Rectangular Drawing Imo Lieberwerth. Content Introduction Rectangular Drawing and Matching...
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Transcript of Rectangular Drawing Imo Lieberwerth. Content Introduction Rectangular Drawing and Matching...
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
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
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
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
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
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
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
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