The intersection of two vertex coloring problems

TL;DR

This paper investigates the complexity of coloring graphs that are free of certain induced subgraphs, specifically focusing on the intersection of two unresolved vertex coloring problems involving even holes and specific forbidden subgraphs.

Contribution

The paper provides partial results on the complexity of coloring {4K1, C4, C6}-free graphs, an intersection of two open problems in graph coloring.

Findings

Partial results on coloring {4K1, C4, C6}-free graphs

Insights into the complexity of coloring even hole-free graphs

Insights into the complexity of coloring {4K1, C4}-free graphs

Abstract

A hole is an induced cycle with at least four vertices. A hole is even if its number of vertices is even. Given a set L of graphs, a graph G is L-free if G does not contain any graph in L as an induced subgraph. Currently, the following two problems are unresolved: the complexity of coloring even hole-free graphs, and the complexity of coloring {4K1, C4}-free graphs. The intersection of these two problems is the problem of coloring {4K1, C4, C6}-free graphs. In this paper we present partial results on this problem.

Equations

$$R=\varnothing$$

$$F\ne \varnothing$$

Full text

Keywords: Graph coloring, perfect graphs

1 Department of Physics and Computer Science, Wilfrid Laurier University,

Waterloo, Ontario, Canada

1 Introduction

There has been recently keen interest in finding polynomial-time algorithms to optimally color graphs $G$ that do not contain any graph in a list $L$ as an induced subgraphs. Particular attention is focused on graphs whose forbidden list $L$ contains graphs with four vertices, and recent papers of Lozin and Malyshev [15], and of Fraser, Hamel, Hoàng, Holmes and LaMantia [10] discuss the state of the art on this problem, identifying three outstanding classes: $L=(4K_{1}$, claw), $L=(4K_{1}$, claw, co-diamond), and $L=(4K_{1},C_{4}$). As a resolution of these cases is likely challenging, a productive approach would be to consider increasing the number of graphs in $L$. But it makes sense to ask, which $L$ is an interesting one to consider? We are particularly interested in cases that are slightly larger than the class $L=(4K_{1},C_{4}$), which is one of the unresolved cases. Before we introduce the problems we need some definitions:

A hole is an induced cycle with at least four vertices. A hole is even if its number of vertices is even. The problem of coloring even-hole-free graphs has been much studied. A theorem of Addario-Berry, Chudnovsky, Havet, Reed, and Seymour [1] shows that for an even-hole-free graph $G$, the chromatic number of $G$ is at most two times its clique number (the number of vertices in a largest clique of $G$). It is currently not known whether even-hole-free graphs can be colored in polynomial time.

We offer the following four problems for consideration:

Problem 1.1

What is the complexity of coloring ($4K_{1},C_{4}$)-free graphs?

This is the original problem, and is likely the most challenging.

Problem 1.2

What is the complexity of coloring even-hole-free graphs?

Problem 1.2 might even be NP-complete. Combining Problems 1.1 and 1.2, we get the following problem:

Problem 1.3

What is the complexity of coloring ($4K_{1}$, even hole)-free graphs?

This problem appears to be more tractable than the previous two. In this paper, we study Problem 1.3. Since a $4K_{1}$-free graph does not contain a hole of length at least 8, Problem 1.3 is equivalent to the following:

Problem 1.4

What is the complexity of coloring ($4K_{1},C_{4},C_{6}$)-free graphs?

Even though we have not been able to solve Problem 1.4, we have succeeded, in some sense, in solving “half” of it, as follows. Consider a ($4K_{1},C_{4},C_{6}$)-free graph $G$. We may assume $G$ is not perfect (there are known algorithms to color perfect graphs). Thus $G$ has to contain a $C_{5}$ or $C_{7}$. If $G$ contains a $C_{7}$, then our result shows that $G$ can be colored in polynomial time. The case where $G$ contains a $C_{5}$ but not a $C_{7}$ is open. Investigation into this problem led us to a proof that there is a polynomial time algorithm to color a ($4K_{1},C_{4},C_{6}$, $C_{5}$-twin)-free graph.

In Section 2, we discuss the background of the problem and state the main results. In Section 3, we study ($4K_{1},C_{4},C_{6}$)-free graphs that contain a $C_{7}$ and show that such graphs can be colored in polynomial time. In Section 4, we study ($4K_{1},C_{4},C_{6}$)-free graphs that contain a $C_{5}$, but no $C_{7}$. In Section 5, we give a polynomial time algorithm to color a ($4K_{1},C_{4},C_{6}$, $C_{5}$-twin)-free graph. Finally, in Section 6, we dicuss open problems related to our work.

2 Background and results

Before discussing our results in more detail, we need introduce a few definitions. Let $G$ be a graph. A colouring of a graph $G=(V,E)$ is a mapping $f:V\rightarrow\{1,\ldots,k\}$ for some nonnegative integer $k$ such that $f(u)\neq f(v)$ whenever $uv\in E$. The chromatic number, denoted $\chi(G)$, is the minimum number of colors needed to colour a graph $G$. VERTEX COLORING is the problem of determining the chromatic number of a graph.

Consider the following operations to build a graph.

(i)

Create a vertex $u$ labeled by integer $\ell$.

(ii)

Disjoint union (i.e., co-join)

(iii)

Join between all vertices with label $i$ and all vertices with label $j$ for $i\not=j$, denoted by $\eta_{i,j}$ (that is, add all edges between vertices of label $i$ and label $j$).

(iv)

Relabeling all vertices of label $i$ by label $j$, denoted by $\rho_{i\rightarrow j}$

The clique width of a graph $G$, denoted by $cwd(G)$, is the minimum number of labels needed to build the graph with the above four operations. It is well-known [8] that if the clique width of a graph is bounded then so is that of its complement. Clique widths have been intensively studied. In Rao [17], the following result is established.

