
Citation 
 Permanent Link:
 http://digital.auraria.edu/AA00001658/00001
Material Information
 Title:
 Competition graphs, Pcompetition graphs, twostep graphs, squares, and domination graphs
 Creator:
 Merz, Sarah Katherine
 Place of Publication:
 Denver, CO
 Publisher:
 University of Colorado Denver
 Publication Date:
 1995
 Language:
 English
 Physical Description:
 x, 119 leaves : illustrations ; 29 cm
Thesis/Dissertation Information
 Degree:
 Doctorate ( Doctor of Philosophy)
 Degree Grantor:
 University of Colorado Denver
 Degree Divisions:
 Department of Mathematical and Statistical Sciences, CU Denver
 Degree Disciplines:
 Applied mathematics
Subjects
 Subjects / Keywords:
 Graphic methods ( lcsh )
Graphic methods ( fast )
 Genre:
 bibliography ( marcgt )
theses ( marcgt ) nonfiction ( marcgt )
Notes
 Bibliography:
 Includes bibliographical references (leaves 113119).
 General Note:
 Submitted in partial fulfillment of the requirements for the degree, Doctor of Philosophy, Applied Mathematics
 General Note:
 Department of Mathematical and Statistical Sciences
 Statement of Responsibility:
 by Sarah Katherine Merz.
Record Information
 Source Institution:
 University of Colorado Denver
 Holding Location:
 Auraria Library
 Rights Management:
 All applicable rights reserved by the source institution and holding location.
 Resource Identifier:
 34267126 ( OCLC )
ocm34267126
 Classification:
 LD1190.L622 1995d .M47 ( lcc )

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COMPETITION GRAPHS, PCOMPETITION
GRAPHS, TWOSTEP GRAPHS, SQUARES, AND
DOMINATION GRAPHS
by
B. A., Whitman College, 1991
M. S., University of Colorado at Denver, 1994
A thesis submitted to the
University of Colorado at Denver
in partial fulfillment
of the requirements for the degree of
Doctor of Philosophy
Applied Mathematics
Sarah Katherine Merz
1995
This thesis for the Doctor of Philosophy
degree by
Sarah Katherine Merz
has been approved
by
David C. Fisher
\ Jennifer Ryan
Date y/l? /9^
Merz, Sarah Katherine (Ph. D., Applied Mathematics)
Competition Graphs, pCompetition Graphs, TwoStep Graphs, Squares, and
Domination Graphs
Thesis directed by Professor J. Richard Lundgren
ABSTRACT
Given a digraph D, the competition graph of D, C(D) is a graph
on the same vertex set with vertices x and y adjacent in C(D) if and only
if there exists a vertex z such that x and y have arcs to z in D. Intro
duced in the context of modeling food webs by Cohen in 1968, competition
graphs have utility in a variety of applications such as radio communication
networks and a certain type of dominating set in tournaments. Can elimina
tion orderings be used to characterize digraphs with interval and/or chordal
competition graphs? We answer this question and consider possible implica
tions in the food web application of competition graphs. We consider several
problems motivated by the use of competition graphs in modeling communi
cation networks. Which symmetric digraphs with a loop at each vertex have
interval competition graphs? Which loopless symmetric digraphs have inter
val competition graphs? For which digraphs can the chromatic number of the
competition graph be bounded? Given a graph G, can we bound the number of
arcs in a digraph D such that C(D) = G? Given a graph G, can we deter
mine canonical forms of a digraph D such that C(D) = G? Which graphs are
the competition graph of a strongly connected or Hamiltonian digraph? These
111
questions are considered. Results pertaining to the last question are general
ized to pcompetition graphs. Lastly, we consider several problems motivated
by tournaments. Given a digraph D, the domination graph of D, Dom(D) is a
graph on the same vertex set as D with vertices and y adjacent in Dom(Z)) if
and only if for every other vertex z in D, x or y has an arc to z (possibly both).
We show a relationship between the competition and domination graphs of a
tournament and consider the following questions: Which graphs are the domi
nation graph of a tournament? What is the maximum number of edges in the
domination graph of a tournament? Which graphs are the domination graph
of an oriented simple graph?
This abstract accurately represents the content of the candidates thesis. I
recommend its publication.
Signed
J. Richard Lundgren
IV
This work is dedicated to
my mother, my father and my sister, Ruth.
Thank you for your encouragement and love.
v
CONTENTS
CHAPTER
1 INTRODUCTION............................................. 1
1.1 Terminology.......................................... 9
2 PROBLEMS MOTIVATED BY FOOD WEBS......................... 11
2.1 Elimination Ordering Characterizations............. 11
2.2 Finding Interval Elimination Orderings ............. 16
2.3 Implications for Food Webs.......................... 18
3 PROBLEMS MOTIVATED BY COMMUNICATION NET
WORKS ..................................................... 20
3.1 Symmetric Digraphs with Loops....................... 20
3.1.1 Using Forbidden Subgraphs.................... 32
3.1.2 Elimination Ordering Characterizations....... 34
3.2 Loopless Symmetric Digraphs......................... 36
3.2.1 Finding Maximal Cliques of the TwoStep Graph 37
3.2.2 Competition Covers........................... 41
3.2.3 Elimination Ordering Characterizations....... 52
3.3 Strongly Connected Digraphs......................... 53
3.3.1 Generalization to pCompetition Graphs....... 57
3.4 Hamiltonian Digraphs................................ 61
vi
3.4.1 Generalization to pCompetition Graphs: Con
structions ................................. 66
3.4.2 Utilizing the Constructions................. 71
3.4.3 Classes of 2Competition Graphs............. 74
3.5 Competition Inverses .............................. 78
3.5.1 Acyclic Competition Inverses................ 79
3.5.2 Strongly Connected Competition Inverses .... 80
3.5.3 Hamiltonian Competition Inverses............ 83
3.5.4 Bounds on the Number of Arcs in a Competition
Inverse..................................... 85
3.6 The Chromatic Numbers of Competition Graphs .... 86
3.6.1 Symmetric Digraphs . ................... 88
3.6.2 Arbitrary Digraphs and Tournaments.......... 90
4 THE COMPETITION GRAPHS OF TOURNAMENTS ... 96
4.1 Necessary Conditions on the Domination Graphs of Tour
naments ................................................ 97
4.2 Sufficient Conditions on the Domination Graphs of Tour
naments ............................................... 103
4.3 The Domination Graphs of Digraphs................. 107
5 A LIST OF OPEN PROBLEMS............................... 109
APPENDIX
A NOTATION.............................................. Ill
BIBLIOGRAPHY ................................................ 113
Vll
FIGURES
FIGURE
1.1 A digraph and its competition graph....................... 1
1.2 An interval graph......................................... 2
1.3 A chordal graph........................................... 3
1.4 The problem with forbidden subgraph characterizations. . . 4
1.5 A properly colored graph.................................. 5
1.6 A 2competition graph..................................... 7
1.7 A tournament and its domination graph..................... 9
2.1 A graph with perfect elimination ordering................. 11
2.2 A digraph with disimplicial elimination ordering......... 12
2.3 A graph with interval elimination ordering................ 14
2.4 A digraph with diinterval elimination ordering........... 15
2.5 Forced incidence diagram for disimplicial elimination ordering. 18
2.6 Forced incidence digraph for diinterval elimination ordering. 19
3.1 Forbidden subgraphs of interval graphs.................... 21
3.2 The smallest tree that is not interval.................... 32
3.3 A graph and an induced subgraph........................... 33
3.4 An interval graph whose twostep graph is not interval. . . 37
3.5 The family of maximal cliques is not S'(G)................ 43
3.6 The set {a;, xi, x2, Â£3} is not in S'(G).................. 48
viii
3.7 A competition graph......................................... 54
3.8 A 2competition graph....................................... 58
3.9 An example.................................................. 64
3.10 Chordal graphs on 5 vertices................................ 76
3.11 Chordal graphs on 5 vertices................................ 77
3.12 A graph and its competition inverse......................... 86
3.13 A graph and its competition inverse......................... 87
3.14 The subgraph of an interval graph........................... 91
3.15 This graph is not an interval graph......................... 92
4.1 A tournament and its domination graph....................... 96
4.2 Regular tournaments on 7 vertices and their domination graphs. 99
4.3 This graph is not the induced subgraph of the domination
graph of a tournament...................................... 100
4.4 A tournament and its domination digraph................. 104
4.5 Orientations of a path.................................. 105
IX
First and foremost, I am grateful to Rich Lundgren for his academic,
financial, personal and professional support. His limitless patience, encour
agement, generosity and capacity as a mentor have been more valuable than I
can express.
I would like to thank Dave Fisher for all the help that he has given
me, particularly in my struggle to do research. In addition, he provided the
extremely valuable templates from which most of the figures in this document
were constructed.
I would also like to thank Kathy Fraughnaugh for the help that she
has given me, particularly with mathematical writing. As a woman in math
ematics, she has been an important role model during my graduate years.
My thanks go to Jenny Ryan and Gary Kochenberger, as well as to
the fine mathematicians with whom I have had the opportunity to collabo
rate: Charlie Anderson, Larry Langley, John Maybee, Patty McKenna, Norm
Pullman, Craig Rasmussen, and Brooks Reid.
I would like to thank the Office of Naval Research and Mark Lipman
for funding my graduate education.
Finally, I would like to thank David Guichard of Whitman College
for encouraging me to go to graduate school.
x
CHAPTER 1
INTRODUCTION
This dissertation addresses a variety of graphtheoretic problems mo
tived by three basic applications of the competition graph of a digraph. Given
a digraph D, the competition graph of D, C(D) is a graph on the same ver
tex set, with x and y adjacent in C(D) if and only if there exists z such that
{x,z) and (y,z) are arcs in D.. Figure 1.1 illustrates a digraph and its com
petition graph. Digraphs can be used in a variety of applications. The three
considered here are in modeling food webs, radio communication networks and
tournaments.
Figure 1.1. A digraph and its competition graph
Competition graphs were introduced in the study of food webs by
Cohen [13] as a method to determine the dimension of ecological phase space.
Let each vertex in a digraph represent a single or collection of similar species.
There is an arc from x to y in the digraph if and only if the species represented
1
by x preys upon the species represented by y. The digraphs in this model
are generally acyclic. Two vertices are adjacent in the competition graph
of the digraph if and only if their corresponding species compete for some
common prey. Section 2 addresses graphtheoretic problems that arise in this
application.
How many variables must be considered in the determination of a
competition graph? One example of such a variable is diet; in the food web
model we suggested, an arc is placed from x to y if x eats y. The boxicity
of a graph, first introduced by Roberts, is defined to be the smallest integer k
such that the graph can be represented as an intersection graph of boxes in k
space. See Roberts [63] for a comprehensive introduction to these concepts. An
interval graph is a graph which is the intersection graph of intervals on the
real line. That is, an interval graph has boxicity one. Figure 1.2 illustrates an
interval graph and the corresponding intervals for each vertex. Cohen observed
that many competition graphs for actual food webs are interval graphs.
8 l
Figure 1.2. An interval graph.
This graph is an intersection graph of intervals on the real line when
the following intervals are assigned to each vertex. 1: [8,11]; 2: [10,14];
3: [12,15]; 4: [9,13]; 5: [1,8]; 6: [4,7]; 7: [3,6]; 8: [2,5].
2
An explanation for this phenomenon has been sought since its discov
ery. Is Cohens empirical observation an artifact of the construction? That is,
perhaps the competition graph of most acyclic digraphs are interval. Roberts
posed this question in 1978 [64]. He showed that this is not the case, since
every graph can be made into the competition graph of an acyclic digraph by
attaching a sufficient number of isolated vertices. So it is interesting to ask
which digraphs have interval competition graphs.
A chordal graph is a graph with no induced cycle on k vertices,
where k > 4 (see Figure 1.3). Equivalently, any fccycle where k > 4 contains
a chord. Interval graphs are chordal. Sugihara [72] suggested the chordal
property was significant with respect to food webs. His analysis was concerned
with holes in niche space. He conjectured that given a collection of species
that exhibit competition in a cyclic structure, it is likely that a chord exists
across the cycle. The lack of such a chord characterizes a hole. For this
reason, it is interesting to ask which digraphs have chordal competition graphs.
We present a characterization of digraphs with chordal and inter
val competition graphs. This characterization is different from previous ap
proaches because it lends a nice interpretation to the structure of the digraph
3
that may be relevant to food webs. Furthermore avoids the difficulty in us
ing forbidden subgraph characterizations. Figure 1.4 illustrates this difficulty.
Even though the competition graph of some digraph is interval, it is possible
that the digraph contains an induced subdigraph with nonchordal competition
graph.
Figure 1.4. The problem with forbidden subgraph characterizations.
From left to right, digraphs and their competition graphs. The digraph
above is an induced sub digraph of the digraph below. Its competi
tion graph is not an induced subgraph of the competition graph of its
superdigraph.
The application of competition graphs in the study of communication
networks was introduced by Raychaudhuri and Roberts [60]. Let each vertex
in a digraph represent a station in a radiocommunication network. There is
an arc from x to y in the digraph if and only if the station represented by y
can receive a signal transmitted by the station represented by x. Two vertices
are adjacent in the competition graph of the digraph if and only if their cor
responding stations are unable to transmit on the same channel because their
signals are received by a common third station. In this setting, competition
graphs are often called conflict graphs. A coloring of a graph is a labeling
4
of its vertices so that adjacent vertices do not have the same label. A coloring
of the competition graph can be translated into a frequency assignment in the
corresponding network. So in the interest of optimally assigning frequencies
in the networks, we would like to efficiently color the competition graph of the
digraph modeling the network. Coloring is generally a hard problem though
there are some classes of graphs which can be easily colored. For example, in
terval graphs can be colored in linear time. Hence the question arises: Which
digraphs have interval competition graphs?
B
Figure 1.5. A properly colored graph.
Digraphs that model communication networks will often have special
structure reflecting the nature of the network. We will take advantage of this
structure as much as possible in considering graphtheoretic problems that
arise in this application. For example, many radio communication networks
can be modeled by a symmetric digraph with a loop at each vertex. That
is, x transmits to y if and only if y transmits to x and each station can
receive its own signal. Such a digraph can be represented by its underlying
5
graph (with or without loops). The competition graphs of digraphs having this
structure have been studied by Raychaudhuri and Roberts [60], Raychaudhuri
[57], Balakrishnan and Paulraja [2, 3], and Laskar and Shier [40]. Given a
symmetric digraph D with a loop at each vertex and underlying interval graph
H, what conditions are necessary and sufficient for the competition graph of
D to be interval? This question was answered by Raychaudhuri and Roberts
in 1985 [60]. Given a symmetric digraph D with a loop at each vertex, what
conditions are necessary and sufficient for the competition graph of D to be
interval? This is the focus of Section 3.1.
A natural extension of this question is to consider symmetric digraphs
without loops. The competition graphs of such digraphs have been studied by
Brigham and Dutton [8], and Lundgren, Maybee, and Rasmussen [45]. The
latter gave necessary and sufficient condition for symmetric digraphs without
loops and having interval underlying graph to have an interval competition
graph. We give necessary and sufficient conditions for symmetric digraphs
without loops to have an interval competition graph in Section 3.2.
A strongly connected digraph is a digraph in which there is a di
rected path from every vertex to every other vertex. Since it is desirable that
a message initiated somewhere in the network be able to reach all stations,
usually the digraphs for communication networks are strongly connected. In a
very large network, certain stations may be equipped with very powerful trans
mitters while other stations transmit much weaker signals. The digraph for
such a network is not necessarily symmetric, but is strongly connected. There
fore it is of interest to determine which graphs are the competition graphs of
6
strongly connected digraphs. In Section 3.3 we will characterize these graphs.
A Hamiltonian digraph is a digraph with a directed cycle that in
cludes every vertex and is therefore an example of a strongly connected di
graph. In Section 3.4 we characterize graphs that are the competition graphs
of Hamiltonian digraphs. The results of Sections 3.3 and 3.4 are extended
to pcompetition graphs, a generalization of the competition graph. The p
competition graph of a digraph D, denoted CP(D) is a graph on the same
vertex set as D, with x and y adjacent in CP(D) if and only if there exist p
vertices z\, z2,..., zp such that
{x,zx), (x, z2), (as, zp) and (y, zt), (y, z2), ...,(y, zp)
are arcs in D. See Figure 1.6 for an example of a 2competition graph.
7
6
8 1

7
2
6
3
5 4
Figure 1.6. A 2competition graph.
The competition graph of this digraph is shown in Figure 1.1. The only
two vertices that compete at least twice are 1 and 8.
Radio frequencies have already been assigned in existing radio com
munication networks thereby establishing a coloring of the competition graph.
Aside from Hefner and Hintz [29], very little work has been done toward devel
oping algorithms that begin with existing communication networks and then
7
optimize, either with respect to the number of communication links (without
changing the competition graph) or with respect to rendering the competi
tion graph easily colored (while preserving qualities of the digraph). We take
preliminary steps in this direction in Section 3.5. Given a graph G, what is
the maximum number of arcs in a digraph having competition graph G? We
answer this question
Given a digraph D and a graph G such that C(D) = G, we say D
is a competition inverse of G. The competition inverse of a graph does
not necessarily exist, nor is it necessarily unique. Characterizing competition
inverses has been studied by Greenberg, Lundgren, and Maybee [24, 25] and
by Roberts [64].
Roberts defined the competition number of a graph G as the small
est integer k(G) such that G, together with k(G) isolated vertices, is the com
petition graph of an acyclic digraph. We first consider the problem of finding a
canonical representative for the class of acyclic competition inverses of chordal
and interval graphs. We next consider the construction of a strongly con
nected competition inverse, as this property is ideally attributed to models of
communication networks. As a special case of strongly connected, we consider
the construction of a Hamiltonian competition inverse for chordal and interval
graphs.
A tournament is an oriented complete graph. The competition
graph of a tournament was first considered by Lundgren, Merz, and Ras
mussen [50] in connection with some problems involving chromatic number.
The structure of the competition graph of a tournament was then examined by
8
Fisher, Lundgren, Merz, and Reid [20]. We discuss the relationship between
competing and dominating pairs in a tournament in Section 4. Two vertices
x and y in a digraph are a dominant pair if and only if every vertex z is
contained in the union of their outsets. Two vertices are adjacent in the dom
ination graph Dom(T) of a tournament T if and only if they are a dominant
pair in T. Figure 1.7 illustrates a tournament and its domination graph.
The domination graph of a tournament T is the complement of the
competition graph of the tournament formed by reversing the arcs of T (this
is not necessarily true for all digraphs). Therefore results on the domination
graphs of tournaments correspond to results on the competition graphs of
tournaments. Since the domination graph of a tournament generally has fewer
edges than the competition graph, it is more convenient to state and prove
results on domination graphs. Section 4 addresses the domination graphs of
tournaments and other oriented graphs.
1.1 Terminology
The open neighborhood of a vertex Â£ in a graph is the set of
vertices adjacent to x. The closed neighborhood of a vertex x is the union
9
of x and its open neighborhood. The inneighbors of a vertex x in a digraph
are the vertices with arcs to x. The outneighbors of a vertex x are the
vertices to which x has arcs. The independence number of a graph G is
the maximum cardinality of a set such that no two vertices in the set are
adjacent. A clique is a set of vertices such that every pair of vertices in the
set is adjacent. The clique number of a graph G is the maximum cardinality
of a clique in G. A proper coloring of a graph G is a labeling of the vertices
such that if two vertices have the same color, then they are not adjacent. A
fecoloring is a proper coloring with k labels. The chromatic number of a
graph G is the smallest integer k such that G has a ^coloring. A clique cover
of a graph G is a labeling of its vertices such that if two vertices have the same
label, then they are adjacent. The clique cover number of a graph G is the
minimum number of labels in a clique cover of G. An edge clique cover of
a graph G is a labeling of its edges such that if two edges have the same label,
then they are contained in a clique. The edge clique cover number of a
graph G is the minimum number of labels in an edge clique cover. An pedge
clique cover of a graph G is a labeling of its edges such that if two edges have
p of the same labels, then they are contained in a clique. The pedge clique
cover number of a graph G is the minimum number of labels in a pedge
clique cover. An isolated vertex is a vertex with no neighbors. A pendant
vertex is a vertex with exactly one neighbor. An internal vertex is a vertex
with at least two neighbors. A simplicial vertex is a vertex whose neighbors
induce a clique. For an explanation of notation used in this document, please
see the appendix.
10
CHAPTER 2
PROBLEMS MOTIVATED BY FOOD WEBS
2.1 Elimination Ordering Characterizations
In this section we explore characterizations of digraphs that have
chordal and/or interval competition graphs. Then we consider the possible im
plications these characterizations may have in answering the question of why
the competition graphs of digraphs modeling food webs are so often interval.
We begin by considering chordal competition graphs. Let G be a graph. A ver
tex v is simplicial in G if and only if the set of vertices adjacent to v induce a
clique. An ordering vi,v2,... ,vn of the vertices of G is a perfect elimination
ordering if and only if u, is simplicial in Gi = G {vi,v2,. ., v;_x}, where
G\ = G. Figure 2.1 illustrates a graph and its perfect elimination ordering.
2
Figure 2.1. A graph with perfect elimination ordering.
Proposition 2.1.1 Rose [67]. A graph is chordal if and only if it has a perfect
elimination ordering.
11
Can we define an elimination ordering of a digraph which corresponds
to a perfect elimination ordering of the competition graph of the digraph, pro
vided one exists? A vertex x is disimplicial in a digraph D if and only if
whenever there are vertices y:z,u,v E V{D) such that
(x,u),(y,u),(x,v),(z,v) A(D)
, then there exists a vertex w E V(D) such that (y,w) and (z,w) E A(D)
An ordering vi,v2,... ,vn of the vertices of D is a disimplicial elimination
ordering if and only if Uj is disimplicial in D{, where A(Di) = A{D) minus
all outarcs of v\,..., Vj_i. Figure 2.2 illustrates a digraph and its disimplicial
elimination ordering.
1
Figure 2.2. A digraph with disimplicial elimination ordering.
Since 1 and 2 compete and 1 and 3 compete, it follows that 2 and 3
compete. Since 2 and 4 compete and 2 and 3 compete, it follows that
3 and 4 compete.
Theorem 2.1.1 Lundgren and Merz [48]. If D is a digraph, then C(D) is
chordal if and only if D has a disimplicial elimination ordering.
12
Proof. Assume that vi,v2,... ,vn is a perfect elimination ordering of C(D).
This is true if and only if given V{ and y,z G
{viiV}i{viiz} Â£ E{C{D)) implies that {y,z} G E(C(D)).
This is true if and only if given V{ and y,z G {vj+i,... ,un}, if there exist u
and v G V(D) such that
(y,u),(vi,u),(z,v),(vi,v) G A(D),
then there exists w G V(D) such that (y,in), (2, w) G A(D). This is true if
and only if Vi,v2,... ,vn is a disimplicial elimination ordering of D.
This characterization is particularly nice because it avoids the for
bidden subgraph problem of previous approaches: It is easy to construct a
digraph whose competition graph is chordal, containing a generated subgraph
whose competition graph is not chordal (see Figure 1.3). Rose, Tarjan, and
Leuker [68] showed that a perfect elimination ordering can be found in linear
time. See Golumbic [23] for an introduction to this algorithm, which finds
a perfect elimination ordering if one exists. This algorithm can be modified
to produce an ordering of the vertices in a digraph which is a disimplicial
elimination ordering if and only if the competition graph is chordal. This
is accomplished by translating the steps of the algorithm to work with the
digraph rather than first constructing its competition graph and then execut
ing the algorithm on the competition graph. Using a disimplicial elimination
ordering in this way to determine if the competition graph of a digraph is
chordal is not computationally faster than first constructing the competition
13
graph and then implementing the algorithm. We now consider digraphs with
interval competition graphs.
Let G be a graph. An ordering vx,v2,... ,vn of the vertices of a graph
G is an interval elimination ordering if and only if the ordering has the
property that for all Vi,Vj,Vk such that i < j < k, {i>, 'i>*} Â£ E(G) implies
that Â£ E{G). Figure 2.3 illustrates a graph and its interval elimination
ordering.
Figure 2.3. A graph with interval elimination ordering.
Proposition 2.1.2 Laskar and Shier [40]. A graph G is interval if and only
if it has an interval elimination ordering.
Can we define an elimination ordering of a digraph which corresponds
to an interval elimination ordering in the competition graph, provided one
exists? We say an ordering vx,v2,... ,vn of the vertices of D is a diinterval
elimination ordering of a digraph D if and only if the ordering has the
property that for all Vi,vj,Vk such that i < j < k, if there exists u such
that (vi,u) and (u*,u) Â£ A{D) then there exists w such that (vk,w) and
(Vj,w) Â£ A{D). Figure 2.4 illustrates a digraph with a diinterval elimination
ordering.
14
5
Figure 2.4. A digraph with diinterval elimination ordering.
Theorem 2.1.2 Lundgren and Merz [48]. If D is a digraph, then C(D) is
interval if and only if D has a diinterval elimination ordering.
Proof. Assume that vi:v2,... ,vn is an interval elimination ordering of C(D).
This is true if and only if for i < j < k, whenever {x>;, x?*} Â£ E(C(D)), then
{vk,Vj} Â£ E(C(D)). And this is true if and only if for i < j < ky whenever
there exists u Â£ V{D) such that (vi,u), (vk,u) Â£ A(D), then there exists
v Â£ V(D) such that (vj,v), (vk,v) Â£ A(D). And this is true if and only if
vi,v2,... ,vn is a diinterval elimination ordering of D.
Again, this characterization also avoids the forbidden subgraph prob
lem of previous characterizations. Can interval elimination orderings be used
to efficiently determine intervality? This question is unanswered. Can di
interval elimination orderings be used to efficiently characterize digraphs with
interval competition graphs? This question is also unanswered. This problem
may be more tractable if we impose certain structures on the digraphs involved.
For example, knowing that the digraph is transitive or acyclic might provide
enough structure so that using diinterval elimination orderings to determine
if the competition graph is interval becomes computationally efficient.
15
2.2 Finding Interval Elimination Orderings
Although the problem of finding an interval elimination ordering di
rectly (if one exists) in an arbitrary graph remains open there are well known
algorithms that can perform intervality testing in linear time. See Golumbic
[23] for an introduction to one of these algorithms. This algorithm uses a char
acterization of interval graphs. A consecutive ranking is an ordering of sets
Ci,C2,..., Cm such that if i < k, then a; Â£ C7; and x Â£ Ck implies x Â£ Cj for
all i < j < k. Fulkerson and Gross [22] showed that a graph is interval if and
only if its maximal cliques have a consecutive ranking. Given this information
we can produce an interval elimination ordering by the following algorithm.
Algorithm 2.2.1 Lundgren and Merz [48]. Given a family of sets of vertices
CuC2t...,Cr corresponding to the consecutively ranked maximal cliques in
an interval graph G, we find an interval elimination ordering for G.
(1) Consider the ri vertices in C\C2. Arbitrarily order them Vi,v2,... ,vT1.
(2) Consider the r2 vertices in C2 C2 that have not yet been ordered.
Arbitrarily order them nPl+i,... ,vTl+r2.
(3) Similarly order the remaining vertices for C2 C4, ..., CT\ CT.
(4) Arbitrarily order the vertices in Cr1 fl Cr.
(5) Complete the ordering by arbitrarily ordering the remaining vertices
in CT resulting in iq,... ,vn.
Theorem 2.2.1 Lundgren and Merz [48]. The ordering produced in Algo
rithm 2.2.1 is an interval elimination ordering for the graph G.
Proof. Consider arbitrary i,j,k such that i < j < k. We claim that if
Â£ E(G), then {vj,Vk} Â£ E(G). Suppose that Â£ E{G). If
Vi,Vk Â£ CT, then since Vi,Vj and Vk were chosen in one of the last two steps of
16
the algorithm we conclude that Vj E Gr and we are done. So assume not. Then
there exists b < r such that V{ E Cf, Cb+i and d > b such that Vk E Cd~ Cd+l
(if d < r) oi Vk E CT. If Vj E CT, then vk E CT. That is, {vj,vk} E E(G). So
assume that Vj E Cf Cf+x for b < f < d, where f < r. Since {ui,Ufc} E E,
Vi and Vk belong to some maximal clique and since V{ E Cb and i>; ^ Cb+1,
then Vi 0 Ca since d > b + 1 and the ranking is consecutive. So Vk E Cb
This consecutive ranking then implies Vk E Cf, since b < f < d. Therefore
{vj,vk} E E(G). U
Given an interval elimination ordering for a graph we can also produce
the consecutively ranked maximal cliques. This is accomplished by consecutive
(with respect to the order of elimination) examination of the neighborhoods
of the vertices. For example, if is the largest integer for which vx and v^
are adjacent, then
{ujjiq and Vj adjacent 1 < j < &i}
is a maximal clique. Furthermore, if k2 is the largest integer for which v2 and
Vk2 are adjacent, then
{vj\v2 and Vj adjacent ,2 < j < k2}
is a maximal clique if and only if either v2 and vx are adjacent or k2 > kx.
A similar approach can be used to find the remaining maximal cliques in a
consecutive ranking.
Thus an interval elimination ordering can be found if you have either
the maximal cliques of the graph or the intervals for which the graph is the
17
intersection graph. These techniques do not use interval elimination order
ings to determine if a graph is interval. Furthermore, they do not translate
to finding a diinterval elimination ordering of a digraph. Short of trying all
possibilities, is there a method for determining whether or not a graph has an
interval elimination ordering or a digraph has a diinterval elimination order
ing? One approach might be to consider the diinterval elimination ordering
for special classes of digraphs such as transitive or acyclic digraphs.
2.3 Implications for Food Webs
As mentioned earlier, the competition graphs of digraphs modeling
cross sections of real food webs are often interval. Do either the disimplicial or
diinterval elimination ordering characterization say anything about the struc
ture of these food webs? Diagrams such as the one in Figure 2.5 indicate
possible interpretations of the situation in which a digraph modeling a food
web has an interval competition graph.
Figure 2.5. Forced incidence diagram for disimplicial elimination ordering.
Assume that v^,V2, .. ,vn is a disimplicial elimination ordering for
a digraph D. The presence of arcs (vk,vn), (vn3,vn), and
(un_2,ni) forces un_2 and un_3 to have common prey.
18
In Figure 2.5 we show a disimplicial elimination ordering for a hypo
thetical food web. Suppose that the species of this food web have been linearly
ordered Vi,V2,... ,vn such that vn is the most adaptable and iq is the least
adaptable. Then if i
they compete and V{ and Vj and have some property in common so that they
compete, then Vj and Vk are very likely to compete as well. This interpreta
tion emphasizes Sugiharas assembly rule for food webs which requires that
species be incorporated into community conservatively by attaching to single
guilds rather than by bridging multiple guilds, where a guild is essentially a
clique in the competition graph [74].
Figure 2.6. Forced incidence digraph for diinterval elimination ordering.
Assume that vi,v2,... ,un is a diinterval elimination ordering for a
digraph D. The presence of arcs (u^ui) and (vi,vi) forces Vk to have
common prey with vertices U{+1,...
Figure 2.6 shows a diinterval elimination ordering for a digraph. The
species of this food web have been linearly ordered Vi,v2,... ,un on some scale
of adaptability where vn is the most adaptable. Let i < j < k. This ordering
implies that if Vk is so adaptable that it competes with V{ then most likely it
competes with Vj.
19
CHAPTER 3
PROBLEMS MOTIVATED BY COMMUNICATION NETWORKS
3.1 Symmetric Digraphs with Loops
In this section we consider the competition graphs of symmetric di
graphs with a loop at each vertex. Such digraphs arise in the model of radio
communication networks. We can represent such a network with a digraph
where a vertex represents each station and there is an arc from a; to 3/ if and
only if the station corresponding to x transmits a signal that the station cor
responding to y can receive. There is a loop at each vertex since each station
may receive its own signal. Given such a digraph, we may refer to its under
lying graph. The underlying graph is a graph on the same vertex set with
an edge between x and y if and only if x has an arc to y and y has an arc to
x. Notice this preserves a loop at each vertex. In this section, we consider the
underlying graph with loops removed.
Given a graph G, the square, G2, is a graph on the same vertex set
such that two vertices x and y are adjacent in the square if and only if there is
a path of length one or two between x and y in G. Raychaudhuri and Roberts
[60] first made the following observation that is stated without proof.
Proposition 3.1.1 Raychaudhuri and Roberts [60]. If D is a symmetric di
graph with a loop at each vertex and G is the underlying graph (loops removed)
of D, then C{D) = G2.
20
Which graphs have interval squares? We begin by observing the fol
lowing result.
Proposition 3.1.2 Raychaudhuri [57]. If G is an interval graph, then G2 is
an interval graph.
Unfortunately the proof given in the work cited is not easily gener
alized to graphs that are not interval. We also observe that there are graphs
having interval squares that are not interval. Consider the forbidden sub
graphs of interval graphs provided by Lekkerkerker and Boland [41] depicted
in Figure 3.1.
Figure 3.1. Forbidden subgraphs of interval graphs.
A graph is interval if and only if it contains no subgraph isomorphic to
Gi,Gf2,C3!3,Gf4,(j!5. Note that the squares of G?i,(n = 4,5),(?3, Â£r4 and
G5 are interval.
Some of these graphs have interval squares. Thus our characterization
cannot be based upon the squares of forbidden subgraphs of interval graphs.
We will employ a characterization of interval graphs due to Fulkerson and
21
Gross [22]. This characterization uses the following definition: An ordering of
a family of sets Gi, G2, ..., Cp is a consecutive ranking if and only if for all
i < j < k, whenever x G Gi and x G G*, then x G Cj.
Proposition 3.1.3 Fulkerson and Gross [22]. A graph G is an interval graph
if and only if the family of maximal cliques of G has a consecutive ranking.
Suppose that we can define a family of sets MC(G2) found in G
corresponding to the maximal cliques in the square. By Proposition 3.1.3, G2
is interval if and only if MC(G2) has a consecutive ranking. Recall that a
vertex is simplicial if its neighborhood induces a clique. A vertex that is not
simplicial is called nonsimplicial. We say that a closed neighborhood of a
nonsimplicial vertex is maximal if it is not properly contained in the closed
neighborhood of any other nonsimplicial vertex in the graph. We define a class
of graphs as having the closed neighborhood property if and only if the
set of maximal closed neighborhoods of nonsimplicial vertices is precisely the
set of maximal cliques in G2. Interval graphs have the closed neighborhood
property as shown by the next theorem.
Theorem 3.1.1 Lundgren, Merz, and Rasmussen [49]. If G ^ Kn is a con
nected interval graph, then G is a maximal clique of G2 if and only if there
exists a nonsimplicial vertex z G V{G) such that G = Ag[z] and Nq[z] is
maximal.
Proof. (<=) Let z be a nonsimplicial vertex in G such that Ng[z\ is maximal.
We claim that the consecutively ranked cliques G;,..., G* containing z form
a maximal clique in the square (since these cliques are Nq[z]. Suppose not.
Then there exists a vertex w adjacent to every vertex in Gi,..., Gj, in G2, but
w is not in G{,..., G*. Without loss of generality, assume that w G Cj where
j > k and j is the smallest such integer. Let x be an arbitrary vertex in G; not
22
equal to z such that x Â£ Ci+i. Since x Cm for m > i and w $ Cm for to < ft
and furthermore i < ft, we conclude that {in,} $ E{G). But {w,x} 6 E(G2)
implies that w and x must be joined by a path of length two and so there
exists t Â£ Ci fl Cj so that x,w 6 Na(t). Since the ranking is consecutive,
t g Ci n n ck n n c5
. That is, Nq[z] C /Vc;[t], a contradiction. Thus no such w exists, i.e., Nq[z]
is a maximal clique in G2.
(=>) Let C be a maximal clique in G2. We will show there exists a
nonsimplicial vertex z (E G such that C = Ng\z\. Let Ci, C2,..., Ci be the
consecutively ranked maximal cliques of G2. Since G is connected, CiC\Ci+\ ^
0, for i = 1,2,...,/ 1. So Ci U Ci+i is a clique in G2 and C % Ci for
i = 1,2,...,/ 1. Similarly C % Ci. Therefore there exists i, the smallest
integer such that Ci PI C ^ 0 and there exists x Â£ Ci fl C such that x $ Ci+i.
Similarly, there exists j, the largest integer such that Cj fl C / 0 and there
exists y Â£ Cj DC such that y ^ Cj1. If j < i, then suppose there exists z Â£ C
such that z 0 Cj fl fl (7,. Then if z Â£ Cm for m > i or z 6 Cm for m < j we
have a contradiction by definition of j and i respectfully.
Therefore i < j. Since the ranking is consecutive, {x,y} ^ E(G) and
so there exists z E Ci and z 6 Cj so that x and y are joined by a path of length
two. Then z Â£ Cp for all i < p < j. Since every vertex of C is contained in a
clique such that i < p < j, we conclude that C C /Vg[z]
Since C is a maximal clique, it follows that C = Nq\z\. If z is
simplicial, then C is a clique in G. Since G is connected and not complete,
23
there exists a vertex x $ C such that x is adjacent to a vertex y Â£ C. But
then {} U C is a clique in G2, a contradiction. So z is nonsimplicial.
Corollary 3.1.1 Lundgren, Merz, and Rasmussen [49]. If G is a connected
interval graph, then G2 is interval.
Proof. If G = Kn, then the statement is clearly true so assume that G ^ Kn.
We claim that since the family of maximal cliques of G has a consecutive
ranking, the family of maximal nonsimplicial closed neighborhoods of G has a
consecutive ranking. Let Ci, C2,..., C* be the consecutively ranked maximal
cliques of G. Let Sij be Ci U U Cj and let S = < j< &}. Let S'
be the sets of S such that no set is properly contained in any other. Since
G is connected, we conclude that (7; D (7;+i 7^ 0. If v Â£ (7, fl (7;+1, then
v is nonsimplicial. Conversely, any nonsimplicial vertex v must be in the
intersection of consecutive cliques. Thus S' is the family of maximal cliques of
G2. Rank S' such that Sab < Sg{ if and only if a < g. Observe a < g implies
c < i. We claim this is a consecutive ranking.
