Ma/CS 6b Class 5: Graph Connectivity

Transcription

1 Ma/CS 6b Class 5: Graph Connectivity By Adam Sheffer Communications Network We are given a set of routers and wish to connect pairs of them to obtain a connected communications network. The network should be reliable a few malfunctioning routers should not disable the entire network. What condition should we require from the network? That after removing any k routers, the network remains connected. 1

2 k-connected Graphs An graph G = (V, E) is said to be k-connected if V > k and we cannot obtain a non-connected graph by removing k 1 vertices from V. Is the graph in the figure 1-connected? Yes. -connected? Yes. 3-connected? No! Connectivity Which graphs are 1-connected? These are the connected graphs ( V > 1). The connectivity of a graph G is the maximum k such that G is k-connected. What is the connectivity of the complete graph K n? n 1. The graph in the figure has a connectivity of.

3 Hypercube 1 A hypercube is a generalization of the cube to any dimension. In d-dimensions, we consider the vertices of the hypercube as the points whose coordinates are 0 s and 1 s. Two vertices are adjacent if they differ in a single coordinate. (0,1) (1,1) 0 (0,0) (1,0) Hypercube Properties 1 The points of a d-dimensional hypercube correspond to the d-dimensional vectors with 0-1 entries. Thus, it has d vertices. Two vertices are adjacent if they have d 1 common coordaintes. Thus, there are d 1 d edges. (0,1) (1,1) 0 (0,0) (1,0) 3

4 Hypercube graphs The hypercube graph Q d corresponds to the d-dimensional hypercube. A vertex for each vertex of the hypercube. An edge for each edge of the hypercube. The graph Q d has d vertices and d 1 d edges. It is d-regular. 4d-hypercube Hypercube graph Properties Problem. Is Q d a bipartite graph? Answer. Yes! On one side we place every vertex with an even number of 1-coordinates. On the other, the vertices with an odd number of 1-coordintes. 4d-hypercube 4

5 More Properties Problem. What is the connectivity of Q d? Answer. We can disconnect Q d by removing the d neighbors of any vertex of Q d. So the connectivity is at most d. We prove by induction that it is exactly d. Q 4 Proof by Induction We prove by induction that the connectivity of Q d is d. Induction basis. Easy to check for d = 1,. Induction step. We can consider Q d as two copies Q, Q of Q d 1 with a perfect matching between their vertices. Q Q Q d 5

6 Completing the Proof S a minimum disconnecting set of Q d. To complete the proof, we claim that S = d. If after removing S both Q and Q connected, it must disconnect Q from Q, and must be of size at least d 1 d. By the induction hypothesis, disconnecting Q (or Q ) requires removing d 1 vertices of Q. To disconnect Q in Q d, we must remove at least one additional vertex from Q. k-edge Connected Graphs A graph G = (V, E) is said to be k-edgeconnected if V > 1 and we cannot obtain a non-connected graph by removing at most k 1 edges from E. Is the graph in the figure 1-edge-connected? Yes. -edge-connected? Yes. 3-edge-connected? Yes. 6

7 k-edge-connectivity Which graphs are 1-edge-connected? These are the connected graphs ( V > 1). The edge-connectivity of a graph G is the maximum k such that G is k-edge-connected. What is the edge-connectivity of the complete graph K n? n 1. The graph in the figure has an edge-connectivity of 3. Edge Connectivity and Minimum Degree What is the relation between the minimum degree and the edge-connectivity of a graph? Can they be equal? Yes Can the edge connectivity be smaller? Yes Can the minimum degree be smaller? 7

8 Edge Connectivity and Minimum Degree (cont.) Claim. In any graph, the edge connectivity is at most the minimum degree. Proof. Consider a vertex v of minimum degree, and denote this degree as d. By removing the d edges that are adjacent to v, we disconnect the graph. Connectivity and Edge-Connectivity What is the relation between the connectivity and the edge-connectivity of a graph? Can they be equal? Yes Can the connectivity be smaller? Yes Can the edge-connectivity be smaller? 8

