Introduction to Graph Theory Lecture 09: Distance and Connectivity

Connectivity Plays an important role in reliability of computer networks Removal of one or more vertices will break the graphs into several components In this lecture, we’ll discuss several connectivity concept

Cut Vertices and Bridges Cut vertex: A vertex of which the deletion disconnects the graph. End vertices cannot be cut vertices A deletion of such a vertex increases the number of components of G Bridge (cut edge): Removal of such an edge increases the number of components. Every edge of a tree is a bridge

Vertex Connectivity Denoted is the minimum number of vertices whose deletion disconnects G or makes G trivial. If G is disconnected then A vertex cutset contains vertices whose removal disconnects the graph. A graph G is called k-connected for some positive integer k if G has a cut vertex if and only if The same terminology applies to the edges too.

Edge Connectivity Denoted is the minimum number of edges whose deletion disconnects G or makes G trivial. An edge cutset contains edges whose removal disconnects the graph. A connected graph has a bridge if and only if

Example 4.3 Find a minimal vertex cutset of order 1 and 2, and minimal edge cutset of size 2 and 4. a d c e b h j f g i

Relation of and Theorem: Given a connected graph G, we have Proof: To see , we can simply remove the edges of the vertex with minimum degree. If S is the edge cutset consisting of k edges, then removal of k suitably chosen vertices removes the edges of S. Thus

Blocks A block is a maximal connected subgraph of G with no cut vertices. What are the blocks for the graph below?

Blocks Theorem: The center of a connected graph G belongs to a single block of G. (We call such a block central block) Proof: By contradiction If G is connected without cut vertex, then the statement is true Assume that G has one cut vertex v, and removal of v results in two components H and J Suppose that , s.t. and then

(cont) Which block is the central block of our previous graph? implies for some z Assume . This implies that v is on every y-z geodesic. Then This implies , so contradiction Therefore x and y must be in the same components of G-v. Which block is the central block of our previous graph?

Menger’s Theorem Menger showed that the connectivity of a graph is related to the number of disjoint paths joining two vertices. Two paths connecting u and v are internally disjoint u-v paths if they have no vertices in common other than u and v. Two paths are edge disjoint if they have no edges in common. A set S of vertices or edges separate u and v if every path connecting u and v passes through S.

Menger’s Theorem Menger’s Theorem: Let u and v be distinct nonadjacent vertices in G. Then the maximum number of internally disjoint paths connecting u and v equals the minimum number of vertices in a set that separate u and v. (Proof omitted)

Example 4.5 Finding the minimum order separating set and a maximum set of internally disjoint u-v path. a f b i g v u e c h d

More Theorem Let u and v be distinct nonadjacent vertices in G. Then the maximum number of edge-disjoint paths connecting u and v equals the minimum number of edges in a set that separate u and v (called minimum cut). Let’s try out this theorem on the previous graph.

Application: Network Reliability You can picture a “network” as a weighted graph, where the weights are probabilities. Network reliability concerned with how well a given network can withstand failure of individual components of a system. There are several reliability models

Edge-Failure Model Assumptions: A common problem: Vertices totally immune to failure All edges fail independently with equal probability Failure happen simultaneously A common problem: K-terminal reliability: to determine the probability that a subset K of terminal vertices remain connected to one another.

Example Find the probability that vertices u and v remain in the same component. u x 0.2 0.2 0.2 0.2 0.2 w v

Vertex-Failure Model Assumptions: A common problem: Edges are perfectly reliable Vertices fail independently with same probability A common problem: The probability that the network remain connected.