Section 3.5 Directed Graphs - University of Maine

Section 3.5 Directed Graphs 1

Section Section Section Section 3333....5555 Di Di Di Directed Graphs : Tournament Graphsrected Graphs : Tournament Graphsrected Graphs : Tournament Graphsrected Graphs : Tournament Graphs

Purpose of Section: Purpose of Section: Purpose of Section: Purpose of Section: We introduce the directed graphdirected graphdirected graphdirected graph and its adjacency matrix,

and then show how directed graphs arise in connection with tournaments of

various types.

IntroductionIntroductionIntroductionIntroduction

Thus far in our in our discussions of graphs we have not associated a

“direction” with the edges of the graph. In the Konigsberg Bridge Problem,

we were allowed to traverse each edge in either direction. In many problems,

however, such as traffic problems with one-way streets, we must restrict the

direction of movement along the edges to only one direction. A directed graphdirected graphdirected graphdirected graph

(or digraphdigraphdigraphdigraph) is defined similarly to a (undirected) graph, except that the edges

of the graph are “directed” from one vertex to another. We call these types

of edges directed edgesdirected edgesdirected edgesdirected edges (sometimes called arcsarcsarcsarcs). If the directed edge goes

from vertex u to vertex v , then u is called the headheadheadhead (or sourcesourcesourcesource) and v is

called the tail tail tail tail (or sinksinksinksink) of the edge, and v is said to be the direct successor direct successor direct successor direct successor of

u and u is the direct predecessordirect predecessordirect predecessordirect predecessor of v . See Figure 1

A directed graph with directed edges joining the vertices

Figure 1

The applications of directed graphs range from transportation problems

in which traffic flow is restricted to one direction, one-way communication

problems, asymmetric social interactions, athletic tournaments, and even the

World-Wide Web. Before talking about applications, however, we introduce

the concept of the adjacency matrix of a directed graph.


Definition Definition Definition Definition The adjacency matrixadjacency matrixadjacency matrixadjacency matrix of a directed graph having n vertices is an

n n× matrix ( )ijM m= where

1 if there is a directed edge from node to node

0 if there is not a directed edge from node to node ij

i jm

i j

=

For example, the adjacency matrix for the digraph in Figure 2 is

0 0 1 1 0

1 0 1 1 0

0 0 0 0 0

0 0 1 0 1

1 1 1 0 0

M

=

Tournament Graphs (Dominance Graphs)Tournament Graphs (Dominance Graphs)Tournament Graphs (Dominance Graphs)Tournament Graphs (Dominance Graphs)

A tournament graphtournament graphtournament graphtournament graph (or dominant graphdominant graphdominant graphdominant graph) is a directed graph where every

pair of vertices is joined by exactly one directed edge. In other words, for

each pair of vertices u and v in a tournament graph, there is a directed edge

from to u v or from to v u , but not both. Figure 2 shows a tournament graph

with five vertices.

Tournament Graph with Five Vertices

Figure 2

Tournament graphs are aptly named since they often are applied to

tournaments of some kind. They are the familiar round-robin tournaments in


tennis, baseball, and so on where every team plays every other team exactly

once (and we assume no ties).

The number of directed edges “leaving” a vertex u is called the outoutoutout----

degreedegreedegreedegree of the vertex and denoted by ( )od u . In the directed graph in Figure 2

we have ( ) ( )od 5 3,od 1 2= = . If the vertices of the graph are athletic teams,

and a directed edge from vertex u to vertex v means team beats team u v ,

then the out-degree of u is the number of wins for the team u . On the other

hand, the number directed edges “entering” a vertex v is called inininin----degreedegreedegreedegree of

v and denoted by ( )id v . In Figure 2 we have ( ) ( )id 5 1, id 1 2= = . In connection

with athletic tournaments, it is the number of losses for team v .

Tournament graphs are even used by sociologists, who call them

dominance graphs, used in the study of social interactions.

Dominance Graphs in SociologyDominance Graphs in SociologyDominance Graphs in SociologyDominance Graphs in Sociology

Many social situations involve people or groups of people (countries,

cultures, universities, and so on) in which some individuals or group

“dominates” others. The word “dominate” can refer to many kinds of

dominance: cultural, physical, political, economic, and so on.

