On the Chromatic Number of (<i>P</i>₅, <i>C</i>₅, Cricket)-Free Graphs

For a graph G, let be the chromatic number of G. It is well-known that holds for any graph G with clique number . For a hereditary graph class , whether there exists a function f such that holds for every has been widely studied. Moreover, the form of minimum such an f is also concerned. A result of Schiermeyer shows that every -free graph G with clique number has . Chudnovsky and Sivaraman proved that every -free with clique number graph is -colorable. In this paper, for any -free graph G with clique number , we prove that . The main methods in the proof are set partition and induction.

• <i>P</i>₅-Free Graphs
• Chromatic Number
• <i>X</i>-Boundedness

1. Introduction

In this paper, we consider undirected, simple graphs. For a given graph H, a graph G is called H-free if G contains no induced subgraphs isomorphic to H. Let H 1 , H 2 , ⋯ , H k ( k ≥ 2 ) be different graphs. If for any 1 ≤ i ≤ k , G is H i -free, then we say that G is ( H 1 , H 2 , ⋯ , H k ) -free. A graph G = ( V , E ) is k-colorable if there exists a function φ : V ( G ) ↦ { 1,2, ⋯ , k } such that for any u v ∈ E ( G ) , there is φ ( u ) ≠ φ ( v ) . The chromatic number of G is the minimum integer k such that G is k-colorable, denoted by χ ( G ) . For a graph G = ( V , E ) , a subset S of V ( G ) is called a clique if S induces a complete subgraph. We use ω ( G ) to denote the maximum size of cliques of G. It is well-known that ω ( G ) ≤ χ ( G ) for every graph G. A graph is perfect if for any induced subgraph G ′ of G, ω ( G ′ ) = χ ( G ′ ) . Chudnovsky et al. [1] gave an equivalent characterization of perfect graphs, which is also called as the Strong Perfect Graph Theorem.

Theorem 1.1. [1] A graph is perfect if and only if it contains neither odd cycles of length at least five nor the complements of these odd cycles.

We say a hereditary graph class G is χ -bounded, if there exists a function f such that for any G ∈ G , χ ( G ) ≤ f ( ω ( G ) ) . Moreover, f is called a χ -binding function of G . Erdös [2] showed that for arbitrary integers k , l ≥ 3 , there exists a graph G with girth at least l and χ ( G ) ≥ k , which implies that the class of H-free graphs is not χ -bounded when H contains a cycle. Gyárfás conjectured that the graph class obtained by forbidding a tree (or forest) is χ -bounded.

Conjecture 1.2. [3] Let T be a tree (or forest), then there exists a function f T such that, for any T-free graph G, χ ( G ) ≤ f T ( ω ( G ) ) .

Moreover, Gyárfás [3] verified this conjecture when T = P k , and showed that f T ≤ ( k − 1 ) ω ( G ) − 1 . When T = P 5 , Esperet et al. [4] gave a χ -binding function of P 5 -free graphs as following.

Theorem 1.3. [4] Suppose G is a P 5 -free graph with clique number ω ≥ 3 . Then χ ( G ) ≤ 5 ⋅ 3 ω − 3 .

As far as we know, for ω ≥ 3 , f ( ω ) = 5 ⋅ 3 ω − 3 is the optimal χ -binding function of P 5 -free graphs at present. Furthermore, determining a polynomial χ -binding function of the class of P 5 -free graphs is an open problem. A result in [5] shows that the class of H-free graphs has a linear χ -binding function f, if and only if f ( ω ) = ω and H is an induced subgraph of P 4 , which means that the class of P 5 -free graphs has no linear χ -binding function.

In this paper, we focus on subclasses of P 5 -free graphs. While the class of P 5 -free graphs has no linear χ -binding function, some subclasses of P 5 -free have linear χ -binding functions.

Theorem 1.4. [6] [7] [8] [9] Suppose H ∈ { d i a m o n d ,   g e m ,   p a r a g l i d e r ,   p a w } , then the class of ( P 5 , H ) -free graphs has a χ -binding function.

More formally, Chudnovsky et al. [6] proved that the class of ( P 5 , gem ) -free graphs has a χ -binding function f ( ω ) ≤ ⌈ 5 4 ω ⌉ . Huang and Karthick [7] showed that ( P 5 , paraglider ) graphs have a χ -binding function f ( ω ) ≤ ⌈ 3 2 ω ⌉ . Karthick and Maffray [8] gave a χ -binding function f ( ω ) = ω + 1 for ( P 5 , diamond ) -free graphs. And Randerath [9] showed that ( P 5 , paw ) -free graphs have a χ -binding function f ( ω ) = ω + 1 (diamond, gem, paraglider and paw are given in Figure 1).

