Graph Structure and Coloring
Matthieu Plumettaz
Submitted in partial fulllment of the
requirements for the degree
of Doctor of Philosophy
in the Graduate School of Arts and Sciences
COLUMBIA UNIVERSITY
2014
c 2014
Matthieu Plumettaz
All Rights Reserved
ABSTRACT
Graph Structure and Coloring
Matthieu Plumettaz
We denote by
G = (V, E)
if no vertex of
G
a graph with vertex set
V
and edge set
E.
A graph
G
is claw-free
has three pairwise nonadjacent neighbours. Claw-free graphs are a natural
generalization of line graphs. This thesis answers several questions about claw-free graphs
and line graphs.
In 1988, Chvátal and Sbihi [15] proved a decomposition theorem for claw-free perfect
graphs. They showed that claw-free perfect graphs either have a clique-cutset or come from
two basic classes of graphs called elementary and peculiar graphs.
In 1999, Maray and
Reed [26] successfully described how elementary graphs can be built using line graphs of
bipartite graphs and local augmentation. However gluing two claw-free perfect graphs on
a clique does not necessarily produce claw-free graphs. The rst result of this thesis is a
complete structural description of claw-free perfect graphs. We also give a construction for
all perfect circular interval graphs. This is joint work with Chudnovsky, and these results
rst appeared in [8].
Erd®s and Lovász conjectured in 1968 that for every graph
such that
G
s + t − 1 = χ(G) > ω(G),
such that
χ(G|S) ≥ s
and
there exists a partition
χ(G|T ) ≥ t.
s
and
t
and all integers
(S, T )
s, t ≥ 2
of the vertex set of
This conjecture is known in the graph theory
community as the Erd®s-Lovász Tihany Conjecture.
cases of the conjecture are when
G
are small.
For general graphs, the only settled
Recently, the conjecture was proved
for a few special classes of graphs: graphs with stability number
2
[2], line graphs [24] and
quasi-line graphs [2]. The second part of this thesis considers the conjecture for claw-free
graphs and presents some progresses on it. This is joint work with Chudnovsky and Fradkin,
and it rst appeared in [5].
Reed's
ω , ∆, χ
conjecture proposes that every graph satises
χ ≤ d 12 (∆ + 1 + ω)e;
it
is known to hold for all claw-free graphs.
The third part of this thesis considers a local
strengthening of this conjecture. We prove the local strengthening for line graphs, then note
that previous results immediately tell us that the local strengthening holds for all quasi-line
graphs. Our proofs lead to polytime algorithms for constructing colorings that achieve our
bounds: The complexity are
O(n2 )
for line graphs and
O(n3 m2 )
for quasi-line graphs. For
line graphs, this is faster than the best known algorithm for constructing a coloring that
achieves the bound of Reed's original conjecture. This is joint work with Chudnovsky, King
and Seymour, and it originally appeared in [7].
Table of Contents
1 Introduction
1
1.1
Perfect graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Claw-free perfect graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.3
Conjectures related to the chromatic number
4
. . . . . . . . . . . . . . . . . .
2 Trigraphs
7
3 The Structure of Claw-Free Perfect Graphs
16
3.1
Preliminary results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
3.2
Essential Triangles
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
3.3
Holes of Length 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
3.4
Long Holes
29
3.5
Some Facts about Linear Interval Join
3.6
Proof of the Main Theorem
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . .
34
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
4 On the Erd®s-Lovász Tihany Conjecture
49
4.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
4.2
Structure Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
4.3
Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
54
4.4
The Icosahedron and Long Circular Interval Graphs
. . . . . . . . . . . . . .
58
4.5
Non-2-substantial and Non-3-substantial Graphs
. . . . . . . . . . . . . . . .
59
4.6
Complements of Orientable Prismatic Graphs . . . . . . . . . . . . . . . . . .
64
4.7
Non-orientable Prismatic Graphs
70
. . . . . . . . . . . . . . . . . . . . . . . . .
i
4.8
Three-cliqued Graphs
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.9
Non-trivial Strip Structures
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
76
4.10 Proof of the Main Theorem
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
78
5 A Local Strengthening of Reed's Conjecture
73
79
5.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
79
5.2
Line Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
81
5.3
Quasi-line Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
90
5.4
Decomposing Quasi-line Graphs . . . . . . . . . . . . . . . . . . . . . . . . . .
93
5.5
Putting the pieces together and Algorithmic Considerations
98
. . . . . . . . . .
Bibliography
99
A Appendix
104
A.1
Orientable prismatic graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
A.2
Non-orientable prismatic graphs . . . . . . . . . . . . . . . . . . . . . . . . . . 107
A.3
Three-cliqued graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
ii
List of Figures
1.1
A claw . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2
Examples of claw-free and quasi-line graphs
. . . . . . . . . . . . . . . . . . .
4
1.3
Example for Reed's Conjecture
. . . . . . . . . . . . . . . . . . . . . . . . . .
6
iii
Acknowledgments
I rst want to express my deepest gratitude to my adviser, Maria. You have mentored
and guided me every step of the way during the last ve years.
You have always made
yourself available whenever I needed help for my research and academic progresses or on
any other topic. It was truly a blessing to be able to work with you.
Merci maman et papa pour votre soutien durant toutes ces années. Vous ne m'avez pas
seulement aidé lors de cette thèse, vos encouragements tout au long de ma vie m'ont permis
d'en arriver ici. Je vous dois tellement. Merci Anaïs et Damien, ça n'a pas été facile tous
les jours d'être loin de la Suisse, mais vous avez transformé chaque visite en un moment
mémorable et relaxant pour reprendre de plus belle. Un grand merci à toute ma famille, vos
messages, pensées et visites me touchent toujours. Muchisimas gracias señora Ana, señor
Alberto, ahijados Helen y Juan. Me hicieron parte de su familia desde el primer día, y tener
el apoyo, el amor y la conanza de su familia ha sido muy importante durante este proceso,
mil gracias.
Thank you Antoine for the countless duels, whether they were in squash, Pentago, TashKalar and for failing to save earth together so many times. Thank you Fanny for all the
delicious meals and all the competitive games we played, and I am really sorry for Carcassonne. Thank you Irene and Nitin for the delicious diners across Manhattan. Thank you
Rodrigo, Daniela, Arseniy and Olga for the many movie nights and celebrations throughout
the years. Thank you Ceci and Yixi for making me part of the CoC by association. Thank
you Rouba and Yori for all the good times and the coees watching the world cup. Thank
you Silvia and Juancho for coming so many times to watch the Patriots lose. Thank you to
the french troop, Laurine, Thibaut, Julia, Justine, Antoine and Anisha, for all the moments
shared together. Thank you Sarah and Bernard, this adventure would have never started
without you. Thank you to all my ocemates during my time here, you brought every day
iv
so many ways not to work, that was awesome. A special thank you to all the people who
read part of this thesis, and gave comments and corrections. And thanks to all my fellow
Ph.D. colleagues for making all of it worth it.
I would like to thank all the sta and faculty members of the IEOR department. Everybody on the sta has always been supportive, helpful and kind and it was amazing to
be able to learn from such knowledgeable faculty. Thank you Alex, Andrew, Gregory, Paul,
Sasha and Yared, for the collaboration on dierent projects.
Thank you Nightwish, Saez, Dido, Queen, Ospring, Nirvana, Ska-P and les Enfoirés for
amusing my ears during the long days and late nights. A big thanks to all the developers
who created so many good games, Blood Bowl, Civilization, Dota, Final Fantasy, Frozen
Synapse, XCOM, in which I lost countless hours of sleep. Thanks the Old World Football
League for the competitive games, the community, the nuing and the whining. I joined
you guys not long after I started this Ph.D., and even during the hardest times, I could not
stop myself from playing my game and let Nue nish the job.
But the most important person is without a doubt my wife, Tulia. Thank you so much
for all you did for me. I would not have been able to make it without your support each
and every day. You made all the hard days easier to live and all the good days a delight to
be part of.
v
To my parents
vi
CHAPTER 1. INTRODUCTION
Chapter 1
Introduction
1.1
Perfect graphs
A graph
G
is a mathematical object used to model pairwise relations among a collection
of entities.
This collection of entities is called the vertex set and is denoted by
The edge set, denoted by
E(G),
represents the relations between pairs of elements of
V (G).
V (G).
Graphs have numerous applications to a wide variety of elds, from nding the shortest path
between two cities on a GPS, to managing inventory in a warehouse or detecting particular
molecules in biology. The major part of the thesis will be about with structural graph theory.
Structural graph theory tries to understand families of graphs.
When someone studies a
particular problem, it is generally possible to characterize some properties of the underlying
family of graphs. One of our main goals is to understand what are the basic graphs in a
given family. In particular, we want to describe a family in terms of well-understood graphs
and construction steps. This description can then lead to a better understanding of how to
approach a problem both from a theoretical and an algorithmic point of view.
For two elements
non-adjacent
members of
to
X
y
if
x, y ∈ V (G),
xy ∈
/ E(G).
are adjacent. A set
are non-adjacent. A set
to at most one vertex of
a
triad
A
A
clique
in
x
G
is
adjacent
is a set
is an
brace
anti-matching
to
y
if
X ⊆ V (G)
X ⊆ V (G) is a stable set
S ⊆ V (G)
S.
we say that
in
xy ∈ E(G)
x
is
such that every two
G if every two members of X
if every vertex in
is a clique of size 2, a
and
triangle
S
is non-adjacent
is a clique of size 3 and
is a stable set of size 3. Note that all the graphs that we consider in this thesis are
1
CHAPTER 1. INTRODUCTION
nite.
We say that
adjacent in
on X
if
H
H
is a
subgraph
G with vertex set X , if every pair of vertices in X
of
are also adjacent in
G.
as the subgraph with vertex set
x is adjacent to y
exists
X ⊆ V (G)
G.
in
For a graph
such that
X ⊆ V (G),
For
X
and such that
G|X = H .
k -coloring
k -coloring
A
ω(G),
denoted by
denoted by
that
x
G
of
G,
G0 .
of
G,
is adjacent to
y
in
G
if and only if
G
x
G
is the graph
of
G if there
For simplicity, we
denoted by
χ(G),
clique number
stability number
of
of
G,
A graph
G is said to be perfect
G0
is equal to the clique
the chromatic number of
The complement of a graph
G,
The
and the
α(G) is the size of the maximum stable set in G.
G0
of
G.
of
if and only
c : V (G) → {1, . . . , k}
is a map
χ(G)-coloring
G|X induced
G|X
induced subgraph
chromatic number
as a coloring. The
is the size of a maximum clique in
if for every induced subgraph
number of
is an
in
v, w ∈ V (G), c(v) 6= c(w).
is the smallest integer such that there exits a
G,
x is adjacent to y
H , we say that H
such that for every pair of adjacent vertices
may also refer to a
we dene the subgraph
that are
with vertex set
is non-adjacent to
y
in
V (G)
and such
G.
Perfect graphs were introduced in 1960 by Claude Berge and are a central family of
graphs because they are the graphs that behave 'perfectly' in terms of coloring.
graph
G, ω(G)
For any
is always a trivial lower bound on the chromatic number. Perfect graphs are
the family of graphs that match this bound and are closed under taking induced subgraph.
When Claude Berge introduced the family of perfect graphs, he also introduced another
family of graphs - that we now call Berge graphs. To give a formal description, we rst need
a few more denitions. A
set can be ordered as
in
G
is a subgraph
C
{c1 , . . . , cn }
such that
that a cycle
C
is a
is non-adjacent to
is
Berge
if
G
path
in
G
{p1 , . . . , pn }
with
ci
n
is a subgraph
such that
pi
vertices for some
is adjacent to
ci+1
for
P
with
n
vertices for
is adjacent to
n ≥ 3,
pi+1
for
n ≥ 1,
whose vertex
1 ≤ i < n.
A
cycle
whose vertex set can be ordered as
1 ≤ i < n,
and
cn
is adjacent to
c1 .
We say
hole, if n > 3 and if for all 1 ≤ i, j ≤ n with i+2 ≤ j and (i, j) 6= (1, n), ci
cj .
The
length
of
C
is the number of vertices of
does not contain any odd holes and
G
C.
We say that a graph
G
does not contain any odd holes. Claude
Berge stated two conjectures when introducing perfect graphs. The rst one, known as the
Weak Perfect Graph Conjecture, states that a graph
G
is perfect if an only if
G
is perfect.
It was proved to be true by Lovász [25]. The second conjecture, known as the Strong Perfect
2
CHAPTER 1. INTRODUCTION
Graph Conjecture, states that a graph is perfect if and only if it is Berge. This conjecture
remained open for more than 40 years before Chudnovsky, Robertson, Seymour and Thomas
proved it in 2002 [9]. Those two results are stated bellow.
1.1.1 (Weak Perfect Graph Theorem.
. A graph G is perfect if and only
Lovász [25])
if G is perfect.
1.1.2 (Strong Perfect Graph Theorem.
Chudnovsky,
Robertson,
Seymour and
. A graph is perfect if and only it is Berge.
Thomas [9])
1.2
The
Claw-free perfect graphs
neighborhood
called
neighbors
of a vertex
of
v.
graph with vertex set
V (L(G)) = E(G)
is
quasi-line
H
G, L(G)
a clique, and a vertex is
G
is the set
N (v) of vertices adjacent to v .
Given a multigraph
corresponding edges in
some multigraph
v
G,
the
line graph of G,
denoted by
is isomorphic to
if every vertex
v
of
N (v) are
L(G),
is the
in which two vertices are adjacent precisely if their
share an endpoint. We say that a graph
bisimplicial
Vertices of
G0 .
A vertex is
G0
simplicial
is a
line graph
if for
if its neighborhood is
if its neighborhood is the union of two cliques. A graph
G
is bisimplicial. A
claw
is the graph with four vertices
and three edges where the edges are all incident to a single vertex (see Figure 1.1). A graph
G
is
claw-free
if it contains no induced claw. It is easy to observe that every line graph is
quasi-line and every quasi-line graph is claw-free. In Figure 1.2, we give two examples that
show that there are quasi-line graphs that are not line graphs and claw-free graphs that are
not quasi-line graphs.
Figure 1.1: This illustration show a claw.
Several attempts have been made to describe claw-free perfect graphs: rst by Chvátal
and Sbihi in 1988 [15], and then by Maray and Reed in 1999 [26]. However, these results
only showed how to decompose claw-free perfect graphs, but did not show how to construct
3
CHAPTER 1. INTRODUCTION
Figure 1.2: The graph on the left is a quasi-line graph but is not a line graph. The graph
on the right is a claw-free graph but is not a quasi-line graph.
them explicitly.
Indeed, Chvátal and Sbihi result uses clique-cutsets to decompose claw-
free perfect graphs into two basic classes of graphs.
But the inverse operation of gluing
two graphs on a clique might produce a claw in the resulting graph. In order to obtain a
structural theorem, understanding how to decompose graphs is only a rst step, but it still
only gives an incomplete picture of the family. We wish to be able to build all graphs in a
family from smaller graphs in such a way that every graph we construct is in the family.
The results presented in Chapter 3 give a complete description of the structure of clawfree perfect graphs. In fact, by the Strong Perfect Graph Theorem 1.1.2, we study claw-free
Berge graphs because in many cases it is easier to prove that a graph is Berge than to prove
that the graph is perfect. Actually we will work with slightly more general objects called
trigraphs which will be dened in Chapter 2. Chudnovsky and Seymour proved a structural
theorem for general claw-free graphs [13] and quasi-line graphs in [14]. Later we will show
that every perfect claw-free graph is a quasi-line graph, however not all quasi-line graphs
are perfect. Our result renes the characterization of quasi-line graphs from [14] to obtain
a precise description of perfect quasi-line graphs.
1.3
Conjectures related to the chromatic number
Finding the exact value of the chromatic number of a graph is a fundamental algorithmic
and theoretical problem in graph theory. Attempt to bound the value of
of graphs have been made since the beginning of graph theory.
example is probably the
Four Color Theorem.
χ(G)
One of the most famous
In the 18th century, the following question
has been raised: Is it was true that the chromatic number of planar graphs is
4
for families
4?
A graph
CHAPTER 1. INTRODUCTION
is
planar
if it can be drawn on a plan with no edge crossing each other. It is easy to build
a planar graph that needs
4
colors, but it took more than a century until Appel and Haken
proved in 1976 the following:
1.3.1 (Four Color Theorem.
. Let G be a planar graph, then
Appel and Haken [1])
χ(G) ≤ 4.
In the last 50 years, many conjectures and many theorems related to the chromatic
number have been stated. We present now one of them, a conjecture that Erd®s and Lovász
made in 1968.
Conjecture 1
. For every graph G with χ(G) > ω(G) and for every
(Erd®s-Lovász Tihany)
two integers s, t ≥ 2 with s + t = χ(G) + 1, there is a partition (S, T ) of the vertex set V (G)
such that χ(G|S) ≥ s and χ(G|T ) ≥ t.
Currently, the only settled cases of the conjecture are
(3, 4), (3, 5)} [3; 28; 33; 34].
(s, t) ∈ {(2, 2), (2, 3), (2, 4), (3, 3),
Recently, Balogh, Kostochka, Prince and Stiebitz proved Conjec-
ture 1 for quasi-line graphs. In Chapter 4, we consider the Erd®s-Lovász Tihany Conjecture
for claw-free graphs.
We prove a slightly weakened version of Conjecture 1 for this class
of graphs. Our proof relies on a structure theorem of claw-free graphs by Chudnovsky and
Seymour [13]
The
degree d(v)
For a graph
number of
G,
G
of a vertex
we dene the
v ∈ V (G)
is the number of vertices adjacent to
maximal degree
is trivially bounded above by
∆(G) = maxv∈V (G) {d(v)}.
∆(G) + 1
Conjecture proposes, roughly speaking, that
Conjecture 2
by
χ(G)
and below by
ω(G).
v
in
G.
The chromatic
Reed's
ω , ∆, χ
falls in the lower half of this range:
. For any graph G,
(Reed)
χ(G) ≤
1
2 (∆(G)
+ 1 + ω(G)) .
One of the rst classes of graphs for which this conjecture was proved is the class of line
graphs [23]. Already for line graph the conjecture is tight. We show in Figure 1.3 examples
of line graphs for which the conjecture holds with equality.
The proof of Conjecture 2 for line graphs was later extended to quasi-line graphs [21; 22]
and claw-free graphs [21]. In his thesis, King proposed a strengthening of Reed's Conjecture,
5
CHAPTER 1. INTRODUCTION
Figure 1.3: Example of a line graph for which Conjecture 2 is tight. The graph on the right
is the line graph of the graph on the left.
ω(v)
denote the size of the
In Chapter 5 we prove that Conjecture 3 holds for line graphs.
Then using methods
giving a bound in terms of local parameters. For a vertex
largest clique containing
Conjecture 3
v,
let
v.
(King [21])
. For any graph G,
χ(G) ≤ max
v∈V (G)
1
2 (d(v)
+ 1 + ω(v)) .
similar to [22], we extend the result to quasi-line graphs.
Furthermore our proofs yield
polytime algorithms for constructing a proper coloring achieving the bound of the theorem:
O(n2 )
time for a line graph on
vertices and
m
n
vertices, and
O(n3 m2 )
time for a quasi-line graph on
n
edges.
The thesis is organized as follow.
In Chapter 2, we introduce trigraphs and notions
associated with them. In Chapter 3, we present and prove our structural theorem for clawfree perfect graphs. In Chapter 4, we explore the Erd®s-Lovász Tihany for claw-free graphs.
Finally in Chapter 5, we prove Conjecture 3 for quasi-line graphs.
6
CHAPTER 2. TRIGRAPHS
Chapter 2
Trigraphs
Trigraphs are a generalization of graphs that are useful for studying problems about forbidden induced subgraphs. Trigraphs will be extensively used in Chapter 3. The majority of
graph notions can be directly extended to trigraphs, and will be described in this chapter.
All graphs and trigraphs considered in this thesis are nite.
A
trigraph G consists of a nite set V (G) of vertices, and a map θG : V (G)2 → {−1, 0, 1},
satisfying:
•
for all
•
for all distinct
u, v ∈ V (G), θG (u, v) = θG (v, u)
•
for all distinct
u, v, w ∈ V (G),
For distinct
antiadjacent
adjacent
v ∈ V (G), θG (v, v) = 0.
u, v ∈ V (G),
if
at most one of
we say that
θG (u, v) = −1,
and
u, v
are
B ⊆ V (G)\{a},
u
is
antiadjacent
we say that
a
is
(resp. antiadjacent) to every vertex in
to
B.
if
if
(resp.
anticomplete )
Similarly, we say that
member of
B,
a
v
if
complete
B.
u, v
antiadjacent
adjacent
u, v
are
if they are
if
u, v
are antiadjacent. For a vertex
a
and
(resp.
is
anticomplete )
For two disjoint
to
B
A, B ⊂ V (G),
if
a
to
is adjacent
we say that
A
is
if every vertex in
A
is complete (resp. anticomplete)
is strongly complete to
B
if
to
B
to
We say that
v
Also, we say that
u
equals 0.
θG (u, v) = 1, strongly
θG (u, v) = 0.
if they are either strongly adjacent or semiadjacent, and
are adjacent, and that
complete
strongly adjacent
semiadjacent
either strongly antiadjacent or semiadjacent.
a set
θG (u, v), θG (u, w)
and so on.
7
a
is strongly adjacent to every
CHAPTER 2. TRIGRAPHS
G
Let
X
are adjacent and
set
X ⊆ V (G)
stable set
3, and a
triad
X ⊆ V (G)
if every two members of
if every two members of
X
X
such that every two members of
X
X
are strongly adjacent. A
are antiadjacent and
are strongly antiadjacent. A
triangle
X
is a
strong
is a clique of size
is a stable set of size 3.
G
X,
and
X ⊆ V (G),
G|X induced on X
we dene the trigraph
and its adjacency function is the restriction of
contains H , and H
is a
subtrigraph
of
θG
to
X 2.
as follows.
G
We say that
G if there exists X ⊆ V (G) such that H
is isomorphic
G|X .
A
claw
{a, b, c}
A
path
in
such that
P
is from
p1
for
C
(i, j) 6= (1, n), ci
ci
and
A trigraph
pn .
to
pi+1
and
cn
cj
G
G
For a path
subpath P 0
of
P
{c1 , . . . , cn }
such that
dened by
with
ci
n
pi
is antiadjacent to
p1 − p2 − . . . − pn
P = p1 − . . . − pn
and
c1 − c2 − . . . − cn − c1 .
is said to be
The
consecutive
Berge
if
cj .
length
is adjacent (resp.
c1 .
n ≥ 3,
i+1=j
if no subtrigraph of
C
or
G
if
with
whose
antiadjacent) to
We say that a cycle (resp.
We will generally denote
of
pj
P 0 = pi − pi+1 − . . . − pj .
antihole ), if n > 3 and if for all 1 ≤ i, j ≤ n with i + 2 ≤ j
are said to be
and
and say
i, j ∈ {1, . . . , n}
vertices for some
is adjacent (resp. antiadjacent) to
(resp.
and
with the following sequence
the
{a, b, c}
does not contains a claw.
1≤i<n
for
is antiadjacent (resp. adjacent) to
following sequence
Vertices
P
if
is complete to
n vertices for n ≥ 1, whose vertex set can be ordered
is adjacent to
pi − P − pj
hole
is a
with
claw-free
anticycle ) in G is a subtrigraph C
(resp.
1 ≤ i < n,
anticycle)
V (H) = {x, a, b, c}, x
is said to be
We generally denote
we denote by
cycle
pi
vertex set can be ordered as
ci+1
G
G is a subtrigraph P
that the path
A
such that
is a triad. A trigraph
{p1 , . . . , pn }
i < j,
H
is a trigraph
|i − j| > 1.
G
is a set
strong clique
if every two members of
Its vertex set is
as
is a
stable set
is a
For a trigraph
to
clique
be a trigraph. A
C
and
with the
is the number of vertices of
C.
{i, j} = {1, n}.
is a hole, and no subtrigraph of
is an antihole. In Chapter 3, we study perfect graphs, which by the strong perfect graph
theorem [9], is equivalent to studying Berge graphs. We will in fact work with the slightly
more general Berge trigraphs. Since it is easier in many cases to prove that a trigraph is
Berge than to prove that the trigraph is perfect, we will only deal with Berge trigraphs.
A trigraph
and
Y
G
is
cobipartite
are strong cliques and
if there exist nonempty subsets
X ∪ Y = V (G).
8
X, Y ⊆ V (G)
such that
X
CHAPTER 2. TRIGRAPHS
X, A, B, C ⊆ V (G),
For
b∈B
for
and
c∈C
such that
X1 , . . . , Xn ⊆ V (G),
there exist
xi ∈ Xi
{X|A, B, C} is a claw
we say that
G|{x, a, b, c}
we say that
hole
is a
x ∈ X , a ∈ A,
{a, b, c}.
is complete to
X1 − X2 − . . . − Xn − X1
x1 − x2 − . . . − xn − x1
such that
x
is a claw and
if there exist
Similarly,
antihole )
(resp.
if
is a hole (resp. antihole). To simplify
notation, we will generally forget the bracket delimiting a singleton, i.e. instead of writing
{{x}|A, {y}, B}
A, B
Let
we will simply denote it by
be disjoint subsets of
V (G).
is a strong clique, and every vertex in
anticomplete to
A.
(A, B)
The pair
strong cliques, and for every vertex
or strongly anticomplete to
to
A,
{x|A, y, B}.
The set
V (G)\A
is called a
A
is called a
homogeneous set
if
A
is either strongly complete or strongly
homogeneous pair
v ∈ V (G)\(A ∪ B), v
in
G
if
A, B
are nonempty
is either strongly complete to
and either strongly complete to
B
A
or strongly anticomplete
B.
Let
V1 , V2
be a partition of
there is a subset
• Ai
Vi \Ai
and
• A1 ∪ A2
• V1 \A1
Ai ⊆ V i
V (G) such that V1 ∪ V2 = V (G), V1 ∩ V2 = ∅,
i = 1, 2
such that:
are not empty for
i = 1, 2,
is a strong clique,
is strongly anticomplete to
The partition
and for
(V1 , V2 ) is called a 1-join
V2 ,
and
V1
is strongly anticomplete to
and we say that
G admits a 1-join
V2 \A2 .
if such a partition
exists.
Let
A1 , A2 , A3 , B1 , B2 , B3 be disjoint subsets of V (G).
is called a
hex-join
if
A1 , A2 , A3 , B1 , B2 , B3
The
6-tuple (A1 , A2 , A3 |B1 , B2 , B3 )
are strong cliques and
• A1
is strongly complete to
B1 ∪ B2 ,
and strongly anticomplete to
B3 ,
and
• A2
is strongly complete to
B2 ∪ B3 ,
and strongly anticomplete to
B1 ,
and
• A3
is strongly complete to
B1 ∪ B3 ,
and strongly anticomplete to
B2 ,
and
•
S
Let
i (Ai
G
∪ Bi ) = V (G).
be a trigraph. For
N (v) = {x : x
every
is adjacent to
v ∈ V (G), N (v)
v ∈ V (G),
v}.
we dene the
The trigraph
G
neighborhood
is said to be a
is the union of two strong cliques.
9
of
v,
denoted
N (v),
quasi-line trigraph
by
if for
CHAPTER 2. TRIGRAPHS
H
A trigraph
subset
thickening
is a
Xv ⊆ V (H),
of a trigraph
G
all pairwise disjoint and with union
•
for each
•
if
u, v ∈ V (G)
are strongly adjacent in
•
if
u, v ∈ V (G)
are strongly antiadjacent in
if
v ∈ V (G), Xv
is a strong clique of
G
v ∈ V (G)
V (H),
there is a nonempty
satisfying the following:
H,
then
G
Xu
then
is strongly complete to
Xu
Xv
in
H,
is strongly anticomplete to
Xv
H,
in
•
if for every
u, v ∈ V (G) are semiadjacent in G then Xu
Xv
anticomplete to
is neither strongly complete nor strongly
H.
in
Next we present some denitions related to quasi-line graphs that have been introduced
in [14]. To develop our structural results in Chapter 3, we need a few more denitions that
rene and extend the concepts used in [14] and will be presented at the same time.
A
stripe
(G, Z)
is a pair
is a strong stable set,
vertex in
G
Z,
G
and a subset
is a strong clique for all
linear interval trigraph
in such a way that for
vi , vk .
1 ≤ i < j < k ≤ n,
Given such a trigraph
(G, {v1 , vn }) a linear interval stripe
if
there is no vertex adjacent to both
G
Let
[0, 1],
G
Σ
and
vi , vk
if
be a circle in the sphere, and let
be a trigraph with vertex set
•
if
u, v ∈ Fi
for some
the interior of
•
if there is no
i
Fi
i
then
V
then
u, v
such that
|Z| ≤ 2, Z
no vertex is semiadjacent to a
are adjacent then
and
v1 , vn
vj
v1 , . . . , v n
{v1 , . . . , vn }
is strongly adjacent
with
n ≥ 2,
we call
are strongly antiadjacent. By analogy
[vi , vj ] = {vk }i≤k≤j , (vi , vj ) = {vk }i<k<j ,
Moreover we will write
F1 , . . . , F k ⊆ Σ
xi < xj
V ⊆Σ
u, v ∈ V ,
are strongly adjacent,
then
u, v
10
i < j.
be nite, and let
are adjacent, and if also at least one of
u, v ∈ Fi
if
be homeomorphic to the interval
share an end-point. Now let
in which, for distinct
u, v
such that
G is connected, no vertex is semiadjacent to v1 or to vn ,
v1 , vn ,
F1 , . . . , Fk
V (G)
Z.
and numbering
(vi , vj ] = {vk }i<k≤j .
such that no two of
of
if its vertex set can be numbered
with intervals, we will use the following notation,
[vi , vj ) = {vk }i≤k<j
Z
z ∈ Z,
and no vertex is adjacent to two vertices of
is said to be a
to both
N (z)
of a trigraph
are strongly antiadjacent.
u, v
belongs to
CHAPTER 2. TRIGRAPHS
Such a trigraph
G
is called a
circular interval trigraph.
We will denote by
Fi∗
the interior of
Fi .
Let
G have four vertices say w, x, y, z , such that w
adjacent to
the pair
Let
z1 , z2
z, x
is semiadjacent to
(G, {w, z})
G
v, z1 , z2
(G\w, {z})
Let
G
Fi .
and
is a
is a
|Xv | = 1
thickening
G,
where
for each
Σ, F1 , . . . , Fk
of
let
z2 ,
and
spot, the pair (G, {z1 }) is a
Z0
be as in the corresponding
and if
z ∈ Fi
say, let no vertex
bubble.
a
Xv (v ∈ V (G))
v ∈ Z,
and
truncated spot.
(G, {z})
We call the pair
is a thickening of
Z ⊆ V (G)
(H, Z 0 )
is a
z1
is strongly adjacent to
z ∈ V (G) belong to at most one of F1 , . . . , Fk ;
be an endpoint of
is strongly
truncated spring.
is a
(G, {z1 , z2 })
be a circular interval trigraph, and let
denition. Let
H
(G\z2 , {z1 })
v
such that
are strongly antiadjacent. Then the pair
and the pair
x, y
and all other pairs are strongly antiadjacent. Then
and the pair
have three vertices say
one-ended spot
If
spring
is a
y,
is strongly adjacent to
are the corresponding subsets, and
be the union of all
Xv (v ∈ Z );
we say that
(G, Z).
The following construction is slightly dierent from how
linear interval joins
have been
dened for general quasi-line graphs [14], but the resulting graphs are exactly the same. We
may also assume that if
(G, Z)
is a stripe then
constructed in the following manner is called a
•
Let
H = (V, E)
•
Let
η : (E × V ) ∪ E → 2V (G) .
•
For every edge
Let
V (G) 6= Z .
Any trigraph
G
that can be
linear interval join.
be a graph, possibly with multiple edges and loops.
e = x1 x2 ∈ E
(Ge , Ye )
(where
x1 = x2
if
e
is a loop)
be either
∗
a spot or a thickening of a linear interval stripe if
∗
the thickening of a bubble if
Moreover let
φe
e
Let
η(e, xj ) = N (φe (xj ))
Let
η(e) = V (Ge )\Ye .
for
is not a loop, or
is a loop.
be a bijection between
e
j = 1, 2
11
Ye
and
and the endpoints of
η(e, v) = ∅
if
v
e.
is not an endpoint of
e.
CHAPTER 2. TRIGRAPHS
•
Construct
η(f, x)
G
with
V (G) =
S
e∈E
η(g, x)
is strongly complete to
is an endpoint of both
f
g,
and
η(e), G|η(e) = Ge \Ye
for all
f, g ∈ E
then the sets
η(f, x)
for all
and
and
e ∈ E
x∈V
η(g, y)
and such that
x
(in particular if
are nonempty and
strongly complete to each other).
H
Moreover, we call the graph
skeleton
of
G
Let
trigraph
G,
e
and we say that
used in the construction of a linear interval join
has been
replaced
by
strong cliques
X1 , . . . , Xn
(S1)
S
(S2)
Xi 6= ∅ ∀ i.
(S3)
Yi
i (Xi
is
and
n ≥ 4, V (G)
Y1 , . . . , Yn
the
(Ge , Ye ).
be a circular interval trigraph. The trigraph
if, for some even integer
G
G
is a
structured circular interval
can be partitioned into pairwise disjoint
such that (all indices are modulo
n):
∪ Yi ) = V (G).
strongly
complete
to
Xi
Xi+1
and
and
strongly
anticomplete
to
V (G)\ (Xi ∪ Xi+1 ∪ Yi ).
(S4) If
Yi 6= ∅
then
Xi
(S5) Every vertex in
(S6)
Xi
to
is strongly complete to
Xi
Xi+1 .
has at least one neighbor in
is strongly complete to
Xi+1
or
Xi−1
Xi+1
and at least one neighbor in
Xi−1 .
and possibly both, and strongly anticomplete
V (G)\(Xi ∪ Xi−1 ∪ Xi+1 ∪ Yi ∪ Yi−1 ).
A bubble
(G, Z) is said to be a structured bubble
if
G is a structured circular interval trigraph.
