Discrete Wiskunde II. Lecture 6: Planar Graphs

Transcription

1 , 2009 Lecture 6: Planar Graphs University of Twente wwwhome.math.utwente.nl/~uetzm/dw/

2 Planar Graphs Given an undirected graph (or multigraph) G = (V, E). A planar embedding of G is a drawing of G = (V, E) in R 2 such that no two edges e E cross. G is planar : planar embedding of G nonplanar... planar... embedding of one and the same (planar) graph, K 4

3 Planar Graphs In Practice VLSI design Map coloring

4 Planar Graphs Question Given an arbitrary graph G, how to figure out if G is planar or not? Answer Characterize (planar) graphs by means of forbidden sub-structures! Active research area within Graph Theory:

5 Planar Graphs: Examples K 1 K 2 K 3 K 4

6 K 5 planar?

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9 one edge crossing (seems) unavoidable

10 Why K 5 isn t planar: Intuition (but no proof) Jordan Curve Theorem (without proof) Let c be a simple closed curve (Jordan curve) in R 2. Then c separates R 2 into two disjoint components, the interior and the exterior, which both have c as boundary.

11 Bipartite Graphs Definition Graph G = (V, E) is bipartite if vertices can be partitioned into V 1 and V 2 such that no edge has both endpoints in V 1, and no edge has both endpoints in V 2. V 1 V 2

12 Bipartite Graphs Theorem Graph G is bipartite if and only if G has no cycle of odd length. Proof. G bipartite odd cycle in G (Exercise!) G has no odd cycle G bipartite: Pick arbitrary v 0, make BFS from v 0 to all other v V Define V 1 = {v V d(v) even} and V 2 = {v V d(v) odd} Because odd cycle, edge with both endpoints in V 1 or V 2 Characterization of bipartite graphs via forbidden substructure!

13 Complete Bipartite Graphs K n,m : bipartite graph ((V 1, V 2 ), E) with V 1 = n and V 2 = m and all edges are present V 1 K 2,3 V 2

14 K 2,3

15 K 2,3 is planar

16 K 3,3 planar? V 1 K 3,3 V 2

17 K 3,3 planar? V 1 K 3,3 V 2

18 K 3,3 planar? V 1 K 3,3 V 2

19 K 3,3 planar? V 1 K 3,3 V 2 again, Jordan Curve Theorem suggests K 3,3 is non-planar but this is intuition, not a proof.

20 Kuratowski s Theorem Minimal non-planar graphs (not yet proved) Complete graphs: K 5 Complete bipartite graphs: K 3,3 Theorem (Kuratowski, 1930) A graph is planar if and only if it contains no subgraph that is homeomorphic to either K 5 of K 3,3. forbidding substructures K 5 and K 3,3 characterizes planar graphs Proof. We ll only prove that the condition is necessary: Any graph homoeomorphic to K 5 or K 3,3 isn t planar.

21 Homeomorphic Graphs Definition A subdivision of G results from successively removing one edge {u, w} and replacing it by two edges {u, v} and {v, w}. Definition Two graphs G 1 and G 2 are homeomorphic if they are both subdivisions of one and the same graph H. Homeomorphic graphs are either both planar or both non-planar.

22 Kuratowski s Theorem Let us assume that neither K 5 nor K 3,3 are planar Necessity G contains subgraph homeomorphic to K 5 or K 3,3 G itself can t be planar (this is clear) Sufficiency G contains no subgraph homeomorphic to K 5 or K 3,3 G is planar (this is harder to prove) But: We still need a formal proof of first claim

23 Regions Definition Given a planar embedding of a planar graph G in R 2. Let r be the number of (topologically) connected components, or regions of G. Figure: embedding K 4 with 4 regions (R 4 is infinite)

24 Euler Formula for Planar Graphs Theorem 11.6 Let G be a planar, connected graph and V = n and E = m, then n m + r = 2 Proof. By induction on m, the number of edges of G. Inductive start: look at m = 0 (case a) or m = 1 (case c, or if loops are allowed, case b).

25 Euler Formula: Proof n m + r = 2 Inductive step: assume formula holds for all planar, connected graphs with m 1 edges. Consider arbitrary graph G with m edges. case 1: G has a cut with only one edge e (a bridge ), consider two connected components G 1 and G 2 of G e, ind. hypothesis for G 1 and G 2 : n 1 m 1 + r 1 = 2 and n 2 m 2 + r 2 = 2, and we know: n = n 1 + n 2, m = m 1 + m 2 + 1, r = r 1 + r 2 1 case 2: all edges e of G are on some cycle, ind. hypothesis for connected graph G 1 = G e (any e): n 1 m 1 + r 1 = 2 G e has exactly one edge and one region less than G, so n = n 1, m = m 1 + 1, r = r 1 + 1

26 Euler Formula Corollary Let G be a planar, connected multi-graph and V = n and E = m, then n m + r = 2 Proof. Each parallel edge increases both m and r by exactly 1 Corollary 11.3 Let G be a planar, connected graph without loops, V = n and E = m 2, then m 3n 6 planar graphs have few edges, namely only O( n ) edges (arbitrary graphs can have as many as Ω( n 2 ) edges)

27 Finally, K 5 and K 3,3 Theorem K 5 is not planar. G planar n m + r = 2 and m 3n 6 Proof. If K 5 was planar, m 15 6 = 9, but m = 10 Theorem K 3,3 is not planar. Proof. Assume K 3,3 planar. Each region in embedding of K 3,3 is bounded by 4 edges, each edge has 2 neighboring regions Thus we conclude 4r 2 m, but r = m n + 2 = = 5, so 10 9, contradiction

28 Dual of a Planar Graph Define dual (multi-) graph G d for G vertex u in G d : Region R u in G edge e = {u, v} in G d : edge between regions in R u and R v in G ( regions share an edge )

29 Map Coloring Feasible coloring of the regions of planar graph G (the map ) Coloring of vertices of G d s.t. adjacent vertices get different colors. Theorem (Appel & Haken 1976) Any map can be feasibly colored with 4 colors, i.e., the dual of any planar graph is 4-colorable (also called 4-partite)

30 4 Colors