It is worth noting that a result in [10] shows that when H contains an independent set with size at least 3, the class of ( P 5 , H ) -free graphs has no linear χ -binding function.

Theorem 1.5. [10] The class of ( 2 K 2 ,3 K 1 ) -free graphs has no linear χ -binding function.

Obviously, when H is a graph with independent number at least 3, ( P 5 , H ) -free graphs is a superclass of ( 2 K 2 ,3 K 1 ) -free graphs. Thus the class of ( P 5 , H ) -free graphs has no χ -binding function.

The following theorem shows that some subclasses of P 5 -free graphs have a χ -binding function f ( ω ) = ( ω + 1 2 ) (The addition forbidden subgraphs are given in Figure 2).

Theorem 1.6. [10] [11] [12] The class of ( P 5 , H ) -free graphs has a χ -binding function f ( ω ) = ( ω + 1 2 ) when H ∈ { b u l l ,   h o u s e ,   h a m m e r } .

In [13], Schiermeyer proved that the class of ( P 5 , H ) -free graphs has a χ -binding function f ( ω ) = ω 2 for H ∈ { claw ,   cricket ,   dart ,   gem + } (see Figure 3).

In addition to the subclasses of P 5 -free graphs we mentioned above, there are many subclasses had been proved that admit a polynomial χ -binding function, which is given in [14] and [15]. More results on χ -binding function, see [16].

The class of ( P 5 , C 5 ) -free graphs, which is a superclass of ( P 5 , C 5 , cricket ) -free graphs, has been studied by Chudnovsky and Sivaraman [11]. They showed that every ( P 5 , C 5 ) -free graph with clique number ω is 2 ω − 1 -colorable. In this paper, we obtain the following result. In the next section, we will give the proof.

Theorem 1.7. Every ( P 5 , C 5 , cricket ) -free graph G with clique number ω has χ ( G ) ≤ ⌈ ω 2 2 ⌉ + ω .

2. The Proof of Main Result

For two vertex sets A and B, let E ( A , B ) = { u v ∈ E ( G ) : u ∈ A   and   v ∈ B } . We say that A is complete to B, if for any x ∈ A and y ∈ B , x y ∈ E ( G ) . For a given graph G = ( V , E ) , let N ( v ) denote the neighborhood of v ∈ V ( G ) , and for a subset S of V ( G ) , set N ( S ) = ∪ v ∈ S     N ( v ) . An induced subgraph D of G is called a dominating D, if there is V ( G ) \ V ( D ) ⊆ N ( V ( D ) ) . In this paper, for an induced P 4 : P = v 1 v 2 v 3 v 4 , we simply write V ( P ) as P. First, we give a lemma based on the structure of a ( P 5 , C 5 ) -free graph.

Lemma 2.1. If P = v 1 v 2 v 3 v 4 is a dominating P 4 of a ( P 5 , C 5 ) -free graph G, then v 2 v 3 is a dominating edge of G.

Suppose, to the contrary, that there exists a vertex u ∉ N ( v 2 ) ∪ N ( v 3 ) . Since P is a dominating P 4 , u ∈ N ( v 1 ) ∪ N ( v 4 ) . By symmetry, we may assume that u v 1 ∈ E ( G ) . If u v 4 ∈ E ( G ) , then u v 1 v 2 v 3 v 4 u would be an induced C 5 . If u v 4 ∉ E ( G ) , then u v 1 v 2 v 3 v 4 would be an induced P 5 . Either deduces a contradiction.

Next, we show that a subclass of ( P 5 , C 5 , cricket ) -free graphs has a χ -binding function f ( ω ) = ⌈ ω 2 2 ⌉ . Let i K 1 + K 2 be the graph consisted of one edge and i isolated vertices.

Lemma 2.2. Every ( P 5 , C 5 ,2 K 1 + K 2 ) -free graph G with clique number ω has χ ( G ) ≤ ⌈ ω 2 2 ⌉ .

