9.7 Planar Graphs

1. 9.7 Planar Graphs

2. Intro problem- 3 houses and 3 utilities  K3,3 problem: Can 3 houses be connected to 3 utilities so that no 2 lines cross? Similarly, can an isomorphic version of K3,3 be drawn in the plane so that no two edges cross? T X Y houses U V W utilities

3. Planar Def Def: A graph is called planar if it can be drawn in the plane without any edges drawing.

4. Sketchpad examples Check examples-- See Fig01 for K4, K5, K2,3, K3,3

5. Q3, Q4 Check examples— Fig02 for Q3, Q4

6. Sketch05

7. Sketch06

8. Sketch07

9. Proof that K3,3 is not planar see Fig 01 and Sketch 08 and Math Teacher article

10. Proof Consider a subgraph T 2 U 1 V X

11. …proof Pf. 1: Case 1: W is outside the graph (region 2). This forms region 2a and 2b. Y must be adjacent to U, V, W…. T   U V W   X

12. Case 2 Case 2: W is inside the graph (region 1). This forms regions 1a and 1b Again, Y is adjacent to U, V, and W…. T U W V X

13. Is K5 planar? (see Fig01)

14. Claim: K5 is nonplanar. Proof: By contradiction… Suppose there is a planar representation of K5. 2 So v1, v2, v3, v4, v5 form a pentagon. 1 3 5 4 {v1,v3} must be present. WLOG, let it be on the inside. Then construct {v2,v4} and {v2,v5} on the outside. So __________ are on the __________  

15. Find # of regions, edges, vertices to discover Euler’s formula

16. Theorem 1: Euler’s Formula Thm: Let G be a connected planar simple graph with e edges and v vertices. Let r be the number of regions in a planar representation of G. Then r = _______  Proof: First, specify a planar representation of G. We will prove by specifying a sequence of subgroups G1, G2, … Ge =G, adding an edge at each step. This is possible because G is connected. Arbitrarily pick an edge of G to obtain G1. Obtain Gn from Gn-1 by arbitrarily adding an edge that is incident with a vertex in Gn-1, adding the other vertex if necessary. 

17. …proof outline By induction: Basis: e=1 G1 r 1 = ___ e 1 = ___ v 1 = ___ So_________  Inductive step: Assume n and show n+1. This means: Assume r n = e n – v n + 2 and add {a n+1, b n+1 } to G n to obtain G n+1 and show ___________…  Case 1: an+1, bn+1 Gn R is split into 2 regoins. r n+1 = ___ e n+1 = ___ v n+1 = ___  So ___________ Case 2: an+1  Gn but bn+1  Gn r n+1 = ___ e n+1 = ___ v n+1 = ___

18. Question: How do you prove a graph is either planar or not planar? • To prove it is… • To prove it isn’t…

19. Corollary 1: e≤3v-6 Corollary 1: If G is a connected planar simple graph with e edges and v vertices where v≥3, then e≤3v-6. Def: deg(R )= number of edges on the boundary of region R Proof: Assume G is simple. Therefore it has no loops or multiple edges. So it has no regions of degree 1 or 2. A planar representation of G has r regions, each of degree at least 3. Note: 2e = ≥ 3r So r ≤ (2/3)e Using Euler’s Theorerm, r = ______ ≤ _____ …

20. Ex: Use the contrapositive of Corollary 1 to prove that K5 is nonplanar.

21. Q: Can you use Corollary 1 to show K3,3 is nonplanar?

22. Cor. 2– region degree ≤ 5 A Corollary of Cor. 1 is the following: Cor. 2: If G is a connected planar simple graph, then G has a vertex of degree not exceeding 5. Proof: • Case 1: G has 1 or 2 vertices: result _______ • Case 2: G has at least 3 vertices. By ___, we know e ≤ ____ so 2e ≤ ____ • To show result, assume degree of every vertex is ______. Then because 2e = _____ by _______,we have 2e ≥ 6v (why?). • But this contradicts ___. So there must be a vertex with degree ≤ 5.

23. Corollary 3: e ≤ 2v – 4 Corollary 3: If G is a connected planar simple graph with e edges, v vertices, v≥3, and no circuits of length 3, then e≤2v – 4. Proof . Assume G is simple. Consider a planar representative of G. Therefore it has no loops or multiple edges, which would create regions of degree 1 or 2. With no circuits of length 3, there are no regions of degree 3. Therefore, all regions are at least degree 4. So 2e = ________ ≥ 4r Solving for r… By Euler’s Formula…

24. K3,3 Ex: Use Corollary 2 to prove that K3,3 is nonplanar.

25. Is Q4 planar or not? Prove.

26. Kuratowski’s Theory Def: Replacing {u,v} with {u,w} and {w,v} is an elementary subdivision. Def: G1=(V1,E1) and G2=(V2,E2) are homeomorphic if they can be obtained from the same graph by a sequence of elementary subdivisions. Kuratowski’s Theorem: A graph is nonplanariff it contains a subgraphhomeomorphic to K3,3 or K5. Proof: clear beyond scope of class

27. examples • Are the following planar or not? Why?

28. Peterson ex   Ex: Use the two Euler Corollaries on the Peterson example (See examples in notes) Cor. 1: e ≤ 3v - 6 Cor. 3: e ≤ 2v - 4

29. Kuratowski’s Theory Ex: Use Kuratowski’s Theory on the Peterson example.   see written handout

30. More ex • See handout for more examples using Euler and Kuratowski • See sketch06 and sketch07 on sketchpad to the right • See attached Sketchpad handout with 7 more ex

31. More – handout ex 1-2

32. Handout 3-4

33. Handout 5-7

34. Ex- book 7, 23