Suppose that x Â£ Sac and x Â£ Sgi. We claim that x Â£ Sdf for all
a < d < g such that Sdf Â£ S'. Then x Â£ C& for some b such that a < b < c
and x Â£ Ch for some h such that g < h
Then d < g. Furthermore c < / and d < /. Suppose that Ce % Sdf for all
b < e < h. Then either d > h or / < b, but d < g < h and b < c < f. So
Ge C Sdf for some e such that b < e < h and we conclude that x Â£ Sdf
Therefore b > h. Then x Â£ Ch fl fl Cb But since d < g < h and
b < c < /, we conclude that Shb ^ Sdf So x Â£ Sdf.
Though this result has been previously established, this approach to
24
the problem may be more easily generalized to other classes of graphs. For
example, trees also have the closed neighborhood property.
Theorem 3.1.2 Lundgren, Merz, and Rasmussen [49]. If T is a tree such
that V(T) > 3, then C is a maximal clique in T2 if and only if there exists a
nonsimplicial vertex v such that !Vg[u] = G and Ng[v] is maximal.
Proof. (<=) Let v be a nonsimplicial vertex in T such that iVr[u] is maximal.
Clearly iVr[v] is a clique in T2. Suppose that it is not maximal. Then there
exists w $ Wr[u] such that w is joined to all x E iVr[i>] by a path of length
one or two. Since {w,v} 0 E[T) there exists x such that w,v E Nt(x).
Since IVrM ^ [sc] there exists y E IV^fu] such that {y,x} Â£ E(T). If
{iw,y} E E(T), then we are done since wxvyw is a 4cycle. So w and y are
joined by a path of length two. That is, there exists a (o ^ x,a ^ v) such that
w, y Â£ Nt(cl). Then wxvyaw is a 5cycle, a contradiction since T is a tree.
Thus no such w can exist. That is, iVr[u] is a maximal clique in T2.
(=>) Let C be a maximal clique in T2. We claim there exists z such
that C C Nx[z] in T. Let R be a subset of C. The proof is by induction on i?.
For the base case, let R = {x,y}. Then either {x,y} E E{T) so R C iVr[a;] or
x, y E Nt(z) for some z, so R C. Nt[z\.
Let R = {s,y,z}. Assume that the claim is false. Suppose {x,y} E
E(T). Then {z,x} and {z,y} 0 E(T), so there exist a,b (possibly a = b) such
that x,z E Nt(cl) and y,z E Nr(b). If a = b, then we are done since xya is a
3cycle. If a ^ b, then we are done since xyazbx is a 5cycle. So assume that
no two vertices in R are adjacent. Then there exists a such that i,j/E IVr(a),
b (b ^ a) such that y,z E ATr(6) and c (c ^ a,c ^ b) such that x,z E ATr(c).
Then xaybzcx is a 6cycle, a contradiction.
25
Assume that the claim is true for all R such that i2 < \C\. Let
R = C. The established base cases imply that \C\ > 4. Let w be an arbitrary
vertex in C. Consider R' = R {to}. By the induction hypothesis there
exists zq such that R' C Nt\zq\. Suppose that the claim is false for R. Then
{w, z0} E(T). Since i2' > 3 there exists x,y G R' such that x ^ z0 and
y ^ z0. If {w,x} and {w,y} 0 E(T), then since there exist a,b (possibly
a = b) such that w,x G Nt(cl) and w,y G Nx(b). Furthermore if a ^ 6, then
waxz0ybw is a 6cycle and if a = b, then axzoya is a 4cycle, a contradiction
in either case. So {in,} C R' for some x. Since R % there exists
V Â£ Nt[zq] such that {y,x} ^ E(T). If {w,y} G E{T), then we are done
since wxz0yw is a 4cycle. Therefore there exists v (v ^ z0,v ^ a:) such that
w,y G Nt(v). Then wxzoyvw is a 5cycle, a contradiction. Therefore there
exists z such that R C Nt[z\. Since C is a maximal clique, C Nt[z\. If z is
simplicial then C is a clique in T. T is connected and not complete. Therefore
there exists x (Â£. C such that x is adjacent to some y G C. But then {x} U C
is a clique in T2, a contradiction. So z is nonsimplicial. I
Corollary 3.1.2 Lundgren, Merz, and Rasmussen [49]. If T is a tree such
that V(T) > 3, then T2 is interval if and only if the maximal nonsimplicial
closed neighborhoods of T have a consecutive ranking.
Two vertices u and v are compatible if and only if either Ag[u] C
iVcfu] or iVofu] C iVcfu]. A vertex v is simple if the vertices in iVc[u] are
pairwise compatible. Observe that if v is simple, then v is simplicial. A graph
is strongly chordal if and only if every induced subgraph has a simple vertex.
Theorem 3.1.2 is generalized using the following lemma that is stated without
proof.
26
Lemma 3.1.1 If v is simple in G and y is the neighbor of v with maximal
neighborhood, then iVc[y] = {u : dG(u,v) < 2}.
Theorem 3.1.3 Raychaudhuri [59]. If G ^ Kn is a connected strongly
chordal graph, then C is a maximal clique of G2 if and only if there exists
a nonsimplicial vertex z such that C NG[z\ and NG[z] is maximal.
Proof. (=>) Let G = ... ,u} be a maximal clique of G2. Then
for i,j E {1,2, ...,to} we conclude that dG(ui,Uj) < 2. Let Pij be a path of
shortest length in G between Ui and Uj, where i,j 1,2,..., n. In particular, if
dc(ui,Uj) = 1, then = {u{,Uj} and if = 2, then P^ = {ui,Vk,Uj}
for some Vk E V(G).
Consider the subgraph G' of G induced by the vertices and Vk,
where i = 1,2,... ,n) and k = 1,2,... ,p, where p < . Since Vk has a pair
of nonadjacent neighbors for all A:, namely Ui and uj, Vk is nonsimplicial in
G' for all k. Therefore Vk is not simple in G' for all k. Since G' is strongly
chordal, there exists ut such that ut is simple in G1. Then by Lemma 3.1.1,
there exists w E such that
Ng>[w] = {z\dGI(ut,z) < 2} D {ui,u2,...,iin}.
Since i\Tc<[tu] C iVc[iu], {ui,U2,...,un} ^ iVcfu;]. Suppose that iVofu;] con
tains a vertex a ^ Ui,U2,... ,un. Then {a,iii,U2,... ,} is a maximal clique
of G2, a contradiction. Therefore, {ui,u2,... ,un} = IVg[^] Since C is a
maximal clique, it follows that C = iVc[tw].
Suppose that w is simplicial in G. Then for i,j = l,2,...,n we
conclude that dc(ui,Uj) = 1. Since G is not complete and connected, there
exists a maximal clique C' of G{C ^ C1) such that C' and C share at least one
27
common nonsimplicial vertex z. But C ^ C' implies there exists x Â£ C such
that x $ C' and y Â£ C' such that y G. Thus dG(x,y) = 2, a contradiction.
Therefore w is nonsimplicial.
Let z\ be a nonsimplicial vertex of G such that Nq[z\] is max
imal. Let Ng(zi) = {z2,z3,... ,zn}. Clearly Nq[z\\ is a clique in G2. Sup
pose that it is not a maximal clique. Then there exists u Â£ V{G) such that
da(u,zi) = 2 and dG(u,z,) < 2 for i 2,3,...,rc. Let Pi be a path of
shortest length in G between u and Z{, where i = 1,2,... n. In particular,
if da(u, Z{) = 1, then Pi = {u,Zi} and if dG(u,Zi) = 2, then Pi = {u,Vi,Zi}.
Consider the subgraph G' of G induced by u, Zi and Vk, where i = 1,2,... ,n
and k = 1,2,... ,p, for p < n. Since G' is strongly chordal, G' has a simple
vertex. Since u*. has a pair of nonadjacent neighbors for all k, namely u and
Zi, we conclude that Vk is not simple for all k. Therefore either u is simple
or there exists zt that is simple. Call this simple vertex s. By Lemma 3.1.1,
there is a neighbor w of s in G' such that /Vg<[u;] = {u : dc{v,s) < 2} where
{zi,z2,...,zn,s} C Ng>[w]. But iVG/[it;] C iVG[u;], so {zlt z2,. .., zn, u} C
1Vg[ii;]. Since w has at least two nonadjacent neighbors u and z\, w is nonsim
plicial, a contradiction since iVG[zi] is maximal. So iVG[zi] is a maximal clique
in G2. U
Corollary 3.1.3 Raychaudhuri [59]. If G is a strongly chordal graph, then
G2 is interval if and only if the maximal nonsimplicial closed neighborhoods
of G have a consecutive ranking.
The girth of a graph is the length of the shortest cycle in the graph,
where girth is undefined if the graph has no cycles. Careful examination of the
proof of Theorem 3.1.2 shows that we obtained contradictions by producing
28
racycles, where n < 6. So we may use similar ideas to generalize the result to
graphs with girth p, where p > 7.
Theorem 3.1.4 Lundgren, Merz, and Rasmussen [49]. If G is a graph with
girth p, where p > 7, then C is a maximal clique in G2 if and only if there
exists a nonsimplicial vertex z such that Ng[z\ C and Ng[z\ is maximal.
Corollary 3.1.4 Lundgren, Merz, and Rasmussen [49]. If G is a graph with
girth p, where p > 7, then G2 is interval if and only if the maximal nonsimpli
cial closed neighborhoods have a consecutive ranking.
Observe that if G is a 6cycle, then the maximal nonsimplicial closed
neighborhoods are not the maximal cliques of G2. If we exclude graphs con
taining 6cycles, we are able to characterize the maximal cliques of the squares
of at least one class of graphs.
Theorem 3.1.5 Lundgren, Merz, and Rasmussen [49]. If G ^ Kn is a con
nected 6cyclefree graph such that every edge is contained in a triangle, then
C is a maximal clique in G2 if and only if there exists a nonsimplicial vertex
v such that C = 1Vg[u] and 1Vg[u] is maximal.
Proof. (<=) Let v be a nonsimplicial vertex in G such that the closed neigh
borhood of v is maximal. Clearly 1Vg[u] is a clique in G2. Suppose that it
is not a maximal clique. Then there exists a vertex w 1Vg[u] such that for
all x Â£ 1Vg[u], the vertices w and x are joined by a path of length at most
two. If for some x 6 1Vg[u], we find that {w, x} 6 E(G), then since every edge
is contained in a triangle we conclude that w and x are joined by a path of
length two. Since w and v are joined by a path of length two, there exists x
such that w,v Â£ Nq(x). Since A^g[u] is not properly contained in 1Vg[x], there
exists y Â£ 1Vg[u] such that {x,y} $ E(G). Since w and y are joined by a path
of length two, there exists a vertex u (u ^ x,u ^ v) such that w,y Ng(u).
29
If {if,} .Â£7(6?), then since {iu,a:} must be contained in a triangle we con
clude that there exists a vertex b (b ^ t/,6 ^ v) such that w,x E Nc{b).
Then wbxvyuw is a 6cycle, a contradiction. So assume that {tt,} E E{G).
Similarly if {u,u} $ E(G), then since {sc,t>} must be contained in a triangle
we conclude that there exists c(c ^ y,c^ w) such that x,v E Ng(c). Then
xcvyuwx forms a 6cycle, a contradiction. So assume that {t>,ti} E E(G).
Since is not properly contained in JVc[it], there exists z E Nq[v]
such that {u, z} $ E{G). If {x, z} E E(G), then we are done since wxzvyuw is
a 6cycle. So assume that {a:,z} ^ E(G). If {w,z} E(G), then we are done
since zwxuyvz is a 6cycle. So assume that {w,z} E E(G). If w,z E Na(y),
then we are done since zywxuvz is a 6cycle. So assume that at most one of
w and z is in No(y). Then since w and z are joined by a path of length two
we conclude that there exists a vertex d(d ^ y,d ^ u,d ^ x,d ^ v) such that
w,z E Na(d). Then wdzvyuw is a 6cycle, a contradiction. Therefore no such
w can exist. That is, iVc[^] is a maximal clique in G2.
(=>) Let C be a maximal clique in G2. We claim that there exists
a vertex z such that C C Ng[z]. Let R be a subset of C. The proof is by
induction on i2. Let R = {x,y}. Then x and y are joined by a path of
length one or two. In either case {x,y} C Ng[z] for some z. Assume that
the statement is true for all R such that fÂ£ < \C\. Let R = C. Let w be an
arbitrary vertex in R. Consider R' = R {to}. By the induction hypothesis
there exists Zq such that R' C iVc[zo] Suppose that there does not exist z
such that R C Ng[z\.
Case 1: Assume that w is adjacent to all x E Ng(z0). Then zq E R.
30
Let x E Ng(zo). Then {x,w} E E(G). Since R % IVg[] there exists y E iV[zo]
such that {x,y} ^ E(G). Since {w,y} E E(G) and every edge is contained
in a triangle, w and y are joined by a path of length two so there exists
v (i> ^ x,v ^ zQ) such that w,y E Nq(v). If x and zQ are not adjacent to v,
then since there exists a (a / v, a ^ il>, a ^ y) such that x,z0 E Ng(cl). Then
xaz0yvwx is a 6cycle, a contradiction. So assume that x and z0 are adjacent
to v. Then R $2 Ng[v] implies there exists t E R' such that {u,t} ^ E(G).
Then {w,t} E E(G) and wtz0yvxw is a 6cycle, a contradiction.
Case 2: There is a vertex y E Ng(zq) such that {y,w} ^ E(G). Since
w and y are joined by a path of length two there exists a vertex v(v ^ zq) such
that w,y E Ng(v). Since R $Â£ iVcM there exists a vertex x E R' such that
{*,v} ^ E(G). Ifw and x are joined by a path of length two, then since there
exists a vertex a (a ^ y,a ^ z
that wvyzQxaw is a 6cycle, a contradiction. So w and x are not joined by
a path of length two. Then {to,s} E E(G) and since {w,x} is contained in
a triangle we conclude that w and x are joined by a path of length two, a
contradiction. Therefore there exists z such that C C ./Vg[z]. Since C is a
maximal clique in G2 it follows that C = IVg[.z]. If z is simplicial, then C is
a clique in G. Since G is connected and not complete, there exists a vertex
x Â£ C such that x is adjacent to a vertex y E Ng[z]. But then {$} U C is a
clique in G2, a contradiction. So z is simplicial.
Corollary 3.1.5 Lundgren, Merz, and Rasmussen [49]. If G ^ Kn is a con
nected graph such that every edge is contained in a triangle and G contains
no 6cycle, then G2 is interval if and only if the maximal closed nonsimplicial
neighborhoods have a consecutive ranking.
31
3.1.1 Using Forbidden Subgraphs Steif [71] showed that, for
arbitrary digraphs, there can be no forbidden subgraph characterization of
digraphs with interval competition graphs. However, there may be partial
forbidden subgraph characterizations, provided the digraph is symmetric with
a loop at each vertex.
(i

Figure 3.2. The smallest tree that is not interval.
Can the square of a tree that is not interval be interval? We already
know that the square of an interval tree is interval. Consider the tree in
Figure 3.2. This tree is not interval nor is its square. Recall that a tree is
interval if and only if it does not contain an induced subgraph isomorphic to
the tree in Figure 3.2. IfT is a tree containing an induced subgraph isomorphic
to the square of the tree in Figure 3.2, does T2 contain an induced subgraph
isomorphic to the square of the tree in Figure 3.2? More generally, if H is
an induced subgraph of G, then H2 is not necessarily an induced subgraph of
G2. An example of such a graph is given in Figure 3.3. For this reason, the
forbidden subgraph approach to characterizing graphs with interval squares
does not generally work. We now show that it will work for trees.
Lemma 3.1.2 Lundgren, Merz, and Rasmussen [49]. If T is a tree and H a
connected subgraph of T, then H2 is an induced subgraph of T2.
Proof. Suppose not. Then there are vertices x,y Â£ H such that {x,y} Â£
E(T2) but {,y} ^ E(H2) Since {x,y} ^ E(H2) and {x,y} $ E(H) there
32
Figure 3.3. A graph and an induced subgraph.
exists but z Â£ H such that x, y E Nq(z). Since H is connected there
is a path xv^.. .v^y joining x and y. Then yzxvi is a cycle in T, a
contradiction. Therefore H2 is an induced subgraph of T2.
Theorem 3.1.6 Lundgren, Merz, and Rasmussen [49]. If T is a tree, then T2
is interval if and only if T is interval.
Proof. ($=) This follows from Lemma 3.1.2.
(=>) Assume that T is not interval and suppose that T2 is interval.
Then T contains an induced subgraph isomorphic to the graph H in Figure 3.2.
By the lemma H2 is an induced subgraph of T2, but H2 is not interval, a
contradiction. Thus T2 is not interval.
Observe that the tree in Figure 3.2 is G2, one of the forbidden sub
graphs shown in Figure 3.1. Furthermore, Cr2 and G\ for n > 6 are the only
forbidden subgraphs of an interval graph with noninterval squares. We now
consider graphs containing subgraphs isomorphic to G\ for n > 6. Provided
that the girth of such a graph is sufficiently large, the square of the graph is
not interval.
Lemma 3.1.3 Lundgren, Merz, and Rasmussen [49]. If G is a graph with
girth p, where p > 6 and G contains an induced pcycle C, then C2 is an
induced subgraph of G2.
33
Proof. Suppose not. Then there exist vertices x,y Â£ V(C) such that {x,y} Â£
E(G2) but {x,y} E(C2). Since {x,y} 0 E(C2), {x,y} 0 E(C) so there
exists z Â£ V(G) such that x,y Â£ Nq^z). Since x and y are on a pcycle with
p > 6, the length of the shortest path joining them is q < p3. Let xv1 .. .vq^y
denote this path. Then yzxv\ u9_iy is a cycle of length q + 2 < p, a
contradiction. Hence C2 is an induced subgraph of G2.
Lemma 3.1.4 Lundgren, Merz, and Rasmussen [49]. If G is a graph with
girth p, where p > 6, then G2 contains an induced cycle of length / where
l > 4.
Proof. Case 1: Suppose that p = 6. Then G contains an induced 6cycle,
G = ViV2 v6vi, so C2 contains an induced four cycle, namely the cycle
V1V2V4V5V1. By Lemma 3.1.3 C2 is an induced subgraph of G2. Hence G2
contains an induced four cycle.
Case 2: Suppose that p > 6 and p is even. G contains an induced
pcycle, C = V\V2 'VpVi, so G2 contains an induced cycle, namely the cycle
Viv3v5 Vp^iVi. By Lemma 3.1.3 C2 is an induced subgraph of G2. Hence
G2 contains an induced cycle, where p > 8.
Case 3: Suppose that p > 6 and p is odd. G contains an induced
pcycle, C = vxv2 vpvi, so C2 contains an induced ^^cycle, namely the
cycle V1V3V5 vpv\. By Lemma 3.1.3 C2 is an induced subgraph of G2. Hence
G2 contains an induced ^ilcycle, where p > 7.
Theorem 3.1.7 Lundgren, Merz, and Rasmussen [49]. If G is a connected
graph with girth p, where p > 5, then G2 is not interval.
3.1.2 Elimination Ordering Characterizations We can de
fine elimination orderings analogous to those used in Section 2 to characterize
34
symmetric digraphs with a loop at each vertex whose competition graph are
chordal and interval.
An ordering v\,v2,... ,vn of the vertices of a graph G is a chordal
square elimination ordering if and only if for all y, z Â£ {ui+i,n+2> ,vn},
whenever there exist u and v such that y,V{ Â£ Ng[u\ and z,u; Â£ IVgM, then
there exists w such that y,z Â£ JVg[u;].
Theorem 3.1.8 Lundgren and Merz [48]. A graph G has a chordal square if
and only if G has a chordal square elimination ordering.
Proof. Assume that Vi,v2,... ,vn is a perfect elimination ordering of G2.
This is true if and only if for V{ and y,z Â£ {ui+i,... ,un},
{u,,]/}, {ui, z} Â£ E(G2) implies that {y,z} Â£ E(G2).
This is true if and only if for V{, and y,z Â£ {u,+i,..., vn}, whenever there
exists u,v such that
Vi,y NG[u\,Vi,z Â£ Ag[u],
then there exists w such that y,z Â£ Ag[iu]. This is true if and only if the
ordering of vertices Vi,v2,... ,vn is a chordal square elimination ordering of G.
m
An ordering v1}v2,... ,vn is an interval square elimination order
ing if and only if for all i < j < k, if there exists u such that Â£ 1Vg[u],
then there exists v such that Vj,Vk Â£ Ag[u].
Theorem 3.1.9 Lundgren and Merz [48]. A graph G has an interval square
if and only if G has an interval square elimination ordering.
Proof. Assume that vi,v2,... ,vn is an interval elimination ordering of G2.
This is true if and only if for i < j < k, whenever {u;,i>fc} Â£ E(G2), then
35
{uj,!;*.} Â£ E(G2). This is true if and only if for i < j < k, whenever there
exists u such that Â£ iVc[u], then there exists v such that Vj,Vk Â£ ATcfu].
And this is necessary and sufficient for v\, u2,... ,un to be an interval square
elimination ordering of G.
3.2 Loopless Symmetric Digraphs
In the previous section, we considered the competition graphs of sym
metric digraphs with a loop at each vertex. This motivates the consideration
of the competition graph of symmetric digraphs without loops. As with sym
metric digraphs with a loop at each vertex, given a symmetric digraph without
loops, we may consider the underlying graph of the digraph. Given a graph
G, the twostep graph, 82(G) is a graph on the same vertex set as G with
{*>2/} Â£ ^(^(G)) if and only if there exists z E V(G) such that {x,z} and
{y, z} are in E(G). This definition motivates the following observation that is
stated without proof.
Proposition 3.2.1 If D is a symmetric digraph without loops and H its
underlying graph, then C(D) = 82(H).
We will characterize classes of graphs with interval twostep graphs.
We first observe that the twostep graph of an interval graph is not necessarily
interval (see Figure 3.4), making this problem more difficult than characteriz
ing graphs with interval squares. As in the characterizations of the previous
section, we employ the Fulkerson and Gross [22] characterization of interval
graphs: A graph G is interval if and only if its family of maximal cliques of G
has a consecutive ranking. We will restrict our discussion to connected graphs
that are not complete since disconnected graphs can be examined by connected
36
components and the twostep graph of the complete graph is also complete.
Figure 3.4. An interval graph whose twostep graph is not interval.
3.2.1 Finding Maximal Cliques of the TwoStep Graph
Our approach is very similar to that used in characterizing graphs with interval
squares. We will find a family of sets MC(iS2(C7)), defined in the original graph,
that corresponds to the maximal cliques in the twostep graph. Then Sz(G)
is interval if and only if MC(5,2(G)) has a consecutive ranking. The first class
we consider is trees.
Theorem 3.2.1 Lundgren, Maybee, Merz, and Rasmussen [44]. If T is a
tree, then C such that C > 2 is a maximal clique in Sz(G) if and only if
G = Nt{v), where v is nonpendant.
Proof. (<=) Let C = Nt(v), where v is a nonpendant vertex in T. Clearly
Nt(v) is a clique in Sz(T). Suppose that it is not a maximal clique. Then
there exists a vertex w C that is joined to every vertex in C by a path
of length two. Since C > 2, there exist distinct vertices x and y in Nt(v).
Since T is a tree, x and y ^ E(T). Then there exist vertices t and u such
that x,w & Nx(t) and y,w Â£ Nt(u). If t = u, then vxuyv is a cycle in
T, a contradiction. Therefore t ^ u. Then vxtwuyv forms a cycle in T, a
contradiction. Thus no such w can exist. Therefore Nt(v) is a maximal clique
37
in S^T). Furthermore, if Nt(v) = Nt(z) for two vertices v and z, then there
exist x and y E Nt(v) fl Nt{z) and vxzyv is a cycle, a contradiction.
(=>) Let G be a maximal clique in ^(T). Then G > 2, so there
exist distinct x and y in G. Since C is a maximal clique in ^(T), there exists a
vertex z such that x,y E Nt(z). Suppose that C ^ Nt(z). Then there exists a
vertex iw E G such that w $ Nt(z). Since T is a tree, {x, y} ^ E(T). Since G is
a maximal clique in S2(T) there exist vertices t and u such that w,x Â£ iVj'(t)
and w,y E Nt(u). If t = u, then txzyt is a cycle in T, a contradiction.
Therefore t ^ u. Then wtxzyuw is a cycle in T, a contradiction. Thus no such
w can exist. That is, Nt(z) = C. I
Corollary 3.2.1 Lundgren, Maybee, Merz, and Rasmussen [44]. If T is a
tree, then ^(T) is interval if and only if the maximal open neighborhoods of
the nonpendant vertices in T have a consecutive ranking.
We can extend this characterization to trianglefree graphs that con
tain no 6cycles (induced or otherwise). First, a lemma that is stated without
proof.
Lemma 3.2.1 If G is a graph and x, y, and z are vertices contained in a
maximal clique in ^(G) and there does not exists v such that x,y,z E iVcfw],
then there must exist distinct a,b, and c such that x,y E iVc(a) ,y,z E No(b)
and x,z Â£ Ng(c), that is, xaybzcx is a 6cycle.
Theorem 3.2.2 Lundgren, Maybee, Merz and Rasmussen [44]. If G Kn is
a connected triangle and 6cyclefree graph, then C such that C > 2 is a
maximal clique in S^G) if and only if G = Nq{z) for some z such that Nq{z)
is maximal.
Proof. (=>) Let G be a maximal clique in ^(G). If G] = 2, then the state
ment is clearly true. So assume that G > 3. Let R C G. We prove by induc
tion on i? that there exists z such that R C Ng[z\ in G. By Lemma 3.2.1,
38
if 22 = 3, then the claim is true. So assume that /2 > 4. Furthermore,
assume that the claim is true for all R such that i2 < k < \C\ and consider
the case il = k < C'. Pick arbitrary x Â£ R. Let R' = R {a:}. By
the induction hypothesis there exists z\ such that R' C in G. Pick
arbitrary y ^ x Â£ R. Let R" = R {y}. By induction hypothesis there
exists z2 such that R" C Nq[z2\. Since x and y are in R there exists z such
that i,j Â£ Ng(z). If z is z\ or z2 we are done so assume not. Observe
Z\, z2 0 R since G is trianglefree (for example, if Zi 6 R then y and z\ are
adjacent and joined by a path of length two). Since iE > 4 there exists w Â£ R
(w 7^ z,w x,w y,w / zi,w 7^ z2) such that w is adjacent to z\ and z2.
Then xzyz\wz2x is a 6cycle in G, a contradiction. Therefore without loss
of generality we conclude z = z\. That is, R C Na[zi\ for all R C C. In
particular C C IVg[zi] and so by maximality of C we conclude C = Ng{z\).
(
is not properly contained in the open neighborhood of any other vertex in G.
Clearly Ng(z) is a clique in S2(G). Suppose that it is not a maximal clique.
Then there exists a vertex w $ Ng(z) such that w is joined by a path of length
two to every vertex in Ng(z) in G. Let x 6 Ng(z). Since w and x are joined
by a path of length two in G there exists a such that x,w Â£ Nc(a). But
Ng(z) is not properly contained in Ng(ci) so there exists y Â£ Ng(z) such that
{a>2/} ^ E{G) Then w and y are joined by a path of length two so there
exists a distinct vertex b such that y,w Â£ JVg(6). Since G is trianglefree,
b ^ x. Then zxawbyz is a 6cycle in G, a contradiction. Thus Ng(z) forms a
maximal clique in S2(G).
39
Corollary 3.2.2 Lundgren, Maybee, Merz, and Rasmussen [44]. If G ^ Kn
is a connected trianglefree and 6cyclefree graph, then 52(G) is interval if
and only if the maximal open neighborhoods of G have a consecutive ranking.
We now consider graphs such that every edge is contained in a trian
gle.
Theorem 3.2.3 Lundgren, Maybee, Merz, and Rasmussen [44]. If G ^ Kn is
a connected 6cyclefree graph such that every edge is contained in a triangle,
then C such that G > 2 is a maximal clique in 52(G) if and only if C = Wg[z]
for some z where Nq[z] is maximal.
Proof. (=$) Let C be a maximal clique in 52(G). By an analogous argument
to that in Theorem 3.2.2 we can show that there exists z such that G C Ng[z].
Since every edge is contained in a triangle and G is maximal we conclude that
G = Na[z].
(
in another closed neighborhood in G. Since every edge of G is contained in
a triangle, clearly Ng[z] forms a clique in 52(G). Suppose that it is not a
maximal clique. Then there exists a vertex w such that w is joined to every
vertex in Ng[z] by a path of length two while {u;, z} ^ E{G). Since w and z are
joined by a path of length two, there exists a vertex v such that w,z E Ng{v).
Since Ng[z\ is not properly contained in iVc[u], there exists y E /Vg[2] such
that y JVcf'u]. Then w and y are joined by a path of length two so there
exists u such that w,y Ng(u), where u ^ v. Then v,z E Ng(u) since
otherwise, because the edge {v,z} is contained in a triangle we conclude that
there exists a vertex t such that v,z E iVc(t), creating the 6cycle ztvwuyz.
But then Ng[z] is not properly contained in iV^fu] so there exists x E Ng[z]
such that x (j iVc[u]. If x Nq(v), then since x and w are joined by a
40
path of length two we conclude that there exists s (possibly equal to y) such
that io,i Â£ Ng(s). Then wvuzxsw forms a 6cycle in G and we have a
contradiction. Thus Â£ i?(G). Then wvxzyuw forms a 6cycle in G.
This contradiction proves no such w can exist.
Corollary 3.2.3 Lundgren, Maybee, Merz, and Rasmussen [44]. If G 9= Kn is
a connected 6cyclefree graph such that every edge is contained in a triangle,
then 52(G) is interval if and only if the maximal closed neighborhoods of G
have a consecutive ranking.
3.2.2 Competition Covers To generalize the results of the
previous section, we must employ the techniques of Lundgren, Maybee, and
Rasmussen [46]. A family 5 = {5i,..., 5r} of sets of vertices of a graph G is
called a competition cover of G if the following conditions are satisfied:
(1) i,j E 5m implies there exists a vertex k such that i,j Â£ Ng(k).
(2) if i,j Â£ Nc(k) for some k then i,j Â£ 5m for some m.
This definition motivates the following proposition that is stated
without proof.
Proposition 3.2.2 Lundgren, Maybee, and Rasmussen [46]. If G is a graph,
then 52(G) is interval if and only if G has a competition cover 5 that has a
consecutive ranking.
The difficulty with this result is finding the right competition cover.
Furthermore, it is very difficult to use this characterization to prove that the
twostep graph of a given graph is not interval. The problem then becomes
finding a family of sets in G that is not only a competition cover, but in
addition, determines conclusively whether or not 52(G) is interval. We have
already achieved this for several classes of graphs in the previous section. We
now present the competition cover used by Lundgren, Maybee, and Rasmussen
41
[46] to characterize interval graphs with interval twostep graphs.
Let v be a nonsimplicial vertex in G. We say v is Type I if every
maximal clique containing v contains three or more vertices. We say v is Type
II if every maximal clique containing v contains exactly two vertices. Other
wise, we say v is Type III. Define S(G), the nonsimplicial competition
cover of a graph G, as family of sets {.Si,..., SV}, where Si is one of the
following:
(1) /Vcfn,], the closed neighborhood of Vi, if u, is Type I.
(2) Na(vi), the open neighborhood of V{, if V{ is Type II.
(3) or actually two distinct sets if v, is Type III,
Si, = CVi = \J{C\C Â£C,ViÂ£ C,\C\ > 3} and = No(vi)t
where C is the family of maximal cliques in G.
Define the maximal nonsimplicial competition cover of a graph
G, denoted as S'(G) as the set of all sets in 82(G) such that no set is properly
contained in any other. We first observe the following results.
Proposition 3.2.3 Lundgren, Maybee, and Rasmussen [46]. If G ^ Kn is a
connected interval graph, then S'(G) is a competition cover of G.
Proposition 3.2.4 Lundgren, Maybee, and Rasmussen [46]. If G ^ Kn is
a connected interval graph, then 82(G) is interval if and only if S'(G) has a
consecutive ranking.
It is interesting to note that a competition cover of a graph does
not necessarily correspond to the maximal cliques in the twostep graph. For
example, the open neighborhoods of a 6cycle form a competition cover, but
the twostep graph of a 6cycle is two triangles. So the open neighborhoods are
not the maximal cliques in the twostep graph. Figure 3.5 illustrates another
42
example in which this is the case. These examples motivate the following
question: When does the competition cover S'{G) correspond to the maximal
cliques in S2(G)? To answer this question we need the following result.
3
Figure 3.5. The family of maximal cliques is not S'(G).
Proposition 3.2.5 Lundgren, Maybee and Rasmussen [46]. If G ^ Kn is a
connected interval graph and x Â£ V(G) such that x is connected by a path of
length two to every vertex in some Si Â£ S'(G), then x Â£ S{.
Theorem 3.2.4 Lundgren, Maybee, Merz, and Rasmussen [44]. If G 9= Kn
is a connected interval graph, then G is a maximal clique in S2(G) if and only
if G Â£ S'(G).
Proof. (<=) Let G Â£ S'(G). Clearly G is a clique in S2(G). Suppose that it is
not a maximal clique. Then there exists a vertex w C such that w is joined
to every vertex in G by a path of length two. But Proposition 3.2.5 implies
w must be an element of G. This contradiction proves G must be a maximal
clique in 52(G).
(=>) Let G be a maximal clique in S2(G). Since G is interval, the
maximal cliques of G have a consecutive ranking Ci,. ., C/. We claim that
there exists a nonsimplicial vertex z such that C C Ng[z\. First we will show
there exists a vertex z such that G C Nq{z\. Suppose not. Let i be the
smallest integer such that there exists a vertex x that is an element of both Gi
43
and G, but x ^ Ci+1. This must occur since C % IVg[e]. Let j be the largest
integer such that there exists a vertex y that is an element of both Cj and G,
but y Cj1. This must occur since C % fVc[y]. Note that i must be less
than j, for if not then C C Gj. for all j < k < i. Since x and y are joined by a
path of length two, there exists a vertex z such that x and z are contained in
a maximal clique and y and z are contained in a maximal clique. Since this
ranking is consecutive and x G;+i, we conclude that z must be in a clique
Ck such that k < i. Since y Cj1, we conclude that z must be in a clique Cm
such that m> j. Since this ranking of cliques is consecutive, we conclude that
z Â£ Cp for all p such that i < p < j. Note that every vertex of C is contained
in a clique Cp such that i < p < j. Thus C C IVg[z], a contradiction. Thus
there must exist a vertex z such that C C Nq[z\.
Suppose that z is simplicial. Since G is connected and not complete,
there exists a vertex x 0 C such that x is adjacent to a vertex y Â£ C. If
y is nonsimplicial, then we are done since C C Na[y]. So assume that y is
simplicial. Then {e} U C is a clique in 82(G) containing C, a contradiction.
Therefore z is nonsimplicial.
If z Â£ C, since C is a maximal clique and z is joined to every vertex
in C by a path of length two, it follows that C = Cz. If z C, since C is a
maximal clique it follows that C = Nq(z). In either case, C Â£ S'(G).
Theorem 3.2.5 Lundgren, Maybee, Merz, and Rasmussen [44]. If G ^ Kn
is a connected 6cyclefree graph then C such that \C\ > 2 is a maximal clique
in 52(G) if and only if C G 5'(G).
Proof. (=>) Let C be a maximal clique in 52(G). First we show that there
exists z such that G C Na[z]. We will prove the result by induction on i2,
44
where R C 67. Clearly this is true if 67 = 2. By Lemma 3.2.1 it is true
if \C\ = 3. So assume that 67 > 4. Assume that there exists z such that
R C Ng[z] for R such that i2 < k < 67 and assume that /2 = A: < 1671.
Pick x E R. Let R' = R {a:}. By the induction hypothesis there exists z\
such that RI C jVc[zi]. Pick y E R such that y ^ x. Let R" = R {y}. By
induction hypothesis there exists z2 such that R" C Ng[z2]. If Zi = z2, then we
are done so assume z\ ^ z2. Then x and y are joined by a path of length two,
so we conclude that there exists z such that x,y E Ng(z). If there exists w not
equal to x,y,zi,z2 or z such that w Â£ R, then we are done since w Â£
and w G A7g[z2] implies that xzyz\wz2x is a 6cycle, so assume no such w
exists. Then R C {as, y, z1,z2,z}. Since f2 >4, at least one of the set {zi, z2}
is in R. Without loss of generality assume that Z\ E R Then Zi E ./Vg[z2]. If
z\ Ng[z] then since z\ and y are joined by a path of length two we conclude
that there exists w such that Zi,y E ATg[u;]. But then zwyz^z2xz is a 6cycle,
a contradiction. Therefore assume that {zi,z} E E(G). If z2 ^ R then since
R C iVc[z] and we are done. So assume that z2 E R. Similarly, if z2 E Ng[z],
then R C Ng[z2\ and we are done so assume z2 Ng[z\. Then x and z2 are
joined by a path of length two implies there exists w such that x,z2 E
implies xwz2ziyzx is a 6cycle. Therefore there exists z such that R C Ng[z].
Suppose that z is simplicial. Since G is connected and not complete,
there exists a vertex x ^ C such that x is adjacent to a vertex y E C. If y
is nonsimplicial we are done since C C A^y]. So assume that y is simplicial.
Then {s} U 67 is a clique in .?2(6r) containing 67, a contradiction. Therefore
there exists nonsimplicial z such that C C Ng[z]. If z E 67, since 67 is a
45
maximal clique in ^(G) and z is joined to every vertex in G by a path of length
two, G = Cz. If z $ C, since C is a maximal clique in 82(G), C = Nq(z).
