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II Graph Isomorphism Cayley's Formula Planar Graphs Is K5 Planar?

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II Graph Isomorphism Cayley's Formula Planar Graphs Is K5 Planar? Great Theoretical Ideas In Computer Science Victor Adamchik CS 15-251 Graph Isomorphism Lecture 9 Carnegie Mellon University Graphs - II Definition. Two simple graphs G and H are isomorphic G H if there is a vertex bijection VH->VG that preserves adjacency and non-adjacency structures. Cayley’s Formula Planar Graphs n-2 Theorem: In any connected The number of labeled trees on n nodes is n planar graph with V vertices, E edges and F faces, then Put another way, it counts the number of V – E + F = 2 spanning trees of a complete graph Kn. 4 2 Theorem: In any connected planar graph with at 1 5 least 3 vertices: P = 5, 1, 1, 5 E ≤ 3 V - 6 3 6 Lemma: In any connected planar graph with at We proved it by finding a bijection between the least 3 vertices: set of Prϋfer sequences and the set of labeled trees. 3 F ≤ 2 E Outline Is K5 planar? Bipartite Graphs Kuratowski Theorem K5 has 5 vertices and 10 edges, thus Graph Coloring E = 10 ≤ 3x5 – 6 = 9 Bipartite Matching which is false, therefore K5 is not planar. 1 Bipartite Graphs Is K3,3 planar? Theorem: In any connected A graph is bipartite if the planar graph with at least 3 vertices can be partitioned vertices: into two sets V and V such 1 2 E ≤ 3 V - 6 that all edges go only between V1 and V2 (no edges go from V1 to V1 or from V2 to V2) K3,3 has 5 vertices and 9 edges, The complete bipartite graphs Km,n have the property that two vertices are adjacent if and thus only if they do not belong together in the E = 9 ≤ 3x6 – 6 = 12 bipartition subsets. Not conclusive! Is K planar? 3,3 Planar Bipartite Graphs ∑(edge, face) ≤ 2 E, since each The previous example established two simple edge is associated with at most criteria for testing whether a given planar graph 2 faces is bipartite. ∑(edge, face) ≥ 4 F , since Theorem. In any bipartite planar graph with at graph contains no simple least 3 vertices: triangle regions of 3 edges. It follows, that E ≤ 2 V - 4 4 F ≤ 2 E and for K we have 3,3 Lemma: In any bipartite planar graph with at 4F ≤ 18 least 3 vertices: F ≤ 4.5 From Euler’s theorem: V – E + F = 2 4 F ≤ 2 E F = 2 + 9 – 6 = 5. Contradiction! Kuratowski Theorem (1930) Subdivision and Contraction Definition. Subdividing an edge means inserting a Theorem. A graph is planar if and only if it new vertex (of degree two) into this edge. contains no subgraph isomorphic to a subdivision of K5 or K3,3. A a B b Petersen e E graph For any graph on V vertices there are efficient d algorithms for checking if the graph is planar. c The best one runs in linear time O(V) C D 2 Theorem. A graph is planar if and only if it Theorem. A graph is planar if and only if it contains no subgraph isomorphic to a contains no subgraph isomorphic to a subdivision of K5 or K3,3. subdivision of K5 or K3,3. A is subdividing (a,e) A A Petersen a graph a b is b e subdividing B b (d,e) e E E Remove B b to get a c c d subgraph d C is subdividing C D C D (c,d) Theorem. A graph is planar if and only if it contains no subgraph isomorphic to a Subdivision and Contraction subdivision of K5 or K3,3. e D Definition. Subdividing an edge means inserting a a new vertex (of degree two) into this edge. a e E c Definition. An edge contraction is an operation c which removes an edge from a graph while d d E simultaneously merging the two vertices it used to connect. D Wagner Theorem Coloring Planar Graphs Theorem. Graph G is planar if and only if it A coloring of a graph is an assignment of a contains no subgraph that can be contracted color to each vertex such that no neighboring to one of the two Kuratowski subgraphs. vertices have the same color Is the Petersen graph planar? 