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Introduction to Graph Theory Lecture 19: Digraphs and Networks.

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1 Introduction to Graph Theory Lecture 19: Digraphs and Networks

2 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.

3 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

4 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.

5 Our first simple theorem Theorem 10.1: If D is a digraph with vertex set and having q arcs, then

6 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.

7 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?

8 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.

9 Example Can we convince that every vertex can reach the root?

10 Acyclic Dgraphs A digraph that has no directed cycle is called acyclic. Theorem 10.3: Every acyclic digraph has at least 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 v w

11 Applications of Acyclic Digraphs A partially ordered set (poset) is often modeled 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”

12 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 transitive arcs?  Do you verify theorem 103 with this poset?

13 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.

14 Round-Robin Tournament A tournament is a directed graph. A tournament on n vertices is an orientation of K n. 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

15 (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.

16 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 with k+1 teams. Let’s consider the the T-v for any v. there is a h- path. Let v i be the first vertex for which, then the h-path in T is If no such v i then the h-path is

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