Chromatic bounds for some classes of $2K_2$-free graphs

TL;DR

This paper investigates chromatic bounds for specific classes of $2K_2$-free graphs, establishing linear $ ext{chi}$-binding functions for certain subclasses and exploring $ ext{chi}$-boundedness of superclasses.

Contribution

It identifies new classes of $2K_2$-free graphs that admit linear $ ext{chi}$-binding functions and extends $ ext{chi}$-boundedness results to some superclasses.

Findings

Certain ($2K_2, H$)-free graph classes have linear $ ext{chi}$-binding functions.

Some superclasses of $2K_2$-free graphs are $ ext{chi}$-bounded.

The class of ($2K_2, 3K_1$)-free graphs does not admit a linear $ ext{chi}$-binding function.

Abstract

A hereditary class $\mathcal{G}$ of graphs is $\chi$-bounded if there is a $\chi$-binding function, say $f$ such that $\chi(G) \leq f(\omega(G))$, for every $G \in \cal{G}$, where $\chi(G)$ ($\omega(G)$) denote the chromatic (clique) number of $G$. It is known that for every $2K_2$-free graph $G$, $\chi(G) \leq \binom{\omega(G)+1}{2}$, and the class of ($2K_2, 3K_1$)-free graphs does not admit a linear $\chi$-binding function. In this paper, we are interested in classes of $2K_2$-free graphs that admit a linear $\chi$-binding function. We show that the class of ($2K_2, H$)-free graphs, where $H\in \{K_1+P_4, K_1+C_4, \overline{P_2\cup P_3}, HVN, K_5-e, K_5\}$ admits a linear $\chi$-binding function. Also, we show that some superclasses of $2K_2$-free graphs are $\chi$-bounded.

Tables

Table 1. Table 1: Known chromatic bounds for ( 2 ​ K 2 , H 2 subscript 𝐾 2 𝐻 2K_{2},H )-free graphs, where H 𝐻 H is any 2 ​ K 2 2 subscript 𝐾 2 2K_{2} -free graph on 5 5 5 vertices with α ​ ( H ) = 2 𝛼 𝐻 2 \alpha(H)=2 , and the graph X ∈ { K ​ i ​ t ​ e , K 4 ∪ K 1 , ( K 3 ∪ K 1 ) + K 1 } 𝑋 𝐾 𝑖 𝑡 𝑒 subscript 𝐾 4 subscript 𝐾 1 subscript 𝐾 3 subscript 𝐾 1 subscript 𝐾 1 X\in\{Kite,K_{4}\cup K_{1},(K_{3}\cup K_{1})+K_{1}\} .

Graph class $𝒞$	$\chi$-bound for $G\in 𝒞$
($2K_2,\overline{P_5}$)-free graphs	$\lfloor \frac{3\omega \left(G\right)}{2}\rfloor$	[9]
($2K_2,C_5$)-free graphs	$\omega {\left(G\right)}^{3/2}$	[11]
($2K_2,K_1+P_4$)-free graphs	$\omega \left(G\right)+1$	(Corollary 1)
($2K_2,K_1+C_4$)-free graphs	$\omega \left(G\right)+5$	(Corollary 2)
($2K_2,\overline{P_2\cup P_3}$)-free graphs	$\omega \left(G\right)+1$	(Corollary 3)
($2K_2$, $HVN$)-free graphs	$\omega \left(G\right)+3$	(Corollary 4)
($2K_2,K_5-e$)-free graphs	$\omega \left(G\right)+4$	(Corollary 5)
($2K_2,K_5$)-free graphs	$2\omega \left(G\right)+1\le 9$	(Corollary 6)
($2K_2,X$)-free graphs	$\left(\frac{\omega \left(G\right)+1}{2}\right)$	[19]

Equations

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$$W_i$$

$$X_i$$

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$$Y_2$$

$$Z_i$$

$$X_i$$

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Full text

Keywords. Chromatic number; clique number; graph classes; $2K_{2}$-free graphs.

1 Introduction

All graphs in this paper are simple, finite and undirected. For notation and terminology that are not defined here, we refer to West [20]. Let $P_{n}$, $C_{n}$, $K_{n}$ denote the induced path, induced cycle and complete graph on $n$ vertices respectively. Let $K_{p,q}$ be the complete bipartite graph with classes of size $p$ and $q$. If $\cal{F}$ is a family of graphs, a graph $G$ is said to be $\cal{F}$-free if it contains no induced subgraph isomorphic to any member of $\cal{F}$. If $G_{1}$ and $G_{2}$ are two vertex disjoint graphs, then their union $G_{1}\cup G_{2}$ is the graph with $V(G_{1}\cup G_{2})$ $=V(G_{1})\cup V(G_{2})$ and $E(G_{1}\cup G_{2})$ $=$ $E(G_{1})\cup E(G_{2})$. Similarly, their join $G_{1}+G_{2}$ is the graph with $V(G_{1}+G_{2})$ $=$ $V(G_{1})\cup V(G_{2})$ and $E(G_{1}+G_{2})$ $=$ E(G_{1})\cup E(G_{2})$$\cup\{(x,y)\mid x\in V(G_{1}),~{}y\in V(G_{2})\}. For any positive integer $k$, $kG$ denotes the union of $k$ graphs each isomorphic to $G$. For a graph $G$, the complement of $G$ is denoted by $\overline{G}$.

A proper coloring (or simply coloring) of a graph $G$ is an assignment of colors to the vertices of $G$ such that no two adjacent vertices receive the same color. The minimum number of colors required to color $G$ is called the chromatic number of $G$, and is denoted by $\chi(G)$. A clique in a graph $G$ is a set of vertices that are pairwise adjacent in $G$. The clique number of $G$, denoted by $\omega(G)$, is the size of a maximum clique in $G$. Obviously, for any graph $G$, we have $\chi(G)\geq\omega(G)$. The existence of triangle-free graphs with large chromatic number (see [16] for a construction of such graphs) shows that for a general class of graphs, there is no upper bound on the chromatic number as a function of clique number.

A graph $G$ is called perfect if $\chi(H)=\omega(H)$, for every induced subgraph $H$ of $G$; otherwise it is called imperfect. A hereditary class $\mathcal{G}$ of graphs is said to be $\chi$-bounded [10] if there exists a function $f$ (called a $\chi$-binding function of $\cal{G}$) such that $\chi(G)\leq f(\omega(G))$, for every $G\in\cal{G}$. If $\cal{G}$ is the class of $H$-free graphs for some graph $H$, then $f$ is denoted by $f_{H}$. We refer to [17] for an extensive survey of $\chi$-bounds for various classes of graphs.

