Geometric AlgorithmsSuman Sourav

Paramasiven AppavooAnuja Meetoo Appavoo

Li JingLu Bingxin

Suhendry EffendyDumitrel Loghin

Introduction amp Motivation

Suman Sourav

Introduction

bull Computational geometry is a branch of computer science devoted to the study of algorithms which can be stated in terms of geometry

bull Deep connections with classical mathematics and theoretical computer science on the one hand and many ties with applications on the other

bull Focus is on discrete nature of geometric problems as opposed to continuous issues

Introduction

bull Any problem that is to be solved using a digital computer has to be discrete in form

bull For CG techniques to be applied to areas that involves continuous issues discrete approximations to continuous curves or surfaces are needed

bull People deal more with straight or flat objects or simple curved objects than with high degree algebraic curves

Introduction

bull Programming in CG is a little difficult Libraries like LEDA and CGAL implement various data structures and geometric algorithms

bull CG algorithms suffer from the curse of degeneracies So certain simplifying assumptions at times like no three points are collinear no four points are cocircular etc are needed to be made

Motivation amp ApplicationThere are many areas that give rise to geometric problems

bull Computer graphics

bull Computer vision and image processing

bull Robotics

bull Computer-aided designing (CAD)

bull Geographic information systems (GIS)

bull Telecommunication

bull VLSI design (chip layout)

Example To find the nearest phone booth on campus

Example The motion planning problem

ExampleTo build a model of the terrainsurface we can start with a numberof sample points where we know theheight

How do we interpolate the height atother points

bull Nearest neighbor interpolation

bull Piecewise linear interpolation by a triangulation

bull Moving windows interpolation

bull Natural neighbor interpolation

For interpolation it is good if triangles are not long and skinny

Example

bull Each time you click inside the editing area of a diagram or inside a web page the application must know inside which object the mouse pointer is

bull Given the high frequency of such a query a fast algorithm must be used

Facility locationbull Given a set of houses and farms

in an isolated area Can we placea helicopter ambulance post sothat each house and farm can bereached within 15 minutes

bull Where should we place anantenna so that a number oflocations have maximumReception

Art gallery problemHow many cameras are needed to guard a given art gallery so that every point is seen

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Introduction amp Motivation

Suman Sourav

Introduction

bull Computational geometry is a branch of computer science devoted to the study of algorithms which can be stated in terms of geometry

bull Deep connections with classical mathematics and theoretical computer science on the one hand and many ties with applications on the other

bull Focus is on discrete nature of geometric problems as opposed to continuous issues

Introduction

bull Any problem that is to be solved using a digital computer has to be discrete in form

bull For CG techniques to be applied to areas that involves continuous issues discrete approximations to continuous curves or surfaces are needed

bull People deal more with straight or flat objects or simple curved objects than with high degree algebraic curves

Introduction

bull Programming in CG is a little difficult Libraries like LEDA and CGAL implement various data structures and geometric algorithms

bull CG algorithms suffer from the curse of degeneracies So certain simplifying assumptions at times like no three points are collinear no four points are cocircular etc are needed to be made

Motivation amp ApplicationThere are many areas that give rise to geometric problems

bull Computer graphics

bull Computer vision and image processing

bull Robotics

bull Computer-aided designing (CAD)

bull Geographic information systems (GIS)

bull Telecommunication

bull VLSI design (chip layout)

Example To find the nearest phone booth on campus

Example The motion planning problem

ExampleTo build a model of the terrainsurface we can start with a numberof sample points where we know theheight

How do we interpolate the height atother points

bull Nearest neighbor interpolation

bull Piecewise linear interpolation by a triangulation

bull Moving windows interpolation

bull Natural neighbor interpolation

For interpolation it is good if triangles are not long and skinny

Example

bull Each time you click inside the editing area of a diagram or inside a web page the application must know inside which object the mouse pointer is

bull Given the high frequency of such a query a fast algorithm must be used

Facility locationbull Given a set of houses and farms

in an isolated area Can we placea helicopter ambulance post sothat each house and farm can bereached within 15 minutes

bull Where should we place anantenna so that a number oflocations have maximumReception

Art gallery problemHow many cameras are needed to guard a given art gallery so that every point is seen

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Introduction

bull Computational geometry is a branch of computer science devoted to the study of algorithms which can be stated in terms of geometry

bull Deep connections with classical mathematics and theoretical computer science on the one hand and many ties with applications on the other

bull Focus is on discrete nature of geometric problems as opposed to continuous issues

Introduction

bull Any problem that is to be solved using a digital computer has to be discrete in form

bull For CG techniques to be applied to areas that involves continuous issues discrete approximations to continuous curves or surfaces are needed

bull People deal more with straight or flat objects or simple curved objects than with high degree algebraic curves

