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 G d constructed from G has a perfect matching. G d 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 1 if 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 G d Add a complete bipartite graph K to G d inside each inner face with a label x K (a, b) where a = 4 – n 1 and b = n x n 1 = number of angles with label 1 n 1 ≤ 4 n x = 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 G d is a set of pairwise non- adjacent edges in G d Perfect matching: if an edge in M is incident to each vertex of G d 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 G d of G has a perfect matching. D can be found in time O(n 1.5 ) 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

Sufficiency Lemma 1: Let J 1, J 2, …, J p be the C o (G)-components of a plane graph G, and let G i = C o (G) U J i, 1 ≤ i ≤ p. Then G has a rectangular drawing with corners a, b, c and d if and only if each G i 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 P c and P cc

Lemma & proof Lemma 3: Assume that a cycle C in the C o (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 P N P ends at the second last vertex of P S 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 P c and P cc

Case 1

Case 2

Case 3.1

Illustration case 3.2

Case 3.2

After the break Rectangular drawing algorithm Advanced topics

Questions?