Theorem 2.1

VERTEX COLORING is polynomial time solvable for graphs with bounded clique width. $\Box$

We will need the folowing well known observation that is easy to establish (for example, see [6]).

Observation 2.2

Let $G$ be a graph and $G^{\prime}$ be the graph obtained from $G$ by removing a constant number of vertices. Then $G$ has bounded clique width if and only if $G^{\prime}$ does. $\Box$

The symbol $\omega(G)$ denotes the number of vertices in a largest clique of $G$. A graph $G$ is perfect if for each induced subgraph $H$ of $G$, we have $\chi(H)=\omega(H)$. A hole is an induced cycle of length at least $4$, i.e. $C_{k}$ for $k\geq 4$. A hole is even or odd depending on the parity of the vertices in the hole. An anti-hole is the complement of a hole.

Two important results are known about perfect graphs. The Perfect Graph Theorem, proved by Lovász [14], states that a graph is perfect if and only if its complement is. The Strong Perfect Graph Theorem, proved by Chudnovsky, Robertson, Seymour, and Thomas [7], states that a graph is perfect if and only if it is odd-hole-free and odd-anti-hole-free. Both of the above results were long standing open problems proposed by Berge [3]. Grötschel, Lovász and Schrijver [12] designed a polynomial-time algorithm for finding a largest clique and a minimum coloring of a perfect graph.

Suppose we want to color a ($4K_{1},C_{4},C_{6}$)-free graph $G$. Note that $G$ contains no $C_{\ell}$ for $\ell\geq 8$ because $G$ is $4K_{1}$-free. By the result of Grötschel et al, we may assume $G$ is not perfect. The result of Chudnovsky et al implies $G$ contains a $C_{5}$ or $C_{7}$ as an induced subgraph (note that the anti-hole of length at least six contains a $C_{4}$ and the $C_{5}$ is self-complementary.) If $G$ contains a $C_{7}$, then we will show that $G$ has bounded clique-width.

Theorem 2.3

Let $G$ be ($4K_{1},C_{4},C_{6}$)-free graph that contains a $C_{7}$. Then $G$ has bounded clique width.

We will use Theorem 2.3 to prove the following theorem that is the main result of this paper.

Theorem 2.4

VERTEX COLORING can be solved in polynomial time for the class of ($4K_{1},C_{4},C_{6}$)-free graphs that contain a $C_{7}$.

Two adjacent vertices $x,y$ of a graph $G$ are twins if for any vertex $z$ different from $x$ and $y$, $xz$ is an edge if and only if $yz$ is an edge. A hole-twin is the graph obtained from a hole by adding a vertex that form twins with some vertex of the hole. Figure 1 shows the $C_{5}$-twin. Hole-twins play an interesting role in graph theory. They are among the forbidden induced subgraphs for line-graphs (Beineke [2]).

By adding the $C_{5}$-twin to the list of forbidden induced subgraphs for our graph class, we obtain the following theorem.

Theorem 2.5

VERTEX COLORING can be solved in polynomial time for the class of ($4K_{1},C_{4},C_{6}$, $C_{5}$-twin)-free graphs.

We will rely on a theorem of Fraser et al [10]. To explain this theorem, we will need to introduce a few definitions. Given sets of vertices $X,Y$, we write $X\;\leavevmode\hbox to14.18pt{\vbox to14.18pt{\pgfpicture\makeatletter\hbox{\hskip 7.09111pt\lower-7.09111pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{ {{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{{{}}}{{}}{}{}{}{}{}{}{}{}{}{{}\pgfsys@moveto{6.89111pt}{0.0pt}\pgfsys@curveto{6.89111pt}{3.8059pt}{3.8059pt}{6.89111pt}{0.0pt}{6.89111pt}\pgfsys@curveto{-3.8059pt}{6.89111pt}{-6.89111pt}{3.8059pt}{-6.89111pt}{0.0pt}\pgfsys@curveto{-6.89111pt}{-3.8059pt}{-3.8059pt}{-6.89111pt}{0.0pt}{-6.89111pt}\pgfsys@curveto{3.8059pt}{-6.89111pt}{6.89111pt}{-3.8059pt}{6.89111pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{ } }{{{{}}\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{-3.22221pt}\pgfsys@invoke{ }\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{0}} }}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} } \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{{{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}\;Y$ to mean there is no edge between any vertex in $X$ and any vertex in $Y$ (also called a co-join). Given sets of vertices $X,Y$, we write $X\;\leavevmode\hbox to14.18pt{\vbox to14.18pt{\pgfpicture\makeatletter\hbox{\hskip 7.09111pt\lower-7.09111pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{ {{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{{{}}}{{}}{}{}{}{}{}{}{}{}{}{{}\pgfsys@moveto{6.89111pt}{0.0pt}\pgfsys@curveto{6.89111pt}{3.8059pt}{3.8059pt}{6.89111pt}{0.0pt}{6.89111pt}\pgfsys@curveto{-3.8059pt}{6.89111pt}{-6.89111pt}{3.8059pt}{-6.89111pt}{0.0pt}\pgfsys@curveto{-6.89111pt}{-3.8059pt}{-3.8059pt}{-6.89111pt}{0.0pt}{-6.89111pt}\pgfsys@curveto{3.8059pt}{-6.89111pt}{6.89111pt}{-3.8059pt}{6.89111pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{ } }{{{{}}\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-2.5pt}{-3.22221pt}\pgfsys@invoke{ }\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{1}} }}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} } \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{{{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}\;Y$ to mean there are all edges between $X$ and $Y$ (also called a join).

