1. Introduction to Planarity Test W. L. Hsu

2. 2/21 Plane Graph A plane graph is a graph drawn in the plane in such a way that no two edges intersect –Except at a vertex to which they are both incident A planar graph is one which is isomorphic to a plane graph –Namely, it has a plane embedding

3. 3/21 Planar Graphs

4. 4/21 Planar Graph Embedding Clockwise edge ordering

5. 5/21 Issues in Planarity Test If you can find a planar embedding, then the graph is planar. How do you determine if a graph is not planar? This is the more difficult part of many recognition algorithm, namely, deciding when a graph “does not” belong to a class –Get a certificate for non-planar graphs –Or alternatively, you have tried all possible ways but still fail to embed the graph in the plane (proof by exhaustion) –Use counting argument

6. 6/21 Basic Non-Planar Graphs K5K5 K 3,3

7. 7/21 Euler’s Theorem (1752) Euler’s theorem Let G be a connected plane graph, and let f be the # of faces of G. Then n + f = m + 2 –Prove by induction on the # of edges. Corollary. m  3 n – 6 –First show that 3f  2m since every face is bounded by at least 3 edges (# of edges is at least 3 f / 2)

8. 8/21 K 5 and K 3,3 are non-planar If K 5 is planar, then by previous Corollary, we have 10  9.  K 3,3 is bipartite. Assume it is planar, then every face is even (has at least 4 edges). –Hence 4 f  2 m or 2 f  m. Do not adopt the previous Corollary directly By Euler’s theorem, f = m + 2 – n = 9 + 2 – 6 = 5 –Namely, 10 = 2 f  m = 9. 

9. 9/21 Kuratowski’s Theorem Two graphs are homeomorphic if they can be obtained from the same graph by inserting new vertices of degree 2 into its edges A graph is planar if and only if it contains no subgraph homeomorphic to K 5 or K 3,3 The latter are referred to as Kuratowski subgraphs

10. Planarity Test

11. 11/21 How do you draw a planar graph without regret ? This means that, besides keeping the current embedding planar, your embedding can also keep future options open. You will have to design an embedding “scheme” rather than obtain a “physical” ( 實體的 ) embedding

12. 12/21 Prior Results 1st approach –Hopcroft and Tarjan [1974],first O(m) time. –PATH ADDITION 2nd approach –Lempel, Even and Cederbaum[1967], O(n 2 ) time –VERTEX ADDITION –st-numbering, consecutive ones testing –Booth and Lueker [1976] used PQ-trees to test the consecutive ones property in O(m+n) time 3rd approach –Shih and Hsu [1999] used PC-trees for recognition and embedding. –EDGE ADDITION

13. A Brief Intro. to the Vertex Addition Approach of LEC

14. 14/21 Vertex Addition Approach of LEC 1.Keep the current partial planar graph connected 2.Keep those non-added vertices a connected subgraph (i.e. in the same face). 3.Apply a consecutive ones test every time a new vertex is added

15. 15/21 st-numbering (I) Consider a 2-connected graph G. Pick any two adjacent vertices s and t. Order the vertices of G into s, v(1),..., v(k), t such that s s v(i)v(i) v(i)v(i) v(i+1) v(i+1),…, t must be imbedded in the same face t t v(i+1)

16. 16/21 St-numbering(II) s v(i)v(i) t s v(i)v(i) t v(i+1) Depth-First-Search s v(i)v(i) t

17. 17/21 1=s 6=t 235 6 235 (a) B 1 (a’) (b) B 2 (b’) 1 2 6 3 5 453 6 35453 Bush Form (1)

18. 18/21 (c) B 2 ’ 1 2 6 354 53 (c’) 6 54533 (d) B 3 1 2 6545 (d’) 6 545 64 46 3 Bush Form (2)

19. 19/21 (e) B 3 ’ 2 6 54 5 46 3 (e’) 6 54 564 (f’) 6 54564 (f) B 4 2 6 5 5 6 3 1 1 65 4 Bush Form (3)

20. 20/21 (g) B 4 ’ 2 6 556 3 1 6 5 4 (g’) 655665 (h’) 6666 (h) B 5 2 6 6 3 1 6 5 4 6 Bush Form (4)

21. 21/21 (i) G=G 6 =B 6 2 6 3 1 5 4 (i’) Bush Form (5)