(<=) Let C Â£ S'(G). By definition there exists a nonsimplicial vertex
z such that C C ^[2]. We have two cases.
Case 1: Suppose that there exists nonsimplicial z such that C = Cz.
Observe that z may be either Type I or Type III. Clearly C is a clique in
52(G). Suppose that it is not a maximal clique. Then there exists w 0 G such
that w is joined to every s Â£ C by a path of length two. So there exists x
such that z, w E Ng(x). Observe that w ^ C, C = Cz, and {w,x} Â£ E(G)
implies {w,z} ^ E(G). Since C
that {x,y} E(G). Then y and w are joined by a path of length two so
there exists a vertex u such that w,y Â£ Ng(u), where u ^ z and u ^ x. If
2 Â£ Ng(u), then since y and z are contained in a triangle we conclude that
a vertex t not equal to x and not equal to w such that y,z Â£ Na(t). Then
ztyuwxz is a 6cycle, a contradiction. So 2 Â£ Nq(u). Then u Â£ G. Suppose
that x Â£ G. If x Ng(u), then since x and 2 are contained in a triangle we
conclude that there exists t not equal to u,w nor y such that x,z Â£ iVc(t).
Then ztxwuyz is a 6cycle. Therefore x Â£ G implies that x Â£ Nq(u). But
G iVc[u] so there exists v Â£ C such that {u,u} 0 E(G). If {v,y} Â£ E(G),
we are done since zvyuwxz is a 6cycle. So {v,y} ^ E(G). Then there exists
s such that w,v Â£ Ng(s) where s is possibly x but s ^ u,s y and s ^ 2.
Then zvswuyz is a 6cycle. Therefore 2:'^ G. This implies {x,u} E(G).
But G <(_ ^[u] so there exists a vertex v Â£ G such that {u,u} ^
E(G). If {v,y} Â£ E(G), we are done since zvyuwxz is a 6cycle, so assume
46
not. Then there exists a vertex s that is possibly equal to x but not equal to
y and not equal to u such that w,v Â£ NG(s). Then zvswuyz is a 6cycle, a
contradiction. Therefore C is a maximal clique in S^G).
Case 2: Suppose that there exists nonsimplicial z such that C =
Ng{z). Observe that z may be Type II or Type III. Clearly C is a clique in
52(G). Suppose that it is not a maximal clique. Then there exists a vertex
w Â£ C joined to every s Â£ C by a path of length two.
Let x Â£ C. Then there exists y such that x,w Â£ NG(y). Since
C <Â£_ NG(y) there exists v Â£ NG(z) such that {y,u} ^ E(G). Then w and v
must be adjacent to x since otherwise there exists a distinct vertex t such that
w,v Â£ NG(t) and wtvzxyw is a 6cycle. If z is Type II, then z is contained in
a triangle, namely vxzv, a contradiction. So assume that z is Type III. Then
since C <Â£ JVg[e] we conclude that there exists u Â£ C such that {x,u} Â£ E(G).
If {/u,y} Â£ E(G), then we are done since wxvzuyw is a 6cycle. So assume that
{u,y} E(G). Suppose {w,u} ^ E{G). Then there exists a distinct vertex t
such that u and w are adjacent to t. Then utwyxzu is a 6cycle, a contradiction.
Thus {10,u} Â£ E(G). Then wvuzxyw is a 6cycle, a contradiction. Therefore
G is a maximal clique in 82(G).
Corollary 3.2.4 Lundgren, Maybee, Merz, and Rasmussen [44]. If G 9= Kn
is a connected 6cyclefree graph, then 52(G) is interval if and only if the
maximal nonsimplicial competition cover of G has a consecutive ranking.
As one might guess from the previous results, six cycles pose a prob
lem in this approach to characterizing graphs with interval twostep graphs.
We now deal with the case that 6cycles are relatively sparse in the graph. Let
47
H = abcdefa denote a 6cycle with vertices consecutively labeled. The alter
nating triples of H are {a, c, e} and {b,d,f}. It is easily observed that the
set of alternating triples of a 6cycle correspond to the maximal cliques in the
twostep graph. Figure 3.6 illustrates a graph whose twostep graph contains
a clique that is neither a set in the maximal nonsimplicial competition cover
nor an alternating triple. Notice that there is a pair of 6cycles that share
more than a single edge.
Z
Figure 3.6. The set {x,xi,x2,x3} is not in S'(G).
Theorem 3.2.6 Lundgren, Maybee, Merz, and Rasmussen [44]. If G ^ Kn
is a connected trianglefree graph such that no two 6cycles have more than a
single edge in common, and C is a maximal clique in S2(G) such that \C\ >
2, then either C = Nq(z) for some nonsimplicial vertex z in G or C is an
alternating triple from a 6cycle in G.
Proof. If \C\ = 2, then C must be the open neighborhood of a nonsimplicial
vertex with precisely two neighbors making the statement is true. If \C\ = 3,
then by Lemma 3.1.3 and the maximality of C, we observe that the statement
is true. So assume that \C\ > 4. Let R denote a subset of C. We will prove
by induction on i2 that there exists a vertex z such that C C Nq[z\.
48
Let iÂ£ = 4. Pick arbitrary x Â£ R. Let R' R {$}. By
Lemma 3.1.3 there exists y such that R' C No[y] or R' is the set of alternating
triples from a 6cycle in G. Assume that there exists y such that R' C jVc[y].
Suppose that y Â£ C. Then y is joined by a path of length two to one of its
neighbors. This is a contradiction since G is trianglefree. Thus y 0 C and
hence y R!. If x Â£ Wg[t/], then we are done so assume x 0 Wc[y]. Further
more assume that there does not exist z such that R C 1Vg[.z]. Since .R = 4,
there exists a vertex a Â£ R1. Then there exists t such that x,a Â£ Nc(t). Since
there does not exist z such that R C Nq[z] and i2 = 4 there exists b Â£ R'
not equal to a and u not equal to t such that a:, 6 Â£ Nq(u). Since there are no
triangles in G, we conclude that t and u are not adjacent to y. Furthermore,
y R. Let c denote the remaining vertex in R'. If there exists a distinct
vertex s such that x,c Â£ Ng(s), then xtaybux and xscybux are two 6cycles
with more than a singe common edge, a contradiction. So assume that no
such 3 exists. Then c must be adjacent to t or u. Without loss of generality
assume that {c, u} Â£ E[G). Then xtaybux and xtaycux are two 6cycles with
more than a single common edge, a contradiction. Thus there must exist z
such that R C Wg[z].
Assume that R' is an alternating triple from a 6cycle in G. Let a, 6, c
denote the vertices of R'. Then there exist p,q,r Â£ V(G) such that bpaqcrb
is a 6cycle in G. If there exists a vertex z such that three elements of R are
in the open neighborhood of z, then we can let R' be the set of these vertices
and we are in the former case. So assume that no such z exists. Then x is
not adjacent to p,q nor r. So there exist distinct vertices w and y such that
49
x,b Â£ Nq(w) and i,aÂ£ Na(y) Then wxyapbw and apbrcqa are two 6cycles
with more than a common edge, a contradiction. So there must exist a vertex
z such that R C Ng\z\.
This verifies the statement for 12[ = 4. Assume that the statement is
true for all R such that /2 < k < \C\ and let i2 = k < 1(71 By assumption
\R\ > 4. Pick an arbitrary vertex x Â£ R and let R' = R {a;}. By the
induction hypothesis there exists zq such that R' C iVcf^o] Suppose that there
does not exist z such that R C Nq[z\. Then {x,z0} $ E(G). Furthermore
1
there must exist distinct vertices y,z Â£ R' and a and b such that x, z Â£ iVc(o)
and x,y G Nc(b). Since there are no triangles in G, we conclude that no two
elements of R are adjacent and a,b $ R',Ng(zq). Furthermore, zq R. Since
1*1 > 4 there exists another distinct vertex w Â£ R'. If there exists a distinct
vertex c such that w, x Â£ Ng(c) we are done since xazz^ybx and xcwz0ybx are
two 6cycles with more than a single common edge. Thus w must be adjacent
to a or b. Without loss of generality, assume that {w,b} Â£ E(G). Then
xazzoybx and xbwzozax are two 6cycles with more than a single common
edge, a contradiction. This proves that for all subsets R of C, there must exist
z such that R C Ng[z\. In particular there exists z such that C C Ng[z]. Since
there are no triangles in G, we conclude that z ^ C. Since C is a maximal
clique in 82(G), we conclude that C = Ng(z). I
Define R(G) as the union of S'(G) and the set of all alternating triples
in G. Define R'(G) as the set of all sets in R(G) such that no set is properly
contained in any other.
50
Theorem 3.2.7 Lundgren, Maybee, Merz, and Rasmussen [44]. If G ^ Kn
is a connected trianglefree graph such that no two 6cycles have more than a
single edge in common and G Â£ R'(G), then C is a maximal clique in
Proof. Since G is trianglefree, every nonsimplicial vertex is of Type II. Thus
we need only consider two cases.
Case 1; Suppose that C is the open neighborhood of a nonsimplicial
vertex z. Then \C\ > 2. Clearly C forms a clique in 82(G). Suppose that it is
not a maximal clique. Then there exists w ^ Ng(z) such that w is joined to
every vertex in Nq(z) by a path of length two. Observe that there does not
exist a vertex p such that {u;} U Nq(z) C Ng(p), since Nq(z) is not properly
contained in Nc(p). Thus there exist x,y Â£ Nq(z) and a and b (not in Nq(z)
since G is triangle free) such that x,w Â£ Ng(u) and y,w Â£ No(b). There
must exist another distinct vertex u Â£ Nq(z), since Ng(z) is not properly
contained in an alternating triple. If there exists a distinct vertex c such that
w,u Â£ Nq(c), then we have two 6cycles in G with more than a single edge in
common, a contradiction. Since G is trianglefree x,y Nq(iv). Thus u must
be adjacent to a or b. In either case we have two 6cycles with more than a
single edge in common, a contradiction. Thus no such w can exist. That is,
Ng(z) is a maximal clique in 82(G).
Case 2: Suppose that C is an alternating triple {x,y,z}. Then there
exist vertices a, 6,and c such that xaybzcx is a 6cycle in G. Clearly C is a
clique in 82(G). Suppose that it is not a maximal clique. Then there exists a
vertex w joined to x,y and z by a path of length two. Since G is trianglefree,
we conclude that x,y,z ^ Ng(w). Suppose that w is adjacent to more than
one element of the set {a,b, c}. Then we have two 6cycles with more than a
51
single edge in common. Suppose that w is adjacent to one element of the set
{a, b, c}. Without loss of generality assume that {ik,c} Â£ E(G). Then since w
and y are joined by a path of length two we conclude that there exists a new
vertex d such that w,y Â£ Nc(d). Then we have two 6cycles with more than a
single edge in common. Thus w is not adjacent to a,b or c. Then there must
exist new vertices s and t such that in, a; Â£ Ng(s) and in, z Â£ Na(t). If s = t,
then xaybzsx and xaybzcx are two 6cycles with more than a single common
edge. Therefore s ^ t. But then xczbyax and xcztwsx are two 6cycles with
more than a single common edge. Thus C is a maximal clique in 82(G).
Corollary 3.2.5 Lundgren, Maybee, Merz, and Rasmussen [44]. If G ^ Kn
is a connected trianglefree graph such that no two 6cycles in G share more
than one edge, then 82(G) is interval if and only if R'(G) has a consecutive
ranking.
3.2.3 Elimination Ordering Characterizations We can de
fine elimination orderings analogous to those used in Chapter 1 to characterize
symmetric digraphs without loops that have chordal and interval competition
graphs.
An ordering Vi,v2,... ,vn is a chordal twostep elimination or
dering if and only if for all y,z Â£ {vi,v2,... ,vn}, whenever there exist u
and v such that y,V{ Â£ Ng(u) and z,u; Â£ Nq(v), there exists w such that
y,z Â£ Ng(w).
Theorem 3.2.8 Lundgren and Merz [48]. A graph G has a chordal twostep
graph if and only if G has a chordal twostep elimination ordering.
Proof. Assume that Vi,V2,... ,vn is a perfect elimination ordering of ^(C?).
This is true if and only if for u; and y,z Â£ {ul+1,i;j+2,... ,un}, whenever
52
{vi,y}, {vhz} ^ we have that {y,z} Â£ E(S2(G)). This is true if
and only if for y,z Â£ {uj+i,ui+2j "Un}, whenever there exist u and v such
that y,Vi Â£ Nq(u) and z,Vi Â£ Nq(v), there exists w such that y,z Â£ Nq(w).
This is true if and only if Vi, v2,... ,vn is a chordal twostep elimination order
ing of G.
An ordering v^,v2,... ,vn is an interval twostep elimination or
dering if and only if for all i,j, and k such that i < j < k, whenever there
exists u such that Vi,Vk Â£ Ng{u), there exists v such that vj,Vk Â£ Ng(v).
Theorem 3.2.9 Lundgren and Merz [48]. A graph G has an interval twostep
graph if and only if G has an interval twostep elimination ordering.
Proof. Assume that v\,v2,... ,vn is an interval elimination ordering of S2(G).
This is true if and only if for i < j < k, whenever Â£ E(S2(G)), we
have that {vj,Vk} Â£ E(S2(G)). This is true if and only if for i < 3 < k,
whenever there exists u such that V{,Vk Â£ Ng(u), there exists v such that
Vj,Vk Â£ Ng(v). This is true if and only if vi,v2,. . ,vn is an interval twostep
elimination ordering.
3.3 Strongly Connected Digraphs
Which graphs are the competition graphs of strongly connected di
graphs? While many digraphs modeling communication networks may be
symmetric, it is more important that they be strongly connected.
Let @e(G) denote the edge clique cover number of G, i.e., the
smallest number of cliques that cover all the edges of G. Let i(G) denote
the number of isolated vertices in a graph G. We begin with the following
proposition that we state without proof.
53
Proposition 3.3.1 Roberts and Steif [66]. G is a competition graph of a
digraph that has no loops if and only if G ^ K2 and Qe(G) < P(Gr).
Will the condition given in Proposition 3.3.1 work for strongly con
nected digraphs as well? The graph in Figure 3.7 illustrates that it will not
work. While Qe(G) = 9, unfortunately Qe{G) )i(Cr) > 9. This is important,
as indicated by the next lemma.
<>
Figure 3.7. A competition graph.
This graph is the competition graph of a loopless digraph, but not one
that is strongly connected.
Lemma 3.3.1 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. If
G is the competition graph of a strongly connected digraph, then Qe(&) +
i(G) < V(G).
An ordering Di, D2,..., Dk of the strongly connected components of
a digraph D is topological if and only if for x Â£ V(Di) and y Â£ V(Dj),
whenever (x,y) Â£ A(D) it follows that i < j.
Theorem 3.3.1 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21].
If G is a graph with no isolated vertices such that G ^ K2 and &e{G) < n,
then G is the competition graph of a strongly connected digraph.
Proof. By Proposition 3.3.1, G = C(D) for some (loopless) digraph D.
Among all such digraphs, let D be one with the smallest number k of strong
components and suppose that k > 2.
Let Â£>1, D2,..., Dk be a topological ordering of the strongly con
nected components of D. Observe that since G has no isolated vertices and
54
every vertex in D has an outgoing arc, this ordering implies there are at least
two vertices in Dk
Case 1: Suppose that V(D\) = {x}. Observe that I(x) = 0. Since
G has no isolated vertices, we conclude that O(x) 0. Let (x,y) Â£ A(D)
for some vertex y. Then create a new digraph D' by simply adding the arc
(y,x) to D. Then D' has fewer strong components than D and C(D') = G, a
contradiction.
Case 2: Suppose that V(.Di) > 2 and there is a vertex y Â£ Dk such
that (x,y) ^ A(D) for some x Â£ D\. Since ) > 2 and Dk(is strongly
connected, we conclude that I(y) ^ 0. As in Case 1 we construct a new
digraph D. Let lD'{y) = Id(x) and Id>(x) = Id(v) with all other arcs in D'
staying the same as in D. We claim that D\ U Dk is strongly connected in
D'. To prove this, we must show that given arbitrary vertices u,w Â£ D\ U Dk,
there is a path from u to w.
Let u Â£ V(Di) not equal to x and w Â£ V(Dk) not equal to y. In D,
u reached x, so in D', u reaches y. In D, y reached w and none of those arcs
have changed. Therefore u reaches all w Â£ V(Dk) in D'. In D, x reached u
and none of those arcs have changed. Therefore x reaches w. Therefore every
vertex in D\ reaches every vertex in Dk
In Z), w reached y, so in D', w reaches x. In Z), x reached u and
none of those arcs have changed, thus w reaches u. In D, y reached w and
none of those arcs have changed, thus y reaches u. Therefore every vertex in
Dk reaches every vertex in D\.
Since every vertex in Di reaches every vertex in Dk and every vertex
55
in Dk reaches every vertex in D\, every vertex in D\ reaches every vertex in
D\ and every vertex in Dk reaches every vertex in Dk, completing the proof
that Di U Dk is strongly connected. So D' has fewer strong components than
D and C(D') = G, a contradiction.
Case 3: Suppose that V(.Di) > 2 and every vertex in D\ has an arc
to all vertices in Dk. Let x Â£ V(D\) an^ V Â£ V(Dfc). Since every vertex in D\
has an arc to y, D\ is a clique in C(D) and all arcs in the inset of x can be
removed in a new digraph D' without changing the competition graph. Now
let ID,(x) = {i/}, and let all other arcs in D' remain the same as in D. Then
D\ U Dk is strongly connected in D' by arguments analogous to that in Case 2,
so D' has fewer strong components than D and C(D') = G, a contradiction.
Thus D is strongly connected, since we assumed that k > 2 was the
smallest number of strongly connected components and in all cases reached a
contradiction. fl
Corollary 3.3.1 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21].
If G is a graph such that G ^ K\ or K2 and 0b(G) + i(G) < V(G), then G
is the competition graph of a strongly connected digraph.
Proof. If G is a graph on V(Cr) isolated vertices, then G is the competition
graph of a cycle. So assume that G is not a graph on IV^G)! isolated vertices.
If G has no isolated vertices, then we are done by the previous theorem. So
assume i(G) = k ^ 0. Let /*. denote the subgraph of G induced by the
k isolated vertices. Then there exists a graph G' such that G = G' U Ik
Observe that G' has m = V(G) k vertices, none of which are isolated and
Qe(G') < m. Let am+i, am+2,..., am+k denote the k vertices of Ik
If G' = Ki, then let x and y be the vertices of G'. Let D be the
56
digraph with cycle xpam+iam+2 ..am+k and the arc (,am+1). Then C(D) =
G and D is strongly connected. If G1 ^ K2, then by the previous theorem there
exists a strongly connected digraph D' such that C(D') = G'. Create D as
follows: Let V{D) = V(D') + Ik. Choose x V(D'). Let ID(am+1) = ID,(x),
Id{x) = am+k and (aj, 0{+i) 6 A' for i = m + l,m + 2,... ,m f (A: 1). Leave
all remaining arcs of D' in D. Then none of the competitions in D' have been
changed so C(D) = G. We claim that D is strongly connected.
Since D' is strongly connected x reaches every vertex in V(D'). The
only arcs that have changed are arcs from 7jqi(x). Therefore x reaches every
vertex in V(D') in D. Let u be a vertex in Id>(x). Clearly u reaches every
vertex in Ik in D and since x reaches u, x reaches every vertex in Ik. Therefore
x reaches every vertex in D. Let v be an arbitrary vertex in V(D'). We
claim v reaches x. Let u 6 7d<(). There is a path from u to x, namely
uam+1om+2 ... and v reaches u in D because v reaches u in D'. Therefore
v reaches a: in D. Clearly an arbitrary vertex v in Ik reaches x. Thus every
vertex in Ik and V(D') reaches s in D. Thus D is strongly connected. B
Corollary 3.3.2 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21].
If G is a graph such that ]F(G) = n > 3, then G is the competition graph of
a strongly connected digraph if and only if Qe{G) + i(G) < n.
3.3.1 Generalization to pCompetition Graphs The graph
in Figure 3.7 is not the competition graph of strongly connected digraph. It
is, however, the 2competition graph of a strongly connected digraph as shown
in Figure 3.8. By making p > 2, we can strengthen Theorem 3.3.1.
Theorem 3.3.2 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If
G is the pcompetition graph of a loopless digraph for p > 2, then G is the
pcompetition graph of a strongly connected loopless digraph.
57
I
Figure 3.8. A 2competition graph.
G is the 1competition graph of a loopless digraph, but not one that is
strongly connected. G is the 2competition graph of a loopless strongly
connected digraph. The digraph is not strongly connected but G is its
2competition graph. Adding the arc (9,5) makes the digraph strongly
connected, but does not change its 2competition graph.
Proof. Observe that if G has V(G) isolated vertices and we let
D be a directed cycle on IV'(G)! vertices, then CP(D) = G for D strongly
connected. Therefore we may assume that G has at least one edge.
Let D be a loopless digraph with fewest number of strongly connected
components such that CP(D) = G. Let Di, D2,..., Dk be the topologically
ordered strongly connected components of D.
If D\ is Dk, then we are done so assume that k > 2. Since G has
at least one edge, we can assume that Di contains a vertex x with at least
one outgoing arc (if D\ does not contain such a vertex then D2,... ,Dk,D\ is
a topological ordering of the strongly connected components). We then have
three cases.
Case 1: Suppose that V{Di) = {a:}. Then /(x) = 0 and 0(x) ^ 0.
Let y be a vertex such that there is an arc from x to y. Let Dy denote the
58
strongly connected component containing y. Create D' by adding the arc (y, x)
to A{D). The set of vertices competing with x at least p times is unchanged.
Since y is the only vertex with an arc to x, the set of vertices competing with y
is unchanged. Thus CP(D) = G implies CP(D') = G and D' has fewer strongly
connected components than D, a contradiction.
Case 2: Suppose that V(.Di) > 2 and there exists y E Dk, x E D\
such that there is no arc from x to y. Suppose that  V( jD/t)  < p. If there exists
q ^ Dk such that (g,y) E A(D), create D' by adding (y,q) to A(D). Since
D is loopless, y competes at most p 1 times with any vertex. In particular,
y competes at most p 2 times for a vertex in Dk and once for q. Therefore
the set of vertices competing with y at least p times is unchanged. Then
CP(D) = G implies CP(D') = G. Letting Dq denote the strongly connected
component containing q, we observe that Dk U Dq is strongly connected in D'.
That is, D' has fewer strongly connected components than Â£), a contradiction.
Thus all arcs incoming at y originate in Dk Create D' by adding
(x,y) and (y,x) to A(D). Then the set of vertices competing with x at least
p times in D has not changed since at most p 1 vertices have arcs to y.
In particular, at most p 2 vertices in Dk and x have arcs to y. The set of
vertices competing with y at least p times in D has not changed since y has
at most p 1 outgoing arcs. In particular, y has arcs to at most p 2 vertices
in Dk and x. Thus CP(D) = G implies CP(D') = G and D\ U Dk is strongly
connected in D'. That is, D' has fewer strongly connected components than
D, a contradiction.
59
Thus V(Dfc) > p. Since V(Di) > 2,1(x) ^ 0 and since V(Z?fc) >
p > 2, I(y) jz 0. Create D' from D by switching the insets of x and y and
leaving all other arcs the same. No competitions have changed. Therefore
CP(D) = G implies CP(D') = G. Since F(Di) > 2 and \V(Dk)\ > 2, D\ U Dk
is strongly connected in D', i.e., D' has fewer strongly connected components
than D, a contradiction.
Case 3: Suppose that V(.Di) > 2 and for all x 6 V(Di) and all
y Â£ V(Dk), (x,y) Â£ A{D). Suppose that F(Z>fc) < p. Create D' by adding
arc (y,x) to A(D). Then the set of vertices competing with y at least p times
has not changed since y has arcs to at most {jp 1) vertices (namely at most
(p 2) in Dk and x). Since the set of vertices competing with x at least p times
has not changed, CP(D) = G implies CP(D') = G. Since D\ U Dk is strongly
connected in D', D' has fewer strong components than Z), a contradiction.
Thus y(.Djfc) > p. Since every vertex in D\ has an arc to every
vertex in Dk, we conclude that D\ is a clique in G. Thus we can remove arcs
between vertices strictly in D\ and the pcompetition graph is unchanged.
Pick an arbitrary vertex x Â£ V{D\). Create D' by deleting all arcs incoming
at x and adding arc (y,x) to A(D). The set of vertices competing with y at
least p times is unchanged since y is the only vertex with an arc to x. Thus
CP(D) = G implies CP(D') = G. Since F(I>i) > 2 and  V(Z>*) >2,AU Dk
is strongly connected in D', i.e., D' has fewer strongly connected components
than D, a contradiction.
Since in each case the contradiction implies D has fewer than k
strongly connected components where k > 2, we must have k = 1, i.e., D
60
is strongly connected. Therefore every graph that is the pcompetition graph
of a loopless digraph is the pcompetition graph of a strongly connected loop
less digraph.
3.4 Hamiltonian Digraphs
One special class of the strongly connected digraphs is the Hamilto
nian digraph. In looking at examples of graphs on IV^G)! vertices satisfying
0e(C7) + i(G) < F(G), we discovered that in many cases, not only are the
graphs competition graphs of strongly connected digraphs, but frequently the
digraphs are Hamiltonian. Certainly Qe(G) + i(G) < V(G) is necessary,
but is it sufficient? Our first result is analogous to Brigham and Duttons [7]
characterization of the competition graphs of acyclic digraphs.
Theorem 3.4.1 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21].
A graph G with V(G) = n is the competition graph of a Hamiltonian digraph
if and only if G has a labeling a\, a2,..., an of its vertices and an edge clique
covering {Ci, C2,..., Gn} satisfying a; ^ C, for i = 1,... ,n and a, Â£ Ci+1 for
i = 1,... ,n 1 and an Â£ C\.
Proof. (=>) Suppose that G = C(D) for D Hamiltonian and ai,..., an is a
Hamiltonian cycle in D. Let Ci = /(a*). Then {Gi,..., C} is an edge clique
covering of G. Furthermore 0 Ci since D has no loops and a; Â£ Ci+1 since
(a;,ai+i) is an arc. Finally an Â£ Ci, since (an,ai) is an arc.
(<Â£=) Suppose that G has a labeling ai,a2,... ,an of its vertices and
an edge clique covering {Ci, C2,..., Gn} satisfying ttj ^ Ci for i = 1 ,...,n
and ai Â£ Ci+1 for i = 1,... ,n 1 and an Â£ C\. Construct D with vertices
ai,...,an such that /(a{) = Ci. Then ai Â£ Ci+1 implies that (ai, o;+i) is an
arc and an Â£ C\ since (an>ai) Is an arc. Therefore D is Hamiltonian. From
61
the construction, it is clear that C(D) = G. I
We now give several classes of graphs that satisfy the condition of
Theorem 3.4.1.
Corollary 3.4.1 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21].
If G is a cycle, then G is the competition graph of a Hamiltonian digraph.
Proof. Let oi,..., an be the vertices of the cycle. Let C\ = {an_i, an}, (72 =
{an,ai},C'3 = {ai,a2}, ,
clique covering of G satisfying Theorem 3.4.1.
Corollary 3.4.2 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21].
The complete graph Kn for n > 3 is the competition graph of a Hamiltonian
digraph.
Proof. Let V(Kn) = {ai,..., an}. Let C{ V {a;}. Then {Ci,..., Cn} is
an edge clique cover of Kn satisfying Theorem 3.4.1.
Corollary 3.4.3 If G is a chordal graph on n > 3 vertices, then G is the
competition graph of a Hamiltonian digraph.
Proof. The proof is by induction on the number of vertices. It is easy to
verify the theorem for chordal graphs with three vertices. Assume that the
result is true for all chordal graphs with less than n vertices, and suppose
that G is a chordal graph with n vertices. If G = Kn, then we are done by
Corollary 3.4.2. So suppose that G Kn, and let x be a simplicial vertex in
G and let G' G {a:}. Since G' is chordal, G C(D') where D' has the
Hamiltonian cycle aia2... a_i. Since G Kn, there is a vertex Oj 0 Wg[],
the closed neighborhood of x. Note that ATct] is a clique since x is simplicial.
We define a new digraph D as follows: Jd(x) = Ii)i{aj)\Iu{oj) = iVc[a;]; and
62
/d(o) = lD'(ai) for all i ^ j. Clearly C(D) = G and aia2... a.jixaj ... an_i
is a Hamiltonian cycle in D.
Corollary 3.4.4 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21].
If G is an interval graph on n > 3 vertices, then G is the competition graph
of a Hamiltonian digraph.
Lemma 3.4.1 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. If
G is a connected trianglefree graph with 0js(G) < V(G), then G is a tree
or a tree with an additional edge.
Proof. Since G is trianglefree, the cliques in any clique cover of minimum
cardinality are just the edges of G. Since G is connected and Oe(G) <  V(G?),
then either \E[G)\ = ^(G)! 1 and G is a tree or IÂ£(Cr) = V"(t?) and G is
a tree plus an edge.
Corollary 3.4.5 If G is a connected trianglefree graph on n > 3 vertices
with Qe(G) < n, then G is the competition graph of a Hamiltonian digraph.
Proof. The proof is by induction on n. For n = 3, a path on three vertices
is the only such graph, and by Corollary 3.4.3 a path on three vertices is the
competition graph of a Hamiltonian digraph. Assume that the result is true
for all connected trianglefree graphs with n vertices and suppose that G has
7ifl vertices. By Lemma 3.4.1, G is either a tree or a tree plus an edge.
If G is a tree, we are done by Corollary 3.4.3, so assume that G is a tree
plus an edge. If G has no pendant vertices, then G is a cycle and the result
follows from Corollary 3.4.2, so assume that G has a pendant vertex. Let x
be a pendant vertex in G. Then G {s} satisfies the induction hypothesis.
Let ax,...,an be the consecutively labeled vertices of the Hamiltonian cycle.
Since n > 3 and deg(a:) = 1, there is a vertex a,j such that {x,aj} 6 E{G)
and a vertex a* such that {x,a,k} ^ E(G). We construct a new digraph D' as
63
follows: Id'{z) = lD{ak),lD'{ak) = {a:, aj},lD'(di) = /ij(a,) for i ^ k. Then
C{D') = (? and ... ajjiaa*.... an is a Hamiltonian cycle in D'.
7s the condition Qe{G) + i(G) < F(G) sufficient for G to be the
competition graph of a Hamiltonian digraph? Consider the graph in Figure 3.9.
The reason this graph is not the competition graph of a Hamiltonian digraph
is a consequence of the following theorem.
7
1 2 3
Figure 3.9. An example.
This is not the competition graph of a Hamiltonian digraph. Since the
component on 6 vertices has a minimum of 7 sets in an edge clique
cover, we conclude no edge clique cover for the entire graph will have
a system of distinct representatives.
Theorem 3.4.2 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21].
If G is the competition graph of a Hamiltonian digraph, then G has an edge
clique covering that has a system of distinct representatives among the vertices
of the cliques.
Proof. Assume that G is the competition graph of a Hamiltonian digraph
D. Let ai,a2,...,an denote the consecutively labeled vertices of a Hamilto
nian cycle in D. Then the family of insets of the vertices of D are an edge
clique cover of G. Furthermore, for i 1, 2,..., n 1, choose a; to represent
7(ai+i) and choose an to represent J(ai) and we have a system of distinct
representatives.
Corollary 3.4.6 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21].
If G is the competition graph of a Hamiltonian digraph, then Qe(G{) < ni>
where G has connected components Gi, each with n{ vertices.
64
Proof. Suppose that G is the competition graph of a Hamiltonian digraph.
If 0#(G;) > Tii for some z, then every edge clique cover C of G requires more
than 7i; cliques to cover G;, and each of these cliques contains only vertices
of G;. It follows that C does not have a system of distinct representatives, a
contradiction. I
Corollary 3.4.7 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21].
If G is a graph satisfying 0js(G) = V(G) and G is the competition graph of a
Hamiltonian digraph, then G contains no adjacent pair of simplicial vertices.
Proof. Suppose that G = C{D) for D Hamiltonian and that Qe{G) = V(G).
Suppose that x and y are simplicial vertices in G that are adjacent. Then
1Vg[x] U 1Vg[i/] is a clique, so G requires at least F(G) 1 cliques
in any clique cover. But then no edge clique cover of G contains a system of
distinct representatives, a contradiction.
Theorem 3.4.3 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21].
If G is a trianglefree graph, then G is the competition graph of a Hamiltonian
digraph if and only if 0Â£j(G,) < V(G;) for each connected component G, of
G.
Proof. One direction of the proof follows from Corollary 3.4.6. Suppose
that Qe(Gi) < 7i; for each component G;, 1 < i < k, where k is the number of
components in G. If ti; > 3, then G; is the competition graph of a Hamiltonian
digraph D; with Hamiltonian cycle a\,... ,aln. by Lemma 3.4.1. If ti; = 2,
then G; = K2, and we let D{ be the arc (o^o^) (observe C(L>;) G; in this
case). If 71; = 1, then H; is the isolated vertex a\. Let D = U^=1Â£);. We
can now form a digraph D' such that C(D') = G and D' is Hamiltonian. For
i = l,... k 1, if ti; > 3, let /Â£>'(a j+1) = Id{o\) If nk > 3, let Id'{o\)
For i = 1,..., k 1, if 71; = 2, let /r>f(a\+1) = {a^Oj}. If ti*. = 2, let
65
/d'(oJ) = {ai,aÂ§}. For i = 1,..., fc 1, if ra, = 1, let /Â£><(ai+1) = a\. If
nk = 1, let Id>(cl\) = aj. Let all other insets in D' be the same as in D.
Observe we have not changed any existing competitions from C(D) to C(D').
Thus if rii ^ 2, we get all the edges of Gj in C(D). For i 1,..., k 1,
if rii 2 then = {ai>a2} gives the edge in G, = K2. Similarly,
lD'(a\) does this if n*. = 2. Thus C(D') = G and D' has the Hamiltonian
cycle a\...a}nia\...al2 ...of
Is the converse of Theorem 3.f.2 true? This is a very interesting open
question. The following theorem proves the converse of Theorem 3.4.2 if loops
are allowed.
Theorem 3.4.4 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21].
If G is a graph on n > 3 vertices satisfying 0js(G) < n and i(G) = 0, then
G is the competition graph of a Hamiltonian digraph possibly having loops if
and only if G has an edge clique covering {Gi,..., Gn} that has a system of
distinct representatives.
Proof. One direction of the proof follows from Theorem 3.4.2. So suppose
that G has such an edge clique covering C. Let the representative for clique
Ci be labeled
on. Define a digraph D by I(fli) = Gj. Then G = G{D) and aia2.. .on is a
Hamiltonian cycle in D. M
3.4.1 Generalization to pCompetition Graphs: Construc
tions Can we generalize the results of the previous section to pcompetition
graphs where p > 2? First we observe the generalization of Theorem 3.4.2.
Theorem 3.4.5 If G is the pcompetition graph of a Hamiltonian digraph,
then G has a pedge clique cover with a system of distinct representatives.
66
Proof. Assume that G is the pcompetition graph of a Hamiltonian digraph
D. Let a\ ,a2,...,an denote the consecutively labeled vertices of a Hamiltonian
cycle in D. Then the family of insets of the vertices are a pedge clique cover
of G. Furthermore, for i = 1,2,... ,n 1, choose ai to represent 7(aj+i) and
choose an to represent I(ai) and we have a system of distinct representatives.
The following constructions will be used in the next section to char
acterize classes of graphs that are the pcompetition graphs of loopless Hamil
tonian digraphs.
Lemma 3.4.2 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If
G is a connected graph such that G is the pcompetition graph of a loopless
Hamiltonian digraph, then adding a pendant vertex x to G results in a graph
G' that is the pcompetition graph of a loopless Hamiltonian digraph.
Proof. Let D be a loopless Hamiltonian digraph such that CP(D) = G. Let
v\v2...vn denote a Hamiltonian cycle in D. Create D' from D as follows.
Add vertex x. Let Vi denote the vertex adjacent to x in G'. Let Id'{x)
i). Let = {aj,Ui}. Observe that viv2 .. .Vi2xvi\Vi.. .vn is a
Hamiltonian cycle in D'. Since G is connected u; has outgoing arcs to at least
p other vertices of D. Let x have an outgoing arc to p 1 of these vertices.
Then CP(D') = G', where D' is Hamiltonian.
Corollary 3.4.8 Langley, Lundgren, McKenna, Merz, and Rasmussen [38].
If T is a tree and has a subtree that is the pcompetition graph of a loopless
Hamiltonian digraph, then T is the pcompetition graph of a loopless Hamil
tonian digraph.
Proof. This follows from Lemma 3.4.2, since we may add pendant vertices
successively to the subtree, obtaining T.
67
Lemma 3.4.3 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If
T is a tree that is the pcompetition graph of a loopless Hamiltonian digraph,
then adding a pendant vertex x to an internal vertex results in a tree T' that
is the (p + l)competition graph of a loopless Hamiltonian digraph.
Proof. Let D be a loopless Hamiltonian digraph such that C(D) = T. Let
V\V2 .. .vn denote a Hamiltonian cycle in D. Create D' from D as follows: Add
x to D. Let Ij}>(x) = V. Let vi denote the vertex adjacent to x in T'. Let vj
and Vk denote two vertices adjacent to V{ in T and T'. Since {vi,Vj} G E(T),
we conclude that V{ and Vj have arcs to at least p common vertices. Let x
have an arc to these vertices. Since G E(T), we conclude that V{ and
Vk have arcs to at least p common vertices. Furthermore there exists vm in
this set of p vertices such that vm has no arc from Uj, since T is a tree. Let x
have an arc to this vertex. Then no previous competitions have changed since
all vertices have arcs to while x and u; compete at least p + 1 times. Then
viv2 ... vm_ixvm .. .vnvi is a Hamiltonian cycle in D'. Therefore CP(D') = T',
where D' is Hamiltonian.