The class of $2K_{2}$-free graphs and its related classes have been well studied in various contexts in the literature; see [2]. Here, we would like to focus on showing $\chi$-binding functions for some classes of graphs related to $2K_{2}$-free graphs. Wagon [19] showed that the class of $mK_{2}$-free graphs admits an $O(x^{2m-2})$ $\chi$-binding function for all $m\geq 1$. In particular, he showed that $f_{2K_{2}}(x)=\binom{x+1}{2}$, and the best known lower bound is $\frac{R(C_{4},K_{x+1})}{3}$, where $R(C_{4},K_{x+1})$ denotes the smallest $k$ such that every graph on $k$ vertices contains either a clique of size $x+1$ or the complement of the graph contains a $C_{4}$ [10]. This lower bound is non-linear because Chung [7] showed that $R(C_{4},K_{t})$ is at least $t^{1+\epsilon}$ for some $\epsilon>0$. It is interesting to note that Brause et al. [3] showed that the class of ($2K_{2},3K_{1}$)-free graphs does not admit a linear $\chi$-binding function. It follows that the class of ($2K_{2},H$)-free graphs, where $H$ is any $2K_{2}$-free graph with independence number $\alpha(H)\geq 3$, does not admit a linear $\chi$-binding function.

Here we are interested in classes of $2K_{2}$-free graphs that admit a linear $\chi$-binding function, in particular, some classes of $2K_{2}$-free graphs that admit a ‘special’ linear $\chi$-binding function $f(x)=x+c$, where $c$ is an integer, that is, $2K_{2}$-free graphs ${G}$ such that $\chi(G)\leq\omega(G)+c$. If $c=1$, then this special upper bound is called the Vizing bound for the chromatic number, and is well studied in the literature; see [13, 17] and the references therein. Brause et al. [3] showed that if $G$ is a connected ($2K_{2},K_{1,3}$)-free graph with independence number $\alpha(G)\geq 3$, then $G$ is perfect. It follows from a result of [12] that if $G$ is a ($2K_{2}$, paw)-free graph, then either $G$ is perfect or $\chi(G)=3$ and $\omega(G)=2$ (see also [3]). Nagy and Szentmiklóssy (see [10]) showed that if $G$ is a ($2K_{2},K_{4}$)-free graph, then $\chi(G)\leq 4$. Blaszik et al. [1] and independently Gyárfás [10] showed that if $G$ is ($2K_{2},C_{4}$)-free graph, then $\chi(G)\leq\omega(G)+1$, and the equality holds if and only if $G$ is a split-graph. It follows from a result of [13] that if $G$ is a ($2K_{2},K_{4}-e$)-free graph, then $\chi(G)\leq\omega(G)+1$. Fouquet et al. [9] showed that if $G$ is a ($2K_{2},\overline{P_{5}}$)-free graph, then $\chi(G)\leq\left\lfloor\frac{3\omega(G)}{2}\right\rfloor$, and the bound is tight. Brause et al. [3] showed that if $G$ is a ($2K_{2},K_{1}+P_{4}$)-free graph, then $\chi(G)\leq 2\omega(G)$.

In this paper, by using structural results, we show that the class of ($2K_{2},H$)-free graphs, where $H\in\{K_{1}+P_{4},K_{1}+C_{4},\overline{P_{2}\cup P_{3}},HVN,K_{5}-e\}$ admits a special linear $\chi$-binding function $f(x)=x+c$, where $c$ is an integer; see Figure 1. We also show that the class of ($2K_{2},K_{5}$)-free graphs admits a linear $\chi$-binding function. Table 1 shows the known chromatic bounds for a ($2K_{2},H$)-free graph $G$, where $H$ is any $2K_{2}$-free graph on $5$ vertices with $\alpha(H)=2$. Some of the cited bounds are consequences of much stronger results available in the literature. Finally, we show $\chi$-binding functions for some superclasses of $2K_{2}$-free graphs.

2 Notation, terminology, and preliminaries

Let $G$ be a graph, with vertex-set $V(G)$ and edge-set $E(G)$. For $x\in V(G)$, $N(x)$ denotes the set of all neighbors of $x$ in $G$. For any two disjoint subsets $S,~{}T\subseteq V(G)$, $[S,T]$ denotes the set of edges $\{e\in E(G)\mid e\mbox{ has one end}$ in $S\mbox{ and the other in}~{}T\}$. Also, for $S\subseteq V(G)$, let $G[S]$ denotes the subgraph induced by $S$ in $G$, and for convenience we simply write $[S]$ instead of $G[S]$. Note that if $H_{1}$ and $H_{2}$ are any two graphs, and if $G$ is ($H_{1},H_{2}$)-free, then $\overline{G}$ is ($\overline{H_{1}},\overline{H_{2}}$)-free. For any integer $k$, we write $[k]$ to denote the set $\{1,2,\ldots,k\}$.

A diamond or a $K_{4}-e$ is the graph with vertex set $\{a,b,c,d\}$ and edge set $\{ab,bc,cd,ad,bd\}$. A paw is the graph with vertex set $\{a,b,c,d\}$ and edge set $\{ab,bc,ac,ad\}$. See Figure 1 for some of the other special graphs used in this paper.

A graph $G$ is a split graph if its vertex set $V(G)$ can be partitioned into two sets $V_{1}$ and $V_{2}$ such that $V_{1}$ is a clique and $V_{2}$ is an independent set. In [8], Földes and Hammer showed that a graph $G$ is a split graph if and only if $G$ is ($2K_{2},C_{4},C_{5}$)-free. A graph $G$ is a pseudo-split graph [15] if $G$ is ($2K_{2},C_{4}$)-free. The class of pseudo-split graphs generalizes the class of split graphs.

A $k$-clique covering of a graph $G$ is a partition $(V_{1},V_{2},\ldots,V_{k})$ of $V(G)$ such that $V_{i}$ is a clique, for each $i\in\{1,2,\ldots,k\}$. The clique covering number of the graph $G$, denoted by $\theta(G)$, is the minimum integer $k$ such that $G$ admits a $k$-clique covering. An independent/stable set in a graph $G$ is a set of vertices that are pairwise non-adjacent in $G$. The independence number of $G$, denoted by $\alpha(G)$, is the size of a maximum independent set in $G$. Clearly, for any graph $G$, we have $\chi(G)=\theta(\overline{G})$ and $\omega(G)=\alpha(\overline{G})$.

Let $G$ be a graph on $n$ vertices $v_{1},v_{2},\ldots,v_{n}$, and let $H_{1},H_{2},\ldots,H_{n}$ be any $n$ vertex disjoint graphs. Then an expansion $G(H_{1},H_{2},\ldots,H_{n})$ of $G$ [4] is the graph obtained from $G$ by

(i) replacing the vertex $v_{i}$ of $G$ by $H_{i}$, $i=1,2,\ldots,n$, and

(ii) joining the vertices $x\in H_{i}$, $y\in H_{j}$ iff $v_{i}$ and $v_{j}$ are adjacent in $G$.

An expansion is also called a composition; see [20]. If $H_{i}$’s are complete, it is called a complete expansion of $G$. By a result of Lovász [14], if $G,H_{1},H_{2},\ldots,H_{n}$ are perfect, then $G(H_{1},H_{2},\ldots,H_{n})$ is perfect.