Apply induction on ω . If ω = 1 , it is obviously true. When ω = 2 , it is also true because every ( P 5 , C 5 , K 3 ) -free graph is a bipartite graph. Moreover, when ω = 3 , by Theorem 1.3, χ ( G ) ≤ 5 = ⌈ 9 2 ⌉ . Next, consider the cases ω ≥ 4 . If G is P 4 -free, then G is perfect by Theorem 1.1. So we may suppose that P = v 1 v 2 v 3 v 4 is an induced P 4 . We claim that P is a dominating P 4 of G. Otherwise, there would exist a vertex u ∈ V ( G ) \ N ( P ) . Noting that P ⊆ N ( P ) , { u , v 1 , v 3 , v 4 } induces a 2 K 1 + K 2 , a contradiction. By Lemma 2.1, v 2 v 3 is a dominating edge of G. Next, denote

V 2 = { v : v v 2 ∈ E ( G )   and   v v 3 ∉ E ( G ) } \ { v 3 } ,

V 3 = { v : v v 2 ∉ E ( G )   and   v v 3 ∈ E ( G ) } \ { v 2 } ,

V 2,3 = N ( v 2 ) ∩ N ( v 3 ) .

For clarity, we give this partition in Figure 4. Let G [ S ] denote the subgraph of G induced by S. Clearly, G [ V 2 ] is ( P 5 , C 5 , K 1 + K 2 ) -free. (Otherwise, let { u 1 , u 2 , u 3 } be an induced K 1 + K 2 of G [ V 2 ] . Then { u 1 , u 2 , u 3 , v 3 } would induce a 2 K 1 + K 2 .) By Theorem 1.1, G [ V 2 ] is perfect. Noting that ω ( G [ V 2 ] ) ≤ ω − 1 , we have χ ( G [ V 2 ] ) ≤ ω − 1 . Similarly, χ ( G [ V 3 ] ) ≤ ω − 1 . Moreover, there is ω ( G [ V 2,3 ] ) ≤ ω − 2 . By induction, χ ( G [ V 2,3 ] ) ≤ ⌈ ( ω − 2 ) 2 2 ⌉ .

Now we color G. Let K = { 1,2, ⋯ , ⌈ ω 2 2 ⌉ } be a color set. First, we color v 2 and v 3 by colors 1 and 2, respectively. Noting that E ( V 2 , { v 3 } ) = ∅ , V 2 can be colored by { 2,3, ⋯ , ω } . Similarly, V 3 can be colored by { 1, ω + 1, ⋯ ,2 ω − 2 } . Thus, χ ( G [ V 2 ∪ V 3 ∪ { v 2 , v 3 } ] ) ≤ 2 ω − 2 . Since v 2 v 3 is a dominating edge of G, V ( G ) = { v 2 , v 3 } ∪ V 2 ∪ V 3 ∪ V 2,3 . So we have

χ ( G ) ≤ χ ( G [ V 2 ∪ V 3 ∪ { v 2 , v 3 } ] ) + χ ( G [ V 2,3 ] ) ≤ 2 ω − 2 + ⌈ ( ω − 2 ) 2 2 ⌉ = ⌈ ω 2 2 ⌉ .

Note that the bound given by Lemma 2.2 is tight for ω = 2 , and C 4 is a ( P 5 , C 5 , cricket ) -free graph with clique number 2 and chromatic number 2.

Proof of Theorem 1.7

When ω ≤ 3 , it is obviously true. Next, assume that ω ≥ 4 . If G is P 4 -free, then χ ( G ) = ω by Theorem 1.1. So we may suppose that P = v 1 v 2 v 3 v 4 is an induced P 4 of G. Let N 2 ( P ) = N ( N ( P ) ) \ N ( P ) and N 3 ( P ) = N ( N 2 ( P ) ) \ N ( P ) . Moreover, for arbitrary different i , j , k ∈ { 1,2,3,4 } , denote

U i = { v ∈ N ( P ) \ P : N ( v ) ∩ P = { v i } } ,

U i , j = { v ∈ N ( P ) \ P : N ( v ) ∩ P = { v i , v j } } ,

U i , j , k = { v ∈ N ( P ) \ P : N ( v ) ∩ P = { v i , v j , v k } } ,

A = { v ∈ N ( P ) \ P : N ( v ) ∩ P = P } .

Clearly, U i , j = U j , i and U i , k , j = U i , j , k = U j , i , k . Since G is ( P 5 , C 5 ) -free, U 1 = U 4 = U 1 , 4 = ∅ . So

A ∪ U 2 ∪ U 3 ∪ U 1 , 2 ∪ U 1 , 3 ∪ U 2 , 3 ∪ U 2 , 4 ∪ U 3 , 4 ∪ U 1 , 2 , 3 ∪ U 1 , 2 , 4 ∪ U 1 , 3 , 4 ∪ U 2 , 3 , 4 = N ( P ) \ P .