A branch of a tree is a path of the tree with the vertex at one end
adjacent to an internal vertex. A maximal branch has the further property
that the other end vertex is a pendant vertex.
Lemma 3.4.4 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If
T is a tree that is the pcompetition graph of a loopless Hamiltonian digraph
and T1 is a tree produced from T by adding a branch of l new vertices where
l > 2, then T' is the ^competition graph of a loopless Hamiltonian digraph
for p
Proof. The proof is by induction on l. If l = 2, observe that T' is a p
competition graph of a loopless Hamiltonian digraph by Corollary 3.4.8. To
show T' is a (p+ l)competition graph of a loopless Hamiltonian digraph, add
68
the first vertex of the branch as indicated in Lemma 3.4.3 creating a tree that
is the (p + l)competition graph of a loopless Hamiltonian digraph. Then by
Corollary 3.4.8, T' is the (p + l)competition graph of a loopless Hamiltonian
digraph.
Assume that the statement is true for the addition of a branch on
l < n new vertices and consider the addition of a branch on l = n new vertices.
By the induction hypothesis, the addition of the first / 1 vertices produces
a tree that is the ^competition graph of a loopless Hamiltonian digraph for
p
a loopless Hamiltonian digraph for p < k < p + l 2. It remains to be shown
that T' is a (p + l l)competition graph of a loopless Hamiltonian digraph.
Let Vi,v2,... ,V[ denote the consecutively labeled vertices of the branch
such that Vi is adjacent to an internal vertex, v0, of T. Let D be a loopless
Hamiltonian digraph such that CP(D) = T. Create D' from D as follows.
Direct an arc from all vertices of T to v\,..., vj_i. Observe that this preserves
all adjacencies from CP(T) in Cp+i\(T').
Since vQ is an internal vertex, there exists vertices t and u adjacent to
Vo in T. Let S denote a set of p + 1 vertices, p of which both t and vq have arcs
directed toward in D and 1 of which u and vq have an arc directed toward,
but t does not. Direct an arc from all vertices in the branch to all vertices of
S.
For i = 1,...,Z 1, direct an arc from u; to all vertices Vk, where
k = 0,1,...,/., but k is not equal to i or i 1. Direct an arc from vi to all
vertices Vk, where k = 0,1,...,/ 3. Observe that nonconsecutively labeled
69
vertices of the branch, and Vk, compete at most (p+ 1) + (Z + 1 4) times in
D' In particular, they compete for for (p + 1) vertices of S and all vertices of
the branch except Vi,Vii,Vk and Vk1. Consecutively labeled vertices of the
branch, V{ and Uj+1, compete at least (p+l) + (Z+l 3) times. In particular,
they compete for {jp + 1) vertices of S and all vertices of the branch except
Vii,v, and Wi+1. Observe that v\ and vq compete at least (p + 1) + (Z 2)
times, while for all other vertices Uj of the branch, and v0 compete at most
(p + 1) (Z 3) times, since vq has no arc to vi and vi has no arc to 2.
Now consider an arbitrary vertex v Â£ T other than v0 and a vertex
Vi of the branch. Since T is a tree, v can have at most p arcs to vertices of
S, while Vi has arcs to Z 2 vertices in the set {u0,... ,uj_i}. Therefore v and
Vi compete at most p + Z 2 times. Thus Cp+i\{D') = T'. Furthermore if
X\X2 ... xn denotes a Hamiltonian cycle in D and xt is any vertex of S, then
*12:2 ... Xi\V\V2 ... viXiXi+i... xnx\ is a Hamiltonian cycle in D'.
Lemma 3.4.5 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If
G\ and G2 are the pcompetition graphs of loopless Hamiltonian digraph for
p > 2, then G\ U G2 is the pcompetition graph of a loopless Hamiltonian
digraph.
Proof. Let D\ and Z>2 be loopless Hamiltonian digraphs such that Cp(Di) =
G\ and Cp(D2) = G2. Let V\V2 .. .vni and X\X2 .. .xn3 denote a Hamiltonian
cycle in Di and D2 respectively. Let v; be an arbitrary vertex in D\ and xi
and arbitrary vertex in D2. Create digraph D from D\ and D2 as follows. Let
loiyi) be the inset of Xi in D2] similarly, let be the inset of Vi in D\.
Then CP(D) = G\ U G2 and v^v2 ... ViiXiXi+i ... xn2Xi ... XiiViVi+i ... vnivi
is a Hamiltonian cycle in D.
70
3.4.2 Utilizing the Constructions Before we can employ the
construction of the previous section, we must establish some base graphs as
the pcompetition graphs of loopless Hamiltonian digraphs.
Lemma. 3.4.6 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If
G is a cycle on n > p f 3 vertices where p >2, then G is the pcompetition
graph of a loopless Hamiltonian digraph.
Proof. Let Ui,U2,... ,vn denote the consecutively labeled vertices of G. Cre
ate digraph D as follows. Let
ID(vi) {^i+lmodnj ^i+2modnj j ^i+p+lmodn}*
Then V{ and ui+lmodT1 compete p times, namely for ul+2modn, ,^+Pmodn, and
Vi+P+imodn Consider nonconsecutive vertices V{ and Vk There are 4 vertices
for which and uj. do not compete, namely Vi, Vk, ^{+imodn and wjfc+imodn
Thus Vi and Vk compete for at most p 1 vertices. Therefore CP(D) = G and
^l^n^ni ^2^1 is a Hamiltonian cycle in D.
Lemma 3.4.7 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. The
complete graph Kn on n > p+2 vertices is the pcompetition graph of a loopless
Hamiltonian digraph.
Proof. Let V(Kn) = {ui,U2,... ,t;n} be the vertices of Kn. Create D as
follows. Let Id(vi) = V(Kn) {uj}. Then CP(D) = G and that v\Vi.. .vnv\
is a Hamiltonian cycle in D.
Lemma 3.4.8 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. The
complete graph minus one edge on n > 2p + 1 vertices is the pcompetition
graph of a loopless Hamiltonian digraph.
Proof. Let G be the complete graph minus one edge for n > 2p + 1. Label
the vertices of G, i>i,V2,... ,vn such that {!,!>} is the missing edge. Create
71
D as follows. For 1 < i < [], let Ioiyi) = V{G) {uj,u1}. For [] < i < n,
let Id{v) = V(G) {uj,vn}. Let Id(v^\) = V(G) {vpaj} if n is odd, and
V{G) {npaj,!;!} if n is even.
Then all pairs compete at least 1 > p times except for V\ and
vn which compete at most once and
'ynt,r?vyr?iivr?i2 ^
is a Hamiltonian cycle in D.
Lemma 3.4.9 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If
G is a path on n > p + 3 vertices, then G is the pcompetition graph of a
loopless Hamiltonian digraph.
Proof. We need only verify this result for n = p + 3 by Corollary 3.4.8.
Let Vi,v2,... ,vn be the consecutively labeled vertices of G. Create D as
follows: for i p + 1, let ID(vi) = V(G) {u.,ui+linodri}; let ID{vP+1) =
V(G) {wi,Uj+lmodn,i;;+2modn} Observe that viv2 ... vnvx is a Hamiltonian
cycle in D. Then u, and u,+1 compete at least p times, namely for
^i+2modn) ^i+3modnj ) ,^i((p+l)modT^
Since V{ does not have an arc to V{ or Uj+imodp, the nonconsecutive vertices
compete at most p 1 times. That is, CP(D) = G where D is Hamiltonian.
We would like to generalize this result to all tree. First we will start
with a subclass. A caterpillar is a tree such that the removal of all pendant
vertices yields a path. This path is called the spine of the caterpillar.
Theorem 3.4.6 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If
G is a caterpillar on n > p + 3 vertices, then G is the pcompetition graph of
a loopless Hamiltonian digraph.
72
Proof. We need only verify this result for n = p + 3 by Corollary 3.4.8.
Let vi,v2,... ,vq denote the consecutively labeled vertices of the spine of G.
Observe q > 3. Since a path on 3 vertices is the competition graph of a
loopless Hamiltonian digraph D, if q = 3 create D' by successively adding
all but one of the remaining pendant vertices as in Lemma 3.4.3. Then add
the final pendant vertex as in Lemma 3.4.2. Then D' is Hamiltonian and
CP(D') = G. If q > 3, then the spine of G is the rcompetition graph of a
loopless Hamiltonian digraph D by Lemma 3.4.9, where r = q 3. Create D'
by successively adding the remaining pendant vertices to D' as in Lemma 3.4.3.
Then D' is Hamiltonian and CP(D') = G.
Theorem 3.4.7 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If
T is a tree on n > 2p vertices, where p > 2, then T is the pcompetition graph
of a loopless Hamiltonian digraph.
Proof. By Corollary 3.4.8 we need only consider the case n = 2p. The case
p 2 can be verified by examination of all possible trees so p > 2. Let q be
the length of the longest path P in T. If q > p + 3, T is the pcompetition
graph of a loopless Hamiltonian digraph by Lemma 3.4.9 and Corollary 3.4.8.
If? < 4, T is a caterpillar and we are done by Theorem 3.4.6.
Saving the case q = 5, consider the case that q > 6. Then n =
2p > 2p q + 6. Thus, if we can show that a tree on 2p q + 6 vertices is
the pcompetition graph of a loopless Hamiltonian digraph, we are done by
Corollary 3.4.8. Assume that T is such a tree with maximum path P.
By Lemma 3.4.9, P is the (g 3)competition graph of a loopless
Hamiltonian digraph. Construct a sequence of subtrees To,Ti,...,Tfc where
To is a path, T* is T and T is constructed from T{_i by the addition of a
73
maximal branch. Let rij be the number of vertices added to to get T,.
Theorems 3.4.3 and 3.4.4 guarantee that T is a (q 3 + 5Z}=1 ki)competition
graph where ki = max{n; 1,1}. The worst case occurs when each addi
tional branch adds two new vertices. In this case, T is a (q 3+.7')competition
graph, where j is half the number of vertices added to To. Since there are q
vertices in To, we add 2p q + 6 q = 2(p (q 3)) vertices to obtain T, and
conclude that T is a p competition graph, where p = (q 3) + (p (q 3)).
If q = 5, T must have a maximal branch with one vertex; otherwise
all branches of T are of length 2, i.e., T has an odd number of vertices, a
contradiction since n = 2p. Remove this branch. The resulting tree has
2p 1 = 2p 2 g + 6 vertices and iÂ§ therefore, by the previous case, a (p 1)
competition graph of a loopless Hamiltonian digraph. Using Lemma 3.4.3 we
conclude T is the pcompetition graph of a loopless Hamiltonian digraph.
Corollary 3.4.9 Langley, Lundgren, McKenna, Merz, and Rasmussen [38].
If G is a forest and all maximal subtrees of G have n > 2p vertices, where
p > 2, then G is the pcompetition graph of a loopless Hamiltonian digraph.
3.4.3 Classes of 2Competition Graphs Using induction on
the number of vertices in the graph, we can establish that chordal graph on at
least 5 vertices are the 2competition graph of loopless Hamiltonian digraph.
An alternate approach for dealing with the base case must be employed in
order to generalize this result to pcompetition graphs where p > 3.
Lemma 3.4.10 Langley, Lundgren, McKenna, Merz, and Rasmussen [38], If
G is chordal on n > 5 vertices and not complete, then G is the 2competition
graph of a loopless Hamiltonian digraph D in which every maximal clique of
G is contained in at least one inset of D.
74
Proof, (by induction on n) If n = 5, we get trees from Theorem 3.4.7 and
verify the remaining cases by considering all possible graphs (see Figures 3.10
and 3.11). Assume that the statement is true for chordal graphs that are not
complete on n = k > 5 vertices and let G be such a chordal graph on n = k +1
vertices. Let Â£ be a simplicial vertex in G. Consider G' = G {x}.
Case 1: Suppose that G' is complete. Suppose that there is exactly
one vertex in G that is not adjacent to x in G. Then G is Kn minus one edge
and by Lemma 3.4.8, is the 2competition graph of a loopless Hamiltonian
digraph. Suppose that there are at least two vertices, X{ and Xj of G' that are
not adjacent to x in G. Let C be the maximal clique in G containing x. Create
D as follows. Let J(x) = V(D) {x}. Let /(x,) = I(xj) = C. For all x0 not
equal to x,x; nor Xj, let J(x0) = V(D) {x0, x}. Then G is the 2competition
graph of D. Since the digraph D {xj,Xj} has all possible arcs, x has an
arc to X{. Furthermore x; has an arc to some vertex x0 of V(G) {Â£,Â£,, Â£j}
and Xo has an arc to Xj. Finally since Xj has an arc to some vertex x*. of
V(G) {x, Â£{, Xj, Xo}, we conclude that D is Hamiltonian. Furthermore C C
J(x;) and V(G') C 7(x), so every maximal clique is contained in at least one
inset of D.
Case 2: Suppose that G' is not complete. By the induction hypoth
esis, G' is the 2competition graph of a loopless Hamiltonian digraph D' such
that every maximal clique of G' is contained in at least one inset of D'. Let
X1X2 .. .Â£_i denote a Hamiltonian cycle of D'. Let C be the maximal clique
containing x in G. Let C' = C fl V(G'). By the inductive hypothesis there is a
vertex Xj such that C' is contained in I(xj) in D'. Create digraph D as follows.
75
Figure 3.10. Chordal graphs on 5 vertices.
76
(q)
(r)
()
(t)
(v)
(w)
Figure 3.11. Chordal graphs on 5 vertices.
77
Add arc {x,xf) to A(D'). Observe that since D' is loopless, {xj,x} ^ E{G).
Suppose there is a vertex x; not equal to Xj that is not adjacent to
x in G. Then let 7d(x) = lD>{xi) and lD'{xi) = C. Observe that x competes
with the other vertices of C in D at X{ and Xj, while x competes at most once
with any other vertex in D, namely at Xj. Since no other competitions have
changed, C(D) = G and Xi... Xi_ixx,... x_i is a Hamiltonian cycle in D.
Suppose that Xj is the only vertex that is not adjacent to x in G. Then
V(G') {aij} is a clique and V(G) {x_,} is a clique. Thus xj is simplicial in
G and since G Xj is complete, we are in Case 1.
Theorem 3.4.8 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If
G is a chordal graph on n > 5 vertices, then G is the 2competition graph of
a loopless Hamiltonian digraph.
Corollary 3.4.10 Langley, Lundgren, McKenna, Merz, and Rasmussen [38].
If G is interval on n > 5 vertices, then G is the 2competition graph of a
loopless Hamiltonian digraph.
3.5 Competition Inverses
Given a graph G, can we characterize the class of digraphs "D{G),
where a digraph D Â£ T>(G) if and only if C(D) G? If a digraph D is in
T>{G) we say D is a competition inverse of G. In this section we begin
by constructing a canonical representative D for T>(G) where D is strongly
connected or Hamiltonian and G is chordal or interval. This work extends
canonical constructions by Roberts [64] in connection to the food web problem.
Roberts gave a canonical construction for D Â£ T>(G) where D is acyclic and G
is chordal (and hence interval). We give a canonical construction for D Â£ T){G)
where D is acyclic and G is interval, but where G merely chordal, D T){G).
78
3.5.1 Acyclic Competition Inverses Since an acyclic digraph
must contain a vertex that has no outgoing arcs, the competition graph of an
acyclic digraph must contain an isolated vertex. Thus not every graph has an
acyclic inverse. The problem of characterizing graphs that are the competition
graph of an acyclic digraph has been extensively studied. In this section, we
cite several characterizations and give canonical constructions for an acyclic
competition inverse based upon these characterizations. Roberts [64] proved
the following proposition that we state without proof.
Proposition 3.5.1 If G is a chordal graph, then G has an acyclic competition
inverse if and only if G has an isolated vertex.
We say an ordering vi,v2,... ,vn of the vertices of an acyclic digraph
is topological if and only if a vertex u; has an arc to a vertex u, implies i < k.
Roberts gave a construction for an acyclic competition inverse for a chordal
graph similar to the one that follows.
Algorithm 3.5.1 Given a chordal graph G with at least one isolated vertex,
we construct an acyclic digraph D such that C(D) = G.
(1) Find a perfect elimination ordering u1,u2, ,vn of G so that the iso
lated vertices of G are eliminated first.
(2) Create the acyclic digraph D on the same vertex set as G such that
ID{vi) = NGi+l[vi+1]. Then C(D) = G.
As a corollary to Roberts result for chordal graphs, we have the
following proposition.
Proposition 3.5.2 Roberts [64]. If G is an interval graph, then G has an
acyclic competition inverse if and only if G has an isolated vertex.
79
The following algorithm gives a canonical representative D 6 P(G)
where D is acyclic and G is an interval graph with at least one isolated vertex.
Algorithm 3.5.2 Given an interval graph with at least one isolated vertex,
we construct an acyclic digraph D such that C(D) = G.
(1) Consecutively rank the maximal cliques in G as Ci,..., Cr so the first
i(G) cliques are the isolated vertices of G.
(2) Let Vi be a vertex in Ci Ci+\ for i 1,2, ,r 1. Such a vertex
must exist because the cliques are maximal and distinct.
(3) Let D be the acyclic digraph on the same vertices of G such that
Id(vi) = C{+1. Then C(D) = G.
3.5.2 Strongly Connected Competition Inverses Using the
characterization in Chapter 3.3 of graph that are the competition graphs of
strongly connected digraphs we can construct a canonical representative D E
V(G) where D is strongly connected.
Lemma 3.5.1 If G is a chordal graph, then e(G) < V(G) i(G).
Proof, (by induction on V(G)) The statement is clearly true for chordal
graphs on 1,2, and 3 vertices. Assume that the statement is true for chordal
graphs G such that [V(G) < k. Let G be a chordal graph such that U(G) =
k. Every chordal graph has a simplicial vertex so let v be an arbitrary simplicial
vertex in G. Consider G' G {u}. By the induction hypothesis 0e(G') <
n i(G) 1. Since v is simplicial v is contained in at most one maximal clique.
Therefore QE(G) < U(G) (G).
Corollary 3.5.1 If G is a chordal graph, then G has a strongly connected
competition inverse.
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Proof. This follows from Lemma 3.5.1 and Corollary 3.3.1.
Corollary 3.5.2 If G is an interval graph, then G has a strongly connected
competition inverse.
In order to devise an algorithm, we must employ techniques to deal
with isolated vertices.
Proposition 3.5.3 Gyarafas [26]. If G is a graph with isolated vertex v, then
0.E(Cr) Qe(G v).
Proof, (by induction on n = F(G)) The statement is clearly true for chordal
graphs on 1, 2, and 3 vertices. Assume that the statement is true for chordal
graphs G such that n < k. Let G be a chordal graph such that n = k. Every
chordal graph has a simplicial vertex so let v be an arbitrary simplicial vertex in
G. Consider G' = G {v}. By the induction hypothesis Qe(G') < n i(G) 1.
Since v is simplicial v is contained in at most one maximal clique. Therefore
e{G)
Let G denote a graph G with isolated vertices removed. Observe that
if a graph G satisfies the conditions of Corollary 3.3.1, then G does as well.
This will be crucial to our algorithm.
Algorithm 3.5.3 Given a graph G such that 0e(G) < y(G) i(G), we
find a strongly connected digraph D such that C(D) = G.
(1) Let G = G without isolated vertices.
(2) Find an edge clique covering of G of size < V(Cr) i(G).
(3) Find a system of distinct representatives for the complements of the
cliques in this edge covering. Such a system of distinct representatives
exists by a result of Roberts and Steif [66].
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(4) Create D by adding an arc from each vertex in a clique in the edge
covering to its representative in the system of distinct representatives
found above. Observe that C(D) = G.
(5) Find the strong components in D using depthfirst search. Let k de
note the number of strong components.
(6) If k = 1 go to step 11.
(7) Topologically order the strong components of D = {Dx, D2,..., Dk}.
(8) If V{DX) = {e} then choose an arbitrary vertex y G V{Dk). Define a
new digraph D' by In> (x) = MlOi lD'{y) [*], and all other insets
in D1 are the same as in D. By the proof of Corollary 3.3.1 in Case
1, Dx U Dk is strongly connected. We have decreased the number of
strong components in the digraph without changing the competition
graph. Go to step 5.
(9) If V(.Di) > 2 and there is a vertex y G V(Dk) such that (x,y) ^ A(D)
for some x G V(DX) then define a new digraph D' by Jd<(i/) = /Â£>(*),
Id'{x) with all other arcs in D' staying the same as in D. By
the proof of Corollary 3.3.1 in Case 2, Dx U Dk is strongly connected.
We have decreased the number of strong components in the digraph
without changing the competition graph. Go to step 5.
(10) If F(Di) > 2 and every vertex in Dx has an arc to every vertex in
Dk then let x G V{D\) and let y G V(Dk). Create a new digraph
D' by letting lD>{y) = [*]> Id>(x) = Id{v) and all other arcs in D'
staying the same as in D. By the proof of Corollary 3.3.1 in Case
3, Dx U Dk is strongly connected. We have decreased the number of
82
strong components in the digraph without changing the competition
graph. Go to step 5.
(11) If G 7^ G, then let oi, 02,..., ol^g) be the isolated vertices in G. Create
a path through the isolated vertices 010203 ... a,(G)iai(G) and let x be
an arbitrary vertex in V(G). Create a new digraph D' by adding
Oi, 02,..., ai(G) to the vertex set, adding the arcs of the path through
these vertices, letting /d<(ui) = Iu(x), Id>{x) {a;(G)} and all other
arcs and vertices of D' the same as in D. Then C(D) = G implies
that C(D') = G and D' is strongly connected.
3.5.3 Hamiltonian Competition Inverses We now present
algorithms to construct canonical representatives for Hamiltonian competition
inverses of certain classes of graphs. The following algorithm constructs a
canonical representative D Â£ X>(Cr) where D is Hamiltonian and G is chordal.
This construction is based upon the existence of a perfect elimination ordering
for G.
Algorithm 3.5.4 Given a chordal graph G on n > 3 vertices, we construct a
Hamiltonian digraph D such that C(D) = G.
(1) If G is not complete then G has two nonadjacent simplicial vertices
(Dirac [17]). Label one Vi and the other vn.
(2) For i = 2,3,... ,n 1, each G{ is chordal and has a simplicial vertex
Vi different from vn. Then the sequence vi,v2,... ,vn is a perfect elim
ination ordering. For i = l,2,...,n, let Ci = Observe that
Vi Ci+1 since u, is not a vertex of vn ^ Ci since iq and vn were
chosen as nonadjacent and {C1,. , Cn} is an edge clique cover of G.
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(3) Construct D by letting I(u{) = C;+1 and I(vn) = C7i. Then D is a
loopless Hamiltonian digraph with Hamiltonian path
VlVnVn_L V2V!
and C(D) = G.
The following algorithm constructs a canonical representative D in
V(G) where D is Hamiltonian and G is interval. This construction employs the
consecutive ranking of the maximal cliques of an interval graph. Observe that
while the digraph produced by Algorithm 3.5.4 is a member of D(G) where
G is interval, the digraph produced by Algorithm 3.5.5 is not necessarily a
member of T>(G) where G is chordal.
Algorithm 3.5.5 Given an interval graph G on n > 3 vertices, we construct
a Hamiltonian digraph D such that C(D) = G.
(1) Find the consecutively ranked maximal cliques C\,C2,... ,CT of G.
Observe r
(2) For each (7;, where i = l,2,...,r 1, choose a vertex in Ci (7;+i and
label it V{. Such a vertex must exists because the cliques are maximal
and distinct. Observe that ^ Ck, where k > i.
(3) For Cr, choose a vertex in CT C>_i and label it vn.
(4) If r < n, for i = 7 + l,7, + 2,...,n 1 choose an unlabeled vertex,
label it Vi, and let Ci =
(5) Construct D by letting J(v,) = Ci+i, I(vn) = C\. Observe that D is a
loopless Hamiltonian digraph with Hamiltonian cycle
V\VnVn! V2Vl
and C(D) = G.
3.5.4 Bounds on the Number of Arcs in a Competition
Inverse The canonical construction of the previous sections may provide
useful in developing algorithms to either maximize or minimize the number of
arcs in a digraph without changing the competition graph. We now present
some elementary bounds that may also be useful. The proof of the first bound
is left to the reader.
Theorem 3.5.1 If G is a graph with competition inverse D and Ci,..., CT is
an edge clique cover of minimum cardinality, then
l4(B) > E C..
i=l
Theorem 3.5.2 If G is a graph with competition graph D and {C\,..., Cr}
is an edge clique cover of minimum cardinality such that each Ci is chosen to
be maximal, then
\A(D)\ < E C, + (y(G) rMG).
i=l
Proof. There must be at least J2 Ci arcs in D to account for the \Ci\ com
petitions in each clique C{ of the edge clique cover. Thus there will be vertices
Vi,i>2,... ,vT with insets C\, C2,..., Cr respectively. There are  V(C?) r ver
tices remaining in the graph. The size of the largest clique in the minimum
edge clique cover is at most u>(G). Thus we can add at most w(G) incoming
arcs at each of the IV^G)! r remaining vertices.
The graph in Figure 3.12 illustrates that this bound is tight. Fig
ure 3.13 illustrates that it cannot be achieved for all graphs. Observe that v2
85
Figure 3.12. A graph and its competition inverse.
For this graph G we can construct a competition inverse D such that
4(D) =210,1 + (\V(G)\ k)w(G) = 26.
in this graph cannot have more than two incoming arcs in any competition
inverse.
Theorem 3.5.3 If G is a graph with competition inverse D and V^G) =
{i,.then
/l(fl) < f>(0 {,}).
1=1
Proof. Recall that D is a loopless digraph. Suppose that a vertex v G V(D)
has more than u>(G {v}) incoming arcs in D. Then either there is a loop at
v in D or there is a clique in G {u} of size greater than uj(G {v}), both
contradictions. Thus A < {,}) H
Figure 3.13 illustrates that this bound is tight. It cannot be achieved
for the graph in Figure 3.12 because there is no vertex Vi for which w{G
{v}) = 2, but in any competition inverse some vertex must have inset {^2^4}
since this is a maximal clique.
3.6 The Chromatic Numbers of Competition Graphs
One of the primary parameters of the competition graph of a digraph
that is of interest in problems motivated by communication networks is the
86
4 5 4 U6
Figure 3.13. A graph and its competition inverse.
For this graph G we can construct a competition inverse D such that
WB)l = Â£w(G{.}) = 14.
87
chromatic number. Recall that the chromatic number of a graph is the
smallest integer x(G) such that G has a proper coloring with x(G) colors. A
proper coloring of the competition graph of a digraph corresponds to a channel
assignment in the radio network. In this section we explore bounds of and in
some cases the exact value of the chromatic number of the competition graph
of a digraph.
3.6.1 Symmetric Digraphs Since G2 = G U ^(G), begin by
making the observation that
A(G) < X(S2(G)) < X(G2).
It is left to the reader to verify the following lemma which yields a preliminary
bound for the chromatic number of a symmetric digraph (possibly having some
loops).
Lemma 3.6.1 Lundgren, Merz, and Rasmussen [50]. If D is a symmetric
digraph with underlying graph (loops removed) G, then S2(G) C G(D) C G2.
Corollary 3.6.1 Lundgren, Merz, and Rasmussen [50]. If D is a symmetric
digraph with underlying graph (loops removed) G, then ^(^(G)) < %(G(D)) <
*(
This bound is not particularly good since given any integer n, we
may construct a symmetric digraph D such that
X(G2)x(S2(G))>n.
For example, let G be the bipartite graph KniTl. Then G2 is K2n and S2(G) is
a graph with two connected components, each of which is Kn.
Recall that u>(G) denotes the size of the largest clique in a graph G.
A graph is perfect if and only if w(G') = %(G') for all induced subgraphs G'
88
of G. This definition gives yields the following theorem.
Theorem 3.6.1 Lundgren, Merz, and Rasmussen [50]. If D is a symmetric
digraph with underlying graph (loops removed) G and both G2 and S2(G) are
perfect, then
w(&(
What conditions are sufficient and necessary for the twostep graph
and the square of the underlying graph of a symmetric digraph (loops removed)
to be perfect? We can answer this question for certain classes of perfect graphs.
For example, in Chapters 3.1 and 3.2 we characterize symmetric digraphs with
interval (and hence perfect) squares and twostep graphs. Which symmetric
digraphs have chordal squares and twostep graphs? The following necessary
condition partially answer this question.
Proposition 3.6.1 Lundgren and Rasmussen, 1993. If T is a tree, then S2(T)
is chordal, and hence perfect.
Proposition 3.6.2 Chang and Nemhauser [12]. If T is a tree, then T2 is
chordal, and hence perfect.
These propositions combined with Theorem 3.6.1 yield the following
corollary.
Corollary 3.6.2 Lundgren, Merz, and Rasmussen [50]. If D is a symmetric
digraph such that the underlying graph (loops removed) of D is a tree T, then
U(S2(T)) < x(C(D)) < u,(T2).
We can make this bound even more precise with the following estab
lished results.
Proposition 3.6.3 Lundgren, Maybee, and Rasmussen [45]. If T is a tree,
then w(S2(T)) = A(T).
89
Proposition 3.6.4 Raychaudhuri and Roberts [60]. If T is a tree, then
w(T2) = 1 + A (T).
Corollary 3.6.3 Lundgren, Merz, and Rasmussen [50]. If D is a symmetric
digraph such that the underlying graph of D with loops removed is a tree T,
then
A(T) < X(0(D)) < 1 + A(T).
We have already seen an example of a chordal graph with a two
step graph that is not chordal in Figure 3.4. Figure 3.1 illustrates a chordal
graph with a square that is not chordal. Rather than consider necessary and
sufficient conditions for the twostep and square of certain classes of perfect
graphs (such as chordal and interval) to be perfect, we will consider other
approaches to bounding the chromatic number of the competition graph of a
digraph.
3.6.2 Arbitrary Digraphs and Tournaments While the in
set of a particular vertex in a digraph is a clique in the competition graph,
it is not necessarily maximal. However, certainly u>(C(D)) > A~(.D), so it is
certainly clear that
A~(D) < x(C(D)) < .
Are there any classes of digraphs such that A~(D) equals x((7(D))?
We now answer this question. A digraph is an interval digraph if and only
if two intervals S(x) and T(x) on the real line can be assigned to vertex x
such that (x,y) G A(D) if and only if S(x) fl T(y) ^ 0. For a comprehen
sive introduction to interval digraphs see Sen, Das, Roy, and West [70]. To
prove results about the competition graph of an interval digraph, we need the
following lemma.
