Abstract

In the early 1980s, Gyárfás introduced the concept of the -bound with -binding functions thereby extending the notion of perfectness. There are a number of challenging conjectures about the -bound. Let , , and be the chromatic number, clique number, and maximum degree of a graph , respectively. In this paper, we prove that if is a triangle-free and -free graph, then unless is one of eight graphs with and , where the eight graphs are extended from the Grötzsch graph as a Mycielskian of a 5-cycle graph. Moreover, we also show that if is a -free graph.

1. Introduction

We consider finite graphs only and refer to [1] for undefined terminology and notation. Let be a graph. A -colouring of is an assignment of colours to its vertices such that no adjacent vertices are assigned the same colour. The minimum for which has a -colouring is called its chromatic number and denoted . For a subset , we call a clique of if all two vertices of are adjacent in . The cardinality of a maximum clique of is called its clique number and denoted .

Let be a family of graphs. A graph is called -free if does not contain an induced copy of any . In particular, we simply write -free instead of -free if . A family of graphs is called -bound with -binding function if holds whenever is an induced subgraph of , which is initiated by Gyárfás [2] and extended the notion of perfectness. Gyárfás [2] conjectured that for each forest , there exists an integer function such that where is a -free graph.

Let and be disjoint graphs, their union graph is the graph with and . The union graph is simply denoted by . Wagon [3] proved that . Randerath and Schiermeyer [4] established that . For more results on the chromatic number of graphs with some forbidden subgraphs, we refer to [59].

The graph is the union graph of paths and . Recently, Bharathi and Choudum [10] have shown that and obtained the sharper bounds for -free graphs. Here, we give the following result for triangle-free and -free graphs.

Theorem 1. Let be a triangle-free and -free graph. Then, unless is the one of eight graphs with and , as shown in Figure 1, where the eight graphs are extended from the Grötzsch graph as a Mycielskian of a 5-cycle graph.

In paper [11], Choudum and Karthick gave the -bounding for -free graphs.

Theorem 2 (Choudum and Karthick [11]). Let be a -free graph. Then, .

Let be a wheel with an order , which is obtained from the star by connecting all leaves in a circuit . In this paper, we study the -bounding for -free graphs and obtain the following result.

Theorem 3. Let be a -free graph. Then, .

2. Proof of Theorem 1

Let be a graph. For a vertex subset , the neighborhood of is defined to be the set

The closed neighborhood of is , and the induced subgraph of is the subgraph of whose vertex set is and whose edge set consists of all edges of which have both ends in . If there is no confusion, , , and are simply denoted by , , and , respectively.

We need the following lemma to prove Theorem 1.

Lemma 1. Let be a triangle-free and -free graph. Then, .

Proof. The result is trivial for . Let and , where and are the neighborhood of and , respectively. The triangle-freeness implies that and are disjoint independent sets. The -freeness implies that is a disjoint union of complete graphs. Since is triangle-free, we have that is a disjoint union of independent vertices and independent edges. Therefore,A connected graph is called a near-bipartite graph if there exists a vertex such that is a star and is a bipartite graph where is the closed neighborhood of in . Clearly, for every near-bipartite graph .

Proof of Theorem 1. Without loss of generality, suppose that is connected. Let be the maximal degree of and be a vertex of such that . The triangle-freeness implies that is a star.
Let . If is -free, it follows from -freeness that is a bipartite graph. Since is a star, is near-bipartite. Thus, . Suppose that has a as it is an induced subgraph. LetThe -freeness implies that there exist at least two vertices of which are adjacent to , and the triangle-freeness implies that there exist at most two nonadjacent vertices of which are adjacent to , where . SoIf there exist such that , without loss of generality, say by (4), then , a contradiction. Thus,By (4) again,Then, , and we have the following three cases to prove Theorem 1.

Case 1. In this case, . By (4) and (5), we have thatBy (4), (5), and (7), it is checked that , as shown in Figure 2. Without loss of generality, let .
If there exists a vertex such that and , then , a contradiction. Thus,Since , it follows from (8) that there are possible three cases: (i) , (ii) , and (iii) by extending from ( as shown in Figure 2) to .For (i), , as shown in Figure 2For (ii), without loss of generality, suppose that and . Let be the neighbour of . Since , . The triangle-freeness forces that . Then,If there exists , then since and (8). Thus, . And, then , a contradiction. So, , as shown in Figure 2.For (iii), if have the same neighbour in , say , then -freeness implies that . By (8) and , , as shown in Figure 2. Otherwise, have different neighbours in , say , respectively. The -freeness implies that . Thus, or else . Therefore, is a 4-regular graph. That is, , as shown in Figure 2.Now, it is sufficient to prove that each member of Figure 2 has a 3-colouring. Since is a star and is a bipartite graph, is near-bipartite for . And, then they have a 3-colouring. For the graph of Figure 2, we check that directly, see Figure 2.