Consider a partition ${\cal P}$ of the vertices of $G$ into sets $S_{1},S_{2},\ldots,S_{k}$ such that each $S_{i}$ induces a clique. A set $S_{i}$ is uniform to a set $S_{j}$ if $S_{i}$
0 $S_{j}$ or $S_{i}$
1 $S_{j}$. The set $S_{i}$ is uniform in ${\cal P}$ if every set $S_{j},i\not=j$, is uniform to it. The partition ${\cal P}$ is uniform if every set $S_{i}$ is uniform in ${\cal P}$. A set $S_{i}$ is near-uniform if there is at most one set $S_{j},i\not=j,$ that is not uniform to $S_{i}$. The partition ${\cal P}$ is near-uniform if every set $S_{i}$ is near-uniform. Thus, a uniform partition is near-uniform. A $k$-near-uniform partition of $G$ is a partition of the vertices of $G$ into $k$ near-uniform sets. In such a partition, the pair of sets $S_{i}$, $S_{j}$ such that each set is not uniform to the other is call a uniform-pair. The following is proved in Fraser et al [10].

Theorem 2.6

[10]** Let $G$ be a graph admitting a $k$-near-uniform partition222We note that the definition of near-uniform partition in [10] is incomplete. The sets $S_{i}$’s must be cliques for the theorem to hold. such that the uniform pairs are $C_{4}$-free. Then we have $cwd(G)\leq 2k$. $\Box$

We will establish the following theorem.

Theorem 2.7

Let $G$ be ($4K_{1},C_{4},C_{6}$)-free graph that contains a $C_{7}$. Then $G-C_{7}$ admits a $k$-uniform partition, for some constant $k$.

By the above discussion, Theorem 2.7 implies Theorem 2.4. We will prove Theorem 2.7 in the next section.

3 When the graphs contain a $C_{7}$

Assume that the graph $G$ is ($4K_{1},C_{4},C_{6}$)-free and contains a $C_{7}$. In this section, we examine the structure of the neighborhood of the $C_{7}$. Then we will prove Theorem 2.7.

We will need first to establish a number of preliminary results. Given a hole, $H$, and a vertex $x$ not in $H$, we say $x$ is a $k$-vertex (for $H$) if $x$ has exactly $k$ neighbours in $H$.

For all the claims below, we shall now assume that $G$ contains an induced $C_{7}$ with vertices $(1,2,3,4,5,6,7)$. The vertex numbers of the $C_{7}$ are taken modulo 7.

Let

• •

$X_{i}$ denote the set of $3$-vertices adjacent to $(i,i+1,i+2)$ in the $C_{7}$,

• •

$Y_{i}$ denote the set of $3$-vertices on $(i,i+1,i+4)$ in the $C_{7}$,

• •

$Z_{i}$ denote the set of $5$-vertices on $(i,i+1,i+2,i+3,i+4)$ in the $C_{7}$, and

• •

$W$ denote the set of $7$-vertices in the $C_{7}$.

We will show that the sets $X_{i},Y_{i},Z_{i},W$ form a partition of $V(G)-C_{7}$.

Observation 3.1

The $C_{7}$ has no $k$-vertex in $G$ for $k=0,1,2,4,6$.

Proof. Suppose that there is a [math]-vertex $v$ for $C_{7}$. This creates an induced $4K1$ $(v,i,i+2,i+4)$, which is forbidden. Therefore, there is no [math]-vertex. Now suppose that there is a $1$-vertex $v$ that is adjacent to $i$. This creates an induced $4K1$ $(v,i+1,i+3,i+5)$, which is forbidden. Therefore, there is no $1$-vertex. Next, suppose that there is a $2$ vertex $v$ for $C_{7}$. It is easy to see that $v$ and some three vertices in the $C_{7}$ form a $4K_{1}$, a contradiction. Consequently, there is no $2$-vertex. Next, suppose that there is a $4$-vertex $v$ for $C_{7}$. Then $G$ contains a $C_{4}$ or $C_{6}$, both of which are forbidden. Therefore, there is no $4$-vertex. Finally, suppose that there exists a $6$-vertex $v$ for $C_{7}$. Then, $G$ contains an induced $C_{4}$, which is forbidden. Therefore, there is no $6$-vertex for $C_{7}$. $\Box$

It is a routine matter to verify the two observations below.

Observation 3.2

Let $v$ be a 3-vertex for the $C_{7}$. Then $v\in X_{i}\cup Y_{i}$ for some $i$. $\Box$

Observation 3.3

Let $v$ be a 5-vertex for the $C_{7}$. Then $v\in Z_{i}$ for some $i$. $\Box$

The above three observations imply the following observation.

Observation 3.4

Let $G^{\prime}$ be the graph obtained from $G$ by removing the $C_{7}$. Then the sets $X_{i},Y_{i},Z_{i},W$ form a partition ${\cal P}$ of $G^{\prime}$.

Our aim is to show that $G^{\prime}$ has bounded clique width. In fact, we will show that ${\cal P}$ is an uniform partition of $G^{\prime}$.

We now examine the adjacencies between the sets of the partition ${\cal P}$.

Observation 3.5

Each of the sets $X_{i},Y_{i},Z_{i},W$ of the partition ${\cal P}$ is a clique.

Proof. It is easy to see that if a set of ${\cal P}$ is not a clique then there is a $C_{4}$. $\Box$

The next sequence of observations will imply that $X_{i}$ is near-uniform in ${\cal P}$.

