Rectangular Drawing Imo Lieberwerth. Content Introduction Rectangular Drawing and Matching Thomassen’s Theorem Rectangular drawing algorithm Advanced.

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?