Introduction

bull Programming in CG is a little difficult Libraries like LEDA and CGAL implement various data structures and geometric algorithms

bull CG algorithms suffer from the curse of degeneracies So certain simplifying assumptions at times like no three points are collinear no four points are cocircular etc are needed to be made

Motivation amp ApplicationThere are many areas that give rise to geometric problems

bull Computer graphics

bull Computer vision and image processing

bull Robotics

bull Computer-aided designing (CAD)

bull Geographic information systems (GIS)

bull Telecommunication

bull VLSI design (chip layout)

Example To find the nearest phone booth on campus

Example The motion planning problem

ExampleTo build a model of the terrainsurface we can start with a numberof sample points where we know theheight

How do we interpolate the height atother points

bull Nearest neighbor interpolation

bull Piecewise linear interpolation by a triangulation

bull Moving windows interpolation

bull Natural neighbor interpolation

For interpolation it is good if triangles are not long and skinny

Example

bull Each time you click inside the editing area of a diagram or inside a web page the application must know inside which object the mouse pointer is

bull Given the high frequency of such a query a fast algorithm must be used

Facility locationbull Given a set of houses and farms

in an isolated area Can we placea helicopter ambulance post sothat each house and farm can bereached within 15 minutes

bull Where should we place anantenna so that a number oflocations have maximumReception

Art gallery problemHow many cameras are needed to guard a given art gallery so that every point is seen

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Introduction

bull Any problem that is to be solved using a digital computer has to be discrete in form

bull For CG techniques to be applied to areas that involves continuous issues discrete approximations to continuous curves or surfaces are needed

bull People deal more with straight or flat objects or simple curved objects than with high degree algebraic curves

Introduction

bull Programming in CG is a little difficult Libraries like LEDA and CGAL implement various data structures and geometric algorithms

bull CG algorithms suffer from the curse of degeneracies So certain simplifying assumptions at times like no three points are collinear no four points are cocircular etc are needed to be made

Motivation amp ApplicationThere are many areas that give rise to geometric problems

bull Computer graphics

bull Computer vision and image processing

bull Robotics

bull Computer-aided designing (CAD)

bull Geographic information systems (GIS)

bull Telecommunication

bull VLSI design (chip layout)

Example To find the nearest phone booth on campus

Example The motion planning problem

ExampleTo build a model of the terrainsurface we can start with a numberof sample points where we know theheight

How do we interpolate the height atother points

bull Nearest neighbor interpolation

bull Piecewise linear interpolation by a triangulation

bull Moving windows interpolation

bull Natural neighbor interpolation

For interpolation it is good if triangles are not long and skinny

Example

bull Each time you click inside the editing area of a diagram or inside a web page the application must know inside which object the mouse pointer is

bull Given the high frequency of such a query a fast algorithm must be used

Facility locationbull Given a set of houses and farms

in an isolated area Can we placea helicopter ambulance post sothat each house and farm can bereached within 15 minutes

bull Where should we place anantenna so that a number oflocations have maximumReception

Art gallery problemHow many cameras are needed to guard a given art gallery so that every point is seen

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Introduction

bull Programming in CG is a little difficult Libraries like LEDA and CGAL implement various data structures and geometric algorithms

bull CG algorithms suffer from the curse of degeneracies So certain simplifying assumptions at times like no three points are collinear no four points are cocircular etc are needed to be made

Motivation amp ApplicationThere are many areas that give rise to geometric problems

bull Computer graphics

bull Computer vision and image processing

bull Robotics

bull Computer-aided designing (CAD)

bull Geographic information systems (GIS)

bull Telecommunication

bull VLSI design (chip layout)

Example To find the nearest phone booth on campus

Example The motion planning problem

ExampleTo build a model of the terrainsurface we can start with a numberof sample points where we know theheight

How do we interpolate the height atother points

bull Nearest neighbor interpolation

bull Piecewise linear interpolation by a triangulation

bull Moving windows interpolation

bull Natural neighbor interpolation

For interpolation it is good if triangles are not long and skinny

Example

bull Each time you click inside the editing area of a diagram or inside a web page the application must know inside which object the mouse pointer is

bull Given the high frequency of such a query a fast algorithm must be used

Facility locationbull Given a set of houses and farms

in an isolated area Can we placea helicopter ambulance post sothat each house and farm can bereached within 15 minutes

bull Where should we place anantenna so that a number oflocations have maximumReception

Art gallery problemHow many cameras are needed to guard a given art gallery so that every point is seen

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Motivation amp ApplicationThere are many areas that give rise to geometric problems

bull Computer graphics

bull Computer vision and image processing

bull Robotics

bull Computer-aided designing (CAD)

bull Geographic information systems (GIS)

bull Telecommunication

bull VLSI design (chip layout)