Observation 3.6

$X_{i}$*
1 $X_{i+1}\cup X_{i+6}$*

Proof. Suppose there are vertices $x_{1}\in X_{i}$ and $x_{2}\in X_{i+1}$ such that $x_{1}x_{2}\notin E$. Then there exists a $4K_{1}$ $(x_{1},x_{2},i+4,i+6)$, which is forbidden. Therefore, $X_{i}$
1 $X_{i+1}$. By symmetry, we have $X_{i}$
1 $X_{i+6}$. $\Box$

Observation 3.7

$X_{i}$*
0 $X_{i+2}\cup X_{i+3}\cup X_{i+4}\cup X_{i+5}$.*

Proof. Consider a vertex $x_{1}\in X_{i}$. Suppose there is a vertex $x_{2}\in X_{i+2}$ such that $x_{1}x_{2}\in E$. This creates a $C_{6}$ $(x_{1},x_{2},i+4,i+5,i+6,i)$, which is forbidden. Therefore, we have $X_{i}$
0 $X_{i+2}$, and by symmetry $X_{i}$
0 $X_{i+5}$. Now suppose that there is a vertex $x_{3}\in X_{i+3}$ such that $x_{1}x_{3}\in E$. This creates a $C_{4}$ $(x_{3},x_{1},i+2,i+3)$, which is forbidden. Therefore, we have $X_{i}$
0 $X_{i+3}$, and by symmetry, $X_{i}$
0 $X_{i+4}$. $\Box$

Observation 3.8

$X_{i}$*
1 $Y_{i}\cup Y_{i+1}\cup Y_{i+4}$.*

Proof. Consider a vertex $x\in X_{i}$. Suppose there is a vertex $y\in Y_{i}$ such that $xy\notin E$. This creates a $4K1$ $(x,y,i+3,i+5)$, which is forbidden. Therefore, $X_{i}$
1 $Y_{i}$. Now suppose there is a vertex $y\in Y_{i+1}$ with $xy\notin E$. This creates a $4K_{1}$ $(x,y,i+3,i+6)$, which is forbidden. Therefore $X_{i}$
1 $Y_{i+1}$. Finally, suppose there is a vertex $y\in Y_{i+4}$ with $xy\notin E$. This creates a $4K_{1}$ $(x,y,i+3,i+6)$, which is forbidden. Therefore $X_{i}$
1 $Y_{i+4}$. Consequently, $X_{i}$
1 $Y_{i}\cup Y_{i+1}\cup Y_{i+4}$. $\Box$

Observation 3.9

$X_{i}$*
0 $Y_{i+2}\cup Y_{i+3}\cup Y_{i+5}\cup Y_{i+6}$*

Proof. Consider a vertex $x\in X_{i}$. Suppose there is a vertex $y\in Y_{i+2}$ with $xy\in E$. This creates a $C_{4}$ $(x,y,i+6,i)$, which is forbidden. Therefore, we have $X_{i}$
0 $Y_{i+2}$, and by symmetry, $X_{i}$
0 $Y_{i+6}$. Now suppose that there is a vertex $y\in Y_{i+3}$ with $xy\in E$. This creates a $C_{4}$ $(x,y,i+3,i+2)$, which is forbidden. Therefore, $X_{i}$
0 $Y_{i+3}$, and by symmetry, $X_{i}$
0 $Y_{i+5}$. $\Box$

Observation 3.10

$X_{i}$*
1 $Z_{i}\cup Z_{i+1}\cup Z_{i+4}\cup Z_{i+5}\cup Z_{i+6}$*

Proof. Consider a vertex $x\in X_{i}$. Suppose there is a vertex $z\in Z_{i}$ such that $xz\notin E$. This creates a $C_{4}$ $(x,i,z,i+2)$, which is forbidden. Therefore, we have $X_{i}$
1 $Z_{i}$, and by symmetry, $X_{i}$
1 $Z_{i+5}$. Now suppose there is a vertex $z\in Z_{i+1}$ such that $xz\notin E$. This creates a $C_{6}$ $(i,i+6,i+5,z,i+2,x)$, which is forbidden. Therefore, we have $X_{i}$
1 $Z_{i+1}$, and by symmetry, $X_{i}$
1 $Z_{i+4}$. Finally, suppose that there is a vertex $z\in Z_{i+6}$ such that $xz\notin E$. This creates a $C_{4}$ $(x,i,z,i+2)$, which is forbidden. Therefore, we have $X_{i}$
1 $Z_{i+6}$, and we are done. $\Box$

Observation 3.11

$X_{i}$*
0 $Z_{i+2}\cup Z_{i+3}$*

Proof. Consider a vertex $x\in X_{i}$. Suppose there is a vertex $z\in Z_{i+2}$ such that $xz\in E$. This creates a $C_{4}$ $(x,z,i+6,i)$, which is forbidden. Therefore, $X_{i}$
0 $Z_{i+2}$. Next, suppose that there is a vertex $z\in Z_{i+3}$ such that $xz\in E$. This also creates a $C_{4}$ $(x,z,i+3,i+2)$, which is forbidden. Therefore, $X_{i}$
0 $Z_{i+3}$. $\Box$

Observation 3.12

$X_{i}$*
1 $W$*

Proof. Suppose that there are vertices $x\in X_{i}$ and $w\in W$ such that $xw\notin E$. This creates a $C_{4}$ $(i,x,i+2,w)$, which is forbidden. Therefore we have $X_{i}$
1 $W$. $\Box$

Observations 3.6–3.12 together imply the following lemma.

Lemma 3.13

For every $i$, the set $X_{i}$ is uniform in the partition ${\cal P}$. $\Box$

Next, we examine the sets $Y_{i}$.

Observation 3.14

At most two of the sets $Y_{1},Y_{2},\ldots,Y_{7}$ can be non-empty. In particular, if $Y_{i}\not=\emptyset$, then $Y_{i+1}=Y_{i+2}=Y_{i+5}=Y_{i+6}=\emptyset$, and we have $Y_{i+3}\neq\emptyset$ or $Y_{i+4}\neq\emptyset$, but not both.

