Ma/CS 6b Class 4: Matchings in General Graphs

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1 Ma/CS 6b Class 4: Matchings in General Graphs By Adam Sheffer Reminder: Hall's Marriage Theorem Theorem. Let G = V 1 V 2, E be a bipartite graph. There exists a matching of size V 1 in G if and only if for every A V 1, we have A N A. We say that G satisfies Hall s condition if for every A V 1, we have A N A. Philip Hall 1

2 k-regular Graphs A graph G = V, E is k-regular if each vertex of V is of degree k. The Petersen graph is a 3-regular graph with ten vertices: Example: k-regular Graphs For any positive integer n, what graph do we know that is n-regular? K n+1 the complete graph with n + 1 vertices (every two vertices are connected). K 7 2

3 Example: 2-regular Graphs For any positive integer n, what graph do we know that is 2-regular and has n vertices? C n a cycle of size n. C 6 1-regular Graphs Problem. Characterize the graphs that are 1-regular. Answer. These are the graphs whose edges form a perfect matching (with no additional edges). 3

4 2-regular Graphs Problem. Characterize the graphs that are 2-regular. Answer. These are the graphs that consist only of vertex-disjoint cycles. k-factors Consider a graph G = V, E. A spanning subgraph of G is a subgraph that contains all of the vertices of V. For a positive integer k, a k-factor of G is a spanning subgraph of G that is k-regular. A graph contains a 1-factor iff it contains a perfect matching. 1-factor 2-factor 4

5 Bipartite k-regular Graphs Claim. Every bipartite k-regular graph G = V 1 V 2, E contains a 1-factor. Proof. We have V 1 = V 2, by double counting the size of E. We complete the proof by showing that G satisfies Hall s condition. For any A V 1, there are k A edges that leave A. Since every vertex of N A is incident to at most k edges from A, then N A A. 2-factor Theorem Theorem. Every 2k-regular graph G = V, G has a 2-factor (where k is a positive integer). One of the earliest works on graph theory by Petersen in Julius Petersen 5

6 Recall: Eulerian Cycles An Eulerian path in a graph G is a path that passes through every edge of G exactly once. An Eulerian cycle is an Eulerian path that starts and ends at the same vertex. a b h g f e c d E = a b c d e f g h a d h e b g c f a Reminder: Graphs Containing Eulerian Cycles Theorem. A connected (possibly not simple) graph G = V, E contains an Eulerian cycle if and only if every vertex of V has an even degree. 6

7 Proof (of 2-factor Theorem) We assume that G is connected. (otherwise, we can consider separately each component). Since every degree in G is 2k, it contains an Eulerian cycle C. We arbitrarily choose a direction for C. We split every vertex v V into v + and v. An edge in C that went from v to u now goes from v + to u. Illustration a d a d b c e b c e a + b + c + a b c d + e + d e * Unlike our case, this graph is not 2k-regular. 7

8 Proof (cont.) We obtain a graph G with 2 V vertices and E edges. We ignore the directions in G and notice that it is a bipartite k-regular graph. a + b + a b d + e + d e c + c Completing the Proof We started with a 2k-regular graph G. There is an Eulerian cycle in G. Splitting every vertex v V into v + and v, we obtained a bipartite k-regular graph B with the same set of edges. B contains a perfect matching M. By merging every pair v + and v back to v, M becomes a 2-factor! 8

9 Another Illustration a + b a c + a b c d + f + b + e d c d e + f e f Merging the vertices of the perfect matching results in a 2-factor. Matchings in General Graphs In bipartite graphs, Hall s condition characterizes the graphs that contain a perfect matching. We now present a similar condition for general graphs, which we denote as Tutte s condition. W. T. Tutte 9

10 Today: Mathematical Facts About Nicolas Cage Odd Components Given a graph G = (V, E), we denote by q G the number of components in G that consists of an odd number of vertices. q G = 4 10

11 Tutte s Condition We say that a graph G = V, E satisfies Tutte s condition if for every S V we have q G S S. S = 4 q G S = 3 The Easy Direction Claim. If a graph G contains a 1-factor M, then it satisfies Tutte s condition. Proof. For any S V, consider the components of G S. Every component with an odd number of vertices must have at least one edge of M connected to a vertex of S. Thus, q G S S. S 11

12 The Other Direction Claim. If a graph G = V, E does not contain a 1-factor, then it does not satisfy Tutte s condition. That is, we need to prove that there exists S V such that q G S > S. Given such a bad set S, if we remove edges from E, then S remains a bad set: If we split an odd sized component into several components, at least one of them is odd-sized. Illustration Removing edges can only increase q(g S). S = 3 q G S = 4 S = 3 q G S = 4 12

13 Rephrasing the Claim We know: if a graph G contains a bad set S, then S remains bad after removing any number of edges. We say that a graph G is edge-maximal without a 1-factor if it does not contain a 1-factor, but adding any one edge to it results in a 1-factor. Thus, it suffices to prove: Claim. If a graph G = V, E is edge-maximal without a 1-factor, then it does not satisfy Tutte s condition. Drowning in a Swimming Pools Correlation: * Taken from 13

14 Odd Number of Vertices If G has an odd number of vertices, it obviously contains no 1-factor. In this case, G does not satisfy Tutte s condition: By taking S = we have 0 = S < q G S = 1. It remains to consider graphs with an even number of vertices. Finding a Bad Set We take S V to be the set of vertices that are connected to every vertex of G (it is possible that S = ). We partition the analysis into two cases: Every component of G S is a complete graph (that is, a complete induced subgraph). There is a component of G S that is not a complete graph. 14

15 The First Case S the set of vertices that are connected to every vertex of G. Consider the case where every connected component of G S is complete. If S q G S, we get a contradiction to G not containing a 1-factor, so S is a bad set! The 1-factor has one edge between every odd component and a distinct vertex of S. G S S Dying in a Helicopter Accident Correlation:

16 The Second Case Assume that a component C of G S that is not a complete graph. There exist two vertices x, z in C such that x, z E but there exists y V S such that x, y, y, z E. Since y S, there exists a vertex w S such that y, w E. x y w S z Building a New Graph Recall that adding any edge to G results in a 1-factor contained in it. M xz 1-factor in G after adding x, z. M yw 1-factor in G after adding y, w. F the set of edges that are either in M xz or in M yw (but not in both). In the graph G F = V, F, every vertex is of degree 0 or 2. That is, G F consists of isolated vertices and disjoint even length cycles. 16

17 Union of Two Perfect Matchings z w z w x y x y M xz M yw z w x y F Different Cycles If x, z and y, w are in different cycles of F, then we can find a 1-factor in G: Take every edge that is in M xz M yw. From the each cycle of F, we take every other edge. We choose those such that we avoid taking x, z and y, w. This contradicts G not containing a 1-factor. z x 17

18 The Same Cycle If x, z and y, w are on the same cycle: If we can take every other edge of the cycle without taking x, z and y, w, we get a contradiction as before. Otherwise, we can use x, y or y, z to obtain a 1-factor containing neither x, z nor y, w. y w z x Concluding the Proof We proved that either S is a bad set, or we get a contradiction to G not containing a 1-factor. That is, we proved that if G is edgemaximal without a 1-factor, then there exists S V such that S < q(g S). Thus, in every graph G without a 1-factor there exists S V such that S < q(g S). 18

19 The End 19