Rectangular Drawing (continue)

Harald Scheper

Overview

• algorithm (directions)

• algorithm in linear time

• outline of algorithm (placement)

• Rect. Drawings without Designated Corners

Algorithm• To find directions of edges in

rectangular drawing of G (vertical, horizontal)

• Later decide the integer coordinates of vertices

Algorithm (outline)• Assume plane graph G has no bad

cycle.• Each C0(G) component treated

independently.• If there exists a boundary NS-, SN-,

WE-, or EW-path then choose it as a partition path

• Otherwise find partition

path PC and PCC from westmost NS-path, and recurse over subgraphs

Algorithm Main• Algorithm Rectangular-Draw(G)

1. Draw the outer cycle C0(G) as a rectangle by 2 horizontal line segments PN, PS and 2 vertical line segments PE and PW

2. Find all C0(G) components J1,..,JP

3. For each component Ji

Gi = C0(G) Ji

Draw(Gi,Ji)

Draw(G,J) if part• If G has boundary NS-, SN-, WE-, or

EW-path P– assume without loss of generality that P

is a boundary NS-path– Draw all edges of P on vertical

line (directions are vertical)– If |E(P)| ≥ 2 then

• F1,..,Fq are C0 components of GP

• For each Fi, i ≤ 1 ≤ q do

– Draw (C0(GP ) Fi, Fi)

Draw(Gi,Ji) else part• No boundary NS-,SN-,EW-, or WE path

• Find westmost NS-path P• Find partition-pair Pc and Pcc from P

• If Pc = Pcc then

– Draw all edges of Pc

on a vertical line segment

• Recursive– Draw(C0(Gi) Fi,Fi)

Draw(Gi,Ji) else, else part• If Pc ≠ Pcc then

– Draw all edges of Pc and Pcc on

alternating sequences of horizontal,

vertical line segments

– G1 is graph obtained form GW by

contracting all edges of PCC that are

on horizontal sides of rectangular

embeddings of C1,C2,..Ck

G2 = of Pc

G3 = G(C1), G4 = G(C2)…G(Ck)

– Recursive: Draw(C0(Gj) Fj,Fj),

1 ≤ j ≤ q

for each graph Gi, 1 ≤ i ≤ k+2

Pcc

Linear time• Find all C0 components

• For each C0 we find boundary

NS-,SN-,EW- and WE-paths if exist, by traversing all boundary faces of G by cc depth-first search.

• Each boundary path gets a label, NN, NE,…, WW (start/end points). So NS-, SN-, EW-, WE-paths are found in constant time

Linear time• no boundary NS-, SN-,

EW-, WE paths

• find partition-pair Pc and Pcc

• give labels to the newly created paths by traversing them (once)

• problem if a subpath of the westmost NS-path is chosen as westmost NS-path P’ in a later recursive stage, which is not on Pc or Pcc.

Linear time• then again have to

traverse the facial cycles

attached to P’, so time

complexity not linear

• so store:– list edges ei elemof E(P) contained in

boundary NN- and EN-paths

– list edges ei elemof E(P) contained in boundary SS- and SE-paths

• to find Pst and Pen in later recursive stages

Linear time• store array of length n =

whether the vertex is a

head or a tail vertex of a clock

wise critical cycle C attached to P and whether ncc(C) = 1 or ncc(C) >1

• indicate existence of critical cycles attached to P’ (not to find critical cycles again)

• every face become boundary face, and not again. so traversed constant time. => linear time.