3 Graph Coloring Graph Coloring Theorem: Any simple planar graph can be colored with 6 colors. Theorem : Every simple planar graph has a vertex of degree at most 5. Proof. (by induction on the number of vertices). Proof. If G has six or less vertices, then the result is ∑deg(vk) = 2 E ≤ 2 (3 V – 6) obvious. Suppose that all such graphs with V-1 vertices are 6-colorable Average degree: Remove a vertex of degree less than 6, use IH. 1/V ∑deg(vk) ≤ 6 – 12/V < 6 Put it back, since it has at most 5 adjacent Thus, there exists a vertex of degree at most 5. vertices, we have enough colors. QED Graph Coloring Theorem: Any simple planar graph can be colored with less than or equal to 5 colors. Proof. (repeat the 6-colors proof) Pick a vertex v of degree 5. Label the Remove edges (v, x1), (v, x2) and (v, x3). vertices adjacent to v as x1, x2, x3, x4 and x5. Contract edges (v, x4), (v, x5). Vertices v, x4, x5 Assume that x4 and x5 are not adjacent to each will be replaced by y, so neighbors of v, x4, x5 other. Why we can will be neighbors of y. assume this? We obtain a new graph H with two less vertices. By IH the graph H can be colored with 5 colors. If they all Next, we assign y-color to x and x adjacent, 4 5 We give v a color different from all colors used on we get K5. the four vertices x1, x2, x3 and y. QED 4 Color Theorem (1976) Bipartite Matching Theorem: Any simple planar graph can be A graph is bipartite if the vertices can be colored with less than or equal to 4 colors. partitioned into two disjoint (also called independent) sets V1 and V2 such that all edges go only between V1 and V2 (no edges go from V1 to V1 It was proven in 1976 by K. Appel and W. Haken. or from V to V ) They used a special-purpose computer program. 2 2 Since that time computer scientists have Personnel Problem. You are the boss been working on developing a formal program proof of a company. The company has M of correctness. The idea is to write code that workers and N jobs. Each worker is describes not only what the machine should do, but qualified to do some jobs, but not also why it should be doing it. others. How will you assign jobs to each worker? In 2005 such a proof has been developed by Gonthier, using the Coq proof system. 4 Bipartite Graphs Bipartite Graphs Theorem. A graph is bipartite iff it does not have Theorem. A graph is bipartite iff it does not have an odd length cycle. an odd length cycle. ) Fix a vertex v. Define two sets of vertices A ={w V | even length Proof. ) shortest path from v to w} If it’s bipartite and B ={w V | odd length has a cycle, its shortest path from v to w} length must be even. If x and y from A, they cannot be adjacent. By contradiction. There will be an odd length cycle. The same argument for B. These sets provide a bipartition. Bipartite Matching Definition. A subset of edges is a matching if no two edges have a common vertex (mutually disjoint). Is a tree always a bipartite graph? Definition. A maximum matching is a matching with the largest possible number of edges Bipartite Matching Hall’s (marriage) Theorem Definition. A perfect matching is a matching in Theorem. (without proof) which each node has exactly one edge incident on it. Let G be bipartite with V1 and V2. For any set SV1, let N(S) denote the set of A perfect matching is like a bijection, which vertices adjacent to vertices in S. requires that |V1| = |V2 | and in which case its inverse is also a bijection. Then, G has a perfect matching if and only if |S| ≤ |N(S)| for every SV1. 5 Alternating Path Augmenting Matching A matching M has some matched and some If a matching M (in green) has an augmenting path, unmatched vertices. (y1,x2),(y3,x4) then we get a larger matching by swapping the edges on the augmenting path. Alternating path has edges alternating between M and E - M. Path x1, y1, x2, y3, x4 is alternating. An alternating path is augmenting if both of x1 x2 x3 x4 x1 x2 x3 x4 its endpoints are free vertices. x1 x2 x3 x4 y1 y2 y3 y4 y1 y2 y3 y4 Path x1, y1, x2, y3, x4, y4 is augmenting. y1 y2 y3 y4 Hungarian Algorithm The algorithm starts with any matching and constructs a tree via a breadth-first search to What is the runtime find an augmenting path. complexity of the Hungarian algorithm? If the search succeeds, then it yields a matching having one more edge than the original. Then we search again (it most it happens is V/2) Complexity of BFS – O(V+E) for a new augmenting path. If the search is unsuccessful, then the algorithm terminates and must be the largest-size matching We run it V/2 times that exists. This, the runtime is O(V E).
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