We need to dene one important class of Berge circular interval trigraphs. Let
graph with vertex set the disjoint union of sets
such that
|Bij | ≤ 1
for
1 ≤ i, j ≤ 3
•
For
i = 1, 2, 3, Bi1 ∪ Bi2 ∪ Bi3
•
For
i = 1, 2, 3, Bii
•
For
1 ≤ i, j ≤ 3
•
For
i = 1, 2, 3, Bii+1
{a1 , a2 , a3 }, B11 , B12 , B13 , B21 , B22 , B23 , B31 , B32 , B33
with adjacency as follows (all additions are modulo 3):
is a strong clique.
is strongly complete to
with
i 6= j , Bij
and
are both nonempty then
G be a tri-
i+1
Bi+2
Bii+1
S3
k
k=1 (Bi+1
k ).
∪ Bi+2
is strongly complete to
S3
k
k=1 Bj .
are either both empty or both nonempty, and if they
is not strongly complete to
12
i+1
Bi+2
.
CHAPTER 2. TRIGRAPHS
•
i = 1, 2, 3, ai is strongly
S3
k
plete to
k=1 Bi+2 .
For
• a1
is antiadjacent to
•
If
•
There exist
a1
a3 ,
is semiadjacent to
complete to
and
a3
a2
then
S3
k
k=1 (Bi
k )
∪ Bi+1
and
ai
is strongly anticom-
{a1 , a3 }.
is strongly anticomplete to
B31 ∪ B21 = ∅.
xi ∈ V (G) ∩ (Bi1 ∪ Bi2 ∪ Bi3 )
for
i = 1, 2, 3,
such that
{x1 , x2 , x3 }
is a
triangle.
We dene
in
C
C
to be the class of all such trigraphs
G.
We will prove in 3.2.7 that all trigraphs
are Berge and circular inteveral. Moreover we dene
i ∈ {1, 2, 3}
such that there exists
i+2
Bi+1
∪ Bii+2 = ∅
and such that
with
z ∈ Xai , H
C0
to be the set of all pairs
(H, {z})
is a thickening of a trigraph in
N (z) ∩ (Xai+1 ∪ Xai+2 ) = ∅
(with
Xai
C
with
as in the denition of
a thickening).
A
cycle
be
signing
C
is
of a graph
v(C) =
evenly signed
(G, s)
P
by
e∈C
s
G = (V, E)
s(e).
s : E → {0, 1}.
The value
v(C)
of a
A graph, possibly with multiple edges and loops, is said to
C
if for all cycles
is said to be an
is a function
in
G, C
has an even value, and in that case the pair
evenly signed graph.
We need to dene three classes of graphs that are going to play an important role in the
structure of claw-free perfect graphs.
F1 :
Let
s
of
(G, s) be a pair of a graph G (possibly with multiple edges and loops) and a signing
G
such that:
• V (G) = {x1 , x2 , x3 },
•
there is an edge
with
•
if
e
F2 :
Let
s
of
all
between
xi
and
xj
with
s(eij ) = 1
for all
{i, j} ⊂ {1, 2, 3}
i 6= j ,
and
We dene
eij
F1
f
are such that
s(e) = s(f ) = 0,
to be the class of all such pairs
then
e
is parallel to
f.
(G, s).
(G, s) be a pair of a graph G (possibly with multiple edges and loops) and a signing
G
such that
e ∈ E(G).
|V (G)| = 4|,
We dene
F2
all pairs of vertices of
G
are adjacent and
to be the class of all such pairs
13
(G, s).
s(e) = 1
for
CHAPTER 2. TRIGRAPHS
F3 :
Let
s
of
(G, s) be a pair of a graph G (possibly with multiple edges and loops) and a signing
G
such that:
• V (G) = {x1 , x2 , . . . , xn }
•
there is an edge
• {x1 , x2 }
if
s(e) = 0,
We dene
An
A
of
F3
and
x2
with
s(e) = 1,
{x3 , . . . , xn },
is a stable set,
then
e
is an edge between
x1
is a pair
(G, s)
of a graph
is either a member of
Here is a construction; a trigraph
G
and
x2 .
(G, s).
to be the class of all such pairs
even structure
G, (A, s|E(A) )
x1
between
is complete to
• {x3 , . . . , xn }
•
e12
n ≥ 4,
with
G
and a signing
F1 ∪ F 2 ∪ F 3
s
such that for all blocks
or an evenly signed graph.
that can be constructed in this manner is called an
evenly structured linear interval join.
•
Let
H = (V, E)
•
Let
η : (E × V ) ∪ E → 2V (G) .
•
For every edge
•
and the signing
e = x1 x2 ∈ E
be an even structure.
(where
x1 = x2
if
e
is a loop),
(Ge , Ye )
be:
∗
a spot if
e
∗
a thickening of a spring if
∗
a trigraph in
∗
either a spot or a thickening of a linear interval stripe if
Let
is in a cycle,
C0
if
e
Let
φe
Let
η(e, xj ) = N (φe (xj ))
Let
η(e) = V (Ge )\Ye .
Construct
x1 6= x2
e
and
s(e) = 1,
is in a cycle,
x1 6= x2 ,
be a bijection between the endpoints of
G
with
V (G) =
is complete to
and
s(e) = 0,
is a loop,
η(f, x)
s
S
η(g, x)
for
e∈E
j = 1, 2
and
e
and
η(e, v) = ∅
η(e), G|η(e) = Ge \Ye
for all
f, g ∈ E
14
and
e
is not in a cycle.
Ye .
if
v
is not an endpoint of
for all
x ∈ V
e ∈ E
e.
and such that
(in particular if
x
is an
CHAPTER 2. TRIGRAPHS
endpoint of both
f
and
g,
then the sets
η(f, x)
and
η(g, y)
are nonempty and strongly
complete to each other).
As for the linear interval join, we call the graph
structured linear interval join
(Ge , Ye )
G
the
skeleton
of
.
15
H
G,
used in the construction of an evenly
and we say that
e
has been
replaced
by
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
Chapter 3
The Structure of Claw-Free Perfect
Graphs
The class of claw-free perfect graphs has been extensively studied in the past.
structural result for this class was obtained by Chvátal and Sbihi [15].
The rst
In particular, in
1998, they proved that every claw-free Berge graph can be decomposed via clique-cutsets
into two types of graphs: 'elementary' and 'peculiar'.
clique-cutset A, B
if
A and B
and there is no edge between
are subset of
A\B
classical technique to decompose
and
We say that a graph
V (G) such that A ∩ B
B\A.
If a graph
G into G|A and G|B .
is a clique,
G
admits a
A ∪ B = V (G)
G admits a clique-cutset A, B , it is a
The structure of peculiar graphs was
determined precisely by their denition, but that was not the case for elementary graphs.
In 1999, Maray and Reed [26] proved that an elementary graph is an augmentation of the
line graph of a bipartite multigraph, thereby giving a precise description of all elementary
graphs. Their result, together with the result of Chvátal and Sbihi, gave an alternative proof
of Berge's Strong Perfect Graph Conjecture for claw-free Berge graphs (the rst proof was
due to Parthasarathy and Ravindra [30]).
However, this still does not describe the class
of claw-free perfect graphs completely, as gluing two claw-free Berge graphs together via a
clique-cutset may introduce a claw. In this chapter we use trigraphs and a previous result
on claw-free graphs by Chudnovsky and Seymour [14] to obtain a full characterization of
claw-free perfect graphs. That is, we give an explicit construction describing all claw-free
16
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
perfect graphs, using a technique that generalizes the construction of line graphs.
This chapter is organized as follows. In Section 3.1, we prove a few introductory results
and state our main theorem 3.1.4.
The proof of 3.1.4 is broken down in several cases
depending on the underlying structure of the graph. Each section analyzes a dierent case
of the main theorem. In Section 3.2, we study circular interval trigraphs that contain special
triangles. Section 3.3 examines circular interval trigraphs that contain a hole of length 4
while Section 3.4 covers the case when a circular interval trigraph contains a long hole. In
Section 3.5, we analyze linear interval joins. Finally, in Section 3.6, we gather our results
and prove 3.1.4.
3.1
Preliminary results
We start by proving two easy facts.
3.1.1. Every claw-free Berge trigraph is a quasi-line trigraph.
Proof.
Let
G
deduce that
v ∈ V (G).
be a claw-free Berge trigraph and let
G|N (v)
does not contain a triad.
does not contain a odd antihole. Thus
G|N (v)
Since
G
Since
G
is claw-free, we
is Berge, we deduce that
is cobipartite and it follows that
G|N (v)
N (v)
is the
union of two strong cliques. This proves 3.1.1.
3.1.2. Let G be a trigraph and H be a thickening of G. If F is a thickening of H then F is
a thickening of G.
Proof.
For
For
v ∈ V (H),
v ∈ V (G),
v ∈ V (G),
let
let
let
XvH
XvF
be the strong clique in
be the strong clique in
Yv ⊆ V (F )
be dened as
H
Yv =
nonempty, pairwise disjoint and their union is
deduce that
(resp.
H
Yv
is a strong clique for all
antiadjacent) in
and thus
Yu
semiadjacent in
G,
then
XuH
F
as in the denition of a thickening.
as in the denition of a thickening. For
S
y∈XvH
V (F ).
v ∈ V (G).
If
XyF .
Clearly, the sets
Since
XvH
is strongly complete (resp. anticomplete) to
G,
then
XuH
Yv
are strongly adjacent
anticomplete) to
in
are all
is a strong clique, we
u, v ∈ V (G)
is strongly complete (resp.
Yv
F.
If
XvH
u, v ∈ V (G)
is neither strongly complete nor strongly anticomplete to
17
in
are
XvH
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
in
H
and hence
Yu
is neither strongly complete nor strongly anticomplete to
Yv
in
F.
This
proves 3.1.2.
The following theorem is the main characterization of quasi-line graphs [14]. It is the
starting point of our structure theorem for claw-free perfect graphs.
3.1.3. Every connected quasi-line trigraph G is either a linear interval join or a thickening
of a circular interval trigraph.
We can now state our main theorem:
3.1.4. Every connected Berge claw-free trigraph is either an evenly structured linear interval
join or a thickening of a trigraph in C .
The goal of this chapter is to prove 3.1.4, but rst we can prove an easy result about
evenly signed graphs. Here is an algorithm that will produce an even signing for a graph:
Algorithm 1
•
Let
•
Arbitrarily assign a value from
•
For every
T
be a spanning tree of
G
e = xy ∈ E(G)\T ,
is the path from
r
to
i
in
and root
{0, 1}
let
T
at some
s(e)
to
P
s(e) =
r ∈ V (G).
for all
f ∈Px
e ∈ T.
s(f ) +
P
f ∈Py
s(f ) (mod 2)
where
Pi
T.
3.1.5. Algorithm 1 produces an evenly signed graph (G, s).
Proof.
Let
C
G.
be a cycle in
First, we notice that for an edge
e in T , s(e) can be expressed
with the same formula used to calculate the signing of an edge outside of
that for all
e ∈ E(G), s(e) =
P
f ∈Px
s(f ) +
P
f ∈Py
s(f ) (mod 2).

X
X
s(e) =
e=xy∈E(C)

X

s(e) +
e∈Px
xy∈E(C)
X
s(e) =
e∈Py
!
=2·
X
X
x∈V (C)
e∈Px
s(e)
which concludes the proof of 3.1.5.
18
Thus,
=0
(mod 2)
T.
In fact we have
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
The result of 3.1.5 shows that if we have a graph
(G, s)
we can nd all signings
s
such that
is an evenly signed graph by using Algorithm 1 with all possible assignments for
on the tree
3.2
G,
s(e)
T.
Essential Triangles
In order to prove 3.1.4, we rst prove the following:
3.2.1. Let G be a Berge circular interval trigraph. Then either G is a linear interval trigraph,
or a cobipartite trigraph, or a thickening of a member of C , or G is a structured circular
interval trigraph.
Before going further, more denitions are needed. Let
dened by
Σ
essential
is
and
F1 , . . . , Fk ⊆ Σ.
if there exist
Let
T = {c1 , c2 , c3 }
i1 , i2 , i3 ∈ {1, . . . , k}
c3 , c1 ∈ Fi3 , and such that Fi1 ∪ Fi2 ∪ Fi3 = Σ.
Σqx,y
the subset of
and
φ(y) = 1
G
be a triangle.
such that
Let
x, y, q
be a circular interval trigraph
We say that
c1 , c2 ∈ Fi1 , c2 , c3 ∈ Fi2
be three points of
Σ.
T
and
We denote by
Σ such that there exists a homeomorphism φ : Σqx,y → [0, 1] with φ(x) = 0
and such that
q ∈ Σqx,y .
Moreover let
q
Σx,y = (Σ\Σqx,y ) ∪ {x, y}.
The following two lemmas are basic facts that will be extensively used to prove 3.2.1.
3.2.2. Let G be a circular interval trigraph dened by Σ and F1 , . . . , Fk . Let x, y, a, b ∈ V (G)
y
such that x ∈ Σa,b . If x is antiadjacent to a and b, then y is strongly antiadjacent to x.
Proof.
Assume not.
x, y ∈ Fi .
Since
x
is adjacent to
It follows that at least one of
but it implies that
a
y,
a, b ∈ Fi∗ .
is strongly adjacent to
x,
we deduce that there exists
Fi
such that
By symmetry we may assume that
a ∈ Fi∗ ,
a contradiction. This proves 3.2.2.
3.2.3. Let G be a circular interval trigraph dened by Σ and F1 , . . . , Fk . Let x, y, z ∈ V (G)
such that x is adjacent to y and x is antiadjacent to z . Then there exists Fi such that
z
Σx,y ⊆ Fi .
Proof.
Since
x
we deduce that
is adjacent to
z∈
/
y
there is
Fi
such that
z
Fi∗ . Thus we conclude that Σx,y
19
x, y ∈ Fi .
⊆ Fi .
Since
z
is antiadjacent to
This proves 3.2.3.
x,
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
3.2.4. Let G be a circular interval trigraph dened by Σ and F1 , . . . , Fk , and let C = c1 −
c2 − . . . − cn − c1 be a hole. Then the vertices of C are in order on Σ.
Proof.
Assume not. By symmetry, we may assume that
and thus we may assume that
{c1 , c3 },
complete to
c
Σc31 ,c2
c4 ∈ Σcc21 ,c3 .
c3
Since
we deduce that there exist
Fi
c1 , c2 , c3 , c4
are not in order on
is antiadjacent to
and
Fj ,
possibly
c1
and since
Fi = Fj ,
c2
Σ,
is
such that
c
Fi and Σc12 ,c3
c
⊆
⊆ Fj . If c4 ∈ Σc31 ,c2 , then since c4 ∈ Fi∗ , we deduce that c4 is strongly
c1
∗
complete to {c1 , c2 }, a contradiction. If c4 ∈ Σc ,c , then since c4 ∈ Fj , we deduce that c4
3 2
{c2 , c3 },
is strongly complete to
a contradiction. This proves 3.2.4.
3.2.5. Let G be a circular interval trigraph dened by Σ and F1 , . . . , Fk . If G is not a linear
interval trigraph, then there exists an essential triangle or a hole in G.
Proof.
Fi1
Let
y
that Σx ,x
0 1
Let
x2
∩ Fi1
k>1
be such that
j=3
such that
Hence
to
x1 .
x0 .
k = 3.
k > 3.
C
xj−1 ∈
/ Fi1 ,
Clearly
Clearly
By the choice of
By the choice of
by 3.2.2,
and while
T = {x0 , x1 , x2 }
Assume now that
adjacent to
Let
x0 , x1 ∈ Fi1
such
let
G
and
xj
x
Σx01 ,x2
and
F ij
is maximal.
be such that
xj ∈ Fij , xj ∈
/ Fik ,
is not a linear interval trigraph, there exists
xk ∈ Fi1 .
Assume rst that
x0 , x2 ∈ Fi3 .
x 2 ∈ F i2 , x 2 ∈
/ Fi1
x1
is maximal. Since
and Σx
j−1 ,xj
k<j
y∈
/ Fi1 .
is maximal and let
is maximal.
Fi2
and
Starting with
for any
Fi1 ∩ V (G)
be such that
Fij , xj−1
Fi1
Fi1 ∪ Fi2 ∪ Fi3 = Σ, x0 , x1 ∈ Fi1 , x1 , x2 ∈ Fi2
and
is an essential triangle.
xj−1
and
is adjacent to
xj
for
j = 1, . . . , k − 1
x0 , x1 , we deduce that xk−1
is antiadjacent to
xj+1
mod k for all
and
xk−1
is
is strongly antiadjacent
j = 1, . . . , k − 1.
Hence
is a hole. This concludes the proof of 3.2.5.
3.2.6. Let G be a circular interval trigraph and C a hole. Let x ∈ V (G)\V (C), then x is
strongly adjacent to two consecutive vertices of C .
Proof.
Let
G
there exists
j
be dened by
such that
Σ
c
cj
and
cj+1 .
F1 , . . . , Fk
x ∈ Σcj+2
.
j ,cj+1
we deduce that there exists
adjacent to
and
Since
i ∈ {1, . . . , k}
cj
and let
C = c1 − c2 − . . . − cl − c1 .
is adjacent to
such that
This proves 3.2.6.
20
c
cj+1
By 3.2.4,
and antiadjacent to
Σcjj+2
,cj+1 ⊆ Fi .
Hence
x
cj+2 ,
is strongly
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
In the remainder of this section, we focus on circular interval trigraphs that contain an
essential triangle. For the rest of the section, addition is modulo 3.
3.2.7. Every trigraph in C is a Berge circular interval trigraph.
Proof.
Let
G
be in
C.
We let the reader check that
G
is indeed a circular interval trigraph,
it can easily be done using the following order of the vertices on a circle:
B13 , B11 , B12 , a1 , B21 , B22 , B23 , a2 , B32 , B33 , B31 , a3
There is no odd hole in G.
(1)
Assume there is an odd hole
to
C = c1 −c2 −. . .−cn −c1
V (G)\{ai+1 }, it follows that V (C)∩Bii = ∅ for all i.
is a cobipartite trigraph, we deduce that
Assume rst that
c2 = a3 .
and
Since
Symmetrically,
Hence,
c3
cn
a1 , a3
is adjacent to
c3 ∈ B31 ∪ B32 .
cn ,
and
As
is strongly adjacent to
c1
is complete to
of generality, let
c2 ∈ Bii+2
that
cn−1 = ai+1 .
with
ai+1
in
C,
and antiadjacent to
a1
is semiadjacent to
and
and
is strongly complete
G|(B12 ∪B13 ∪B21 ∪B23 ∪B31 ∪B32 )
C.
We may assume that
c1 = a1
c2 , we deduce that cn ∈ B21 ∪ B23 .
a3 ,
it follows that
{c2 , cn }∩{a1 , a2 , a3 } = ∅.
we deduce that
i+2
cn ∈ Bi+1
.
Since
Symmetrically, we deduce that
we deduce that
Bii
B21 ∪ B31 = ∅.
a contradiction.
c1 = ai
{c2 , cn },
Since
|{a1 , a2 , a3 } ∩ V (C)| ≥ 1.
c1
cn ,
G.
Since
are two consecutive vertices of
Thus, we may assume that
to
in
n > 5.
But
Since
c2
is antiadjacent
i+2
{c2 , cn } = Bii+2 ∪ Bi+1
.
cn−1
is antiadjacent to
c3 = ai+2 .
Since
|{x ∈ V (G) : x
ai+2
Without loss
c2 ,
we deduce
is not consecutive
antiadjacent to
c2 }| ≤ 2,
a
contradiction. This proves (1).
(2)
There is no odd antihole in G.
Assume there is an odd antihole
that
C
has length at least
V (C) ∩ Bii = ∅
for all
Assume rst that
B33 )| = 7,
7.
Bii
in
is strongly complete to
G.
By (1), we may assume
V (G)\{ai+1 },
it follows that
i.
a1
is semiadjacent to
we deduce that
two neighbors in
Since
C = c1 − c2 − . . . − cn
a3 .
Then
Since
|V (G)\(B11 ∪ B22 ∪
V (C) = ({a1 , a2 , a3 } ∪ B12 ∪ B13 ∪ B32 ∪ B23 ).
({a1 , a2 , a3 } ∪ B12 ∪ B13 ∪ B32 ∪ B23 ),
strongly antiadjacent to
B31 ∪ B21 = ∅.
a3 .
21
But
a2
has only
a contradiction. This proves that
a1
is
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
|V (C) ∩ {a1 , a2 , a3 }| = 1. We may assume that a1 ∈ V (C) and
S
S
k
k
follows that V (C) = {a1 } ∪
j6=k Bj . But G|({ai } j6=k Bj ) is not an antihole of length
S
2
k
since the vertex of B1 has 5 strong neighbors in ({ai }
j6=k Bj ), a contradiction.
Assume now that
Hence we may assume that
deduce that
and
|V (C) ∩ {a1 , a2 , a3 }| ≥ 2.
|C ∩ {a1 , a2 , a3 }| = 2
a3 ∈
/ C.
B12 ∪ B13
But since
a1 ,
anticomplete to
Since there is no triad in
we deduce that
{c4 , c5 , c6 } ⊆ B21 ∪ B23 ,
a2
and
7,
we
c1 = a1 , c2 = a2
and by symmetry we may assume that
is strongly anticomplete to
C,
it
B31 ∪ B32
is strongly
a contradiction. This proves (2).
Now by (1) and (2), we deduce 3.2.7.
3.2.8. Let G be a Berge circular interval trigraph such that G is not cobipartite. If G has
an essential triangle, then G is a thickening of a trigraph in C .
Proof.
Let
{x1 , x2 , x3 }
be an essential triangle and let
x2 ∈ F1 ∩ F2 , x3 ∈ F2 ∩ F3
(1)
and
F1 , F2 , F3
be such that
x1 ∈ F1 ∩ F3 ,
F1 ∪ F2 ∪ F3 = Σ.
xi is not in a triad for i = 1, 2, 3.
x1
Assume
x1 ∈ F1 ∩ F3 ,
is in a triad.
y, z
Then there exist
we deduce that
y, z ∈ F2∗
and so
y
such that
{x1 , y, z}
is strongly adjacent to
is a triad.
z,
Since
a contradiction.
This proves (1).
By (1) and as
assume that
G is not a cobipartite trigraph, there exists a triad {a∗1 , a∗2 , a∗3 } and we may
a∗i ∈ Fi \(Fi+1 ∪ Fi+2 ), i = 1, 2, 3.
0
such that ai , ai are in triads and
Ai+2 ), Ai = V (G) ∩ Ai
vertex in
(2)
B1 ∪ B2 ∪ B3
ai ∈ Fi ∩ Σxa∗i ,a∗
a∗
Σaii ,a0 is maximal. Let
i
Bi = V (G) ∩ Bi .
i
and
i+2
a∗i
Ai = Σai ,a0 , Bi =
i
By the denition of
x
a0i ∈ Fi ∩ Σa∗i+1
,a∗
i
i+1
Σxa∗i ,a∗ \(Ai
i i+2
a1 , a2 , a3 , a01 , a02 , a03 ,
∪
no
is in a triad.
{a1 , a2 , a3 } and {a01 , a02 , a03 } are triads.
By the denition,
ai ∈ Ai , i = 2, 3.
{a1 , a2 , a3 }
(3)
and
Let
and
a1
is in a triad.
By 3.2.2,
{a01 , a02 , a03 }
a1
Let
{a1 , a2 , a3 }
is non adjacent to
a3 .
be a triad, then we assume that
Now, using symmetry, we deduce that
are triads. This proves (2).
For all x ∈ Ai there exist y ∈ Ai+1 , z ∈ Ai+2 such that {x, y, z} is a triad.
22
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
x ∈ A1 .
By symmetry, we may assume that
If
|A1 | = 1,
x = a∗1
then
and
{a∗1 , a∗2 , a∗3 }
is
a triad.
Therefore, we may assume that
a3 .
5,
length
a2
is strongly antiadjacent to
we deduce that
{x, a2 , a3 }
x
a03 .
x
By (2) and 3.2.2,
{x, a02 , a3 } is not a triad, then a02
We may assume that
and 3.2.2,
a1 6= a01 .
Since
is antiadjacent to
is strongly adjacent to
x − a2 − a02 − a3 − a03 − x
{a2 , a03 }.
is not strongly complete to
a3 .
a02
and
By (2)
is not a hole of
But now one of
{x, a02 , a03 },
is a triad. This proves (3).
{x1 , x2 , x3 } is a triangle such that xi ∈ Bi for i = 1, 2, 3.
(4)
By (3),
xi ∈
/ A1 ∪ A2 ∪ A3
for
i = 1, 2, 3.
{x1 , x2 , x3 }
for
i = 1, 2, 3.
(5)
(A1 , A2 , A3 |B1 , B2 , B3 ) is a hex-join.
Moreover,
By the denition of
Bi ) = V (G).
Clearly
If there exist
Ai
such that
Bi
yi
is a strong clique for
a∗2
a∗1 ,a∗3 ,
B1 ⊂ Σ
S
i (Ai ∪
Ai ⊂ Fi , i = 1, 2, 3.
is antiadjacent to
By symmetry, it remains to show that
Since
xi ∈ Bi
is an essential triangle. This proves (4).
is a strong clique as
yi , yi0 ∈ Bi
A1 .
it follows that
A1 , A2 , A3 , B1 , B2 , B3 , they are clearly pairwise disjoint and
by 3.2.2, a contradiction. Thus
complete to
Bi ,
By the denition of
B1
yi0 ,
then
{yi , yi0 , a∗i+1 }
i = 1, 2, 3.
is strongly anticomplete to
we deduce that
is a triad
B1
A2
and strongly
is strongly anticomplete to
A2
by 3.2.2 and (3).
Suppose there is
a2 ∈ A2
and
b1 ∈
and
a 3 ∈ A3
a 1 ∈ A1
be such that
such that
a1
{a1 , a2 , a3 } is a triad.
b1 ∈ B1 .
b1
is antiadjacent to
Since
a2
b1 .
By (3), let
is anticomplete to
is strongly antiadjacent to
a2 .
Thus
{a1 , a3 },
{a1 , a2 , b1 }
This concludes the proof of (5).
There is no triangle {a1 , a2 , a3 } with ai ∈ Ai , i = 1, 2, 3
Let
ai ∈ Ai , i = 1, 2, 3
ci ∈ Ai , i = 2, 3
c3 ∈
to
b1 ∈ B1
a
Σa21 ,a3 , we deduce by 3.2.2 that
is a triad, a contradiction as
(6)
and
a
Σa12 ,a3 . Since
a2 .
such that
be such that
{a1 , c2 , c3 }
{a2 |a1 , c2 , c3 }
By (2) and as
a2 ∈
a0
Σa30 ,a0 ,
2 1
a1
ai , i = 2, 3.
is adjacent to
is a triad.
By 3.2.3,
c2 ∈
is not a claw, we deduce that
a03
is antiadjacent
23
a2 .
Since
c3
a
Σa12 ,a3 .
By (3), let
By symmetry,
is strongly antiadjacent
a
a3 ∈ Σc32,a03
and by (2),
a3
is
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
a2 .
strongly antiadjacent to
This proves (6).
For the rest of the proof of 3.2.8, let
{j, k, l} = {1, 2, 3}.
There is no induced 3-edge path w − x − y − z such that w ∈ Aj , x, y ∈ Ak , z ∈ Al .
(7)
Assume that
z ∈ A3 .
w−x−y−z
Now by (5),
is an induced 3-edge path such that
w − x − y − z − x1 − w
is a hole of length
5,
w ∈ A1 , x, y ∈ A2 ,
a contradiction. This
proves (7).
For i = 1, 2, 3, let yi ∈ Ai . Then yk is strongly antiadjacent to at least one of yj , yl .
(8)
Suppose there exist
yi ∈ Ai , i = 1, 2, 3 such that y2
strongly antiadjacent to
z3
and
that
is antiadjacent to
y1
y3 .
By (3), there exist
y3 .
Since
is strongly adjacent to
z3 ,
is adjacent to
z1 , z3 ∈ A2
such that
y1
and
y3 .
z1
is antiadjacent to
By (6),
y1
is
y1
{y2 |y1 , y3 , z3 } and {y2 |y1 , y3 , z1 } are not claws, we deduce
and
y3
is strongly adjacent to
z1 .
But
y1 − z − 3 − z1 − y3
is a 3-edge path, contrary to (7). This proves (8).
(9)
Aj is strongly anticomplete to at least one of Ak , Al .
Assume not. By symmetry, we may assume there are
such that
y
x
is adjacent to
y
and
is strongly antiadjacent to
But now
(10)
x−y−z−w
w,
z
is adjacent to
and
z
w.
x ∈ A1 , y, z ∈ A2
By (8),
x
and
w ∈ A3
is strongly antiadjacent to
is strongly antiadjacent to
x;
and in particular
w,
y 6= z .
is am induced 3-edge path, contrary to (7). This proves (9).
For i = 1, 2, 3, let bi ∈ Bi such that bk is adjacent to bl . Then bj is strongly adjacent
to at least one of bk , bl .
By symmetry, we may assume that
is not a hole of length
b2 , b3 .
(11)
z,
by (5) we deduce that
b1
Since
b1 −a∗3 −b3 −b2 −a∗1 −b1
is strongly adjacent to at least one of
This proves (10).
Let x ∈ Bj , then x is strongly complete to one of Bk , Bl .
Assume there is
to
5,
j = 1, k = 2 and l = 3.
then
x
y ∈ Bk
such that
is strongly adjacent to
z
x
is antiadjacent to
since
{x, y, z}
24
y.
Let
z ∈ Bl .
If
y
is antiadjacent
is not a triad. By (10), if
y
is strongly
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
adjacent to
z,
then
x
is strongly adjacent to
z.
Thus
x
Bl .
is strongly complete to
This
proves (11).
By (9) and symmetry, we may assume that
Let
Bii
be the set of all vertices of
Bi
j 6= i, let Bij be the set of all vertices of
S3
j
we deduce that Bi =
j=1 Bi .
(12)
A2
is strongly anticomplete to
Bi+1 ∪ Bi+2 .
that are strongly complete to
Bi \Bii
A1 ∪ A3 .
For
Bj .
By (11),
Bj ∪ Bk ,
contrary
that are strongly complete to
If Bjk = ∅, then Blk = ∅.
Assume that
Bjk
of the denition of
Bll
Blk .
and
Now, we observe that
G
A2 , B11 , B22 , B33
is a thickening of a member of
anticomplete to
that either
Since
G
are homogeneous sets and
If
A1
C.
(A1 , A3 ), (B12 , B32 ),
is strongly anticomplete to
By (12), it follows that
is a thickening of a member of
C.
A3 ,
A1
Thus, we may assume that
A1 − A3 − B31 − A2 − B21 − A1
B21 = ∅ or B31 = ∅.
we deduce that
3.3
A3 .
is strongly complete to
This proves (12).
(B23 , B13 ), (B31 , B21 ) are homogeneous pairs.
and (12),
Blk
is empty. It implies that
is not strongly
is not a hole of length
B21 ∪ B31
then by (4)
5,
we deduce
is empty. Using (4) and (12),
This concludes the proof of 3.2.8.
Holes of Length 4
Next we examine circular interval trigraphs that contain a hole of length 4.
3.3.1. Let G be a Berge circular interval trigraph. If G has a hole of length 4 and no
essential triangle, then G is a structured circular interval trigraph.
Proof.
In the following proof, the addition is modulo 4.
F1 , . . . , Fk .
Let
x∗1 − x∗2 − x∗3 − x∗4 − x∗1
Let
be a hole of length
G
4.
be dened by
Σ
and
We may assume that
x∗i , x∗i+1 ∈ Fi , i = 1, 2, 3, 4.
(1)
x∗i is strongly antiadjacent to x∗i+2 .
Assume not.
By symmetry we may assume that
may assume that there exists
i ∈ {1, . . . , k}
x∗1
such that
25
is adjacent to
Σ
x∗2
x∗1 ,x∗3
⊆ Fi .
x∗3 .
If
Moreover, we
i = 4,
it implies
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
that
{x∗1 , x∗2 , x∗3 , x∗4 } ⊂ F4 ,
and thus
Symmetrically, we may assume that
Fi ∪ F3 ∪ F4 = Σ,
For
i 6= 3.
But now
{x∗1 , x∗3 , x∗4 } is an essential triangle since
Xi , Yi
let
Xi , Yi ⊂ Σ
Xi , Yi ⊂ V (G)
and
is homeomorphic to
Xi ⊆ V (G) ∩ Xi , Yi ⊆ V (G) ∩ Yi , i = 1, 2, 3, 4,
(H3)
S
(H4)
X1 , X2 , X3 , X4 , Y1 , Y2 , Y3 , Y4
(H5)
Yi ⊆ Σx∗i+2
,x∗ , i = 1, 2, 3, 4,
(H6)
x∗i ∈ Xi , i = 1, 2, 3, 4,
(H7)
X1 , X2 , X3 , X4 , Y1 , Y2 , Y3 , Y4
(H8)
S
∪ Yi ) = Σ ,
i (Xi
are pairwise disjoint,
x∗
i
i (Xi
i+1
∪ Yi )
are disjoints strong cliques satisfying (S2)-(S6),
is maximal.
By (1), such a structure exists. We may assume that
x ∈ V (G)\
S
complete to
x,
For
i (Xi
∪ Yi ).
and by
i = 1, 2, 3, 4,
such that
Σ
x∗i+1
xli ,xri
let
S ⊆ V (G)\{x},
the subset of
xli , xri ∈ Xi
S
V (G)\
S
we denote by
i (Xi
∪ Yi )
SC
the subset of
that is anticomplete to
be such that
x∗i−1 , xli , xri , x∗i+1
is not empty. Let
S
that is
x.
are in this order on
Σ
and
is maximal.
{xri , xli+1 } is complete to Xi ∪ Xi+1 .
exists
Fl
such that
is complete to
Xi .
a ∈ Xi
such that
{a, xri } ∪ Xi+1 ⊆ Fl
a
is adjacent to
and thus
xri
xri+1 .
is complete to
By 3.2.3 and (S6), there
Xi+1 .
By symmetry,
xli+1
This proves (2) by (H7).
If Xi is not complete to Xi+1 , then xli is strongly antiadjacent to xri+1 .
Let
a
For
SA
By (S5), there exists
(3)
be such that:
[0, 1),
(H2)
(2)
is not a hole, a contradiction.
a contradiction. This proves (1).
i = 1, 2, 3, 4,
(H1) each of
x∗1 − x∗2 − x∗3 − x∗4 − x∗1
a ∈ Xi
and
b ∈ Xi+1
be such that
is strongly antiadjacent to
xri+1 .
a
is strongly antiadjacent to
By 3.2.2 and (S6),
This proves (3).