Example To find the nearest phone booth on campus

Example The motion planning problem

ExampleTo build a model of the terrainsurface we can start with a numberof sample points where we know theheight

How do we interpolate the height atother points

bull Nearest neighbor interpolation

bull Piecewise linear interpolation by a triangulation

bull Moving windows interpolation

bull Natural neighbor interpolation

For interpolation it is good if triangles are not long and skinny

Example

bull Each time you click inside the editing area of a diagram or inside a web page the application must know inside which object the mouse pointer is

bull Given the high frequency of such a query a fast algorithm must be used

Facility locationbull Given a set of houses and farms

in an isolated area Can we placea helicopter ambulance post sothat each house and farm can bereached within 15 minutes

bull Where should we place anantenna so that a number oflocations have maximumReception

Art gallery problemHow many cameras are needed to guard a given art gallery so that every point is seen

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example To find the nearest phone booth on campus

Example The motion planning problem

ExampleTo build a model of the terrainsurface we can start with a numberof sample points where we know theheight

How do we interpolate the height atother points

bull Nearest neighbor interpolation

bull Piecewise linear interpolation by a triangulation

bull Moving windows interpolation

bull Natural neighbor interpolation

For interpolation it is good if triangles are not long and skinny

Example

bull Each time you click inside the editing area of a diagram or inside a web page the application must know inside which object the mouse pointer is

bull Given the high frequency of such a query a fast algorithm must be used

Facility locationbull Given a set of houses and farms

in an isolated area Can we placea helicopter ambulance post sothat each house and farm can bereached within 15 minutes

bull Where should we place anantenna so that a number oflocations have maximumReception

Art gallery problemHow many cameras are needed to guard a given art gallery so that every point is seen

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example The motion planning problem

ExampleTo build a model of the terrainsurface we can start with a numberof sample points where we know theheight

How do we interpolate the height atother points

bull Nearest neighbor interpolation

bull Piecewise linear interpolation by a triangulation

bull Moving windows interpolation

bull Natural neighbor interpolation

For interpolation it is good if triangles are not long and skinny

Example

bull Each time you click inside the editing area of a diagram or inside a web page the application must know inside which object the mouse pointer is

bull Given the high frequency of such a query a fast algorithm must be used

Facility locationbull Given a set of houses and farms

in an isolated area Can we placea helicopter ambulance post sothat each house and farm can bereached within 15 minutes

bull Where should we place anantenna so that a number oflocations have maximumReception

Art gallery problemHow many cameras are needed to guard a given art gallery so that every point is seen

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

ExampleTo build a model of the terrainsurface we can start with a numberof sample points where we know theheight

How do we interpolate the height atother points

bull Nearest neighbor interpolation

bull Piecewise linear interpolation by a triangulation

bull Moving windows interpolation

bull Natural neighbor interpolation

For interpolation it is good if triangles are not long and skinny

Example

bull Each time you click inside the editing area of a diagram or inside a web page the application must know inside which object the mouse pointer is

bull Given the high frequency of such a query a fast algorithm must be used

Facility locationbull Given a set of houses and farms

in an isolated area Can we placea helicopter ambulance post sothat each house and farm can bereached within 15 minutes

bull Where should we place anantenna so that a number oflocations have maximumReception

Art gallery problemHow many cameras are needed to guard a given art gallery so that every point is seen

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

How do we interpolate the height atother points

bull Nearest neighbor interpolation

bull Piecewise linear interpolation by a triangulation

bull Moving windows interpolation

bull Natural neighbor interpolation

For interpolation it is good if triangles are not long and skinny

Example

bull Each time you click inside the editing area of a diagram or inside a web page the application must know inside which object the mouse pointer is

bull Given the high frequency of such a query a fast algorithm must be used

Facility locationbull Given a set of houses and farms

in an isolated area Can we placea helicopter ambulance post sothat each house and farm can bereached within 15 minutes

bull Where should we place anantenna so that a number oflocations have maximumReception

Art gallery problemHow many cameras are needed to guard a given art gallery so that every point is seen

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

For interpolation it is good if triangles are not long and skinny

Example

bull Each time you click inside the editing area of a diagram or inside a web page the application must know inside which object the mouse pointer is

bull Given the high frequency of such a query a fast algorithm must be used

Facility locationbull Given a set of houses and farms

in an isolated area Can we placea helicopter ambulance post sothat each house and farm can bereached within 15 minutes

bull Where should we place anantenna so that a number oflocations have maximumReception

Art gallery problemHow many cameras are needed to guard a given art gallery so that every point is seen