Suppose a sociologist wishes to study dominance patterns in a close-knit

group of college students. The group consists of five woman: Amy (A), Betty

(B), Carol (C), Denise (D), and Elaine (E). The sociologist begins by

conducting interviews with each pair of students to determine the pair wise

dominance1 of the students. If person A dominates person B, we denote this

by writing A B→ . (We assume that for each pair of students, one person

dominates the other.) After conducting the interviews, the sociologist draws

the dominance graph that represents pair wise dominations of the entire

group, which is shown in Figure 3. Note that Amy dominates Betty and that

Denise dominates Carol.

1 The determination of one-on-one dominance can be carried out by a series of questions, although it can be

subjective in some instances.


Dominance Graph for Amy, Betty, Carol, Denise, and Elaine

Figure 3

The adjacency matrix for the dominance graph is

0 1 1 0 1

0 0 0 0 0

0 1 0 0 1

1 1 1 0 0

0 1 0 1 0

A B C D E

A

B

M C

D

E

=

The number of 1’s in each row is the out-degree of the row vertex and

represents the number of firstfirstfirstfirst----stage stage stage stage dominancesdominancesdominancesdominances of that individual. Note that

vertices and A D each have a “score” of 3, vertices and C E each have a

score of 2, and B has a score of 0. In other words, Amy dominates three

people in first stage dominances (Betty, Carol, and Elaine), whereas Carol

dominates two people in the first stage (Betty and Elaine), and Betty

dominates no one.

The goal is to find the group leadergroup leadergroup leadergroup leader2222. If one person dominates every

person in the group, then obviously we would call that person the group

leader. If no one person dominates every other person, then the person with

the most first-stage dominances is declared group leader. If two or more

people tie with the most first-stage dominances, then we must resort to

ssssecondecondecondecond----stage stage stage stage dominancesdominancesdominancesdominances. In our example, Amy and Denise are tied, each

with 3 first-stage dominances.

2 If these were athletic teams playing in a conference round-robin tournament, we would want to know who

should be declared the conference winner.


To understand second-stage dominances, note that Elaine dominates

Denise and Denise dominates Amy. Hence, we say that Elaine has a second-

stage dominance (or second-stage influence) over Amy and we denote this by

E D A→ → . To find the number of second-stage dominances, consider the

square of the adjacency matrix 2M , which is

2

0 2 0 1 1

0 0 0 0 0

0 1 0 1 0

0 2 1 0 2

1 1 1 0 0

A B C D E

A

B

M C

D

E

=

To understand the meaning of 2M (See Problem 15), observe in the dominance

graph in Figure 3 that A dominates E and E dominates D , which we write as

A E D→ → , which is the only second-stage dominance of A over D . This

fact is indicated by the 1 in row A , column D of 2M . Also note that A has

two second-stage dominances over B ( A C B→ → and A E B→ → ), which is

indicated by a 2 in row A , column B of 2M . By examining 2M we see that

Denise has a total of five second-state dominances compared to four second-

state dominances for Amy. Thus, we declare Denise the group leader.3

We could continue our analysis of group dominance by computing the

sum 2M M+ , whose elements give the total number of first-stage and

second-stage dominances of one person over another.

2

2

0 1 1 0 1 0 2 0 1 1 0 3 1 1 2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 1 0 0 1 0 1 0 1 0 0 2 0 1 1

1 1 1 0 0 0 2 1 0 2 1 3 2 0 2

0 1 0 1 0 1 1 1 0 0 1 2 1 1 0

A B C D E A B C D E A B C D E

A A A

B B B

M M C C C

D D D

E E E

M M

+ = + =

+

3 Since Amy and Denise both have at least three second-stage dominances, we could compute ( )3

M G ,

which would give the number of third-stage dominances. One suspects, however, that although we might

define third-order dominances in theory, it is difficult to observe them in reality.


For example, Amy ( )A has 3 first- and second-stage dominances over Betty

( )B . The sum of the entries in row A represent the total number of ways

Amy dominates some person in the group in the first two stages. Here, Amy

has 7 first- and second-stage dominances over the members in the group.

(Can you find them?) However, Denise has a total of 8 first- and second-

stage dominances, so we would call Denise the leader of the group.

Tournament graphs also arise in conjunction with paired comparison paired comparison paired comparison paired comparison

experimentsexperimentsexperimentsexperiments in consumer testing, where a subject is given pairs of alternatives

and asked to determine which is preferable.