26
xri+1
b.
By 3.2.2 and (S6),
is strongly antiadjacent to
xli .
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
xl
x∈
/ Σxli+1
for all i.
,xr
(4)
i
i
xl
x ∈ Σx2l ,xr .
Assume not. We may assume that
1
let
Xi0 = Xi
and let
X10 = X1 ∪ {x}.
Since
by (S3) and (S6), we deduce by 3.2.2 that
xr1
xr4
is adjacent to
X4 .
X1
clique. If
X2 .
X2 .
Since
X1
x
X2 ,
above arguments show that
x
X10 , . . . , X40 , Y10 , . . . , Y40
then
Xi , Yi i = 1, 2, 3, 4 clearly satisfy (H1)-(H5)
S
maximality of
i (Xi ∪ Yi ). This proves (4).
By (4) and by symmetry, we may assume that
x
is strongly anticomplete to
x4 .
adjacent to
exists
is strongly complete to
Xi0 , Yi0 i = 1, 2, 3, 4,
with
x∗
x ∈ Σx3r ,xl
Since
We may assume there exist
By (S6),
Fi , i ∈ {1, . . . , k},
j ∈ {1, . . . , k}
Fi 6= Fj .
such that
By (S6),
x3
and therefore
2
x ∈ F1 ,
{x, x3 , x4 }
⊆
x3 ∈ X3C
is strongly antiadjacent to
such that
x, x3 ∈ Fi
x, x4 ∈ Fj
and
The
contrary to
x ∈ F1 .
we deduce that
and
x∗2 ∈
/ Fj .
and
x∗1
x
By 3.2.2
is strongly
x∗1 ∈
/ Fi .
x3
Fi , Σx,x
4
⊆
x
Fj and Σx3 ,x4
Fl
such that
x3
By symmetry, there exists
is
we deduce that
Now, since
x3 is adjacent to
x3 , x4 ∈ Fl
we deduce that
Fj ,
x∗2 ∈ Fi ,
Moreover, as
such that
⊆ Fk ,
x4 ∈ X4C
and therefore by 3.2.3 there
x∗i is strongly anticomplete to xi+2 for i = 1, 2.
we deduce from 3.2.3 that there exists
x4
Since Σx,x
3
(6)
is a strong
X3C is strongly anticomplete to X4C .
Assume not.
x4 ,
X10
Y1 .
complete to
(5)
Y3 .
and not strongly
we deduce that
x
and not
x is strongly complete to X4 .
1
and (S3),
Y1
Y4
Since
are disjoint cliques satisfying (S2)-(S6).
Moreover,
the
xr1 ,
it follows from 3.2.3 that
X4 ,
Y2 ∪ Y3 ∪ X3 .
is strongly complete to
is strongly complete to
is strongly complete to
{xr1 , xl1 }
is strongly anticomplete to
is strongly anticomplete to
is strongly adjacent to
is strongly complete to
By symmetry, if
x
By symmetry,
xl1
i = 1, 2, 3, 4, let Yi0 = Yi , for i = 2, 3, 4,
Y2 ∪ Y3 ∪ X3
by (2), we deduce by 3.2.3 that
strongly anticomplete to
anticomplete to
For
1
and
l ∈ {1, . . . , k}\{i, j}.
Fi ∪ Fj ∪ Fk = Σ.
Hence,
is an essential triangle, a contradiction. This proves (5).
At least one of X3C , X4C is empty.
Assume not. Let
that there is
a ∈ X4C .
By 3.2.3 and since
Fi , i ∈ {1, . . . , k},
such that
a is strongly anticomplete to X2 , we deduce
{a, xr4 , x} ∈ Fi
27
and thus
xr4 ∈ X4C .
Symmetrically,
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
xl3 ∈ X3C .
and
X2
By (5),
xr4
is strongly antiadjacent to
is strongly complete to
by (2) and (S6),
X3 .
xl3 .
X1
By (S6),
x
By (2) and (5),
is strongly complete to
{xr3 , xl4 }.
is anticomplete to
X4 ,
But now
x − xl4 − xl2 − xr4 − xl3 − xr1 − xr3 − x is an antihole of length 7, a contradiction.
This proves (6).
By symmetry, we may assume that
x
is strongly complete to
is strongly anticomplete to
and
x is strongly anticomplete to X3 .
x − Y4 − xl4 − xl3 − X2 − x
anticomplete to
deduce that
Yi0 = Yi ,
X1
Y2 ∪ Y4 .
Since
x − X2 − X3 − X4 − X1 − x
is strongly complete to
and let
Y10 = Y1 ∪ {x}.
X2 .
For
Yi0 = Yi ,
show that
let
x
let
X10 = X1 ∪ Y4C
X10 , . . . , Xn0 , Y10 , . . . , Yn0
strongly complete to
X4 ,
then
X20
is complete to
X30 .
•
If
Y1 6= ∅
then
X10
is complete to
X20 .
For the former, we observe that if
X3C
x
i = 2, 3, 4,
X10 , . . . , Xn0 , Y10 , . . . , Yn0
Xi0 , Yi0 , i = 1, 2, 3, 4,
This proves (7).
let
Xi0 = Xi ,
X20 = X2 ∪ {x}.
for
i = 1, 2, 3,
The above arguments
and
satisfy (S4) and (S6).
Y30 = Y3 ,
and since
is not strongly complete to
is not a hole of length
is strongly complete to
is empty. Since
i = 3, 4,
for
X10 \X1 ⊂ Y4
is
it is enough to check the following:
Y2 6= ∅
x
and let
X30 = X3 , X40 = X4
If
since
For
∪ Yi ).
X10 , . . . , Xn0 , Y10 , . . . , Yn0
•
X3 − X4 − X1 − x
Y2 .
i (Xi
is strongly
are disjoint cliques satisfying (S2), (S3) and (S5). To get
a contradiction, it remains to show that
First we check (S4). Since
Xi0 = Xi ,
Moreover, it is easy to nd
is strongly complete to
Y40 = Y4A ,
let
The above arguments show that
S
x
is not a cycle of length 5, we
i = 1, 2, 3, 4,
satisfying (H1)-(H5), contrary to the maximality of
By 3.2.3 and (7),
x − Y2 − xr3 − xr4 − X1 − x
Since
are not holes of length 5, we deduce that
are disjoint cliques satisfying (S2)-(S6).
Y1
By (2) and 3.2.3,
X1 ∪ X2 .
Assume not. By 3.2.2,
let
X4 .
x is adjacent to xl3 .
(7)
let
x
X1 ,
5,
we deduce that
Y2
it is enought to show that if
is not empty, it follows that
is empty by 3.2.3 and (S4).
28
x∗
2
Y1 ⊆ Σx,x
∗.
1
X3 ,
then since
x − Y2 −
is empty. For the latter,
Y1
Now if
is not empty, then
Y4C
Y4C
is not empty, then
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
To check (S6), we need to prove the following:
(i) If
X10
is not strongly complete to
X20
then
X20
is strongly complete to
X30 .
(ii) If
X20
is not strongly complete to
X30
then
X30
is strongly complete to
X40 .
(iii) If
X30
is not strongly complete to
X40
then
X40
is strongly complete to
X10 .
(iv) If
X40
is not strongly complete to
X10
then
X10
is strongly complete to
X20 .
For (i), rst assume that
strongly anticomplete to
xr3 .
x is not strongly complete to X3 .
Since
x − xr2 − xr3 − X4 − Y4 − x
Y4C
are not cycles of length 5, we deduce that
to
X2 .
Thus
complete to
X10 = X1
X20 .
and since
x
x
X1 ,
is strongly complete to
X3
X30
is not strongly complete to
and thus
X20
x is
x − xr2 − xr3 − X4 − X1 − x
X1
is strongly complete
X10
it follow that
is strongly complete to
X2
it follows that
and
is empty and that
is strongly complete to
So we may assume that
By 3.2.2, we deduce that
X3 .
is strongly
By 3.2.3 and (S6),
is strongly complete to
X30 .
This
proves (i).
For (ii), if
antiadjacent to
xr4 .
X2
Moreover by (S4),
X40 ,
then by (3) it follows that
is strongly complete to
xr4 − X1 − x
is not a cycle of length 5, we deduce, using (2), that
X3
X30
and thus
is strongly complete to
X20 .
For (iii) and (iv), we may assume that
strongly complete to
implies that
X4
Y4 ,
we deduce that
is strongly complete to
follows. Also by (S6), we deduce that
follows that
Y4
complete to
X20 ,
is empty. Since
x
X1
x − xl3 − xr3 −
is strongly complete to
is not strongly complete to
is not strongly complete to
and thus
X40
X10 .
X1 .
is strongly complete to
X1 ,
X2 .
Since
X4
is
But by (S6), it
is strongly complete to
is strongly complete to
X30 , and (iii)
Moreover by (S4), it
we deduce that
X10
is strongly
and (iv) follows.
The above arguments show that
(S2)-(S6). Moreover, it is easy to nd
the maximality of
3.4
X3
x
Since
is strongly
This proves (ii).
X40
X4
X3 .
xl3
S
i (Xi
∪ Yi ).
X10 , . . . , Xn0 , Y10 , . . . , Yn0
Xi0 , Yi0 , i = 1, 2, 3, 4,
are disjoint cliques satisfying
satisfying (H1)-(H5), contrary to
This concludes the proof of 3.3.1
Long Holes
In this section, we study circular interval trigraphs that contain a hole of length at least 6.
29
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
A result equivalent to 3.4.1 has been proved independently by Kennedy and King [20].
The following was proved in joint work with Varun Jalan.
3.4.1. Let G be a circular interval trigraph dened by Σ and F1 , . . . , Fk ⊆ Σ. Let P =
p0 − p1 − . . . − pn−1 − p0 and Q = q0 − q1 − . . . − qm−1 − q0 be holes. If n + 1 < m then there
is a hole of length l for all n < l < m. In particular, if G is Berge then all holes of G have
the same length.
Proof.
We start by proving the rst assertion of 3.4.1. We may assume that the vertices of
P
Q
and
m > 5.
Σ.
are ordered clockwise on
Since
P
and
Q
are holes, it follows that
n≥4
and
We are going to prove the following claim which directly implies the rst assertion
of 3.4.1 by induction.
(1)
There exists a hole of length m − 1.
We may assume that
(2)
Q
and
P
are chosen such that
|V (Q) ∩ V (P )|
is maximal.
If there are i ∈ {0, . . . , m − 1}, j ∈ {0, . . . , n − 1} such that
p
qi , qi+1 ∈ Σpjj+2
,pj+1 \{pj , pj+1 }
with qm = q0 , pn = p1 and pn−1 = p0 , then there is a hole of length m − 1 in G.
We may assume that
that
p
q3 ∈
/ Σp31 ,p2 .
Since
p
q1 , q2 ∈ Σp31 ,p2 \{p1 , p2 }.
q
p2 ∈ Σq12 ,q3 ,
Since
q1
is antiadjacent to
p2
we deduce by 3.2.3 that
q3 ,
we deduce
is strongly anticomplete to
{q0 , q5 }.
If
p2
is adjacent to
we may assume that
is a hole of length
p2
it follows that
Q − q1 − p 2 − q4 − Q
is strongly antiadjacent to
q4 .
is a hole of length
But then
{x0 , x1 , . . . , xs−1 }.
i ∈ {0, . . . , s − 1},
m > n+1, we may assume that |V (P )∩V (Q)| > 1.
We may assume that
let
Ai =
x0 , . . . , xs−1
x
mod s
Σxi+2
. Since
i ,xi+1 mod s
|Ak ∩ V (P )| < |Ak ∩ V (Q)|.
m > n + 1,
Thus
Let
This proves (2).
V (P )∩V (Q) =
are in clockwise order on
By (2), it follows that
30
q − 1.
Q0 = Q − q1 − p2 − q3 − Q
m with |V (Q0 ) ∩ V (P )| > |V (Q) ∩ V (P )|, a contradiction.
By (2) and since
such that
q4 ,
there exists
Σ.
For
k ∈ {0, . . . , s − 1}
|Ak ∩ V (P )| = |Ak ∩ V (Q)| − 1.
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
Let
P0
that
V (P 0 ) = V (P ) ∩ Ak .
be the subpath of P such that
V (Q0 ) ∩ Ak = {xi , xi+1 }.
Then
Let
x1 − P 0 − x2 − Q0 − x1
Q0
be the subpath of
is a hole of length
Q
such
m − 1.
This proves (1) and the rst assertion of 3.4.1. Since every hole in a Berge trigraph has
even length, the second assertion of 3.4.1 follows immediately from the rst. This concludes
the proof of 3.4.1.
3.4.2. Let G be a Berge circular interval trigraph. If G has a hole of length n with n ≥ 6,
then G is a structured circular interval trigraph.
Proof.
G be a Berge circular interval trigraph. Let X1 , . . . , Xn and Y1 , . . . , Yn be pairwise
S
disjoint cliques satisfying (S2)−(S6) and with |
i (Xi ∪Yi )| maximum. Such sets exist since
Let
n in G. Moreover since G is Berge, it follows that n is
S
S
V (G)\ i (Xi ∪ Yi ) is not empty. Let x ∈ V (G)\ i (Xi ∪ Yi ).
there is a hole of length
may assume that
For
S ⊆ V (G)\{x},
the subset of
(1)
S
we denote by
that is anticomplete to
S
the subset of
that is complete to
x,
SA
and by
x.
C then y is strongly adjacent to z .
If y ∈ XiC and z ∈ Xi+1
Assume not.
By (S4),
plete to
We may assume
Y1 = ∅.
X1 .
Since
By (S6),
n + 1,
we deduce that
x
Since
is not a hole of length
For
X10 = X1 ∪ YnC
(S2) − (S6)
but
is not a claw,
is strongly anticomplete to
Xi0 = Xi ,
y
but
X3 ,
x
for
x
is antiadjacent to
and
Xn
let
is strongly com-
X3
is not a hole of length
or
Xn .
Without loss of
Since
x−X3 −X4 −. . .−Xn −x
Xn .
Since
is strongly complete to
i = 1, . . . , n − 1,
z.
is strongly anticomplete to
x − z − X3 − . . . − Xn−1 − y − x
x is strongly complete to X3 .
n − 1, x
let
z ∈ X2C
is strongly complete to
are not claws, we deduce that
i = 3, . . . , n,
and
is strongly complete to at least one of
generality, we may assume that
{X3 |X2 , X4 , x}
X2
y ∈ X1C
n−1
{x|y, z, ∪n−1
i=4 Xi ∪i=3 Yi }
X4 , . . . , Xn−1 , Y3 , . . . , Yn−1 .
(2)
SC
even. We
Yi0 = Yi .
{X3 |X4 , Y2 , x}
and
Y2 ∪ X2 .
Let
X20 = X2 ∪ {x},
Yn0 = YnA . Then X10 , . . . , Xn0 , Y10 , . . . , Yn0 are disjoint cliques satisfying
S
S
0
0
with |
i (Xi ∪ Yi )| < | i (Xi ∪ Yi )|, a contradiction. This proves (1).
and
C 6= ∅ then X A = ∅.
If XiC 6= ∅ and Xi+2
i+1
Assume not. We may assume
is not a claw by (S6),
x
y ∈ XnC
and
z ∈ X2C
and
is strongly anticomplete to
31
w ∈ X1A .
Since
X4 , . . . , Xn−2 .
n−2
{x|y, z, ∪i=4
Xi }
Assume that
C =
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
x − X3 − . . . − Xn−1 − x
to
V (C)\{x},
or
Xn−1 .
is a hole. Then
contrary to 3.2.6.
x
is strongly anticomplete to at least one of
x
but
w
is strongly anticomplete
is strongly anticomplete to
is not a hole length
By (S6) and symmetry, we may assume that
{z|X3 , x, w}
(3)
has length
By symmetry, we may assume that
x − X2 − X3 − . . . − Xn−1 − x
Xn−1 .
Thus
n − 2,
C
n − 1, x
X1
X3 .
X3
Since
is strongly anticomplete to
is strongly complete to
X2 .
But now
is a claw, a contradiction. This proves (2).
C = ∅.
If XiC 6= ∅, then Xi+2
Assume not. We may assume there exist
complete to
X1 .
n−2
{x|y, z, ∪i=4
Xi ∪n−2
j=3 Yj }
Since
X4 , . . . , Xn−2
strongly anticomplete to
If
y ∈ XnC
and
and
z ∈ X2C .
By (2),
x
is strongly
is not a claw by (S6), it follows that
x
is
Y3 , . . . , Yn−2 .
X3C 6= ∅, then either {x|X1 , X3 , Xn−1 } is a claw or x − X3 − X4 − . . . − Xn − x is a hole
of length
n−1
and therefore odd, hence
is strongly anticomplete to
is strongly complete to
For
Since
X3 .
By symmetry,
x
{y|Xn−1 , x, Yn }
are not claws,
x
is strongly anticomplete to
{z|X3 , x, Y1 }
and
Y1 ∪ Yn .
i = 3, . . . , n − 1,
X20 = X2 ∪ Y2C ,
Xn−1 .
x
let
Xi0 = Xi
The above arguments show that
let
Yi0 = Yi .
Let
{x|Xn , Y1 , Y2C }
cases, it implies that
to
0
Xn−1
.
Y10
let
X10 , . . . , Xn0 , Y10 , . . . , Yn0
contradiction, we need to show that
Since
Y20 = Y2A ,
let
C and let Y 0
A
Xn0 = Xn ∪ Yn−1
n−1 = Yn−1 .
S
S
0
0
0
0
0
0
Clearly X1 , . . . , Xn , Y1 , . . . , Yn are disjoint cliques such that |
i (Xi ∪Yi )| < | i (Xi ∪Yi )|.
let
X10 = X1 ∪ {x},
i = 1, 3, 4, . . . , n − 2, n,
and for
satisfy (S2) and (S5).
X10 , . . . , Xn0 , Y10 , . . . , Yn0
satisfy (S3), (S4) and (S6).
is not a claw, we deduce that either
X20 .
is strongly complete to
It remains to prove the following.
(i) If
Y1 6= ∅,
then
X10
is strongly complete to
X20
(ii) If
Yn 6= ∅,
then
Xn0
is strongly complete to
X10
(iii)
X20
is strongly complete to at least one of
X30 , X10 .
(iv)
Xn0
is strongly complete to at least one of
0
0
Xn−1
, X2 .
(v)
X10
is strongly complete to at least one of
Xn0 , X20 .
32
Y1 = ∅
Symmetrically,
Hence, (S3) is satised.
To get a
Yn0
or
Y2C = ∅.
In both
is strongly complete
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
Y1 6= ∅.
Assume that
X1
It implies by (S4), that
{x|Yn , Y1 , Y2C } is not a claw, we deduce that Y2C = ∅.
is not a hole of length
X20 .
n + 1,
Y2C 6= ∅,
x − Y1 − X2A − X3 − . . . − Xn − x
and thus
X10
is strongly complete to
is strongly complete to
X30
and iii) holds. Thus we
we deduce that
Y2C
to at least one of
X1 , X3 ,
is empty. If
n + 1,
X2A 6= ∅.
we deduce that
Assume that there exist
rst that
w ∈ X2C .
we deduce that
But
v
Since
X2A = ∅
q ∈ X2A .
{r|p, q, x}
Since
x − Y1 − X2A − X3 − . . . − Xn − x
w ∈ X2
and
v ∈ X3
such that
But
x
X2A .
is not a hole of
{z|x, v, w}
v.
is antiadjacent to
Suppose
is not a cycle of length
By (S5), there exists
a ∈ X2C
n + 1,
adjacent to
w ∈ X2A
X2
is a claw, a contradiction. Hence
is strongly complete to at least one of
By (S5) and (S6), there is
sume that
x
C
Yn−1
= ∅,
it follows that
is strongly complete to
C 6= ∅.
Yn−1
X10
Since
r ∈ X1
Xn .
v
and
is
is strongly
X20 .
This proves v). This concludes the proof of (3).
strongly anticomplete to
induced a cycle of length
{x1 , x2 }.
p
and
Xn0
p ∈ XnA
q.
But
X1 .
If
and v) holds.
Thus we may
we deduce that
Y2C = ∅.
and thus
X10
n + 1,
Since
we deduce that
is strongly complete to
n with xi ∈ Xi .
By 3.2.6,
x is strongly
Without loss of generality, we may assume that
By (1),
x1
X3 ∪ X4 ∪ Xn−1 ∪ Xn .
p 6= n
X2
be a hole of length
C.
Suppose that
By symmetry we may as-
is not a hole of length
is strongly complete to
adjacent to two consecutive vertices of
Xn .
is strongly complete to
C , Y C } is not a claw,
{x|X1 , Yn−1
2
is empty. By (1),
is strongly complete to
Xn
By (1),
is strongly complete to
C = x1 − x2 − . . . − xn − x1
or
that is adjacent to both
X2A
X1
X2
This proves the claim.
C −X
A
x − Yn−1
n−1 − . . . − X3 − X2 − X1 − x
Let
w
x − w − X2A − v − X4 − . . . − Xn − x
X2C .
X10 , X30 .
is strongly complete to at least one of
Y1 = ∅.
is a claw, a contradiction.
assume that
is strongly complete
This proves iii) and by symmetry iv) holds.
We claim that
and
X20
X2
is a claw, a contradiction. Thus we may assume that
strongly complete to
X3 .
is empty, and since by (S6),
is strongly anticomplete to
{a|x, v, X2A }
complete to
X2A
it follows that
Thus we may assume that
length
X20
it follows by (S4) that
may assume that
x
Since
Since
This proves i) and by symmetry ii) holds.
If
v.
X2 .
is strongly complete to
is strongly adjacent to
Since
G|({x}
by 3.4.1, we deduce that
33
x
S
x2 .
By (3),
x
is
i Xi ) does not contain an
is strongly anticomplete to
Xi
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
i = 5, . . . , n − 2.
for
is a hole of length
Since
length
in
otherwise there
x
and
x − Yn − Xn − . . . − X2 − x
are not holes of
is strongly anticomplete to
Y2 ∪ Yn .
{X2C |X1A , x, X3 } and {X1C |X2A , x, Xn } are not claws, it follows that X1A
X2C
anticomplete to
complete to
and
b ∈ X2A
By (S5), there is
X4 ,
X2 .
X1C
is strongly anticomplete to
adjacent to
a.
contrary to 3.2.6.
Since
{X1 |x, Y1 , Xn }
But
X2A .
G|({x1 , x2 , a, b}
Thus
is strongly
Suppose there is
a ∈ X1A .
is a hole of length 4 strongly
X1A = X2A = ∅
and by (1),
is not a claw, we deduce that
x
X1
is strongly
is strongly complete
Y1 .
For
i = 1, . . . , n,
Xi0 = Xi ,
let
above arguments show that
|
Y3 ∪ . . . ∪ Yn−1
is strongly anticomplete to
G.
we deduce that
anticomplete to
to
p 6= n
x
x − Y2 − X3 − . . . − Xn − X1 − x
n + 1,
Since
Similarly,
S
i (Xi
∪ Yi )| < |
0
i (Xi
S
for
i = 2, . . . , n,
let
X10 , . . . , Xn0 , Y10 , . . . , Yn0
∪ Yi0 )|,
Yi0 = Yi
and let
Y10 = Y1 ∪ {x}.
are cliques satisfying
The
(S2) − (S6)
but
a contradiction. This concludes the proof of 3.4.2.
We now have all the tools to prove 3.2.1.
Proof of 3.2.1.
We may assume that
G is not a linear interval trigraph and not a cobipartite
trigraph. By 3.2.5, there is an essential triangle or a hole in
G
G.
Then by 3.2.8, 3.3.1 and 3.4.2,
is either a structured circular interval trigraph or is a thickening of a trigraph in
C.
This
proves 3.2.1.
3.5
Some Facts about Linear Interval Join
In this section we prove some lemmas about paths in linear interval stripes.
3.5.1. Let G be a linear interval join with skeleton H such that G is Berge. Let e be an edge
of H that is in a cycle. Let η(e) = V (T )\Z where (T, Z) is a thickening of a linear interval
stripe (S, {x1 , xn }). Then the lengths of all paths from x1 to xn in (S, {x1 , xn }) have the
same parity.
Proof.
For
Assume not. Let
i = 0, . . . , n − 1,
φei (ci ) = x1i
and
let
C = c0 − c1 − . . . − cn − c0
ci ci+1 = ei , (Gei , {x1i , x2i })
φei (ci+1 ) = x2i
be a cycle in
be such that
H
such that
e = c0 cn .
η(ei ) = V (Gei )\{x1i , x2i },
as in the denition of a linear interval join. We may assume
34
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
that
x1
φe (cn ) = x1
to
xn
in
S
and
and
i = 0, 1, . . . , n − 1,
φe (c0 ) = xn .
Let
O = x1 − o1 − . . . − ol−1 − xn
P = x1 − p1 − . . . − pl0 −1 − xn
let
Qi
be a path in
Gei
from
be an odd path from
x1
be an even path from
x1i
to
x2i .
Q0i
Let
to
xn
in
S.
Qi
be the subpath of
For
with
V (Q0i ) = V (Qi )\{x1i , x2i }.
Let
C1 = Xo1 − . . . − Xol−1 − Q00 − Q01 − . . . − Q0n−1 − Xo1
Q00 − Q01 − . . . − Q0n−1 − Xp1 .
Then one of
C1 , C2
and
C2 = Xp1 − . . . − Xpl0 −1 −
is an odd hole in
G,
a contradiction. This
proves 3.5.1.
Before the next lemma, we need some additional denitions. Let
ear interval stripe.
and
rp = xn )
The
right path
of
G
is the path
max{j|xj
is adjacent to
ri−1 }
ilarly the
left path
is the path
of
G
inductively starting with
ri
(i.e. from
R = r0 − . . . − rp
i = 1
dened inductively starting with
(G, {x1 , xn })
(where
ri = xi∗
such that
take a maximal edge on the right to
L = l0 − . . . − lp
i = p−1 such that li = xi∗
(where
with
l0 = x1
and
i∗ = min{j|xj
be a lin-
r0 = x1
with
i∗ =
ri+1 ).
Sim-
lp = xn )
dened
is adjacent to li+1 }.
3.5.2. Let (G, {x1 , xn }) be a linear interval stripe and R be the right path of G. If x, y ∈
V (R), then x − R − y is a shortest path between x and y .
Proof.
Let
P = x − p1 − . . . − pt−1 − y
x − rl − . . . − rs+l−2 − y = x − R − y .
stripe, we deduce that
rl+i−1 ≥ pi
be a path between
By the denition of
for
i = 1, . . . , s − 1.
R
x
and
y
and since
of length
G
t
and let
is a linear interval
Hence it follows that
s ≤ t.
This
proves 3.5.2.
3.5.3. Every linear interval trigraph is Berge.
Proof.
Let
G
be a linear interval trigraph with
induction on the number of vertices.
trigraph, so inductively
N (vn )
H
is Berge.
is a strong clique. But if
a ∈ V (A),
it follows that
consequently
G
A
Clearly
Since
G
V (G) = {v1 , . . . , vn }.
H = G|{v1 , . . . , vn−1 }
is a linear interval
is a linear interval trigraph, it follows that
is an odd hole or an odd antihole in
N (a) ∩ V (A)
The proof is by
G,
is not a strong clique. Therefore
is Berge. This proves 3.5.3.
35
then for every
vn ∈
/ V (A)
and
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
3.5.4. Let (G, {x1 , xn }) be a linear interval stripe. Let S and Q be two paths from x1 to xn
of length s and q such that s < q . Then there exists a path of length m from x1 to xn in G
for all s < m < q .
Proof.
Let
G0
be a circular interval trigraph obtained from
G
x
as
and
C2
by adding a new vertex
follows:
• V (G0 ) = V (G) ∪ {x},
• G0 |V (G) = G,
• x
is strongly anticomplete to
• x
is strongly complete to
V (G)\{x1 , xn },
{x1 , xn }.
s < m < q , C1 = x 1 − S − x n − x − x 1
Let
are holes of length
s+2 and q +2 in G0 .
and
C2 = x 1 − Q − x n − x − x 1 .
By 3.4.1, there exists a hole
Clearly,
C1
Cm of length m+2 in G0 .
G,
Since it is easily seen from the denition of linear interval trigraph that there is no hole in
we deduce that
x ∈ V (Cm ).
we may assume that
length
m
from
x1
to
Let
c1 = x1
xn
in
Cm = x − c1 − c2 − . . . − cm+1 − x.
cm+1 = xn .
and
G.
N (x) = {x1 , xn },
x1 − c2 − . . . − cm − xn
is a path of
This proves 3.5.4.
We say that a linear interval stripe
have length
But now
Since
(G, {x1 , xn })
has
length p if all paths from x1
to
xn
p.
3.5.5. Let (G, {x1 , xn }) be a linear interval stripe of length p. Let L = l0 − . . . − lp and
R = r0 − . . . − rp be the left and right paths. Then r0 < l1 ≤ r1 < l2 ≤ r2 < . . . < lp−1 ≤
rp−1 < lp .
Proof.
that
Since
G
is a linear interval trigraph and by the denition of right path, it follows
r0 < r1 < r2 < . . . < rp .
We claim that if li
ri−1
is adjacent to
li−1 ≤ ri−1 .
Since
∈ (ri−1 , ri ],
then li−1
ri , we deduce that li
ri−1 < li
∈ (ri−2 , ri−1 ].
is adjacent to
ri−1 .
Assume that li
∈ (ri−1 , ri ].
By the denition of the left path,
and by the denition of the right path, we deduce that
strongly antiadjacent to li . Since
G
is a linear interval trigraph, we deduce that li−1
This proves the claim.
36
Since
ri−2
is
> ri−2 .
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
Now, since lp
for
i = 1, . . . , p.
∈ (rp−1 , rp ]
and using the claim inductively, we deduce that
ri−1 < li ≤ ri
This proves 3.5.5.
3.5.6. Let (G, {x1 , xn }) be a linear interval stripe of length p. Let L = l0 − . . . − lp and
R = r0 − . . . − rp be the left and right paths. Then [r0 , li ) is strongly anticomplete to [li+1 , lp ]
and [r0 , ri ] is strongly anticomplete to (ri+1 , lp ] for i = 0, . . . , p.
Proof.
Assume not.
b ∈ [li+1 , lp ]
By symmetry, we may assume that there exist
such that
a
is adjacent to
b.
Since li+1
trigraph, we deduce that li+1 is adjacent to
a.
∈ (a, b]
a < li ,
But
and since
i, a ∈ [r0 , li )
G
and
is a linear interval
contrary to the denition of the
left path. This proves 3.5.6.
3.5.7. Let (G, {x1 , xn }) be a linear interval stripe of length p ≥ 3. Let L = l0 − . . . − lp and
R = r0 − . . . − rp be the left and right paths. If li and ri+1 are strongly adjacent for some
0 < i < p, then G admits a 1-join.
Proof.
Let
i
be such that
trigraph, we deduce that
strongly anticomplete to
Suppose there exists
(ri+1 , rp ].
[li , li+1 ).
li
and
ri+1
[li , ri+1 ]
is a strong clique.
x ∈ [li , ri+1 ]
a ∈ (ri−1 , li ).
We claim that
a
x ∈ (ri , ri+1 ].
is strongly anticomplete to
li+3 .
By symmetry,
a
is adjacent to
we deduce that
P
x
a ∈ [r0 , li )
[r0 , li )
is
[li+1 , lp ]
and
b ∈
and thus
x ∈
x ∈ (ri , li+1 ).
we deduce that
By
a∈
/ [r0 , ri−1 ].
b ∈ (ri+1 , li+2 ).
ri−2
is a path. Since
ri−1 < a and
is strongly antiadjacent to
and by the denition of the left path, we deduce that
b
a.
Since
is strongly antiadjacent to
a ∈ (ri−1 , li ) and b ∈ (ri+1 , li+2 ), it follows that a and b are strongly
antiadjacent. Moreover since
But
x,
P = r0 − R − ri−1 − a − x − b − li+2 − L − lp
By 3.5.6 and since
is a linear interval
Hence by 3.5.5, we deduce that
by the denition of the right path, we deduce that
b < li+2
G
By 3.5.6, it follows that
that is adjacent to a vertex
the denition of the right path and since
Hence
Since
(ri+1 , rp ].
By 3.5.6, it follows that
Symmetrically,
are strongly adjacent.
x ∈ (ri , li+1 )
and by the denition of the left and right path,
is strongly anticomplete to
is an path of length
{ri−1 , li+2 }.
p + 1, a contradiction.
anticomplete to at least one of
[r0 , li ), (ri+1 , rp ].
37
This proves the claim.
Hence for all
x ∈ [li , ri+1 ], x is strongly
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
Let
V1 = {x ∈ [li , ri+1 ] : x
is strongly anticomplete to
The above arguments shows that
(ri+1 , rp ]}
and
([r0 , li )∪V1 , (ri+1 , rp ]∪V2 ) is a 1-join.
V2 = [li , ri+1 ]\V1 .
This proves 3.5.7.
3.5.8. Let (G, {x1 , xn }) be a linear interval stripe of length p with p > 3, then G admits a
1-join.
Proof.
If
Assume not. Let
r2 = l2 ,
L = l0 − . . . − lp
it follows that
r2
By 3.5.7, we may assume that
is adjacent to
r2 .
Hence l0
be the left and right paths.
is strongly adjacent to at least one of
Thus by 3.5.5, we may assume that l2
By 3.5.5, it follows that l2
R = r0 − . . . − rp
and
∈ (r1 , r2 ).
l1 , r3 ,
contrary to 3.5.7.
< r2 .
l1
is antiadjacent to
Since
G
r2
l2
and
r3 .
is antiadjacent to
is a linear interval trigraph, we deduce that l2
− l1 − l2 − r2 − R − rp
is a path of length
p + 1,
a contradiction.
This proves 3.5.8.
3.5.9. Let (G, {x1 , xn }) be a linear interval stripe of length three, and (H, Z) a thickening
of (G, {x1 , xn }). Then either H admits a 1-join or (H, Z) is the thickening of a spring.
Proof.