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example

bull Each time you click inside the editing area of a diagram or inside a web page the application must know inside which object the mouse pointer is

bull Given the high frequency of such a query a fast algorithm must be used

Facility locationbull Given a set of houses and farms

in an isolated area Can we placea helicopter ambulance post sothat each house and farm can bereached within 15 minutes

bull Where should we place anantenna so that a number oflocations have maximumReception

Art gallery problemHow many cameras are needed to guard a given art gallery so that every point is seen

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Facility locationbull Given a set of houses and farms

in an isolated area Can we placea helicopter ambulance post sothat each house and farm can bereached within 15 minutes

bull Where should we place anantenna so that a number oflocations have maximumReception

Art gallery problemHow many cameras are needed to guard a given art gallery so that every point is seen

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Art gallery problemHow many cameras are needed to guard a given art gallery so that every point is seen

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

A Real Life Application

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Voronoi

Voronoi diagram- Properties- Construction methods

Fortunersquos Algorithm- How does it work- Implementation aspects

- Data structures- Pseudocode

- Complexity (time amp space)

Paramasiven Appavoo amp Anuja Meetoo Appavoo

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

bull Voronoi edges Each point on an edge of the Voronoi diagram is equidistant from its two nearest neighbors pi and pj

bull Voronoi vertices It is the center of the circle passing through these sites and this circle contains no other sites in its interior

bull Degree If we assume that no four points are cocircular then the vertices of the Voronoi diagram all have degree three

Voronoi diagram Properties

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

bull Convex hull A cell of the Voronoi diagram is unbounded if and only if the corresponding site lies on the convex hull

bull Size If n denotes the number of sites then the Voronoi diagram is a planar graph (if we imagine all the unbounded edges as going to a common vertex infinity) with exactly n faces

Voronoi diagram Properties

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

VoronoiConstruction methods

bull Perpendicular bisector

bull Divide and conquer

bull Fortunersquos algo

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

VoronoiFortunersquos algo

bull Time complexity of ldquoperpendicular bisectorrdquo

bull Extra computation at the merge of ldquodivide and conquerrdquo

bull Steven Fortune developed a plane sweep algorithm called the ldquofortunersquos algorithmrdquo that useso a sweep line o a beach line o site event ando circle eventto generate the diagrams

Steven FortuneBell Laboratories

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos Algo

>

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algo - ConceptsSweep line

bull The idea behind algorithms of this type is to imagine that a line is swept or moved across the plane stopping at some points

bull A sweep line splits the problem domain into two regions an explored region and an unexplored region

bull Operations are restricted to objects that are either intersecting or are in the immediate vicinity of the sweep line whenever it stops

bull The complete solution is available once the line has passed over all objects

bull It also applies an ordering

bull to the problem

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algo - ConceptsSite events

bull A site event happens when the sweep line hits a site

bull Property of the line perpendicular to the sweep line at the site evento All points are equidistant to

bull The sitebull The closest point on the sweep line (the site

itself at this stage) site event

What happens when the sweep line moves further away

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algo - ConceptsWhat happenshellip

bull The set of points that are equidistant to both the site and the closest point on the sweep line degenerate from a line to form a parabola

The parabola defines the set of points

Set of points that belong to this sitersquos voronoi cell

Important property implying that the cross section of such parabolas from a number of sites will define the edges of a voronoi cell

Why are we doing this

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algo - Concepts2 sites - Crossing parabolashellip

bull p1 and p2 are two siteshellip

bull At the site event of p2 point q is equidistant from p1 and p2

bull As the sweep line goes down the intersection of the parabolas breakpoints are the endpoints for a partial edge

P1

P2

P1

P2 site event

q

P1

breakpoints

part of voronoi diagramrsquos edge

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algo - ConceptsMore siteshellip amp beach line

bull This dividing line is termed the beach line

bull The algorithm for computing the Voronoi diagram of a set of points depends entirely on how this beach line changes as the sweep moves through the space

bull The beach linersquos topology changes when a new arc is added or another is absorbed

breakpoints

parts of the edges of the voronoi diagrams

What is the stopping condition for an edgersquos growth

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

bull One endpoint is formed with a circle evento a vertex in the voronoi diagram

bull Each relating site is equidistant from the vertexbull Vertex is the centre of a circle on which the three

points lieEither by cross sections of parabolas or simply by using the intersection of each or the perpendicular bisector of the inner triangle

Fortunersquos algo - ConceptsCircle event

vertex

circle event

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algo - ConceptsFrom the vertex

vertex q If a circumcircle is empty in its interior then in a Voronoi diagrambull a b c would be Voronoi sitesbull q would be a Voronoi vertexbull The perpendicular bisectors of abc would be Voronoi edges

a

bc

Nowhellip from the vertex

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algo Implementation Aspectsbull Parabolas defining beach line not computed as sweep

line moves through the problem spaceo Computationally expensive and unnecessaryo Calculations required only when the beach line

changes topology Site events Circle events

bull Sweep line makes discrete steps rather than a continuous sweep

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoData Structures

bull Priority queueo State of sweep lineo Indicates when topology of beach line can change