90

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COMPETITION GRAPHS, PCOMPETITION GRAPHS, TWOSTEP GRAPHS, SQUARES, AND DOMINATION GRAPHS by Sarah Katherine Merz B. A., Whitman College, 1991 M.S., University of Colorado at Denver, 1994 A thesis submitted to the University of Colorado at Denver in partial fulfillment of the requirements for the degree of Doctor of Philosophy Applied Mathematics 1995
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This thesis for the Doctor of Philosophy degree by Sarah Katherine Merz has been approved by
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Merz, Sarah Katherine (Ph. D., Applied Mathematics) Competition Graphs, pCompetition Graphs, TwoStep Graphs, Squares, and Domination Graphs Thesis directed by Professor J. Richard Lundgren ABSTRACT Given a digraph D, the competition graph of D, C(D) is a graph on the same vertex set with vertices x and y adjacent in C(D) if and only if there exists a vertex z such that x and y have arcs to z in D. Intro duced in the context of modeling food webs by Cohen in 1968, competition graphs have utility in a variety of applications such as radio communication networks and a certain type of dominating set in tournaments. Can elimination orderings be used to characterize digraphs with interval and/ or chordal competition graphs? We answer this question and consider possible implica tions in the food web application of competition graphs. We consider several problems motivated by the use of competition graphs in modeling communi cation networks. Which symmetric digraphs with a loop at each vertex have interval competition graphs? Which loopless symmetric digraphs have inter val competition graphs? For which digraphs can the chromatic number of the competition graph be bounded? Given a graph G, can we bound the number of arcs in a digraph D such that C(D) = G? Given a gmph G, can we determine canonical forms of a digraph D such that C(D) = G? Which graphs are the competition graph of a strongly connected or Hamiltonian digmph? These 111
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questions are considered. Results pertaining to the last question are general ized to pcompetition graphs. Lastly, we consider several problems motivated by tournaments. Given a digraph D, the domination graph of D, Dom(D) is a graph on the same vertex set as D with vertices x andy adjacent in Dom(D) if and only iffor every other vertex z in D, x or y has an arc to z (possibly both). We show a relationship between the competition and domination graphs of a tournament and consider the following questions: Which graphs are the domination graph of a tournament? What is the maximum number of edges in the domination graph of a tournament? Which graphs are the domination graph of an oriented simple graph? This abstract accurately represents the content of the candidate's thesis. I recommend its publication. Signed IV
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This work is dedicated to my mother, my father and my sister, Ruth. Thank you for your encouragement and love. v
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CONTENTS CHAPTER 1 INTRODUCTION 1.1 Terminology 2 PROBLEMS MOTIVATED BY FOOD WEBS 2.1 Elimination Ordering Characterizations 2.2 Finding Interval Elimination Orderings 2.3 Implications for Food Webs 3 PROBLEMS MOTIVATED BY COMMUNICATION NET1 9 11 11 16 18 WORKS . . . . . . . . 20 3.1 Symmetric Digraphs with Loops 20 3.1.1 Using Forbidden Subgraphs 32 3.1.2 Elimination Ordering Characterizations 34 3.2 Loopless Symmetric Digraphs . . . . . 36 3.2.1 Finding Maximal Cliques of the TwoStep Graph 37 3.2.2 Competition Covers .......... 3.2.3 Elimination Ordering Characterizations 3.3 Strongly Connected Digraphs . . . . . 3.3.1 Generalization topCompetition Graphs 3.4 Hamiltonian Digraphs .............. VI 41 52 53 57 61
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3.4.1 Generalization to pCompetition Graphs: Con structions . . . . . 3.4.2 Utilizing the Constructions 3.4.3 Classes of 2Competition Graphs 3.5 Competition Inverses ....... 3.5.1 3.5.2 3.5.3 3.5.4 Acyclic Competition Inverses Strongly Connected Competition Inverses Hamiltonian Competition Inverses .... Bounds on the Number of Arcs in a Competition Inverse .................. 3.6 The Chromatic Numbers of Competition Graphs 3.6.1 Symmetric Digraphs ......... 3.6.2 Arbitrary Digraphs and Tournaments 4 THE COMPETITION GRAPHS OF TOURNAMENTS 4.1 Necessary Conditions on the Domination Graphs of Tour66 71 74 78 79 80 83 85 86 88 90 96 naments. . . . . . . . . . . . . . 97 4.2 Sufficient Conditions on the Domination Graphs of Tournaments ............... 4.3 The Domination Graphs of Digraphs 5 A LIST OF OPEN PROBLEMS ..... APPENDIX A NOTATION BIBLIOGRAPHY VII 103 107 109 111 113
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FIGURES FIGURE 1.1 A digraph and its cqmpetition graph. 1 1.2 An interval graph. 2 1.3 A chordal graph. 3 1.4 The problem with forbidden subgraph characterizations. 4 1.5 A properly colored graph. 5 1.6 A 2competition graph. 7 1.7 A tournament and its domination graph. 9 2.1 A graph with perfect elimination ordering. 11 2.2 A digraph with disimplicial elimination ordering. 12 2.3 A graph with interval elimination ordering. . 14 2.4 A digraph with diinterval elimination ordering. 15 2.5 Forced incidence diagram for disimplicial elimination ordering. 18 2.6 Forced incidence digraph for diinterval elimination ordering. 19 3.1 Forbidden subgraphs of interval graphs. 21 3.2 The smallest tree that is not interval. 32 3.3 A graph and an induced subgraph. . 33 3.4 An interval graph whose twostep graph is not interval. 37 3.5 The family of maximal cliques is not S'(G). 43 3.6 The set {x,x1,x2,x3 } is not in S'(G). . 48 Vlll
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3.7 A competition graph. 54 3.8 A 2competition graph. 58 3.9 An example. 0 0 0 64 3.10 Chordal graphs on 5 vertices. 76 3.11 Chordal graphs on 5 vertices. 77 3.12 A graph and its competition inverse. 86 3.13 A graph and its competition inverse. 87 3.14 The subgraph of an interval graph .. 91 3.15 This graph is not an interval graph. 92 4.1 A tournament and its domination graph. 96 4.2 Regular tournaments on 7 vertices and their domination graphs. 99 4.3 This graph is not the induced subgraph of the domination graph of a tournament. ........... 100 4.4 A tournament and its domination digraph. 104 4.5 Orientations of a path. .......... 105 IX
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First and foremost, I am grateful to Rich Lundgren for his academic, financial, personal and professional support. His limitless patience, encour agement, generosity and capacity as a mentor have been more valuable than I can express. I would like to thank Dave Fisher for all the help that he has given me, particularly in my struggle to do research. In addition, he provided the extremely valuable templates from which most of the figures in this document were constructed. I would also like to thank Kathy Fraughnaugh for the help that she has given me, particularly with mathematical writing. As a woman in mathematics, she has been an important role model during my graduate years. My thanks go to Jenny Ryan and Gary Kochenberger, as well as to the fine mathematicians with whom I have had the opportunity to collaborate: Charlie Anderson, Larry Langley, John Maybee, Patty McKenna, Norm Pullman, Craig Rasmussen, and Brooks Reid. I would like to thank the Office of Naval Research and Mark Lipman for funding my graduate education. Finally, I would like to thank David Guichard of Whitman College for encouraging me to go to graduate school. X
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CHAPTER 1 INTRODUCTION This dissertation addresses a variety of graphtheoretic problems motived by three basic applications of the competition graph of a digraph. Given a digraph D, the competition graph of D, C(D) is a graph on the same vertex set, with :c andy adjacent in C(D) if and only if there exists z such that (x,z) and (y,z) are arcs in D .. Figure 1.1 illustrates a digraph and its competition graph. Digraphs can be used in a variety of applications. The three considered here are in modeling food webs, radio communication networks and tournaments. 8 1 8 5 1 4 Figure 1.1. A digraph and its competition graph. Competition graphs were introduced in the study of food webs by Cohen [13] as a method to determine the dimension of ecological phase space. Let each vertex in a digraph represent a single or collection of similar species. There is an arc from :c toy in the digraph if and only if the species represented 1
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by x preys upon the species represented by y. The digraphs in this model are generally acyclic. Two vertices are adjacent in the competition graph of the digraph if and only if their corresponding species compete for some common prey. Section 2 addresses graphtheoretic problems that arise in this application. How many variables must be considered in the determination of a competition graph? One example of such a variable is diet; in the food web model we suggested, an arc is placed from x to y if x eats y. The boxicity of a graph, first introduced by Roberts, is defined to be the smallest integer k such that the graph can be represented as an intersection graph of boxes in kspace. See Roberts [63] for a comprehensive introduction to these concepts. An interval graph is a graph which is the intersection graph of intervals on the real line. That is, an interval graph has boxicity one. Figure 1.2 illustrates an interval graph and the corresponding intervals for each vertex. Cohen observed that many competition graphs for actual food webs are interval graphs. 8 1 7 2 6 3 5 4 Figure 1.2. An interval graph. This graph is an intersection graph of intervals on the real line when the following intervals are assigned to each vertex. 1: [8,11]; 2: [10,14]; 3: [12,15]; 4: [9,13]; 5: [1,8]; 6: [4,7]; 7: [3,6]; 8: [2,5]. 2
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An explanation for this phenomenon has been sought since its discov ery. Is Cohen's empirical observation an artifact of the construction? That is, perhaps the competition graph of most acyclic digraphs are interval. Roberts posed this question in 1978 [64]. He showed that this is not the case, since every graph can be made into the competition graph of an acyclic digraph by attaching a sufficient number of isolated vertices. So it is interesting to ask which digraphs have interval competition graphs. Figure 1.3. A chordal graph. A chordal graph is a graph with no induced cycle on k vertices, where k 2: 4 (see Figure 1.3). Equivalently, any kcycle where k 2: 4 contains a chord. Interval graphs are chordal. Sugihara [72] suggested the chordal property was significant with respect to food webs. His analysis was concerned with "holes in niche space." He conjectured that given a collection of species that exhibit competition in a cyclic structure, it is likely that a chord exists across the cycle. The lack of such a chord characterizes a "hole". For this reason, it is interesting to ask which digraphs have chordal competition graphs. We present a characterization of digraphs with chordal and inter val competition graphs. This characterization is different from previous ap proaches because it lends a nice interpretation to the structure of the digraph 3
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that may be relevant to food webs. Furthermore avoids the difficulty in using forbidden subgraph characterizations. Figure 1.4 illustrates this difficulty. Even though the competition graph of some digraph is interval, it is possible that the digraph contains an induced sub digraph with nonchordal competition graph. 2rvr 203 1 4 1 4 5 5 1 4 1 4 Figure 1.4. The problem with forbidden subgraph characterizations. From left to right, digraphs and their competition graphs. The digraph above is an induced subdigraph of the digraph below. Its competition graph is not an induced subgraph of the competition graph of its superdigraph. The application of competition graphs in the study of communication networks was introduced by Raychaudhuri and Roberts [60]. Let each vertex in a digraph represent a station in a radiocommunication network. There is an arc from x to y in the digraph if and only if the station represented by y can receive a signal transmitted by the station represented by x. Two vertices are adjacent in the competition graph of the digraph if and only if their corresponding stations are unable to transmit on the same channel because their signals are received by a common third station. In this setting, competition graphs are often called conflict graphs. A coloring of a graph is a labeling 4
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of its vertices so that adjacent vertices do not have the same label. A coloring of the competition graph can be translated into a frequency assignment in the corresponding network. So in the interest of optimally assigning frequencies in the networks, we would like to efficiently color the competition graph of the digraph modeling the network. Coloring is generally a hard problem though there are some classes of graphs which can be easily colored. For example, in terval graphs can be colored in linear time. Hence the question arises: Which digraphs have interval competition graphs? B G R R B Figure 1.5. A properly colored graph. Digraphs that model communication networks will often have special structure reflecting the nature of the network. We will take advantage of this structure as much as possible in considering graphtheoretic problems that arise in this application. For example, many radio communication networks can be modeled by a symmetric digraph with a loop at each vertex. That is, x transmits to y if and only if y transmits to x and each station can receive its own signal. Such a digraph can be represented by its underlying 5
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graph (with or without loops). The competition graphs of digraphs having this structure have been studied by Raychaudhuri and Roberts [60], Raychaudhuri [57], Balakrishnan and Paulraja [2, 3], and Laskar and Shier [40]. Given a symmetric digraph D with a loop at each vertex and underlying interval graph H, what conditions are necessary and sufficient for the competition graph of D to be interval? This question was answered by Raychaudhuri and Roberts m 1985 [60]. Given a symmetric digraph D with a loop at each vertex, what conditions are necessary and sufficient for the competition graph of D to be interval? This is the focus of Section 3.1. A natural extension of this question is to consider symmetric digraphs without loops. The competition graphs of such digraphs have been studied by Brigham and Dutton [8], and Lundgren, Maybee, and Rasmussen [45]. The latter gave necessary and sufficient condition for symmetric digraphs without loops and having interval underlying graph to have an interval competition graph. We give necessary and sufficient conditions for symmetric digraphs without loops to have an interval competition graph in Section 3.2. A strongly connected digraph is a digraph in which there is a di rected path from every vertex to every other vertex. Since it is desirable that a message initiated somewhere in the network be able to reach all stations, usually the digraphs for communication networks are strongly connected. In a very large network, certain stations may be equipped with very powerful transmitters while other stations transmit much weaker signals. The digraph for such a network is not necessarily symmetric, but is strongly connected. There fore it is of interest to determine which graphs are the competition graphs of 6
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strongly connected digraphs. In Section 3.3 we will characterize these graphs. A Hamiltonian digraph is a digraph with a directed cycle that ineludes every vertex and is therefore an example of a strongly connected digraph. In Section 3.4 we characterize graphs that are the competition graphs of Hamiltonian digraphs. The results of Sections 3.3 and 3.4 are extended to pcompetition graphs, a generalization of the competition graph. The pcompetition graph of a digraph D, denoted Cp(D) is a graph on the same vertex set as D, with x and y adjacent in Cp(D) if and only if there exist p vertices zl> z2 Zp such that (x,z1),(x,z2), . ,(x,zp) and (y,z1),(y,z2), ,(y,zp) are arcs in D. See Figure 1.6 for an example of a 2competition graph. 8 1 7 6 8 5 Figure 1.6. A 2competition graph. 4 2 3 The competition graph of this digraph is shown in Figure 1.1. The only two vertices that compete at least twice are 1 and 8. Radio frequencies have already been assigned in existing radio communication networks thereby establishing a coloring of the competition graph. Aside from Hefner and Hintz [29], very little work has been done toward developing algorithms that begin with existing communication networks and then 7
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optimize, either with respect to the number of communication links (without changing the competition graph) or with respect to rendering the competition graph easily colored (while preserving qualities of the digraph). We take preliminary steps in this direction in Section 3.5. Given a graph G, what is the maximum number of arcs in a digraph having competition graph G? We answer this question Given a digraph D and a graph G such that C(D) = G, we say D IS a competition inverse of G. The competition inverse of a graph does not necessarily exist, nor is it necessarily unique. Characterizing competition inverses has been studied by Greenberg, Lundgren, and Maybee [24, 25] and by Roberts [64]. Roberts defined the competition number of a graph G as the small est integer "(G) such that G, together with "(G) isolated vertices, is the competition graph of an acyclic digraph. We first consider the problem of finding a canonical representative for the class of acyclic competition inverses of chordal and interval graphs. We next consider the construction of a strongly con nected competition inverse, as this property is ideally attributed to models of. communication networks. As a special case of strongly connected, we consider the construction of a Hamiltonian competition inverse for chordal and interval graphs. A tournament is an oriented complete graph. The competition graph of a tournament was first considered by Lundgren, Merz, and Ras mussen [50] in connection with some problems involving chromatic number. The structure of the competition graph of a tournament was then examined by 8
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Fisher, Lundgren, Merz, and Reid [20]. We discuss the relationship between competing and dominating pairs in a tournament in Section 4. Two vertices x and y in a digraph are a dominant pair if and only if every vertex z is contained in the union of their outsets. Two vertices are adjacent in the domination graph Dom(T) of a tournament T if and only if they are a dominant pair in T. Figure 1.7 illustrates a tournament and its domination graph. 8 1 7 6 5 4 8 5 1 4 Figure 1. 7. A tournament and its domination graph. 2 The domination graph of a tournament T is the complement of the competition graph of the tournament formed by reversing the arcs ofT (this is not necessarily true for all digraphs). Therefore results on the domination graphs of tournaments correspond to results on the competition graphs of tournaments. Since the domination graph of a tournament generally has fewer edges than the competition graph, it is more convenient to state and prove results on domination graphs. Section 4 addresses the domination graphs of tournaments and other oriented graphs. 1.1 Terminology The open neighborhood of a vertex x in a graph is the set of vertices adjacent to x. The closed neighborhood of a vertex x is the union 9
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of"' and its open neighborhood. The inneighbors of a vertex"' in a digraph are the vertices with arcs to "' The outneighbors of a vertex "' are the vertices to which "' has arcs. The independence number of a graph G is the maximum cardinality of a set such that no two vertices in the set are adjacent. A clique is a set of vertices such that every pair of vertices in the set is adjacent. The clique number of a graph G is the maximum cardinality of a clique in G. A proper coloring of a graph G is a labeling of the vertices such that if two vertices have the same color, then they are not adjacent. A kcoloring is a proper coloring with k labels. The chromatic number of a graph G is the smallest integer k such that G has a kcoloring. A clique cover of a graph G is a labeling of its vertices such that if two vertices have the same label, then they are adjacent. The clique cover number of a graph G is the minimum number of labels in a clique cover of G. An edge clique cover of a graph G is a labeling of its edges such that if two edges have the same label, then they are contained in a clique. The edge clique cover number of a graph G is the minimum number of labels in an edge clique cover. An pedge clique cover of a graph G is a labeling of its edges such that if two edges have p of the same labels, then they are contained in a clique. The pedge clique cover number of a graph G is the minimum number of labels in a pedge clique cover. An isolated vertex is a vertex with no neighbors. A pendant vertex is a vertex with exactly one neighbor. An internal vertex is a vertex with at least two neighbors. A simplicial vertex is a vertex whose neighbors induce a clique. For an explanation of notation used in this document, please see the appendix. 10
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CHAPTER 2 PROBLEMS MOTIVATED BY FOOD WEBS 2.1 Elimination Ordering Characterizations In this section we explore characterizations of digraphs that have chordal and/or interval competition graphs. Then we consider the possible implications these characterizations may have in answering the question of why the competition graphs of digraphs modeling food webs are so often interval. We begin by considering chordal competition graphs. Let G be a graph. Avertex vis simplicial in G if and only if the set of vertices adjacent to v induce a clique. An ordering v1 v2 vn of the vertices of G is a perfect elimination ordering if and only if v; is simplicial in G, = G{v1,v2 ,v,_1}, where G1 = G. Figure 2.1 illustrates a graph and its perfect elimination ordering. 2 1 6 5 Figure 2.1. A graph with perfect elimination ordering. Proposition 2.1.1 Rose [67]. A graph is chordal if and only if it has a perfect elimination ordering. 11
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Can we define an elimination ordering of a digraph which corresponds to a perfect elimination ordering of the competition graph of the digraph, pro vided one exists? A vertex x is disimplicial in a digraph D if and only if whenever there are vertices y, z, u, v E V(D) such that (x,u),(y,u),(x,v),(z,v) E A(D) then there exists a vertex wE V(D) such that (y,w) and (z,w) E A(D) An ordering v1 v2 Vn of the vertices of D is a disimplicial elimination ordering if and only if v; is disimplicial in D;, where A(D;) =A( D) minus all outarcs of v1 ... v;_1 Figure 2.2 illustrates a digraph and its disimplicial elimination ordering. Figure 2.2. A digraph with disimplicial elimination ordering. Since 1 and 2 compete and 1 and 3 compete, it follows that 2 and 3 compete. Since 2 and 4 compete and 2 and 3 compete, it follows that 3 and 4 compete. Theorem 2.1.1 Lundgren and Merz [48]. If D is a digraph, then C(D) is chordal if and only if D has a disimplicial elimination ordering. 12
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Proof. Assume that v1, v2 Vn is a perfect elimination ordering of C(D). This is true if and only if given v; andy, z E { Vi+ll ... vn}, {v;,y},{v;,z} E E(C(D)) implies that {y,z} E E(C(D)). This is true if and only if given v; and y, z E { vi+l, ... vn}, if there exist u and v E V(D) such that (y,u),(vi,u),(z,v),(vi,v) E A(D), then there exists wE V(D) such that (y,w),(z,w) E A(D). This is true if and only if vh v2 Vn is a disimplicial elimination ordering of D. This characterization is particularly nice because it avoids the for bidden subgraph problem of previous approaches: It is easy to construct a digraph whose competition graph is chordal, containing a generated subgraph whose competition graph is not chordal (see Figure 1.3). Rose, Tarjan, and Leuker [68] showed that a perfect elimination ordering can be found in linear time. See Golumbic [23] for an introduction to this algorithm, which finds a perfect elimination ordering if one exists. This algorithm can be modified to produce an ordering of the vertices in a digraph which is a disimplicial elimination ordering if and only if the competition graph is chordal. This is accomplished by translating the steps of the algorithm to work with the digraph rather than first constructing its competition graph and then execut ing the algorithm on the competition graph. Using a disimplicial elimination ordering in this way to determine if the competition graph of a digraph is chordal is not computationally faster than first constructing the competition 13
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graph and then implementing the algorithm. We now consider digraphs with interval competition graphs. Let G be a graph. An ordering v1 v2 Vn of the vertices of a graph G is an interval elimination ordering if and only if the ordering has the property that for all v;,v;,vk such that i :S j :S k, {v;,vk} E E(G) implies that { v;, vk} E E( G). Figure 2.3 illustrates a graph and its interval elimination ordering. 5 6 3 7 3 I 2 Figure 2.3. A graph with interval elimination ordering. Proposition 2.1.2 Laskar and Shier [40]. A graph G is interval if and only if it has an interval elimination ordering. Can we define an elimination ordering of a digraph which corresponds to an interval elimination ordering in the competition graph, provided one exists? We say an ordering v1,v2 ,vn of the vertices of Dis a diinterval elimination ordering of a digraph D if and only if the ordering has the property that for all v;, v;, vk such that i :S j :S k, if there exists u such that (v;,u) and (vk,u) E A(D) then there exists w such that (vk,w) and (v;,w) E A(D). Figure 2.4 illustrates a digraph with a diinterval elimination ordering. 14
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5 70 3 1 2 Figure 2.4. A digraph with diinterval elimination ordering. Theorem 2.1.2 Lundgren and Merz [48]. If D is a digraph, then C(D) is interval if and only if D has a diinterval elimination ordering. Proof. Assume that v1 v2 . Vn is an interval elimination ordering of C(D). This is true if and only if fori< j < k, whenever {v;,vk} E E(C(D)), then { Vk, v;} E E( C(D)). And this is true if and only if for i < j < k, whenever there exists u E V( D) such that ( v,, u ), ( vk> u) E A( D), then there exists v E V(D) such that (v;,v),(vk,v) E A(D). And this is true if and only if v1 v2 Vn is a diinterval elimination ordering of D. Again, this characterization also avoids the forbidden subgraph problem of previous characterizations. Can interval elimination orderings be used to efficiently determine intervality? This question is unanswered. Can diinterval elimination orderings be used to efficiently characterize digraphs with interval competition graphs? This question is also unanswered. This problem may be more tractable if we impose certain structures on the digraphs involved. For example, knowing that the digraph is transitive or acyclic might provide enough structure so that using diinterval elimination orderings to determine if the competition graph is interval becomes computationally efficient. 15
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2.2 Finding Interval Elimination Orderings Although the problem of finding an interval elimination ordering directly (if one exists) in an arbitrary graph remains open there are well known algorithms that can perform intervality testing in linear time. See Golumbic [23] for an introduction to one of these algorithms. This algorithm uses a characterization of interval graphs. A consecutive ranking is an ordering of sets cl, c2, ... em such that if i < k, then X E C; and X E ck implies X E C; for all i < j < k. Fulkerson and Gross [22] showed that a graph is interval if and only if its maximal cliques have a consecutive ranking. Given this information we can produce an interval elimination ordering by the following algorithm. Algorithm 2.2.1 Lundgren and Merz [48]. Given a family of sets of vertices Cl> C2 ... C, corresponding to the consecutively ranked maximal cliques in an interval graph G, we find an interval elimination ordering for G. (1) Consider the r1 vertices in C1C2 Arbitrarily order them v1 v2, ... v,,. (2) Consider the r2 vertices in C2 C3 that have not yet been ordered. Arbitrarily order them Vr,+1, ... ,vr,+r, (3) Similarly order the remaining vertices for C3 C4 C.1 C,. (4) Arbitrarily order the vertices in C,_1 n C,. (5) Complete the ordering by arbitrarily ordering the remaining vertices in C, resulting in V1, ... Vn Theorem 2.2.1 Lundgren and Merz [48]. The ordering produced in Algorithm 2.2.1 is an interval elimination ordering for the graph G. Proof. Consider arbitrary i, j, k such that i < j < k. We claim that if { v;, vk} E E( G), then { v;, vk} E E( G). Suppose that { v;, vk} E E( G). If v;, vk E C., then since v;, v; and Vk were chosen in one of the last two steps of 16
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the algorithm we conclude that v; E 0, and we are done. So assume not. Then there exists b < r such that v; E 0&0&+1 and d :::0: b such that vk E OdOd+1 (if d < r) or vk E 0,. Ifv; E 0., then vk E 0,. That is, {v;,vk} E E(G). So assume that v; E OJ01+1 forb:<; f :<; d, where f < r. Since {v;,vk} E E, v; and Vk belong to some maximal clique and since v; E 0& and v; f/. Cb+1> then v; f/. Od since d :::0: b + 1 and the ranking is consecutive. So vk E 0&. This consecutive ranking then implies vk E Of> since b :<; f :<; d. Therefore {v;,vk} E E(G). Given an interval elimination ordering for a graph we can also produce the consecutively ranked maximal cliques. This is accomplished by consecutive (with respect to the order of elimination) examination of the neighborhoods of the vertices. For example, if k1 is the largest integer for which v 1 and vk, are adjacent, then is a maximal clique. Furthermore, if k2 is the largest integer for which v2 and vk, are adjacent, then {v;fv2 and v; adjacent ,2 :<; j :<; k2 } JS a maximal clique if and only if either v2 and v1 are adjacent or k2 > k1 A similar approach can be used to find the remaining maximal cliques in a consecutive ranking. Thus an interval elimination ordering can be found if you have either the maximal cliques of the graph or the intervals for which the graph is the 17
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intersection graph. These techniques do not use interval elimination orderings to determine if a graph is interval. Furthermore, they do not translate to finding a diinterval elimination ordering of a digraph. Short of trying all possibilities, is there a method for determining whether or not a graph has an interval elimination ordering or a digraph has a diinterval elimination ordering? One approach might be to consider the diinterval elimination ordering for special classes of digraphs such as transitive or acyclic digraphs. 2.3 Implications for Food Webs As mentioned earlier, the competition graphs of digraphs modeling cross sections of real food webs are often interval. Do either the disimplicial or diinterval elimination ordering characterization say anything about the structure of these food webs? Diagrams such as the one in Figure 2.5 indicate possible interpretations of the situation in which a digraph modeling a food web has an interval competition graph. Figure 2.5. Forced incidence diagram for disimplicial elimination ordering. Assume that v1 v2 v,. is a disimplicial elimination ordering for a digraph D. The presence of arcs (vk,vn), (vn3,vn), (vk,v,), and ( Vn2, v1 ) forces Vn2 and Vn3 to have common prey. 18
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In Figure 2.5 we show a disimplicial elimination ordering for a hypothetical food web. Suppose that the species of this food web have been linearly ordered v1 v2 Vn such that Vn is the most adaptable and v1 is the least adaptable. Then if i < j < k and v; and Vk have some property in common so they compete and v; and v; and have some property in common so that they compete, then v; and Vk are very likely to compete as well. This interpretation emphasizes Sugihara's assembly rule for food webs which requires that "species be incorporated into community conservatively by attaching to single guilds rather than by bridging multiple guilds," where a guild is essentially a clique in the competition graph [74]. I I v, v; I Vn1 I I Vn Figure 2.6. Forced incidence digraph for diinterval elimination ordering. Assume that v1 v2 Vn is a diinterval elimination ordering for a digraph D. The presence of arcs (vk,v1 ) and (v;,v1 ) forces vk to have common prey with vertices v;+l, ... ,vk1 Figure 2.6 shows a diinterval elimination ordering for a digraph. The species of this food web have been linearly ordered v1 v2 Vn on some scale of adaptability where v,. is the most adaptable. Let i < j < k. This ordering implies that if vk is so adaptable that it competes with v; then most likely it competes with v;. 19
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CHAPTER 3 PROBLEMS MOTIVATED BY COMMUNICATION NETWORKS 3.1 Symmetric Digraphs with Loops In this section we consider the competition graphs of symmetric digraphs with a loop at each vertex. Such digraphs arise in the model of radiocommunication networks. We can represent such a network with a digraph where a vertex represents each station and there is an arc from x to y if and only if the station corresponding to x transmits a signal that the station corresponding to y can receive. There is a loop at each vertex since each station may receive its own signal. Given such a digraph, we may refer to its underlying graph. The underlying graph is a graph on the same vertex set with an edge between x and y if and only if x has an arc to y and y has an arc to x. Notice this preserves a loop at each vertex. In this section, we consider the underlying graph with loops removed. Given a graph G, the square, G2 is a graph on the same vertex set such that two vertices x andy are adjacent in the square if and only if there is a path of length one or two between x andy in G. Raychaudhuri and Roberts [60] first made the following observation that is stated without proof. Proposition 3.1.1 Raychaudhuri and Roberts [60]. If D is a symmetric digraph with a loop at each vertex and G is the underlying graph (loops removed) of D, then C(D) = G'. 20
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Which graphs have interval squares? We begin by observing the fol lowing result. Proposition 3.1.2 Raychaudhuri [57]. If a is an interval graph, then a' is an interval graph. Unfortunately the proof given in the work cited is not easily generalized to graphs that are not interval. We also observe that there are graphs having interval squares that are not interval. Consider the forbidden subgraphs of interval graphs provided by Lekkerkerker and Boland [41] depicted in Figure 3 .1. G, Ga Figure 3.1. Forbidden subgraphs of interval graphs. A graph is interval if and only if it contains no subgraph isomorphic to a1 a,, a3 a4 a,. Note that the squares of a1 (n = 4, 5), a3 a. and a, are interval. Some of these graphs have interval squares. Thus our characterization cannot be based upon the squares of forbidden subgraphs of interval graphs. We will employ a characterization of interval graphs due to Fulkerson and 21
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Gross [22]. This characterization uses the following definition: An ordering of a family of sets C1 C2 Cp is a consecutive ranking if and only if for all i < j < k, whenever X E c, and X E ck, then X E Cj. Proposition 3.1.3 Fulkerson and Gross [22]. A graph G is an interval graph if and only if the family of maximal cliques of G has a consecutive ranking. Suppose that we can define a family of sets MC( G') found in G corresponding to the maximal cliques in the square. By Proposition 3.1.3, G' is interval if and only if MC(G2 ) has a consecutive ranking. Recall that a vertex is simplicial if its neighborhood induces a clique. A vertex that is not simplicial is called nonsimplicial. We say that a closed neighborhood of a nonsimplicial vertex is maximal if it is not properly contained in the closed neighborhood of any other nonsimplicial vertex in the graph. We define a class of graphs as having the closed neighborhood property if and only if the set of maximal closed neighborhoods of nonsimplicial vertices is precisely the set of maximal cliques in G2 Interval graphs have the closed neighborhood property as shown by the next theorem. Theorem 3.1.1 Lundgren, Merz, and Rasmussen [49]. If G 'J! Kn is a con nected interval graph, then C is a maximal clique of G' if and only if there exists a nonsimplicial vertex z E V( G) such that C = Na[z] and Na[z] is maximal. Proof. (<=)Let z be a nonsimplicial vertex in G such that Na[z] is maximal. We claim that the consecutively ranked cliques C;, ... Ck containing z form a maximal clique in the square (since these cliques are Na[z]. Suppose not. Then there exists a vertex w adjacent to every vertex in C,, ... Ck in G', but w is not in C;, ... Ck. Without loss of generality, assume that w E Cj where j > k and j is the smallest such integer. Let x be an arbitrary vertex inC; not 22
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equal to z such that x i and w i or z E em form< j we have a contradiction by definition of j and i respectfully. Therefore i < j. Since the ranking is consecutive, {:v,y}
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there exists a vertex x rf_ C such that x is adjacent to a vertex y E C. But then { x} U C is a clique in G2 a contradiction. So z is nonsimplicial. Corollary 3.1.1 Lundgren, Merz, and Rasmussen [49]. If G is a connected interval graph, then G2 is interval. Proof. If G S"' Kn, then the statement is clearly true so assume that G '/. Kn. We claim that since the family of maximal cliques of G has a consecutive ranking, the family of maximal nonsimplicial closed neighborhoods of G has a consecutive ranking. Let C1 C2 Ck be the consecutively ranked maximal cliques of G. Let Sii be C, U U Ci and let S = {Sij[i :::; j :::; k}. Let S' be the sets of S such that no set is properly contained in any other. Since G is connected, we conclude that C, n C,+l f 0. If v E C, n Ci+1 then v is nonsimplicial. Conversely, any nonsimplicial vertex v must be in the intersection of consecutive cliques. Thus S' is the family of maximal cliques of G2 Rank S' such that S.b < S9 if and only if a < g. Observe a < g implies c < i. We claim this is a consecutive ranking. Suppose that X E s., and X E s.,. We claim that X E sdf for all a < d < g such that sd/ E S'. Then X E cb for some b such that a < b < c and X E ch for some h such that g < h < i. If b :::; h, then X E cb n ... n ch. Then d < g. Furthermore c < f and d < f. Suppose that C, 1= Sdf for all b < e < h. Then either d > h or f < b, but d < g :::; h and b :::; c < f. So C, <;;: Sd/ for some e such that b < e < h and we conclude that x E Sdf Therefore b > h. Then X E ch n ... n cb. But since d < g :::; h and b:::; c < f, we conclude that shb <;;: sdf So X E sd/. Though this result has been previously established, this approach to 24
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the problem may be more easily generalized to other classes of graphs. For example, trees also have the closed neighborhood property. Theorem 3.1.2 Lundgren, Merz, and Rasmussen [49]. If T is a tree such that \V(T)[ 3, then Cis a maximal clique in T2 if and only if there exists a nonsimplicial vertex v such that Na[v] = C and Na[v] is maximal. Proof. ( <=) Let v be a nonsimplicial vertex in T such that Nr[v] is maximal. Clearly Nr[v] is a clique in T2 Suppose that it is not maximal. Then there exists w (/; Nr[v] such that w is joined to all x E Nr[v] by a path of length one or two. Since {w,v} (/; E(T) there exists x such that w,v E Nr(x). Since Nr[v] rf_ Nr[x] there exists y E Nr[v] such that {y,x} (/; E(T). If { w, y} E E(T), then we are done since wxvyw is a 4cycle. So w and y are joined by a path of length two. That is, there exists a (a I x,a I v) such that w, y E Nr( a). Then wxvyaw is a 5cycle, a contradiction since T is a tree. Thus no such w can exist. That is, Nr[v] is a maximal clique in T2 ( =:) Let C be a maximal clique in T2 We claim there exists z such that C c:;; Nr [ z] in T. Let R be a subset of C. The proof is by induction on [ R[. For the base case, let R = {x,y}. Then either {x,y} E E(T) so R c:;; Nr[x] or x,y E Nr(z) for some z, so R c:;; Nr[z]. Let R = {x,y, z}. Assume that the claim is false. Suppose {x,y} E E(T). Then {z, x} and {z, y} (/; E(T), so there exist a, b (possibly a= b) such that x,z E Nr(a) and y,z E Nr(b). If a= b, then we are done since xya is a 3cycle. If a I b, then we are done since xyazbx is a 5cycle. So assume that no two vertices in Rare adjacent. Then there exists a such that x,y E Nr(a), b (b I a) such that y, z E Nr(b) and c (c I a, c I b) such that x, z E Nr(c). Then xaybzcx is a 6cycle, a contradiction. 25