Paired Comparison Experiments (Cycles in Digraphs)Paired Comparison Experiments (Cycles in Digraphs)Paired Comparison Experiments (Cycles in Digraphs)Paired Comparison Experiments (Cycles in Digraphs)

A psychologist presents a subject with sounds of five musical

instruments, two at a time, and asks the subject which of the two is more

soothing. The instruments are : an alto horn ( )A , a bassoon ( )B , a clarinet

( )C , a dudek ( )D , and an English horn ( )E . The results of the experiment

given to one person4 are displayed in Figure 4.

Dominance Experiment in a Paired Comparison Experiment

Figure 4

It is natural to think that if instrument i is more soothing than j , and that j is

more soothing than k then i should be more soothing than k . When this

happens we call the three sounds ( ), ,i j k a consistent transitive tripleconsistent transitive tripleconsistent transitive tripleconsistent transitive triple. In

Figure 4 the subject says the clarinet ( )C is more soothing than the bassoon

4 The results displayed in the dominance graph in Figure 4 could be the result of a single subject or the

consensus results from a group of people.


( )B , and the bassoon ( )B is more soothing than the dudek ( )D . Since the

subject also said the clarinet ( )C was more soothing than the dudek ( )D , the

vertices ( ), ,C B D form a consistent transitive triple in the dominance graph.

In the dominance graph in Figure 4 there are 8 transitive triples:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ){ }, , , , , , , , , , , , , , , , , , , , , , ,C B D A C B A C D A B D B E D E C D E A C E A D

In some cases, however, the individual may be “inconsistent.” For

instance, the dominance graph indicates the subject says the clarinet ( )C is

more soothing than the bassoon ( )B and that the bassoon ( )B is more

soothing than the English horn ( )E , but that the English horn ( )E is more

soothing than the clarinet ( )C . We call this triple ( ), ,C B E a triple cycletriple cycletriple cycletriple cycle (or 3333----

cyclecyclecyclecycle). The dominance graph in Figure 4 has 2 triple cycles:

( ) ( ){ }, , , , ,C B E A B E

From the point of view of a psychologist a triple cycle represents an

inconsistency in preferences. To determine the number of transitive triples

and triple cycles5 in a dominance graph in Figure

Error! Reference source not found.Error! Reference source not found.Error! Reference source not found.Error! Reference source not found., we first compute the number of out-

degrees of the vertices, which are

( ) ( ) ( ) ( ) ( )od 3, od 2, od 2, od 0, od 3A B C D E= = = = =

One can show6 that if the vertices of a directed graph are numbered 1,2,…, n ,

then the number of transitive triples ( )TT and triple cycles ( )TC is given by

( ) ( )( )

( ) ( )

1

1od od 1 (number of transitive triples)

2

11 2 (number of triple cycles)

6

n

i

TT i i

TC n n n TT

=

= −

= − − −

∑

In our example, we find

5 We may also wish to find the number of transitive paths and cycles of different lengths other than just

paths of length three. 6 The proof of this result can be found in more advanced book in graph theory.


( ) ( )( ) ( ) ( )( )

( )

( )( )

( )

1od 1 od 1 1 od 5 od 5 1

2

13 2+ 2 1+ 2 1+ 0 1 + 3 2

2

8

11 2

6

15 4 3 8

6

2

TT

TC n n n TT

= − + + −

= ⋅ ⋅ ⋅ ⋅ − ⋅

=

= − − −

= ⋅ ⋅ −

=

�

RRRRanking Vertices anking Vertices anking Vertices anking Vertices of a of a of a of a Directed Directed Directed Directed GraphGraphGraphGraph

The challenging task in athletics is ranking athletic teams that have not

played each other. There are over 100 teams in NCAA Division 1 football,

and every year the National Champion is lambasted by teams who were

ranked 2,3, … . The following strategy is an attempt to “order” teams in a

directed graph when teams do not play every other team. (In other words,

only a few vertices in the graph are connected by a directed edge.) The

strategy takes into account the level of competition each team competes

against. (If Team A beats a good team, then it is rewarded more than beating

a poor team, etc.) To make the numbers manageable, we consider only five

universities, Maine (M), New Hampshire (N), Rhode Island (R), Connecticut

(C), and Delaware (D), where Figure 5 shows the season results between

these teams. The direction of the edges indicates what team beat what team,

and the numbers on the edges indicate by how many points. For example

Maine beat New Hampshire by 21 points, and lost to Rhode Island by 7. We

have also indicated the won-lost record of each team, which would give

ammunition to fans from Delaware and New Hamphire that their team should

be ranked #1. On the other hand Maine beat New Hampshire, which was a

good team, by 21 so they might make an argument for their team too.