Proof. Suppose $Y_{i}\not=\emptyset$. Let $y$ be a vertex in $Y_{i}$. Suppose the set $Y_{i+1}$ contains a vertex $y_{1}$. If $yy_{1}\notin E$, then there exists a $4K_{1}$ $(y,y_{1},i+3,i+6)$, which is forbidden. If $yy_{1}\in E$, then there is a $C_{4}$ ($y,y_{1},i+5,i+4$), which is forbidden. So we have $Y_{i+1}=\emptyset$, and by symmetry, $Y_{i+6}=\emptyset$. Now, suppose $Y_{i+2}$ is non-empty and contains a vertex $y_{2}$. If $yy_{2}\in E$, then there is a $C_{4}$ ($y,i+4,i+3,y_{2}$), a contradiction. But if $yy_{2}\notin E$, then there exists a $C_{6}$ $(y,i+4,i+3,y_{2},i+6,i)$, a contradiction. So, we have $Y_{i+2}=\emptyset$, and by symmetry, $Y_{i+5}=\emptyset$.

The first part of this proof shows that if $Y_{i+3}\not=\emptyset$, then $Y_{i+4}=\emptyset$. So only one of the two sets $Y_{i+3},Y_{i+4}$ can be non-empty. $\Box$

Observation 3.15

$Y_{i}$*
1 $Y_{i+3}\cup Y_{i+4}$*

Proof. Consider a vertex $y\in Y_{i}$. Suppose there is a vertex $y_{3}\in Y_{i+3}$ such that $yy_{3}\notin E$. There is a $C_{4}$ $(y,i,y_{3},i+4)$, a contradiction. So, we have $Y_{i}$
1 $Y_{i+3}$, and by symmetry, $Y_{i}$
1 $Y_{i+4}$. $\Box$

Observation 3.16

If $Y_{i}\neq\emptyset$, then $Z_{i+5}=Z_{i+6}=\emptyset$

Proof. Assume that $Y_{i}\neq\emptyset$ and $Z_{i+5}\neq\emptyset$. Consiser vertices $y\in Y_{i}$ and $z\in Z_{i+5}$. If $yz\in E$, then there exists a $C_{4}$ $(y,z,i+5,i+4)$, which is forbidden. However, if $yz\notin E$, then there exists a $C_{6}$ $(y,i,z,i+2,i+3,i+4)$, which is also forbidden. Therefore if $Y_{i}\neq\emptyset$, then $Z_{i+5}=\emptyset$, and by symmetry, $Z_{i+6}=\emptyset$. $\Box$

Observation 3.17

$Y_{i}$*
1 $W\cup Z_{i}\cup Z_{i+1}\cup Z_{i+3}\cup Z_{i+4}$*

Proof. Consider a vertex $y\in Y_{i}$. Let $w\in W$. If $yw\notin E$, then there is a $C_{4}$ $(i,w,i+4,y)$, which is forbidden. Therefore, $Y_{i}$
1 $W$.

Consider a vertex $z\in Z_{i}\cup Z_{i+1}\cup Z_{i+3}\cup Z_{i+4}$. Then $z(i+4)\in E$, and $z$ is adjacent to a vertex $i^{\prime}\in\{i,i+1\}$. If $zy\notin E$, there there is a $C_{4}$ $(i^{\prime},y,i+4,z)$, which is forbidden. Therefore, $Y_{i}$
1 $Z_{i}\cup Z_{i+1}\cup Z_{i+3}\cup Z_{i+4}$. $\Box$

Observation 3.18

$Y_{i}$*
0 $Z_{i+2}$*

Proof.Consider vertices $y\in Y_{i}$ and $z\in Z_{i+2}$ such that $yz\in E$. This creates a $C_{4}$ $(y,z,i+6,i)$, which is forbidden. Therefore $Y_{i}$
0 $Z_{i+2}$. $\Box$

Observations 3.14–3.18 together imply the following lemma.

Lemma 3.19

For every $i$, the set $Y_{i}$ is uniform in the partition ${\cal P}$. $\Box$

Next, we examine the sets $Z_{i}$.

Observation 3.20

If $Z_{i}\neq\emptyset$, then $Z_{i+2}=\emptyset$ and $Z_{i+5}=\emptyset$.

Proof. Suppose that $Z_{i}\neq\emptyset$. Also, suppose $Z_{i+2}\neq\emptyset$. Consider vertices $z_{i}\in Z_{i}$ and $z_{i+2}\in Z_{i+2}$. If $z_{i}z_{i+2}\in E$, then there is a $C_{4}$ ($z_{i},z_{i+2},i+6,i$), a contradiction. If $z_{i}z_{i+2}\not\in E$, then there is a $C_{4}$ ($z_{i},i+4,z_{i+2},i+2$), a contradiction. So $Z_{i+2}$ is empty, and by symmetry, $Z_{i+5}$ is empty. $\Box$

Observation 3.21

There can exist at most $3$ distinct sets of $5$-vertices for $C_{7}$.

Proof. Follows from Observation 3.20.

Observation 3.22

$Z_{i}$*
1 $W\cup Z_{i+1}\cup Z_{i+3}\cup Z_{i+4}\cup Z_{i+6}$*

Proof. Consider a vertex $z_{i}\in Z_{i}$ and a vertex $z\in V(G)-C_{7}$ with $z\not=z_{i}$. If $z$ is adjacent to two non-adjacent vertices, say $a$ and $b$, of the set $\{i,i+1,i+2,i+3,i+4\}$, then $zz_{i}\in E$, for otherwise there is a $C_{4}$ ($z,a,z_{i},b$). Observe that any vertex in $W\cup Z_{i+1}\cup Z_{i+3}\cup Z_{i+4}\cup Z_{i+6}$ is adjacent to two non-adjacent vertices of $\{i,i+1,i+2,i+3,i+4\}$. The Observation follows. $\Box$

Observations 3.20–3.22 together imply the following lemma.

Lemma 3.23

For every $i$, the set $Z_{i}$ is uniform in the partition ${\cal P}$. $\Box$

We can now prove our main results.