Let
L = l0 − l1 − l2 − l3
If l1 is strongly adjacent to
r2
and
R = r0 − r1 − r2 − r3
G
then by 3.5.7,
be the left and right paths of
admits a 1-join, and so does
Thus, we may assume that l1 is not strongly adjacent to
a ∈ (r1 , l2 ).
a
Since
a > r1 ,
we deduce that
is strongly antiadjacent to
l3 .
r0 − l1 − a − r2 − l3
length
3.
r0
is a path of length
4,
l1 .
Symmetrically,
Symmetrically,
Since
a
contrary to the fact that
G
is a linear
is adjacent to
r2 .
(G, {x1 , xn })
has
(r1 , l2 ) = ∅.
G
is a linear interval trigraph, we deduce that
is a strong clique, and moreover, that
r0
is strongly complete to
follows that
is strongly adjacent to
r0
[l2 , l3 ].
l1 ∈ (r0 , r1 ]
and
r2 ∈ [l2 , l3 ).
is not strongly complete to
the right path, we deduce that
(r0 , r1 ]
[l2 , l3 ).
((r0 , r1 ], [l2 , l3 ))
Since
Since
l1
V (G) = {r0 , l3 } ∪
is antiadjacent to
r2 ∈ [l2 , l3 )
By 3.5.6, it
is a homogeneous pair.
r2 ,
we deduce
and by the denition of
is not strongly anticomplete to
38
(r0 , r1 ].
By symmetry and since
the above arguments show that
Moreover by 3.5.5,
(r0 , r1 ]
r1
is strongly anticomplete to
(r0 , r1 ] ∪ [l2 , l3 ),
that
is adjacent to
a ∈ (l1 , l2 ).
r0 .
and as
Since
(r0 , r1 ]
Thus
a
H.
Suppose that there exists
is strongly antiadjacent to
By 3.5.5, it follows that
interval trigraph, we deduce that
Hence
a
r2 .
G.
[l2 , l3 ).
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
Xw = {l0 }, Xx = (r0 , r1 ], Xy = [l2 , l3 )
Now setting
Xz = {r3 },
and
we observe that
(G, {x1 , xn }) is the thickening of a spring, and therefore (H, Z) is the thickening of a spring.
This proves 3.5.9.
3.6
Proof of the Main Theorem
In this section we collect the results we have proved so far, and nish the proof of the main
theorem.
3.6.1. Let (G, {x}) be a connected cobipartite bubble. Then (G, {x}) is a thickening of a
truncated spot, a thickening of a truncated spring or a thickening of a one-ended spot.
Proof.
Let
X
assume that
Hence
and
Y
{x} ⊆ X .
(G, {x})
{x} ∪ N (x) = V (G),
If
it follows that
N (x)
We may
is a homogeneous set.
is the thickening of a truncated spot.
Thus we may assume that
x
X ∪ Y = V (G).
be two disjoint strong cliques such that
{x}∪N (x) 6= V (G).
Y1 = Y ∩N (x) and Y2 = Y \Y1 .
Let
is strongly complete to
Y1
and strongly anticomplete to
homogeneous pair. Since
G
is connected, we deduce that
strongly anticomplete to
Y2 .
If
N (x)
Y2 .
Observe that
|N (x)| ≥ 1
is strongly complete to
Y2 ,
(N (x), Y2 )
and that
N (x)
we observe that
a thickening of a one-ended spot. And otherwise, we observe that
(G, {x})
Then
is a
is not
(G, {x})
is
is a thickening
of a truncated spring. This concludes the proof of 3.6.1.
3.6.2. Let (G, {z}) be a stripe such that G is a thickening of a trigraph in C . Then (G, {z})
is in C 0 .
Proof.
Let
H
Bij ⊆ V (H)
be a trigraph in
and
ai ∈ V (H)
C
such that
such that
b∈
/ {z}.
Xai ∪ Xai+1 ⊆ N (b),
Thus there exists
and since there exists no
we deduce that
is a thickening of
be as in the denition of
as in the denition of a thickening. For
i
G
and
For
is not strongly complete to
such that
c ∈ Xak+1 ∪ Xak+2
with
N (z) ∩ (Xak+1 ∪ Xak+2 ) = ∅.
39
z ∈ Xak .
c
For
i = 1, 2, 3,
b ∈ V (G)\(Xa1 ∪ Xa2 ∪ Xa3 )
Xai
k ∈ {1, 2, 3}
C.
H.
Since
k+2
Bk+1
let
Xai ⊂ V (G)
be
let
and since there exists
Xai+1 ,
S3
we deduce that
1
i=1 (Bk
strongly complete to
Since
i, j = 1, 2, 3,
i
∪ Bk+1
) ⊆ N (z)
S3
1
i
i=1 (Bk ∪ Bk+1 ),
is anticomplete to
Bkk+2
and
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
k+2
Bk+1
∪ Bkk+2 ⊆ N (z),
deduce that
(G, {z})
we deduce from the denition of
is in
C0.
C
k+2
Bk+1
∪ Bkk+2 = ∅.
that
Hence we
This proves 3.6.2.
3.6.3. Let G be a trigraph and let H be a thickening of G. For v ∈ V (G), let Xv be as in
the denition of thickening of a trigraph. Let C = c1 − c2 − . . . − cn − c1 be an odd hole or
an odd antihole of H . Then |V (C) ∩ Xv | ≤ 1 for all v ∈ V (G).
Proof.
Assume not. We may assume that
C
Assume rst that
is antiadjacent to
c1
is semiadjacent to
y
vertex in
G,
c3 ∈ Xy .
and
we deduce that
k ∈ {3, . . . , n − 1}
with
x ∈ V (G).
is a hole. By symmetry, we may assume that
c2 ,
and adjacent to
Assume now that
|V (C) ∩ Xx | ≥ 2
C
By symmetry, and since
cn ∈ Xy ,
a contradiction since
is an antihole.
such that
we deduce that there exists
c1 , ck ∈ Xx .
x
c1 , c2 ∈ Xx .
y ∈ V (G)
c3
Since
such that
x
is semiadjacent to at most one
Xy
is a strong clique.
By symmetry, we may assume that there exists
Moreover we may assume by symmetry that
k
is
even.
(1)
For i ∈ {1, . . . , k/2}, if i is odd then ci , ck−i+1 ∈ Xx , and there exists y ∈ V (G) such
that if i is even then ci , ck−i+1 ∈ Xy .
By induction on
cent to
c1 ,
c2 ∈ Xy .
that
i.
By assumption,
we deduce that there exists
By symmetry, and since
x
c1 , ck ∈ Xx .
y ∈ V (G)
Since
c2
such that
is adjacent to
x
ck
and antiadja-
is semiadjacent to
is semiadjacent to at most one vertex in
G,
y
in
G
and
we deduce
ck−1 ∈ Xy .
Now let
sume that
i ∈ {3, . . . , k/2}
ci−1 , ck−i+2 ∈ Xy .
deduce that
ci ∈ Xx
ck−i+1 ∈ Xx .
Now if
since
i
y
and assume rst that
Since
ci
i
is adjacent to
is semiadjacent only to
x
is odd.
ck−i+2
in
G.
By induction, we may as-
and antiadjacent to
ci−1 ,
we
By symmetry, we deduce that
is even, the same argument holds by symmetry. This proves (1).
By (1), there exists
z ∈ {x, y}
such that
ck/2 , ck/2+1 ∈ Xz ,
a contradiction. This con-
cludes the proof of 3.6.3.
3.6.4. Let G be a trigraph and let H be a thickening of G. Then G is Berge if and only if
H is Berge.
40
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
Proof.
If
C = c1 − c2 − . . . − cn − c1
Xc1 − Xc2 − . . . − Xcn − Xc1
Now assume that
By 3.6.3, there is
for all
i 6= j .
is an odd hole (resp.
is an odd hole (resp. antihole) in
C = c1 − c2 − . . . − cn − c1
xi ∈ V (G)
such that
ci ∈ Xxi
x1 − x2 − . . . − xn − x1
But
antihole) in
G
then
C0 =
H.
is an odd hole or an odd antihole in
for
i = 1, . . . , n
and such that
is an odd hole or an odd antihole in
H.
xi 6= xj
G.
This
proves 3.6.4.
3.6.5. Let G be a structured circular interval trigraph. Then G is Berge.
Proof.
Yi
be as in the denition of structured
circular interval trigraph. Let
be an odd hole or an odd antihole in
G.
C = c1 − . . . − cn − c1
Sn
clique for all y ∈
i=1 Yi ,
we deduce that
Assume not.
Since
N (y)
For
i = 1, . . . , n,
is a strong
let
Xi
and
V (C) ∩
Sn
i=1 Yi
= ∅.
But by 3.6.3 and (S1)-(S6), we get a contradiction. This proves 3.6.5.
3.6.6. Let G be a structured circular interval trigraph. Then G is a thickening of an evenly
structured linear interval join.
Proof.
Let
X1 , . . . , Xn , Y1 , . . . , Yn
and
n be as in the denition of structured circular interval
trigraph. Throughout this proof, the addition is modulo
Let
H = (V, E)
be a graph and
s
n.
be a signing such that:
|Y |
|Y |
• V ⊆ {h1 , h2 , . . . , hn } ∪ {l11 , . . . , l1 1 } ∪ . . . ∪ {ln1 , . . . , ln n },
•
•
if
Xi
ei
between
if
if
hi
and
hi+2 ,
and
Xi+1 ,
then
hi+1 ∈
/ V,
and there is exactly one edge
s(ei ) = 0,
|Xi |
Xi is strongly complete to Xi−1 ∪Xi+1 , then there are |Xi | edges e1i , . . . , ei
hi
•
is not strongly complete to
and
hi+1 ,
and
s(eki ) = 1
for
k = 1, . . . , |Xi |,
hi ∈ V , there is one edge between hi
Then
G
k
and li−1 with
k ) = 1 for k = 1, . . . , |Y
s(hi li−1
i−1 |.
is an evenly structured linear interval join with skeleton
stripe associated with an edge
e
between
with
s(e) = 1
We can now prove the following.
41
H
and such that each
is a spot. This proves 3.6.6.
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
3.6.7. Let G be a linear interval join. Then G is Berge if and only if G is an evenly
structured linear interval join.
Proof.
⇐
Let
G
be an evenly structured linear interval join. We have to show that
By 3.5.3, linear interval stripes are Berge.
G
is Berge.
C0
By 3.2.7 and 3.6.4, trigraphs in
are
Berge. By 3.6.5, structured bubbles are Berge. Clearly spots, truncated spots, oneended spots and truncated springs are Berge. By 3.6.4 and due to the construction of
evenly structured linear interval join, the only holes created are of even length due to
G
the signing. Thus
⇒
Let
H
G
is Berge.
be a Berge linear interval join. Let
is chosen among all skeletons of
that with
|E(H)|
φe : V (e) → Ye
(1)
maximum. Let
G
H
be a skeleton of
such that
|V (H)|
(Ge , Ye ), e = x1 x2
be associated with
H
G.
We may assume that
is maximum and subject to
(with
x1 = x2
if
e
is a loop) and
as in the denition of linear interval join.
If (Ge , Ye ) is a thickening of a linear interval stripe such that e is in a cycle in H
but e is not a loop, then Ge does not admit a 1-join.
Assume not. Let
Ye = {y, z}
and
e = x1 x2 .
We may assume that
φe (x1 ) = y
and
φe (x2 ) = z .
Let
H0
be the graph obtained from
H
V (H) ∪ {a0 }, H 0 |V (H) = H\e and a0
Let
(Fe , Ze )
Fe
admits a 1-join. Let
Moreover let
is a thickening of
that
is adjacent to
be a linear interval stripe such that
and such that
of 1-join.
by adding a new vertex
W1 , W2
Fe |Wk
for
(Ge , Ye )
V1 \A1 .
G1e \Ye1 = Ge |(W1 \Ye ).
Let
Let
k = 1, 2
(G1e , Ye1 )
Fi2
x2 ,
and
(W1 , W2 )
and
42
V (Ge )
0
vk+1
(Fe , Ze )
be as in the denition
Ge |Wk
such that
We may assume
V2 = {vk+1 , . . . , vn }.
and
V (H 0 ) =
and to no other vertex.
is a 1-join.
be the thickening of
be such that
as follows:
is a thickening of
be the natural partition of
0
V (Fe1 ) = {v1 , . . . , vk , vk+1
}, Fe1 |V1 = Fe
anticomplete to
and
V1 , V2 , A1 , A2 ⊂ V (Fe )
V (Fe ) = {v1 , . . . , vn }, V1 = {v1 , . . . , vk }
such that
x1
a0
Let
is complete to
0
(Fe1 , {v1 , vk+1
})
Fe1
A1
be
and
such that
V (Fe2 ) = {vk0 , vk+1 , . . . , vn }, Fe2 |V2 = Fe
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
vk0
and
is complete to
(Fe2 , {vk0 , vn })
of
G
Now
A2
such that
a0 x1
a0 x2 ,
and
H
V2 \A2 .
Let
(G2e , Ye2 )
except for stripe
H0
using the same stripes as the con-
(G1e , Ye1 )
and
(G2e , Ye2 )
|V (H)|.
contrary to the maximality of
associated with the
This proves (1).
s be a signing of G such that s(e) = 1 if (Ge , Ye ) is a spot,
Let
be the thickening
G2e \Ye2 = Ge |(W2 \Ye ).
is a linear interval join with skeleton
struction with skeleton
edges
and anticomplete to
and
s(e) = 0 if (Ge , Ye )
is not a spot.
It remains to prove that:
(P1)
if e is not a loop and is in a cycle and s(e) = 0, then (Ge , Ye ) is a thickening of
a spring, and
(P2)
(H, s) is an even structure,
(P3)
if e is a loop, then (Ge , Ye ) is a trigraph in C 0 .
First we prove (P1).
not a loop.
Let
been replaced by
φe (x1 ) = y
from
y
to
z
and
Let
e = x1 x2
be in a cycle and such that
s(e) = 0
and
(Ge , Ye )
be a thickening of a linear interval stripe such that
(Ge , Ye )
in the construction. Let
φe (x2 ) = z .
by 3.5.8 and 3.5.9,
(Ge , Ye )
e∈H
By 3.5.1 and 3.5.4, if
have the same length. By (1),
Ye = {y, z}.
(Ge , Ye )
e
e
is
has
We may assume that
is in a cycle, then all paths
does not admit a 1-join, and thus
is the thickening of a spring. This proves (P1).
Before proving (P2). We need the following claims.
(2)
Let C = c1 − c2 − c3 − c1 be a cycle in H with edge set E(C) = {e1 , e2 , e3 }. If
s(e1 ) = s(e2 ) = 0 and s(e3 ) = 1, then there is an odd hole in G.
By (P1),
and
(Ge1 , Ye1 )
(Ge2 , Ye2 )
and
(Ge2 , Ye2 )
are springs. It follows that the springs
together with the spot
(Ge3 , Ye3 )
induce a hole of length
(Ge1 , Ye1 )
5
in
G,
a
contradiction. This proves (2).
(3)
P
Let C = c1 − c2 − . . . − cn − c1 be a cycle in H such that n > 3 and such that
e∈E(C) s(e)
is odd; then there is an odd hole in G.
The proof of (3) is similar to the proof of (2) and is omitted.
43
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
Let {z1 , z2 , z3 } be a triangle in H . For i = 1, 2, 3, let ei be an edge between zi and
(4)
zi+1
such that s(ei ) = 1. If y ∈ V (H)\{z1 , z2 , z3 } is adjacent to at least two
mod 3
vertices in {z1 , z2 , z3 }, then s(f ) = 1 for every edge f with one end y and the other
end in {z1 , z2 , z3 }.
Assume that there is an edge
s(e4 ) = 0.
e4
By symmetry, we may assume that
e5
may also assume that there is an edge
by (2) using
s(e5 ) = 1,
y
with one end
y − z1 − z2 − y
and the other end in
z1
between
is an end of
e4 .
y
s(e5 ) = 0,
and
z2 .
that there is an odd hole in
we deduce by (2) using
{z1 , z2 , z3 }
y − z 1 − z3 − z2 − y
G,
If
with
By symmetry, we
we deduce
a contradiction. But if
that there is an odd hole in
G,
a contradiction. This proves (4).
Let A be a block of H . Assume that there is a cycle C = c1 − c2 − c3 − c1 in H
(5)
such that s(e) = 1 for all e ∈ E(C). Then all connected components of A\V (C) have
size 1.
Let
B be a connected components of A\V (C) such that |B| > 1.
2-connected, there are at least 2 vertices in
i 6= j .
bi , bj ∈ B
bi
P
from
b1
to
C2 = c1 − b1 − P − b2 − c2 − c1
b2
i=1
in
B.
{c1 , c2 , c3 }.
that are not anticomplete to
is adjacent to
By symmetry, we may assume that
deduce that there is a path
and
such that
B∪{c1 , c2 , c3 } is
that are not anticomplete to
{c1 , c2 , c3 }
Similarly, there are at least 2 vertices in
Hence, we can nd
B
Since
ci
and
bj
is adjacent to
with
and
j = 2.
But
C1 = c3 −c1 −b1 −P −b2 −c2 −c3
Since
B
cj
B.
is connected, we
are cycles of length greater than
them has an odd value, thus by (3) there is an odd hole in
G,
3
and one of
a contradiction. This
proves (5).
Now we prove (P2).
of
F1 ∪ F 2 ∪ F 3
that
(A, s|A )
or an evenly signed graph.
and
s(e) = 1
By (2), if
Let
A
H
is either a member
be such a block and assume
is not an evenly signed graph. It follows that there exists a cycle
c1 − c2 − . . . − cn − c1
3
We need to prove that every block of
in
A
of odd value. By (3) and (2), we deduce that
for all edges
|V (A)| = 3
C
C =
has length
e ∈ E(C).
we deduce that
A
44
is a member of
F1 .
Hence we may assume
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
that there is
c1
and
c2 .
c4 ∈ A.
s(e) = 1
By (4), we deduce that
Assume rst that
there is
By (5) and by symmetry, we deduce that
c4
is adjacent to
y ∈ A\{c1 , c2 , c3 , c4 }
c3 .
for all edges
Assume that
such that
y
Since there is a cycle
s(e) = 1
e ∈ E(Cijk ),
Hence
y
is anticomplete to
Assume now that there is an edge
e
that
c1
is between
and
c2 .
Now
value. By (3), it follows that
all edges
e
in
A
G
|V (A)| > 4.
y
f
{c1 , c2 }
and
is a member of
E(A\{c1 , c2 , c4 }) = ∅.
V (A)\{c1 , c2 }.
Finally we prove (P3). Let
that
e
(Ge , Ye )
{c1 , c2 , c3 , c4 }.
of length
be a loop. Let
A
Since
4
of odd
s(e) = 1
for
E(A\{c1 , c2 , c3 }) = ∅.
A is 2-connected, it follows
s(f ) = 1 for all edges
F3 .
This proves (P2).
be a thickening of a bubble such
(F, W ).
is a thickening of
cl .
|V (A)| = 4.
is a cycle of length
is a member of
(Ge , Ye )
with
is not adjacent to
By (4), we deduce that
Hence
3
Let
F2 .
e has been replaced by (Ge , Ye ) in the construction.
a bubble such that
is connected,
has an odd hole, a contradiction. Hence
A
c4 .
and
By symmetry, we may assume
c1 − c2 − c3 − c4 − c1 ,
{c1 , c2 } is complete to V (A)\{c1 , c2 }.
between
A
a contradiction. It follows that
Assume now that c4 is not adjacent to c3 . By (5), we deduce that
that
Since
Cijk = ci − cj − ck − ci
e in A with s(e) = 0.
and we deduce that
Similarly by (5), it follows that
{c1 , c2 , c3 }
between
we deduce by (5) that
{c1 , c2 , c3 , c4 },
is adjacent to both
is not anticomplete to
{i, j, k, l} = {1, 2, 3, 4}.
for all edges
e
c4
Let
Ye = {y}.
By 3.2.1,
F
Let
(F, W ) be
is a linear interval
trigraph, a cobipartite trigraph, a structured circular interval trigraph or a thickening
of a trigraph in
C.
Assume rst that
of
F.
Let
F
is a linear interval trigraph. Let
k ∈ {1, . . . , n} be such that {vk } = W .
in the denition of a thickening. Let
that
1 < l
vertices
and
a1 , a2
adjacent to
r < n.
as follows:
to
vn
H0
vi ∈ V (F ), let Xvi ⊂ V (Gi ) be as
be such that
N (vk ) = {vl , . . . , vr }.
be the graph obtained from
H
and
v0
is adjacent to
Let
(Gle , Yel )
45
and
a2
are
Fl be such that V (Fl ) = {v0 , v1 , . . . , vk },
Let
v1
and to no other vertex.
V (Fr ) = {vk , . . . , vn , vn+1 }, Fr \vn+1 = F |{vk , . . . , vn }
and to no other vertex.
Assume
by adding two new
V (H 0 ) = V (H) ∪ {a1 , a2 }, H 0 |V (H) = H\e, a1
φ−1
e (y) and to no other vertex.
Fl \v0 = F |{v1 , . . . , vk }
such that
Let
l<r
For
{v1 , . . . , vn } be the set of vertices
and
be the thickening of
vn+1
Let
Fr
be
is adjacent
(Fl , {v0 , vk })
such
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
that
that
Gle \Yel = Ge |
Sk−1
Gre \Yer = Ge |
Sn
j=1
Xvj .
Let
(Gre , Yer )
j=k+1 Xvj . Now
G
be the thickening of
(Ge , Ye ),
a0
new vertex
φ−1
e (y)
to
as follows:
thickening of
and
(G0e , Ye0 )
vn+1
H0
except for
|V (H)|.
H0
(Gle , Yel )
V (H 0 ) = V (H) ∪ {a0 }, H 0 |V (H) = H\e
Let
is adjacent to
F0
vn
be such that
and
H
(Ge , Ye ),
using
(Gre , Yer )
and
by adding a
a0
is adjacent
V (F 0 ) = {v1 , . . . , vn , vn+1 },
and to no other vertex. Let
Now
(G0e , Ye0 )
be the
G is a linear interval join
using the same stripes as the construction with skeleton
instead of
such
Hence by symmetry, we
be the graph obtained from
(F 0 , {v1 , vn+1 }) such that G0e \Ye0 = Ge \Ye .
with skeleton
for
Now let
and to no other vertex.
F 0 |V (F ) = F
H
contrary to the maximality of
l = 1.
may assume that
H0
is a linear interval join with skeleton
the same stripes as the construction with skeleton
instead of
(Fr , {vk , vn+1 })
|V (H)|.
contrary to the maximality of
H
Hence
except
F
is not
a linear interval trigraph.
F
Assume now that
is a structured circular interval trigraph. Using the same construc-
tion as in the proof of 3.6.6, it is easy to see that there exist
and a set of stripes
stripes of
S,
such that
G
H0
with
|V (H 0 )| > |V (H)|
is a linear interval join with skeleton
S , contrary to the maximality of |V (H)|.
Hence
F
H0
using the
is not a structured circular
interval trigraph.
Assume now that
F
is a cobipartite trigraph. Clearly any thickening of a cobipartite
trigraph is a cobipartite trigraph.
By 3.6.1,
(Ge , Ye )
is a thickening of a truncated
spot, a thickening of a truncated spring or a thickening of a one-ended spot.
Assume that
(Ge , Ye )
is a thickening of a one-ended spot.
in the denition of a thickening. Let
new vertex
a0
as follows:
H0
V (H 0 ) = V (H) ∪ {a0 }, H 0 |V (H) = H\e,
a0
φ−1
e (y).
Let the stripes associated with the edges between
0
φ−1
e (y), there is a loop l on a
let the stripe associated with the loop on
G
is a linear interval join with skeleton
with skeleton
Hence
(Gi , Yi )
H
Xv ⊂ V (Ge )
be the graph obtained from
between
and
Let
a0
H0
and
a0
H
by adding a
there is
|Xv |
edges
is adjacent to no other vertex than
a0
and
φ−1
e (y)
be spots and
be a thickening of a truncated spot. Now
using the same stripes as the construction
except for additional edges, contrary to the maximality of
is not a thickening of a one-ended spot.
46
be as
|V (H)|.
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
(Ge , Ye )
Assume now that
obtained from
H
is a thickening of a truncated spot. Let
by adding
|V (Ge )| − 1
1}, aj
is adjacent to
with skeleton
H0
φ−1
e (y)
be the graph
a1 , . . . , a|V (Ge )|−1
new vertices
V (H 0 ) = V (H) ∪ {a1 , . . . , a|V (Ge )|−1 }, H 0 |V (H) = H\e,
H0
and for
and to no other vertex. Now
G
as follows:
j ∈ {1, . . . , |V (Ge )| −
is a linear interval join
using the same stripes as the construction with skeleton
H
and such
that the stripes associated with the added edges are spots, contrary to the maximality
|V (H)|.
of
Hence
Assume that
(Ge , Ye )
(Ge , Ye )
obtained from
H
is not a thickening of a truncated spot.
is a thickening of a truncated spring.
by adding a new vertex
H 0 |V (H) = H\e,
a0
and
is adjacent to
linear interval join with skeleton
H
skeleton
H0
a0
as follows:
φ−1
e (y)
|V (H)|.
H0
be the graph
V (H 0 ) = V (H) ∪ {a0 },
and no other vertex.
Now
G
is a
using the same stripes as the construction with
and such that the stripe associated with the edge
trary to the maximality of
Let
Hence
(Ge , Ye )
a0 φ−1
e (y) is a spring, con-
is not a thickening of a truncated
spring.
Finally assume that
(Ge , Ye )
Hence
is in
G
C0.
Ge
is a thickening of a trigraph in
C.
By 3.6.2, it follows that
This concludes the proof of (P3).
is an evenly structured linear interval join.
This concludes the proof of 3.6.7.
A last lemma is needed for the proof of 3.1.4.
3.6.8.
3.6.9. Let G be a cobipartite trigraph. Then G is a thickening of a linear interval trigraph.
Proof.
Let
Y, Z
be two disjoint strong cliques such that
homogeneous pair. Let
H
be the trigraph such that
• y
is strongly adjacent to
if
Y
• y
is strongly antiadjacent to
z
• y
is semiadjacent to
z
if
z
Y
Y ∪ Z = V (G).
V (H) = {y, z}
is strongly complete to
if
Y
Clearly
(Y, Z)
is a
and
Z,
is strongly anticomplete to
Z,
is neither strongly complete nor strongly anticomplete to
Z.
47
CHAPTER 3. THE STRUCTURE OF CLAW-FREE PERFECT GRAPHS
Now setting
Y = Xy
a nd
Z = Xz , we observe that G is a thickening of H .
a linear interval trigraph, it follows that
G
Since
H
is clearly
is a thickening of a linear interval trigraph. This
proves 3.6.9.
Proof of 3.1.4.
Let
G be a Berge claw-free connected trigraph.
By 3.1.3,
interval join or a thickening of a circular interval trigraph. By 3.2.1, if
circular interval trigraph, then
G
C,
or
G
is a structured circular interval trigraph.
is a structured circular interval trigraph, then
linear interval join. By 3.6.9, if
G is a thickening of a
G is a thickening of a linear interval trigraph, or a cobipartite
trigraph, or a thickening of a member of
But by 3.6.6, if
G is either a linear
G
is an evenly structured
G is a cobipartite trigraph, then G is a thickening of a linear
interval trigraph. Moreover, any thickening of a linear interval trigraph is clearly an evenly
structured linear interval join. Finally by 3.6.7, if
G
is a linear interval join, then
evenly structured linear interval join. This proves 3.1.4.
48
G
is an
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
Chapter 4
On the Erd®s-Lovász Tihany
Conjecture
4.1
Introduction
In 1968, Erd®s and Lovász made the following conjecture:
Conjecture 1
(Erd®s-Lovász Tihany)
. For every graph G with χ(G) > ω(G) and any two
integers s, t ≥ 2 with s + t = χ(G) + 1, there is a partition (S, T ) of the vertex set V (G)
such that χ(G|S) ≥ s and χ(G|T ) ≥ t.
Let
G
be a graph such that
χ(G\{u, v}) ≥ χ(G) − 1.
is
Tihany
if
χ(G) > ω(G).
More generally, if
K
We say that a brace
is a clique of size
k
in
G,
Tihany
if
then we say that
K
{u, v}
is
χ(G \ K) ≥ χ(G) − k + 1.
The following theorem is the main result of this chapter:
4.1.1. Let G be a claw-free graph with χ(G) > ω(G). Then there exists a clique K with
|K| ≤ 5 such that χ(G\K) > χ(G) − |K|.
To prove 4.1.1 we use a structure theorem for claw-free graphs due to Chudnovsky and
Seymour that appears in [13] and is described in the next section. Section 4.3 contains some
lemmas that serve as 'tools' for later proofs. The next six sections are devoted to dealing
with the dierent outcomes of the structure theorem, proving that a minimal counterexample
49
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
to 4.1.1 does not fall into any of those classes. In Section 4.4 we deal with the icosahedron
and long circular interval graphs, in Section 4.5 with non-2-substantial and non-3-substantial
graphs, in Section 4.6 with orientable prismatic graphs, in Section 4.7 with non-orientable
prismatic graphs, in Section 4.8 with three-cliqued graphs and nally in Section 4.9 with
strip structures. In Section 4.10 all of these results are collected to prove 4.1.1.
4.2
Structure Theorem
The goal of this section is to state and describe the structure theorem for claw-free graphs
appearing in [13] (or, more precisely, its corollary). We begin with some denitions which
are modied from [13].
Let
X, Y
be two subsets of
V (G)
to each other if every vertex of
anticomplete
are
A ⊆ V (G)
and
and
Let
X ∩ Y = ∅.
to
F ⊆ V (G)2
A
if
v
then
v
is
X
Y,
to
A
if
v
there is a nonempty subset
H
are
complete
and we say that they
Y.
Similarly, if
is adjacent to every vertex in
A,
A.
be a set of unordered pairs of distinct vertices of
vertex appears in at most one pair. Then
Y
and
is adjacent to a member of
complete
has no neighbor in
X
We say that
is adjacent to every vertex of
to each other if no vertex of
v ∈ V (G) \ A,
anticomplete
X
with
is a
thickening
of
(G, F )
G
such that every
if for every
v ∈ V (G)
Xv ⊆ V (H), all pairwise disjoint and with union V (H) satisfying
the following:
•
for each
v ∈ V (G), Xv
•
if
u, v ∈ V (G)
are adjacent in
•
if
u, v ∈ V (G)
are nonadjacent in
H
is a clique of
G
and
G
{u, v} 6∈ F ,
and
then
{u, v} 6∈ F ,
Xu
then
is complete to
Xu
Xv
in
is anticomplete to
H
Xv
in
H
•
if
{u, v} ∈ F
then
Xu
is neither complete nor anticomplete to
In this denition of graph thickening, elements of
F
Xv
in
H.
have the role of pair of vertices
semiadjacent in the description of thickening for trigraphs. Here are some classes of clawfree graphs that come up in the structure theorem.
50
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
• Graphs from the icosahedron.
icosahedron
The
is the unique planar graph with
twelve vertices all of degree ve. Let it have vertices
10, vi
v1 , v3 , v5 , v7 , v9 ,
and
be obtained from
G ∈ T1
if
G
v11
G0
(reading subscripts modulo 10), and
is adjacent to
by deleting
v11
v2 , v4 , v6 , v8 , v10 .
G2
and let
v0
G1
1≤i≤
is adjacent to
Let this graph be
be obtained from
(G0 , ∅), (G1 , ∅),
is a thickening of
• Fuzzy long circular interval graphs.
homeomorphic to the interval
[0, 1],
and no three of them have union
with vertex set
where for
G0 .
Let
by deleting
G1
v10 .
F 0 = {{v1 , v4 }, {v6 , v9 }}.
Furthermore, let
Let
vi+1 , vi+2
is adjacent to
v0 , v1 , . . . , v11 ,
V
(G2 , F )
V ⊆Σ
Now let
u, v ∈ V
for some
F1 , . . . , Fk
F ⊆ F 0.
F1 , . . . , Fk ⊆ Σ
be a circle, and let
such that no two of
Σ.
in which distinct
Σ
Let
or
be
share an endpoint,
be nite, and let
are adjacent precisely if
H
be a graph
u, v ∈ Fi
for some
i.
Let
F0
Let
F ⊆ F0
G
{u, v}
G ∈ T2
such that every vertex of
if
G
|X| = 4,
{u, v}
X
no triad of
there is a triad of
or
F
if
G
A graph
H
X
contains
H
u
H
is
antiprismatic
Let
F.
i.
Then
if for every
H
and no triad of
u
F
be a graph and let
is in at most one member of
containing both
X ⊆ V (K)
H
and
contains
v,
v,
F
be a set of pairs
and
or
and no other triad of
H
contains
is
antipris-
v.
is the set of changeable edges discussed in [11]. The pair
the vertex pairs in
if
for some
is not a claw and there are at least two
matic if for every F 0 ⊆ F , the graph obtained from H
graph
Fi
(H, F ).
is a thickening of
K
that are adjacent. Let
such that every vertex of
u
are distinct endpoints of
appears in at most one member of
the subgraph induced by
pairs of vertices in
Thus
G
u, v
is a fuzzy long circular interval graph.
• Fuzzy antiprismatic graphs.
with
such that
fuzzy long circular interval graph
is a
Let
be the set of pairs
G
G ∈ T3
F0
if
G
by changing the adjacency of all
is antiprismatic. We say that a graph
is a thickening of
(H, F )
G
is a
for some antiprismatic pair
is a fuzzy antiprismatic graph.
51
(H, F )
fuzzy antiprismatic
(H, F ).
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
Next, we dene what it means for a claw-free graph to admit a strip-structure. For a
H
multigraph
G
Let
be a graph. A
and a function
(F, h)
pair
F ∈ E(H),
and
η
with
we denote by
F
the set of all
h ∈ V (H)
incident with
strip-structure (H, η) of G consists of a multigraph H
mapping each
F ∈ E(H)
and
F ∈ E(H)
h∈F
η(F )
to a subset
η(F, h)
to a subset
of
of
V (G),
η(F ),
F.
E(H) 6= ∅,
with
and mapping each
satisfying the following
conditions.
(SD1) The sets η(F ) (F ∈ E(H)) are nonempty and pairwise disjoint and have union V (G).
(SD2)
For each
clique of
h ∈ V (H),
the union of the sets
η(F, h)
for all
F ∈ E(H)
with
h∈F
is a
G.