Case 2. . In this case, . We claim that is isomorphic to one member of Figure 1.
Suppose there exists such that . Without loss of generality, let . By (4), where each indices of are taken modulo 5. On the other hand, by (4) and (5),A contradiction. Therefore, for . It follows from (4) that . Thus,By (4), (5), and (11), it is checked that , as shown in Figure 1 ( is the Grötzsch graph as a Mycielskian of a 5-cycle graph). Without loss of generality, suppose that .
By a similar argument in (8) of Case 1, we obtain thatSuppose that there exists such that . The triangle-freeness implies with . If , then the triangle-freeness forces and for or else . Thus, , a contradiction so and by the same argument. And we have thatFor , by the same argument, we obtain thatLet . By (14), . If there exists , then by (12). That is, there exists such that . Thus, , a contradiction. Therefore,If there exist such that , then (or else by (14), and if there exist such that , then or else by (14), where the indices are taken modulo 5. By (14) and (15) again, as shown in Figure 1, we have thatIt is well known that the Grötzsch graph as a Mycielskian of a 5-cycle graph (see in Figure 1) satisfies one’s chromatic number is 4. By Lemma 1, where is one member of Figure 1.

Case 3. . Suppose that , without loss of generality, let . The triangle-freeness forces that for . And then the -freeness forces that there exist at least two vertices of which are adjacent to , say . Thus, since is triangle-free. So, , a contradiction. Therefore, , and similarly. Thus,Since , there exist two vertices in , say and , which are adjacent to neither nor . Then, , a contradiction.
This completes Proof of Theorem 1.

3. Proof of Theorem 3

Let and be disjoint subsets of . The set of all edges in with one end in and another end in is denoted by , and if every vertex in is adjacent to every vertex in , then is said to be complete. We call an independent set of if no two vertices of are adjacent in . For the sake of convenience, if is an induced subgraph of , we write .

Proof of Theorem 3. If is -free, by Theorem 2, then . So we suppose that contains a as it is an induced subgraph, let , , andThen, it is easily verified that . For convenience, we further partition , and as follows:Then, , , and . Unless otherwise stated, we assume that the indices are taken modulo 4 in this section. For each , we have the following results.R 3.1. is an independent set (or else ).R 3.2. is an independent set (or else ).R 3.3. is the disjoint union of complete graphs (or else ).R 3.4. is the disjoint union of complete graphs (or else ).R 3.5. is complete (or else ).R 3.6. (or else ).R 3.7. (or else ).By R 3.3, let is the number of components of , then where is complete and with .R 3.8. If , then and .Proof of R 3.8. Suppose that , there exist , , and such that is an independent set. Let , the -freeness forces that , say . Thus, , a contradiction. So . By the similar argument, one get .R 3.9. If , then and are both independent sets with order two, and .Proof of R 3.9. Since , let , , and , . The -freeness forces that and . The -freeness forces that and . So . Without loss of generality, let , then or else . Therefore, . Thus, .Suppose there exists , then or else . And then, , a contradiction. So . By the similar argument, . Therefore, and .R 3.10. If , then or .Proof of R 3.10. Since , and are complete. Suppose that there exist , such that , it suffices to show that and are both complete. If there exists such that , then , a contradiction. So is complete. Let , such that . If , then , a contradiction. Otherwise, , the -freeness forces that , and thus, , a contradiction. So is complete.R 3.11. If for each , then .Proof of R 3.11. By R 3.8, we have for each . Suppose , without loss of generality, let or .If , by R 3.9, let , , and , say . Let , then -freeness forces and the -freeness forces . Without loss of generality, let and . The -freeness and force . Thus, , a contradiction.Now, suppose that . Let , and . The -freeness forces that and , say and . Thus,Let . The -freeness forces that . If , then (or else ) and . And then , a contradiction. So, , and similarly. Thus, , a contradiction.R 3.12. If for each and , then .Proof of R 3.12. Without loss of generality, let , , and . By R 3.3, and are both complete. Suppose that and . Let , , and . The -freeness forces , say . Then, (or else ), say . And thus, , a contradiction. So, there is at least one with order 1 of .Let , and , . The -freeness forces . Suppose that , without loss of generality, let . Then, or else . And thus, , a contradiction. So , without loss of generality, let and . Then, or else .The -freeness forces . Without loss of generality, let , then or else . And then, or else . And or else . Therefore, .Next, it suffices to show that .If there exists , then the -freeness forces and . Thus,