Rectangular Drawing

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Rectangular Drawing

• Imo Lieberwerth

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Content

• Introduction
• Rectangular Drawing and Matching
• Thomassens Theorem
• Rectangular drawing algorithm
• Advanced topics

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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

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Example
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Rectangular Drawing and Matching

• A graph G with ? 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

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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

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Example of Labeling

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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

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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

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Example labeling
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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 Tuttes
  transformation for finding an f-factor of a
  graph

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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 a with label x and his corresponding
  edge is contained in a perfect matching, then a
  label 2 otherwise a label 1

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Example labeling
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Theorem

• Let G be a plane graph with ? 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.

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Thomassens Theorem

• Assume that G is a 2-connected plane graph with
  ? 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

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Definitions

• leg of cycle
• k-legged cycle
• good cycle, bad cycle
• Thomassens Theorem G has a rectangular drawing
  if and only if G has no bad cycle

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Number of corners
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Sufficiency

• Lemma 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

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Critical cycle
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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
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Partition-pair Pc and Pcc
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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.

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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

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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

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Case 1
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Case 2
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Case 3.1
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Illustration case 3.2
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Case 3.2

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After the break

• Rectangular drawing algorithm
• Advanced topics

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Questions?