1. Introduction
All graphs considered in this paper are finite and simple. We use
and
to denote a
path and a
cycle on
k vertices, respectively, and follow [
1] for undefined notations and terminology. Let
G be a graph, and
X be a subset of
. We use
to denote the subgraph of
G induced by
X, and call
X a
clique (
independent set) if
is a complete graph (has no edge). The
clique number of
G is the maximum size taken over all cliques of
G. Let
denote the minimum degree of
G.
For , let be the set of vertices adjacent to v, , , . For , let has a neighbor in and . If it does not cause any confusion, we will omit the subscript G and simply write and .
For positive integer i, let }, where is the distance between u and v in G. Moreover, let . Notice that is the neighborhood of X in G. We write for .
Let G and H be two vertex disjoint graphs. Join is the graph with and }. Union is the graph with and . The union of k copies of the same graph G will be denoted by . We say that G induces H if G has an induced subgraph isomorphic to H, and say that G is H-free otherwise. Analogously, for a family of graphs, we say that G is -free if G induces no member of . The complement of a graph G will be denoted by .
Let k be a positive integer, and let . A k-coloring of G is a mapping , such that whenever in G. The chromatic number of G is the minimum integer k, such that G admits a k-coloring. It is well known that . A perfect graph is one such that for all of its induced subgraphs H. A family of graphs is said to be -bounded if there is a function f such that for every , and if such a function does exist for , then f is said to be a χ-binding function of .
Let G be a graph, , and let X and Y be two subsets of . We say that v is complete to X if v is adjacent to all vertices of X, and say that v is anticomplete to X if v is not adjacent to any vertex of X. We say that X is complete (respectively, anticomplete) to Y if each vertex of X is complete (resp. anticomplete) to Y. Particularly, we say that X is almost complete to Y if at most one vertex of X is not complete to Y. For , we simply write if , and write if .
A hole of G is an induced cycle of length at least 4, and a k-hole is a hole of length k. A k-hole is called an odd hole if k is odd, and is called an even hole otherwise. An antihole is the complement of some hole. An odd (respectively, even) antihole is defined analogously. The famous Strong Perfect Graph Theorem states that
Theorem 1 ([
2]).
A graph is perfect if and only if it induces neither an odd hole nor an odd antihole. Erdös [
3] proved that for any positive integers
, there exists a graph
G with
and no cycles of length less than
l. This result motivates us to study the chromatic number of
F-free graphs, where
F is a forest (a disjoint union of trees). Gyárfás [
4] and Sumner [
5] independently conjectured that if
F is a forest, then
F-free graphs are
-bounded.
The class of
-free graphs has attracted a great deal of interest in recent years. It is known that for every
-free graph
G,
[
6]. Up to now, the best known
-binding function for
-free graphs is
[
7]. We refer the interested readers to [
6,
8,
9,
10] for results of
-free graphs, and to [
11,
12,
13] for more results and problems about the
-bounded problem. In particular, Brause et al. [
8] proved that the class of
-free graphs does not admit a linear
-binding function. In 2022, Brause et al. [
14] gave a more general theorem.
Theorem 2 (Lemma 1 of [
1]).
Let be a set of graphs and ℓ be an integer such that has girth at most ℓ for each . If the class of -free graphs is χ-bounded, then the class of -free graphs does not admit a linear χ-binding function. Many scholars began to show interest in
-free graphs as
-free graphs is a superclass of
-free graphs. The best known
-binding function for
-free graphs is
[
15]. In [
16,
17,
18,
19,
20,
21], the authors proved that the class of (
)-free graphs has a linear
-binding function when
diamond, house, gem, paraglider,
,
}. Recently, Wu and Xu [
22] proved that
if
G is (
, crown)-free. (see
Figure 1 for the illustration of
and some forbidden configurations.)
In this paper, we prove that
Theorem 3. if G is (, kite)-free.
Let G be a graph on n-vertices and let be n vertex-disjoint graphs. An expansion of G is a graph obtained from G by
(i) Replacing each of G by , ;
(ii) Joining every vertex in with every vertex in , whenever and are adjacent in G.
In addition, and , denoted by .
Let
H be the Mycielski–Gröstzsch graph (see
Figure 2). Then,
and
. It is clear that
is (
, kite)-free, and
. Let
F be the complement of Schläfli graph (see
https://houseofgraphs.org/graphs/19273) (accessed on 1 September 2023). Then,
and
. Obviously,
is (
, kite)-free. Hence,
and
. This implies that our bound is optimal when
.
By Theorem 2, we have that the class of (, hammer)-free graphs has no linear -binding functions. In this paper, we prove that
Theorem 4. if G is (, hammer)-free.
Notice that H is also (, hammer)-free. Therefore, the -binding function for (, hammer)-free graphs is tight when .
In [
23], Choudum and Karthick proved that
if
G is
-free. By Theorem 2, we have that the class of
-free graphs does not admit a linear
-binding function. In this paper, we prove that
Theorem 5. if G is ()-free.
In [
24], Wang and Zhang proved that
if
G is a
-free graph. In this paper, we also prove that
Theorem 6. if G is ()-free.
Notice that is a -free graph with and . In reality, as mentioned above, if G is -free, where diamond, paraglider, gem, house, , kite}. So, we may ask a question as follows.