We also use the following known results:

1. (R1)

Seinsche ([18]): If $G_{1}$ and $G_{2}$ are $P_{4}$-free, then $G_{1}\cup G_{2}$ and $G_{1}+G_{2}$ are $P_{4}$-free. 2. (R2)

Seinsche ([18]): Every $P_{4}$-free graph is perfect. 3. (R3)

Chudnovsky et al. ([6]) (The Strong Perfect Graph Theorem (SPGT)): A graph is perfect if and only if it contains no odd hole (chordless cycle) of length at least $5$ and no odd antihole (complement graph of a hole) of length at least $5$. 4. (R4)

Choudum et al. ([5]): Let $\cal{G}$ and $\cal{F}$ be hereditary classes of graphs where $\cal{F}$ admits a linear $\chi$-binding function. If there exists a constant $k$ such that for any $G\in\cal{G}$, $V(G)$ can be partitioned into $k$ subsets $V_{1},V_{2},\ldots,V_{k}$, where $[V_{i}]\in\cal{F}$ for each $i\in\{1,\ldots,k\}$, then $\cal{G}$ has a linear $\chi$-binding function. 5. (R5)

Blazsik et al. ([1]): For every pseudo-split graph $G$, $\chi(G)\leq\omega(G)+1$. 6. (R6)

Brause et al. ([3]): For every ($2K_{2}$, paw)-free graph $G$, $\chi(G)\leq\omega(G)+1$. 7. (R7)

Karthick and Maffray ([13]): For every ($2K_{2}$, diamond)-free graph $G$, $\chi(G)\leq\omega(G)+1$.

3 Linearly $\chi$-bounded $2K_{2}$-free graphs

In this section, we show that the class of ($2K_{2},H$)-free graphs, where $H\in\{K_{1}+P_{4},K_{1}+C_{4},\overline{P_{2}\cup P_{3}},K_{5}-e,HVN,K_{5}\}$ is linearly $\chi$-bounded. Note that the class of ($2K_{2},K_{1}+C_{4}$)-free graphs and the class of ($2K_{2},\overline{P_{2}\cup P_{3}}$)-free graphs generalize the class of ($2K_{2},C_{4}$)-free graphs or pseudo-split graphs. Also the class of ($2K_{2},K_{5}-e$)-free graphs, the class of ($2K_{2},\overline{P_{2}\cup P_{3}}$)-free graphs, and the class of ($2K_{2},HVN$)-free graphs generalize the class of ($2K_{2},K_{4}-e$)-free graphs.

3.1 The class of ($2K_{2},K_{1}+P_{4}$)-free graphs

First we prove a structure theorem for the complement graph of a ($2K_{2},K_{1}+P_{4}$)-free graph.

Theorem 1

Let $G$ be an imperfect $(P_{4}\cup K_{1},C_{4})$-free graph. Then $G$ is connected and there exists a partition $(V_{1},V_{2})$ of $V(G)$ such that $V_{1}$ induces a perfect subgraph of $G$, and $V_{2}$ is a clique.

Proof. Let $G$ be an imperfect $(P_{4}\cup K_{1},C_{4})$-free graph. Since $G$ is ($P_{4}\cup K_{1}$)-free, $G$ contains no hole of length at least $7$, and since $G$ is $C_{4}$-free, $G$ contain no anti-hole of length at least $7$. Thus, it follows from SPGT [6] that $G$ contains a $5$-hole (hole of length $5$), say $C$ with vertex-set $\{v_{1},v_{2},v_{3},v_{4},v_{5}\}$, and edge-set $\{v_{1}v_{2},v_{2}v_{3},v_{3}v_{4},v_{4}v_{5},v_{5}v_{1}\}$. Throughout this proof, we take all the subscripts of $v_{i}$ to be modulo 5.

Claim 1

Any vertex $x\in V(G)\setminus V(C)$ is adjacent to at least two vertices in $C$.

Proof of Claim 1. Suppose not. If $x$ is not adjacent to any of the vertices in $C$, or if $x$ is adjacent to exactly one vertex in $C$, say to $v_{1}$, then $\{v_{2},v_{3},v_{4},v_{5},x\}$ induces a $P_{4}\cup K_{1}$ in $G$, which is a contradiction. $\Diamond$

By Claim 1, $G$ is connected.

Claim 2

If $x\in V(G)\setminus V(C)$, then $[N(x)\cap V(C)]$ is isomorphic to a member of $\{K_{2},P_{3},C_{5}\}$.

Proof of Claim 2. Suppose not. Then by Claim 1, $N(x)\cap V(C)$ is either $\{v_{i},v_{i+3}\}$ or $\{v_{i},v_{i+1},v_{i+3}\}$ or $\{v_{i},v_{i+1},v_{i+2},v_{i+3}\}$, for some $i$. But then in all the cases, $\{v_{i},x,v_{i+3},v_{i+4}\}$ induces a $C_{4}$ in $G$, which is a contradiction. $\Diamond$

For $i\in[5]$, let:

[TABLE]

Moreover, let $A=A_{1}\cup\cdots\cup A_{5}$, and $B=B_{1}\cup\cdots\cup B_{5}$. Then by Claims 1 and 2, we have $V(G)=V(C)\cup A\cup B\cup D$.

Claim 3

For each $i\in[5]$ ($i$ mod 5), the following hold:

• (i)

$A_{i}$* induces a $P_{4}$-free subgraph of $G$.*

• (ii)

$[A_{i},A_{i+1}]$* is complete.*

• (iii)

If $A_{i}\neq\emptyset$ and $A_{i+1}\neq\emptyset$, then $A_{i}$ and $A_{i+1}$ are cliques in $G$.

• (iv)

If $A_{i}\neq\emptyset$, then $A_{i+2}=\emptyset=A_{i-2}$.

Proof of Claim 3. (i) Suppose to the contrary that $[A_{i}]$ contains an induced $P_{4}$, say $P$. Then by the definition of $A_{i}$, $V(P)\cup\{v_{i+3}\}$ induces a $P_{4}\cup K_{1}$ in $G$, a contradiction. So (i) holds.

Suppose that (ii) does not hold. Then there exist vertices $x\in A_{i}$ and $y\in A_{i+1}$ such that $xy\notin E(G)$. But, then $\{v_{i+3},v_{i+4},v_{i},x,y\}$ induces a $P_{4}\cup K_{1}$ in $G$, a contradiction. So (ii) holds.

Suppose that (iii) does not hold. Then up to symmetry, there exist two non-adjacent vertices $a$ and $b$ in $A_{i}$, and let $x\in A_{i+1}$. Then by (ii), $ax,bx\in E(G)$. But then $\{a,b,x,v_{i}\}$ induces a $C_{4}$ in $G$, a contradiction. So (iii) holds.

Suppose that (iv) does not hold. Then there exist vertices $x\in A_{i}$ and $y\in A_{i+2}\cup A_{i-2}$. By symmetry, we may assume that $y\in A_{i+2}$. But then $\{x,v_{i+1},v_{i+2},y\}$ induces a $C_{4}$ in $G$, if $xy\in E(G)$, and $\{x,v_{i+1},v_{i+2},y,v_{i+4}\}$ induces a $P_{4}\cup K_{1}$ in $G$, if $xy\notin E(G)$, a contradiction. So (iv) holds. $\Diamond$

Claim 4

For each $i\in[5]$ ($i$ mod 5), the following hold:

• (i)

$\{v_{i}\}\cup B_{i}\cup D$* is a clique.*

• (ii)

$[B_{i},B_{i+2}]=\emptyset=[B_{i},B_{i-2}]$.