(SD3) For all distinct F1 , F2 ∈ E(H), if v1 ∈ η(F1 ) and v2 ∈ η(F2 ) are adjacent in G, then
there exists
h ∈ F1 ∩ F2
such that
v1 ∈ η(F1 , h)
v2 ∈ η(F2 , h).
and
There is also a fourth condition, but it is technical and we will not need it in this thesis.
Let
Let
(H, η) be a strip-structure of a graph G, and let F ∈ E(H), where F = {h1 , . . . , hk }.
v1 , . . . , v k
v1 , . . . , v k ,
be new vertices, and let
where
(J, {v1 , . . . , vk })
vi
is complete to
is called the
J
be the graph obtained from
G|η(F )
by adding
η(F, hi ) and anticomplete to all other vertices of J .
strip of (H, η) at F .
A strip-structure
(H, η)
is
Then
nontrivial
if
|E(H)| ≥ 2.
We now describe some strips that we will need for the structure theorem of claw-free
graph.
Z1 :
Let
if
H
be a graph with vertex set
vi , vk
are adjacent then
vj
{v1 , . . . , vn },
such that for
is adjacent to both
vi , vk .
1 ≤ i < j < k ≤ n,
Let
n ≥ 2,
be nonadjacent, and let there be no vertex adjacent to both
F 0 ⊆ V (H)2
be the set of pairs
is nonadjacent to
vj+1 ,
and
such that every vertex of
fuzzy linear interval graph
H
vj
{vi , vj }
such that
is nonadjacent to
i < j , vi 6= v1
vi−1 .
H
and
52
and
and
F.
v1 , vn
vn .
Let
vj 6= vn , vi
Furthermore, let
appears in at most one member of
if for some
v1
let
F ⊆ F0
Then
G
is a
F , G is a thickening of (H, F ) with
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
|Xv1 | = |Xvn | = 1.
set of all pairs
Z2 :
Let
n ≥ 2.
Xv1 = {u1 }, Xvn = {un },
Let
Construct a graph
A, B, C ,
H
where
as follows.
A, B, C
are cliques. For
i = j,
adjacent if and only if
aj , bj
are nonadjacent.
H0 = H \ X
F ⊆ V (H 0 )2
Now let
G
and let
0 ≤ i, j ≤ n
i 6= j 6= 0.
X ⊆ A ∪ B ∪ C \ {a0 , b0 }
(H 0 , F )
with
Z3 :
Let
H
H
is the set of all pairs
be a graph, and let
in order, such that
is incident with one of
by making the edges
{{h2 h3 , h3 h4 }}
Z4 :
h2 h3
and let
G
h3 h4
Xh1 h2 = {u}, Xh4 h5 = {v},
Let
H
b2 , c1
H
G
and
H
(vertices of
(H 0 , F )
H,
H 0)
with
nonadjacent. Let
is the set of all pairs
and
v1 − · · · − v6 − v1
v1 , v2 ; v8
to
v3 , v4 , v5 , v6 , v9 ; v11
is adjacent to
with
{a1 , b1 , c2 }
v4 , v5 ; v9
is adjacent to
G
are cliques;
F = {{b2 , c2 },
|Xa0 | = |Xb0 | = 1.
Let
(G, Z).
and with adjacency as follows.
v7
is adjacent
v6 , v1 , v2 , v3 ; v10
is adjacent
of length 6.
is adjacent to
v3 , v4 , v6 , v1 , v9 , v10 ;
No other pairs are adjacent. Let
53
Let
(G, Z).
and adjacency
is the set of all pairs
{v1 , . . . , v12 },
is an induced cycle in
to
v2 , v3 , v5 , v6 , v9 , v10 .
(H, F )
F ⊆
|Xh1 h2 | = |Xh4 h5 | = 1.
{a0 , a1 , a2 , b0 , b1 , b2 , b3 , c1 , c2 }
Z = {a00 , b00 }. Z4
be the graph with vertex set
because they are
be obtained from the line graph of
Z = {u, v}. Z3
be a thickening of
and
Xa0 = {a00 }, Xb0 = {b00 },
are adjacent; and all other pairs are nonadjacent.
and let
F
Let
H
{a0 , a1 , a2 }, {b0 , b1 , b2 , b3 }, {a2 , c1 , c2 },
Xa0 = {a00 }, Xb0 = {b00 },
H
and
be the graph with vertex set
{b3 , c1 }}
Let
of
H0
|C \ X| ≥ 2.
and every edge of
be a thickening of
Let
as follows:
Z5 :
and
Let
be
be the vertices of a path of
both have degree one in
h2 , h3 , h4 .
ci
let
(G, Z).
h1 − h2 − h3 − h4 − h5
h1 , h5
be
|Xa0 | = |Xb0 | = 1
with
technical and we will not need them in our proof ). Let
Z = {a00 , b00 }. Z2
ai , bj
All other pairs not specied so far
(we will not specify the possibilities for the set
and
let
0 ≤ j ≤ n,
and
A =
say
Adjacency is as
(i, j) 6= (0, 0),
with
1 ≤ i ≤ n
be a thickening of
|C| = n,
and
C = {c1 , . . . , cn }.
and
and for
if and only if
Its vertex set is the disjoint
|A| = |B| = n + 1
{a0 , a1 , . . . , an }, B = {b0 , b1 , . . . , bn },
adjacent to
is the
(G, Z).
union of three sets
follows.
Z = {u1 , un }. Z1
and
Next,
and
v12
is adjacent to
H 0 be a graph isomorphic to
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
H \X
for some
(H 0 , F )
Z5
with
X ⊆ {v11 , v12 } and let F ⊆ {{v9 , v10 }}.
|Xa0 | = |Xb0 | = 1.
is the set of all pairs
Let
Let
G be a thickening of
Xv7 = {v70 }, Xv8 = {v80 },
and
Z = {v70 , v80 }.
(G, Z).
We are now ready to state a structure theorem for claw-free graphs that is an easy
corollary of the main result of [13].
4.2.1. Let G be a connected claw-free graph. Then either
• G is a member of T1 ∪ T2 ∪ T3 , or
• V (G) is the union of three cliques, or
• G admits a nontrivial strip-structure such that for each strip (J, Z), 1 ≤ |Z| ≤ 2, and
if |Z| = 2, then either
|V (J)| = 3 and Z is complete to V (J) \ Z , or
(J, Z) is a member of Z1 ∪ Z2 ∪ Z3 ∪ Z4 ∪ Z5 .
4.3
Tools
In this section we present a few lemmas that will then be used extensively in the following
sections to prove results on the dierent graphs used in 4.2.1.Let
denote by
C(K)
the set of common neighbors of the members of
their common non-neighbors, and by
K.
K
M (K)
be a clique in
K,
by
A(K)
G.
We
the set of
the set of vertices that are mixed on the clique
Formally,
C(K) ={x ∈ V (G) \ K : ux ∈ E(G)
A(K) ={x ∈ V (G) : ux ∈
/ E(G)
for all
for all
u ∈ K}
u ∈ K}
M (K) =V (G) \ (C(K) ∪ A(K)).
We say that a clique
K
is
dense
if
C(K)
is a clique and we say that it is
good
if
C(K)
is an
anti-matching.
The following result is taken from [34]. Because it is fundamental to many of our results,
we include its proof here for completeness.
54
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
4.3.1. Let G be a graph with chromatic number χ and let K be a clique of size k in G. If
K is not Tihany, then every color class of a (χ − k)-coloring of G \ K contains a vertex
complete to K .
Proof.
C
K
Suppose not. Since
be a color class of a (χ
K ∪C
coloring of
is not Tihany, it follows that
− k )-coloring
G \ (K ∪ C)
is
K.
This denes a
(χ − k − 1)-colorable.
is
(χ − k)-colorable.
with no vertex complete to
by giving a distinct color to each vertex of
a color of one of its non-neighbors in
that
G\K
of
G\K
K
K.
Let
Dene a
and giving each vertex of
k -coloring
G|(K ∪ C).
of
However, this implies that
G
C
Note also
is (χ − 1)-colorable,
a contradiction. This proves 4.3.1.
The next lemma is one of our most important and basic tool.
4.3.2. Let G be a graph such that χ(G) > ω(G). Let K be a clique of G. If K is dense,
then it is Tihany.
Proof.
Suppose that
color class of
of
C(K)
Let
G
if
C
A, B
is not Tihany. Let
be disjoint subsets of
A
reduced
(A, B)
is
into
B1, B2
such that
A1
Note that since
A
B1
Let
A
H
if we can partition
is complete to
reduced
G \ K.
A
By 4.3.1, every
Hence, every color class contains a member
a contradiction. This proves 4.3.2.
(A, B)
v ∈ V (G) \ (A ∪ B), v
B
homogeneous pair
in
is either complete to
A
is called a
or anticomplete to
into two sets
B 1 , A2
A1
B.
A2
and
A2 , B 2
A
B.
W -join (A, B)
is a
We say that a
W-
and we can partition
B , and B 2
is anticomplete to
is neither complete nor anticomplete to
B,
is anticomplete
it follows that both
is non-empty. We call a
B
W -join
A1
that is
non-reduced W -join.
be a thickening of
notice that
of
is neither complete nor anticomplete to
are non-empty and at least one of
not reduced a
K.
(χ − k)-coloring
The pair
and either complete to
join
and
V (G).
are cliques, and for every vertex
homogeneous pair in which
A.
be a
|C(K) ∪ K| ≥ χ(G) > ω(G),
or anticomplete to
to
C
contains a vertex complete to
and so
(A, B)
K
(Xu , Xv )
W -join
(G, F )
for some valid
F ⊆ V (G)2
{u, v} ∈ F
W -join
in
H.
If for every
then we say that
H
is a
reduced thickening
is a
The following result appears in [4].
55
and let
of
{u, v} ∈ F .
we have that
G.
Then we
(Xu , Xv )
is a
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
4.3.3. Let G be a claw-free graph and suppose that G admits a non-reduced W -join. Then
there exists a subgraph H of G with the following properties:
1. H is a claw-free graph, |V (H)| = |V (G)| and |E(H)| < |E(G)|.
2. χ(H) = χ(G).
The result of 4.3.3 implies the following:
4.3.4. Assume that G be a claw-free graph with χ(G) > ω(G) that is a minimal counterexample to 4.1.1. Assume also that G is a thickening of (H, F ) for some claw-free graph H
and F ⊆ V (H)2 . Then G is a reduced thickening of (H, F ).
For a clique
K
s.t.
K ⊆ V (G)
{x, k} ∈ F
and
and
F ⊆ V (G)2 ,
we dene
SF (K) = {x ∈ V (G) : ∃k ∈
x ∈ C(K\k)}.
4.3.5. Let G be a reduced thickening of (H, F ) for some claw-free graph H and F ⊆ V (H)2 .
Let K be a clique in H such that for all x, y ∈ C(K) ⊆ V (H), {x, y} 6∈ F . If C(K) ∪ SF (K)
is a clique, then there exists a dense clique of size |K| in G.
Proof.
Let
K0
be a clique of size
|K|
in
G
such that
K 0 ∩ Xv 6= ∅
denition of a thickening such a clique exists. Moreover since
follows that
K0
for all
v ∈ K.
C(K) ∪ SF (K)
By the
is a clique, it
is dense. This proves 4.3.5.
The following lemma is a direct corollary of 4.3.2 and 4.3.5.
4.3.6. Let G be a reduced thickening of (H, F ) for some claw-free graph H and F ⊆ V (H)2 .
Let K be a dense clique in H such that for all x, y ∈ C(K), {x, y} 6∈ F . If C(K) ∪ SF (K)
is a clique, then there exists a Tihany clique of size |K| in G.
The following result helps us handle the case when
vertex
C(x)
is an antimatching for some
x ∈ V (G).
4.3.7. Let G be a graph with χ(G) > ω(G). Let u, x, y ∈ V (G) such that ux, uy ∈ E(G),
xy 6∈ E(G) and x 6= y . Let E = {u, x} and E 0 = {u, y}. If C(E) = C(E 0 ) then E, E 0 are
Tihany.
56
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
Proof.
Suppose that
C∈C
be the color class such that
E,
complete to
E
and so
is not Tihany.
z ∈ C(E).
Tihany and by symmetry, so is
y ∈ C.
But
C
Let
y
be a
(χ(G) − 2)-coloring
By 4.3.1, there is a vertex
is complete to
C(E),
of
G \ {u, x}.
z∈C
Let
such that
a contradiction. Hence
z
is
E
is
E0.
In particular, if we have a vertex
x
such that
C(x)
is an antimatching, we can nd a
Tihany edge either by 4.3.2 or by 4.3.7.
4.3.8. Let H be a graph, G a thickening of (H, F ) for some valid F ⊆ H(V )2 such that
χ(G) > ω(G). Let K be a clique of H . Assume that for all {x, y} ∈ F such that x ∈ K ,
y is complete to C(K)\{y}. Let u, v ∈ C(K) with u 6= v be such that u is not adjacent to
v and {u, v} is complete to C(K)\{u, v}. Moreover assume that if there exists E ∈ F with
{u, v} ∩ E 6= ∅, then E = {u, v}. Then there exists a Tihany clique of size |K| + 1 in G.
Proof.
If
Assume not. Let
{u, v} ∈
/ F,
Xv2
Xv2
a ∈ Xu , A = Xu , b ∈ Xv
is empty, let
|K| in G such that K 0 ∩ Xy 6= ∅ for all y ∈ K .
and
B = Xv .
a ∈ Xu2 , A = Xu2 , b ∈ Xv1
a ∈ Xu2 , A = Xu , b ∈ Xv2
Now let
Ta = K 0 ∪ {a}
χ(G) − |K| − 1.
neighbor of
to
be a clique of size
If
{u, v} ∈ F ,
Xu1 , Xu2 , Xv1
let
be as in the denition of reduced W-join. By symmetry, we may assume that
empty. If
let
let
K0
and
B = Xv1 ;
Tb = K 0 ∪ {b}.
Since no vertex of
it follows that
and if
Xv2
is not
is not empty,
B = Xv .
We may assume that
By 4.3.1, we may assume that every color class of
Ta .
C(Ta )\A,
and
and
Xu2
and
B
is complete to
|A| > |B|.
Ta ,
But similarly,
χ(G\Ta ) = χ(G\Tb ) =
G\Ta
and since
|B| > |A|,
B
contains a common
is a clique complete
a contradiction.
This
proves 4.3.8.
We need an additional denition before proving the next lemma. Let
denote by
C(K)
the closed neighborhood of
K,
i.e.
K
be a clique; we
C(K) := C(K) ∪ K .
4.3.9. Let G be a graph such that χ(G) > ω(G). Let A and B be cliques such that 2 ≤
|A|, |B| ≤ 3 (i.e., each one is a brace or a triangle). If C(A) ∩ C(B) = ∅ and C(A) ∪ C(B)
contains no triads then at least one of A, B is Tihany.
Proof.
Assume not and let
class must have a vertex in
k = χ(G) − |A|.
C(A).
By 4.3.1, in every
As there is no triad in
57
k -coloring
C(A) ∪ C(B),
of
G\A every
color
it follows that every
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
vertex of
C(A)
C(B),
is in a color class with at most one vertex of
symmetry, it follows that
C(A) < C(B),
thus
C(A) > C(B).
By
a contradiction. This proves 4.3.9.
4.3.10. Let G be a claw-free graph such that χ(G) > ω(G). If G admits a clique cutset, then
there is a Tihany brace in G.
Proof.
Let
V (G).
Let
that
(1)
A, B ⊂ V (G)\K
be a clique cutset. Let
χA = χ(G|(A ∪ K))
such that
χB = χ(G|(B ∪ K)).
and
A ∩ B = ∅ and A ∪ B ∪ K =
By symmetry, we may assume
χA ≥ χB .
χ(G) = χA
SA = (A1 , A2 , . . . , AχA )
Let
K)
K
and
G|(B ∪ K).
assume that
Let
Now let
x ∈ B
is a
for all
K\{y}) ≥ χA − 1 ≥ χ(G) − 1.
4.4
of
G.
{x, y}
G|(A ∪
Then
S = (A1 ∪ B1 , A2 ∪ B2 , . . . , AχB ∪
This proves (1).
be such that
Hence
be optimal colorings of
Up to renaming the stable sets, we may
i = 1, 2, . . . , |K|.
χA -coloring
y ∈ K
and
SB = (B1 , B2 , . . . , BχB )
K = {k1 , k2 , . . . , k|K| }.
Ai ∩ Bi = {ki }
BχB , AχB +1 , . . . , AχA }
and
xy ∈ E(G).
Then
χ(G\{x, y}) ≥ χ(G|(A ∪
is a Tihany brace. This proves 4.3.10.
The Icosahedron and Long Circular Interval Graphs
4.4.1. Let G ∈ T1 . If χ(G) > ω(G), then there exists a Tihany brace in G.
Proof.
Let
v0 , v1 , . . . , v11
be as in the denition of
F ⊆ {(v1 , v4 ), (v6 , v9 )}.
Xv11
is empty when
thickening of
G
(G2 , F )).
First suppose that
Xv5 ∪ Xv6
be as in the denition of the icosahedron. Let
T1 .
Then
is a thickening of either
0 ≤ i ≤ 11,
For
Let
G
xi ∈ Xvi
Xvi
let
is a thickening of
and
(G1 , ∅)
So we may assume that
G
E = {x1 , x3 }.
{x4 , x6 }
contains at least one vertex from
or
(G2 , F )
F
for
be as in the denition of thickening (where
(G2 , F ),
or
(G1 , ∅)
and
Xv10
is empty when
G
Then
C({x4 , x6 }) = Xv4 ∪
is a
G\E
or
(G2 , F ).
is a Tihany brace by 4.3.2.
is a thickening of
Then
(G0 , ∅), (G1 , ∅),
and
wi = |Xvi |.
is a thickening of
is a clique. Therefore,
Tihany and let
G
G0 , G1 , G2 ,
is
(G0 , ∅).
Suppose that no brace of
(χ − 2)-colorable.
is
By 4.3.1, every color class
C(E) = (Xv1 ∪ Xv2 ∪ Xv3 ∪ Xv0 ) \ {x1 , x3 }.
58
G
Since
α(G) = 3,
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
it follows that every color class has at most two vertices from
S11
i=4 Xvi . Hence we conclude
that
w4 + w5 + w6 + w7 + w8 + w9 + w10 + w11 ≤ 2 · (w1 + w2 + w3 + w0 − 2)
{xi , xj }. Summing these
P11
P
( i=0 20wi ) ≤ ( 11
i=0 20wi ) − 120, a
A similar inequality exists for every brace
{xi , xj },
braces
it follows that
inequalities over all
contradiction.
This
proves 4.4.1.
4.4.2. Let G ∈ T2 . If χ(G) > ω(G), then there exists a Tihany brace in G.
Proof.
Let
(H, F ).
H, F, Σ, F1 , . . . , Fk
Let
Fi
be as in the denition of
be such that there exists no
j
with
Fi ⊂ Fj .
and without loss of generality, we may assume that
C({xk , xl }) = {xk+1 , . . . , xl−1 },
a Tihany brace in
4.5
G.
it follows that
T2
such that
Let
{xk , . . . , xl }
{xk , xl }
G
is a thickening of
{xk , . . . , xl } = V (H) ∩ Fi
are in order on
Σ.
Since
is dense. Hence by 4.3.6 there exists
This proves 4.4.2.
Non-2-substantial and Non-3-substantial Graphs
In this section we study graphs where a few vertices cover all the triads. An antiprismatic
graph
The
G is k-substantial
matching number
matching in
G.
if for every
of a graph
S ⊆ V (G) with |S| < k there is a triad T
with
S ∩T = ∅.
G, denoted by µ(G), is the number of edges in a maximum
Balogh et al. [2] proved the following theorem.
4.5.1. Let G be a graph such that α(G) = 2 and χ(G) > ω(G). For any two integers s, t ≥ 2
such that s + t = χ(G) + 1 there exists a partition (S, T ) of V (G) such that χ(G|S) ≥ s and
χ(G|T ) ≥ t.
The following theorem is a result of Gallai and Edmonds on matchings and it will be
used in the study of non-2-substantial and non-3-substantial graphs.
4.5.2
. Let G = (V, E) be a graph. Let D
(Gallai-Edmonds Structure Theorem [17], [19])
denote the set of nodes which are not covered by at least one maximum matching of G. Let
A be the set of nodes in V \D adjacent to at least one node in D. Let C = V \(A ∪ D). Then:
59
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
i) The number of covered nodes by a maximum matching in G equals |V | + |A| − c(D),
where c(D) denotes the number of components of the graph spanned by D.
ii) If M is a maximum matching of G, then for every component F of G|D, E(D) ∩ M
covers all but one of the nodes of F , E(C) ∩ M is a perfect matching of G|C and M
matches all the nodes of A with nodes in distinct components of D.
4.5.3. Let G be an antiprismatic graph. Let K be a clique and assume that u, v ∈ V (G)\C(K)
are non-adjacent. If α(G|(C(K) ∪ {u, v})) = 2 and α(G|K ∪ {u, v}) = 3, then G|C(K) is
cobipartite.
Proof.
Since there is no triad in
C(K) ∪ {u, v},
{u, v}.
is claw-free and
anticomplete to
is no vertex in
and for all
v
to
C(K)
Since
G
complete to
x ∈ C(K), x
{u, v}.
is adjacent to
and non-adjacent to
u
a clique and by symmetry
if
Cu
u
x ∈ Cv .
Let
we deduce that there is no vertex in
α(G|K ∪ {u, v}) = 3,
Cu , Cv ⊆ C(K)
and non-adjacent to
Since
if
x ∈ Cu ,
α(G|(Cv ∪ {u})) = 2,
is a clique. Hence
C(K)
it follows that there
be such that
v
C(K)
Cu ∪ Cv = C(K)
and
x
is adjacent
Cv
we deduce that
is
is the union of two cliques. This
proves 4.5.3.
4.5.4. Let G be a claw-free graph such that χ(G) > ω(G). Let K be a clique such that
α(G\K) ≤ 2. Then there exists a Tihany clique of size at most |K| + 1 in G.
Proof.
Assume not. Let
n = |V (G)|, w ∈ C(K)
and
K 0 = K ∪ {w}
(such a vertex
w
exists
by 4.3.1).
(1)
χ(G) = n − µ(Gc ).
K0
is not Tihany, it follows that
deduce that
n−|K 0 |
, and thus
2
Since
of
G
χ(G\K 0 ) ≥
χ(G\K 0 ) = χ(G) − |K 0 |.
χ(G) ≥
follows that
χ(G) ≥ n − µ(Gc ).
But clearly
G
α(G\K 0 ) ≤ 2,
2,
and since
G
is claw-free,
where all color classes have size
1
or
χ(G) ≤ n − µ(Gc ), thus χ(G) = n − µ(Gc ).
proves (1).
(2)
Let T be a clique of size |K| + 1 in G, then χ(G\T ) = n − |T | − µ(Gc \T ).
60
we
n+|K 0 |
. Hence in every optimal coloring
2
the color classes have an average size strictly smaller than
we deduce that there is an optimal coloring of
Since
2.
It
This
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
Since
T
is not Tihany, it follows that
|V (G\T )|
. Hence in every optimal coloring of
2
smaller than
2,
and since
G is claw-free,
G\T ,
1 or 2.
χ(G\T ) = n − |T | − µ(Gc \T ).
This proves (2).
A, D, C
be as in 4.5.2.
n−|K 0 |
that
µ(Gc ) ≤
that
c(D) ≥ |K 0 |.
2
.
Let
Since
− |T | =
n−|T |
2
It follows that
χ(G) ≥
D1 , D2 , . . . , Dc(D)
χ(G\T ) ≥ |V (G\T |−µ(Gc \T ).
n+|K 0 |
and
2
By 4.5.2 i), we deduce that
=
the color classes have an average size
we deduce that there is an optimal coloring of
where all color classes have size
Let
n+|K 0 |
2
χ(G\T ) = χ(G) − |T | ≥
χ(G) = n − µ(Gc ),
µ(Gc ) =
n+|A|−c(D)
.
2
be the anticomponents of
G\T
Hence
we deduce
Thus, it follows
G|D.
Let
di ∈ Di
for
i = 1, . . . , c(D).
|Di | = 1 for all i.
(3)
|D1 | > 1.
Assume not, and by symmetry assume
α(G|D1 ) = 2.
Thus there exist
x, y ∈ D1
Since
such that
x
G
is claw-free, we deduce
is adjacent to
y.
Now
T =
{x, y, d2 , . . . , d|K| } is a clique of size |K|+1 and by 4.5.2 ii), it follows that µ(Gc \T ) < µ(Gc ).
By (1) and (2), it follows that
χ(G\T ) + |T | = n − µ(Gc \T ) > n − µ(Gc ) = χ(G),
a contra-
diction. This proves (3).
Let
T = {d1 , . . . , d|K|+1 }.
deduce that
C(T ) ∩ A 6= ∅.
C(T ) ∩ D
By (3), it follows that
Let
x ∈ C(T ) ∩ A.
Now
is a clique.
By 4.3.2, we
S = {d1 , . . . , d|K| , x}
is a clique of
µ(Gc \S) < µ(Gc ).
size
|K| + 1
that
χ(G\S) + |S| = n − µ(Gc \S) > n − µ(Gc ) = χ(G),
and by 4.5.2 ii), it follows that
By (1) and (2), it follows
a contradiction. This concludes the
proof of 4.5.4.
4.5.5. Let H be an antiprismatic graph such that there exists x ∈ V (H) with α(H\x) = 2.
Let G be a reduced thickening of (H, F ) for some valid F ⊆ V (G)2 such that χ(G) >
ω(G) and |Xx | > 1. Then for all {u, v} ∈ Xx , χ(G\{u, v}) ≥ χ(G) − 1.
Proof.
and
Let
u, v ∈ Xx .
We may assume that
S = (S1 , S2 , . . . , Sk )
Ij = {i : |Si | = j}
and let
be a
k -coloring
is not Tihany.
G\{u, v}.
of
O = C({u, v}) ∩
{u, v}
S
i∈I1 ∪I2
61
Si
k = χ(G\{u, v})
Si ∩ C({u, v}) 6= ∅.
S
P = C({u, v}) ∩ i∈I3 Si .
By 4.3.1,
and
Let
Let
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
α(H\x) = 2,
Since
complete to
O
and thus
ω(G|O) < |I1 ∪ I2 |.
coloring of
O.
G|O.
Now let
such that
ends.
Si ∩ Xx 6= ∅
If
Ti = {p2i−1 , p2i }
|Ss | = 2,
we deduce that
Thus the coloring
S
i = 1, . . . , l.
Let
s
P
is a clique
we deduce that
G|O
is cobipartite.
does not induce an optimal
P = p1 − p2 − . . . − p2l
be such that
p1 ∈ Ss
and
e
be
They are the color classes where the augmenting antipath starts and
Tl+1 = ({u} ∪ Ss \p1 ),
let
Tl+2 = ({v} ∪ Se \p2l ),
Now
for
Hence,
χ(G) > ω(G),
It follows that there exists an augmenting antipath
p2l ∈ Se .
|J| = l + 1.
Since
O ⊆ C(Xx ),
By 4.5.3 and since
i ∈ I3 .
for all
ω(G|O ∪ P ) = ω(G|O) + |I3 |.
χ(G|O) = ω(G|O) < |I1 ∪ I2 |.
Hence
in
it follows that
otherwise let
(T1 , T2 , . . . , Tl+2 )
with the color classes
Si
for
Tl+2 = {v}.
is a
i ∈
/ J
otherwise let
Let
(l + 2)-coloring
create a
Tl+1 = {u}.
If
|Se | = 2,
J = {i|Si ∩ V (P ) 6= ∅}.
S
of
i∈J Si ∪ {u, v}, which
k + 1-coloring
of
G,
let
Clearly
together
a contradiction.
This
proves 4.5.5.
The next lemma is a direct corollary of 4.5.4 and 4.5.5.
4.5.6. Let H be a non-2-substantial antiprismatic graph. Let G be a reduced thickening of
(H, F ) for some valid F ⊆ V (G)2 such that χ(G) > ω(G). Then there exists a Tihany brace
in G.
Now we look at non-3-substantial graphs.
4.5.7. Let H be a non-3-substantial antiprismatic graph. Assume that u, v ∈ H satisfy
α(H\{u, v}) = 2. Let G be a reduced thickening of H such that χ(G) > ω(G). If u is not
adjacent to v , then there exists a Tihany brace or triangle in G.
Proof.
Assume not.
Let
Nu = C(u)\C({u, v})
and
Nv = C(v)\C({u, v}).
antiprismatic, it follows that
Nu
and
Nv
are antimatchings.
By 4.5.6, we deduce that
Nu
and
Nv
are not cliques. Let
and
xv , yv ∈ Nv
be non-adjacent.
may assume by symmetry that
α(H\{u, v}) = 2
and
strongly complete to
H
Since
xu xv , yu yv
are edges, and
and strongly anticomplete to
and strongly anticomplete to
xu xv .
x ∈ Nx
and and all
xu xv
Let
(Nx , Ny )
y ∈ Ny
62
H
is
be non-adjacent,
H
is antiprismatic, we
xu yv , yu xv
are non-edges. Since
and
is antiprismatic, it follows that every vertex in
xu xv
are complete to
α(H\{u, v}) = 2
xu , yu ∈ Nu
Since
yu yv ,
C({u, v})
is either
or strongly complete to
be the partition of
are complete to
C({u, v})
yu yv .
yu yv
such that all
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
Assume rst that
Tv = {v, xv , nx }.
and
Nx 6= ∅ and Ny 6= ∅.
Clearly
Tu
Tv
and
Let
nx ∈ Nx
and
ny ∈ Ny
and let
Tu = {u, yu , ny }
are triangles.
α(G|(C(Tu ) ∪ C(Tv )) = 2 and C(Tu ) ∩ C(Tv ) = ∅.
(1)
Assume not. Since
C(Tu ) ∩ C(Tv ) = ∅.
u ∈ T.
hence
Clearly,
Let
T ∈ C(Tu ) ∪ C(Tv )
T \u ∈ Nv .
T \u ∈
/ C(Tu ) ∪ C(Tv ),
Now let
and
C(Tu ) ⊆ Ny ∪ Nu ∪ {u}
Su , Sv ∈ G
But since
H
and
H,
we deduce that
be a triad. By symmetry, we may assume that
is antiprismatic, we deduce that
T \u ⊆ C(nx ),
a contradiction. This proves (1).
be triangles such that
|Sv ∩ Xv | = |Sv ∩ Xxv | = |Sv ∩ Xnx | = 1.
thickening of
C(Tv ) ⊆ Nx ∪ Nv ∪ {v},
|Su ∩ Xu | = |Su ∩ Xyu | = |Su ∩ Xny | = 1
By (1) and 4.3.9 and since
Nx , Ny
C({u, xu })
Nx
is empty. Since
G.
This concludes the proof of 4.5.7.
is a reduced
G.
we deduce that there is a Tihany triangle in
Now assume that at least one of
G
is empty. By symmetry, we may assume that
is an antimatching, by 4.3.8 there exists a Tihany triangle in
4.5.8. Let H be a non-3-substantial antiprismatic graph. Let u, v ∈ H be such that
α(G\{u, v}) = 2. Let G be a reduced thickening of (H, F ) for some valid F ⊆ V (H)2
such that χ(G) > ω(G). If u is adjacent to v , then there exists a Tihany clique K in G with
|K| ≤ 4.
Proof.
Assume not.
assume that
By 4.5.4, we may assume that
|Xu | > 0
or
|Xv | > 0.
If
|Xu | = 1,
|Xu ∪ Xv | > 2.
then
G\Xu
By 4.5.6, we may
is a reduced thickening of a
non-2-substantial antiprismatic graph. By 4.5.5, there exists a brace
{x, y}
in
χ(G\({x, y} ∪ Xu )) ≥ χ(G\Xu ) − 1.
hence
{x, y} ∪ Xu
But
Tihany triangle, a contradiction. Thus
Let
x1 , y1 ∈ Xu
Let
k = χ(G\C)
that
and
and
x2 , y2 ∈ Xv ,
and
|Xu | > 1,
thus
α(H\{u, v}) = 2,
let
Since
be a
of
G\C .
and let
Si ∩ (Xu ∪ Xv ) 6= ∅
χ(G) > ω(G),
63
such that
is a
|Xv | > 1.
is a clique of size
k -coloring
Ij = {i : |Si | = j}
it follows that
ω(G|O ∪ P ) = ω(G|O) + |I3 |.
and by symmetry
C = {x1 , x2 , y1 , y2 }
S = (S1 , S2 , . . . , Sk )
Si ∩ N (C) 6= ∅. For j = 1, 2, 3
S
P = N (C) ∩ i∈I3 Si .
Since
χ(G\Xu ) − 1 ≥ χ(G) − 2,
Xv
4.
By 4.3.1, it follows
O = N (C) ∩
for all
we deduce that
S
i∈I1 ∪I2
i ∈ I3 .
Si
Hence,
ω(G|O) < |I1 ∪ I2 |.
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
By 4.5.3, we deduce that
the coloring
S
is cobipartite. Hence
P = p1 − p2 − . . . − p2l
s be such that p1 ∈ Ss
and
in
that there exists
ŝ ∈ {1, 2}
Tl+2 = {x1 , x2 }\xŝ .
exists
G|O.
ê ∈ {1, 2}
such that
xê
xŝ
Se \p2l
Since
such that
O.
Ss \p1
for
i = 1, . . . , l.
They are the color classes where the
{x1 , y1 },
is not complete to
Ss \p1 .
is antiadjacent to
is not complete to
is antiadjacent to
Thus
It follows that there exists an
Ti = {p2i−1 , p2i }
Now let
e be such that p2l ∈ Se .
augmenting antipath starts and ends. Since
and
χ(G|O) = ω(G|O) < |I1 | + |I2 |.
does not induce an optimal coloring of
augmenting antipath
Let
G|O
Se \p2l .
Let
{x2 , y2 },
Let
we deduce
Tl+1 = {xŝ } ∪ Ss \p1
we deduce that there
Tl+3 = {xê } ∪ Se \p2l
and
Tl+4 = {y1 , y2 }\xê .