The partition is shown in Figure 5. Since G is P 5 -free, there is no vertex with a distance of 4 to P. So we can partition V ( G ) into N ( P ) , N 2 ( P ) , N 3 ( P ) , and color these sets respectively. Next, we give two claims based on N 3 ( P ) and N 2 ( P ) .

Claim 1 N 3 ( P ) = ∅ .

Otherwise, suppose there are vertices x 3 ∈ N 3 ( P ) and x 2 ∈ N 2 ( P ) such that x 2 x 3 ∈ E ( G ) . Let u ∈ N ( P ) \ P be a neighbor of x 2 . If u ∈ A , then { x 2 , u , v 1 , v 2 , v 4 } would induce a cricket, a contradiction. So there exists v i and v j ( i , j ∈ { 1,2,3,4 } ) such that v i v j ∈ E ( G ) , u v i ∈ E ( G ) and u v j ∉ E ( G ) . Now x 3 x 2 u v i v j is an induced P 5 , a contradiction.

Claim 2 Let T be a connected component of G [ N 2 ( P ) ] with | V ( T ) | ≥ 2 , then then at least one vertex of U 2,3 is complete to V ( T ) .

First, we show that every edge x y in T has N ( x ) ∩ N ( P ) = N ( y ) ∩ N ( P ) . Suppose, to the contrary, that there exists a vertex u ∈ ( N ( x ) ∩ N ( P ) ) \ ( N ( y ) ∩ N ( P ) ) . Similar to the proof of Claim 1, there is an induced cricket or induced P 5 , a contradiction. So, for each x y ∈ E ( T ) , x and y have same neighborhood in N ( P ) . By connectivity and transitivity, all vertices in T have same neighborhood in N ( P ) . Then there is at least one vertex, say u, in N ( P ) \ P such that V ( T ) is complete to { u } .

Next, we pick an arbitrary edge x y in T. Then x u y is a triangle. If u ∈ U 2 ∪ U 1,2 , then x u v 2 v 3 v 4 would be an induced P 5 . And if u ∈ A ∪ U 1,3 ∪ U 1,2,3 ∪ U 1,3,4 , then { x , y , u , v 1 , v 3 } would induce a cricket. Up to symmetry, there must be u ∈ U 2,3 .

By Claim 2, for an arbitrary connected component T of G [ N 2 ( P ) ] , there exists a vertex u ∈ U 2,3 such that { u } is complete to V ( T ) . If there exists x , y ∈ V ( T ) such that x y ∉ E ( G ) , then { x , y , u , v 2 , v 3 } would induce a cricket. Thus V ( T ) is a clique with size at most ω − 1 , which implies that

χ ( G [ N 2 ( P ) ] ) ≤ ω − 1. (1)

Let G ′ = G [ N ( P ) ] . Note that P is a dominating P 4 of G ′ . By Lemma 2.1, v 2 v 3 is a dominating edge of G ′ . Thus V ( G ′ ) \ { v 2 , v 3 } can be partitioned into { V 2 , V 3 , V 2,3 } , which is defined as in Lemma 2.2. Since G ′ is ( P 5 , C 5 , cricket ) -free, both G [ V 2 ] and G [ V 3 ] are ( P 5 , C 5 , K 1 + K 2 ) -free. Thus, by the coloring described in Lemma 2.2, there is

χ ( G [ V 2 ∪ V 3 ∪ { v 2 , v 3 } ] ) ≤ 2 ω − 2 . Moreover, noting that G [ V 2,3 ] is complete to { v 2 , v 3 } , we have that G [ V 2,3 ] is ( P 5 , C 5 ,2 K 1 + K 2 ) -free and ω ( G [ V 2,3 ] ) ≤ ω − 2 . By Lemma 2.2, χ ( G [ V 2,3 ] ) ≤ ⌈ ( ω − 2 ) 2 2 ⌉ . Thus,

χ ( G ′ ) ≤ χ ( G [ V 2 ∪ V 3 ∪ { v 2 , v 3 } ] ) + χ ( G [ V 2 , 3 ] ) ≤ 2 ω − 2 + ⌈ ( ω − 2 ) 2 2 ⌉ ≤ ⌈ ω 2 2 ⌉ . (2)

By Claim 1, V ( G ) = N ( P ) ∪ N 2 ( P ) . Hence, by Inequality (1) and (2), there is

χ ( G ) ≤ χ ( G ′ ) + χ ( G [ N 2 ( P ) ] ) ≤ ⌈ ω 2 2 ⌉ + ω .

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

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