Ranking Athletic Teams

Figure 5

The strategy in ranking these teams is to give Team A x more points

than Team B is Team A beat Team B by x points. In other words, we want to

give Maine 21 more points than New Hampshire, and we want to give

Connecticut 14 more points than Rhode Island, and so on. In other words we

want the following seven equations to hold:

7 New Hampshire over Conn by 7

3 New Hampshire over Delaware by 3

14 Connecticut over Rhode Island by 14

21 Delaware over Connecticut by 21

7 Delaware over Rhode Isl

N C

N D

C R

D C

D R

= +

= +

= +

= +

= + and by 7

7 Rhode Island over Maine by 7

21 Maine over New Hampshire by 21

R M

M N

= +

= +

What we have is 7 equations in 5 unknowns , , , ,N M C D R , and like most over-

determined linear systems, this system has no solution. In other words, we

can’t give Maine 21 more points than New Hampshire, since one could argue

that New Hampshire beat Connecticut (by 7), and Connecticut beat Rhode

Island (by 14) and Rhode Island beat Maine (by 7), so New Hampshire fans

could (and would) argue that they indirectly beat Maine by 21. So the


argument goes, how can we assign numbers to N,C,D,R,M so each team is

satisfied? (As much as possible anyway.) The idea is to pick the variables

that minimizes the sum of squares (a democratic choice) of the difference of

between left and right hand sides of the equations. That is pick , , , ,N C D R M

so it minimizes

( ) ( ) ( ) ( )

( ) ( ) ( )

2 2 2 2

2 2 2

Sum of Squares 7 3 14 21

7 7 21

C N D N R C C D

R D M R N M

= + − + + − + + − + + −

+ + − + + − + + −

This problem is a standard problem in statistics and the answer gives the final

rankings of the five teams as

12.1 Maine ranaked #1

9.7 Delaware ranked #2

3.6 New Hampshire ranked #3

0.6 Rhode Island ranked #4

0.0 Connecticut ranked #5

M

D

N

R

C

=

=

=

=

=

(No doubt there would be some complaining of fans from teams ranked 2 and

3. Also note that if each of the five variables , , , ,M D N R C were increased by

1 in the sum of squares equation, so its not a actual value of the variables that

is important but the difference between them. So after obtaining the least

squares solution for the variables we rescaled them so the lowest value was 0,

which in this case was Connecticut.

The reason Maine came out on top having only a 1-1 record was due to

the fact they beat a 2-1 team by 21 points. This system takes into

consideration the level of teams played and the difference in scores between

winners and losers.


ProblemsProblemsProblemsProblems

For the digraphs in Problems 1-6, find the adjacency matrix M . Then

compute 2M and 2M M+ and verify that the elements of these matrices

agree with the number of dominations in the graphs.

1.

2.

3.

4.


5.

6.

7. Group DominanceGroup DominanceGroup DominanceGroup Dominance

The graph shown in Figure 5 shows the dominance of a group of four

classmates: Andy ( )A , Betty ( )B , Charlie ( )C , and Denise ( )D .

Figure 5

a) Construct the adjacency matrix M for this graph.

b) Is there a consensus leader for the group?

c) Compute 2M and interpret its elements.

d) Who is the group leader?

e) Which person is dominated by the most other people?


8. RoundRoundRoundRound----Robin TournamentsRobin TournamentsRobin TournamentsRobin Tournaments

The graph shown in Figure 6 shows the results of a round-robin

tournament for baseball teams in the Yankee Conference.

Round-robin tournament graph

Figure 6


b) Is there a consensus leader for the group?

c) Compute 2M and interpret its elements.

d) Which team is the conference winner?

e) Which team is dominated by the most other teams?