J = {i|Si ∩ V (P ) 6= ∅}. Clearly |J| = l + 1. Now (T1 , T2 , . . . , Tl+2 , Tl+3 , Tl+4 ) is
S
(l + 4)-coloring of i∈J Si ∪ {x1 , x2 , y1 , y2 }, which together with the color classes Si , for
Let
a
i∈
/ J,
create a
k + 3-coloring
of
G,
a contradiction. This proves 4.5.8.
The following lemma is a direct corollary of 4.5.7 and 4.5.8.
4.5.9. Let H be a non-3-substantial antiprismatic graph. Let G be a reduced thickening of
H such that χ(G) > ω(G). Then there exists a Tihany clique K ⊂ V (G) with |K| ≤ 4.
4.6
Complements of Orientable Prismatic Graphs
In this section we study the complements of orientable prismatic graphs. A graph is
prismatic
if its complement is antiprismatic. We can also dene also prismatic graph in a direct way.
A graph
G
Let
G is prismatic if for every triangle T ⊆ V (G) and x ∈ V (G)\T , then |N (x)∩T | = 1.
be prismatic and let
S, T
be two disjoint triangles in
exists a perfect matching between
orientation
O(T )
for every triangle
are disjoint triangles with
si ti ∈ E(G) i = 1, 2, 3.
non-orientable
The
core
S
T
and
of
T.
G
An
orientation O
such that if
O(S) = s1 → s2 → s3 → s1
We say that
G
is
G.
orientable
By denition of
of
G
there
is a choice of a cyclic
S = {s1 , s2 , s3 }
and
G
and
T = {t1 , t2 , t3 }
O(T ) = t1 → t2 → t3 → t1 ,
then
G
is
is a triangle in
G
if it admits an orientation, and
otherwise.
of a graph
G
is the union of all the triangles in
weak.
strong
core
is the subset of the core such that no vertex in the strong core is weak. As proved
of
G
64
and
c
{a, b, c}
b, c
then
b
If
and both
only belong to one triangle in
G,
G.
are said to be
The
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
in [11], if
y
H
is a thickening of
(G, F )
for some valid
F ⊆ V (G)2
{x, y} ∈ F ,
and
then
x
and
are not in the strong core.
A
three-cliqued claw-free graph (G, A, B, C)
cliques
A, B, C
of
cliqued graph is a
G,
pairwise disjoint and with union
3-colored graph.
three-cliqued graph, where
Let
n ≥ 0,
with vertex set
V (G1 ) ∪ · · · ∪ V (Gn )
•
for
1 ≤ i ≤ n, G|V (Gi ) = Gi ;
•
for
1 ≤ i < j ≤ n , Ai
•
for
1 ≤ i < j ≤ n,
if
(Gc , A, B, C)
A, B, C
is a
is a
and
worn hex-chain
(Gc1 , A1 , B1 , C1 )
G1 , . . . , G n .
If
If each
(Gi , Ai , Bi , Ci )
let
C = C1 ∪ · · · ∪ Cn ,
V (Gj ) \ Bj ; Bi
and let
G
be a
be the graph
V (Gj ) \ Cj ;
is complete to
are adjacent then
v ∈ Cj ,
and if
u ∈ Ci
(G, A, B, C)
and so
When
(Gc , A, B, C),
(Gc2 , A2 , B2 , C2 ).
Gi
1 ≤ i ≤ n,
graph; we call the sequence
(G2 , A2 , B2 , C2 ).
for
and
and
are cliques,
3-colored
and
v ∈ Bj
and
u ∈ Bi
worn hex-chain for (G, A, B, C).
(G1 , A1 , B1 , C1 )
The complement of a tree-
and
Ci
and
u ∈ Ai
and the same applies if
In particular,
and three
and with adjacency as follows:
is complete to
V (Gj ) \ Aj ;
is complete to
and for
V (G).
G
V (G1 ), . . . , V (Gn ) are all nonempty and pairwise vertex-disjoint.
A = A1 ∪ · · · ∪ An , B = B1 ∪ · · · ∪ Bn ,
Let
consists of a claw-free graph
n=2
we say that
and
v ∈ Aj .
(Gi , Ai , Bi , Ci ) (i = 1, . . . , n)
(G, A, B, C)
G
n = 2, (Gc , A, B, C)
and if each
is a
a
worn hex-join of
(Gci , Ai , Bi , Ci ) (i = 1, . . . , n)
Note also that every triad of
is claw-free then so is
are both in no triads;
is a three-cliqued graph and
Similarly, the sequence
and when
u, v
Gci
is a
G
worn hex-join
of
is a triad of one of
is prismatic then so is
Gc .
(G, A, B, C) is a three-cliqued graph, and {A0 , B 0 , C 0 } = {A, B, C}, then (G, A0 , B 0 , C 0 )
is also a three-cliqued graph, that we say is a
permutation of (G, A, B, C).
A list of the denitions needed for the study of orientable prismatic graphs can be found
in appendix A.1. The structure of prismatic graphs has been extensively studied in [11] and
[12]; the resulting two main theorems are the following.
4.6.1. Every orientable prismatic graph that is not 3-colorable is either not 3-substantial, or
a cycle of triangles graph, or a ring of ve graph, or a mantled L(K3,3 ).
65
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
4.6.2. Every 3-colored prismatic graph admits a worn chain decomposition with all terms in
Q0 ∪ Q1 ∪ Q2 .
In the remainder of the section, we use these two results to prove our main theorem for
complements of orientable prismatic graphs.
We begins with some results that deal with
the various outcomes of 4.6.1.
4.6.3. Let H be a prismatic cycle of triangles and G be a reduced thickening of (H, F ) for
some valid F ⊆ V (H)2 such that χ(G) > ω(G). Then there exists a Tihany brace or triangle
in G.
Proof.
Let the set
we may assume
Xi
be as in the denition of a cycle of triangles. Up to renaming the sets,
|X̂2n | = |X̂4 | = 1.
Let
u ∈ X̂2i
j=1
|X̂2 | > 1,
then
|R1 | = |L3 | = ∅
there is a Tihany brace in
anti-matching between
G.
R1
If
and
hence
uv
is an edge. We have
Xj ∪ R1 ∪ L3 .
mod 3,j≥4
and so
|X̂2 | = 1,
L3 .
v ∈ X̂4 ;
[
CH ({u, v}) =
If
and
CH ({u, v})
is a clique.
the only non-edges in
Therefore by 4.3.6,
G|CH ({u, v})
are a perfect
Hence by 4.3.8, there is a Tihany triangle in
G.
This
proves 4.6.3
4.6.4. Let H be a ring of ve graph. Let G be a reduced thickening of (H, F ) for some valid
F ⊆ V (H)2 such that χ(G) > ω(G). Then there is a Tihany triangle in G.
Proof.
Let
and thus
a2 , b3 , a4
{a2 , b3 , a4 }
{a2 , b3 , a4 } ∩ E = ∅
C({a2 , b3 , a4 }) = V2 ∪ V4
be as in the denition of a ring of ve.
is a dense triangle.
for all
E ∈ F.
By the denitions of
H
and
F,
it follows that
Hence by 4.3.6, there exists a Tihany triangle in
G.
This
proves 4.6.4.
4.6.5. Let H be a mantled L(K3,3 ) and G be a reduced thickening of (H, F ) for some valid
F ⊆ V (H)2 . If χ(G) > ω(G), then there exists a Tihany brace in G.
Proof.
Assume not. Let
W, aij , V i , Vi
the clique corresponding to
aij
be as in the denition of mantled
in the thickening and
66
W
(resp.
Vi , V i )
L(K3,3 ).
Let
Xji
be
be the set of vertices
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
corresponding to
W
(resp.
Vi , V i )
in the thickening. Let
xji ∈ Xij , V = ∪3i=1 Vi ∪ V i
and
k = χ(G).
Recall that for a clique
K,
M (K) = V (G) \ (C(K) ∪ A(K)).
and
A(K) = {x ∈ V (G) : ux ∈
/ E(G)
we dene
MV (E) := M (E) ∩ V , AW (E) := A(E) ∩ W
and let
(1)
S
be a color class in a
E = {x11 , x12 }
and
of
G,
X11 ∪ X21 ∪ X31
we deduce that
V 2 ∪ V 3,
Since
AV (E) := A(E) ∩ V .
0
Let
E = {xji , xji0 }
G\E .
x
is complete to
y, z ∈
/ V1 ∪ V2 ∪ V3
V 1 ∪ V 2 ∪ V 3,
and
it follows that
is a clique, we deduce that
assume by symmetry that
y, z ∈
/
V1 ∪ V2 ∪ V3
Xij , for
and
Xij ,
|{y, z} ∩ (V 1 ∪ V 2 ∪ V 3 )| ≤ 1.
z ∈ V 2 ∪ V 3.
But
for
i = 1, 2, 3 j = 2, 3.
|{y, z} ∩ (X11 ∪ X21 ∪ X31 )| ≤ 1.
y ∈ X11 ∪ X21 ∪ X31
and
Since
Hence, we may
X11 ∪ X21 ∪ X31
is complete
a contradiction. This proves (1).
If S ∩ MV (E) 6= ∅, then |S| ≤ 2.
Assume not.
E = {x11 , x12 }
and
we deduce that
V2 ∪ V3 ,
that
S = {x, y, z}
Let
x ∈ V1 .
Since
x
y, z ∈
/ V1 ∪ V2 ∪ V3
it follows that
But
and without loss of generality we may assume that
is complete to
and
V1 ∪ V2 ∪ V3
V2 ∪V3
X2j ∪ X3j ,
for
X11 ∪ X12 ∪ X13
is a clique, we deduce
As
is complete to
X11 ∪X12 ∪X13 , a contradiction.
(k − 2)-coloring
of
G\E
By (1) and (2), it follows that color classes with vertices in
y ∈ V2 ∪ V3
E=
0
{xji , xji0 }
i, j = 1, 2, 3,
and
This proves (2).
must have a vertex in
AV (E) ∪ MV (E)
C(E).
have size 2.
|AV (E) + |MV (E)| + 21 |AW (E)| + 12 |MW (E)| ≤ |C(E)| − 2.
this inequality on all braces
j = 1, 2, 3,
Since there is no triad
Hence we may assume by symmetry that
By 4.3.1, every color class of a
Hence we deduce that
and
y, z ∈
/ X2j ∪ X3j , for j = 1, 2, 3.
|{y, z} ∩ (V2 ∪ V3 )| ≤ 1.
|{y, z} ∩ (X11 ∪ X12 ∪ X13 )| ≤ 1.
z ∈ X11 ∪X12 ∪X13 .
3
MW (E) := M (E) ∩ W ,
let
and without loss of generality we may assume that
x ∈ AV (E) = V 1 .
Since there is no triad in
in
(k − 2)-coloring
S = {x, y, z}
Let
i = 1, 2, 3 j = 2, 3,
(2)
and
in
x ∈ K}
If S ∩ AV (E) 6= ∅, then |S| ≤ 2.
Assume not.
to
E
For a brace
for all
Summing
it follows that
X
X
X
X j
4X j
8X j
(|Vi | + |V i |) + 6
(|Vi | + |V i |) +
|Xi | +
|Xi | < 9
(|Vi | + |V i |) + 6
|Xi |,
2
2
i
i
i,j
i,j
which is a contradiction. This proves 4.6.5.
67
i
i,j
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
4.6.6. Let (H, H1 , H2 , H3 )c be a path of triangle s and (I, I1 , I2 , I3 ) an antiprismatic threecliqued graph. Let G be a worn hex-join of (H, H1 , H2 , H3 ) and (I, I1 , I2 , I3 ), and G0 be a
reduced thickening of (G, F ) for some valid F ∈ V (G)2 such that χ(G0 ) > ω(G0 ). Then there
exists a Tihany clique K in G0 , with |K| ≤ 4.
Proof.
we may assume that
Assume rst that
in
G.
Xj
Assume not. Let the set
Moreover
Hi = ∪j=i
|X̂2i | > 1
{u, v}
H
of
be as in the denition of a path of triangles and
mod 3 Xj .
for some
i.
Let
u ∈ X2i−2
[
[
Xj ∪
mod 3,
j=2i+2
j=2i−2
j≥2i+2
in
k = 2i + 1 mod 3.
G0 ,
so
uv
is an edge
Hence
CG ({u, v})
n≥3
and let
Xj ∪ Ik
mod 3,
j≤2i−2
is a clique and so by 4.3.6, there is a Tihany brace
a contradiction. Hence we may assume that
Assume that
v ∈ X2i+2 ,
is in the strong core. Thus
CG ({u, v}) =
for
and
u ∈ X̂2 , v ∈ X̂6 .
|X̂2i | = 1
Then
uv
for all
i.
is an edge in
G.
Moreover
{u, v}
is in the strong core. Thus
[
CG ({u, v}) =
j=0
Hence
CG ({u, v})
is an antimatching, and by 4.3.8, there exists a Tihany triangle in
a contradiction. It follows that
Assume now that
X2 ∪ R3 ∪ L5 ∪ H3 .
n = 2.
Let
u ∈ X̂2 , v ∈ L5 .
G|C({u, v})
Thus
Thus we deduce that
be a neighbor of
n = 1.
v.
CG ({u, v}) ⊆ X2 ∪ L3 ∪ H3
G0 ,
G0 ,
uv
is an edge in
G and CG ({u, v}) =
R3
and
L5 .
Hence
a contradiction.
Assume that
|R1 | = |L3 | = 1.
Let
u ∈ X2
Without loss of generality, we may assume that
and
v ∈ L3 .
v ∈
Since
is a clique, it follows by 4.3.6 that there is a Tihany brace in
be adjacent. By 4.5.6, we may assume that
x ∈ I2
Then
is a perfect anti-matching between
a contradiction. Hence we deduce that
that there exists
G0 ,
n ≤ 2.
by 4.3.8, there is a Tihany triangle in
R1 ∪ L3
Xj ∪ X2 ∪ R3 ∪ L5 ∪ H3 .
mod 3, j≥6
such that
x
|R1 | = |L3 | > 1.
G
is not a
u ∈ R1
2-non-substantial
is in a triad. Thus
68
Now, let
CG ({u, v, x})
and
v ∈ L3
graph. If follows
is an antimatching,
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
and by 4.3.8, there exists a Tihany clique
K
in
G0
with
|K| ≤ 4,
a contradiction.
This
proves 4.6.6.
4.6.7. Let (G, A, B, C) be an antiprismatic graph that admit a worn chain decomposition
(Gi , Ai , Bi , Ci ). Suppose that there exists k such that (Gk , Ak , Bk , Ck ) is the line graph of
K3,3 . Let G0 be a reduced thickening of (G, F ) for some valid F ∈ V (G)2 . If χ(G0 ) > ω(G0 ),
then there is a Tihany brace in G0 .
Proof.
Assume not. Let
Xji = Xai
j
{aij }i,j=1,2,3
be the vertices of
be the clique corresponding to
aij
Gk
using the standard notation. Let
in the thickening.
Moreover, let
xji ∈ Xij ,
wij = |Xij |.
Since all of the vertices in the thickening of
Gk
are in triads,
Gk
is linked to the rest of
the graph by a hex-join.
Note that
G\{x11 , x12 }
containing a vertex in
is
(χ(G) − 2)-colorable.
X12 ∪ X13
w12 + w13 ≤ w21 + w31 − 1
By 4.3.1, it follows that every color class
must have a vertex in
and by symmetry
X21 ∪ X31 .
Hence we deduce that
w22 + w23 ≤ w11 + w31 − 1.
Summing these two
inequalities, it follows that
w12 + w13 + w22 + w23 < w21 + w11 + 2w31 .
A similar inequality can be obtained for all edges
4
P
ij
wij < 2
P
ij
wij + 2
P
ij
wij ,
xji xji0 .
Summing them all, we deduce that
a contradiction. This proves 4.6.7
4.6.8. Let H be a 3-colored prismatic graph. Let G be a reduced thickening of (H, F ) for
some valid F ⊆ V (H)2 such that χ(G) > ω(G). Then there exists a Tihany brace or triangle
in G.
Proof.
By 4.6.2,
H
admits a worn chain decomposition with all terms in
one term of the decomposition is in
K
in
G
with
|K| ≤ 4 G.
G.
If
then by 4.6.6, it follows that there is a Tihany clique
If one term of the decomposition is in
that there is a Tihany brace in
there are no triads in
Q2
Q0 ∪ Q 1 ∪ Q 2 .
Q1 ,
then by 4.6.7, it follows
Hence we may assume that all terms are in
G and thus by 4.5.1,
Q0 .
Therefore
it follows that there is a Tihany brace in
proves 4.6.8.
69
G.
This
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
We can now prove the main result of this section.
4.6.9. Let H be an orientable prismatic graph. Let G be a reduced thickening of (H, F ) for
some valid F ⊆ V (H)2 such that χ(G) > ω(G). Then there exists a Tihany clique K in G
with |K| ≤ 4.
Proof.
G
If
H
admits a worn chain decomposition with all terms in
admits a Tihany brace or triangle. Otherwise, by 4.6.1,
cycle of triangles, a ring of ve graph, or a mantled
If
H
is not
3-substantial,
H
Q0 ∪Q1 ∪Q2 , then by 4.6.8,
is either not
K
in
G
with
cycle of triangles, then by 4.6.3, there is a Tihany brace or triangle in
ve graph, then by 4.6.4, there is a Tihany triangle in
4.7
G.
a
L(K3,3 ).
then by 4.5.7, there is a clique
then by 4.6.5, there is a Tihany brace in
3-substantial,
G.
Finally, if
H
|K| ≤ 4.
G.
If
H
If
H
is a
is a ring of
is a mantled
L(K3,3 ),
This proves 4.6.9.
Non-orientable Prismatic Graphs
The denitions needed to understand this section can be found in appendix A.2.
The
following is a result from [12].
4.7.1. Let G be prismatic. Then G is orientable if and only if no induced subgraph of G is
a twister or rotator.
In the following two lemmas, we study complements of orientable prismatic graphs. We
split our analysis based on whether the graph contains a twister or a rotator as an induced
subgraph.
4.7.2. Let H be an non-orientable prismatic graph. Assume that there exists D ⊆ V (H) such
that G|D is a rotator. Let G be a reduced thickening of (H, F ) such that χ(G) > ω(G) for
some valid F ⊆ V (H)2 . Then there exists a Tihany clique K in G with |K| ≤ 5.
Proof.
let
Ai
Assume not. Let
D = {v1 , . . . , v9 } be as in the denition of a rotator.
be the set of vertices of
{v1 , v2 , v3 }
V (H)\D
is a triangle, it follows that
that are adjacent to
vi .
A1 ∪ A2 ∪ A3 = V (H)\D.
70
Since
H
For
i = 1, 2, 3,
is prismatic and
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
I1
Let
{{5, 6}, {5, 9}, {6, 8}, {8, 9}}, I2
=
I3 = {{4, 5}, {4, 8}, {5, 7}, {7, 8}}.
tices of
V (H)\D
i = 1, 2, 3
For
H
and
{k, l} ∈ Ii
and
let
Ak,l
i
and
be the set of ver-
{v1 , v2 , v3 } and {vi , vi+3 , vi+6 } are
S
k,l
that Ai =
{k,l}∈Ii Ai for i = 1, 2, 3.
Since
is prismatic, we deduce
and since
{k, l} ∈ Ii ,
and
{vi , vk , vl }.
that are complete to
i = 1, 2, 3
triangles for
i = 1, 2, 3
For
{{4, 6}, {4, 9}, {6, 7}, {7, 9}}
=
{v1 , v4 , v7 }, {v2 , v5 , v8 }, {v3 , v6 , v9 }
are triangles and
H is prismatic, it follows that Ak,l
i is anticomplete to vm for all m ∈ {4, 5, 6, 7, 8, 9}\{i, k, l}.
4,9
4,8
4,9
Assume that A2 and A3 are not empty. Since H is prismatic, we deduce that A2 is
4,8
4,9
4,8
anticomplete to A3 in H . Let x ∈ A2 and y ∈ A3 . Then CH ({v1 , v5 , v6 , x, y}) is a clique
and
{v1 , v5 , v6 , x, y}
size 5 in
is in the strong core. Hence by 4.3.6, there exists a Tihany clique of
G.
A4,9
2
Assume now that
CH ({v1 , v5 , v6 , x})
A34,8
is empty.
is a clique and
{v1 , v5 , v6 , x}
is in the core.
{v2 , v5 , v8 }
is a triad and
v2
the strong core. Since
there exists
is not empty, but
E ∈ F
with
v5 ∈ E ,
then
We may now assume that
K
But
of size 4 in
= ∅.
A24,9 = A4,8
3
Since
v8
H
is an anti-matching.
strong core. For
i = 2, 3, since {vi , vi+3 , vi+6 } is a triad and vi
that if there exists
v6
and
v9
E∈F
with
is not adjacent to
vi+3 ∈ E ,
v5 .
then
Moreover
Then
{v1 , v6 , x}
is in
is not adjacent to
v6
in
H.
G.
CH ({v1 , v5 , v6 })
Moreover
x ∈ A4,9
2 .
is in the strong core, it follows that if
E = {v5 , v8 }.
Hence by 4.3.6, there exists a Tihany clique
Let
{v1 , v5 , v6 }
is prismatic, it follows that
is in the core and
v1
is in the
is in the strong core, it follows
E = {vi+3 , vi+6 }.
But
v8
is not adjacent to
Hence by 4.3.6, there exists a Tihany triangle in
G.
This
concludes the proof of 4.7.2.
4.7.3. Let H be a non-orientable prismatic graph. Assume that there exists W ⊆ V (H)
such that H|W is a twister. Further, assume that there is no induced rotator in H . If G is
a reduced thickening of (H, F ) such that χ(G) > ω(G), then there exists a Tihany clique K
in G with |K| ≤ 4.
Proof.
Assume not.
Let
W = {v1 , v2 , . . . , v8 , u1 , u2 }
Throughout the proof, all addition is modulo 8. For
tices in
V \W
that are adjacent to
are adjacent to
vi
and
vi+2 .
be as in the denition of a twister.
i = 1, . . . , 8,
let
Ai,i+1
be the set of ver-
vi and vi+1 and let Bi,i+2 be the set of vertices in V \W
Moreover, let
C ⊆ V \W
71
that
be the set of vertices that are anticom-
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
plete to
over
Ai,i+1
Since
to
W.
H
Since
is prismatic, we deduce that
{vi , vi+1 , vi+3 , vi+6 }
is complete to
is prismatic, it follows also that
W \{vi , vi+2 , ui
(1)
H
mod 2 }. Moreover,
C
S8
i=1 (Ai,i+1
∪ Bi,i+2 ) ∪ C = V \W .
and anticomplete to
Bi,i+2
is complete to
is anticomplete to
More-
W \{vi , vi+1 , vi+3 , vi+6 }.
mod 2 } and anticomplete
ui
{v1 , v2 , . . . , v8 }.
There exists i ∈ {1, . . . , 8}, such that Ai,i+1 and Ai+3,j+4 are either both empty or both
non-empty.
Assume not. By symmetry we may assume that
Since
A1,2
is not empty, we deduce that
follows that
A7,8
v6 , v3 , v7 , v2 , y}
(2)
A3,4
and
A6,7
are not empty. Let
A1,2
is not empty and
is empty. Since
x ∈ A7,8
and
A4,5
and
y ∈ A3,4 .
A4,5
A6,7
Then
is empty.
are empty, it
G|{v8 , u1 , v4 , x,
is a rotator, a contradiction. This proves (1).
If Ai,i+1 and Ai+3,i+4 are both non-empty for some i ∈ {1, . . . , 8}, then there exists a
Tihany clique of size 5 in G.
Assume that
neighborhood of
A2,3
and
A5,6
are not empty and let
{v1 , v7 , u2 , x, y}
in
H
x ∈ A2,3
is a stable set. Moreover,
strong core and hence by 4.3.6 there is a Tihany clique of size
(3)
5
and
y ∈ A5,6 .
The anti-
{v1 , v7 , u2 , x, y}
in
G.
is in the
This proves (2).
If Ai,i+1 and Ai+3,i+4 are both empty for some i ∈ {1, . . . , 8}, then there exists a Tihany
clique of size 4 in G.
Assume that
H
is
A2,3
and
A5,6
are both empty. Then the anti-neighborhood of
A8,2 ∪A2,4 ∪A4,6 ∪A6,8 which is a matching.
is in the core. Possibly
Moreover
u2 is in the strong core and {v1 , v7 }
{v1 , v5 } and {v3 , v7 } are in F , but A2,8 ∪A2,4 ∪A4,6 ∪A6,8 ∪ {v3 , v7 } is
also an anti-matching. Hence by 4.3.8, there is a Tihany clique of size
Now by (1), there exists
non-empty. If
of size 5 in
size 4 in
Ai,i+1
G.
G.
If
and
Ai,i+1
{v1 , v7 , u2 } in
i
such that
Ai+3,i+4
and
Ai,i+1
and
Ai+3,i+4
4 in G.
This proves (3).
are either both empty or both
are both non-empty, then by (2) there is a Tihany clique
Ai+3,i+4
are both empty, then by (3) there is a Tihany clique of
This concludes the proof of 4.7.3.
4.7.4. Let H be a non-orientable prismatic graph. Let G be a reduced thickening of (H, F )
for some valid F ⊆ V (G)2 such that χ(G) > ω(G); then there exists a Tihany clique K in
G with K ≤ 5.
72
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
Proof.
By 4.7.1, it follows that there is an induced twister or an induced rotator in
there is an induced rotator in
size
5
in
G.
H,
If
then by 4.7.2, it follows that there is a Tihany clique of
If there is an induced twister and no induced rotator in
G.
follows that there is a Tihany clique of size 4 in
4.8
H.
H,
then by 4.7.3, it
This proves 4.7.4.
Three-cliqued Graphs
In this section we prove 4.1.1 for those claw-free graphs
into three cliques.
Section 4.6.
G for which V (G) can be partitioned
The denition of three-cliqued graphs has been given at the start of
A list of three-cliqued claw-free graphs that are needed for the statement of
the structure theorem can be found in appendix A.3. We begin with a structure theorem
from [13].
4.8.1. Every three-cliqued claw-free graph admits a worn hex-chain into terms each of which
is a reduced thickening of a permutation of a member of one of T C 1 , . . . , T C 5 .
(G, A, B, C)
Let
strongly Tihany
hex-join
be a three-cliqued graph and
if for all three-cliqued graphs
K
be a clique of
(H, A0 , B 0 , C 0 ), K
G.
We say that
K
is
is Tihany in every worn
(I, A ∪ A0 , B ∪ B 0 , C ∪ C 0 ) of (G, A, B, C) and (H, A0 , B 0 , C 0 ) such that χ(I) > ω(I).
A clique
and every
K
is said to be
v∈K
bi-cliqued
if exactly two of
is in a triad. A clique
are all non-empty and every
v∈K
K
K ∩ A, K ∩ B, K ∩ C
is said to be
tri-cliqued
if
are non-empty
K ∩ A, K ∩ B, K ∩ C
is in a triad.
4.8.2. Let K be a dense clique in (G, A1 , A2 , A3 ). If both K and C(K) are bi-cliqued, then
K is strongly Tihany.
Proof.
Let
hex-join of
(G0 , A0 , B 0 , C 0 ) be a three-cliqued
(G, A, B, C)
complete to
and
C(K) ∩ V (G).
(G0 , A0 , B 0 , C 0 ).
claw-free graph and let
Then in
Hence, by 4.3.2,
K
(H, D, E, F ) be a worn
H , C(K) ∩ V (G0 )
is Tihany in
H
is a clique that is
and hence
H
is strongly
Tihany.
4.8.3. Let K be a dense clique of a three-cliqued graph (G, A, B, C). If K is tri-cliqued, then
K is strongly Tihany.
73
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
Proof.
Let
join of
(G, A, B, C)
(G0 , A0 , B 0 , C 0 )
H.
is a clique in
be a three-cliqued claw-free graph and let
(G0 , A0 , B 0 , C 0 ).
and
Hence, by 4.3.2,
K
(H, D, E, F )
H , CH (K) ∩ V (G0 ) = ∅
Then in
be a hex-
and thus
CH (K)
is strongly Tihany.
4.8.4. Let (G, A, B, C) be an element of T C 1 and G0 be a reduced thickening of (G, F ) for
some valid F ⊆ V (G)2 . Then there is either a strongly Tihany brace or a strongly Tihany
triangle in G0 .
Proof.
H, v1 , v2 , v3
Let
of vertices of
H
similarly. Let
V123
that are adjacent to
Suppose that
in
H.
Vij 6= ∅
Then
Tihany brace in
for some
vij vi
V123
v2
and
H
xj
L(H) = G.
but not to
Then let
and
so
v3
V12
be the set
V13 , V23
be dened
Let
and let
{v1 , v2 , v3 }.
vij ∈ Vij ,
and let
be the vertex in
G
xi
be the vertex in
G
corresponding to the edge
and thus by 4.3.5 and 4.8.2, there exists a strongly
G0 .
is not empty. Let
to the edges
in
i, j .
CG ({xi , xj }) = ∅,
So we may assume that
that
v1
be the set of vertices complete to
corresponding to the edge
vij vj
T C1;
be as in the denition of
vv1 , vv2 , vv3
of
Vij = ∅
for all
v ∈ V123
H,
i, j .
and let
Then from the denition of
x1 , x2 , x3
respectively. Then
be the vertices in
CG ({x1 , x2 , x3 }) = ∅
and 4.8.3, there exists a strongly Tihany triangle in
G0 .
G
T C1,
it follows
corresponding
and hence by 4.3.5
This proves 4.8.4.
4.8.5. Let (G, A, B, C) be an element of T C 2 and let (G0 , A0 , B 0 , C 0 ) be a reduced thickening
of (G, F ) for some valid F ⊆ V (G)2 . Then there is either a strongly Tihany brace or a
strongly Tihany triangle in G0 .
Proof.
Let
Σ, F1 , . . . , Fk , L1 , L2 , L3 be as in the denition of T C 2 .
we may assume that
there exists
Fi
A
such that
is not anticomplete to
Fi ∩A and Fi ∩B
B.
It follows from the denition of
are both not empty. Let
and without loss of generality, we may assume that
Let
Fi
be such that there exists no
j
with
{xk , . . . , xl }
Fi ⊂ Fj .
and without loss of generality, we may assume that
C({xk , xl }) = {xk+1 , . . . , xl−1 },
of
Fi ,
Without loss of generality,
it follows that
Let
G.
This proves 4.4.2.
74
Σ.
{xk , . . . , xl } = V (H) ∩ Fi
are in order on
is dense. If
it follows by 4.3.1 and 4.3.5 that there is a Tihany brace in
there exists a Tihany brace in
that
{xk , . . . , xl } = V (H)∩Fi
are in order on
{xk , . . . , xl }
{xk , xl }
G
xk , xl
G.
Σ.
Since
are the endpoints
Otherwise, by 4.3.6
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
4.8.6. Let (G, A, B, C) be an element of T C 3 and let (G0 , A0 , B 0 , C 0 ) be a reduced thickening
of (G, F ) for some valid F ∈ V (G)2 . Then there is either a strongly Tihany brace or a
strongly Tihany triangle in G0 .
Proof.
Let
H, A = {a0 , a1 , . . . , an }, B = {b0 , b1 , . . . , bn }, C = {c1 , . . . , cn },
the denition of near-antiprismatic graphs. Suppose that for some
since
in
G,
X
be as in
i, ai , bi ∈ V (G).
|C \ X| ≥ 2, it follows that there exists j 6= i such that cj ∈ V (G).
is dense and tri-cliqued in
and
Now
Then
T = {ai , bi , cj }
and so by 4.3.5 and 4.8.3 there is a strongly Tihany triangle
G0 .
So we may assume that for all i, if
ai ∈ V (G), then bi 6∈ V (G).
every vertex is in a triad, it follows that
ai , aj ∈ V (G)
G.
i 6= j .
Then
C(E)
i 6= j
j 6= i we have cj ∈ V (G).
T C3
Now suppose that
is a non-reduced homogeneous pair in
ai , aj
at most one of
Now
ai ∈ V (G).
whenever
({ai , aj }, {ci , cj })
Hence we may assume that for all
then for some
G0 .
for some
ci ∈ V (G)
Since by denition of
is in
V (G).
Let
ai ∈ V (G),;
E = {ai , cj } is dense and bi-cliqued.
is bi-cliqued, hence by 4.3.5 and 4.8.2, it follows that
E
Moreover
is a strongly Tihany brace in
This proves 4.8.6.
4.8.7. Let G be an element of T C 5 and G0 be a reduced thickening of (G, F ) for some valid
F ⊆ V (G)2 . Then there exists either a brace E ∈ V (G0 ) that is strongly Tihany or a triangle
T ∈ V (G0 ) that is strongly Tihany in G0 .
Proof.
G ∈ T C 15 .
First suppose that
v4 ∈ V (G)
then
{v2 , v4 }
Let
H, {v1 , . . . , v8 }
is dense and bi-cliqued.
be as in the denition of
Moreover
thus by 4.3.5 and 4.8.2, there is a strongly Tihany brace in
dense and bi-cliqued. Moreover
a strongly Tihany brace in
T = {v1 , v6 , v7 }
G0 .
G0 .
If
If
is bi-cliqued and
v 3 ∈ G,
then
{v3 , v5 }
is
is bi-cliqued and so by 4.3.5 and 4.8.2, there is
So we may assume that
v4 , v3 6∈ V (G).
But then the triangle
is dense and tri-cliqued and thus by 4.3.5 and 4.8.3, there exists a strongly
Tihany triangle in
G0 .
We may assume now that
C({v2 , v3 })
C({v3 , v5 })
C({v2 , v4 })
T C 15 .
is bi-cliqued.
G ∈ T C 25 .
Otherwise,
If
v3 ∈ G
{v2 , v4 }
then
{v2 , v3 }
is dense, bi-cliqued and
is dense, bi-cliqued and
C({v2 , v4 })
is bi-
cliqued. In both cases, it follows from 4.3.5 and 4.8.2 that there exists a strongly Tihany
brace in
G0 .
This proves 4.8.7.
75
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
We are now ready to prove the main result of this section.
4.8.8. Let G be a three-cliqued claw-free graph such that χ(G) > ω(G). Then G contains
either a Tihany brace or a Tihany triangle in G.
Proof.
By 4.8.1, there exist
(Gi , Ai , Bi , Ci ),
(Gi , Ai , Bi , Ci ) (i = 1, . . . , n)
is
a
worn
for
i = 1, . . . , n,
hex-chain
for
such that the sequence
(G, A, B, C)
and
such
that
(Gi , Ai , Bi , Ci ) is a reduced thickening of a permutation of a member of one of T C 1 , . . . , T C 5 .