Rectangular Grid Drawing 1

• Algorithm finds only directions of all edges in G

• Now coordinates of vertices in G determined in linear time

• Assume for simplicity not-corner vertices have degree 3

Rectangular Grid Drawing 2• Graph Ty spanning tree obtained from

G

delete

not deleted

Rectangular Grid Drawing 3

16 maximal vertical lseg

15 maximal horizontal lseg

Rectangular Grid Drawing 4• to each maximal horizontal line

segment L, we assign y(L) as y-coordinate of every vertex on L– PS is lowermost, PN is topmost– y(PS)=0– compute y(L) bottom to top– For each vertex v assign temp(v) as

temporary y-coordinate of v– For every vertex v on L two cases:

• v has neighbor u below v (temp(v) = y(L’)+1)• v has no neighbor u below v (temp(v) = 0)

– y(L) = max{temp(v)} v

Rectangular Grid Drawing 5• All maximal horizontal line segments with

depth-first search = linear time• Upperbounds on area of grid, W+H ≤ n/2

and W · H ≤ n2/16– coordinate of south-west corner = (0,0),

northeast = (W,H), at least one hor., vert. line segment

– lv = #vertical linesegments, lh = #horizontal linesegments, l = lv + lh

– each vertex (- 4) = one of the l-4 max. line seg (PN…) so n-4 = 2(l-4) => l-2 = n/2 because compact: H ≤ lh -1, W ≤ lv -1

= W+H ≤ lh + lv – 2 = l – 2 = n/2 => W · H ≤ n2/4

RD no designated corners 1• until now considered rect. plane

graph G with Δ ≤ 3 with 4 outer vertices of degree 2 as corners

• now corners not designated

• efficiently find whether G has 4 outer vertices (degree 2) such that there is a rect. drawing

RD no designated corners 2• independent: no common vertex

S1 = {C1,C2}, S2 = {C2,C3,C4}

RD no designated corners 3• Theorem 6.4.1:

G is 2 connected graph, Δ ≤ 3, has min. 4 outer vertices of degree 2, then rect. draw with 4 corners desig. corners if G satisfies:– every 2-legged cycle in G contains at

least 2 outer vertices of degree 2– every 3-legged cycle in G contains at

least 1 outer vertex of degree 2– if an independent set S of cycles in G

consists of c2 2 legged cycles and c3 3-legged cycles then 2c2 + c3 ≤ 4

RD no designated corners 2• independent: no common vertex

S1 = {C1, C2}, c2 = 2, c3 = 0, 2*2+0 = 4

S2 = {C2, C3, C4}, c2 = 1, c3 = 2, 2*1+2 = 4

RD no designated corners 3• Necessity of Th. 6.4.1 :

– assume G has rectangular drawing D with 4 corners. by fact 6.3.1:

an indep. set S has c2 2-legged C contains at least 2 corners, c3 3-legged C contains at least 1 corner. All cycles independent, so at least 2c2 + c3 corners in D. Since there are 4 corners in D, 2c2+c3 ≤ 4

In any rectangular drawing D of G, every 2-legged cycle of G contains 2 or more corners, every 3-legged cycle of G contains 1 or more corner, more than 3-legged has no corner

RD no designated corners 4• prove of 6.4.1:

– show if the 3 conditions hold, then can choose 4 outer vertices of degree 2 as corners a,b,c,d, such that (theorem):

• any 2-legged cycle contains 2 or more corners

• any 3-legged cycle contains 1 or more corners

RD no designated corners 5• outline = (complete in article)

• let J1,..,Jp p ≥ 1 be C0(G) components of G in series

• C1 and C2 in J1 and Jp

• if 4 corners not been chosen, we choose a corner from each of innermost 3-legged cycles.

RD no designated corners 6• example

RD of Planar Graphs• until now considered rectangular

drawings of plane graphs (with fixed embedding), now of planar graphs with Δ ≤ 3 (without fixed embedding)

• say planar graph G has rect. drawing if at least one of the plane embeddings has rect. drawing

RD of Planar Graphs• b, c, d are embeddings of planar

graph b = rd, c and d not (3-legged cycle with no outer vertex of degree 2)

RD of Planar Graphs• finding not trivial because

exponential number of plane embeddings by algorithm as before.

• now linear algorithm to examine if there is a plane embedding with rect. drawing

• Questions?