9. Paired Comparison ExperimentPaired Comparison ExperimentPaired Comparison ExperimentPaired Comparison Experiment

The dominance graph shown in Figure 7 shows the preference of a group

of voters over political candidates.


b) Find the out-degrees of the vertices.

c) Find the number of transitive triples and triple cycles7.

d) Find the transitive triples and triple cycles in the graph.

7 It was interesting in the 2008 Democratic primary, one poll showed a majority of voters preferring Barack

Obama over Hillary Clinton, and Hillary Clinton over John Edwards in person-to-person face-offs.

However, these same voters preferred John Edwards over Barack Obama.


Political Polling

Figure 7

10. Transitive GraphsTransitive GraphsTransitive GraphsTransitive Graphs

A dominance graph is called transitive transitive transitive transitive if the graph has no triple cycles.

Transitive graphs have the property that each vertex has a different out-

degree and thus possible to rank the vertices (teams) from first to last.

a) Show the dominance graph in Figure 8 is a transitive graph.

b) For transitive graphs, it is possible to find a Hamiltonian path (not a

closed tour) by moving from vertex to vertex in decreasing order of their

out-degree. Find the Hamiltonian path of this graph.

c) Draw a transitive graph with six vertices. Do you see the general

strategy for constructing transitive graphs of any size?

d) Draw a digraph with five vertices with one triple cycle.

Transitive graph

Figure 8


11. Landau’s TheoremLandau’s TheoremLandau’s TheoremLandau’s Theorem

A theorem by Landau states that if some vertex u in a dominance graph

has a larger out-degree than all other vertices, then either u dominates all

other vertices v , or if it does not dominate a given vertex v , then u

dominates a third vertex w which in turn dominates v . What does the

theorem say in the language of round-robin tournaments? Verify this theorem

fro the dominance graphs in Problems 1-6.

12. Landau’s Theorem in Landau’s Theorem in Landau’s Theorem in Landau’s Theorem in the Yankee Conference the Yankee Conference the Yankee Conference the Yankee Conference

Suppose the football teams in the Yankee Conference play every other

team exactly once during the season. At the end of the season Maine has won

more games than any other team. However, Maine lost to Vermont. What

does Landau’s theorem say in the language of the Yankee Conference?

13. Food WebFood WebFood WebFood Web

On a certain island, the only living things are snails, birds, goats, thistles,

lions, people, and lettuce. The snails eat lettuce, the birds eat snails, the

goats eat lettuce and thistles, the people eat goats and lettuce, and the lion

eats goats and people. A directed graph where the vertices represent forms

of life and a directed edge from vertex u to vertex v means that u eats v is

called a food webfood webfood webfood web. Construct the food web for this ecological system.

14. SecondSecondSecondSecond----Stage DominancesStage DominancesStage DominancesStage Dominances

Let M be the adjacency matrix of 0s and 1s for a given directed graph.

Show that the thij element of the square 2M of the adjacency matrix

represents the number of second-stage dominances of vertex i over vertex j .

Hint: Recall how matrices are multiplied and note that the product 1 1 1⋅ =

indicates a second stage dominance, and that 0 1 1 0 0⋅ = ⋅ = does not.

15. OutOutOutOut----degree/Indegree/Indegree/Indegree/In----DegreeDegreeDegreeDegree Show that the sum of the out-degrees and in-

degrees of vertices in a directed graph are equal. Verify this fact for the

directed graph in Figure Error! Reference source not found.Error! Reference source not found.Error! Reference source not found.Error! Reference source not found.. Hint: Very easy.

16. Directed Acyclic GraphDirected Acyclic GraphDirected Acyclic GraphDirected Acyclic Graph A directed acdirected acdirected acdirected acyclic graphyclic graphyclic graphyclic graph (denoted a DAGDAGDAGDAG) is a

directed graph that has no cycles of any length. That is, for any vertex u

there is no directed path that ends with itself. Roughly speaking, it means the

graph “flows” in one direction. Show that a DAG gives rise to a partial order

on the set of vertices, where u v≤ is defined as a directed path going from

vertex to u v .


17. Directed Acyclic GraphDirected Acyclic GraphDirected Acyclic GraphDirected Acyclic Graph Verify by inspection that the graph in Figure 9 is

a DAG.

Directed Acyclic Graph

Figure 9

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Section 3.5 Directed Graphs - University of Maine