Proof of Theorem 2.7. Let $G$ be a $(4K_{1},C_{4},C_{6}$)-free graph with a $C_{7}$. Define the sets $X_{i},Y_{i},Z_{i},W$ as above. Let $G^{\prime}=G-C_{7}$. Observations 3.4 implies that the sets $X_{i},Y_{i},Z_{i},W$ form a partition of $G^{\prime}$. Lemmas 3.13, 3.19, and 3.23 show that each of the sets $X_{i},Y_{i},Z_{i}$ is uniform. Now the set $W$ is also uniform because every other set is uniform to $W$. Thus, the partition is uniform. $\Box$

Proof of Theorem 2.3. Let $G$ be a $(4K_{1},C_{4},C_{6}$)-free graph with a $C_{7}$. The graph $G^{\prime}=G-C_{7}$ has bounded clique width by Theorem 2.7 and Theorem 2.6, $G^{\prime}$ has bounded clique width. Thus, $G$ has bounded clique width by Observation 2.2. $\Box$

Proof of Theorem 2.4. Let $G$ be a $(4K_{1}$, even hole)-free graph with a $C_{7}$. By Theorem 2.3, $G$ has bounded clique width. By Theorem 2.1, $G$ can be optimally colored in polynomial time. $\Box$

So, if our graphs contain a $C_{7}$, we know how to color them. If they do not contain a $C_{7}$, then we know they must contain a $C_{5}$, for otherwise they are perfect and we would know how to color them. In the next section, we discuss the case of the $C_{5}$.

4 When the graphs contains a $C_{5}$

In this section, we assume $G$ is ($4K_{1}$, $C_{4}$, $C_{6}$)-free. For the all the claims below, we will also assume that $G$ contains an induced $C_{5}$ with vertices $(i,i+1,i+2,i+3,i+4)$. Let $R$ denote the set of [math]-vertices for this $C_{5}$, let $F_{i}$ be the set of $1$-vertices adjacent to $i$, let $T_{i}$ be the set of $2$-vertices with neighbors $(i,i+1)$, let $X_{i}$ be the set of $3$-vertices with neighbors $(i,i+1,i+2)$ and let $W$ denote the set of $5$-vertices.

The following observation is immediate.

Observation 4.1

The sets $F_{i},T_{i},X_{i},R,W$ form a partition of the vertex set of $G-C_{5}$ $\Box$

Observation 4.2

Each of $F_{i},T_{i},X_{i},R,W$ form a clique.

Proof. Consider two non-adjacent vertices $x,y$ of $G$. If both $x,y$ belong to $F_{i}$, then $x$ and $y$ and some two non-adjacent vertices of the $C_{5}$ form a $4K_{1}$, a contradiction. Similarly, we can see that $x,y$ cannot both belong to $T_{i}$, or to $R$. If $x,y$ both belong to $X_{i}$ or to $W$, then $x,y$ and some two non-adjacent vertices of $C_{5}$ form a $C_{4}$. $\Box$

Observation 4.3

$R$*
1 $F_{i}\cup T_{i}$.*

Proof. Consider a vertex $r\in R$ and a vertex $s\in F_{i}\cup T_{i}$. If $rs\notin E$, then $r,s$ and some two non-adjacent vertices in the $C_{5}$ form a $4K_{1}$. $\Box$

Observation 4.4

If $F_{i}\neq\emptyset$ then $F_{j}=\emptyset$ for all $j\neq i$.

Proof. Consider vertices $f_{i}\in F_{i}$, $f_{j}\in F_{j}$, with $i\not=j$. We must have $f_{i}f_{j}\in E$, for otherwise $f_{i},f_{j}$ and some two vertices in the $C_{5}$ form a $4K_{1}$. If $j=i+1$, then there is a $C_{4}$ ($f_{i},f_{j},j,i$). So we have $j\not=i+1$, and by symmetry, $j\not=i-1$. If $j=i+2$, then there is a $C_{6}$ ($f_{i},f_{j},j,j+1,j+2,i$). So we have $j\not=i+2$, and by symmetry, $j\not=i-2$. $\Box$

Observation 4.5

$F_{i}$*
1 $T_{i}\cup T_{i+2}\cup T_{i+4}$.*

Proof. Let $f\in F_{i}$, and let $t\in T_{i}\cup T_{i+2}$. If $ft_{i}\notin E$, then $f,t$ and some two non-adjacent vertices of the $C_{5}$ form a $4K_{1}$. So $F_{i}$
1 $T_{i}\cup T_{i+2}$. By symmetry (with the case $T_{i}$), we have $F_{i}$
1 $T_{i+4}$. $\Box$

Observation 4.6

$F_{i}$*
0 $T_{i+1}\cup T_{i+3}$.*

Proof. Let $f\in F_{i}$, and let $t\in T_{i+1}$. If $ft\in E$, then there is a $C_{4}$ ($f,t,i+1,i$). So we have $F_{i}$
0 $T_{i+1}$, and by symmetry, $F_{i}$
0 $T_{i+3}$. $\Box$

Observation 4.7

$F_{i}$*
0 $X_{i+1}$.*

Proof. Let $f\in F_{i}$, and let $x\in X_{i+1}$. If $fx\in E$, then there is a $C_{4}$ ($f,t,i+1,i$). So we have $F_{i}$
0 $X_{i+1}$. $\Box$

We note that vertices of $F_{i}$ may have neighbors and non-neighbors in $X_{i}\cup X_{i+3}\cup X_{i+4}$.

