Introduction to Graph Theory Lecture 09: Distance and Connectivity

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 di

sconnects the graph. A graph G is called k-connected for some positive i

nteger k if G has a cut vertex if and only if The same terminology applies to the edges too.

G

0G

kG kG

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

G

1

Example 4.3

Find a minimal vertex cutset of order 1 and 2, and minimal edge cutset of size 2 and 4.

e

hb

i

da

c

jf g

Relation of and

Theorem: Given a connected graph G, we have

Proof: To see , we can simply remove the ed

ges 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

GGG

G G

GG

GG

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 state

ment 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 the

n

GCyx , HVx JVy Gyexe rad

(cont)

implies for some z Assume . This implies that v is on every

y-z geodesic. Then This implies , so contradictio

n Therefore x and y must be in the same componen

ts of G-v. Which block is the central block of our previo

us graph?

Gve rad Gvezvd rad, JVz

GGzvdvydzyd radrad1,,, Gradye GCy

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.

i

g

fa

b

ec

u

h

d

v

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: 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.

w v

ux

0.2 0.20.2

0.2

0.2

Vertex-Failure Model

Assumptions: Edges are perfectly reliable Vertices fail independently with same probability

A common problem: The probability that the network remain

connected.