• (iii)

$[B_{i}\cup B_{i+1}\cup B_{i+2}\cup B_{i+3}]$* is a perfect subgraph of $G$.*

Proof of Claim 4. Suppose that (i) does not hold. Then there exist vertices $x,y\in B_{i}\cup D$ such that $xy\notin E(G)$. But, then $\{x,v_{i-1},y,v_{i+1}\}$ induces a $C_{4}$ in $G$, a contradiction. So (i) holds.

Suppose that (ii) does not hold. Then there exist vertices $x\in B_{i}$ and $y\in B_{i+2}\cup B_{i-2}$ such that $xy\in E(G)$. By symmetry, we may assume that $y\in B_{i+2}$. But, then $\{x,v_{i+4},v_{i+3},y\}$ induces a $C_{4}$ in $G$, a contradiction. So (ii) holds. $\Diamond$

It is clear that (iii) follows from (i), (ii), and by SPGT [6].

Claim 5

For each $i\in[5]$ ($i$ mod 5), the following hold:

• (i)

$[A_{i},B_{i}\cup B_{i+1}]$* are complete.*

• (ii)

$[A_{i},B_{i+3}]=\emptyset$.

• (iii)

If $x\in B_{i+2}\cup B_{i-1}$, then either $[\{x\},A_{i}]$ is complete or $[\{x\},A_{i}]=\emptyset$.

Proof of Claim 5. Suppose that (i) does not hold. Then there exist vertices $x\in A_{i}$ and $y\in B_{i}\cup B_{i+1}$ such that $xy\notin E(G)$. But, then $\{v_{i+2},v_{i+3},v_{i+4},y,x\}$ induces a $P_{4}\cup K_{1}$ in $G$, a contradiction. So (i) holds.

Suppose that (ii) does not hold. Then there exist vertices $x\in A_{i}$ and $y\in B_{i+3}$ such that $xy\in E(G)$. But, then $\{x,v_{i},v_{i+4},y\}$ induces a $C_{4}$ in $G$, a contradiction. So (ii) holds.

By symmetry, we may assume that $x\in B_{i+2}$. Suppose that (iii) does not hold. Then there exist vertices $a$ and $b$ in $A_{i}$ such that $ax\in E(G)$ and $bx\notin E(G)$. Then since $\{v_{i+4},v_{i+3},a,x,b\}$ does not induce a $P_{4}\cup K_{1}$, we have $ab\in E(G)$. But, then $\{v_{i+2},a,x,b,v_{i+4}\}$ induces a $P_{4}\cup K_{1}$, a contradiction. So (iii) holds. $\Diamond$

By Claim 3(iv), we may assume that $A\setminus(A_{1}\cup A_{2})=\emptyset$. If $A_{1}\neq\emptyset$ and $A_{2}\neq\emptyset$ or if $A_{1}\neq\emptyset$ is a clique and $A_{2}=\emptyset$ or if $A_{1}\cup A_{2}=\emptyset$, then we define $V_{1}:=\{v_{1},v_{3},v_{4},v_{5}\}\cup B_{1}\cup B_{3}\cup B_{4}\cup B_{5}$ and $V_{2}:=\{v_{2}\}\cup A_{1}\cup A_{2}\cup B_{2}$. Then by the definitions of $B_{i}$ and by Claim 4(iii), $V_{1}$ induces a perfect subgraph of $G$. Also, by Claim 3(iii) and by Claim 5(i), $V_{2}$ is a clique in $G$. So $(V_{1},V_{2})$ is a required partition of $G$ and the theorem holds.

So, suppose that $A_{1}$ is not a clique. Let $a$ and $b$ be two vertices in $A_{1}$ that are non-adjacent. First, note that by Claim 5(i), $[A_{1},B_{1}\cup B_{2}]$ is complete. Moreover:

Claim 6

We have the following:

• (i)

$[A_{1},B_{5}]=\emptyset$.

• (ii)

$[B_{1},B_{2}]$, $[B_{1},B_{5}]$ and $[B_{3},B_{4}]$ are complete.

Proof of Claim 6. $(i)$: Suppose that (i) does not hold. Then there exists a vertex $x$ in $B_{5}$ and a vertex in $A_{1}$ that are adjacent. Then by Claim 5(iii), $[\{x\},A_{1}]$ is complete. In particular, $ax,ay\in E(G)$. But, then $\{x,a,b,v_{2}\}$ induces a $C_{4}$ in $G$, a contradiction. So (i) holds.

$(ii)$: If $[B_{1},B_{2}]$ is not complete, then there exist vertices $x\in B_{1}$ and $y\in B_{2}$ such that $xy\notin E(G)$. But then $\{x,y,a,b\}$ induces a $C_{4}$ in $G$, a contradiction. So, $[B_{1},B_{2}]$ is complete.

If $[B_{1},B_{5}]$ is not complete, then there exist vertices $x\in B_{1}$ and $y\in B_{5}$ such that $xy\notin E(G)$. Then since $\{y,v_{5},x,a,v_{3}\}$ or $\{y,v_{5},x,b,v_{3}\}$ do not induce a $P_{4}\cup K_{1}$, we have $ya,yb\in E(G)$. But then $\{y,a,b,v_{2}\}$ induces a $C_{4}$ in $G$, a contradiction. So, $[B_{1},B_{5}]$ is complete.

If $[B_{3},B_{4}]$ is not complete, then there exist vertices $x\in B_{3}$ and $y\in B_{4}$ such that $xy\notin E(G)$. Then by Claim 5(ii), $ya,yb\notin E(G)$. Then since $\{v_{5},y,v_{3},x,a\}$ or $\{v_{5},y,v_{3},x,b\}$ do not induce a $P_{4}\cup K_{1}$, we have $xa,xb\in E(G)$. But then $\{x,a,b,v_{1}\}$ induces a $C_{4}$ in $G$, a contradiction. So, $[B_{3},B_{4}]$ is complete. $\Diamond$

Now, we define $V_{1}:=\{v_{1},v_{2},v_{5}\}\cup A_{1}\cup B_{1}\cup B_{2}\cup B_{5}$ and $V_{2}:=\{v_{3},v_{4}\}\cup B_{3}\cup B_{4}$. Then by above claims, we see that $V_{1}$ induces a perfect subgraph of $G$ as it is a join of two perfect subgraphs induced by $\{v_{1}\}\cup B_{1}$ and $\{v_{2},v_{5}\}\cup A_{1}\cup B_{2}\cup B_{5}$, and $V_{2}$ is a clique. Hence the theorem is proved. $\Box$

The following corollary is an improvement over that in [3], where it is shown that for every ($2K_{2},K_{1}+P_{4}$)-free graph $G$, $\chi(G)\leq 2\omega(G)$.

Corollary 1

Let $G$ be a ($2K_{2},K_{1}+P_{4}$)-free graph. Then $\chi(G)\leq\omega(G)+1$.

Proof. Consider the complement $H$ of $G$. Then $H$ is a ($P_{4}\cup K_{1},C_{4}$)-free graph.

If $H$ is perfect, then $\theta(H)=\alpha(H)$, and the corollary holds.