Site events Circle events

bull Doubly connected edge list (DCEL)o State of Voronoi diagramo Stores

Vertex records Edge records Face records

bull Balanced binary search tree ( AVL or red-black tree)o State of beach line

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoPriority Queue of Events

bull Priority based on their y-coordinates

bull Site events are represented by their coordinates

bull Circle events o Computed on the fly o Represented by coordinates of

lowest point of empty circle touching 3 sites

o ldquoAnticipatedrdquo and may be false

s1 (x1 y1) s2 (x2 y2) s3 (x3 y3) s5 (x5 y5) s4 (x4 y4)

c1 (x y)circle event added to queue

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoDCEL

Vertex Coordinates Incident edge

v1 (x1 y1) e12

v2 (x2 y2) e23

v8 (x8 y8) e87

Vertex records

Half edge

Source vertex

Twin Incident face

Prev edge

Next edge

e23 v2 e32 s3 e52 e37

Edge recordsFace Incident edge

s1 e32

s2 e56

Face records

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algo Balanced Binary Search Tree

bullInternal nodes o Represent breakpoints between two arcs (tuple)o Contains pointer to record of edge being traced

bullLeaf nodeso Stores site defining arc on beach lineo Contains a pointer to a potential circle event

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoPseudocode

1 Initialise data structures Event queue Q larr all site events

Binary search tree TlarremptyDCEL D larrempty

2 while(queue not empty)3 event = pop first event from queue4 process(event T D)5 finish all edges in binary search tree

How to process site event- Add the event and breakpoints to BST- Add new edge in DCEL- Delete false alarms for circle events- Add potential circle event(s) to queue

How to process circle event- Update breakpoints and leaf in BST- Record new vertex- Delete false alarms - Add possible circle event to queue

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoProcessing Site Event

bull Locate existing arc (if any) that is above the new siteo x-coordinate of new site is used for binary searcho x-coordinate of each breakpoint along root to leaf

path is computed on the fly

s5 s4 Q

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoProcessing Site Event

bull Break the arco Replace the leaf node with a sub tree representing

the new arc and its breakpoints

Different arcs can be identified by the same sites4 Q

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoProcessing Site Event

bull Add new edge record in DCEL

bull Create new face record with pointer to new edge

New half-edge record

s4 Q

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoProcessing Site Event

bull Check for potential circle eventso Scan for triple of consecutive arcs and determine if

breakpoints converge

Potential circle event

Add potential circle event to to priority queue

Store a pointer to circle event in leaf

node for s1

c1 s4 Q

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoProcessing Site Event

bull Converging breakpoints may not always yield a circle evento Appearance of a new site before the circle event

makes the potential circle non-empty

Original circle event becomes a false alarm

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoProcessing Circle Event

bull Create new vertex record

bull Add vertex to corresponding edge record

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoProcessing Circle Event

bull Delete disappearing arc

s4 Q

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoProcessing Circle Event

s4 Q

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoProcessing Circle Event

New edge being traced by new breakpointlts5 s3gt

New half-edge record

Create new edge record

Check new triple edges for potential circle events

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoAlgorithm Termination

bull When Q is empty beach line and its breakpoints continue to trace edgeso Terminate ldquohalf-infiniterdquo edges via a bounding box

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algo Complexity

bull Locate leaf representing existing arc above new siteo Delete potential circle event from queue

bull Break arc by replacing leaf node with a sub tree representing new arc and break points

bull Add new edge and face records to DCEL

bull Check for potential circle event(s) o Add event to queue if they existo Store pointer to event in proper leaf

Running Time

O(log n)

O(1)

Steps in handling site events

O(1)

O(log n)

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoComplexity

bull Delete disappearing BST leaf node and its associated circle events from event queue

bull Add vertex record in DCEL

bull Create new edge record in DCEL

bull Check for potential circle event(s)

Running TimeSteps in handling circle events

O(log n)

O(1)

O(1)

O(1)

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Fortunersquos algoComplexity

bull Time complexity for each event O(log n)bull How many events are there

o Number of site events + Number of circle eventso How many site events

no How many circle events

Each circle event corresponds to a vertex bull 2n - 5 (Eulerrsquos formula)

False alarms deleted before they are processed Total number of events = 3n - 5

bull Overall running time O(n log n)

bull Storage complexity O(n)

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Delaunay Triangulation

Li Jing amp Lu Bingxin

bullDelaunay and Voronoi

bullDelaunay properties

bullRandomized Algorithm

- Idea

- Implementation aspects- Pseudocode- Data structure

- Complexity (time amp space)