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Assume that the claim is true for all R such that IRI < ICI. Let R =C. The established base cases imply that ICI 2: 4. Let w be an arbitrary vertex in C. Consider R' = R { w }. By the induction hypothesis there exists z0 such that R' c:;; NT[z0]. Suppose that the claim is false for R. Then { w, z0 }
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Lemma 3.1.1 If v is simple in G and y is the neighbor of v with maximal neighborhood, then Na[y] = {u: da(u,v) 2}. Theorem 3.1.3 Raychaudhuri [59]. If G "/ Kn is a connected strongly chordal graph, then C is a maximal clique of G2 if and only if there exists a nonsimplicial vertex z such that C = Na[z] and Na[z] is maximal. Proof. ( =>) Let C = { u1 u1 un} be a maximal clique of G2 Then for i,j E {1,2, ... ,n} we conclude that da(u;,u;) 2. Let P;; be a path of shortest length in G between u; and u;, where i, j = 1, 2, ... n. In particular, if da(u;,u;) = 1, then P;; = {u;,u;} and if da(u;,u;) = 2, then P;; = {u;,vk,u;} for some vk E V(G). Consider the subgraph G' of G induced by the vertices u; and vk, where i = 1,2, ... ,n) and k = 1,2, ... ,p, where p Since Vk has a pair of nonadjacent neighbors for all k, namely u; and u;, Vk is nonsimplicial in G' for all k. Therefore Vk is not simple in G' for all k. Since G' is strongly chordal, there exists u, such that u, is simple in G'. Then by Lemma 3.1.1, there exists w E Na[ut] such that Since Na[w] Na[w], { u1 u2 ... un} Na[w]. Suppose that Na[w] contains a vertex a f. U1, U2, ... Uno Then {a, u1, u2, ... un} is a maximal clique of G2 a contradiction. Therefore, {u1,u2 ... ,un} = Na[w]. Since C is a maximal clique, it follows that C = Na[w]. Suppose that w is simplicial in G. Then for i, j = 1, 2, ... n we conclude that da(u;,u;) = 1. Since G is not complete and connected, there exists a maximal clique C' of G ( C /c C') such that C' and C share at least one 27
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common nonsimplicial vertex z. But C =J C' implies there exists x E C such that x rf C' andy E G' such that y rf C. Thus da(x,y) = 2, a contradiction. Therefore w is nonsimplicial. ( <=) Let z1 be a nonsimplicial vertex of G such that Na[z1 ] is maximal. Let Na(z1 ) = {z2,z3 ,zn} Clearly Na[z1 ] is a clique in G2 Sup pose that it is not a maximal clique. Then there exists u E V( G) such that da(u,z1 ) = 2 and da(u,z;) 2 fori= 2,3, ... ,n. Let P; be a path of shortest length in G between u and z;, where i = 1, 2, ... n. In particular, if da(u,z;) = 1, then P; = {u,z;} and if da(u,z;) = 2, then P; = {u,v;,z;}. Consider the subgraph G' of G induced by u, z; and vk, where i = 1, 2, ... n and k = 1, 2, ... p, for p n. Since G' is strongly chordal, G' has a simple vertex. Since vk has a pair of nonadjacent neighbors for all k, namely u and z;, we conclude that Vk is not simple for all k. Therefore either u is simple or there exists z, that is simple. Call this simple vertex s. By Lemma 3.1.1, there is a neighbor w of sinG' such that Na[w] = {v: da(v,s) 2} where {z,,z2, ... ,zn,s} <;:; Na[w]. But Na[w] <;:; Na[w], so {z1,z2 ,zn,u} <;:; Na[w]. Since w has at least two nonadjacent neighbors u and z1 w is nonsimplicial, a contradiction since Na[z1 ] is maximal. So Na[z1 ] is a maximal clique in G2 Corollary 3.1.3 Raychaudhuri [59]. If G is a strongly chordal graph, then G2 is interval if and only if the maximal nonsimplicial closed neighborhoods of G have a consecutive ranking. The girth of a graph is the length of the shortest cycle in the graph, where girth is undefined if the graph has no cycles. Careful examination of the proof of Theorem 3.1.2 shows that we obtained contradictions by producing 28
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ncycles, where n :::; 6. So we may use similar ideas to generalize the result to graphs with girth p, where p :::0: 7. Theorem 3.1.4 Lundgren, Merz, and Rasmussen [49]. If G is a graph with girth p, where p :::0: 7, then C is a maximal clique in G2 if and only if there exists a nonsimplicial vertex z such that Na[z] = C and Na[z] is maximal. Corollary 3.1.4 Lundgren, Merz, and Rasmussen [49]. If G is a graph with girth p, where p :::0: 7, then G2 is interval if and only if the maximal nonsimpli cial closed neighborhoods have a consecutive ranking'. Observe that if G is a 6cycle, then the maximal nonsimplicial closed neighborhoods are not the maximal cliques of G2 If we exclude graphs containing 6cycles, we are able to characterize the maximal cliques of the squares of at least one class of graphs. Theorem 3.1.5 Lundgren, Merz, and Rasmussen [49]. If G ':p Kn is a con nected 6cyclefree graph such that every edge is contained in a triangle, then C is a maximal clique in G2 if and only if there exists a nonsimplicial vertex v such that C = Na[v] and Na[v] is maximal. Proof. ( <=) Let v be a nonsimplicial vertex in G such that the closed neighborhood of v is maximal. Clearly Na[v] is a clique in G2 Suppose that it is not a maximal clique. Then there exists a vertex w fj Na[v] such that for all x E Na[v], the vertices w and x are joined by a path of length at most two. If for some x E Na[v], we find that {w,x} E E(G), then since every edge is contained in a triangle we conclude that w and x are joined by a path of length two. Since w and v are joined by a path of length two, there exists x such that w,v E Na(x). Since Na[v] is not properly contained in Na[x], there exists y E Na[v] such that {x,y} fj E(G). Since wand yare joined by a path of length two, there exists a vertex u (u f x,u f v) such that w,y E Na(u). 29
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If { u, x}
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Let x E Na(z0). Then {x,w} E E(G). Since R Cl Na[x) there exists y E N[zo) such that { x, y} !f. E( G). Since { w, y} E E( G) and every edge is contained in a triangle, w and y are joined by a path of length two so there exists v (v of x,v of z0 ) such that w,y E Na(v). If x and z0 are not adjacent to v, then since there exists a (a of v, a of w, a of y) such that x, z0 E Na(a). Then xaz0yvwx is a 6cycle, a contradiction. So assume that x and z0 are adjacent to v. Then R Cl Na[v) implies there exists t E R' .such that {v,t} !f. E(G). Then { w, t} E E( G) and wtz0yvxw is a 6cycle, a contradiction. Case 2: There is a vertex y E N a( z0 ) such that {y, w} !f. E( G). Since w andy are joined by a path of length two there exists a vertex v ( v f z0 ) such that w,y E Na(v). Since R Cl Na[v) there exists a vertex x E R' such that {x,v} !f. E(G). Ifw and x are joined by a path of length two, then since there exists a vertex a (a of y,a of z0,a of v) such that w,x E Na(a) we conclude that wvyz0xaw is a 6cycle, a contradiction. So w and x are not joined by a path of length two. Then { w, x} E E( G) and since { w, x} is contained in a triangle we conclude that w and x are joined by a path of length two, a contradiction. Therefore there exists z such that C c;:: Na[z]. Since C is a maximal clique in G2 it follows that C = Na[z). If z is simplicial, then Cis a clique in G. Since G is connected and not complete, there exists a vertex x !f. C such that xis adjacent to a vertex y E N0[z]. But then {x} U Cis a clique in G2 a contradiction. So z is simplicial. Corollary 3.1.5 Lundgren, Merz., and Rasmussen [49]. If G 'jt K,. is a con nected graph such that every edge is contained in a triangle and G contains no 6cycle, then G2 is interval if and only if the maximal closed nonsimplicial neighborhoods have a consecutive ranking. 31
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3.1.1 Using Forbidden Subgraphs Steif [71] showed that, for arbitrary digraphs, there can be no forbidden subgraph characterization of digraphs with interval competition graphs. However, there may be partial forbidden subgraph characterizations, provided the digraph is symmetric with a loop at each vertex. Figure 3.2. The smallest tree that is not interval. Can the square of a tree that is not interval be interval? We already know that the square of an interval tree is intervaL Consider the tree in Figure 3.2. This tree is not interval nor is its square. Recall that a tree is interval if and only if it does not contain an induced subgraph isomorphic to the tree in Figure 3.2. JfT is a tree containing an induced subgraph isomorphic to the square of the tree in Figure 3.2, does T2 contain an induced subgraph isomorphic to the square of the tree in Figure 3.2? More generally, if H is an induced subgraph of G, then H2 is not necessarily an induced subgraph of G2 An example of such a graph is given in Figure 3.3. For this reason, the forbidden subgraph approach to characterizing graphs with interval squares does not generally work. We now show that it will work for trees. Lemma 3.1.2 Lundgren, Merz, and Rasmussen [49]. If T is a tree and H a connected subgraph ofT, then H2 is an induced subgraph of T2 Proof. Suppose not. Then there are vertices x,y E H such that {x,y} E E(T2 ) but {x,y} tJ E(H2 ) Since {x,y} tJ E(H2 ) and {x,y} tJ E(H) there 32
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Figure 3.3. A graph and an induced subgraph. exists z E T but z \l H such that x,y E Na(z). Since His connected there is a path xv1 VkY joining x and y. Then yzxv1 VkY is a cycle in T, a contradiction. Therefore H2 is an induced subgraph of T2 Theorem 3.1.6 Lundgren, Merz, and Rasmussen [49]. If Tis a tree, then T2 is interval if and only if T is interval. Proof. (<=)This follows from Lemma 3.1.2. ( =}) Assume that T is not interval and suppose that T2 is interval. Then T contains an induced subgraph isomorphic to the graph H in Figure 3.2. By the lemma H2 is an induced subgraph of T2 but H2 is not interval, a contradiction. Thus T2 is not interval. Observe that the tree in Figure 3.2 is G2 one of the forbidden subgraphs shown in Figure 3 .1. Furthermore, G2 and G1 for n 2 6 are the only forbidden subgraphs of an interval graph with noninterval squares. We now consider graphs containing subgraphs isomorphic to G1 for n 2 6. Provided that the girth of such a graph is sufficiently large, the square of the graph is not interval. Lemma 3.1.3 Lundgren, Merz, and Rasmussen [49]. If G is a graph with girth p, where p 2 6 and G contains an induced pcycle C, then 02 is an induced subgraph of G2 33
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Proof. Suppose not. Then there exist vertices x,y E V(C) such that {x,y} E E(G2 ) but {x,y} cj_ E(C2). Since {x,y} cj_ E(C2), {x,y} cj_ E(C) so there exists z E V(G) such that x,y E Na(z). Since x andy are on a pcycle with p 2': 6, the length ofthe shortest path joining them is q :S p3. Let xv1 ... VqIY denote this path. Then yzxv1 vq_1 y is a cycle of length q + 2 < p, a contradiction. Hence C2 is an induced subgraph of G2 Lemma 3.1.4 Lundgren, Merz, and Rasmussen [49]. If G is a graph with girth p, where p 2': 6, then G2 contains an induced cycle of length l where l 2': 4. Proof. Case 1: Suppose that p = 6. Then G contains an induced 6cycle, C = V] V2 VsVJ, SO C2 contains an induced four cycle, namely the cycle v 1 v 2 v 4 v 5 v 1 By Lemma 3.1.3 C2 is an induced subgraph of G2 Hence G2 contains an induced four cycle. Case 2: Suppose that p > 6 and p is even. G contains an induced pcycle, C = v 1 v 2 vPv1 so G2 contains an induced namely the cycle v 1 v 3 v 5 vP_1 v 1 By Lemma 3.1.3 C2 is an induced subgraph of G2 Hence G2 contains an induced where p 2': 8. Case 3: Suppose that p > 6 and p is odd. G contains an induced pcycle, C = v 1 v 2 "vpv1 so C2 contains an induced P!1cycle, namely the cycle v 1 v 3 v 5 vpv1 By Lemma 3.1.3 C2 is an induced subgraph of G2 Hence G2 contains an induced where p 2': 7. Theorem 3.1.7 Lundgren, Merz, and Rasmussen [49]. If G is a connected graph with girth p, where p > 5, then G2 is not interval. 3.1.2 Elimination Ordering Characterizations We can define elimination orderings analogous to those used in Section 2 to characterize 34
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symmetric digraphs with a loop at each vertex whose competition graph are chordal and interval. An ordering v1 v2 . Vn of the vertices of a graph G is a chordal square elimination ordering if and only if for ally, z E { v;+J, v;+2 vn}, whenever there exist u and v such that y,v; E Na[u] and z,v; E Na[v], then there exists w such that y, z E Na[w]. Theorem 3.1.8 Lundgren and Merz [48]. A graph G has a chordal square if and only if G has a chordal square elimination ordering. Proof. Assume that v1 v2 Vn is a perfect elimination ordering of G2 This is true if and only if for v; andy, z E { Vi+J, ... vn}, {v;,y},{v;,z} E E(G2 ) implies that {y,z} E E(G2). This is true if and only if for v;, and y, z E { v;+J, ... vn}, whenever there exists u, v such that v;, y E Na[u], v;, z E Na[v], then there exists w such that y, z E Na[w]. This is true if and only if the ordering of vertices v1,v2 ,vn is a chordal square elimination ordering of G An ordering v1 v2 Vn is an interval square elimination ordering if and only if for all i < j < k, if there exists u such that v,, Vk E Na[u], then there exists v such that v;,vk E Na[v]. Theorem 3.1.9 Lundgren and Merz [48]. A graph G has an interval square if and only if G has an interval square elimination ordering. Proof. Assume that v1,v2 . ,vn is an interval elimination ordering of G2 This is true if and only if for i < j < k, whenever { v;, vk} E E( G2), then 35
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{v;,vk} E E(G2 ). This is true if and only if fori < j < k, whenever there exists u such that v,, Vk E NG[u], then there exists v such that vi, Vk E NG[v]. And this is necessary and sufficient for v1 v2 Vn to be an interval square elimination ordering of G. 3.2 Loopless Symmetric Digraphs In the previous section, we considered the competition graphs of symmetric digraphs with a loop at each vertex. This motivates the consideration of the competition graph of symmetric digraphs without loops. As with symmetric digraphs with a loop at each vertex, given a symmetric digraph without loops, we may consider the underlying graph of the digraph. Given a graph G, the twostep graph, S2(G) is a graph on the same vertex set as G with {x,y} E E(S2(G)) if and only if there exists z E V(G) such that {x,z} and {y, z} are in E( G). This definition motivates the following observation that is stated without proof. Proposition 3.2.1 If D 1s a symmetric digraph without loops and H its underlying graph, then C(D) = S2(H). We will characterize classes of graphs with interval twostep graphs. We first observe that the twostep graph of an interval graph is not necessarily interval (see Figure 3.4), making this problem more difficult than characterizing graphs with interval squares. As in the characterizations of the previous section, we employ the Fulkerson and Gross [22] characterization of interval graphs: A graph G is interval if and only if its family of maximal cliques of G has a consecutive ranking. We will restrict our discussion to connected graphs that are not complete since disconnected graphs can be examined by connected 36
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components and the twostep graph of the complete graph is also complete. Figure 3.4. An interval graph whose twostep graph is not interval. 3.2.1 Finding Maximal Cliques of the TwoStep Graph Our approach is very similar to that used in characterizing graphs with interval squares. We will find a family of sets MC(S2(G)), defined in the original graph, that corresponds to the maximal cliques in the twostep graph. Then S2(G) is interval if and only if MC(S2(G)) has a consecutive ranking. The first class we consider is trees. Theorem 3.2.1 Lundgren, Maybee, Merz, and Rasmussen (44]. If T is a tree, then C such that ICI 2:: 2 is a maximal clique in S2(G) if and only if C == Nr( v ), where v is nonpendant. Proof. ( <=) Let C == Nr( v ), where v is a non pendant vertex in T. Clearly Nr( v) is a clique in S2(T). Suppose that it is not a maximal clique. Then there exists a vertex w 'I C that is joined to every vertex in C by a path of length two. Since ICI 2:: 2, there exist distinct vertices x andy in Nr(v). Since T is a tree, x and y 'I E(T). Then there exist vertices t and u such that x,w E Nr(t) and y,w E Nr(u). If t == u, then vxuyv is a cycle in T, a contradiction. Therefore t =f u. Then vxtwuyv forms a cycle in T, a contradiction. Thus no such w can exist. Therefore Nr( v) is a maximal clique 37
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in S2(T). Furthermore, if Nr( v) = Nr( z) for two vertices v and z, then there exist :v andy E Nr(v) n Nr(z) and v:vzyv is a cycle, a contradiction. ( =;.) Let C be a maximal clique in S2(T). Then /C/ 2 2, so there exist distinct :v andy in C. Since Cis a maximal clique in S2(T), there exists a vertex z such that :v,y E Nr(z). Suppose that C f Nr(z). Then there exists a vertexw E C such that w tj Nr(z). Since Tis a tree, {:v,y} t/. E(T). Since Cis a maximal clique in S2(T) there exist vertices t and u such that w, :v E Nr(t) and w, y E Nr( u ). If t = u, then t:vzyt is a cycle in T, a contradiction. Therefore t f u. Then wt:vzyuw is a cycle in T, a contradiction. Thus no such w can exist. That is, Nr(z) =C. Corollary 3.2.1 Lundgren, Maybee, Merz, and Rasmussen (44]. If T is a tree, then S2(T) is interval if and only if the maximal open neighborhoods of the nonpendant vertices in T have a consecutive ranking. We can extend this characterization to trianglefree graphs that conlain no 6cycles (induced or otherwise). First, a lemma that is stated without proof. Lemma 3.2.1 If G is a graph and :v, y, and z are vertices contained in a maximal clique in S2(G) and there does not exists v such that :v,y,z E Na[v], then there must exist distinct a,b, and c such that :v,y E Na(a), y,z E Na(b) and :v,z E Na(c), that is, :vaybzc:v is a 6cycle. Theorem 3.2.2 Lundgren, Maybee, Merz and Rasmussen [44]. If Gop Kn is a connected triangleand 6cyclefree graph, then C such that /C/ ::0: 2 is a maximal clique in S2(G) if and only if C = Na(z) for some z such that Na(z) is maximal. Proof. ( =;.) Let C be a maximal clique in S2 ( G). If /C/ = 2, then the state ment is clearly true. So assume that /C/2 3. Let R C::: C. We prove by induc tion on /R/ that there exists z such that R C::: Na[z] in G. By Lemma 3.2.1, 38
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if IRI = 3, then the claim is true. So assume that IRI 2 4. Furthermore, assume that the claim is true for all R such that IRI < k :S ICI and consider the case IRI = k :S ICI. Pick arbitrary x E R. Let R' = R{x}. By the induction hypothesis there exists z1 such that R' S:::: N0[z1 ] in G. Pick arbitrary y x E R. Let R" = R{y }. By induction hypothesis there exists z2 such that R" S:::: N a [ z2]. Since x and y are in R there exists z such that x,y E Na(z). If z is z1 or z2 we are done so assume not. Observe z1 z2 if R since G is trianglefree (for example, if z1 E R then y and z1 are adjacent and joined by a path of length two). Since IRI 2 4 there exists w E R (w z,w x,w y,w z1,w z2 ) such that w is adjacent to z1 and z2 Then xzyz1wz2x is a 6cycle in G, a contradiction. Therefore without loss of generality we conclude z = z1 That is, R S:::: N a[ z,] for all R S:::: C. In particular C S:::: Na[z1 ] and so by maximality of C we conclude C = N0(z1). ( <=) Let z be a vertex in G such that the open neighborhood of z is not properly contained in the open neighborhood of any other vertex in G. Clearly Na(z) is a clique in S2(G). Suppose that it is not a maximal clique. Then there exists a vertex w if Na(z) such that w is joined by a path of length two to every vertex in Na(z) in G. Let x E N0(z). Since w and x are joined by a path of length two in G there exists a such that x, w E Na(a). But Na(z) is not properly contained in Na(a) so there exists y E Na(z) such that {a,y} if E(G). Then w and y are joined by a path of length two so there exists a distinct vertex b such that y,w E Na(b). Since G is trianglefree, b x. Then zxawbyz is a 6cycle in G, a contradiction. Thus Na(z) forms a maximal clique in S2 ( G). 39
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Corollary 3.2.2 Lundgren, Maybee, Merz, and Rasmussen [44]. If G ft Kn is a connected trianglefree and 6cyclefree graph, then S2 ( G) is interval if and only if the maximal open neighborhoods of G have a consecutive ranking. We now consider graphs such that every edge is contained in a triangle. Theorem 3.2.3 Lundgren, Maybee, Merz, and Rasmussen [44]. If G ft Kn is a connected 6cyclefree graph such that every edge is contained in a triangle, then C such that ICI 2 is a maximal clique in S2(G) if and only if C = Na[z] for some z where Na[z] is maximal. Proof. ( =?) Let C be a maximal clique in S2 ( G). By an analogous argument to that in Theorem 3.2.2 we can show that there exists z such that C c::; Na[z]. Since every edge is contained in a triangle and C is maximal we conclude that C = Na[z]. (=)Let z be a vertex in G such that Na[z] is not properly contained in another closed neighborhood in G. Since every edge of G is contained in a triangle, clearly Na[z] forms a clique in S2 ( G). Suppose that it is not a maximal clique. Then there exists a vertex w such that w is joined to every vertex in Na[z] by a path oflength two while { w, z} 'f. E( G). Since wand z are joined by a path of length two, there exists a vertex v such that w,z E Na(v). Since Na[z] is not properly contained in Na[v], there exists y E Na[z] such that y 'f. Na[v]. Then w andy are joined by a path of length two so there exists u such that w,y E Na(u), where u jc v. Then v,z E Na(u) since otherwise, because the edge { v, z} is contained in a triangle we conclude that there exists a vertex t such that v, z E Na(t), creating the 6cycle ztvwuyz. But then Na[z] is not properly contained in Na[u] so there exists x E Na[z] such that x 'f. Na[u]. If x 'f. Na(v), then since x and w are joined by a 40
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path of length two we conclude that there exists s (possibly equal to y) such that w,x E Na(s). Then wvuzxsw forms a 6cycle in G and we have a contradiction. Thus {x,v} E E(G). Then wvxzyuw forms a 6cycle in G. This contradiction proves no such w can exist. Corollary 3.2.3 Lundgren, Maybee, Merz, and Rasmussen [44]. If G '/! Kn is a connected 6cyclefree graph such that every edge is contained in a triangle, then S2 ( G) is interval if and only if the maximal closed neighborhoods of G have a consecutive ranking. 3.2.2 Competition Covers To generalize the results of the previous section, we must employ the techniques of Lundgren, Maybee, and Rasmussen [46]. A family S = {S1 ... Sr} of sets of vertices of a graph G is called a competition cover of G if the following conditions are satisfied: (1) i,j E Sm implies there exists a vertex k such that i,j E Na(k). (2) if i,j E Na(k) for some k then i,j E Sm for some m. This definition motivates the following proposition that 1s stated without proof. Proposition 3.2.2 Lundgren, Maybee, and Rasmussen [46]. If G is a graph, then S2 ( G) is interval if and only if G has a competition cover S that has a consecutive ranking. The difficulty with this result is finding the right competition cover. Furthermore, it is very difficult to use this characterization to prove that the twostep graph of a given graph is not interval. The problem then becomes finding a family of sets in G that is not only a competition cover, but in addition, determines conclusively whether or not S2 ( G) is interval. We have already achieved this for several classes of graphs in the previous section. We now present the competition cover used by Lundgren, Maybee, and Rasmussen 41
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[46] to characterize interval graphs with interval twostep graphs. Let v be a nonsimplicial vertex in G. We say v is Type I if every maximal clique containing v contains three or more vertices. We say vis Type II if every maximal clique containing v contains exactly two vertices. Otherwise, we say v is Type III. Define S( G), the nonsimplicial competition cover of a graph G, as family of sets {St, ... S,}, where S; is one of the following: (1) Na[v;], the closed neighborhood of vi, if vi is Type I. (2) Na(vi), the open neighborhood of vi, if Vi is Type II. (3) or actually two distinct sets if Vi is Type III, S;, = Cv; = U{CJC E C,v; E C, JCJ 3} and S;, = Na(v;), where C is the family of maximal cliques in G. Define the maximal nonsimplicial competition cover of a graph G, denoted as S'( G) as the set of all sets in S2 ( G) such that no set is properly contained in any other. We first observe the following results. Proposition 3.2.3 Lundgren, Maybee, and Rasmussen [46]. If G ?/! Kn is a connected interval graph, then S'( G) is a competition cover of G. Proposition 3.2.4 Lundgren, Maybee, and Rasmussen [46]. If G ?/! Kn is a connected interval graph, then S2 ( G) is interval if and only if S'( G) has a consecutive ranking. It is interesting to note that a competition cover of a graph does not necessarily correspond to the maximal cliques in the twostep graph. For example, the open neighborhoods of a 6cycle form a competition cover, but the twostep graph of a 6cycle is two triangles. So the open neighborhoods are not the maximal cliques in the twostep graph. Figure 3.5 illustrates another 42
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example in which this is the case. These examples motivate the following question: When does the competition cover S'( G) correspond to the maximal cliques in S2 ( G)? To answer this question we need the following result. 3 1 6 5 Figure 3.5. The family of maximal cliques is not S'( G). Proposition 3.2.5 Lundgren, Maybee and Rasmussen [46]. If G 'f Kn is a connected interval graph and x E V (G) such that x is connected by a path of length two to every vertex in some S; E S'( G), then x E S;. Theorem 3.2.4 Lundgren, Maybee, Merz, and Rasmussen [44]. If G 'f Kn is a connected interval graph, then C is a maximal clique in S2 ( G) if and only if C E S'(G). Proof. ( <=) Let C E S'( G). Clearly Cis a clique in S2 ( G). Suppose that it is not a maximal clique. Then there exists a vertex w tf_ C such that w is joined to every vertex in C by a path of length two. But Proposition 3.2.5 implies w must be an element of C. This contradiction proves C must be a maximal clique in S2 ( G). (=?)Let C be a maximal clique in S2(G). Since G is interval, the maximal cliques of G have a consecutive ranking C1 C1. We claim that there exists a nonsimplicial vertex z such that C <;;; Na[z]. First we will show there exists a vertex z such that C <;;; Na[z]. Suppose not. Let i be the smallest integer such that there exists a vertex x that is an element of both C; 43
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and c, but X tj ci+1 This must occur since c% Na[x]. Let j be the largest integer such that there exists a vertex y that is an element of both C; and C, but y tj C;1 This must occur since C % Na[y]. Note that i must be less than j' for if not then c c;; ck for all j ::; k ::; i. Since X and y are joined by a path of length two, there exists a vertex z such that x and z are contained in a maximal clique and y and z are contained in a maximal clique. Since this ranking is consecutive and x tj C;+l, we conclude that z must be in a clique Ck such that k =:; i. Since y tj C;_1 we conclude that z must be in a clique Cm such that m 2 j. Since this ranking of cliques is consecutive, we conclude that z E Cp for all p such that i::; p =:; j. Note that every vertex of Cis contained in a clique Cp such that i ::; p::; j. Thus C c;; Na[z], a contradiction. Thus there must exist a vertex z such that C c;; Na[z]. Suppose that z is simplicial. Since G is connected and not complete, there exists a vertex x r/ C such that x is adjacent to a vertex y E C. If y is nonsimplicial, then we are done since C c;; Na[y]. So assume that y is simplicial. Then { x} U C is a clique in 52 ( G) containing C, a contradiction. Therefore z is nonsimplicial. If z E C, since C is a maximal clique and z is joined to every vertex in C by a path of length two, it follows that C = Cz. If z tj C, since C is a maximal clique it follows that C = Na(z). In either case, C E S'(G). Theorem 3.2.5 Lundgren, Maybee, Merz, and Rasmussen [44]. If G cp Kn is a connected 6cyclefree graph then C such that ICI 2 2 is a maximal clique in S2(G) if and only if C E S'(G). Proof. ( =?) Let C be a maximal clique in 52 ( G). First we show that there exists z such that C c;; Na[z]. We will prove the result by induction on jRj, 44
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where R 5;:; C. Clearly this is true if ICI = 2. By Lemma 3.2.1 it is true if \CI = 3. So assume that \CI ::::: 4. Assume that there exists z such that R 5;:; Na[z] for R such that IRI < k :S ICI and assume that IRI = k :S IC\. Pick x E R. Let R' = R{x}. By the induction hypothesis there exists z1 such that R' 5;:; Na[z1]. Picky E R such that y f x. Let R" = R{y}. By induction hypothesis there exists z2 such that R" 5;:; Na[z2]. If z1 = z2 then we are done so assume z1 f z2 Then x and y are joined by a path of length two, so we conclude that there exists z such that x,y E Na(z). If there exists w not equal to x, y, z1 z2 or z such that w E R, then we are done since w E Na[z1 ] and w E Na[z2 ] implies that xzyz1wz2x is a 6cycle, so assume no such w exists. Then R 5;:; { x, y, z, z2 z }. Since \RI ::::: 4, at least one of the set { z, z2 } is in R. Without loss of generality assume that z1 E R. Then z1 E Na[z2]. If z1 '/. Na[z] then since z1 andy are joined by a path of length two we conclude that there exists w such that z1,y E Na[w]. But then zwyz1z2xz is a 6cycle, a contradiction. Therefore assume that { z1 z} E E( G). If z2 'f. R then since R 5;:; Na[z] and we are done. So assume that z2 E R. Similarly, if z2 E Na[z], then R 5;:; Na[z2 ] and we are done so assume z2 f/c Na[z]. Then x and z2 are joined by a path of length two implies there exists w such that x, z2 E Na[w] implies xwz2z1yzx is a 6cycle. Therefore there exists z such that R 5;:; Na[z]. Suppose that z is simplicial. Since G is connected and not complete, there exists a vertex x f/c C such that x is adjacent to a vertex y E C. If y is nonsimplicial we are done since C 5;:; Na[y]. So assume that y is simplicial. Then {x} U Cis a clique in S2(G) containing C, a contradiction. Therefore there exists nonsimplicial z such that C 5;:; Na[z]. If z E C, since C is a 45
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maximal clique in S2 ( G) and z is joined to every vertex in C by a path of length two, C = C,. If z f/; C, since Cis a maximal clique in S2(G), C = No(z). ((=) Let C E S'( G). By definition there exists a nonsimplicial vertex z such that C <;:; N0[z]. We have two cases. Case 1: Suppose that there exists nonsimplicial z such that C = C,. Observe that z may be either Type I or Type III. Clearly C is a clique in S2(G). Suppose that it is not a maximal clique. Then, there exists w f/; C such that w is joined to every 8 E C by a path of length two. So there exists x such that z,w E Na(x). Observe that w f/; C, C =C., and {w,x} E E(G) implies {w,z} f/; E(G). Since C rj_ Na[x], there exists a vertex y E C such that { x, y} f/; E( G). Then y and w are joined by a path of length two so there exists a vertex u such that w,y E Na(u), where u I z and u I x. If z f/; Na(u), then since y and z are contained in a triangle we conclude that a vertex t not equal to x and not equal to w such that y, z E Na(t). Then ztyuwxz is a 6cycle, a contradiction. So z E Na(u). Then u E C. Suppose that x E C. If x f/; N0(u), then since x and z are contained in a triangle we conclude that there exists t not equal to u, w nor y such that x, z E No( t ). Then ztxwuyz is a 6cycle. Therefore x E C implies that x E Na(u). But C rj_ Na[u] so there exists v E C such that {u,v} f/; E(G). If {v,y} E E(G), we are done since zvyuwxz is a 6cycle. So {v,y} f/; E(G). Then there exists 8 such that w,v E No(8) where 8 is possibly x but 8 I u,s I y and 8 I z. Then zvswuyz is a 6cycle. Therefore x f/; C. This implies {x, u} f/; E(G). But C rj_ Nc[u] so there exists a vertex v E C such that {u,v} f/; E(G). If {v,y} E E(G), we are done since zvyuwxz is a 6cycle, so assume 46
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not. Then there exists a vertex 8 that is possibly equal to x but not equal to y and not equal to u such that w,v E Na(8). Then zv8wuyz is a 6cycle, a contradiction. Therefore C is a maximal clique in S2 ( G). Case 2: Suppose that there exists nonsimplicial z such that C = Na(z). Observe that z may be Type II or Type III. Clearly C is a clique in S2 ( G). Suppose that it is not a maximal clique. Then there exists a vertex w f/_ C joined to every 8 E C by a path of length two. Let x E C. Then there exists y such that x,w E Na(y). Since C r:t Na(Y) there exists v E Na(z) such that {y,v} f/_ E(G). Then wand v must be adjacent to x since otherwise there exists a distinct vertex t such that w,v E Na(t) and wtvzxyw is a 6cycle. If z is Type II, then z is contained in a triangle, namely vxzv, a contradiction. So assume that z is Type III. Then since C r:t Na[x] we conclude that there exists u E C such that {x,u} f/_ E(G). If { u, y} E E( G), then we are done since wxvzuyw is a 6cycle. So assume that { u, y} f/_ E( G). Suppose { w, u} f/_ E( G). Then there exists a distinct vertex t such that u and w are adjacent tot. Then utwyxzu is a 6cycle, a contradiction. Thus {w,u} E E(G). Then wvuzxyw is a 6cycle, a contradiction. Therefore Cis a maximal clique in S2 ( G). Corollary 3.2.4 Lundgren, Maybee, Merz, and Rasmussen [44]. If G '"F Kn is a connected 6cyclefree graph, then S2(G) is interval if and only if the maximal nonsimplicial competition cover of G has a consecutive ranking. As one might guess from the previous results, six cycles pose a problem in this approach to characterizing graphs with interval twostep graphs. We now deal with the case that 6cycles are relatively sparse in the graph. Let 47
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H = abcdefa denote a 6cycle with vertices consecutively labeled. The alternating triples of Hare {a,c,e} and {b,d,f}. It is easily observed that the set of alternating triples of a 6cycle correspond to the maximal cliques in the twostep graph. Figure 3.6 illustrates a graph whose twostep graph contains a clique that is neither a set in the maximal nonsimplicial competition cover nor an alternating triple. Notice that there is a pair of 6cycles that share more than a single edge. z z, y Figure 3.6. The set {x,x1,x2,x3 } is not in S'(G). Theorem 3.2.6 Lundgren, Maybee, Merz, and Rasmussen [44]. If G '/' Kn is a connected trianglefree graph such that no two 6cycles have more than a single edge in common, and Cis a maximal clique in S2(G) such that ICI 2: 2, then either C = Na(z) for some nonsimplicial vertex z in G or C is an alternating triple from a 6cycle in G. Proof. If ICI = 2, then C must be the open neighborhood of a nonsimplicial vertex with precisely two neighbors making the statement is true. If ICI = 3, then by Lemma 3.1.3 and the maximality of C, we observe that the statement is true. So assume that ICI 2: 4. Let R denote a subset of C. We will prove by induction on IRI that there exists a vertex z such that C c:;: N0[z]. 48
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Let IRI = 4. Pick arbitrary x E R. Let R' = R{x}. By Lemma 3.1.3 there exists y such that R' <;;; Na[Y] orR' is the set of alternating triples from a 6cycle in G. Assume that there exists y such that R' <;;; Na[y]. Suppose that y E C. Then y is joined by a path of length two to one of its neighbors. This is a contradiction since G is trianglefree. Thus y C and hence y 1/ R'. If x E Na[y], then we are done so assume x Na[y]. Further more assume that there does not exist z such that R <;;; N G [ z ]. Since I Rl = 4, there exists a vertex a E R'. Then there exists t such that x, a E Na(t). Since there does not exist z such that R <;;; Na[z] and IRI = 4 there exists b E R' not equal to a and u not equal tot such that x,b E Na(u). Since there are no triangles in G, we conclude that t and u are not adjacent to y. Furthermore, y 1/ R. Let c denote the remaining vertex in R'. If there exists a distinct vertex s such that x,c E Na(s), then xtaybux and xscybux are two 6cycles with more than a singe common edge, a contradiction. So assume that no such s exists. Then c must be adjacent to t or u. Without loss of generality assume that { c, u} E E( G). Then xtaybux and xtaycux are two 6cycles with more than a single common edge, a contradiction. Thus there must exist z such that R <;;; Na[z]. Assume that R' is an alternating triple from a 6cycle in G. Let a, b, c denote the vertices of R'. Then there exist p,q,r E V(G) such that bpaqcrb is a 6cycle in G. If there exists a vertex z such that three elements of R are in the open neighborhood of z, then we can let R' be the set of these vertices and we are in the former case. So assume that no such z exists. Then x is not adjacent to p, q nor r. So there exist distinct vertices w and y such that 49
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x,b E Na(w) and x,a E Na(y). Then wxyapbw and apbrcqa are two 6cycles with more than a common edge, a contradiction. So there must exist a vertex z such that R t:;; Na[z]. This verifies the statement for IRI = 4. Assume that the statement is true for all R such that IRI < k :S: [C[ and let IRI = k :S: [C[. By assumption IRI > 4. Pick an arbitrary vertex"' E R and let R' = R{x}. By the induction hypothesis there exists z0 such that R' t:;; Na[z0]. Suppose that there does not exist z such that R t:;; Na[z]. Then {x,z0 } if_ E(G). Furthermore there must exist distinct vertices y, z E R' and a and b such that x, z E Na(a) and x,y E Na(b). Since there are no triangles in G, we conclude that no two elements of Rare adjacent and a, b if_ R', Na(z0). Furthermore, z0 if_ R. Since I Rl > 4 there exists another distinct vertex w E R'. If there exists a distinct vertex c such that w,"' E Na(c) we are done since xazz0ybx and xcwz0ybx are two 6cycles with more than a single common edge. Thus w must be adjacent to a or b. Without loss of generality, assume that { w, b} E E( G). Then xazz0ybx and xbwz0zax are two 6cycles with more than a single common edge, a contradiction. This proves that for all subsets R of C, there must exist z such that R t:;; Na[z]. In particular there exists z such that C t:;; Na[z]. Since there are no triangles in G, we conclude that z tj C. Since C is a maximal clique in S2(G), we conclude that C = Na(z). DefineR( G) as the union of S'( G) and the set of all alternating triples in G. Define R'( G) as the set of all sets in R( G) such that no set is properly contained in any other. 50
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Theorem 3.2.7 Lundgren, Maybee, Merz, and Rasmussen [44]. If G '/! Kn is a connected trianglefree graph such that no two 6cycles have more than a single edge in common and C E R'( G), then Cis a maximal clique in S2 ( G). Proof. Since G is trianglefree, every nonsimplicial vertex is of Type II. Thus we need only consider two cases. Case 1: Suppose that Cis the open neighborhood of a nonsimplicial vertex z. Then ICI 2': 2. Clearly C forms a clique in S2 ( G). Suppose that it is not a maximal clique. Then there exists w e) Na(z) such that w is joined to every vertex in Na(z) by a path of length two. Observe that there does not exist a vertex p such that {w} U Na(z)!:;; Na(p), since Na(z) is not properly contained in Na(p). Thus there exist x,y E Na(z) and a and b (not in Na(z) since G is trianglefree) such that x,w E Na(a) and y,w E Na(b). There must exist another distinct vertex u E Na(z), since Na(z) is not properly contained in an alternating triple. If there exists a distinct vertex c such that w,u E Na(c), then we have two 6cycles in G with more than a single edge in common, a contradiction. Since G is trianglefree x, y e) Na( w ). Thus u must be adjacent to a or b. In either case we have two 6cycles with more than a single edge in common, a contradiction. Thus no such w can exist. That is, Na(z) is a maximal clique in S2(G). Case 2: Suppose that Cis an alternating triple {x,y, z}. Then there exist vertices a, b,and c such that xaybzcx is a 6cycle in G. Clearly C is a clique in S2(G). Suppose that it is not a maximal clique. Then there exists a vertex w joined to x,y and z by a path of length two. Since G is trianglefree, we conclude that x,y,z e) Na(w). Suppose that w is adjacent to more than one element of the set {a, b, c }. Then we have two 6cycles with more than a 51
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single edge in common. Suppose that w is adjacent to one element of the set {a, b, c }. Without loss of generality assume that { w, c} E E( G). Then since w and y are joined by a path of length two we conclude that there exists a new vertex d such that w,y E NG(d). Then we have two 6cycles with more than a single edge in common. Thus w is not adjacent to a, b or c. Then there must exist new vertices 8 and t such that w,x E NG(s) and w,z E NG(t). If 8 = t, then xaybzsx and xaybzcx are two 6cycles with more than a single common edge. Therefore s fc t. But then xczbyax and xcztwsx are two 6cycles with more than a single common edge. Thus C is a maximal clique in S2 ( G). Corollary 3.2.5 Lundgren, Maybee, Merz, and Rasmussen [44]. If G ';/! Kn is a connected trianglefree graph such that no two 6cycles in G share more than one edge, then S2 ( G) is interval if and only if R'( G) has a consecutive ranking. 3.2.3 Elimination Ordering Characterizations We can define elimination orderings analogous to those used in Chapter 1 to characterize symmetric digraphs without loops that have chordal and interval competition graphs. An ordering v1,v2 ,vn is a chordal twostep elimination ordering if and only if for all y,z E {v1,v2 ,vn}, whenever there exist u and v such that y,v; E NG(u) and z,v; E NG(v), there exists w such that y,z E NG(w). Theorem 3.2.8 Lundgren and Merz [48]. A graph G has a chordal twostep graph if and only if G has a chordal twostep elimination ordering. Proof. Assume that v1,v2 ,vn is a perfect elimination ordering of S2(G). This is true if and only if for v; and y, z E {Vi+!, v;n, ... vn}, whenever 52