If $H$ is imperfect, then by Theorem 1, $H$ is connected and there exists a partition ($V_{1},V_{2}$) of $H$ such that $V_{1}$ induces a perfect subgraph of $H$, and $V_{2}$ is a clique in $H$. So, $\theta(H)\leq\theta([V_{1}])+\theta([V_{2}])=\alpha([V_{1}])+1\leq\alpha(H)+1$, and the corollary follows. $\Box$

The bound in Corollary 1 is tight. For example, consider the graph $G$ isomorphic to $C_{5}[K_{t}^{c},K_{t}^{c},K_{t}^{c},K_{t}^{c},K_{t}^{c}]$. Then $G$ is $(2K_{2},K_{1}+P_{4})$-free with $\omega(G)=2$ and $\chi(G)=3$.

3.2 The class of ($2K_{2},K_{1}+C_{4}$)-free graphs

First we prove a structure theorem for the class of ($2K_{2},K_{1}+C_{4}$)-free graphs.

Theorem 2

Let $G$ be a connected ($2K_{2},K_{1}+C_{4}$)-free graph. Then $G$ is either a pseudo-split graph or there exists a partition $(V_{1},\ldots,V_{6})$ of $V(G)$ such that

1. (i)

$[V_{1}]$* is either a pseudo-split graph of $G$ with $\omega([V_{1}])\leq\omega(G)-1$ or the complement of a bipartite graph of $G$, and* 2. (ii)

$V_{i}$* is an independent set, for each $i\in\{2,\ldots,6\}$. Moreover, if $V_{1}$ induces a pseudo-split graph of $G$, then $V_{5}=\emptyset=V_{6}$.*

Proof. Let $G$ be a connected ($2K_{2},K_{1}+C_{4}$)-free graph.

If $G$ is $C_{4}$-free, then $G$ is a pseudo-split graph, and the theorem holds.

Suppose that $G$ contains an induced $C_{4}$, say $C$ with vertex-set $L_{0}:=\{v_{1},v_{2},v_{3},v_{4}\}$, and edge-set $\{v_{1}v_{2},v_{2}v_{3},v_{3}v_{4},v_{4}v_{1}\}$. Define sets $L_{1}:=\{y\in V(G)\setminus L_{0}\mid y\mbox{ has a neighbor in }L_{0}\}$ and $L_{2}:=V(G)\setminus(L_{0}\cup L_{1})$. Throughout this proof, we take all the subscripts of $v_{i}$ to be modulo 4.

Claim 1

If $x\in L_{1}$, then $|N(x)\cap L_{0}|\in\{1,2,3\}$.

Proof of Claim 1. Otherwise, $L_{0}\cup\{x\}$ induces a $K_{1}+C_{4}$ in $G$, a contradiction. $\Diamond$

So, for any $x\in L_{1}$, there exists an index $j\in[4]$ such that $xv_{j}\in E(G)$ and $xv_{j+1}\notin E(G)$. For $i\in[4]$, let:

[TABLE]

Moreover, let $W=W_{1}\cup\cdots\cup W_{4}$, $X=X_{1}\cup\cdots\cup X_{4}$, and $Z=Z_{1}\cup\cdots\cup Z_{4}$. Then, by Claim 1, $V(G)=L_{0}\cup W\cup X\cup Y_{1}\cup Y_{2}\cup Z\cup L_{2}$. Now:

Claim 2

The following hold:

• (i)

If $x\in W\cup X$, then $N(x)\cap L_{2}=\emptyset$.

• (ii)

$L_{2}$* is an independent set.*

Proof of Claim 2. (i) We may assume that $x\in W_{1}\cup X_{1}$, and suppose to the contrary that $y\in N(x)\cap L_{2}$. Then $\{y,x,v_{3},v_{4}\}$ induces a $2K_{2}$ in $G$, a contradiction. So (i) holds.

Suppose that (ii) does not hold. Then there exist two adjacent vertices, say $x$ and $y$ in $L_{2}$. But, then $\{x,y,v_{1},v_{2}\}$ induces a $2K_{2}$ in $G$, a contradiction. So (ii) holds. $\Diamond$

Claim 3

For each $i\in[4]$ ($i$ mod 4), the following hold:

• (i)

$W_{i}\cup W_{i+1}\cup X_{i}$* is an independent set.*

• (ii)

If $X_{i}\neq\emptyset$, then either $X_{i+1}=\emptyset$ or $X_{i+2}=\emptyset$.

• (iii)

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

Proof of Claim 3. We prove the claim for $i=1$.

(i) Suppose to the contrary that there exist two adjacent vertices, say $x$ and $y$ in $W_{1}\cup W_{2}\cup X_{1}$. Then $\{x,y,v_{3},v_{4}\}$ induces a $2K_{2}$ in $G$, a contradiction. So (i) holds.

Suppose that (ii) does not hold. Then there exist vertices $x_{1}\in X_{1}$, $x_{2}\in X_{2}$ and $x_{3}\in X_{3}$. Then since $\{x_{1},v_{1},x_{2},v_{3}\}$ or $\{x_{1},v_{1},x_{3},v_{3}\}$ or $\{x_{2},v_{2},x_{3},v_{4}\}$ do not induce a $2K_{2}$ in $G$, $\{x_{1},x_{2},x_{3}\}$ induces a triangle in $G$. But, then $\{x_{1},v_{2},v_{3},x_{3},x_{2}\}$ induces a $K_{1}+C_{4}$ in $G$, a contradiction. So (ii) holds.

Suppose that (iii) does not hold. Then there exist vertices $x\in Z_{1}$ and $y\in Z_{3}$. But then $\{v_{1},v_{2},y,v_{4},x\}$ induces a $K_{1}+C_{4}$ in $G$, if $xy\in E(G)$, or $\{v_{1},x,y,v_{3}\}$ induces a $2K_{2}$ in $G$, if $xy\notin E(G)$, a contradiction. So (iii) holds. $\Diamond$

Claim 4

For each $i\in\{1,2\}$, $Y_{i}$ is a union of a clique and an independent set.

Proof of Claim 4. We prove the claim for $i=1$. First, we show that $[Y_{1}]$ is $P_{3}$-free. Suppose to the contrary that $[Y_{1}]$ contains an induced $P_{3}$, say $P$. Then by the definition of $Y_{1}$, $V(P)\cup\{v_{1},v_{3}\}$ induces a $K_{1}+C_{4}$ in $G$, a contradiction. So, $[Y_{1}]$ is $P_{3}$-free, and hence it is a union of cliques. Then since $G$ is $2K_{2}$-free, it follows that $Y_{1}$ is a union of a clique and an independent set, and the claim holds. $\Diamond$

By Claim 4, for each $i\in\{1,2\}$, we define $Y_{i}:=Y_{i}^{\prime}\cup Y_{i}^{\prime\prime}$, where $Y_{i}^{\prime}$ is a clique, and $Y_{i}^{\prime\prime}$ is an independent set.

Claim 5

For each $i\in\{1,2\}$, $[Z_{i}\cup Z_{i+2},Y_{3-i}]=\emptyset$.

Proof of Claim 5. We prove the claim for $i=1$. Suppose to the contrary that there exist vertices, say $z\in Z_{1}\cup Z_{3}$ and $y\in Y_{2}$ such that $zy\in E(G)$. But, then $\{v_{1},v_{2},y,v_{4},z\}$ or $\{v_{2},v_{3},v_{4},y,z\}$ induces a $K_{1}+C_{4}$ in $G$, a contradiction. So the claim holds. $\Diamond$

Claim 6

For each $i\in\{1,2\}$, if $Z_{i}\cup Z_{i+2}\neq\emptyset$, then $Y_{3-i}$ is an independent set.

