Introduction to Graph Theory

Lecture 19: Digraphs and Networks

Introduction

Diagraph: A directed graph with directions on the edges.

Digraph are used to model problems where the direction of flow of some quantity is important.

Network: a digraph with limits placed on the quantity flown through a particular directed edge.

Directed Graph

A digraph consists of a finite nonempty set of vertices V(D) and a set of ordered pairs of distinct vertices called arcs.

For the graph below, the arc xy goes from x to y. x is adjacent to y y is adjacent from x

x y

More Terminology

od(v): outdegree. The number of vertices that v is adjacent to.

id(v): indegree. The number of vertices that v is adjacent from.

Transmitter: a vertex v with id(v)=0. (only sending information, but receiving none)

Receiver: a vertex v with in(v)=0.

Our first simple theorem

Theorem 10.1: If D is a digraph with vertex set and having q arcs, then nvvvDV ,,, 21

qvvn

ii

n

ii

11

odid

Connectivity

Weakly connected: A connected digraph with pairs of vertices not accessible from each other.

Unilateral: for every pair of vertices (u,v) there is either a directed u-v or v-u path.

Strongly connected: for every pair of vertices (u,v) there are both directed u-v and v-u paths.

Strong Orientation

Assigning a direction to each edge is orienting the graph.

If the resulting graph becomes strongly connected, it is strong orientation.

Q1: Is there any way that we could make a graph with a bridge strongly oriented?

Q2: Is strong orientation always possible if G contains no bridge?

Achieving Strong Orientation

Theorem 10.2: A connected graph G has a strong orientation if and only if it contains no bridges, i.e. every edge is in some cycle.

To obtain a strong orientation, we use our favorite search strategy --- DFS Obtain a tree T using DFS Orient each edge of T toward the vertex with

higher number Orient the remaining edges of G toward the vertex

with lower number.

Example

Can we convince that every vertex can reach the root?

Acyclic Dgraphs A digraph that has no directed cycle is called

acyclic. Theorem 10.3: Every acyclic digraph has at l

east one vertex of outdegree zero and at least one vertex of indegree zero. Proof:

Consider the last vertex v in any longest path in the digraph, od(v)=0

Consider the first vertex u of a longest path P, id(v)=0

u x y vw

Applications of Acyclic Digraphs A partially ordered set (poset) is often modele

d by using an acyclic diagraph. A partial order on a set is a relation that is

Reflexive (a~a, “a is related to a”) Antisymmetric (a~b and b~a implies a=b) Transitive (a~b and b~c implies a~c)

Examples of such relations are “less equal” and “a subset of”

Example

Consider the set A={2,3,5,6,10,12,15,39} with the relation | (divide) Is the relation a partial order If so, draw the associated acyclic digraph What is the acyclic digraph if we omit the transitiv

e arcs? Do you verify theorem 103 with this poset?

Tournaments

Since some of you are volleyball player, this topic might interest you.

There are two kinds of tournament: Elimination tournament --- once a team loses a

game, it is out of the competition Round-robin tournament --- each team plays each

other team exactly once. We’ll focus out discussion on round-robin

tournaments.

Round-Robin Tournament

A tournament is a directed graph. A tournament on n vertices is an orientation of Kn. An arc from u to v indicates that vertex u defeated

vertex v. How many possible outcome for a tournament of 3

teams and 4 teams? What we would like to do is to rank the players from

best to worst, which is a hard task Consider a tournament of 5 teams

(cont)

However, it is still possible to arrange the players on a list so that player i beats player i+1 for .

The next theorem should convince us the statement.

Theorem 10.4: Every Tournament contains a directed hamiltonian path.

nppp ,,21,

ni1

Proof of Theorem 10.4

Proof by induction Basic case: True for Hypothesis: True for every tournament with n=k We want to prove that it is true for a tournament w

ith k+1 teams. Let’s consider the the T-v for any v. there is a h-path

. Let vi be the first vertex for which , then the h-path

in T is If no such vi then the h-path is

3n

kvvvP ,,, 21 Tvvi

kiiT vvvvvvP ,,,,,,, 1121 vvvvP k ,,,, 21