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Delaunay and VoronoiAn intuitive conception

General position assumption no 4 points are co-circular

(a) Voronoi diagram

(b) Delaunay triangulation

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Delaunay and VoronoiDelaunay and Voronoi complexes are dual to each other

Dual correspondence

Voronoi complexes Delaunay triangulation

cells (regions) vertices

edges edges

vertices faces

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Delaunay Triangulation (DT) Whatrsquos the difference

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

PropertiesCircumcircle propertyThe circumcircle of any triangle in DTS) is empty (It contains no points of S in its interior)

Proof

By general position assumption the degree of all Voronoi vertex is 3 edge vu exists

Consider S4 S1S4 is perpendicular to l and divided half by l S4 is outside the circle

In fact one definition of Delaunay Triangulation isDelaunay Triangulation is a triangulation that circumcircle of each triangle is empty

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

PropertiesEmpty circle propertyTwo points are connected by an edge in the Delanuay triangulation There is an empty circle passing through these two points

Proof Trivial from the circumcircle property

There are a series of circumcircles that pass through Si and Sj

Note down the centres of these circumcircles until they pass through another point

The track is the bound for both cell(Si) and cell(Sj)

Si Sj is an edge in the DT(S)

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Locally Delaunay edge An edge ab is locally Delaunay if

it belongs to only one triangle or it belongs to two triangles abc and abd and d lies

outside the circumcircle of abc

If an edge is locally Delaunaywe also call it legal otherwise illegal

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Delaunay LemmaIf every edge in Ts is locally Delaunay then Ts is the Delaunay triangulation of S

Proof Let x be an arbitrary point in abc

Let abc=Α0 A1 A2 hellip Ak be the sequence of triangles that intersect xp

Let di(p) = |p ndash ai|2 ndash ri2

Because the edges along xp are locally delaunay d0(p)gt d1(p)gt hellip gt dk(p)

dk(p) = 0 so d0(p)gt0

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Edge Flipping Flip all edges in a triangulation until they are all locally Delaunay edges

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Randomized Incremental Algorithm

Randomized the points as p1 p2hellip pn

Find a sufficiently large triangle that contains P

Insert p1 then p2 and finally pn

suppose we have computed DT(Pi-1)

insert pi which splits a triangle into three

perform edge flips until no illegal edge remains we have just computed DT(Pi)

Repeat the process until i = n

Discard the initial large triangle

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

First stepFind a sufficiently large triangle that contains P

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

First step

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

First step

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example

insert p

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example

split abc into abp bcp and acp

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example

check edges ab bc and ac

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example

edge ab is illegal flip it

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example

edge ab is flipped into pd edge ad and bd are to be checked edge ad is legal keep it

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example

edge bd is illegal flip it

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example

edge bd is flipped into pe edge ed and be are legal keep them

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example

edge bc is illegal flip it

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example

edge bc is flipped into fp edge bf and cf are legal keep them

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example

edge ac is illegal flip it

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example

edge ac is flipped into pg no more edge to flip we are done

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

PseudocodeAlgorithm DelaunayTriangulation(P)Input a suitably shuffled (permuted uniformly at random) set of points P = (p1

p2 p3⋯ pn)Output DT(P) ( use a global DCEL to store DT(P) )1 Find a sufficiently large triangle T(p-3p-2p-1) containing P2 for i = 1 to n do3 Insert(pi)4 Endfor5 Discard the triangle T(p-3p-2p-1)

Algorithm Insert(p)Input a point p a set of point P and T = DT(P)Output DT(P u p)1 Find the triangle T(abc) of DT(P) containing p

( use conflict lists )2 Insert edges papb and pc( update conflict lists )3 SwapTest(ab)4 SwapTest(bc)5 SwapTest(ca)

Algorithm SwapTest(ab)1 if ab is an edge of the exterior face of DT(P)2 do return3 d lt- the vertex (other than ab) of the triangle adjacent to triangle T(pab) along edge ab4 if inCircle(p a b d) lt 05 do Flip edge ab for pd

( update conflict lists )6 SwapTest(ad)7 SwapTest(db)

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Conflict list Conflict --- a non-inserted point is inside a triangle in the

current triangulation

non-inserted points

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Current triangle Non-inserted points

T(p1p2p3) p7 p8

T(p4p5p6) p9

Triangles in the current Delaunay Triangulation

bi-directional pointer

Each triangle of the current triangulation --- Bucket

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Time complexity

Major steps in the algorithm

Find a sufficiently large triangle

Find the triangle containing a non-inserted point

Update the triangulation

Update conflict lists

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Find a sufficiently large triangle M maximum absolute value of either x or y coordinate

of all the points in P

Time costO(1)