Proof of Claim 6. We prove the claim for $i=1$. Let $z\in Z_{1}\cup Z_{3}$. Up to symmetry, we may assume that $z\in Z_{1}$. By Claim 5, $[\{z\},Y_{2}]=\emptyset$. Now, we show that $Y_{2}$ is an independent set. Suppose to the contrary that there exist adjacent vertices, say $p$ and $q$ in $Y_{2}$. Then since $[\{z\},Y_{2}]=\emptyset$, we have $zp\notin E(G)$ and $zq\notin E(G)$. But, then $\{z,v_{1},p,q\}$ induces a $2K_{2}$ in $G$, a contradiction. So the claim holds. $\Diamond$

Now, by using Claim 3(iii), we prove the theorem in two cases.

Case 1. Suppose that $Z_{i}=\emptyset$, for every $i\in[4]$.

By Claim 3(ii) and by symmetry, we may assume that either $X_{2}\cup X_{4}=\emptyset$ or $X_{3}\cup X_{4}=\emptyset$. Then we define $V_{1}:=Y_{1}^{\prime}\cup Y_{2}^{\prime}\cup\{v_{1},v_{2}\}$, $V_{2}:=Y_{1}^{\prime\prime}\cup\{v_{4}\}$, $V_{3}:=Y_{2}^{\prime\prime}\cup\{v_{3}\}$, $V_{4}:=W_{1}\cup W_{2}\cup X_{1}\cup L_{2}$. Further: If $X_{2}\cup X_{4}=\emptyset$, then we define $V_{5}:=W_{3}\cup W_{4}\cup X_{3}$ and $V_{6}:=\emptyset$; and if $X_{3}\cup X_{4}=\emptyset$, then we define $V_{5}:=W_{3}\cup X_{2}$ and $V_{6}:=W_{4}$.

Now, by Claims 2 and 3(i), and by the definition of $Y_{i}^{\prime}$’s and $Y_{i}^{\prime\prime}$’s, we see that $[V_{1}]$ is isomorphic to the complement of a bipartite graph, and $V_{i}$’s are independent sets, for each $i\in\{2,\ldots,6\}$. So, $(V_{1},\ldots,V_{6})$ is a required partition of $V(G)$.

Case 2. Suppose that $Z_{i}\cup Z_{i+1}\neq\emptyset$, for exactly one $i\in[4]$.

We may assume up to symmetry that $i=1$ and $Z_{1}\neq\emptyset$. Then by Claim 6, $Y_{2}$ is an independent set. Then, we define $V_{1}:=N(v_{1})$, $V_{2}:=W_{2}\cup X_{2}\cup L_{2}$, $V_{3}:=W_{3}\cup W_{4}\cup X_{3}$, $V_{4}:=Y_{2}\cup\{v_{1},v_{3}\}$, $V_{5}:=\emptyset$ and $V_{6}:=\emptyset$. Then since $G$ is ($K_{1}+C_{4}$)-free, $V_{1}$ induces a pseudo-split graph in $G$. Also, $\omega([V_{1}])\leq\omega(G)-1$. So, by Claims 2 and 3(i), we see that $V_{i}$’s are independent sets, for each $i\in\{2,3,4\}$, and hence $(V_{1},\ldots,V_{6})$ is a required partition of $V(G)$.

This completes the proof of the theorem. $\Box$

Corollary 2

Let $G$ be a ($2K_{2},K_{1}+C_{4}$)-free graph. Then $\chi(G)\leq\omega(G)+5$.

Proof. Let $G$ be a ($2K_{2},K_{1}+C_{4}$)-free graph. We may assume that $G$ is connected. We use Theorem 2. If $G$ is a pseudo-split graph, then, by (R5), $\chi(G)\leq\omega(G)+1$. So, suppose that $V(G)$ admits a partition as in Theorem 2. Now:

(a) Suppose that $V_{1}$ induces a pseudo-split graph with $\omega([V_{1}])\leq\omega(G)-1$. Then $V_{5}=\emptyset=V_{6}$. So, $\chi(G)\leq\chi([V_{1}])+3$. Then by (R4) and (R5), $\chi(G)\leq\omega([V_{1}])+1+3\leq\omega(G)+3$.

(b) Suppose that $[V_{1}]$ is isomorphic to the complement of a bipartite graph, and $V_{i}$’s are independent sets, for each $i\in\{2,\ldots,6\}$. Then since $[V_{1}]$ is perfect, it follows by (R4) that $\chi(G)\leq\omega([V_{1}])+5\leq\omega(G)+5$. Hence the corollary is proved. $\Box$

3.3 The class of ($2K_{2},\overline{P_{2}\cup P_{3}}$)-free graphs

We use the following structure theorem for $(P_{2}\cup P_{3},C_{4})$-free graphs proved in [4].

Theorem 3** ([4])**

If $G$ is a connected $(P_{2}\cup P_{3},C_{4})$-free graph, then $G$ is either chordal or there exists a partition $(V_{1},V_{2},V_{3})$ of $V(G)$ such that (1) $[V_{1}]\cong K_{m}^{c}$, for some $m\geq 0$, (2) $[V_{2}]\cong K_{t}$, for some $t\geq 0$, (3) $[V_{3}]$ is isomorphic to a graph obtained from one of the basic graphs $G_{t}$ ($1\leq t\leq 17$) shown in Figure 2 by expanding each vertex indicated in circle by a complete graph (of order $\geq 1$), (4) $[V_{1},V_{3}]=\emptyset$, and (5) $[V_{2},V_{3}\setminus S]$ is complete (see Figure 2 for the set $S$).

For $t\in[17]$, let ${\cal{G}}_{t}$ denote the class of graphs obtained from $G_{t}$ (see Figure 2) by the operations stated in Theorem 3.

Corollary 3

Let $G$ be a ($2K_{2},\overline{P_{2}\cup P_{3}}$)-free graph. Then $\chi(G)\leq\omega(G)+1$.

Proof. Consider the complement $H$ of $G$. Then $H$ is a $(P_{2}\cup P_{3},C_{4})$-free graph.

If $H$ is chordal, then $H$ is perfect and so $\theta(H)=\alpha(H)$, and the corollary holds.