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{v;,y},{v;,z} E E(S2(G)), we have that {y,z} E E(S2(G)). This is true if and only if for y, z E { Vi+h v;+2 vn}, whenever there exist u and v such that y,v; E Na(u) and z,v; E Na(v), there exists w such that y,z E Na(w). This is true if and only if v1 v2 Vn is a chordal twostep elimination ordering of G. An ordering v1 v2 Vn is an interval twostep elimination ordering if and only if for all i,j, and k such that i < j < k, whenever there exists u such that v;,vk E Na(u), there exists v such that Vj,vk E Na(v). Theorem 3.2.9 Lundgren and Merz [48). A graph G has an interval twostep graph if and only if G has an interval twostep elimination ordering. Proof. Assume that Vj, v,, ... 'Vn is an interval elimination ordering of s,( G). This is true if and only if fori < j < k, whenever {v;,vk} E E(S2(G)), we have that {vj,vk} E E(S2(G)). This is true if and only if fori < j < k, whenever there exists u such that v;,vk E Na(u), there exists v such that Vj,Vk E Na(v). This is true if and only if v 1,v2 ... ,vn is an interval twostep elimination ordering. 3.3 Strongly Connected Digraphs Which graphs are the competition graphs of strongly connected digraphs? While many digraphs modeling communication networks may be symmetric, it is more important that they be strongly connected. Let GE(G) denote the edge clique cover number of G, i.e., the smallest number of cliques that cover all the edges of G. Let i( G) denote the number of isolated vertices in a graph G. We begin with the following proposition that we state without proof. 53
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Proposition 3.3.1 Roberts and Steif [66]. G is a competition graph of a digraph that has no loops if and only if G of K2 and 0E(G) :S IV( G) I. Will the condition given in Proposition 3. 3.1 work for strongly connected digraphs as well? The graph in Figure 3.7 illustrates that it will not work. While 0E(G) = 9, unfortunately 0E(G) + i(G) > 9. This is important, as indicated by the next lemma. j Figure 3. 7. A competition graph. This graph is the competition graph of a loopless digraph, but not one that is strongly connected. Lemma 3.3.1 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. If G is the competition graph of a strongly connected digraph, then 0 E( G) + i(G)::: IV(G)I. An ordering D1, D2, ... Dk of the strongly connected components of a digraph D is topological if and only if for x E V(D;) and y E V(Dj), whenever (x,y) E A(D) it follows that i :<; j. Theorem 3.3.1 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. If G is a graph with no isolated vertices such that G of K2 and 0 E( G) :<; n, then G is the competition graph of a strongly connected digraph. Proof. By Proposition 3.3.1, G = C(D) for some (loopless) digraph D. Among all such digraphs, let D be one with the smallest number k of strong components and suppose that k 2. Let Dl> D2 Dk be a topological ordering of the strongly connected components of D. Observe that since G has no isolated vertices and 54
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every vertex in D has an outgoing arc, this ordering implies there are at least two vertices in Dk. Case 1: Suppose that V(D1 ) = {x}. Observe that I(x) = 0. Since G has no isolated vertices, we conclude that O(x) f 0. Let (x,y) E A(D) for some vertex y. Then create a new digraph D' by simply adding the arc (y,x) to D. Then D' has fewer strong components than D and C(D') = G, a contradiction. Case 2: Suppose that IV(D1)12: 2 and there is a vertex y E Dk such that (x,y) if A(D) for some x E D1 Since IV(Dk)l 2: 2 and Dk is strongly connected, we conclude that l(y) f 0. As in Case 1 we construct a new digraph D'. Let ln(Y) = In(x) and ln(x) = In(y) with all other arcs in D' staying the same as in D. We claim that D1 U Dk is strongly connected in D'. To prove this, we must show that given arbitrary vertices u, w E D1 U Dk, there is a path from u tow. Let u E V(D,) not equal to x and wE V(Dk) not equal toy. In D, u reached x, so in D', u reaches y. In D, y reached w and none of those arcs have changed. Therefore u reaches all w E V(Dk) in D'. In D, x reached u and none of those arcs have changed. Therefore x reaches w. Therefore every vertex in D1 reaches every vertex in Dk. In D, w reached y, so in D', w reaches x. In D, x reached u and none of those arcs have changed, thus w reaches u. In D, y reached w and none of those arcs have changed, thus y reaches u. Therefore every vertex in Dk reaches every vertex in D1 Since every vertex in D1 reaches every vertex in Dk and every vertex 55
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in Dk reaches every vertex in D 1 every vertex in D 1 reaches every vertex in D 1 and every vertex in Dk reaches every vertex in Dk, completing the proof that D1 U Dk is strongly connected. So D' has fewer strong components than D and C(D') = G, a contradiction. Case 3: Suppose that IV(D1)1 2': 2 and every vertex in D1 has an arc to all vertices in Dk. Let x E V(D!) andy E V(Dk) Since every vertex in D 1 has an arc to y, D 1 is a clique in C(D) and all arcs. in the inset of x can be removed in a new digraph D' without changing the competition graph. Now let ID(x) = {y}, and let all other arcs in D' remain the same as in D. Then D 1 U Dk is strongly connected in D' by arguments analogous to that in Case 2, so D' has fewer strong components than D and C( D') = G, a contradiction. Thus D is strongly connected, since we assumed that k 2': 2 was the smallest number of strongly connected components and in all cases reached a contradiction. Corollary 3.3.1 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. If G is a graph such that G f K1 or K2 and GE(G) + i(G) :0:: IV( G) I, then G is the competition graph of a strongly connected digraph. Proof. If G is a graph on IV( G) I isolated vertices, then G is the competition graph of a cycle. So assume that G is not a graph on IV( G) I isolated vertices. If G has no isolated vertices, then we are done by the previous theorem. So assume i( G) = k f 0. Let h denote the subgraph of G induced by the k isolated vertices. Then there exists a graph G' such that G = G' U h. Observe that G' has m = IV( G)lk vertices, none of which are isolated and If G' = K2 then let x and y be the vertices of G'. Let D be the 56
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digraph with cycle xyam+lam+2 ... am+k and the arc (x, am+l) Then C(D) = G and D is strongly connected. If G1 f K2 then by the previous theorem there exists a strongly connected digraph D1 such that C(D1 ) = G1 Create D as follows: Let V(D) = V(D1 ) +h. Choose x E V(D1). Let lv(am+,) = lv(x), Iv(x) = am+k and (a;,a;+1 ) E A1 fori= m+ 1,m + 2, ... ,m+ (k 1). Leave all remaining arcs of D1 in D. Then none of the competitions in D1 have been changed so C(D) =G. We claim that Dis strongly connected. Since D1 is strongly connected x reaches every vertex in V( D1). The only arcs that have changed are arcs from lv(x). Therefore x reaches every vertex in V(D1 ) in D. Let u be a vertex in lv(x). Clearly u reaches every vertex in h in D and since x reaches u, x reaches every vertex in h Therefore x reaches every vertex in D. Let v be an arbitrary vertex in V(D1). We claim v reaches x. Let u E lv(x). There is a path from u to x, namely uam+lam+2 ... am+k"' and v reaches u in D because v reaches u in D1 Therefore v reaches x in D. Clearly an arbitrary vertex v in h reaches x. Thus every vertex in hand V(D1 ) reaches x in D. Thus Dis strongly connected. Corollary 3.3.2 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman (21]. If G is a graph such that IV( G) I= n?: 3, then G is the competition graph of a strongly connected digraph if and only if 8 E( G) + i( G) ::; n. 3.3.1 Generalization to pCompetition Graphs The graph in Figure 3.7 is not the competition graph of strongly connected digraph. It is, however, the 2competition graph of a strongly connected digraph as shown in Figure 3.8. By making p?: 2, we can strengthen Theorem 3.3.1. Theorem 3.3.2 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If G is the pcompetition graph of a loopless digraph for p ?: 2, then G is the pcompetition graph of a strongly connected loopless digraph. 57
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. 6 1 5 2 4 3 4 3 Figure 3.8. A 2competition graph. G is the 1competition graph of a loopless digraph, but not one that is strongly connected. G is the 2competition graph of a loopless strongly connected digraph. The digraph is not strongly connected but G is its 2competition graph. Adding the arc (9, 5) makes the digraph strongly connected, but does not change its 2competition graph. Proof. Observe that if G has !V(G)I isolated vertices and we let D be a directed cycle on !V(G)I vertices, then CP(D) = G for D strongly connected. Therefore we may assume that G has at least one edge. Let D be a loopless digraph with fewest number of strongly connected components such that Cp(D) = G. Let D1 D2 .. Dk be the topologically ordered strongly connected components of D. If D1 is Dk, then we are done so assume that k 2: 2. Since G has at least one edge, we can assume that D1 contains a vertex x with at least one outgoing arc (if D1 does not contain such a vertex then D2 Dk, D1 is a topological ordering of the strongly connected components). We then have three cases. Case 1: Suppose that V(D1 ) = {x}. Then I(x) = 0 and O(x) fc 0. Let y be a vertex such that there is an arc from x to y. Let Dy denote the 58
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strongly connected component containing y. CreateD' by adding the arc (y,x) to A( D). The set of vertices competing with x at least p times is unchanged. Since y is the only vertex with an arc to x, the set of vertices competing with y is unchanged. Thus Cv(D) = G implies Cv(D') = G and D' has fewer strongly connected components than D, a contradiction. Case 2: Suppose that IV(D1)j 2 and there exists y E Dk, x E D1 such that there is no arc from x toy. Suppose that jV(Dk)j < p. If there exists q rj Dk such that (q,y) E A(D), createD' by adding (y,q) to A(D). Since Dis loopless, y competes at most p1 times with any vertex. In particular, y competes at most p2 times for a vertex in Dk and once for q. Therefore the set of vertices competing with y at least p times is unchanged. Then Cp(D) = G implies Cp(D') = G. Letting Dq denote the strongly connected component containing q, we observe that Dk U Dq is strongly connected in D'. That is, D' has fewer strongly connected components than D, a contradiction. Thus all arcs incoming at y originate in Dk. Create D' by adding ( x, y) and (y, x) to A( D). Then the set of vertices competing with x at least p times in D has not changed since at most p1 vertices have arcs to y. In particular, at most p2 vertices in Dk and x have arcs to y. The set of vertices competing with y at least p times in D has not changed since y has at most p l outgoing arcs. In particular, y has arcs to at most p2 vertices in Dk and x. Thus Cp(D) = G implies CP(D') = G and D1 U Dk is strongly connected in D'. That is, D' has fewer strongly connected components than D, a contradiction. 59
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Thus IV(Dk)J 2': p. Since JV(D,)J 2': 2, I(x) f 0 and since IV(Dk)J 2': p ::0: 2, l(y) f 0. Create D' from D by switching the insets of x and y and leaving all other arcs the same. No competitions have changed. Therefore Cv(D) == G implies Cv(D') ==G. Since JV(D,)J ::0:2 and IV(Dk)i ::0:2, D1 UDk is strongly connected in D', i.e., D' has fewer strongly connected components than D, a contradiction. Case 3: Suppose that IV(D1)J 2': 2 and for all x E V(D1 ) and all y E V(Dk), (x,y) E A(D). Suppose that IV(Dk)/ < p. CreateD' by adding arc (y,x) to A( D). Then the set of vertices competing withy at least p times has not changed since y has arcs to at most (p 1) vertices (namely at most (p2) in Dk and x). Since the set of vertices competing with x at least p times has not changed, Cv(D) == G implies Cv(D') =G. Since D1 U Dk is strongly connected in D', D' has fewer strong components than D, a contradiction. Thus IV(Dk)! ::0: p. Since every vertex in D1 has an arc to every vertex in Dk, we conclude that D1 is a clique in G. Thus we can remove arcs between vertices strictly in D1 and the pcompetition graph is unchanged. Pick an arbitrary vertex x E V( D1 ). Create D' by deleting all arcs incoming at x and adding arc (y, x) to A( D). The set of vertices competing with y at least p times is unchanged since y is the only vertex with an arc to x. Thus Cp(D) == G implies Cp(D') ==G. Since IV(D,)J ::0: 2 and JV(Dk)l ::0: 2, D1 U Dk is strongly connected in D', i.e., D' has fewer strongly connected components than D, a contradiction. Since in each case the contradiction implies D has fewer than k strongly connected components where k ::0: 2, we must have k = 1, i.e., D 60
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is strongly connected. Therefore every graph that is the pcompetition graph of a loopless digraph is the pcompetition graph of a strongly connected loopless digraph. 3.4 Hamiltonian Digraphs One special class of the strongly connected digraphs is the Hamiltonian digraph. In looking at examples of graphs on [V(G)I vertices satisfying 8E(G) + i(G) [V(G)I, we discovered that in many cases, not only are the graphs competition graphs of strongly connected digraphs, but frequently the digraphs are Hamiltonian. Certainly 8E(G) + i(G) [V(G)I is necessary, but is it sufficient? Our first result is analogous to Brigham and Dutton's [7] characterization of the competition graphs of acyclic digraphs. Theorem 3.4.1 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. A graph G with [V( G)l = n is the competition graph of a Hamiltonian digraph if and only if G has a labeling a 1 a 2 ... an of its vertices and an edge clique covering { 01 02 Cn} satisfying a; rf C; for i = 1, ... n and a; E Ci+1 for i = 1, ... n1 and an E C1. Proof. ( =}) Suppose that G = C(D) for D Hamiltonian and a1 .. an is a Hamiltonian cycle in D. Let C; =I( a;). Then {01 Cn} is an edge clique covering of G. Furthermore a; rf C; since D has no loops and a; E 0;+1 since (a;,a;+l) is an arc. Finally an E 01 since (an, a!) is an arc. ( <=) Suppose that G has a labeling a1 a2 an of its vertices and an edge clique covering { 01 02 Cn} satisfying a; rf C; for i = 1, ... n and a; E 0;+1 for i = 1, ... n 1 and an E 01 Construct D with vertices al, ... ,an such that J(a;) = C;. Then a; E ci+l implies that (a;, ai+l) is an arc and an E 01 since (an, a1 ) is an arc. Therefore D is Hamiltonian. From 61
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the construction, it is clear that C(D) = G. We now give several classes of graphs that satisfy the condition of Theorem 3.4.1. Corollary 3.4.1 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. If G is a cycle, then G is the competition graph of a Hamiltonian digraph. Proof. Let a1 ,an be the vertices of the cycle. Let C1 = {an_1,an},C2 = {an,a!},Cs = {a,,a2},,Cn = {an2,anJ} Then {CJ,,Cn} is an edge clique covering of G satisfying Theorem 3.4 .1. Corollary 3.4.2 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. The complete graph Kn for n 3 is the competition graph of a Hamiltonian digraph. Proof. Let V(Kn) = {a1 ,an}. Let C, = V{a,}. Then {C,, ... ,Cn} is an edge clique cover of Kn satisfying Theorem 3.4.1. Corollary 3.4.3 If G is a chordal graph on n > 3 vertices, then G is the competition graph of a Hamiltonian digraph. Proof. The proof is by induction on the number of vertices. It is easy to verify the theorem for chordal graphs with three vertices. Assume that the result is true for all chordal graphs with less than n vertices, and suppose that G is a chordal graph with n vertices. If G = Kn, then we are done by Corollary 3.4.2. So suppose that G f Kn, and let x be a simplicial vertex in G and let G' = G{ x }. Since G' is chordal, G' = C(D') where D' has the Hamiltonian cycle a1a2 an! Since G f Kn, there is a vertex a; !/. NG[x], the closed neighborhood of x. Note that NG[x] is a clique since xis simplicial. We define a new digraph D as follows: Iv(x) = fv,(a;);Iv(a;) = NG[x]; and 62
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Iv(a;) = lv(a;) for all i j j. Clearly C(D) = G and a1a2 . a;_1xa; ... an! is a Hamiltonian cycle in D. Corollary 3.4.4 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. If G is an interval graph on n 2': 3 vertices, then G is the competition graph of a Hamiltonian digraph. Lemma 3.4.1 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. If G is a connected trianglefree graph with E>E(G) :0::: IV(G)I, then G is a tree or a tree with an additional edge. Proof. Since G is trianglefree, the cliques in any clique cover of minimum cardinality are just the edges of G. Since G is connected and E> E( G) :0::: IV( G) I, then either IE(G)I = IV(G)I1 and G is a tree or IE(G)I = IV(G)I and G is a tree plus an edge. Corollary 3.4.5 If G is a connected trianglefree graph on n 2': 3 vertices with e E( G) :::: n, then G is the competition graph of a Hamiltonian digraph. Proof. The proof is by induction on n. For n = 3, a path on three vertices is the only such graph, and by Corollary 3.4.3 a path on three vertices is the competition graph of a Hamiltonian digraph. Assume that the result is true for all connected trianglefree graphs with n vertices and suppose that G has n + 1 vertices. By Lemma 3.4.1, G is either a tree or a tree plus an edge. If G is a tree, we are done by Corollary 3.4.3, so assume that G is a tree plus an edge. If G has no pendant vertices, then G is a cycle and the result follows from Corollary 3.4.2, so assume that G has a pendant vertex. Let x be a pendant vertex in G. Then G{x} satisfies the induction hypothesis. Let a1 . an be the consecutively labeled vertices of the Hamiltonian cycle. Since n 2': 3 and deg(x) = 1, there is a vertex a; such that {x,a;} E E(G) and a vertex ak such that {x,ak} t/ E(G). We construct a new digraph D1 as 63
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C(D') = G and a1a2 ak_1xak an is a Hamiltonian cycle in D'. Is the condition GE(G) + i(G) :S: IV( G) I sufficient for G to be the competition graph of a Hamiltonian digraph? Consider the graph in Figure 3.9. The reason this graph is not the competition graph of a Hamiltonian digraph is a consequence of the following theorem. Figure 3.9. An example. This is not the competition graph of a Hamiltonian digraph. Since the component on 6 vertices has a minimum of 7 sets in an edge clique cover, we conclude no edge clique cover for the entire graph will have a system of distinct representatives. Theorem 3.4.2 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. If G is the competition graph of a Hamiltonian digraph, then G has an edge clique covering that has a system of distinct representatives among the vertices of the cliques. Proof. Assume that G is the competition graph of a Hamiltonian digraph D. Let a1 a2 an denote the consecutively labeled vertices of a Harnilto nian cycle in D. Then the family of insets of the vertices of D are an edge clique cover of G. Furthermore, for i = 1, 2, ... n1, choose a; to represent I(ai+!) and choose an to represent I(a1 ) and we have a system of distinct representatives. Corollary 3.4.6 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. If G is the competition graph of a Hamiltonian digraph, then e E( G;) :s; n,, where G has connected components G;, each with n; vertices. 64
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Proof. Suppose that G is the competition graph of a Hamiltonian digraph. If EJ E( G;) > n; for some i, then every edge clique cover C of G requires more than n; cliques to cover G;, and each of these cliques contains only vertices of G;. It follows that C does not have a system of distinct representatives, a contradiction. Corollary 3.4.7 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. If G is a graph satisfying EJE(G) = IV(G)I and G is the competition graph of a Hamiltonian digraph, then G contains no adjacent pair of simplicial vertices. Proof. Suppose that G = C(D) forD Hamiltonian and that EJE(G) = JV(G)I. Suppose that x and y are simplicial vertices in G that are adjacent. Then NG[x] U NG[y] is a clique, so G{x,y} requires at least JV(G)I1 cliques in any clique cover. But then no edge clique cover of G contains a system of distinct representatives, a contradiction. Theorem 3.4.3 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. If G is a trianglefree graph, then G is the competition graph of a Hamiltonian digraph if and only if EJE(Gi) _:::: JV(Gi)l for each connected component G; of G. Proof. One direction of the proof follows from Corollary 3.4.6. Suppose that ElE(Gi) _:::: n; for each component G;, 1 _:::: i _:::: k, where k is the number of components in G. If n; :;:: 3, then G; is the competition graph of a Hamiltonian digraph D; with Hamiltonian cycle a\, ... by Lemma 3.4.1. If n; = 2, then G; 3" K,, and we let D; be the arc (a\, (observe C(D;) fc G; in this case). If n; = 1, then D; is the isolated vertex a\. Let D = We can now form a digraph D' such that C(D') = G and D' is Hamiltonian. For i = 1, ... k1, if n; :;:: 3, let In.(a\+1 ) = In(a\). If nk :;:: 3, let ln(ai) =In( Fori= 1, ... ,k 1, if n; = 2, let ln(ai+1 ) = {a\,an. If nk = 2, let 65
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I D' (aD = an. For i = 1' ... k 1' if n, = 1' let I D' I) = If nk = 1, let Iv(aD = Let all other insets in D' be the same as in D. Observe we have not changed any existing competitions from C(D) to C(D'). Thus if n; 2, we get all the edges of G, in C(D). For i = 1, ... k 1, if n; = 2 then = {a\, aD gives the edge in G; = K2 Similarly, Iv(aD does this if nk = 2. Thus C(D') = G and D' has the Hamiltonian Is the converse of Theorem 3.42 true? This is a very interesting open question. The following theorem proves the converse of Theorem 3.4 .2 if loops are allowed. Theorem 3.4.4 Fraughnaugh, Lundgren, Maybee, Merz, and Pullman [21]. If G is a graph on n :::>_ 3 vertices satisfying 8E(G) ::; nand i(G) = 0, then G is the competition graph of a Hamiltonian digraph possibly having loops if and only if G has an edge clique covering { C1 Cn} that has a system of distinct representatives. Proof. One direction of the proof follows from Theorem 3.4.2. So suppose that G has such an edge clique covering C. Let the representative for clique c, be labeled ai1 fori = 2, ... ,n and the representative for cl be labeled a,. Define a digraph D by I(ai) = C,. Then G = C(D) and a1a2 ... an is a Hamiltonian cycle in D. 3.4.1 Generalization to pCompetition Graphs: Construetions Can we generalize the results of the previous section topcompetition graphs where p :::>_ 2? First we observe the generalization of Theorem 3.4.2. Theorem 3.4.5 If G is the pcompetition graph of a Hamiltonian digraph, then G has a pedge clique cover with a system of distinct representatives. 66
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Proof. Assume that G is the pcompetition graph of a Hamiltonian digraph D. Let a1 a2 an denote the consecutively labeled vertices of a Hamiltonian cycle in D. Then the family of insets of the vertices are a pedge clique cover of G. Furthermore, for i = 1, 2, ... n 1, choose a; to represent I( a;+1 ) and choose an to represent I( a1 ) and we have a system of distinct representatives The following constructions will be used in the next section to characterize classes of graphs that are the pcompetition graphs of loopless Hamiltonian digraphs. Lemma 3.4.2 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If G is a connected graph such that G is the pcompetition graph of a loopless Hamiltonian digraph, then adding a pendant vertex x to G results in a graph G' that is the pcompetition graph of a loopless Hamiltonian digraph. Proof. Let D be a loopless Hamiltonian digraph such that CP(D) = G. Let v1v2 ... vn denote a Hamiltonian cycle in D. CreateD' from D as follows. Add vertex x. Let v; denote the vertex adjacent toxin G'. Let Iv(x) = Hamiltonian cycle in D'. Since G is connected v; has outgoing arcs to at least p other vertices of D. Let x have an outgoing arc top1 of these vertices. Then Cp(D') = G', where D' is Hamiltonian. Corollary 3.4.8 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If T is a tree and has a subtree that is the pcompetition graph of a loopless Hamiltonian digraph, then T is the pcompetition graph of a loopless Hamiltonian digraph. Proof. This follows from Lemma 3.4.2, since we may add pendant vertices successively to the subtree, obtaining T. 67
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Lemma 3.4.3 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If T is a tree that is the pcompetition graph of a loopless Hamiltonian digraph, then adding a pendant vertex x to an internal vertex results in a tree T' that is the (p + 1 )competition graph of a loopless Hamiltonian digraph. Proof. Let D be a loopless Hamiltonian digraph such that C(D) = T. Let v1v2 Vn denote a Hamiltonian cycle in D. CreateD' from D as follows: Add x to D. Let In(x) = V. Let v; denote the vertex adjacent toxin T'. Let vi and vk denote two vertices adjacent to v; in T and T'. Since {v;,vj} E E(T), we conclude that v; and Vj have arcs to at least p common vertices. Let x have an arc to these vertices. Since { v;, vk} E E(T), we conclude that v; and vk have arcs to at least p common vertices. Furthermore there exists Vm in this set of p vertices such that Vm has no arc from Vj, since T is a tree. Let x have an arc to this vertex. Then no previous competitions have changed since all vertices have arcs to x, while x and v; compete at least p + 1 times. Then v,v, .. vmJXVm ... VnVJ is a Hamiltonian cycle in D'. Therefore Cp(D') = T', where D' is Hamiltonian. A branch of a tree is a path of the tree with the vertex at one end adjacent to an internal vertex. A maximal branch has the further property that the other end vertex is a pendant vertex. Lemma 3.4.4 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If T is a tree that is the pcompetition graph of a loopless Hamiltonian digraph and T' is a tree produced from T by adding a branch of l new vertices where l 2: 2, then T' is the kcompetition graph of a loopless Hamiltonian digraph for p :S: k :S: p + l 1. Proof. The proof is by induction on l. If l = 2, observe that T' is a pcompetition graph of a loopless Hamiltonian digraph by Corollary 3.4.8. To show T' is a (p + 1 )competition graph of a loopless Hamiltonian digraph, add 68
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the first vertex of the branch as indicated in Lemma 3.4 .3 creating a tree that is the (p +I)competition graph of a loopless Hamiltonian digraph. Then by Corollary 3.4.8, T' is the (p +I)competition graph of a loopless Hamiltonian digraph. Assume that the statement is true for the addition of a branch on l < n new vertices and consider the addition of a branch on l = n new vertices. By the induction hypothesis, the addition of the first l 1 vertices produces a tree that is the kcompetition graph of a loopless Hamiltonian digraph for p S: k S: p + l2. Then by Corollary 3.4.8, T' is the kcompetition graph of a loopless Hamiltonian digraph for p :S k :S p + l 2. It remains to be shown that T' is a (p + l 1 )competition graph of a loopless Hamiltonian digraph. Let v1 v2 Vt denote the consecutively labeled vertices of the branch such that v1 is adjacent to an internal vertex, v0 ofT. Let D be a loopless Hamiltonian digraph such that Cp(D) = T. Create D' from D as follows. Direct an arc from all vertices ofT to vi> ... Vtl Observe that this preserves all adjacencies from Cp(T) in Cp+t1 (T'). Since v0 is an internal vertex, there exists vertices t and u adjacent to v0 in T. Let S denote a set of p + 1 vertices, p of which both t and v0 have arcs directed toward in D and 1 of which u and v0 have an arc directed toward, but t does not. Direct an arc from all vertices in the branch to all vertices of s. For i = 1, ... l 1, direct an arc from v; to all vertices vk, where k = 0, 1, ... l., but k is not equal to i or i1. Direct an arc from Vt to all vertices vk, where k = 0, 1, ... l3. Observe that nonconsecutively labeled 69
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vertices of the branch, vi and Vk, compete at most (p + 1) +(I+ 1 4) times in D' In particular, they compete for for (p + 1) vertices of S and all vertices of the branch except Vi, ViI, vk and vkl Consecutively labeled vertices of the branch, Vi and Vi+!> compete at least (p + 1) + (l + 13) times. In particular, they compete for (p + 1) vertices of S and all vertices of the branch except ViJ,Vi and Vi+J Observe that v1 and v0 compete at least (p + 1) + (l 2) times, while for all other vertices Vi of the branch, v; and v0 compete at most (p + 1)(l3) times, since v0 has no arc to vi and Vi has no arc to Vi_2 Now consider an arbitrary vertex v E T other than v0 and a vertex Vi of the branch. Since T is a tree, v can have at most p arcs to vertices of S, while Vi has arcs to l 2 vertices in the set { v 0 ... ViI}. Therefore v and Vi compete at most p + l2 times. Thus Cp+iI(D') = T'. Furthermore if x 1 x2 Xn denotes a Hamiltonian cycle in D and Xi is any vertex of S, then Lemma 3.4.5 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If G1 and G2 are the pcompetition graphs of loopless Hamiltonian digraph for p 2, then G1 U G2 is the pcompetition graph of a loopless Hamiltonian digraph. Proof. Let D1 and D2 be loopless Hamiltonian digraphs such that CP(D1 ) = G1 and CP(D2 ) = G2. Let V1V2 ... Vn1 and x1x2 Xn, denote a Hamiltonian cycle in D1 and D2 respectively. Let Vi be an arbitrary vertex in D1 and Xi and arbitrary vertex in D2 Create digraph D from D1 and D2 as follows. Let Iv(vi) be the inset of x; in D2 ; similarly, let Iv(x;) be the inset of v; in D1 is a Hamiltonian cycle in D. 70
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3.4.2 Utilizing the Constructions Before we can employ the construction of the previous section, we must establish some base graphs as the pcompetition graphs of loopless Hamiltonian digraphs. Lemma 3.4.6 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If G is a cycle on n ;::: p + 3 vertices where p ;::>: 2, then G is the pcompetition graph of a loopless Hamiltonian digraph. Proof. Let v1 v2 Vn denote the consecutively labeled vertices of G. Create digraph D as follows. Let Iv( vi) = { Vi+lmodn, Vi+2modn, .. 'Vi+p+Imodn}. Then Vi and Vi+lmodn compete p times, namely for Vi+2modn, . 'Vi+pmodn, and Vi+v+1modn Consider nonconsecutive vertices v; and vk. There are 4 vertices for which v; and Vk do not compete, namely v;, vk, Vi+lmodn and vk+lmodnThus v; and Vk compete for at most p l vertices. Therefore Cp(D) = G and vlvnVn1 ... v,v1 is a Hamiltonian cycle in D. Lemma 3.4.7 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. The complete graph Kn on n;::: p+2 vertices is the pcompetition graph of a loopless Hamiltonian digraph. Proof. Let V(Kn) = {v1,v2 ... ,vn} be the vertices of Kn. CreateD as follows. Let ln(v;) = V(Kn) {vi} Then Cv(D) = G and that v 1v, ... VnV1 is a Hamiltonian cycle in D. Lemma 3.4.8 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. The complete graph minus one edge on n ;::: 2p + 1 vertices is the pcompetition graph of a loopless Hamiltonian digraph. Proof. Let G be the complete graph minus one edge for n ;::: 2p + 1. Label the vertices of G, vi> v 2 ... Vn such that {vi> vn} is the missing edge. Create 71
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D as follows. For 1::; i < let In(v;) = V(G){v;,v1}. For < i::; n, let In( v;) = V( G) { v;, vn}. Let In( = V( G) { if n is odd, and V( G) { v1} if n is even. Then all pairs compete at least l 1 2: p times except for v1 and Vn which compete at most once and is a Hamiltonian cycle in D. Lemma 3.4.9 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If G is a path on n 2: p + 3 vertices, then G is the pcompetition graph of a loopless Hamiltonian digraph. Proof. We need only verify this result for n = p + 3 by Corollary 3.4.8. Let vi> v2 Vn be the consecutively labeled vertices of G. Create D as follows: fori f p + 1, let In(v;) = V(G){v;,vi+lmodn}i let In(vp+l) = V(G) {vi,Vi+1modn,Vi+2modn} Observe that V1V2 ... VnV1 is a Hamiltonian cycle in D. Then v; and Vi+l compete at least p times, namely for Since v; does not have an arc to v; or Vi+lmodp, the nonconsecutive vertices compete at most p 1 times. That is, Gp(D) = G where Dis Hamiltonian. We would like to generalize this result to all tree. First we will start with a subclass. A caterpillar is a tree such that the removal of all pendant vertices yields a path. This path is called the spine of the caterpillar. Theorem 3.4.6 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If G is a caterpillar on n 2: p + 3 vertices, then G is the pcompetition graph of a loopless Hamiltonian digraph. 72
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Proof. We need only verify this result for n = p + 3 by Corollary 3.4.8. Let v, v2 Vq denote the consecutively labeled vertices of the spine of G. Observe q 2': 3. Since a path on 3 vertices is the competition graph of a loopless Hamiltonian digraph D, if q = 3 create D' by successively adding all but one of the remaining pendant vertices as in Lemma 3.4.3. Then add the final pendant vertex as in Lemma 3.4.2. Then D' is Hamiltonian and Cp(D') = G. If q > 3, then the spine of G is the rcompetition graph of a loopless Hamiltonian digraph D by Lemma 3.4.9, where r = q3. CreateD' by successively adding the remaining pendant vertices to D' as in Lemma 3.4.3. Then D' is Hamiltonian and Cp(D') =G. Theorem 3.4.7 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If Tis a tree on n 2': 2p vertices, where p 2': 2, then Tis the pcompetition graph of a loopless Hamiltonian digraph. Proof. By Corollary 3.4.8 we need only consider the case n = 2p. The case p = 2 can be verified by examination of all possible trees so p > 2. Let q be the length of the longest path P in T. If q 2': p + 3, T is the pcompetition graph of a loopless Hamiltonian digraph by Lemma 3.4.9 and Corollary 3.4.8. If q :0::: 4, Tis a caterpillar and we are done by Theorem 3.4.6. Saving the case q = 5, consider the case that q 2': 6. Then n = 2p 2': 2pq + 6. Thus, if we can show that a tree on 2pq + 6 vertices is the pcompetition graph of a loopless Hamiltonian digraph, we are done by Corollary 3.4.8. Assume that T is such a tree with maximum path P. By Lemma 3.4.9, P is the (q3)competition graph of a loopless Hamiltonian digraph. Construct a sequence of subtrees T0 T1 Tk where T0 is a path, Tk is T and T, is constructed from T,_1 by the addition of a 73
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maximal branch. Let n; be the number of vertices added to T;_1 to get Ti. Theorems 3.4.3 and 3.4.4 guarantee that T; is a (q3 + L:;j=1 k!)competition graph where k; = max{n; 1, 1}. The worst case occurs when each addi tional branch adds two new vertices. In this case, T is a ( q3 + j)competition graph, where j is half the number of vertices added to To. Since there are q vertices in T0 we add 2pq + 6q = 2(p(q3)) vertices to obtain T, and conclude that Tis a p competition graph, where p = (q3) + (p(q3)). If q = 5, T must have a maximal branch with one vertex; otherwise all branches of T are of length 2, i.e., T has an odd number of vertices, a contradiction since n = 2p. Remove this branch. The resulting tree has 2p1 = 2p2q + 6 vertices and is therefore, by the previous case, a (p1 ) competition graph of a loopless Hamiltonian digraph. Using Lemma 3.4.3 we conclude T is the pcompetition graph of a loopless Hamiltonian digraph. Corollary 3.4.9 Langley, Lundgren, McKenna, Merz, and Rasmussen (38]. If G is a forest and all maximal subtrees of G have n 2p vertices, where p 2, then G is the pcompetition graph of a loopless Hamiltonian digraph. 3.4.3 Classes of 2Competition Graphs Using induction on the number of vertices in the graph, we can establish that chordal graph on at least 5 vertices are the 2competition graph of loopless Hamiltonian digraph. An alternate approach for dealing with the base case must be employed in order to generalize this result to pcompetition graphs where p 3. Lemma 3.4.10 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If G is chordal on n 5 vertices and not complete, then G is the 2competition graph of a loopless Hamiltonian digraph D in which every maximal clique of G is contained in at least one inset of D. 74
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Proof. (by induction on n) If n = 5, we get trees from Theorem 3.4.7 and verify the remaining cases by considering all possible graphs (see Figures 3.10 and 3.11). Assume that the statement is true for chordal graphs that are not complete on n = k 2: 5 vertices and let G be such a chordal graph on n = k + 1 vertices. Let x be a simplicial vertex in G. Consider G' = G{x}. Case 1: Suppose that G' is complete. Suppose that there is exactly one vertex in G' that is not adjacent toxin G. Then G is Kn minus one edge and by Lemma 3.4.8, is the 2competition graph of a loopless Hamiltonian digraph. Suppose that there are at least two vertices, x; and Xj of G' that are not adjacent to x in G. Let C be the maximal clique in G containing x. Create D as follows. Let I(x) = V(D){x}. Let I(x;) = I(xi) =C. For all x0 not equal to x,x; nor Xj, let I(xo) = V(D){x0,x}. Then G is the 2competition graph of D. Since the digraph D { x;, xi} has all possible arcs, x has an arc to x;. Furthermore x; has an arc to some vertex xo of V(G){x,x;,xj} and x0 has an arc to Xj Finally since Xj has an arc to some vertex Xk of V(G){x,x;,xj,xo}, we conclude that Dis Hamiltonian. Furthermore C I(x;) and V(G') I(x), so every maximal clique is contained in at least one inset of D. Case 2: Suppose that G' is not complete. By the induction hypothesis, G' is the 2competition graph of a loopless Hamiltonian digraph D' such that every maximal clique of G' is contained in at least one inset of D'. Let x1x2 .. Xnl denote a Hamiltonian cycle of D'. Let C be the maximal clique containing x in G. Let C' = Cn V(G'). By the inductive hypothesis there is a vertex Xj such that C' is contained in I( xi) in D'. Create digraph D as follows. 75
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() (<) 11 (d) f4 () \7 (f) (g) v (b) Vl (;) \X (j) \21 '(l)w' Figure 3.10. Chordal graphs on 5 vertices. 76
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lml ffJ 1 fJ:J 1 \jl f 1,1 w (q) / (r) w ( s) ( t) /. (u) C7 1>1 c. (w) Figure 3.11. Chordal graphs on 5 vertices. 77
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Add arc (x,x;) to A(D'). Observe that since D' is loopless, {x;,x} fj E(G). Suppose there is a vertex x; not equal to x; that is not adjacent to x in G. Then let Iv(x) = lv(x;) and lv(x;) =C. Observe that x competes with the other vertices of C in D at x; and x;, while x competes at most once with any other vertex in D, namely at x;. Since no other competitions have changed, C(D) = G and x1 ... x;_1xx; ... Xn1 is a Hamiltonian cycle in D. Suppose that x; is the only vertex that is adjacent toxin G. Then V(G'){x;} is a clique and V(G){x;} is a clique. Thus x; is simplicial in G and since Gx; is complete, we are in Case 1. Theorem 3.4.8 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If G is a chordal graph on n 2: 5 vertices, then G is the 2competition graph of a loopless Hamiltonian digraph. Corollary 3.4.10 Langley, Lundgren, McKenna, Merz, and Rasmussen [38]. If G is interval on n 2: 5 vertices, then G is the 2competition graph of a loopless Hamiltonian digraph. 3.5 Competition Inverses Given a graph G, can we characterize the class of digraphs 'D( G), where a digraph D E 'D(G) if and only if C(D) = G? If a digraph D is in 'D(G) we say D is a competition inverse of G. In this section we begin by constructing a canonical representative D for 'D( G) where D is strongly connected or Hamiltonian and G is chordal or interval. This work extends canonical constructions by Roberts [64] in connection to the food web problem. Roberts gave a canonical construction for D E 'D( G) where D is acyclic and G is chordal (and hence interval). We give a canonical construction forD E 'D( G) where D is acyclic and G is interval, but where G merely chordal, D fj 'D( G). 78
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3.5.1 Acyclic Competition Inverses Since an acyclic digraph must contain a vertex that has no outgoing arcs, the competition graph of an acyclic digraph must contain an isolated vertex. Thus not every graph has an acyclic inverse. The problem of characterizing graphs that are the competition graph of an acyclic digraph has been extensively studied. In this section, we cite several characterizations and give canonical constructions for an acyclic competition inverse based upon these characterizations. Roberts [64] proved the following proposition that we state without proof. Proposition 3.5.1 If G is a chordal graph, then G has an acyclic competition inverse if and only if G has an isolated vertex. We say an ordering v1 v2 . Vn of the vertices of an acyclic digraph is topological if and only if a vertex Vi has an arc to a vertex Vi implies i ::; k. Roberts gave a construction for an acyclic competition inverse for a chordal graph similar to the one that follows. Algorithm 3.5.1 Given a chordal graph G with at least one isolated vertex, we construct an acyclic digraph D such that C(D) =G. (1) Find a perfect elimination ordering v1,v2 ,vn of G so that the isolated vertices of G are eliminated first. (2) Create the acyclic digraph D on the same vertex set as G such that As a corollary to Roberts' result for chordal graphs, we have the following proposition. Proposition 3.5.2 Roberts [64]. If G is an interval graph, then G has an acyclic competition inverse if and only if G has an isolated vertex. 79
