Proving Lower Bounds on Graph Drawing Problems

Rajat AnantharamDepartment of Gaming and Media Technology

Utrecht University

Objective

Technique for proving exponential area lower bounds

The Logic Engine Description Illustration Simulation Extension

Tune ups

Planar acyclic graphs

Exponential variables

Planar straight line Upward drawing

Upper and Lower bounds

s1

s0

t0

t1

G1

Sn

Sn-1

Sn-2

Tn-2

Tn-1

Tn

Gn-1Gn

(a)

(b)

Digraphs G1 and Gn

Theorem 1

Given any resolution rule, a planar straight line upward drawing of digraph Gn (with 2n + 2 vertices) has area 2n)

Approach to Proving

With An as the minimum area of a planar straight line upward drawing Gn We use induction to prove that

An >= 4. An -2

Since A1 >= c for some constant c depending on the resolution rule, this implies the claimed result.

Variable Instantiation

n straight line drawing of Gn with min area

n-1 straight line drawing of Gn-1 by removing vertices and edges sn and tn

n-2 for Gn-2 and so on ..

Variable Instantiation – Part 2

Horizontal line through vertex sn-2

Horizontal line through vertices tn-2

Line extending edge (tn-3, tn-2)

Angle formed by edge (tn-3, tn-2) and the x-axis

Angle fomed by edge (sn-2, sn-3) and the x-axis

Line paralle to through vertex sn-2

Line extending edge (sn-2, sn-3)

Observations Leading to Inference

Area(P) >= 2 * Area(n-2) >= 2 * An-2

An = Area(n) >= Area(P) + Area(Area

An = 2 * Area(P) >= 4 * Area(n-2) = 4 * An-2

Logic Engine

Links in a Chain

Each armature is connected to the shaft by a chain

When the engine lies flat then the chain can be in two possible positions – viz.aj and aj*

If xj = 1, aj = xj

If xj = 0, aj* = xj

For clauses c1, c2, ... cm the chain links are numbered in outward order as 1,2, ... m

Flags and Flag Collision

Two flags attached along the same row across two armatures collide with each other if they point towards each other

Flag connected to the outermost armature An collides with the frame if it points towards it

If literal xj appears in the clause ci then link i of aj is unflagged

If the literal xj * appears in the clause ci then the link i of aj * is unflagged

The Flag

Theorem

An instance of NAE3SAT is a “yes” instance if and only if the corresponding logic engine has a flat collission free configuration.

And .. The proof

Assume “yes” instance of NAE3SAT

Rotate armatures such that if truth assignment t(xj) = 1, aj is on top and if t(xj) = 0, aj is in bottom

Since clause ci contains atleast one literal y with t(y) = 1 and atleast one literal z with t(z) = 0 there’s atleast one unflagged link in each row.

So we allign the chain such that the flaggs from the remaining link point towards the unflagged link leading to “no collission”

A similar analogy can be drawn from a flat collission free configuration thereby implying the validity of the theorem

Logic engine and a graph Drawing Problem

To prove that the following problem is NP-Hard Theorem :

“Unit length planar straight line drawing of a graph is NP-Hard”

The question: - Is there a straight line planar drawing of G such that every edge is of length one ?

Prelude

How to construct the unversal part of a logic graph?

A graph G is uniquely drawable if all the unit length planar drawings f G can be obtained through translation, rotation, scaling, mirroring

Construction of the Logic Engine

Each unit length of the constructed Logic Engine corresponds to the link graph

Armature and Outer frame is necessarily unique

The max Euclidean distance b/w the extremal endpoints to the shaft gives the number of edges in the shaft making its construction unique as well

Hence the Proof

The logic graph corresponding to an instance of NAE3SAT with n variables and m clauses has O((m+n)2) vertices and edges and the time taken to construct the logical graph is linear in the size of the graph

Extending Usage of Logic Graphs Is there a grid drawing of a tree T, such that each edge

has length one ?

Is there a grid drawing of tree T of area at most K ? – where K is an integer.

Is there a drawing of tree T such that T is a minimum spanning tree of the vertex locations?

Is there a drawing of graph G such that G is the mutual nearest neighbour graph of the vertex location?

Thank You