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Rectangular Drawing

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

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Rectangular Drawing

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  1. Rectangular Drawing Imo Lieberwerth

  2. Content Introduction Rectangular Drawing and Matching Thomassen’s Theorem Rectangular drawing algorithm Advanced topics

  3. 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

  4. Example

  5. Rectangular Drawing and Matching A graph G with Δ ≤ 4 has a rectangular drawing D if and only if a new bipartite graph Gdconstructed from G has a perfect matching. Gd is called a decision graph Assumption: G is 2-connected

  6. 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

  7. Example of Labeling

  8. 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

  9. 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

  10. Example labeling

  11. 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 – n1and 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

  12. 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 otherwiseα= label 1

  13. Example labeling

  14. 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.

  15. 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

  16. 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

  17. Number of corners

  18. 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

  19. Critical cycle

  20. 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

  21. Partition-pair Pc and Pcc

  22. 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.

  23. 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 Pis minimum • Counterclockwise depth-first search w

  24. 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

  25. Case 1

  26. Case 2

  27. Case 3.1

  28. Illustration case 3.2

  29. Case 3.2

  30. After the break • Rectangular drawing algorithm • Advanced topics

  31. Questions?

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