Suppose that $H$ is not chordal. Let $H_{1},H_{2},\ldots,H_{k}$ ($k\geq 1$) denote the components of $H$. Then since $H$ is not chordal and since $H$ is $(P_{2}\cup P_{3},C_{4})$-free, by Theorem 3, we may assume that there exists a component, say $H_{1}$ of $G$ such that $V(H_{1})$ admits a partition $(V_{1},V_{2},V_{3})$ as in Theorem 3 where $[V_{3}]$ contains either a $C_{5}$ or a $C_{6}$, and $[V_{3}]\in{\cal{G}}_{t}$, for $t\in[17]$. Then since $H$ is ($P_{2}\cup P_{3}$)-free, $H_{i}\cong K_{1}$, for each $i\in\{2,3,\ldots,k\}$. So $\alpha(H)=\alpha(G_{t})+|V_{1}|+(k-1)$. Now, $\theta(H)\leq\theta([V_{1}])+\theta([V_{2}\cup V_{3}])+(k-1)=|V_{1}|+\theta([V_{2}\cup V_{3}])+(k-1)=\theta([V_{2}\cup V_{3}])+\alpha(H)-\alpha(G_{t})$. It is easily verified that $\theta([V_{2}\cup V_{3}])\leq\alpha(G_{t})+1$. Hence, $\theta(H)\leq\alpha(H)+1$, and the corollary is proved. $\Box$

The graphs $C_{5}(\overline{K}_{n_{1}},\overline{K}_{n_{2}},\overline{K}_{n_{3}},\overline{K}_{n_{4}},\overline{K}_{n_{5}})$ show that the bound in Corollary 3 is tight.

3.4 The class of ($2K_{2},H$)-free graphs, $H\in\{HVN,K_{5}-e\}$

In order to prove our next results, we need the following notation. Let $G$ be a connected graph that contains an induced diamond, say $D$, with vertex set $L_{0}:=\{v_{1},v_{2},v_{3},v_{4}\}$ and edge set $\{v_{1}v_{2},v_{2}v_{3},v_{3}v_{4},v_{4}v_{1},v_{2}v_{4}\}$. Define sets $L_{1}:=\{y\in V(G)\setminus L_{0}\mid y\mbox{ has a neighbor in }L_{0}\}$ and $L_{2}:=V(G)\setminus(L_{0}\cup L_{1})$. Moreover, let:

[TABLE]

Then we have the following lemma, and we leave its proof as it can be routinely verified.

Lemma 1

Let $G$ be a connected $2K_{2}$-free graph that contains an induced diamond $D$. Let $L_{0}$, subsets of $L_{1}$, and $L_{2}$ be defined as above. Then the following hold:

1. (1)

$V(G)=N(v_{4})\cup\{v_{4}\}\cup X_{1}\cup X_{2}\cup X_{3}\cup Y_{1}\cup Y_{2}\cup Z_{1}\cup Z_{2}\cup L_{2}$. 2. (2)

We have either $X_{1}=\emptyset$ or $X_{3}=\emptyset$. 3. (3)

$X_{1}\cup X_{2}\cup Y_{1}$, $Y_{2}$, $Z_{1}$ and $L_{2}$ are independent sets. 4. (4)

$[X_{1}\cup X_{2}\cup X_{3}\cup Y_{1}\cup Y_{2}\cup Z_{1},L_{2}]=\emptyset$. $\Box$

Theorem 4

Let $G$ be a connected ($2K_{2},HVN$)-free graph. Then $G$ is either a ($2K_{2}$, diamond)-free graph or there exists a partition $(V_{1},\ldots,V_{4})$ of $V(G)$ such that

1. (i)

$V_{1}$* induces a ($2K_{2}$, paw)-free graph of $G$ with $\omega([V_{1}])\leq\omega(G)-1$, and* 2. (ii)

$V_{i}$* is an independent set, for each $i\in\{2,3,4\}$.*

Proof. Let $G$ be a connected ($2K_{2},HVN$)-free graph. If $G$ is diamond-free, then the theorem holds. Suppose that $G$ contains an induced diamond, say $D$. We use Lemma 1. By (2) and by symmetry, we may assume that $X_{3}=\emptyset$. Now, since $G$ is $HVN$-free, we have the following:

• •

For any $v\in V(G)$, $N(v)$ induces a paw-free graph with $\omega([N(v)])\leq\omega(G)-1$.

• •

$Y_{2}\cup Z_{2}$ is an independent set (by using (3)).

Define $V_{1}:=N(v_{4})$, $V_{2}:=X_{1}\cup X_{2}\cup Y_{1}$, $V_{3}:=Y_{2}\cup Z_{2}$, and $V_{4}:=Z_{1}\cup L_{2}\cup\{v_{4}\}$. Then by (3) and (4), and by the above properties, we see that $(V_{1},\ldots,V_{4})$ is a required partition of $V(G)$, and the theorem is proved. $\Box$

Corollary 4

Let $G$ be a ($2K_{2},HVN$)-free graph. Then $\chi(G)\leq\omega(G)+3$.

Proof. Let $G$ be a connected ($2K_{2},HVN$)-free graph. We use Theorem 4.

If $G$ is a ($2K_{2}$, diamond)-free graph, then by (R8), $\chi(G)\leq\omega(G)+1$, and the corollary holds. Suppose that $V(G)$ admits a partition as in Theorem 4. So, $\chi(G)\leq\chi([V_{1}])+3$. Since $\chi([V_{1}])\leq\omega([V_{1}])+1$ (by (R7)), we have $\chi(G)\leq\omega([V_{1}])+1+3\leq\omega(G)+3$, as desired. $\Box$

Theorem 5

Let $G$ be a connected ($2K_{2},K_{5}-e$)-free graph. Then $G$ is either a ($2K_{2}$, diamond)-free graph or there exists a partition $(V_{1},\ldots,V_{5})$ of $V(G)$ such that

1. (i)

$V_{1}$* induces a ($2K_{2}$, diamond)-free graph of $G$ with $\omega([V_{1}])\leq\omega(G)-1$, and* 2. (ii)

$V_{i}$* is an independent set, for each $i\in\{2,3,4,5\}$.*

Proof. Let $G$ be a connected ($2K_{2},K_{5}-e$)-free graph. If $G$ is diamond-free, then the theorem holds. Suppose that $G$ contains an induced diamond, say $D$. We use Lemma 1. By (2) and by symmetry, we may assume that $X_{3}=\emptyset$. Now, since $G$ is ($K_{5}-e$)-free, we have the following:

• •

For any $v\in V(G)$, $N(v)$ induces a diamond-free graph with $\omega([N(v)])\leq\omega(G)-1$.

• •

$Z_{2}$ is an independent set.

Define $V_{1}:=N(v_{4})$, $V_{2}:=X_{1}\cup X_{2}\cup Y_{1}$, $V_{3}:=Y_{2}$, $V_{4}:=Z_{1}\cup L_{2}\cup\{v_{4}\}$, and $V_{5}:=Z_{2}$. Then by (3) and (4), and by the above properties, we see that $(V_{1},\ldots,V_{5})$ is a required partition of $V(G)$, and the theorem is proved. $\Box$

Corollary 5

Let $G$ be a ($2K_{2},K_{5}-e$)-free graph. Then $\chi(G)\leq\omega(G)+4$.

Proof. Let $G$ be a connected ($2K_{2},K_{5}-e$)-free graph. We use Theorem 5.

If $G$ is a ($2K_{2}$, diamond)-free graph, then by (R8), $\chi(G)\leq\omega(G)+1$, and the corollary holds. Suppose that $V(G)$ admits a partition as in Theorem 5. So, $\chi(G)\leq\chi([V_{1}])+4$. Since $\chi([V_{1}])\leq\omega([V_{1}])+1$ (by (R7)), we have $\chi(G)\leq\omega([V_{1}])+1+4\leq\omega(G)+4$, as desired. $\Box$

3.5 The class of ($2K_{2},K_{1}+H$)-free graphs, for any graph $H$

Theorem 6

Let $H$ be any graph. Suppose that for every $(2K_{2},H)$-free graph $G^{\prime}$, $\chi(G^{\prime})\leq f(\omega(G^{\prime}))$. Then for every ($2K_{2},K_{1}+H)$-free graph $G$, we have $\chi(G)\leq 2f(\omega(G)-1)+1$.