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Query the conflict list

Find the triangle containing a non-inserted point

Time cost for one iterationO(1)

Time cost for all the n iterationsO(n)

Non-inserted point

Current triangle

p7 T(p1p2p3)

p8 T(p1p2p3)

p9 T(p4p5p6)

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Backward analysis Imagine that the algorithm is run backwards starting

from the delaunay triangulation we have at the end

In analyzing the ith step imagine that we are deleting one of the i points in the current triangulation and then update the triangulation The work done in this case is the same as running

the algorithm forward Assume that each of the i points is equally likely to

be deleted at the ith step since the points were added randomly in the original algorithm

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Time to update triangulation Consider the ith step of the algorithm

Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the whole point set P i gt3 or the set of points in DT(Pi)

Run the ith step backward (deleting a random point p in Pi ) Note that each new edge added in DT(Pi) due to the

insertion of p is incident to p

The total number of edge changes made in the triangulation for the insertion of p is proportional to the degree of p after the insertion is complete

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Time to update triangulation

The total degree of the vertices in Pi is less than 6i There are at most 3i edges in DT(Pi) (planar graph)

The sum of vertex degrees is twice the number of edges in a graph

The expected degree of a random point in Pi is at most 6

The expected number of edge changes for inserting a random point p is O(1)

Summing over all the n steps the time for updating the triangulation is O(n)

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Update conflict lists Each non-inserted point is assigned to a trianglebucket

For a non-inserted point if the triangle which it lies within is destroyed we have to find another new triangle containing this non-inserted point

The expected time to update conflict lists is the expected time to rebucket non-inserted points

is proportional to the expected number of non-

inserted points required to be rebucketed

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Consider the ith step of the algorithm Pi the set of the first i points (p1 p2 p3 ⋯ pi) in the

whole point set P i gt3 or the set of points in DT(Pi) PPi the set of non-inserted points (pi+1 pi+2 pi+3 ⋯

pn)

Run the ith step backward (deleting a random point in Pi) some triangles in DT(Pi) are destroyed some points in PPi are required to be rebucketed

Rebucket points

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Time to rebucket points Assume a random point p in Pi is deleted

For a random point q in PPi suppose that q is bucketed in the triangle T(abc) of DT(Pi)

If p is one of the three vertices of T(abc) T(abc) will be destroyed and q will be rebucketed

Pr (p is deleted) = 1iPr (T(abc) is destroyed) = 3iPr (q needs to be rebucketed) = 3i

The expected number of points in PPi required to be rebucketed is 3(n-i)i

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Time to rebucket points Summing over all the n steps

The total expected number of non-inserted points required to be rebucketed is O(nlogn)

The expected time to update conflict lists is O(nlogn)

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Time complexitybull Major steps in the algorithm Running

time

o Find a sufficiently large triangle O(1)

o Find the triangle containing a point O(n)

o Update the triangulation O(n)

o Update conflict lists O(nlogn)

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Complexity Expected time complexity O(nlogn)

Expected space complexity O(n) DCEL Conflict list

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Trapezoidal Decomposition

Dumitrel Loghin amp Suhendry Effendy

ConceptsRandomized AlgorithmMotivation - Point LocationComplexity Analysis

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Defining the problem

bull Given a set S of n segments in the plane with no two distinct end-points having the same x coordinate (general position)

bull These segments intersect in k pointsbull There is a bounding rectangle R containing all n

segmentsbull If we draw vertical lines through each end-point or

intersection point till they intersect a segment or R we obtain a set of trapezoids T(S) which is the trapezoidal decomposition of S

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Example

p1

p2

p3 q1

q2

q3

R

k1k2

k3

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Details

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Construction Example

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Randomized Algorithm

bull Start with a random permutation S = s1 hellip sn with a bounding rectangle R and with T(S)=Oslash

bull FOR i = 1 TO n DO add segment si and update T(S)ndash Get the trapezoid which contains left end-point of si and

draw a vertical linendash Move to the right on the planar graph and get all the

intersected trapezoidsbull If the segment intersect the bottom or top segment of

the trapezoid the point is one of the k intersection points hence draw a vertical line

bull Draw a vertical line at the right end-pointndash Rescan the intersected vertical lines and trim them based

on their starting point merge the trapezoids which share the deleted line

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Point Locationbull Given a query point q find in which trapezoid it lies

bull Data structurendash DAG

internal nodes are end-points intersection points or segments leaves are trapezoids and they may be shared (in-degree of a leaf

may be more than one)ndash Can be constructed simultaneously with trapezoidal

decompositionbull Algorithm ndash key idea

ndash At step i some of the leaves (trapezoids) become internal nodes (end-points of the segment si or intersection points) and new trapezoids are added