If there exists
i ∈ {1, . . . , n}
tation of a member of
such that
(Gi , Ai , Bi , Ci )
T C1, T C2, T C3,
or
T C5,
is a reduced thickening of a permu-
then by 4.8.4,
4.8.5,
4.8.6, or 4.8.7 (re-
spectively), there is a strongly Tihany brace or a strongly Tihany triangle in
there is a Tihany brace or a Tihany triangle in
reduced thickening of a member of
T C4
for all
G.
Thus it follows that
i = 1, . . . , n.
Hence
Gi ,
and thus
(Gi , Ai , Bi , Ci )
is a
G is a reduced thickening
of a three-cliqued antiprismatic graph. By 4.6.8, there exists a Tihany brace or triangle in
G.
This proves 4.8.8.
4.9
Non-trivial Strip Structures
In this section we prove 4.1.1 for graphs
G that admit non-trivial strip structures and appear
in [13].
Let
Z
(J, Z)
be a strip. We say that
is complete to
(J, Z)
is a
line graph strip
if
|V (J)| = 3, |Z| = 2
and
V (J) \ Z .
The following two lemmas appear in [4].
4.9.1. Suppose that G admits a nontrivial strip-structure such that |Z| = 1 for some strip
(J, Z) of (H, η). Then either G is a clique or G admits a clique cutset.
4.9.2. Let G be a graph that admits a nontrivial strip-structure (H, η) such that for every
F ∈ E(H), the strip of (H, η) at F is a line graph strip. Then G is a line graph.
We now use these lemmas to prove the main result of this section.
4.9.3. Let G be a claw-free graph with χ(G) > ω(G) that is a minimal counterexample
to 4.1.1. Then G does not admit a nontrivial strip-structure (H, η) such that for each strip
76
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
(J, Z) of (H, η), 1 ≤ |Z| ≤ 2, and if |Z| = 2 then either |V (J)| = 3 and Z is complete to
V (J) \ Z , or (J, Z) is a member of Z1 ∪ Z2 ∪ Z3 ∪ Z4 ∪ Z5 .
Proof.
Suppose that
(J, Z)
of
that
G
|Z| = 2
|Z| = 1
Further suppose that
G
is a clique or
χ(G) > ω(G),
(H, η)
admits a nontrivial strip-structure
(H, η), 1 ≤ |Z| ≤ 2.
by 4.9.1 either
satisfy
G
for some strip
admits a clique cutset; in the former case
for all strips
(H, η)
Z1 = {a1 , b1 }.
B2 = NG (B1 ) \ V (J1 ).
Then
(1)
G
Then
does not
(J, Z).
are line graph strips, then by 4.9.2,
the result follows from [2]. So we may assume that some strip
Let
(J, Z).
and in the latter case 4.9.3 follows from 4.3.10. Hence we may assume
If all the strips of
strip.
such that for each strip
Let
is a line graph and
(J1 , Z1 )
is not a line graph
A1 = NJ1 (a1 ), B1 = NJ1 (b1 ), A2 = NG (A1 ) \ V (J1 ),
C1 = V (J1 ) \ (A1 ∪ B1 )
Let
G
and
and
C2 = V (G) \ (V (J1 ) ∪ A2 ∪ B2 ).
V (G) = A1 ∪ B1 ∪ C1 ∪ A2 ∪ B2 ∪ C2 .
If C2 = ∅ and A2 = B2 , then there is a Tihany clique K in G with |K| ≤ 5.
Note that
V (G) = A1 ∪ B1 ∪ C1 ∪ A2 .
(J1 , Z1 )
strip, it follows that
Since
|Z1 | = 2
and
(J1 , Z1 )
Z1 ∪ Z 2 ∪ Z 3 ∪ Z 4 ∪ Z 5 .
is a member of
is not a line graph
We consider the cases
separately:
1. If
(J1 , Z1 )
Z1 ,
is a member of
then
J1
is a fuzzy linear interval graph and so
G
is a
fuzzy long circular interval graph and 4.9.3 follows from [2].
2. If
(J1 , Z1 )
Z2 , Z3 ,
is a member of
cliques and so
V (G)
or
(J1 , Z1 )
and for
to
is a member of
1 ≤ i ≤ 12
let
Xvi
Z5 .
Let
Let
a2 , adjacent to v1 , v2 , v4
no triad of
H 00
thickening of
contains
(H 0 , F ),
K
with
A1 , B1 ,
A1 ∪ A2 , B1 ,
v9
or
and
v10 .
Then
H 00
Thus the pair
Hence,
H0
A2
Z5
is complete
by adding a new
is an antiprismatic graph. Moreover,
(H 0 , F )
is antiprismatic, and
so 4.9.3 follows from 4.6.9 and 4.7.4.
77
C1 .
be as in the denition of
be the graph obtained from
v5 .
are all
|K| ≤ 5.
v1 , . . . , v12 , X, H, H 0 , F
H 00
C1
and
and
be as in the denition of a thickening. Then
Xv1 ∪ Xv2 ∪ Xv4 ∪ Xv5 .
vertex
In all of these cases,
is the union of three cliques, namely
by 4.8.8, there exists a Tihany clique
3. If
Z4 .
G
is a
CHAPTER 4. ON THE ERDŽS-LOVÁSZ TIHANY CONJECTURE
This proves (1).
By (1), we may assume
A2
it follows that
assume that
v ∈ A2 \ B2
4.10
C2 6= ∅
or
A2 6= B2 .
is a clique cutset of
A2 6= B2
and let
G
Suppose that
A2 = B2 .
Then since
and the result follows from 4.3.10. Hence, we may
and without loss of generality we may assume that
w ∈ A1 \ B 1 .
Then
C2 6= ∅
A2 \ B2 6= ∅.
Let
E = {v, w} is dense and 4.9.3 follows from 4.3.2.
Proof of the Main Theorem
We can now prove the main theorem.
Proof of 4.1.1.
Let
not exist a clique
K
G
be a claw-free graph with
in
it follows that either
T2 .
V (G)
and suppose that there does
G with |K| ≤ 5 such that χ(G\K) > χ(G) − |K|.
G
By 4.4.1, it follows that
of
χ(G) > ω(G),
is a member of
T1 ∪ T 2 ∪ T 3
G is not a member of T1 .
By 4.6.9 and 4.7.4, we deduce that
G
or
V (G)
By 4.9.3 and 4.2.1,
is the union of three cliques.
By 4.4.2, it follows that
is not a member
T3 .
G is not a member
Hence, it follows that
is the union of three cliques. But by 4.8.8, it follows that there is a Tihany brace or
triangle in
G,
a contradiction. This proves 4.1.1.
78
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
Chapter 5
A Local Strengthening of Reed's
Conjecture
5.1
Introduction
The chromatic number is a notion of utmost importance in graph theory. Finding its exact
value for a graph is a central problem both from a theoretical and algorithmic point of view.
For general graphs, there is a trivial lower and upper bound on the chromatic number that
we present now. We include the proof for completeness.
5.1.1. Let G be a graph. Then ω(G) ≤ χ(G) ≤ ∆(G) + 1.
Proof.
Let
K
be a clique of size
we need at least
ω(G)
ω(G).
No two vertices of
colors for the vertices of
K.
K
can have the same color, hence
It follows that
χ(G) ≥ ω(G).
For the upper bound we will use induction on the number of vertices in
has one vertex, then
that for all graph
has
n−1
H
χ(G) = 1 ≤ ∆(G) + 1.
with
vertices and so
colors since
coloring of
|V (H)| < n
then
to a coloring of
G
Clearly if
G
G be such that |V (G)| = n and assume
χ(H) ≤ ∆(H) + 1.
χ(G\x) ≤ ∆(G\x) + 1 ≤ ∆(G) + 1.
|N (x)| = d(x) ≤ ∆(G).
G\x
Now let
G.
Let
But
x ∈ V (G).
N (x)
Now
uses at most
G\x
∆(G)
Therefore there is at least one color free to extend the
using at most
79
∆(G) + 1
colors. This proves 5.1.1.
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
In 1998, Reed made the following conjecture.
Conjecture 2
. For any graph G,
(Reed)
χ(G) ≤
1
+ 1 + ω(G)) .
2 (∆(G)
Conjecture 2 has been proved rst for line graphs [23] and was then extended to quasi-line
graphs [21; 22] and later claw-free graphs [21]. Later, King proposed a local strengthening
of Reed's Conjecture.
Conjecture 3
(King)
. For any graph G,
χ(G) ≤ max
1
v∈V (G)
2 (d(v)
+ 1 + ω(v)) .
There are several pieces of evidence that lend credence to Conjecture 3. First is the fact
that the result holds for claw-free graphs with stability number at most three [21]. However,
for the remaining classes of claw-free graphs, which are constructed as a generalization of
line graphs [10], the conjecture has remained open.
The second piece of evidence for Conjecture 3 is that the fractional relaxation holds. The
fractional chromatic number χf (G)
is the optimal value of the following linear program,
which is the linear relaxation of the standard integer program formulation of the graph
coloring problem.
χf (G) =
min
P
xS
S
subject to
P
xS ≥ 1 ∀ v ∈ V (G)
S3v
xS ∈ [0, 1] ∀
stable set
S
It was noted by McDiarmid as an extension of a theorem of Reed [27] that the following
holds.
5.1.2
. For any graph G,
(McDiarmid)
χf (G) ≤ max
v∈V (G)
1
2 (d(v)
The main result of this chapter is:
80
+ 1 + ω(v)) .
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
5.1.3. For any quasi-line graph G,
1
χ(G) ≤ max
2 (d(v)
v∈V (G)
This chapter is organized as follow.
graphs.
+ 1 + ω(v)) .
In Section 5.2, we prove Conjecture 3 for line
In Section 5.3, we introduce quasi-line graphs and some important concepts.
In
Section 5.4, we study how quasi-line graphs can be decomposed into smaller pieces that are
well understood, and nally in Section 5.5 we put the dierent pieces together to prove 5.1.3
and discuss some algorithmic notions.
5.2
Line Graphs
In order to prove Conjecture 3 for line graphs, we prove an equivalent statement in the setting
of edge colorings of multigraphs. Given distinct adjacent vertices
G,
µG (uv)
we let
denote the number of edges between
the maximum, over all vertices
{u, v, w}.
w∈
/ {u, v},
u
and
u
v.
and
v
in a multigraph
We let
tG (uv)
denote
of the number of edges with both endpoints in
That is,
tG (uv) :=
max
w∈N (u)∩N (v)
(µG (uv) + µG (uw) + µG (vw)) .
We omit the subscripts when the multigraph in question is clear.
Observe that given an edge
d(u) + d(v) − µ(uv) − 1.
incident to
see that
u,
ω(v)
e
in
L(G)
with endpoints
And since any clique in
the edges incident to
in
G
is equal to
v,
u
L(G)
and
v,
the degree of
containing
e
in
L(G)
is
comes from the edges
or the edges in a triangle containing
max{d(u), d(v), t(uv)}.
uv
u
and
v,
we can
Therefore we prove the following
theorem, which, aside from the algorithmic claim, is equivalent to proving Conjecture 3 for
line graphs:
5.2.1. Let G be a multigraph on m edges, and let
γl0 (G) := max
uv∈E(G)
max d(u) + 12 (d(v) − µ(vu)), d(v) + 21 (d(u) − µ(uv)),
1
2 (d(u)
+ d(v) − µG (uv) + t(uv)) .
Then χ0 (G) ≤ γl0 (G), and we can nd a γl0 (G)-edge-coloring of G in O(m2 ) time.
81
(5.1)
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
The most intuitive approach to achieving this bound on the chromatic index involves
assuming that
G−e
G
is a minimum counterexample, then characterizing
e.
for an edge
of
We want an algorithmic result, so we will have to be a bit more careful
to ensure that we can modify partial
we can extend to a complete
Our
γl0 (G)-edge-colorings
O(m2 )-time
γl0 (G)-edge-colorings
γl0 (G)-edge-coloring
of
eciently until we nd one that
G.
algorithm requires time-ecient data structures, i.e. a combination of
lists and matrices. Our algorithm will build the multigraph one edge at a time, maintaining
a proper
k -edge-coloring
given vertices
1
to
n
at each step (in this case,
and a multiset of
m
k = γl0 (G)).
We may assume we are
edges, which we rst sort at a cost of
time, then add to the graph in lexicographic order. We may also assume that
O(m log m)
G contains no
isolated vertices.
As we build the multigraph we maintain an
n×n
adjacency matrix, each cell of which
contains a list of edges between the two vertices in question. We also maintain a sorted list
of neighbors for each vertex, and a sorted list of edges incident to each vertex. Further, we
maintain a
k×n
color-vertex incidence matrix, and for each vertex
colors appearing on an edge incident to
v,
v
two lists: a list of the
and a list of the colors not incident to
these structures shall be connected with appropriate links. For example, if color
appear at vertex
v.
If
c
does appear at
will be linked to the corresponding node in the list of colors incident to
v
with color
the multigraph in time
and
t(uv)
structures in
does not
in
O(nm) ⊆ O(m2 ),
O(n)
v,
v,
the cell
as well as to the
c.
To initialize the coloring data structures we rst need to determine
µ(uv),
All
v , the corresponding cell of the color-vertex incidence matrix will be linked
to the corresponding node in the list of colors absent at
edge incident to
c
v.
for each edge
time. So we can determine
O(nm + γl0 (G)n) ⊆ O(m2 )
uv
γl0 (G)
in
γl0 (G).
After building
we can determine
O(nm)
d(u), d(v),
time and initialize the
time.
These structures allow us to update eciently: When we add an edge to the multigraph,
the fact that the edges are presorted allows us to update all lists and matrices in constant
time.
When changing the color of an edge, the interlinkedness of the matrices and lists
allows us to update in constant time.
We begin by dening, for a vertex
v,
a
fan hinged at v .
82
Let
e
be an edge incident
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
to
v,
v1 , . . . , v `
and let
c : E \ {e} → {1, . . . , k}
F = (e; c; v; v1 , . . . , v` )
j
than
vi
u∈
/ {v, v1 , . . . , v` }
fan.
is a
fan
The
size
v
and
missing
(i.e. a color
vj
`).
Then
i
there exists some
less
F
We say that
is
hinged at v .
is a fan, we say that
F
If there is no
is a
maximal
These fans generalize Vizing's fans, originally used in the proof of
incident to w
k -edge-coloring
of
w,
¯
C(w)
and we use
Fans allow us to modify partial
to denote
k ≥ γl0 (G),
k -edge-coloring
or we can easily nd a
of
G.
G−e
w.
w,
we say that a
C(w)
We use
to denote
[k] \ C(w).
then either every maximal fan has
For more general results related to
fans, see [35]. We rst prove that we can construct a
of
and a vertex
k -edge-colorings of a graph (specically those with exactly
one uncolored edge). We will show that if
k -edge-coloring
G
if the color appears on an edge incident to
the set of colors incident to
2
k.
Let
of a fan refers to the number of neighbors of the hinge vertex contained in
Vizing's theorem [37]. Given a partial
size
v1 .
and
is assigned a color that does not appear on any
vi ).
at
v
between
for some xed
2 ≤ j ≤ `,
such that
F 0 = (e; c; v; v1 , . . . , v` , u)
such that
the fan (in this case,
color is
j
if for every
e
with
G \ {e}
be a proper edge coloring of
such that some edge between
edge incident to
v
be a set of distinct neighbors of
k -edge-coloring
of
G
from a partial
whenever we have a fan for which certain sets are not disjoint.
5.2.2. For some edge e in a multigraph G and positive integer k, let c be a k-edge-coloring
¯ ∩ C(v
¯ j ) 6= ∅,
of G − e. If there is a fan F = (e; c; v; v1 , . . . , v` ) such that for some j , C(v)
then we can nd a k-edge-coloring of G in O(k + m) time.
Proof.
Let
j
result is trivial, since we can extend
and we can nd
j
in
O(m)
time.
f : {2, . . . , `} → {1, . . . , ` − 1}
between
v
and
vi
such that
increasing values of
We construct a
to a proper
e1
such that for each
c(ei )
is missing at
vi
k -edge-coloring
to be
i,
is nonempty. If
(1)
vf (i) .
e.
of
G.
j = 1,
k -edge-coloring cj
c(ef (j) )
from
k -edge-coloring cj
We can therefore extend
cj
of
then the
Otherwise
j ≥2
We then construct a function
f (i) < i
and (2) there is an edge
We can nd this function in
of
G − ej
ef (j)
to
G − ej
to a proper
from
ef (f (j)) ,
c
f
for
{ei }`i=2 .
ej
to
and so on, until we shift a color to
e.
by shifting the color
c(ej )
from
such that some color is missing at both
k -edge-coloring
83
ei
O(k + m)
at which each color is missing, and computing
starting at 2. While doing so we also nd the set of edges
shifting the color
We now have a
vj .
i
c
We dene
time by building a list of the earliest
ef (j) ,
¯ ∩ C(v
¯ j)
C(v)
be the minimum index for which
of
G
in
O(k + m)
time.
v
and
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
5.2.3. For some edge e in a multigraph G and positive integer k, let c be a k-edge-coloring
of G − e. If there is a fan F = (e; c; v; v1 , . . . , v` ) such that for some i and j satisfying
¯ i ) ∩ C(v
¯ j ) 6= ∅, then we can nd vi and vj in O(k + m) time, and we can
1 ≤ i < j ≤ `, C(v
nd a k-edge-coloring of G in O(k + m) time.
Proof.
We can easily nd
β
and let
and
be a color in
i
and
j
O(k + m)
in
¯ i ) ∩ C(v
¯ j ).
C(v
α
time if they exist. Let
be a color in
Note that by 5.2.2, we can assume
¯
C(v)
α ∈ C(vi ) ∩ C(vj )
β ∈ C(v).
Gα,β
Let
be the subgraph of
of
Gα,β
containing
of
Gα,β
not containing
k -edge-coloring
of
v , vi , or vj
v.
G−e
to apply 5.2.2 to nd a
G
containing those edges colored
is a path on
≥ 2 vertices.
Exchanging the colors
in which either
k -edge-coloring
Thus either
α and β
G.
or
β.
vi or vj
Every component
is in a component
on this component leaves us with a
¯ ∩ C(v
¯ i ) 6= ∅
C(v)
of
α
or
¯ ∩ C(v
¯ j ) 6= ∅.
C(v)
We can easily do this work in
This allows us
O(m)
time.
The previous two lemmas suggest that we can extend a coloring more easily when we
have a large fan, so we now consider how we can extend a fan that is not maximal. Given
a fan
F = (e; c; v; v1 , . . . , v` ),
we use
d(F )
to denote
d(v) +
P`
i=1 d(vi ).
5.2.4. For some edge e in a multigraph G and integer k ≥ ∆(G), let c be a k-edge-coloring of
G − e and let F be a fan. Then we can extend F to a maximal fan F 0 = (e; c; v; v1 , v2 , . . . , v` )
in O(k + d(F 0 )) time.
Proof.
We proceed by setting
F0 = F
maintain two color sets. The rst,
not between
in
C
v
and extending
C,
and another vertex of
color in
C
F 0.
The second,
but
consists of those colors that are
is maximal if and only if
C¯F 0 = ∅.
We
time by counting the number of times each
F 0 = (e; c; v; v1 , v2 , . . . , v` ),
may assume is not empty.
update
O(k + d(F ))
F0
C¯F 0 ,
v
appears incident to a vertex of the fan.
Now suppose we have
endpoint
until it is maximal. To this end we
consists of those colors appearing incident to
and are missing at some fan vertex. Clearly
can perform this initialization in
F0
v`+1 .
C¯F 0
Take an edge incident to
We now update
C
along with sets
v
C
and
with a color in
C¯F ;
by removing all colors appearing between
by removing all colors appearing between
84
v
and
v`+1 ,
v
C¯F 0 ,
which we
call its other
and
v`+1 .
We
and adding all colors in
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
¯ `+1 ).
C ∩ C(v
Set
F 0 = (e; c; v; v1 , v2 , . . . , v`+1 ).
We can perform this update in
d(v`+1 )
time;
the lemma follows.
We can now prove that if
we can nd a
k -edge-coloring
k ≥ γl0 (G)
of
G
in
and we have a maximal fan of size
O(k + m)
1
or at least
3,
time.
5.2.5. For some edge e in a multigraph G and positive integer k ≥ γl0 (G), let c be a k-edgecoloring of G − e and let F = (e; c; v; v1 ) be a fan. If F is a maximal fan we can nd a
k -edge-coloring of G in O(k + m) time.
Proof.
If
¯ ∩ C(v
¯ 1)
C(v)
k -edge-coloring
of
G.
is nonempty, then we can easily extend the coloring of
¯ ∩ C(v
¯ 1)
C(v)
So assume
nonempty. Therefore there is a color in
endpoint, call it
v2 ,
is not
v1 .
Thus
G−e
to a
¯ 1)
k ≥ γl0 (G) ≥ d(v1 ), C(v
is empty. Since
is
¯ 1 ) appearing on an edge incident to v whose other
C(v
(e; c; v; v1 , v2 )
is a fan, contradicting the maximality of
F.
5.2.6. For some edge e in a multigraph G and positive integer k ≥ γl0 (G), let c be a k-edgecoloring of G − e and let F = (e; c; v; v1 , v2 , . . . , v` ) be a maximal fan with ` ≥ 3. Then we
can nd a k-edge-coloring of G in O(k + m) time.
Proof.
Let
v0
denote
v
for ease of notation.
If the sets
pairwise disjoint, then using 5.2.2 or 5.2.3 we can nd a
¯ 0 ), C(v
¯ 1 ), . . . , C(v
¯ `)
C(v
k -edge-coloring
of
G
We can easily determine whether or not these sets are pairwise disjoint in
are not all
O(m)
time.
O(k + m)
time.
in
Now assume they are all pairwise disjoint; we will exhibit a contradiction, which is enough
to prove the lemma.
The number of missing colors at
if
i ∈ {0, 1}.
outside
Since
{v0 , . . . , v` }
F
vi ,
i.e.
¯ i )|,
|C(v
is
k − d(vi ) if 2 ≤ i ≤ `,
is maximal, any edge with one endpoint
must have a color not appearing in
`
X
!
k − d(vi )
+2+
d(v0 ) −
i=0
`
X
v0
¯ i ).
∪`i=0 C(v
and
k − d(vi ) + 1
and the other endpoint
Therefore
!
µ(v0 vi )
≤ k.
(5.2)
i=1
Thus
`k + 2 −
`
X
µ(v0 vi ) ≤
i=1
`
X
i=1
85
d(vi ).
(5.3)
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
But since
k ≥ γl0 (G),
(5.1) tells us that for all
i ∈ [`],
d(vi ) + 12 (d(v0 ) − µ(v0 vi )) ≤ k
Thus substituting for
k
(5.4)
tells us
`
X
d(v0 ) + 2d(vi ) − µ(v0 vi )
2
i=1
+2−
`
X
µ(v0 vi ) ≤
i=1
`
X
d(vi ).
i=1
So
2+
1
2 `d(v0 )
−
3
2
`
X
µ(v0 vi ) ≤ 0
i=1
2 + 12 `d(v0 ) ≤
`
2 d(v0 )
This is a contradiction, since
<
3
2
`
X
µ(v0 vi )
i=1
3
d(v
0 ).
2
` ≥ 3.
We are now ready to prove the main lemma of this section.
5.2.7. For some edge e0 in a multigraph G and positive integer k ≥ γl0 (G), let c0 be a
k -edge-coloring of G − e. Then we can nd a k -edge-coloring of G in O(k + m) time.
As we will show, this lemma easily implies 5.2.1. We approach this lemma by constructing
a sequence of overlapping fans of size two until we can apply a previous lemma. If we cannot
do this, then our sequence results in a cycle in
G
and a set of partial
k -edge-colorings
of
G
with a very specic structure that leads us to a contradiction.
Proof of 5.2.7.
Let
fan. If
v0
and
We postpone algorithmic considerations until the end of the proof.
v1
be the endpoints of
|{u1 , . . . , u` }| =
6 1,
e0 , and let F0 = (e0 ; c0 ; v1 ; v0 , u1 , . . . , u` ) be a maximal
then we can apply 5.2.5 or 5.2.6. More generally, if at any time
we nd a fan of size three or more we can nish by applying 5.2.6. So assume
is a single vertex; call it
Let
color
C¯0
v2 .
denote the set of colors missing at
α0 ∈ C¯0 .
Note that if
appears between
v1
{u1 , . . . , u` }
α0
and a vertex
v0
in the partial coloring
does not appear on an edge between
u∈
/ {v0 , v1 , v2 },
86
so there is a fan
c0 ,
v1
and take some
and
v2 ,
then
(e0 ; c0 ; v1 ; v0 , v2 , u)
α0
of size
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
3
v1
edge between
e1
Let
coloring
and
G − e1
of
fan. As with
from
F0 ,
v1
c0
α1 ∈ C¯1
that
α1
α1
in
c1 .
color
α1 .
F1
given color
C¯1 .
Now let
v2
and
α0
v3 .
at
vi+2
ci , we make αi
in the coloring
ci+2 )
j
fans until we reach some
c2
of
from
available at
for which
G − e0
the original edge-coloring of
vi
precisely if
of color
cycle
and
i
α1 ,
vi+1
precisely if
αi .
for
C,
nd a
H
j = 3.
A
j
α0
vj = v0
α1
of
be a maximal
v3
may
in
c0
G
α0
C¯0
and
j
e2
v3
we can see
having color
and assigning
e1
i = 0, 1, 2, . . .
for
(the set of colors missing
is odd. To see this, consider
1 ≤ i ≤ j − 1, α0
α1
j ).
v1
vi
and
vi+1
Let
(j − 1)/2
C
denote the
times on
C,
in a
G consisting of those edges between
Let
(and therefore in any
appears between
appears on an edge
F0 = Fj .
both appear
A
be the set of colors missing on
be the sub-multigraph of
87
and
c0 ,
form a matching, and so do the edges
since any two vertices of
We know that every color in
in
appears on an edge between
(with indices modulo
HA
v2
This is possible because when we
odd. Furthermore
and
v2
by uncoloring
H
consisting of
e0
and those
ci ).
If some color is missing on two vertices of
k -edge-coloring
denote the
which will inevitably happen if we never
be the sub-multigraph of
and let
edges receiving a color in
Suppose
is odd, and
In each coloring,
0 ≤ j ≤ j −1
at least one vertex of
C¯1
We continue constructing our sequence of
vj ∈ {vi }j−1
i=0 ,
and
and
vi+2 , so the set C¯i+2
and note that for
vj = v0
we indeed have
near-perfect matching. Let
vi+1
Let
Fi = (ei , ci ; vi+1 ; vi , vi+2 )
is even. Thus since the edges of color
v0 , v1 , . . . , vj−1 .
and
i
c1
αi+2 = αi .
always contains
v1
be the edge between
G − e2
nd a fan of size 3 or greater. We claim that
between
α0 .
F1 = (e1 ; c1 ; v2 ; v1 , v3 )
appears between
e2
Now let
in this way, maintaining the property that
from
color
We construct a new
exists and is indeed maximal. The vertex
We continue to construct a sequence of fans
ci+1
c0 .
in
e0
and assigning
c1 .
Just as
We construct a coloring
construct
α0
v0 .
be a color in
appears between
e1
by uncoloring
we can assume that
v2
and
in the coloring
or may not be the same as
Let
v1
denote the edge between
c1
does appear on an
v2 .
set of colors missing at
vi
α0
and apply 5.2.6 to complete the coloring. So we can assume that
C
C
in
c0 , c1 , or c2 , we can easily
are the endpoints of
and
v2 ,
e0 , e1 ,
and every color in
C¯1
or
e2 .
appears
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
between
v2
and
v3 = v0 .
|E(HA )| = |A| + 1.
Therefore
Therefore
2γl0 (G) ≥ dG (v0 ) + dG (v1 ) + tG (v0 v1 ) − µG (v0 v1 )
= dHA (v0 ) + dHA (v1 ) + 2(k − |A|) + tG (v0 v1 ) − µG (v0 v1 )
≥ dHA (v0 ) + dHA (v1 ) + 2(k − |A|) + tHA (v0 v1 ) − µHA (v0 v1 )
≥ 2|E(HA )| + 2(k − |A|)
> 2|A| + 2(k − |A|) = 2k
This is a contradiction since
Let
β
be a color in
k ≥ γl0 (G).
A \ {α0 , α1 }.
ci
then we can easily extend
we can easily
k -edge-color G,
number of vertices at which
β
β
is missing at two consecutive vertices
k -edge-coloring
to a
a maximal fan, we claim that if
If
β
j ≥ 5.
We can therefore assume that
G.
of
v1 , β
Likewise, since
vj−2 .
Finally, suppose
eβ
β
or the number of edges colored
is missing in any
by giving
is a
e2
color
β
k -edge-coloring
extend
G − e0
to a
k -edge-color G.
at which
β
v1
appears between
and
v2
with color
and giving
eβ
color
be the edge between
of
v1
G − e0
Fi
is
β
in
HA
in which
k -edge-coloring
of
α1
β
G.
and
β
is at least twice the
ci .
v1
vj−1 , β
is not missing at
β
vi+1 ,
is not missing at two consecutive vertices, then either
appears on an edge between
does.
and
Bearing in mind that each
β ∈ C¯0 .
To prove this claim, rst assume without loss of generality that
not missing at
vi
in
and
v2
appears on an edge between
v2 ,
and is missing at
c0 .
β
v0
and
is missing at
We therefore have at least two edges of
v3
in
eβ
HA
v3
colored
β
v1 .
or
and
vj−1
c0 .
is
α0
and
Then let
c00
from
c0
e2 ).
Thus
c00
We construct a coloring
is missing at both
β
for the same reason that
(i.e. we swap the colors of
Thus if
Since
We can therefore
vj−3
we can easily
for every vertex of
C
is missing, and we do not double-count edges. This proves the claim, and the
analogous claim for any color in
A
also holds.
Now we have
j−1
X
µHA (vi vi+1 ) = |E(HA )| > 2
i=0
j−1
X
(k − dG (vi )) .
(5.5)
i=0
Therefore taking indices modulo
j−1
X
j,
we have
dG (vi ) + 21 µHA (vi+1 vi+2 ) > jk.
i=0
88
(5.6)
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
Therefore there exists some index
i
for which
dG (vi ) + 12 µHA (vi+1 vi+2 ) > k.
(5.7)
k ≥ dG (vi ) + 12 µG (vi+1 vi+2 ) > k.
(5.8)
Therefore
This is a contradiction, so we can indeed nd a
that we can do so in
O(k + m)
Given the coloring
ci ,
k -edge-coloring
G.
It remains to prove
time.
we can construct the fan
O(k+d(Fi )) time.
whether or not it is maximal in
of
Fi = (ei , ci ; vi+1 ; vi , vi+2 )
If it is not maximal, we can complete the
Therefore we will either complete the coloring or construct our cycle of fans
O(
i=0 (k
which
Fi ,
in
+ d(Fi )))
k > d(Fi ).
i
in
for
In this case we certainly have two intersecting sets of available colors in
Fi , and nd the k -edge-coloring of G
Pj−1
jk = O( i=0 (d(Fi ))) = O(m), and we indeed
so we can apply 5.2.2 or 5.2.3 when we arrive at
O(k + m)
time. If no such
Since each
Fi
i
exists, then
is a maximal fan, in
two consecutive vertices
we rst check for any
i
guarantees a
vi
vi+1 ,
and
β ∈ C¯i ∩ C¯i+1 ,
O(|C¯i | + |C¯i+1 |)
O(k + m)
c0
time.
there must be some color
which we can easily do in
which we can nd in
time for each
i
i,
β∈
/ {α0 , α1 }
missing at
otherwise we reach a contradiction. To nd
i for which |C¯i | > d(vi+1 ),
exist, we search for a satisfying
in
F0 , . . . , Fj−1
time. This is not the desired bound, so suppose there is an index
complete the construction of all fans in
an
k-
G in O(m) time; this will happen at most once throughout the entire process.
edge-coloring of
Pj−1
and determine
by comparing
O(k)
C¯i
for each
i
i
satises
|C¯i | ≤ d(vi+1 ),
and since each
time in total. Therefore the entire operation takes
O(k + m)
0
to
and
i,
O(m) time such
time. If such a trivial
from
β
j.
i
does not
We can do this
this takes
O(m)
time.
We now complete the proof of 5.2.1.
Proof of 5.2.1.
Let
k = γl0 (G).
As noted in Section 5.2, we can compute
Taking the (lexicographically presorted) edges
the subgraph of
Given a
G
on edges
k -edge-coloring
of
{ej | j ≤ i}.
Gi ,
e1 , . . . , e m
Since
we can nd a
G0
of
in
O(m2 )
G, for i = 0, . . . , m let Gi
is empty it is vacuously
k -edge-coloring
89
k
of
Gi+1
in
time.
denote
k -edge-colored.
O(k + m)
time by
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
applying 5.2.7. Since
k = γl0 (G) = O(m),
total running time of
O(m2 ).
O(m)
each augmentation step takes
time, for a
The theorem follows.
This gives us the following result for line graphs, since for any multigraph
G
we have
|V (L(G))| = |E(G)|:
5.2.8. Given a line graph G on n vertices, we can nd a proper coloring of G using γl (G)
colors in O(n2 ) time.
Proof.
To
γl (G)-color G
we rst nd a multigraph
H
ply 5.2.1. As discussed in [21] Ÿ4.2.3, we can construct
such that
G = L(H),
H
in
from
G
then we ap-
O(|E(G)|)
time using
one of a number of known algorithms.
This is faster than the algorithm of King, Reed, and Vetta [23] for
graphs, which is given an improved complexity bound of
5.3
O(n5/2 )
γ(G)-coloring
line
in [21], Ÿ4.2.3.
Quasi-line Graphs
We now leave the setting of edge colorings of multigraphs and consider vertex colorings of
simple graphs.
As mentioned in the introduction, we can extend Conjecture 3 from line
graphs to quasi-line graphs using the same approach that King and Reed used to extend
Conjecture 2 from line graphs to quasi-line graphs in [22]. We do not require the full power
of Chudnovsky and Seymour's structure theorem for quasi-line graphs [14]. Instead, we use a
simpler decomposition theorem from [10]. Our proof of 5.1.3 yields a polytime
γl (G)-coloring
algorithm; we sketch a bound on its complexity at the end of the section.
