Rectangular Drawing

Imo Lieberwerth

Content Introduction

Rectangular Drawing and Matching

Thomassen’s Theorem

Rectangular drawing algorithm

Advanced topics

Introduction Conditions

Each vertex is drawn as a point Each edge is drawn as a horizontal or vertical line segment, without edge-crossing Each face is drawn as a rectangle

Special case of a convex drawing Not every plane graph has a rectangular drawing Rectangular drawing is used in floorplanning

Example

Rectangular Drawing and Matching

A graph G with Δ ≤ 4 has a rectangular drawing D if and only if a new bipartite graph Gd constructed from G has a perfect matching.

Gd is called a decision graph

Assumption: G is 2-connected

Definitions Angle of vertex v = the angle formed by two edges incident to v

In rectangular drawing alone angles of: 90° = label 1 180° = label 2 270° = label 3

Example of Labeling

Regular labeling A regular labeling of G satisfies:

For each vertex the sum of labels is equal to 4 The label of any inner angle is 1 or 2 Every inner face has exactly four angles with label 1 The label of any outer angle is 2 or 3 The outer face has exactly four angles of label 3

Follows: A non-corner vertex v with degree 2, has two labels 2 if d(v) = 3, then one angle with label 2 and the other 1if d(v) = 4, four angles with label 1

Regular labeling (2) A plane graph G has a rectangular drawing if and only if G has a regular labeling

Have to find a regular labeling

Assumptions: convex corners a, b, c, and d of degree 2 are given

Example labeling

Decision graph All vertices wit a label x are vertices of Gd

Add a complete bipartite graph K to Gd

inside each inner face with a label x K(a, b) where

a = 4 – n1 and b = nx

n1 = number of angles with label 1

n1 ≤ 4 nx = number of angles with x

The idea of adding K originates from Tutte’s transformation for finding an “f-factor” of a graph

Matching A matching M of Gd is a set of pairwise non-adjacent edges in Gd

Perfect matching: if an edge in M is incident

to each vertex of Gd

If an angle α with label x and his

corresponding edge is contained in a perfect matching, then α = label 2 otherwise α = label 1

Example labeling

Theorem Let G be a plane graph with Δ ≤ 4 and four outer vertices a, b, c and d be designated as corners. Then G has a rectangular drawing D with the designated corners if and only if the decision graph Gd of G has a perfect matching. D can be found in time O(n1.5) whenever G has D.

Thomassen’s Theorem Assume that G is a 2-connected plane graph with Δ ≤ 3 and the four outer vertices of degree two are designated as the corners a, b, c and d. Then G has rectangular drawing if and only if:

any 2-legged cycle contains two or more corners

Any 3-legged cycle contains one or more corners

Definitions leg of cycle k-legged cycle good cycle, bad cycle

Thomassen’s Theorem: G has a rectangular drawing if and only if G has no bad cycle

Number of corners

SufficiencyLemma 1: Let J1, J2, …, Jp be the Co(G)-components of a plane graph G , and let Gi = Co(G) U Ji , 1 ≤ i ≤ p. Then G has a rectangular drawing with corners a, b, c and d if and only if each Gi has a rectangular drawing with corners a, b, c and d

Critical cycle

Boundary face

Lemma 2: If G has no bad cycle, then every boundary NS-, SN-, EW- or WE-path P of G is a partitioning path, that is, G can be splitted along P into two subgraphs, each having no bad cycle

Partition-pair Pc and Pcc

Lemma & proof

Lemma 3: Assume that a cycle C in the Co(G)-component J has exactly four legs. Then the subgraph G(C) of G inside C has no bad cycle when the four leg-vertices are designated as corners of G(C).

Proof. If G(C) has a bad cycle, then it is also a bad cycle in G, a contradiction to the assumption that G has no bad cycle.

Westmost NS-path

A path is westmost if: P starts at the second vertex of PN

P ends at the second last vertex of PS

The number of edges in G P is

minimum

Counterclockwise depth-first search

w

Lemma

Lemma 4: If G has no bad cycle and has no boundary NS-, SN-, EW- or WE-path, then G has a partition-pair Pc and Pcc

Case 1

Case 2

Case 3.1

Illustration case 3.2

Case 3.2

After the break

Rectangular drawing algorithm

Advanced topics

Questions?