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The following algorithm gives a canonical representative D E 1J( G) where D is acyclic and G is an interval graph with at least one isolated vertex. Algorithm 3.5.2 Given an interval graph with at least one isolated vertex, we construct an acyclic digraph D such that G(D) =G. (1) Consecutively rank the maximal cliques in Gas 01 .. G, so the first i( G) cliques are the isolated vertices of G. (2) Let v; be a vertex in G;Gi+1 fori = 1, 2, r1. Such a vertex must exist because the cliques are maximal and distinct. (3) Let D be the acyclic digraph on the same vertices of G such that In(v;) = Gi+l Then G(D) =G. 3.5.2 Strongly Connected Competition Inverses Using the characterization in Chapter 3.3 of graph that are the competition graphs of strongly connected digraphs we can construct a canonical representative D E 1J( G) where D is strongly connected. Lemma 3.5.1 If G is a chordal graph, then ElE(G) :S IV(G)Ii(G). Proof. (by induction on IV( G) I) The statement is clearly true for chordal graphs on 1,2, and 3 vertices. Assume that the statement is true for chordal graphs G such that IV( G) I < k. Let G be a chordal graph such that IV( G) I = k. Every chordal graph has a simplicial vertex so let v be an arbitrary simplicial vertex in G. Consider G' = G{ v }. By the induction hypothesis ElE( G') :S ni(G)1. Since vis simplicial vis contained in at most one maximal clique. Therefore ElE(G) :S jV(G)Ii(G). Corollary 3.5.1 If G is a chordal graph, then G has a strongly connected competition inverse. 80
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Proof. This follows from Lemma 3.5.1 and Corollary 3.3.1. Corollary 3.5.2 If G is an interval graph, then G has a strongly connected competition inverse. In order to devise an algorithm, we must employ techniques to deal with isolated vertices. Proposition 3.5.3 Gyarafas [26]. If G is a graph with isolated vertex v, then EJE(G) = EJE(Gv). Proof. (by induction on n =IV( G) I) The statement is clearly true for chordal graphs on 1, 2, and 3 vertices. Assume that the statement is true for chordal graphs G such that n < k. Let G be a chordal graph such that n = k. Every chordal graph has a simplicial vertex so let v be an arbitrary simplicial vertex in G. Consider G' = G{v }. By the induction hypothesis EJE(G') :'::: ni(G) 1. Since v is simplicial v is contained in at most one maximal clique. Therefore EJE(G) :'::: ni(G). Let G denote a graph G with isolated vertices removed. Observe that if a graph G satisfies the conditions of Corollary 3.3.1, then G does as well. This will be crucial to our algorithm. Algorithm 3.5.3 Given a graph G such that EJE(G) :S IV(G)Ii(G), we find a strongly connected digraph D such that C(D) =G. (1) Let G = G without isolated vertices. (2) Find an edge clique covering of G of size :S IV(G)Ii(G). (3) Find a system of distinct representatives for the complements of the cliques in this edge covering. Such a system of distinct representatives exists by a result of Roberts and Steif [66]. 81
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(4) CreateD by adding an arc from each vertex in a clique in the edge covering to its representative in the system of distinct representatives found above. Observe that C(D) =G. (5) Find the strong components in D using depthfirst search. Let k denote the number of strong components. (6) If k = 1 go to step 11. (7) Topologically order the strong components of D = {D1 D2, ... Dk}. (8) If V(D!) = {x} then choose an arbitrary vertex y E V(Dk) Define a new digraph D' by ln(x) = In(y), ln(y) = [x], and all other insets in D' are the same as in D. By the proof of Corollary 3.3 .1 in Case 1, D1 U Dk is strongly connected. We have decreased the number of strong components in the digraph without changing the competition graph. Go to step 5. (9) If IV( D1 )I 2': 2 and there is a vertex y E V( Dk) such that ( x, y) t/ A(D) for some x E V(D!) then define a new digraph D' by ln(y) = In(x), In( x) = In(Y) with all other arcs in D' staying the same as in D. By the proof of Corollary 3.3.1 in Case 2, D1 U Dk is strongly connected. We have decreased the number of strong components in the digraph without changing the competition graph. Go to step 5. (10) If IV(D1)1 2': 2 and every vertex in D1 has an arc to every vertex in Dk then let x E V(D!) and let y E V(Dk) Create a new digraph D' by letting ln(Y) = [x], ln(x) = In(Y) and all other arcs in D' staying the same as in D. By the proof of Corollary 3.3.1 in Case 3, D1 U Dk is strongly connected. We have decreased the number of 82
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strong components in the digraph without changing the competition graph. Go to step 5. (11) If G f G, then let a 1,a2 ... ,a;( G) be the isolated vertices in G. Create a path through the isolated vertices a 1 a 2 a 3 ... ai(G)lai(G) and let x be an arbitrary vertex in V( G). Create a new digraph D' by adding a1 a 2 .. ai(G) to the vertex set, adding the arcs of the path through these vertices, letting lD(a!) = ID(x), ID(x) = {ai(G)} and all other arcs and vertices of D' the same as in D. Then C (D) = G implies that C( D') = G and D' is strongly connected. 3.5.3 Hamiltonian Competition Inverses We now present algorithms to construct canonical representatives for Hamiltonian competition inverses of certain classes of graphs. The following algorithm constructs a canonical representative D E 7J( G) where D is Hamiltonian and G is chordal. This construction is based upon the existence of a perfect elimination ordering for G. Algorithm 3.5.4 Given a chordal graph G on n :::: 3 vertices, we construct a Hamiltonian digraph D such that C(D) = G. (1) If G is not complete then G has two nonadjacent simplicial vertices (Dirac [17]). Label one v1 and the other Vn. (2) Fori= 2,3, ... ,n 1, each G; is chordal and has a simplicial vertex v; different from Vn Then the sequence v1 v2, ... Vn is a perfect elimination ordering. Fori= 1,2, ... ,n, let C; = Na;[v;]. Observe that Vi tJ. ci+l since Vi is not a vertex of Gi+l) Vn tJ. cl since VI and Vn were chosen as nonadjacent and ... Cn} is an edge clique cover of G. 83
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(3) Construct D by letting I(v;) = C;+1 and I(vn) = C1 Then D is a loopless Hamiltonian digraph with Hamiltonian path and C(D) =G. The following algorithm constructs a canonical representative D in 'D(G) where Dis Hamiltonian and G is interval. This.construction employs the consecutive ranking of the maximal cliques of an interval graph. Observe that while the digraph produced by Algorithm 3.5.4 is a member of 'D(G) where G is interval, the digraph produced by Algorithm 3.5.5 is not necessarily a member of 'D( G) where G is chordal. Algorithm 3.5.5 Given an interval graph G on n 2: 3 vertices, we construct a Hamiltonian digraph D such that C(D) = G. (1) Find the consecutively ranked maximal cliques C2 Cr of G. Observer :':: n. (2) For each C;, where i = 1,2, ... ,r1, choose a vertex in C;Ci+1 and label it v;. Such a vertex must exists because the cliques are maximal and distinct. Observe that v; <1Ck, where k 2: i. (3) For Cr, choose a vertex in CrCr! and label it Vn ( 4) If r < n, for i = r + 1, r + 2, ... n1 choose an unlabeled vertex, label it v;, and let C; = { v;}. (5) Construct D by letting I(v;) = C;+l, I(vn) = C1 Observe that Dis a loopless Hamiltonian digraph with Hamiltonian cycle 84
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and C(D) == G. 3.5.4 Bounds on the Number of Arcs in a Competition Inverse The canonical construction of the previous sections may provide useful in developing algorithms to either maximize or minimize the number of arcs in a digraph without changing the competition graph. We now present some elementary bounds that may also be useful. The proof of the first bound is left to the reader. Theorem 3.5.1 If G is a graph with competition inverse D and C1 C, is an edge clique cover of minimum cardinality, then r /A( L IC. i=l Theorem 3.5.2 If G is a graph with competition graph D and { ... C,} is an edge clique cover of minimum cardinality such that each C; is chosen to be maximal, then r /A( D)/ :S L /Cd +(IV( G)/r)w(G). i=l Proof. There must be at least L: /Cd arcs in D to account for the /C;/ competitions in each clique C; of the edge clique cover. Thus there will be vertices v1 v2 v, with insets C1 C2 C, respectively. There are /V( G)/r vertices remaining in the graph. The size of the largest clique in the minimum edge clique cover is at most w( G). Thus we can add at most w( G) incoming arcs at each of the IV( G)/'T' remaining vertices. The graph in Figure 3.12 illustrates that this bound is tight. Figure 3.13 illustrates that it cannot be achieved for all graphs. Observe that v2 85
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Let ID(v,) = {v,,v3, vr}, ID(v,) = {v!,VJ,Vs,vr}, ID(v3) = {v,, v4}, "' ID(v) = {v,,vJ,Vs,vr}, ID(vs) = {v,,v6}, ID(v6) = {v,,v3,vs,vr}, ID(vr) = {v,,v,,vs} and "' lD(vs) = {v,,v3,v,,v7}. Figure 3.12. A graph and its competition inverse. For this graph G we can construct a competition inverse D such that IA(D)I =I: IC;I + (IV(G)Ik)w(G) = 26. in this graph cannot have more than two incoming arcs li1 any competition inverse. Theorem 3.5.3 If G !S a graph with competition inverse D and V( G) { V1, ... Vn}, then n IA(D)I :Ew(G{vi}). i=l Proof. Recall that Dis a loopless digraph. Suppose that a vertex v E V(D) has more than w( G { v}) incoming arcs in D. Then either there is a loop at v in D or there is a clique in G { v} of size greater than w( G { v} ), both contradictions. Thus IAI Z:w(G{vi}). Figure 3.13 illustrates that this bound is tight. It cannot be achieved for the graph in Figure 3.12 because there is no vertex Vi for which w( G { v;}) = 2, but in any competition inverse some vertex must have inset { v2 v4 } since this is a maximal clique. 3.6 The Chromatic Numbers of Competition Graphs One of the primary parameters of the competition graph of a digraph that is of interest in problems motivated by communication networks is the 86
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"' Figure 3.13. A graph and its competition inverse. For this graph G we can construct a competition inverse D such that IA(D)I = I;w(G{vi})= 14. 87
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chromatic number. Recall that the chromatic number of a graph is the smallest integer x( G) such that G has a proper coloring with x( G) colors. A proper coloring of the competition graph of a digraph corresponds to a channel assignment in the radio network. In this section we explore bounds of and in some cases the exact value of the chromatic number of the competition graph of a digraph. 3.6.1 Symmetric Digraphs Since G2 = G u S2 ( G), begin by making the observation that t!.(G)::; x(S,(G)) :S x(G'). It is left to the reader to verify the following lemma which yields a preliminary bound for the chromatic number of a symmetric digraph (possibly having some loops). Lemma 3.6.1 Lundgren, Merz, and Rasmussen [50]. If D is a symmetric digraph with underlying graph (loops removed) G, then S2(G) c:;; C(D) c:;; G2 Corollary 3.6.1 Lundgren, Merz, and Rasmussen [50]. If D is a symmetric digraph with underlying graph (loops removed) G, then x(S2(G)) ::; x(C(D))::; x(G'). This bound is not particularly good since given any integer n, we may construct a symmetric digraph D such that For example, let G be the bipartite graph Kn,n Then G2 is K2n and S,(G) is a graph with two connected components, each of which is Kn Recall that w(G) denotes the size of the largest clique in a graph G. A graph is perfect if and only if w( G') = x( G') for all induced subgraphs G' 88
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of G. This definition gives yields the following theorem. Theorem 3.6.1 Lundgren, Merz, and Rasmussen [50]. If D is a symmetric digraph with underlying graph (loops removed) G and both G2 and S2 ( G) are perfect, then w(S,(G)) :<; x(C(D)) :<; w(G2). What conditions are sufficient and necessary for the twostep graph and the square of the underlying graph of a symmetric digraph (loops removed} to be perfect? We can answer this question for certain classes of perfect graphs. For example, in Chapters 3.1 and 3.2 we characterize symmetric digraphs with interval (and hence perfect) squares and twostep graphs. Which symmetric digraphs have chordal squares and twostep graphs? The following necessary condition partially answer this question. Proposition 3.6.1 Lundgren and Rasmussen, 1993. If Tis a tree, then S2(T) is chordal, and hence perfect. Proposition 3.6.2 Chang and Nemhauser [12]. If T is a tree, then T2 is chordal, and hence perfect. These propositions combined with Theorem 3.6.1 yield the following corollary. Corollary 3.6.2 Lundgren, Merz, and Rasmussen [50]. If Dis a symmetric digraph such that the underlying graph (loops removed) of Dis a tree T, then w(S2(T)) :<; x(C(D)) :<; w(T2). We can make this bound even more precise with the following established results. Proposition 3.6.3 Lundgren, Maybee, and Rasmussen [45]. If T is a tree, then w(S2(T)) = 6.(T). 89
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Proposition 3.6.4 Raychaudhuri and Roberts [60]. If T Js a tree, then w(T2 ) = 1 + .6.(T). Corollary 3.6.3 Lundgren, Merz, and Rasmussen [50]. If D is a symmetric digraph such that the underlying graph of D with loops removed is a tree T, then .6.(T) x(C(D)) 1 + .6.(T). We have already seen an example of a chordal graph with a twostep graph that is not chordal in Figure 3.4. Figure 3.1 illustrates a chordal graph with a square that is not chordal. Rather than consider necessary and sufficient conditions for the twostep and square of certain classes of perfect graphs (such as chordal and interval) to be perfect, we will consider other approaches to bounding the chromatic number of the competition graph of a digraph. 3.6.2 Arbitrary Digraphs and Tournaments While the inset of a particular vertex in a digraph is a clique in the competition graph, it is not necessarily maximal. However, certainly w( C(D)) 2 .6. (D), so it is certainly clear that .6.(D) x(C(D)) n. Are there any classes of digraphs such that .6_(D) equals x(C(D))? We now answer this question. A digraph is an interval digraph if and only if two intervals S( x) and T( x) on the real line can be assigned to vertex x such that (x,y) E A(D) if and only if S(x) n T(y) f 0. For a camprehensive introduction to interval digraphs see Sen, Das, Roy, and West [70]. To prove results about the competition graph of an interval digraph, we need the following lemma. 90
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Lemma 3.6.2 Langley, Lundgren, and Merz (39]. If Dis an interval digraph and C is a maximal clique of C(D) then there is some x E V(D) such that I(x) =C. Proof. Consider the vertices of C. Let x be a vertex such that I(x) n C is not properly contained in J(v) n C for any other vertex v. Suppose that I(x) n C of C. Then there exists a vertex a E C such that a rf_ I(x). Since I(x) is not properly contained in the inset of any other vertex there must exist b,c E I(x) n C and y,z such that (a,y),(c,y),(a,z),(b,z) E A(D) but (b,y),(c,z) rf_ A(D). Since Dis an interval digraph we have intervals on the real line S(a), S(b), S(c) and T(x), T(y), T(z) such that the intersection graph G of these six intervals contains graph H in Figure 3.14 as a subgraph. c y a b z Figure 3.14. The subgraph of an interval graph. Observe that G of H since H is not interval. Further observe that {x,a},{y,b}, and {z,c} rf_ E(G). Without loss of generality assume that { x, y} E E( G). Then xyazbx is a five cycle. Since G is interval this five cycle must have a chord. Suppose that {b, a} E E(G). Then G has the chordless four cycle xyabx. So {b,a} rf_ E(G) and therefore {y,z} or {x,z} must be in E(G). If 91
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only one of these edges isinG, G has a chordless four cycle, so both {y,z} and { x, z} E E( G). Suppose that { c, b} E E( G). Then cbzy is a chordless four cycle, so { c, b} fj E( G). Suppose that {a, c} E E( G). Then cazx is a chordless four cycle, so {a, c} fj E( G). Thus E( G) is completely determined and G is shown in Figure 3.15. c b z a Figure 3.15. This graph is not an interval graph. Since the vertices a, b and c form an asteroidal triple, we have a contradiction, i.e., no such z can exist. Theorem 3.6.2 Langley, Lundgren, and Merz [39]. If D 1s an interval di graph, then C(D) is an interval graph. Proof. By Lemma 3.6.2, every maximal clique of C(D) is the inset of some vertex x. Choose a unique such vertex for each maximal clique of C(D). Consider the set of terminal intervals for these vertices. If T(x) c;; T(y) then I(x) c;; I(y). Thus no interval in this set may be properly contained in another interval and we may order the maximal cliques C1 C2 CP, where C; :": C; implies the left endpoint of the interval corresponding to C; is strictly less that the left endpoint for C;. We claim this ranking is consecutive. Let Xi denote the vertex chosen such that Inset( Xi) = C;. If x E C; then S( x) n T( xi) =J 0. Suppose that x E C,, x E Ck and i < j < k. Then 92
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S(x)nT(x;) rf 0 and S(x)nT(xk) rf 0. Ifx is not inc, then S(x)nT(x;) is to the left of T(x,). But S(x) n T(xk) is to the right of T(x,) and since S(x) is an interval, this is impossible. Thus C1 C2 . Cn is a consecutive ranking of the maximal cliques of C(D), i.e., C(D) is an interval graph. Theorem 3.6.3 Langley, Lundgren, and Merz [39]. If D is an interval digraph, then x(C(D)). Proof. Since C(D) is interval it is perfect and so ,x(C(D)) is the size of the largest clique. By Lemma 3.6.2 we conclude that x(C(D)) Can we determine the chromatic number of the competition graphs of any other arbitrary digraphs? Recall from Theorem 2.1.1 that the competition graph of a digraph D is chordal if and only if D has a disimplicial elimination ordering. Lemma 3.6.3 Lundgren, Merz, and Rasmussen [50]. If D has a disimplicial elimination ordering v1 v2 Vn and X;= {vi} U {vkJk > i and O(v;) n O(vk) rf 0} and Cis a maximal clique in C(D), then C =X; where i = min {kJvk E C}. Proof. Suppose that z = Vk E C. If k = i, then Vk E X;. So assume that k rf i. By choice of v;, it follows that k > i. Since z, v; E C, we conclude that O(z) n O(v;) rf 0, but then z E X;. Consequently, C <::;: X;. Since D has a disimplicial elimination ordering, it follows that X; induces a clique in G. If X; rf C, then X; must properly contain C, but this is impossible since Cis maximal. Theorem 3.6.4 Lundgren, Merz, and Rasmussen [50]. If Dis a digraph with a disimplicial elimination ordering v1 v2 Vn and X; is defined as in the preceding lemma, then x(C(D)) = m11x JX;J. ls=;ts=;n 93
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Proof. Since D has a disimplicial elimination ordering, by the preceding result, G = C(D) is chordal. Since chordal graphs are perfect, x(G) = w(G) and the result follows from the previous lemma. A circulant tournament is a tournament that can be labeled such that row i of the adjacency matrix has 1's in columns i + k mod n where k = 1, ... ln;'1 and O's everywhere else, unless n is even, in which case the only exception is that the matrix has a zero in the [n/2 + 1, 1]position. Observe that if n is odd the circulant ntournament is regular. Lemma 3.6.4 Lundgren, Merz, and Rasmussen [50]. If Hn is the circulant tournament of order n, then a(C(Hn)) = 2. Proof. A simple calculation shows that rows n and of the adjacency matrix of Hn are orthogonal, so vertices n and have no common neighbors, i.e., these vertices are nonadjacent in C(Hn)). Thus a(C(Hn)) 2: 2. Since each row of the adjacency matrix of Hn has at least l n;l J ones, then if we choose any three rows i,j, and k, there must exist some column p such that at least two of rows i,j, and k have ones in column p. Thus a(C(Hn)) :<:; 2, and the result follows. Theorem 3.6.5 Lundgren, Merz, and Rasmussen [50]. If Hn is the circulant tournament of order n, then x(C(Hn)) = m. Proof. We first show that C(Hn) can be properly colored using colors. To each vertex i, where i = 1,2, ... ,n, assign color i (mod Since the adjacency matrix of Hn has the consecutive ones property for columns, and since each column sum is either l n;l J or ln;'l, then for any i # j we have 94
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that if both a;,k = 1 and aj,k = 1 then i j (mod fil ), so vertices i and j receive distinct colors. Thus x(C(Hn)) fil If fewer than fil colors are used, then at least one color must be used on at least three vertices, but this is impossible since, by the preceding lemma, each color class can have at most two members. Thus x(C(Hn)) fil, and the result follows. The reader is referred to the next chapter for more on the chromatic number of the competition graphs, specifically, the co,mpetition graph of tournaments. 95
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CHAPTER 4 THE COMPETITION GRAPHS OF TOURNAMENTS An ndigraph is an oriented graph on n vertices. In an ndigraph, we say vertex x beats vertex y if and only if there is an arc from x toy. An ntournament is an oriented complete graph. See Moon [55] and Reid and Beineke [61] for more about tournaments. Two vertices x and y dominate an ndigraph D if x and y beat all other vertices i.e., {x,y} U O(x) U O(y) = V(D). Two such vertices are called a dominant pair. The domination graph of D, denoted Dom(D), is a graph on V(D) with {x,y} E E(Dom(D) if and only if x andy dominate D. Figure 4.1 shows the domination graph of a tournament. For example, {1,8} is an edge in the domination graph because 0(1) = {2,3,5, 7} and 0(8) = {1,4,5,6}, but {4,5} is not an edge because 6 0(4) U 0(5). 8 Dom(T) = 5 4 7 6 8 5 1 4 Figure 4.1. A tournament and its domination graph. 2 The proof of the following proposition is left to the reader. This 96
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observation summarizes the relationship between the domination and competition graphs of a tournament. Proposition 4.0.5 If Tis a tournament, then Dom(T) is the complement of the competition graph of the reversal of T. Thus results pertaining to the domination graphs of tournaments pertain to the competition graphs of tournaments. Since tournaments are arc dense it is not surprising that the competition graphs of tournaments are edge dense. This means that the domination graphs of tournaments are much easier to work with than the competition graphs because they generally have fewer edges. 4.1 Necessary Conditions on the Domination Graphs of Tournaments Which graphs can be the domination graph of a tournament? We begin to answer this question by finding graphs that are not subgraphs of Dom(T) for all tournaments T. Proposition 4.1.1 Fisher, Lundgren, Merz, and Reid [20]. If Tis a tournament and z E V(T), then O(z) is an independent set of Dom(T). Proof. Let x,y E O(z). Then z tj O(x) U O(y) U {x,y}. That is, x andy do not dominate T. Therefore {x,y} E E(Dom(T)). Corollary 4.1.1 Fisher, Lundgren, Merz, and Reid [20]. IfT is a tournament, then .6. +(T) ::; a(Dom(T)). Let Ck denote the undirected cycle on k vertices. Lemma 4.1.1 Fisher, Lundgren, Merz, and Reid [20]. If k 2: 4 is an even number and T is a ktournament, then Ck is not a subgraph of Dom(T). 97
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Proof. Suppose that Ck is a subgraph of Dom(T). Let 1, 2, ... k be the con secutively labeled vertices of Ck. Since k is even, ,6. +(T) 2': Corollary 4.1.1 shows Therefore ,6. +(T) = The only two possible independent sets of order in Dom(T) are the two maximal independent sets of Ck given by A = {1,3, ... ,k1} and B = {2,4, ... ,k}. Say a+(1) by Proposi tion 4.1.1, 0(1) =B. Then for any i E B, we have O(i) of. A because 1 beats i, and O(i) of. B because i E B. For any i E A{1}, we have O(i) f A because i E A, and 0( i) of. B because i beats 1. Thus a+ ( i) < for all i f 1. So by symmetry, at most one vertex of T can have outdegree Thus the sum of the outdegrees of the vertices ofT is at most + ( k 1) ( 1). For k 2': 3, this is less than the k(k;l) arcs in T, a contradiction, completing the proof. Let k 2': 3 be an odd integer. Let S be a ( k;l )set contained in zk (the integers mod k) where 0 rf s and SJ + 82 lc 0 for all SJ, 82 E S. For such sets, we can form a regular tournament T(S) called the rotational tournament with symbol S whose vertices are labeled by the elements of Zk and with arcs (i,j) ifji := s where s E S. Let Uk = T({1,3, ... ,k2}). Figure 4.2(a) shows U7 Figure 4.2(b) shows T({l,2,4}). The regular 7tournament in Figure 4.2( c) is not rotational as 0(0) and 0( 4) induce distinct 3tournaments. Lemma 4.1.2 Fisher, Lundgren, Merz, and Reid [20]. If k is odd and Tis a ktournament, then Ck is a subgraph of Dom(T) if and only if Tis isomorphic 98
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0 0 6 5 2 4 3 4 3 0 0 6. ol 5 2 5. 02 4 3 4 3 0 0 6 1 5 2 4 3 4 3 Figure 4.2. Regular tournaments on 7 vertices and their dom ination graphs. 99
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Proof. ( <=) Since the only dominating pairs in Uk are i and j where j i = 1 or k 1, Dom(Uk) is the kcycle with vertices labeled consecutively by 0,1,2, ... ,k 1. ( =>) Assume that Gk is a sub graph of Dom(T). Consecutively label the vertices of Gk by 0, 1, 2, ... k1. Corollary 4.1.1 shows k1 t.+(T) :S a(Dom(T)) :S a(Gk) = 2. Since the average outdegree in a ktournament is k;t, we have that d+(i) = k;t for all i. Without loss of generality, assume that 0 beats 1. The only independent set of Gk of order k;t without either 0 or 1 is {2, 4, ... k 1 }, so 0(1) is this set. In particular, 1 beats 2. The only independent set of Gk of order k;t without either 1 or 2 is {3, 5, ... k}, so 0(2) is this set. By similar arguments we can show that fori= 3,4, ... ,k, O(i) = {kjk i s,s E {1,3, ... ,k 2}}, completing the proof. Tg if a b c d Figure 4.3. This graph is not the induced subgraph of the domination graph of a tournament. As defined in Section 3, a caterpillar is a tree such that the removal of all pendant vertices yields a path. Figure 4.3 shows the smallest tree that is 100
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not a caterpillar. This tree will be called NC7 (noncaterpillar on 7 vertices). It is left to the reader to verify that any tree that is not a caterpillar must contain a copy of NC7. Lemma 4.1.3 Fisher, Lundgren, Merz, and Reid [20]. If Tis a ?tournament, then NC7 is not a subgraph of Dom(T). Proof. Suppose that NC7 is a subgraph of Dom(T) with the vertices of T labeled as in Figure 4.3. Corollary 4.1.1 shows t:,.+(T) o:(Dom(T)) o:(NC7) = 4. Since the average outdegree in a tournament with 7 vertices is 3, either Tis regular or T has a vertex with outdegree 4. Figure 1.7 shows the domination graphs of the only three nonisomorphic regular ?tournaments. None of these graphs has NC7 as a subgraph. Thus, T has a vertex with out degree 4. Since {a,c, e,g} is the only 4 vertex independent set in NC7, by Proposition 4.1.1 only b, d, or f can have outdegree 4 in T. By symmetry, without loss of generality assume that b has outdegree 4. Then b beats a, c, e and g, and loses to d and f. Since b and c dominate T, it follows that c beats d and f. Without loss of generality, assume that d beats f. Since f and g dominate T and c and d both beat f, we see that g beats c and d. But then c and d do not dominate T because neither beats g, a contradiction. Thus NC7 is not a subgraph of the domination graph of a ?tournament. Lemma 4.1.4 Fisher, Lundgren, Merz, and Reid [20]. If S is an induced subdigraph of a digraph D, then the induced subgraph of Dom(D) on the vertices of Sis a subgraph of Dom(S). Proof. Letx,y E Swhere{:z:,y}isanedgeinDom(D). ThenOn(x)UOn(y)U {x,y} = V(D). As V(S) <;;; V(D), we have Os(x) U Os(y) U {x,y} = V(S). Thus {:z:,y} E E(Dom(S)). 101
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A spiked cycle is a connected graph such that the removal of all pendant vertices yields a cycle. Theorem 4.1.1 Fisher, Lundgren, Merz, and Reid [20]. IfT is anntournament, then Dom(T) is either a spiked odd cycle, with or without isolated vertices, or a forest of caterpillars. Proof. Lemmas 4.1.1 and 4.1.4 show that Dom(T) has no even cycles. First assume that Dom(T) has a kcycle C where k is odd. Lemmas 4.1.2 and 4.1.4 show that the subtournament ofT on the vertices of Cis Uk. By Lemma 4.1.4, the induced subgraph of Dom(T) on Cis a subgraph of Dom(Uk) = Ck. So C = Ck is an induced subgraph of Dom(T). By Proposition 4.1.1, if :vis not on C, then 0( :v) n V( C) is an independent set. Since the independent sets in a kcycle have at most vertices, two vertices not in C cannot beat all k vertices in C. So the subgraph induced on the vertices that are not on C has no edges. If some vertex :v that is not on C is adjacent in Dom(T) to at least two vertices on C, say y and z, then edges {:v,y} and {:v,z} together with one of the two path connecting y and z form an even cycle in Dom(T), a contradiction. Thus Dom(T) is a spiked odd cycle possibly with isolated vertices. Otherwise, assume that Dom(T) is cyclefree. Then by Lemmas 4.1.3 and 4.1.4, each component must be a caterpillar. So, Dom(T) is a forest of caterpillars. From Theorem 4.1.1, we deduce results about various graph parameters for domination and competition graphs of tournaments. The bounds in Corollaries 4.1.2 to 4.1.7 are achieved for all allowed values of n. 102
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Corollary 4.1.2 Fisher, Lundgren, Merz, and Reid [20]. If n 2 2, then the maximum possible number of edges in the domination graph of an n tournament is n. Corollary 4.1.3 Fisher, Lundgren, Merz, and Reid [20]. If n 2: 2, then the minimum possible number of edges in the competition graph of an n tournament is (;) n. Corollary 4.1.4 Fisher, Lundgren, Merz, and Reid [20]. If n 2 3 and T is an ntournament, then the clique number and chromatic number of the domination graph ofT are at most 3. Corollary 4.1.5 Fisher, Lundgren, Merz, and Reid [20]. If n 2 3 and Tis anntournament, then the independence number and the clique cover number of the domination graph ofT are at least ln/2J. Since the independence number of a graph equals the clique number of its complement and the clique cover number of a graph equals the chromatic number of its complement, we have the following corollaries. Corollary 4.1.6 Fisher, Lundgren, Merz, and Reid [20]. If n 2 3 and Tis anntournament, then the independence number and the clique cover number of the competition graph ofT are at most 3. Corollary 4.1.7 Fisher, Lundgren, Merz, and Reid [20]. If n 2 3 and Tis an ntournament, then the clique number and the chromatic number of the competition graph ofT are at least l n/2 J. 4.2 Sufficient Conditions on the Domination Graphs of Tournaments Which graphs are the domination graph of a tournament? We begin with spiked cycles. Theorem 4.2.1 Fisher, Lundgren, Merz, and Reid [20]. If G is a spiked odd cycle with possible isolated vertices, then G is the domination graph of a tournament. 103
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Proof. Let G be such a graph on n vertices with a cycle, C, of length k. Consecutively label the vertices of Cas {0, 1, ... k1}. Let N; be the set of vertices pendant to i, where i E {0, 1, ... k1}. Let J be the set of isolated vertices. We then construct a tournament T with Dom(T) = G as follows. Create arcs between vertices in V( C) so that the induced subgraph on V( C) is Uk. Let i beat all vertices in N;. Let all vertices in N; beat all vertices in V( C) that dominate i and let all vertices in V( C) that are dominated by i beat all vertices in N;. For all i,j E {0, 1, ... k1} with i =f j, if i beats j, then let each vertex in N; beat all vertices in Ni. Let each vertex not in J beat all vertices in J. Pairs of vertices in N; and pairs of vertices in J are joined by arcs in an arbitrary manner. Then Dom(T) = G. I 3 I li 0 I 3 I 9 / I li 0 0 Figure 4.4. A tournament and its domination digraph. Given a tournament T, the domination digraph ofT, D(T) is a digraph on V(T) with (x,y) E A(D(T)) if and only if x andy dominate T and x beats yin T. Thus D(T) is the orientation of Dom(T) induced by T. Figure 4.4 shows the domination digraph of the tournament in Figure 4.1. Theorem 4.2.1 shows that vertices in D(T) can be beaten at most 104
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once. Further, it shows that a vertex can beat only one other vertex that has beaten other vertices. Lemma 4.2.1 Fisher, Lundgren, Merz, and Reid [19]. If Tis a tournament, then in 1J(T), a vertex can have at most one inneighbor and at most one of its outneighbors can have outneighbors. Proof. First suppose that vertices x and z both beat y in 1J(T). Without loss of generality, assume that x beats z in T. So x beats both y and z in T, and y and z do not dominate T, a contradiction. Thus a vertex has at most one inneighbor in 7J(T). Suppose that in D(T) that vertex x beats both w and y where w beats v andy beats z. Without loss of generality, assume that w beats yin T. Since w and x both beat y in T and since z and y dominate T, z must beat both w and x. But then w and x do not dominate T, a contradiction. So, a vertex has at most one outneighbor in D(T) that has an outneighbor. (a) (b) 1 2 3 n1 n n 1 n Figure 4.5. Orientations of a path. Theorem 4.2.2 Fisher, Lundgren, Merz, and Reid [19]. A path on four or more vertices is not the domination graph of a tournament. Proof. Suppose that there exists some ntournament with n 2: 4 such that Dom(T) is a path. Up to symmetry, Figure 4.5 shows the only orientations of 1J(T) allowed by Theorem 4.2.1. First assume that D(T) is oriented as in Figure 4.5(a). Then for 1 ::; i::; n1, vertex i beats vertex i + 1 in T, and 105
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vertices i and i + 1 dominate T. So for 1 i n2, since i beats i + 1 and since i + 1 and i + 2 dominate T, i + 2 must beat i. Next for 1 i n3, since i + 3 beats i + 1 and since i and i + 1 dominate T, i must beat i + 3. Then for 1 i n4, since i beats i + 3 and since i + 3 and i + 4 dominate T, i + 4 must beat i. Continuing this way we can show that when 1 i < j n, vertex i beats vertex j if j i is odd, and vertex j beats vertex i if j i is even. So the outneighbors of vertex 1 are the ev,en numbered vertices, and the outneighbors of either vertex n (if n is odd), or vertex n1 (if n is even) are the odd numbered vertices except nor nl. So either {1, n} E Dom(T) or {1,n1} E Dom(T). This is a contradiction when n :::0: 3. Assume that D(T) is oriented as in Figure 4.5(b ). Similar arguments show that when 2 i < j n, vertex i beats vertex j if j i is odd, and vertex j beats vertex i if j i is even. So the outneighbors of vertex 2 are the odd numbered vertices (including 1 because of the orientation), and the outneighbors of either vertex n (if n is even), or vertex n1 (if n is odd) are the even numbered vertices except n or n 1. Thus either {2, n} E Dom(T) or {2, n 1} E Dom(T). This is a contradiction when n :::0: 4, completing the proof. Which forests of caterpillars are the domination graph of a tourna ment? Which forests of caterpillars are the domination graph of an oriented graph? Which graphs are the domination graph of an arbitrary digraph? These questions are open. 106
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4.3 The Domination Graphs of Digraphs Tournaments record the result of contests where each team plays every other team exactly once and there are no ties. A natural generalization is to digraphs without cycles of length 1 or 2, called oriented (simple) graphs. For these, the lack of an arc between two vertices indicates that the teams either tied or did not play. Alternatively, the digraph represents a double round robin tournament where ( x, y) E A(D) if and only if x beats y twice. That is, (x, y) and (y, x) rf_ A( D) implies that x beat y once and y beat x once. We begin by extending Theorem 4.1.1 to oriented graphs. Theorem 4.3.1 Fisher, Lundgren, Merz, and Reid [19]. If Dis an oriented graph, then Dom(D) is either an odd cycle with or without isolated and/or pendant vertices, or a forest of caterpillars. Proof. Arbitrarily add arcs to D so as to form a tournament T. Since all dominant pairs in D are dominant pairs in T, we have that Dom( D) is a subgraph of Dom(T). The result then follows from Theorem 4.1.1. Theorem 4.2.1 showed that a graph consisting of an odd cycle, with or without isolated and/or pendant vertices is the domination graph of some tournament. Since tournaments are oriented graphs, any graph consisting of an odd cycle, with or without isolated and/or pendant vertices is the domination graph of some oriented graph. However, there are some graphs that are the domination graph of an oriented graph, but not of a tournament. Theorem 4.3.2 Fisher, Lundgren, Merz, and Reid [19]. If G is a caterpillar, then G is the domination graph of an oriented graph. Proof. Let 1, 2, ... k be the consecutively labeled vertices of a longest path in G. Observe that both 1 and k have degree one. Let j = k if k is odd and 107
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j = k 1 if k is even. Let G' be the graph formed by adding edge {l,j} to G. Since G' has an odd cycle with pendant edges, Theorem 4.2.1 shows there is a tournament T with Dom(T) = G', and that the subtournament ofT on the vertices 1, 2, ... j is UJ. Let D be the digraph formed by removing arc (1, 2) from T. Then 1 and k do not dominateD because neither beats 2. However, 1 and 2 dominate D. All other dominant pairs of T are dominant pairs of D, because no other dominant pairs ofT includes 1, completing the proof. In particular, observe that while paths are not the domination graph of a tournament, they are the domination graph of some oriented graph, since paths are caterpillars. Theorem 4.3.3 If G is a caterpillar plus one isolated vertex, then G is the domination graph of an oriented graph. Proof. Let G' be the same graph as prescribed in the proof of the previous theorem. Theorem 4.2.1 shows that there is a tournament T with Dom(T) = G' and that the subtournament ofT on 1,2, ... ,j is Uj Add a vertex toT that beats only 1 and j. Then {i,j} is removed and all other arcs remam, completing the proof. 108
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CHAPTER 5 A LIST OF OPEN PROBLEMS ( 1) Can disimplicial/ diinterval elimination orderings be used to deter mine if a digraph has a chordal/interval competition graph computationally faster than constructing the competition graph? (2) Characterize classes of graphs, other than strongly chordal, trees, graphs with girth at least 7, and 6cycle free graphs such that that every edge is in a triangle, that have interval squares. (3) Characterize classes of graphs, other than trees, 6cycle free and cactus, that have interval twostep graphs. ( 4) Prove or disprove: If G has an edge clique cover that has a system of distinct representatives, then G is the competition graph of a loopless Hamiltonian digraph. ( 5) Prove or disprove: If G is a chordal graph on 2p + 1 vertices, then G is the pcompetition graph of a loopless Hamiltonian digraph. (6) Prove or disprove: If G is an interval graph on 2p + 1 vertices, then G is the pcompetition graph of a loopless Hamiltonian digraph. (7) Prove or disprove: If T is a tree on p + 3 vertices, then T IS the pcompetition graph of a loopless Hamiltonian digraph. (8) Generalize the bounds of Chapter 3.5.4 topcompetition graphs where p ?:. 2. 109
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(9) Which forests are the domination graph of an oriented graph? (10) Which forests are the domination graph of a tournament? 110
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APPENDIX A NOTATION C(D) the competition graph of a digraph D Cp(D) the pcompetition graph of a digraph D E( G) the edge set of a graph G V( G) the vertex set of a graph G { x, y} E E( G) x and y are adjacent in G A( D) the arc set of a digraph D (x,y) E A(D) there is an arc from x toy in D Na(x) the open neighborhood of x in G Na[x] the closed neighborhood of x in G On(x) the set of outneighbors of x in D In( x) the set of inneighbors of x in D b.+(D) maxxEV(D) IO(x)l b.(D) maxxEV(D) II(:z:)l o+(D) minxEV(D) IO(x)l o(D) minxEV(D) II(x)l b.( G) maxxEV(G) INa(x)l o(G) minxEV(G) INa(:z:)l a( G) the independence number of a graph G w( G) the clique number of a graph G 111
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x( G) the chromatic number of a graphG X1X2 Xn a path from X1 to Xn (directed if in a digraph) XJ X2 XnX1 a cycle through x1 x2 Xn (directed if in a digraph) GE(G) the edge clique cover number of G G) the pedge clique cover number of G i( G) the number of isolated vertices in G da(x,y) the minimum number of vertices in a path from x toyinG Cz the family of maximal cliques 0 such that /0 I 2: 3 and x E 0 112
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