We wish to describe the structure of quasi-line graphs. If a quasi-line graph does not
contain a certain type of homogeneous pair of cliques, then it is either a circular interval
graph or built as a generalization of a line graph where in a line graph we would replace each
edge with a vertex, we now replace each edge with a linear interval graph. We now describe
this structure more formally, which is equivalent to the quasi-line trigraph decomposition
that we used in Chapter 3.
We restate here the decomposition in term of graphs and
reintroduce some denition for completeness. It is important to notice that in this chapter,
we take the view of gluing strips together into 'composition of linear interval strips', where
90
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
in Chapter 3 and Chapter 4 we took the opposite approach of decomposing 'linear interval
joins' and 'strip structures' into strips.
A
linear interval graph is a graph G = (V, E) with a linear interval representation, which
u
is a point on the real line for each vertex and a set of intervals, such that vertices
G
are adjacent in
the real line. If
and
Y
are specied cliques in
G
|X|
leftmost and
respectively, we say that
X
and
Y
|Y |
are
These cliques may be empty.
Accordingly, a
|V |
consisting of the
G
rightmost vertices (with respect to the real line) of
i.e.
v
precisely if there is an interval containing both corresponding points on
X
end-cliques of G.
and
circular interval graph
is a graph with a
circular interval representation,
points on the unit circle and a set of intervals (arcs) on the unit circle such that two
vertices of
G are adjacent precisely if some arc contains both corresponding points.
Circular
interval graphs are the rst of two fundamental types of quasi-line graph. Deng, Hell, and
Huang proved that we can identify and nd a representation of a circular or linear interval
graph in
O(m)
time [16].
We now describe the second fundamental type of quasi-line graph.
A
X
linear interval strip (S, X, Y )
and
Y.
multigraph
strip
We compose a set of strips as follows.
H,
Cv =
We construct
For
v ∈ V (H)
[
{Xe | e
G
is an edge out of
from the disjoint union of
{Se | e ∈ E(H)}
Let
Gh
we take a linear interval
as
[
v} ∪
{Ye | e
composition of linear interval strips.
with specied end-cliques
e of H
hub clique Cv
we dene the
S
We begin with an underlying directed
possibly with loops, and for every every edge
(Se , Xe , Ye ).
is then a
is a linear interval graph
is an edge into
by making each
v} .
Cv
denote the subgraph of
G
a clique;
G
induced on
the union of all hub cliques. That is,
Gh = G[∪v∈V (H) Cv ] = G[∪e∈E(H) (Xe ∪ Ye )].
Compositions of linear interval strips generalize line graphs: note that if each
|Se | = |Xe | = |Ye | = 1
then
|A| + |B| ≥ 3,
satises
G = Gh = L(H).
A pair of disjoint nonempty cliques
if
Se
every vertex outside
homogeneous pair of cliques
(A, B)
in a graph is a
A∪B
is adjacent to either all or none of
91
A,
and
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
every vertex outside
A∪B
is adjacent to either all or none of
nonlinear if G contains an induced C4 in A ∪ B
the subgraph of
G
induced by
A∪B
B.
Furthermore
(A, B)
is
(this condition is equivalent to insisting that
is a linear interval graph).
Chudnovsky and Seymour's structure theorem for quasi-line graphs [10] tells us that any
quasi-line graph not containing a clique- cutset is made from the building blocks we just
described.
5.3.1. Any quasi-line graph containing no clique-cutset and no nonlinear homogeneous pair
of cliques is either a circular interval graph or a composition of linear interval strips.
To prove 5.1.3, we rst explain how to deal with circular interval graphs and nonlinear
homogeneous pairs of cliques, then move on to considering how to decompose a composition
of linear interval strips.
We can easily prove Conjecture 3 for circular interval graphs by combining previously
known results. Niessen and Kind proved that every circular interval graph
dχf (G)e
G satises χ(G) =
[29], so 5.1.2 immediately implies that Conjecture 3 holds for circular interval
graphs. Furthermore Shih and Hsu [32] proved that we can optimally color circular interval
graphs in
O(n3/2 )
time, which gives us the following result:
5.3.2. Given a circular interval graph G on n vertices, we can γl (G)-color G in O(n3/2 )
time.
There are many lemmas of varying generality that tell us we can easily deal with nonlinear
homogeneous pairs of cliques; we use the version used by King and Reed [22] in their proof
of Conjecture 2 for quasi-line graphs:
5.3.3. Let G be a quasi-line graph on n vertices containing a nonlinear homogeneous pair
of cliques (A, B). In O(n5/2 ) time we can nd a proper subgraph G0 of G such that G0 is
quasi-line, χ(G0 ) = χ(G), and given a k-coloring of G0 we can nd a k-coloring of G in
O(n5/2 ) time.
It follows immediately that no minimum counterexample to 5.1.3 contains a nonlinear
homogeneous pair of cliques.
92
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
5.4
Decomposing Quasi-line Graphs
Decomposing graphs on clique-cutsets for the purpose of nding vertex colorings is straightforward and well understood.
For any monotone bound on the chromatic number for a hereditary class of graphs, no
minimum counterexample can contain a clique-cutset, since we can simply paste together
two partial colorings on a clique-cutset. Tarjan [36] gave an
O(nm)-time
algorithm for con-
structing a clique-cutset decomposition tree of any graph, and noted that given
of the leaves of this decomposition tree, we can construct a
in
O(n2 )
cutset in
time. Therefore if we can
O(f (n, m))
O(f (n, m) + nm)
γl (G)-color
time for some function
f,
k -coloring
k -colorings
of the original graph
any quasi-line graph containing no cliquewe can
γl (G)-color
any quasi-line graph in
time.
If the multigraph
H
contains a loop or a vertex of degree 1, then as long as
G
is not a
clique, it will contain a clique-cutset.
A canonical interval 2-join is a composition by which a linear interval graph is attached
to another graph. Canonical interval 2-joins arise from compositions of strips, and can be
viewed as a local decomposition rather than one that requires knowledge of a graph's global
structure as a composition of strips.
Given four cliques
X1 , Y1 , X2 ,
and
Y2 ,
we say that
((V1 , X1 , Y1 ), (V2 , X2 , Y2 ))
is an
interval 2-join if it satises the following:
• V (G) can be partitioned into nonempty V1
such that for
v1 ∈ V 1
and
v2 ∈ V2 , v1 v2
and
V2
with
X1 ∪ Y1 ⊆ V1
is an edge precisely if
and
{v1 , v2 }
X2 ∪ Y2 ⊆ V2
is in
X1 ∪ X2
or
Y1 ∪ Y2 .
• G|V2
is a linear interval graph with end-cliques
and
Y2 .
X2
interval 2-join.
The following decomposition theorem is a straightforward consequence of
disjoint, then we say
((V1 , X1 , Y1 ), (V2 , X2 , Y2 ))
canonical
If we also have
and
Y2
X2
is a
the structure theorem for quasi-line graphs:
5.4.1. Let G be a quasi-line graph containing no nonlinear homogeneous pair of cliques.
Then one of the following holds.
93
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
• G is a line graph
• G is a circular interval graph
• G contains a clique-cutset
• G admits a canonical interval 2-join.
Therefore to prove 5.1.3 it only remains to prove that a minimum counterexample cannot contain a canonical interval 2-join.
Before doing so we must give some notation and
denitions.
We actually need to bound a renement of
((V1 , X1 , Y1 ), (V2 , X2 , Y2 ))
in
G|V1 ,
and let
let
G2
denote
G|V2
G
γl (G).
Given a canonical interval 2-join
with an appropriate partitioning
H2
as the size of the largest clique in
G|(V2 ∪ X1 ∪ Y1 ).
denote
H2
containing
v
V1
For
and
V2 ,
v ∈ H2
let
G1
we dene
and not intersecting both
Observe that
γlj (H2 ) ≤ γl (G).
If
v ∈ X1 ∪ Y1 ,
then
ω 0 (v)
is
ω 0 (v)
X1 \ Y1
Y1 \ X1 , and we dene γlj (H2 ) as maxv∈H2 ddG (v) + 1 + ω 0 (v)e (here the superscript j
join).
denote
and
denotes
|X1 | + |X2 |, |Y1 | + |Y2 |,
or
|X1 ∩ Y1 | + ω(G|(X2 ∪ Y2 )).
The following lemma is due to King and Reed and rst appeared in [21]; we include the
proof for the sake of completeness.
5.4.2. Let G be a graph on n vertices and suppose G admits a canonical interval 2-join
((V1 , X1 , Y1 ), (V2 , X2 , Y2 )). Then given a proper l-coloring of G1 for any l ≥ γlj (H2 ), we can
nd a proper l-coloring of G in O(nm) time.
Since
γlj (H2 ) ≤ γl (G),
this lemma implies that no minimum counterexample to 5.1.3
contains a canonical interval 2-join.
It is easy to see that a minimum counterexample cannot contain a simplicial vertex
(i.e. a vertex whose neighborhood is a clique).
((V1 , X1 , Y1 ), (V2 , X2 , Y2 ))
Y2
Therefore in a canonical interval 2-join
in a minimum counterexample, all four cliques
X1 , Y1 , X2 ,
and
must be nonempty.
Proof.
We proceed by induction on
l,
observing that the case
by modifying the coloring so that the number
94
k
l = 1
is trivial.
of colors used in both
X1
and
We begin
Y1
in the
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
l-coloring
G1
of
is maximal. That is, if a vertex
Y1
then every color appearing in
exceeds
v ∈ X1
N (v).
appears in
l = γlj (H2 )
l
set whose removal lowers both
and
O(n2 )
This can be done in
γlj (H2 ) we can just remove a color class in G1
Thus we can assume that
gets a color that is not seen in
Y1 ,
time. If
l
and apply induction on what remains.
and so if we apply induction we must remove a stable
γlj (H2 ).
We use case analysis; when considering a case we may assume no previous case applies.
In some cases we extend the coloring of
we remove a color class in
G1
G1
to an l-coloring of
is a stable set, and when we remove it we reduce
apply induction on l. Notice that if
may have a large clique in
of
Y1 ;
Case 1.
G2
together with vertices in
H2
X1 ∩ Y1 6= ∅
γlj (v)
in one step. In other cases
such that everything we remove
for every
v ∈ H2 ;
after doing this we
and there are edges between
which contains some but not all of
X1
X2
and
Y2
we
and some but not all
this is not necessarily obvious but we deal with it in every applicable case.
Y1 ⊆ X1 .
H2
in
is a circular interval graph and
O(n3/2 )
X1
is a clique-cutset.
G1 .
In this case
done. By symmetry, this covers the case in which
k=0
and
Y1
are disjoint. Take a stable set
this we mean that we start with
along the vertices of
leave
S
to see that
Y2 ,
S
γl (H2 ) ≤ γlj (H2 ) ≤ l
so we are
X1 ⊆ Y1 .
S = {v1 },
S
greedily from left to right in
the leftmost vertex of
in linear order, adding a vertex to
hits
X2 .
If it hits
Y2 , remove S
and
Y1
S
S
X2 ,
G2
γlj (v),
By
and we move
along with a color class in
v ∈ G2
G1
it is easy
either loses two neighbors or
intersects every maximal clique containing
are disjoint,
G2 .
whenever doing so will
these vertices together make a stable set. If
will drop: every remaining vertex in
and therefore
γlj (H2 )
If
γlj (v)
X1
S
X1 ∪ Y1 ;
in which case
then since
ω 0 (v),
G2
a stable set. So
not intersecting
is in
γl (H2 )-color H2
l > |X1 | + |Y1 |.
and
X1
Here
We can
time using 5.3.2. By permuting the color classes we can ensure that this
coloring agrees with the coloring of
Case 2.
G
v.
If
v ∈ X1 ∪ Y1 ,
ω 0 (v) is either |X1 |+|X2 | or |Y1 |+|Y2 |; in either case
drops when
S
and the color class are removed. Therefore
drops, and we can proceed by induction.
does not hit
Y2
we remove
S
along with a color class from
95
G1
that hits
Y1
(and
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
X1 ).
therefore not
S ∩ Y2 = ∅
Since
the vertices together make a stable set. Using
the same argument as before we can see that removing these vertices drops both
γlj (H2 ),
Case 3.
k=0
l−1
and
X1
l = |X1 | + |Y1 |.
and
Y1
neighbors in
are disjoint. By maximality of
G1 .
Since
l = |X1 | + |Y1 |
ω 0 (Y1 ) ≤ |X1 | + |Y1 | − |Y2 |.
without loss of generality that
We rst attempt to l-color
appears in
X1
k , every vertex in X1 ∪ Y1
we know that
|Y1 | ≥ 2|X2 |
Thus
G.
then use
and
X1
Dene
of
H3
|Y2 |
|Y2 | ≤ |X2 |.
Y2
to get a proper l-coloring
the vertices of
vertices of
X1 .
b < |Y2 |.
We want an
H3 ,
This is possible since
Y2
have no edges between them.
b
as
l − ω(H3 );
we can assume that
H3 ;
but not
X1 .
this clique does not intersect
Denote by
vj
There is some clique
X1
because
the leftmost neighbor of
vi .
|X1 ∪ X2 | ≤
Since
γlj (vi ) ≤ l,
we
colors appear in
Y2
then we are done, otherwise we will roll back the coloring, starting at
vi .
vi
has at most
can be assured that
We now color
X1
Y2
colors appear in
ω(H3 )-coloring
b < |Y2 | ≤ 12 |X1 |
it is clear that
but not
b
in
l − 21 |Y1 | ≤ l − |Y2 | < l − b.
H3
That is, for every
vi ∈
/ X2 .
2b
neighbors outside
Since
p ≥ i,
vi ∈
/ Y2
ω(H3 ).
we modify the coloring of
ω(H3 ).
we can roll back the coloring at least
proper colorings of
C,
and since
ω(H3 ) > |Y2 |, vi ∈
/ Y2 .
from left to right, modulo
one that it currently has, modulo
b
|Y2 |
and
G1
ω(H3 )-color
then this is easy: we can
new colors to recolor
such that at most
ω(H3 )
Assume
this is clearly sucient to prove the lemma since we can permute the
C = {vi , . . . , vi+ω(H3 )−1 }
Since
|X1 | ≥ 2|Y2 |.
H2 −Y1 , which we denote by H3 , such that every color in Y2
ω(H3 ) ≤ l − |Y2 |,
If
has at least
ω 0 (X1 ) ≤ |X1 | + |Y1 | − |X2 |
and similarly
color classes and paste this coloring onto the coloring of
of
and
so we can proceed by induction.
and
Again,
l
Since
vi
If at most
H3
by giving
has at most
ω(H3 ) − 2b − 1
b
vp
2b
the color after the
neighbors behind it,
times for a total of
ω(H3 ) − 2b
H3 .
the colors on
Y2
will appear in order modulo
possible sets of colors appearing on
colors appearing in
Y2
but not
X1 .
Y2 ,
and in
2b + 1
ω(H3 ).
Thus there are
of them there are at most
It follows that as we roll back the coloring of
96
H3
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
we will nd an acceptable coloring.
Henceforth we will assume that
Case 4.
|X1 | ≥ |Y1 |.
0 < k < |X1 |.
Take a stable set
G,
from
remove
S
in
G2 − X2
along with a color class from
S
S
greedily from left to right. If
along with a color class from
it is a simple matter to conrm that
G1
intersecting
G1
intersecting both
γlj (v)
X1
Y1 .
but not
drops for every
Y2 ,
hits
X1
and
v ∈ H2
we remove
S
Otherwise, we
Y1 .
In either case
as we did in Case
2. We proceed by induction.
Case 5.
k = |Y1 | = |X1 | = 1.
In this case
|X1 | = k = 1.
If
G2
is not connected, then
cutsets and we can proceed as in Case 1. If
v ∈ V2
then there is some
G2
l ≥ γlj (H2 ).
l
Case 6.
G1
to a proper l-coloring of
in the
So
in linear time using only colors not appearing in
of
and
Y1
are both clique-
is connected and contains an l-clique,
of degree at least
contradicting our assumption that
X1
l-clique.
ω(G2 ) < l.
Thus
We can
γlj (H2 ) > l,
ω(G2 )-color G2
X1 ∪ Y1 , thus extending the l-coloring
G.
k = |Y1 | = |X1 | > 1.
Suppose that
k
is not minimal. That is, suppose there is a vertex
closed neighborhood does not contain all
change the color of
v
X1
has degree at least
γlj (H2 ) ≥ l ≥ 21 (l + |X2 | + |X1 | + |X2 |),
|X2 | + |Y2 | ≤ l − k .
can
ω(G2 )-color G2 ,
X1 ∪ Y1
and
colors in the coloring of
and apply Case 4. So assume
Therefore every vertex in
so
l
Since there are
v ∈ X1 ∪ Y1
so
k
Then we can
is minimal.
l + |X2 | − 1.
2|X2 | ≤ l − k .
l−k
G1 .
whose
Since
X1 ∪ X2
Similarly,
is a clique,
2|Y2 | ≤ l − k ,
colors not appearing in
X1 ∪ Y1 ,
we
then permute the color classes so that no color appears in both
X2 ∪ Y2 .
Thus we can extend the l-coloring of
G1
to an l-coloring of
G.
These cases cover every possibility, so we need only prove that the coloring can be found
in
O(nm)
time. If
every vertex in
k
has been maximized and we apply induction,
X1 ∪ Y1
k
will stay maximized:
will have every remaining color in its closed neighborhood except
97
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
possibly if we recolor a vertex in Case 6. In this case the overlap in what remains is
which is the most possible since we remove a vertex from
k.
Hence we only need to maximize
k
X1
Y1 ,
or
k − 1,
each of which has size
once. We can determine which case applies in
O(m)
time, and it is not hard to conrm that whenever we extend the coloring in one step our
work can be done in
O(nm) time.
6, all our work can be done in
can be completed in
5.5
O(nm)
When we apply induction, i.e. in Cases 2, 4, and possibly
O(m)
time. Since
l<n
it follows that the entire
l-coloring
time.
Putting the pieces together and Algorithmic Considerations
We are now ready to prove 5.1.3.
Proof of 5.1.3.
Let
G
be a minimum counterexample. By 5.3.3, it follows that
no nonlinear homogeneous pair of cliques. By 5.2.1, we deduce that
and 5.3.2 implies that
G
G
G
contains
is not a line graph
G
is not a circular interval graph. By 5.4.2, it follows that
not admit a canonical interval 2-join. Therefore by 5.4.1,
G
does
cannot exist.
It is fairly clear that our proof of 5.1.3 gives us a polytime coloring algorithm. Here we
sketch a bound of
O(n3 m2 )
on its running time.
We proceed by induction on
n.
We reduce to the case containing no nonlinear homoge-
O(m)
neous pair of cliques by applying 5.3.3
of
G
such that
O(n2 m2 )
χ(G) = χ(G0 ),
and given a
time. We must now color
containing no clique-cutsets since
If
G0
time.
If
G0 .
times in order to nd a quasi-line subgraph
k -coloring
of
G0 ,
we can nd a
G
in
n3 m2 ≥ nm.
is a line graph we can determine this in
Roussopoulos [31], then
of
Following Section 5.4, we need only consider graphs
is a circular interval graph we can determine this and
G0
k -coloring
G0
γl (G)-color it in O(n2 ) time.
O(m)
γl (G)-color
it in
O(n3/2 )
time using an algorithm of
Otherwise,
G0
must admit a canonical
interval 2-join. In this case Lemma 6.18 in [21], due to King and Reed, tells us that we can
nd such a decomposition in
O(n2 m)
This canonical interval 2-join
time.
((V1 , X1 , Y1 ), (V2 , X2 , Y2 ))
98
leaves us to color the induced
CHAPTER 5. A LOCAL STRENGTHENING OF REED'S CONJECTURE
subgraph
G1
coloring of
G
in
of
G1
O(n2 m2 )
G0 ,
we can
time.
n−1
vertices and is quasi-line.
O(nm)
time, then reconstruct the
which has at most
γl (G)-color G0
in
The induction step takes
O(n2 m2 )
vertices, so the total running time of the algorithm is
Remark:
Given a
γl (G)-
γl (G)-coloring
of
time and reduces the number of
O(n3 m2 ).
This bound does not use recent, more sophisticated results on decomposing
quasi-line graphs, such as those found in [6] and [18]. We suspect that by applying these
results carefully, one should be able to reduce the running time of the entire
algorithm to
O(m2 ).
99
γl (G)-coloring
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103
p-graph. Diskret. Analiz,
APPENDIX A. APPENDIX
Appendix A
Appendix
A.1
Orientable prismatic graphs
• Q0
is the class of all 3-coloured graphs
(G, A, B, C)
such that
• Q1
is the class of all 3-coloured graphs
(G, A, B, C)
where
graph of
• Q2
G
G
has no triangle.
is isomorphic to the line
K3,3 .
is the class of all canonically-coloured path of triangles graphs.
• Path of triangles.
A graph
G
is a
path of triangles
graph if for some integer
X1 , . . . , X2n+1
there are pairwise disjoint stable subsets
of
V (G)
with union
n≥1
V (G),
satisfying the following conditions (P1)-(P7).
(P1) For
1 ≤ i ≤ n,
0 < i < n,
(P2) For
at least one of
j−i=2
either
(2) if
i, j
and for
has cardinality 1.
modulo 3 and there exist
are odd and
j − i 6= 2
j = i + 2,
modulo 3 then either
1 ≤ i ≤ n + 1, X2i−1
M2i−1 , R2i−1 ,
(P4) If
X̂2i , X̂2i+2
X̂2i ⊆ X2i ; |X̂2 | = |X̂2n | = 1,
1 ≤ i < j ≤ 2n + 1
(1) if
(P3) For
there is a nonempty subset
where
or
u ∈ Xi
i, j
and
v ∈ Xj ,
are even and
j =i+1
or
Xi
nonadjacent, then
u∈
/ X̂i
and
v∈
/ X̂j ;
is anticomplete to
Xj .
is the union of three pairwise disjoint sets
L2i−1 ,
L1 = M1 = M2n+1 = R2n+1 = ∅.
R1 = ∅ then n ≥ 2 and |X̂4 | > 1, and if L2n+1 = ∅ then n ≥ 2 and |X̂2n−2 | > 1.
104
APPENDIX A. APPENDIX
(P5) For
1 ≤ i ≤ n, X2i
M2i−1 ∪ M2i+1 ;
is anticomplete to
(P6) For
1 ≤ i ≤ n,
M2i+1
L2i−1
(4) if
is between
L2i+1 .
then
X̂2i
R2i−1
then
and
X2i+1
and
M2i−1 , X̂2i−2
M2i−1 ∪R2i−1
and
L2i+1 ∪
L2i+1 ;
R2i−1 ∪ M2i−1 ∪ L2i+1 ∪ M2i+1 ;
is complete to
is complete to
i>1
and
is anticomplete to
is adjacent to exactly one end of
are matched, and every edge between
(2) the vertex in
(3)
R2i−1
|X̂2i | = 1,
if
R2i−1 , L2i+1
(1)
X2i \X̂2i
and every vertex in
every edge between
L2i−1 ∪ R2i+1 ; X2i \X̂2i
X2i−1
is complete to
are matched, and if
R2i+1
i<n
then
M2i+1 , X̂2i+2
are
matched.
(P7) For
1 < i < n,
(1)
|X̂2i | > 1
if
then
R2i−1 = L2i+1 = ∅;
(2) if
u ∈ X2i−1
and
v ∈ X2i+1 ,
have the same neighbour in
S
(Xi : 1 ≤ i ≤ 2n + 1
then
u, v
X̂2i .
Let
Ak =
is a
canonically-coloured path of triangles graphs.
• Cycle of triangles.
with
n=2
with union
modulo
A graph
i = k mod 3) (k = 0, 1, 2).
and
G
are nonadjacent if and only if they
is a
cycle of triangles
Then
graph if for some integer
X1 , . . . , X2n
modulo 3, there are pairwise disjoint stable subsets
V (G),
(G, A1 , A2 , A3 )
n≥5
of
V (G)
satisfying the following conditions (C1)-(C6) (reading subscripts
2n):
(C1) For
1 ≤ i ≤ n,
X̂2i , X̂2i+2
(C2) For
there is a nonempty subset
(1) if
and all
k
with
2 ≤ k ≤ 2n − 2,
let
j ∈ {1, . . . , 2n}
with
modulo 2n:
k =2
either
modulo 3 and there exist
i, j
are odd and
u ∈ Xi
k ∈ {2, 2n − 2}, or i, j
and
v ∈ Xj
k 6= 2
modulo 3 then
Xi
is anticomplete to
105
, nonadjacent, then
are even and
;
(2) if
and at least one of
has cardinality 1.
i ∈ {1, . . . , 2n}
j =i+k
X̂2i ⊆ X2i ,
Xj .
u∈
/ X̂i
and
v∈
/ X̂j
APPENDIX A. APPENDIX
(Note that
k = 2 modulo 3 if and only if 2n−k = 2 modulo 3, so these statements
are symmetric between
(C3) For
i
1 ≤ i ≤ n + 1, X2i−1
and
j .)
is the union of three pairwise disjoint sets
L2i−1 , M2i−1 ,
R2i−1 .
(C4) For
1 ≤ i ≤ n, X2i
M2i−1 ∪ M2i+1 ;
is anticomplete to
(C5) For
(1)
1 ≤ i ≤ n,
is between
(3)
L2i−1
(4)
M2i−1 , X̂2i−2
(1)
L2i+1 .
then
X̂2i
R2i−1
X2i+1
and
u ∈ X2i−1
• Ring of ve.
and
G
Let
{a1 , . . . , a5 , b1 , . . . , b5 }
R2i−1 ∪ M2i−1 ∪ L2i+1 ∪ M2i+1 ;
X2i−1
is complete to
M2i+1 , X̂2i+2
v ∈ X2i+1 ,
is complete to
all stable; for
remainder of
R2i+1
are matched.
are nonadjacent if and only if they
X̂2i .
V (G)
V0 , V1 , . . . , V5 .
the union of the disjoint sets
{b1 , . . . , b5 }
is a triangle, and
and anticomplete to
is complete to
is anticomplete to
and the adjacency between
Vi , Vi+1
is adjacent to
{a1 , . . . , a5 }; V0 , V1 , . . . , V5
{ai−1 , bi , ai+1 }
V1 ∪ · · · ∪ V5 ;
ai
for
Let
G
be a graph with
1 ≤ i ≤ 5 Vi
is anticomplete
is arbitrary. We call such a graph a
V (G)
the union of seven sets
W = {aji : 1 ≤ i, j ≤ 3}, V 1 , V 2 , V 3 , V1 , V2 , V3 ,
106
are
and anticomplete to the
of ve.
• Mantled L(K3,3 ).
W =
Let adjacency be as follows (reading sub-
1 ≤ i ≤ 5, {ai , ai+1 ; bi+3 }
i = 1, . . . , 5, Vi
W ; V0
u, v
then
be a graph with
and
scripts modulo 5). For
Vi+2 ;
L2i+1 ∪
then
have the same neighbour in
to
and
R2i−1 = L2i+1 = ∅;
(2) if
bi ; V0
M2i−1 ∪R2i−1
L2i+1 ;
are matched and
|X̂2i | > 1
if
and
is complete to
is complete to
1 ≤ i ≤ n,
and
is anticomplete to
is adjacent to exactly one end of
are matched, and every edge between
(2) the vertex in
(C6) For
R2i−1
|X̂2i | = 1,
if
R2i−1 , L2i+1
M2i+1
X2i \X̂2i
and every vertex in
every edge between
L2i−1 ∪ R2i+1 ; X2i \X̂2i
ring
APPENDIX A. APPENDIX
1 ≤ i, j, i0 , j 0 ≤ 3, aji
with adjacency as follows. For
i0 6= i
and
j 0 6= j .
For
i = 1, 2, 3, V i , Vi
anticomplete to the remainder of
to the remainder of
W.
G a
A.2
Vi
aji0
are adjacent if and only if
is complete to
{a1i , a2i , a3i },
and
W ; and Vi is complete to {ai1 , ai2 , ai3 } and anticomplete
V1∪V2∪V3
Moreover,
there is no triangle included in
are stable;
0
and
V1∪V2∪V3
is anticomplete to
or in
V1 ∪ V2 ∪ V3 .
V1 ∪ V2 ∪ V3 ,
and
We call such a graph
mantled L(K3,3 ).
Non-orientable prismatic graphs
• A rotator.
{v4 , v5 , v6 }
Let
G
have nine vertices
{v7 , v8 , v9 },
is complete to
and there are no other edges. We call
• A twister.
Let G have ten vertices
i = 1, . . . , 8 vi
i = 1, 2, ui
a
twister
A.3
is adjacent to
is adjacent to
and
u1 , u2
v1 , v2 , . . . , v9 ,
and for
G
a
u1 , u2 , v1 , . . . , v8
vi−1 , vi+1 , vi+4
axis
i = 1, 2, 3, vi
{v1 , v2 , v3 }
is a triangle,
is adjacent to
vi+3 , vi+6 ,
rotator.
vi , vi+2 , vi+4 , vi+6 ,
is the
where
, where
u1 , u2
are adjacent, for
(reading subscripts modulo
8),
and for
and there are no other edges. We call
G
of the twister.
Three-cliqued graphs
• A type of line trigraph.
such that every edge of
H
Let
v1 , v2 , v3
be distinct nonadjacent vertices of a graph
is incident with one of
at least three, and let all other vertices of
all distinct
i, j ∈ {1, 2, 3},
is adjacent to
with
v1 , v2 , v3
vi
H
v1 , v2 , v3 .
Let
v1 , v2 , v3
all have degree
have degree at least one. Moreover, for
let there be at most one vertex dierent from
and not to
vj
in
H.
respectively, and let
G
three-cliqued claw-free trigraph; let
Let
A, B, C
v1 , v2 , v3
be the sets of edges of
be a line trigraph of
T C1
H,
H.
Then
H
that
incident
(G, A, B, C)
is a
be the class of all such three-cliqued trigraphs
such that every vertex is in a triad.
• Long circular interval trigraphs.
let
Σ
be a circle with
V (G) ⊆ Σ,
circular interval trigraph. By a
line
Let
and
G
be a long circular interval trigraph, and
F1 , . . . , Fk ⊆ Σ,
as in the denition of long
we mean either a subset
107
X ⊆ V (G)
with
|X| ≤ 1,
APPENDIX A. APPENDIX
or a subset of some
in
V (G).
Let
Fi
homeomorphic to the closed unit interval, with both end-points
L1 , L2 , L3
be pairwise disjoint lines with
V (G) ⊆ L1 ∪ L2 ∪ L3 ;
(G, V (G) ∩ L1 , V (G) ∩ L2 , V (G) ∩ L3 ) is a three-cliqued claw-free trigraph.
by
T C2
then
We denote
the class of such three-cliqued trigraphs with the additional property that
every vertex is in a triad.
• Near-antiprismatic trigraphs.
A, B, C, X
dene
Let
H
be a near-antiprismatic trigraph, and let
be as in the denition of near-antiprismatic trigraph. Let
B0, C 0
T C3
denote by
(H, A0 , B 0 , C 0 )
similarly; then
A0 = A\X
and
is a three-cliqued claw-free trigraph. We
the class of all three-cliqued trigraphs with the additional property
that every vertex is in a triad.
• Antiprismatic trigraphs.
partition of
V (G)
Let
G
be an antiprismatic trigraph and let
into three strong cliques; then
(G, A, B, C)
A, B, C
be a
is a three-cliqued claw-
free trigraph. We denote the class of all such three-cliqued trigraphs by
T C4.
(In [11]
Chudnovsky and Seymour described explicitly all three-cliqued antiprismatic graphs,
and their "changeable" edges; and this therefore provides a description of the threecliqued antiprismatic trigraphs.) Note that in this case there may be vertices that are
in no triads.
• Sporadic exceptions.
Let
H
vi , vj
and
v7
be the trigraph with vertex set
are strongly adjacent for
v2 v6
strongly antiadjacent to
v2
is antiadjacent to
Let
1 ≤ i < j ≤ 6
are strongly antiadjacent;
is strongly antiadjacent to
v5 .
v1 , v6 , v7
the pairs
with
and adjacency as follows:
j − i ≤ 2;
the pairs
v1 v5
are pairwise strongly adjacent, and
v2 , v3 , v4 , v5 ; v7 , v8
v1 , . . . , v 6 ;
Let
{v1 , . . . , v8 }
v1 v4
are strongly adjacent, and
and
v3 v6
v8
is
are semiadjacent, and
A = {v1 , v2 , v3 }, B = {v4 , v5 , v6 }
and
C = {v7 , v8 }.
X ⊆ {v3 , v4 }; then (H\X, A\X, B\X, C) is a three-cliqued claw-free trigraph,
and all its vertices are in triads.
Let
sets
H
be the trigraph with vertex set
{v1 , . . . , v9 },
A = {v1 , v2 }, B = {v3 , v4 , v5 , v6 , v9 }
108
and
and adjacency as follows: the
C = {v7 , v8 }
are strong cliques;
v9
APPENDIX A. APPENDIX
is strongly adjacent to
antiadjacent to
v1 , v8
v4 , v5 , v6 , v7 ,
is strongly antiadjacent to
strongly antiadjacent to
to
v8
v5 v7
and strongly antiadjacent to
semiadjacent to
v5 , v6 , v7 , v8
and strongly adjacent to
and strongly adjacent to
v7 ;
X ⊆ {v3 , v4 , v5 , v6 },
{v3 , v4 }\X
∗ v7
is not strongly anticomplete to
{v5 , v6 }\X
Then
v4 , v5 ∈
/X
then
v2
(H\X, A, B\X, C)
We denote by
T C5
are
is adjacent to
v4
v2 v4
and
such that
is not strongly anticomplete to
if
v8 ; v2
v3 ; v3 , v4
and the adjacency between the pairs
∗ v2
∗
is strongly
v7 , v8 ; v5 is strongly antiadjacent to v8 ; v6 is semiadjacent
and strongly adjacent to
is arbitrary. Let
v3
v2 , v7 ; v1
and
v5
is adjacent to
v7 .
is a three-cliqued claw-free trigraph.
the class of such three-cliqued trigraphs (given by one of these two
constructions) with the additional property that every vertex is in a triad.
109