Observation 4.8

$T_{i}$*
0 $T_{j}$ for all $j\not=i$.*

Proof. Consider vertices $t_{i}\in T_{i},t_{i+1}\in T_{i+1}$. If $t_{i}t_{i+1}\in E$, then there is a $C_{6}$ ($t_{i},t_{i+1},i+2,i+3,i+4,i$). So we have $T_{i}$
0 $T_{i+1}$, and by symmetry, $T_{i}$
0 $T_{i+4}$. Now consider a vertex $t_{i+2}\in T_{i+2}$. If $t_{i}t_{i+2}\in E$, then there is a $C_{4}$ ($t_{i},i+1,i+2,t_{i+2}$). So we have $T_{i}$
0 $T_{i+2}$, and by symmetry, $T_{i}$
0 $T_{i+3}$. $\Box$

Observation 4.9

$T_{i}$*
0 $X_{i+2}$.*

Proof. Consider vertices $t_{i}\in T_{i},x_{i+2}\in X_{i+2}$. If $t_{i}x_{i+2}\in E$, then there is a $C_{4}$ ($t_{i},x_{i+2},i+2,i+1$). $\Box$

Observation 4.10

$X_{i}$*
0 $X_{i+2}$.*

Proof. Consider vertices $x_{i}\in X_{i},x_{i+2}\in X_{i+2}$. If $x_{i}x_{i+2}\in E$, then there is a $C_{4}$ ($x_{i},x_{i+2},i+4,i$). $\Box$

In the next section, we will use the results of this section to prove Theorem 2.5.

5 Clique cutset decomposition

In this section, we present a proof of Theorem 2.5. We will need to introduce definitions and background for the problem.

Consider a graph $G$. A clique cutset of $G$ is a set of vertices $S$ such that $S$ is a clique and $G-S$ is disconnected. Consider the following procedure to decompose $G$. If $G$ has a clique cutset $C$, then $G$ is decomposed into subgraphs $G_{1}=G[V_{1}]$ and $G_{2}=G[V_{2}]$ where $V=V_{1}\cup V_{2}$ and $C=V_{1}\cap V_{2}$ ($G[X]$ denotes the subgraph of $G$ induced by $X$ for a subset $X$ of vertices of $V(G)$). Given optimal colourings of $G_{1},G_{2}$, we can obtain an optimal colouring of $G$ by identifying the colouring of $C$ in $G_{1}$ with that of $C$ in $G_{2}$. In particular, we have $\chi(G)=\max(\chi(G_{1}),\chi(G_{2}))$. If $G_{i}$ ($i\in\{1,2\}$) has a clique cutset, then we can recursively decompose $G_{i}$ in the same way. This decomposition can be represented by a binary tree $T(G)$ whose root is $G$ and the two children of $G$ are $G_{1}$ and $G_{2}$, which are in turn the roots of subtrees representing the decompositions of $G_{1}$ and $G_{2}$. Each leaf of $T(G)$ corresponds to an induced subgraph of G that contains no clique cutset; we will call such graph an atom. Algorithmic aspects of the clique cutset decomposition are studied in Tarjan [18] and Whiteside [19]. In particular, the decomposition tree $T(G)$ can be constructed in $O(nm)$ time such that the total number atoms is at most $(n-1)$ [18] (Here, as usual, $n$, resp., $m$, denotes the number of vertices, resp., edges, of the graph $G$). This discussion can be summarized by the theorem below.

Theorem 5.1** ([18], [19])**

Let $G$ be a graph. If every atom of $G$ can be colored in polynomial time, then $G$ can be colored in polynomial time. $\Box$

We will need the following theorem that illustrates the structure of ($4K_{1},C_{4},C_{6},C_{5}$-twin)-free graphs.

Theorem 5.2

Let $G$ be a ($4K_{1},C_{4},C_{6},C_{5}$-twin)-free graph. If $G$ contains a $C_{5}$, then one of the following holds:

(i)

$G$* contains a clique cutset.*

(ii)

$G$* contains a $C_{7}$.*

(iii)

$G$* has bounded clique width.*

(iv)

$G$* is the join of a (possibly empty) clique and a $C_{5}$. In this case, (iii) is also satisfied.*

Proof of Theorem 5.2. Let $G$ be a ($4K_{1},C_{4},C_{6},C_{5}-twin$)-free graph and suppose $G$ contains a $C_{5}$. Assume that $G$ contains no clique cutset and no $C_{7}$, for otherwise we are done. Define the sets $F_{i},T_{i},X_{i},W,R$ as above. Note that $X_{i}=\emptyset$ for all $i$ because $G$ contains no $C_{5}$-twin. By Observation 4.4, at most one set $F_{i}$ can be non-empty. We will assume that this one set, it it exists, is $F_{1}$. Define $T=T_{1}\cup\ldots\cup T_{5}$.

We will show that

[TABLE]

Suppose $R\not=\emptyset$. Note that $W\cup T\cup F_{1}$ is a cutset, separating $R$ from the $C_{5}$. Let $C$ be a minimal ($R,C_{5}$)-separator of $G$ that is contained in $W\cup T\cup F_{1}$. By assumption, $C$ is not a clique. Consider two non-adjacent vertices $a,b$ in $C$. By the minimality of $C$, there is a chordless path $P$ with endpoints being $a,b$, and interior vertices belonging to $R$. Since $R$ is a clique, $P$ has at most three edges.

Suppose first that $P$ has three edges. Enumerate the vertices of $P$ as $x,r_{1},r_{2},y$ with $r_{i}\in R$. Since $r_{2}$ is not adjacent to $a$, by Observation 4.3, we have $a\in W$. Similarily, we have $b\in W$. But by Observation 4.2, $ab$ is an edge, a contradiction.

So $P$ has two edges. Enumerate the vertices of $P$ as $x,r,y$ with $r\in R$. Note that both $a,b$ have neighbors in the $C_{5}$. No vertex $c\in C_{5}$ can be adjacent to both $a,b$, for otherwise, there is a $C_{4}$ ($c,a,r,b$). So we have $a,b\in F_{1}\cup T$, in particular, $a,b\notin W$. Since $ab$ is not an edge, by Observation 4.2, either $a$ or $b$, or both, belongs to some $T_{s}$. We may assume $b\in T_{s}$, that is, $b$ is a 2-vertex.