Proof. Let $G$ be a ($2K_{2},K_{1}+H)$-free graph. If $G$ is an edgeless graph, then the theorem is obvious. So we may assume that there exist adjacent vertices, say $v_{1}$ and $v_{2}$ in $V(G)$. For each $i\in\{1,2\}$, let $A_{i}:=\{x\in V(G)\setminus\{v_{1},v_{2}\}\mid N(x)\cap\{v_{1},v_{2}\}=\{v_{i}\}\}$. Also, let $B:=\{x\in V(G)\setminus\{v_{1},v_{2}\}\mid N(x)\cap\{v_{1},v_{2}\}=\{v_{1},v_{2}\}\}$ and $C:=V(G)\setminus(\{v_{1},v_{2}\}\cup A_{1}\cup A_{2}\cup B)$. Then, we have the following:

1. (i)

Since $G$ does not induce a $K_{1}+H$, we have: for any $v\in V(G)$, $N(v)$ induces a $H$-free graph. So, for each $i\in\{1,2\}$, $[A_{i}\cup B]$ is a $H$-free graph with $\omega([A_{i}\cup B])\leq\omega(G)-1$. 2. (ii)

Since $G$ does not induce a $2K_{2}$, we see that $C$ is an independent set.

Now, $\chi(G)\leq\chi([N(v_{1})])+\chi([A_{2}])+\chi([C\cup\{v_{1}\}])=\chi([A_{1}\cup B\cup\{v_{2}\}])+\chi([A_{2}])+\chi([C\cup\{v_{1}\}])$. Since for every ($2K_{2},H$)-free graph $G^{\prime}$, $\chi(G^{\prime})\leq f(\omega(G^{\prime}))$, and since $C\cup\{v_{2}\}$ is an independent set (by (ii)), we have, by (R4), $\chi(G)\leq f(\omega(G)-1)+f(\omega(G)-1)+1=2f(\omega(G)-1)+1$ (by (i)), as desired. $\Box$

Corollary 6

Let $G$ be a ($2K_{2},K_{5}$)-free graph. Then $\chi(G)\leq 2\omega(G)+1\leq 9$.

Proof. Since $G$ is a ($2K_{2}$, $K_{1}+K_{4}$)-free graph, and since for every ($2K_{2},K_{4}$)-free graph $G^{\prime}$, $\chi(G^{\prime})\leq\omega(G^{\prime})+1\leq 4$ (see [10]), the corollary follows by Theorem 6. $\Box$

4 Superclasses of $2K_{2}$-free graphs

In this section, we show that some superclasses of $2K_{2}$-free graphs are $\chi$-bounded.

If $G$ is a graph and if $e:=uv$ is an edge in $G$, then we simply write $A(e)$ to denote the set of all vertices in $G$ that are not adjacent to both $u$ and $v$ in $G$. The proof of the following theorem is very similar to the proof of Wagon [19] for the class of $2K_{2}$-free graphs, and we give it here for completeness.

Theorem 7

Let $\cal{H}$ be a class of graphs and let $G$ be any graph. Suppose that $\cal{H}$ is $\chi$-bounded with $\chi$-binding function $f$. Suppose that for every edge $e$ in $G$, $[A(e)]\in\cal{H}$. Then $\chi(G)\leq\binom{\omega(G)}{2}\cdot f(\omega(G))+\omega(G)$.

Proof. Let $\omega:=\omega(G)$ and let $K$ be a complete subgraph of $G$ with $|K|=\omega$, and $V(K)=\{v_{1},v_{2},\ldots,v_{\omega}\}$. Then every vertex in $x\in V(G)\setminus V(K)$ is not adjacent to at least one vertex in $K$. Otherwise, $\{x\}\cup V(K)$ induces a clique of size larger than $\omega$ which is a contradiction. For each $i,j\in[\omega]$, $i\neq j$, let $A_{ij}:=A(e_{ij})$, where $e_{ij}$ is the edge $v_{i}v_{j}$, and let $B_{i}:=\{x\in V(G)\setminus V(K)\mid[\{x\},V(K)\setminus\{v_{i}\}]~{}\mbox{is~{}complete}\}$. Moreover, let $A:=\cup A_{ij}$ and $B:=\cup B_{i}$. Then $V(G)=V(K)\cup A\cup B$.

Now, for each $i,j\in[\omega]$, $i\neq j$, we have:

1. (i)

Since for every edge $e$ in $G$, $[A(e)]\in\cal{H}$, we have $[A_{ij}]\in\cal{H}$. 2. (ii)

$B_{i}\cup\{v_{i}\}$ is an independent set. If not, then there exist adjacent vertices, say $x$ and $y$ in $B_{i}$. But, then $\{x,y\}\cup(V(K)\setminus\{v_{i}\})$ induces a clique of size $\omega+1$, a contradiction.

So, $\chi(G)\leq\sum_{\{i,j\}\subseteq[\omega]}\chi([A(e_{ij})])+\sum_{i=1}^{\omega}\chi([B_{i}\cup\{v_{i}\}])$. Then by (i) and (ii), and by (R4), $\chi(G)\leq\sum_{\{i,j\}\subseteq[\omega]}f(\omega([A(e_{ij})]))+\sum_{i=1}^{\omega}\chi([B_{i}\cup\{v_{i}\}])$. Then since $\omega([A(e_{ij})])\leq\omega$ and since $B_{i}\cup\{v_{i}\}$ is an independent set, for each $i,j\in[\omega]$, $i\neq j$, we have $\chi(G)\leq\binom{\omega}{2}\cdot f(\omega)+\omega$, and the theorem is proved. $\Box$

Then we immediately have the following.

Corollary 7** ([19])**

Let $G$ be a $2K_{2}$-free graph. Then $\chi(G)\leq\binom{\omega(G)+1}{2}$.

Proof. Since $G$ is $2K_{2}$-free, for each $i,j\in[\omega]$, $i\neq j$, $A(e_{ij})$ is an independent set in $G$. So, $\omega([A(e_{ij})])\leq 1$, and hence the corollary follows from the proof of Theorem 7. $\Box$

Corollary 8

Let $G$ be any graph. If for every edge $e$ in $G$, $A(e)$ induces a perfect graph, then $\chi(G)\leq\frac{\omega(G)^{3}-\omega(G)^{2}+2\omega(G)}{2}$. $\Box$

Corollary 9

Let $G$ be a $(P_{2}\cup P_{4})$-free graph. Then $\chi(G)\leq\frac{\omega(G)^{3}-\omega(G)^{2}+2\omega(G)}{2}$.

Proof. Since every $P_{4}$-free is perfect (by (R2)), the corollary follows from Corollary 8. $\Box$

Acknowledgement. The first author sincerely thanks Prof. Ingo Schiermeyer for the fruitful discussions.