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Point Location Example

s1

s2

p1

q2

q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7

Δ8

Δ9

Δ10

k1

q

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Point Search in DAGp1

q1

p2

Δ1

Δ4s1

q2

Δ10k1k1Δ3

s2 Δ9 Δ3 s2

Δ6 Δ5 Δ8 Δ7

s1

s2p1

q2q1

p2

Δ1 Δ4

Δ3

Δ2

Δ5

Δ6

Δ7Δ8

Δ9

Δ10

k1

q

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Complexity Analysis

Assumption

points (end of segment) are in ldquogeneral positionrdquo= no two points lies in a same vertical line= all x-coordinate are different

Later we will see how to lift this assumption

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Complexity Analysis

What is the complexity of adding one segment s

bull find the left end point of segment s

bull trace (and update) all trapezoids intersected by s

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Adding 1 Segment

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Adding 1 Segment

How to find this trapezoid

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Finding Trapezoid

How to find trapezoid which contain the left end-pointof segment s

Trapezoid Segment

T1 s5 s7

T2 s8

T3 s6 s9 s10

Segment Trapezoid

s5 T1

s6 T3

s7 T1

s8 T2

s9 T3

s10 T3

bi-directional pointercouple of ways

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Bi-directional pointer

bi-directional pointer

bull query find in O(1)

bull need to update the pointer for each changed trapezoid (only update trapezoids which are intersected by new segment s)

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Adding 1 Segment

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Adding 1 Segment

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Adding 1 Segment

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Complexity Analysis

For each segment s we need to1 find the left end-point of s2 trace intersected trapezoids3 update the trapezoids4 update the bi-directional pointers

For each trapezoid Overall

t(f) the ldquocomplexityrdquo of trapezoid fp(f) update the bi-directional pointer for trapezoid f

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Backward Analysis

Imagine the algorithm run backwards deleting segment one at a time

When we delete a segment s from Hi only trapezoids in Hi which adjacent to s will be affected

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Complexity Analysis

Since we insert (or delete in backward analysis) segment s in random order then every remaining segment is equally likely to be chosen

E(i) the expected complexity of deleting the ith segment(in backward analysis)

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Observation

Each trapezoid is adjacent to at most 4 segments(using general position assumption)

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Observation

Trapezoid can have an arbitrary number ofadjacent segments in non general position

We will dealwith this caselater

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Complexity Analysis

Each trapezoid is adjacent to at most 4 segments=

Each trapezoid appears in at most 4 segmentsrsquo adjacency list

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Complexity Analysis

= all the left endpoints of segment that have yet to be added (or have been deleted in backward analysis)

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Complexity Analysis

= proportional to the number of vertices in all decomposition

k = the number of intersection

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Complexity Analysis

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Complexity Analysis

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Randomized vs Deterministic

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Non General Position

How to handle non general position

bull Rotationor

bull Transformation

Shear Transformationxrsquo = x + ε y

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

SummaryVoronoi Diagrams -- Use Fortune Algorithm

Delaunay Triangulation -- Randomized Incremental Construction ndash Dual of Voronoi

Trapezoidal Decomposition -- Randomized Incremental Construction

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Conclusion

bull Widely used in various other areas

bull We use it sometimes without even realising it

bull Lot of potential of further development

bull Numerous interesting open problemshttpcompgeomcsuiucedu~jeffeopen

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

Thank You

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf

ReferencesVoronoibull Derek Johns An Optimal Algorithm for Computing 2D Voronoi Diagrams Fortunes Sweep Algorithm

Available at httpcgmcsmcgillca~mcleish644ProjectsDerekJohnsSweephtm [Accessed February 2014]

bull Dheeraj Kumar Singh Lecture 20 Voronoi Diagrams and Fortunes Algorithm Available at httpintinnoiitkgpernetincourses91wfiles37906 [Accessed February 2014]

bull Voronoi Diagrams Available at httpimaudgedu~sellaresComGeoVor2D_1ppt [Accessed February 2014]

Delaunay Triangulationbull httpwwwcsumdedu~mount754Lects754lectspdfbull httpwwwcsuunlgeobookinterpolationpdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes04pdfbull httpwwwcompnusedusg~hchengacademiccoursescs5237notes05pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl9cs4235pdfbull httpwwwcompnusedusg~tantcioi_trainingCGl10cs4235pdfbull httpgroupscsailmitedugraphicsclasses6838F01lecturesDelaunayDelaunay2Dppt

Trapezoidal Decompositionbull Rajeev Motwani Prabhakar Raghavan ldquoRandomized Algorithmsrdquo 1995bull Subhash Suri ldquoPoint Locationrdquo Available at httpwwwcsucsbedu~surics235Locationpdf