Let $i$ be a vertex in the $C_{5}$ that is adjacent to $a$. Let $j$ be a the vertex in $C_{5}$ that is adjacent to $b$ and is closest to $i$ in the $C_{5}$. Then $P^{\prime}=(a,i,i+1,\ldots,j$) is an induced path. If $P^{\prime}$ has length at least four, the $P\cup P^{\prime}$ induces a chordless cycle of length at least six, a contradiction. So we know $j=i+1$. Since $b\in T_{s}$ and $bi$ is an edge, we know $b(i+3),b(i-1)\notin E$. It follows that $b(i+2)\in E$. The vertex $a$ may or may not be adjacent to $i-1$. If $a(i-1)\in E$, let $P^{\prime\prime}$ be the chordless path $b,i+2,i+3,i-1,a$; otherwise, let $P^{\prime\prime}$ be the chordless path $b,i+2,i+3,i-1,i,a$. Then $P^{\prime\prime}$ and $r$ together induces a $C_{6}$ or $C_{7}$, a contradiction. We have establised (1).

Next, we claim that

[TABLE]

Suppose $F=\emptyset$. If $T=\emptyset$, then $G$ is the join of $W$ and the $C_{5}$, and we are done. (Note that in this case $G$ has clique width three). We may assume some $T_{i}$ is not empty. By Observation 4.8, the vertices in $T_{i}$ have no neighbors in $T_{j}$ with $i\not=j$. So $W\cup\{i,i+1\}$ is a clique cutset separating $T_{i}$ from $\{i+2,i+3,i+4\}$, a contradiction. We have established (2).

Now, we may assume $F=F_{1}$ is not empty. Suppose $T_{2}\not=\emptyset$. By Observations 4.8 and 4.6, $T_{2}$’s neighbors belong to $W\cup\{2,3\}$. But then $W\cup\{2,3\}$ is a clique cutset of $G$, a contradiction. So we have $T_{2}=\emptyset$, and by symmetry, $T_{4}=\emptyset$.

Suppose that $T_{3}\not=\emptyset$. If $|T_{3}|\geq 2$, then there is a $C_{5}$-twin ($f,1,2,3,t,t^{\prime}$) for any $f\in F_{1}$ and $t,t^{\prime}\in T_{3}$ (by Observation 4.5, $f$ is adjacent to $t,t^{\prime}$). So we have $|T_{3}|=1$. If $T_{1}\not=\emptyset$, then there is a $C_{5}$-twin ($f,t_{3},3,2,t_{1},1$) for any $f\in F_{1}$, $t_{1}\in T_{1}$, and $t_{3}\in T_{3}$. So we have $T_{1}=\emptyset$, and by symmetry, $T_{5}=\emptyset$. Consider the graph $G^{\prime}$ obtained from $G$ by removing the six vertices of $C_{5}\cup T_{3}$. The partition $W,F_{1}$ is a near-uniform partition of $G^{\prime}$. By Theorem 2.6, $G^{\prime}$ has bounded clique width. By Observation 2.2, $G$ has bounded clique width.

So, we may assume that $T_{3}=\emptyset$. If $T_{1}=\emptyset$, then $\{1,5\}\cup W$ is a clique cutset separation $F_{1}\cup T_{5}$ from $\{2,3,4\}$, a contradictionl. So we have $T_{1}\not=\emptyset$, and by symmetry, $T_{5}\not=\emptyset$. Consider vertices $t_{5}\in T_{5},f\in F_{1},t_{1}\in T_{1}$. By Observations 4.8 and 4.5, we have $ft_{5},ft_{1}\in E$, and $t_{1}t_{5}\notin E$. Now, there is a $C_{7}$ ($t_{5},f,t_{1},2,3,4,5$) and so (ii) holds. $\Box$

We are now in position to prove Theorem 2.5.

Proof of Thereom 2.5. Let $G$ be a ($4K_{1},C_{4},C_{6},C_{5}$-twin)-free graph. We may assume that $G$ is not perfect, for otherwise, we may use the algorithm of [12] to color $G$. Since $G$ is $C_{4}$-free, $G$ contains no anti-hole of length at least six. So $G$ must contain a $C_{5}$ or $C_{7}$. If $G$ contains a $C_{7}$, then we are done by Theorem 2.4. So, we may assume that $G$ contains a $C_{5}$, but no $C_{7}$. By Theorem 5.1, we only need to show that every atom of $G$ can be colored in polynomial time. Let $A$ be an atom of $G$ (an induced subgraph with no clique cutset). By Theorem 5.2, $A$ has bounded clique width, thus it can be colored in poylynomial time by Theorem 2.1. $\Box$

6 Conclusions

In this paper, we studied the complexity of VERTEX COLORING for ($4K_{1},C_{4},C_{6}$)-free graphs. We showed the problem admits a polynomial time algorithm when the graph in our class has a $C_{7}$. We have not solved the problem when the graph contains a $C_{5}$. We leave this as an open problem. In addition, we designed a polynomial time algorithm for VERTEX COLORING for ($4K_{1},C_{4},C_{6}$, $C_{5}$-twin)-free graphs. We note that the more general problem to color a $(4K_{1},C_{4}$)-free graph in polynomial time is still open.

Acknowledgement

This work was supported by the Canadian Tri-Council Research Support Fund. The authors A.M.H. and C.T.H. were each supported by individual NSERC Discovery Grants. Author T.P.M was supported by an NSERC Undergraduate Student Research Award (USRA). This work was done by authors D.J.F. and K.H. in partial fulfillment of the course requirements for CP493: Directed Research Project I in the Department of Physics and Computer Science at Wilfrid Laurier University.