9 Connectivity and Edge-Connectivity Claim. In any graph G = V, E, the connectivity is at most the edge-connectivity. Proof. Let F E be a minimum set of edges whose removal disconnects G. We need to find a set of at most F vertices that disconnects G. We divide the analysis into two cases: A vertex v V not adjacent to any edge of F. Every vertex of V is adjacent to an edge of F. Analysis of Case 1 Assume that a vertex v V is not adjacent to any edge of F. By definition, the removal of F disconnects G. Let C be the connected component in G F that contains v. Removing the vertices V in C that are adjacent to an edge of F disconnects v from the vertices of G C. Since F is minimal, each edge of F has at most F v one endpoint in C, so V F. 9

10 Analysis of Case Assume that every vertex of V is adjacent to an edge of F. If G is a complete graph, the connectivity is V 1 and edge-connectivity is V 1. Analysis of Case (cont.) Assume that every vertex of V is adjacent to an edge of F. If G is not complete, there exists v V such that deg v < V 1. Removing the set V of neighbors of v disconnects G. C the connected component of v in G F. Every vertex of V is either in C or connected to v by an edge of F. Since no edge of F is between two vertices of C, we have F V F. v 10

11 Connectivity in 3-regular Graphs Claim. If G = V, E is a connected 3-regular graph, then the connectivity and the edge-connectivity of G are equal. Proof k v (vertex) connectivity. k e edge connectivity. We already know that k v k e. Thus, it suffices to find a set of k v edges that disconnects G. S a minimum set of k v vertices whose removal disconnects G into C 1 and C (which might not be connected themselves) 11

12 Proof (cont.) Every vertex of v S is connected to both C 1 and C, since S is a minimum disconnecting set. Due to 3-regularity, either v is adjacent to a single vertex of C 1, or to a single vertex of C. That is, by removing a single edge, we can disconnect v from either C 1 or from C. By removing such an edge from each v S, we obtain a disconnecting set of k v edges. Recap We proved that in every graph: Edge Connectivity connectivity Minimum degree This implies that small minimum degree implies small connectivity. Does large minimum degree imply large connectivity. No! 1

13 Highly Connected Subgraphs Recall. The average degree of a graph G = V, E is deg G = 1 V v V deg v = E V. Theorem (Mader `7). Let k be a positive integer and let G = V, E be a graph such that deg G 4k. Then there exists a k + 1 -connected subgraph H G with minimum degree > deg(g)/. Proof Set d = deg G 4k. Let H = (V, E ) be a subgraph of G such that V k and E > d V k. Among the subgraphs that satisfy the above, we take as H one that minimizes V. Such subgraphs exist since G is one: Since d 4k, there exists a vertex of degree at least 4k, and thus V 4k + 1. E = E V V = d V > d V k. 13

14 Proof () Set d = deg G 4k. H = (V, E ) subgraph of G such that V k and E > d V k. If V = k, we get the contradiction E > d V k = d k k > k = V E. Proof (3) H = (V, E ) the minimal subgraph of G such that V k and E > d V k. We actually have V k + 1. Removing a vertex of degree d/ decreases d V k by d and E by at most d. That is, both V k and E > d k remains valid, contradicting the minimality of H. V 14

15 Proof (4) H = (V, E ) the minimal subgraph of G such that V k and E > d V k. We proved that every vertex of V is of degree larger than d/ in H. We need to show that H is (k + 1)-connected. Assume for contradiction that the removal of a set S of k vertices disconnects H. We can thus partition the vertices of V S into the subsets V 1, V with no edges between them. Proof (5) Assume for contradiction that the removal of a set S of k vertices whose removal disconnects H. Partition the vertices of V S into subsets V 1, V with no edges between them. H 1 the subgraph induced by V 1 S. Consider a vertex v V 1 and notice that all the edges that are adjacent to it remain in H 1. Since deg v > d k, we have V 1 S > k. S V 1 V 15

16 Illustration V 1 V S H 1 H S S Proof (6) The subgraph H 1 has more than k vertices. In order not to contradict the minimality of H, the number of edges of H 1 must be d V 1 S k. Similarly, the number of edges of H must be d V S k. E d V 1 S k + d V S k = d V k 16

17 The End 17