Problem 1. Is it true that if G is ()-free?
2. (, Kite)-Free Graphs
In this section, we consider (, kite)-free graphs. We may always assume that G is a (, kite)-free graph, such that for every induced subgraph of G different from G, and . The following lemmas will be used in our proof.
Lemma 1 (Lemma 2.1 of [
16]).
Let u and v be two nonadjacent vertices in G. Then, and . Proof. Suppose to its contrary that by symmetry. By assumption, . Since we can color u by the color of v, it follows that , a contradiction. □
Lemma 2 ([
24]).
If G is a -free graph, then . Lemma 3 (1.8 of [
25]).
Let G be a ()-free graph. If G contains a , then . We will complete the proof of Theorem 3 by the following three propositions.
Proposition 1. G is -free.
Proof. Suppose not. Let
Q be an induced
in
G with
such that
is a triangle. We will prove that
Suppose
has two adjacent vertices
and
. If
, we may by symmetry assume that
and
, then
induces a
, a contradiction. So,
. By symmetry,
. Therefore, there must exist a vertex in
, which is complete to
, say
. Now,
induces a kite, a contradiction. So,
is independent and, by symmetry,
is independent. This proves (
1).
Let . Since G is -free, we have that is -free. Choose a maximum clique in , say . Without loss of generality, we may assume that . Let , which is complete to , and . So, D is a clique, and for each vertex y in C, y is not complete to .
Suppose there exists a vertex
such that
is anticomplete to
. Since
is not complete to
, it follows that
C has a vertex
, such that
. If
is anticomplete to
, then
induces a
, a contradiction. So,
has a neighbor in
, say
. Then,
induces a kite, a contradiction. So,
Suppose the contrary. We may assume that there exists a vertex
in
C, such that
. By (
2),
. If
and
, then
induces a kite, a contradiction. So,
and, by symmetry,
. Similarly, for each vertex
, if
y is not complete to
, then
.
By the definition of
C, we have that there exists a vertex
in
C such that
. Suppose
is not complete to
. Then,
. If
, then
induces a
, a contradiction. If
, then
induces a kite, a contradiction. So,
is complete to
. Now,
induces a kite, a contradiction. This proves (
3).
Let
. By (
3) and
is a maximum clique in
, we have that
. So,
. It is clear that
D is complete to
C. Since
, we have that
.
Note that
and
. By (
1),
, a contradiction.
This proves Proposition 1. □
Proposition 2. G is hammer-free.
Proof. Suppose not. Let Q be an induced hammer in G with such that is a triangle, and . Notice that induces a . For a subset , we define and if and only if . Let and . By Proposition 1, we have that G is -free.
If , let , then induces a , a contradiction. So, . Similarly, . Since G is -free, we have that if . Therefore, . Moreover, since G is -free, we have that is -free, which implies that each component of is a vertex or an edge.
If
has two adjacent vertices
and
, then
induces a kite, a contradiction. So,
is independent if
and
. Similarly,
is independent if
and
. Let
and
. We will prove that
Suppose not. Without loss of generality, there exists a vertex
such that
. If
, then
or
induces a kite, a contradiction. So,
. But now,
induces a kite, a contradiction. So,
is anticomplete to
. This proves (
4).
By (
4), we have that
. If
is not complete to
, let
such that
, then
induces a kite, a contradiction. So,
is complete to
. In particular,
is complete to
, and thus
.
Suppose . Then, , which implies that , a contradiction. So, . Now, , a contradiction.
This proves Proposition 2. □
Proposition 3. G is -free.
Proof. Suppose not. Let Q be an induced in G with , such that is a triangle. By Lemma 1, there exists a vertex , such that and . By Proposition 1 and 2, G is (, hammer)-free.
If is anticomplete to , then induces a , a contradiction. If is complete to , then induces a kite, a contradiction. So, . But now, induces a hammer, a contradiction. This proves Proposition 3. □
Proof of Theorem 3. By Lemma 2, we may assume that . Now, by Lemma 3 and Proposition 3, we have that , a contradiction. This completes the proof of Theorem 3. □
Actually, by the proof above, we have the following proposition.
Proposition 4. if F is a -free graph.
5. ()-Free Graphs
In this section, we consider ()-free graphs. By Lemma 2, we may assume that G contains a triangle. Let G be a ()-free graph; we will complete the proof of Theorem 6 by the two following propositions.
Proposition 7. If G contains an induced , then .
Proof. Let Q be an induced in G with such that is a triangle. For a subset , we define and if and only if . Note that , as G is -free.
Suppose . Let . If , let and , then induces a , a contradiction. So, , which implies that , as G is -free. By symmetry, we may assume that , and . Now, induces a , a contradiction. Therefore, and, by symmetry, .
Since G is -free, it follows that , and are all independent. Moreover, is -free, as G is -free, and thus , as G is -free. Therefore, . This proves Proposition 7. □
Proposition 8. If G is -free and contains an induced , then .
Proof. Let Q be an induced in G with , such that is a triangle. For a subset , we define and if and only if . Note that as G is -free.
Let . If is anticomplete to , then induces a , a contradiction. If is adjacent to exactly one element of , say , then induces a , a contradiction. So, is complete to , and thus is complete to . By symmetry, is complete to . Since G is -free, we have that is independent.
Since G is -free, it follows that , and are all independent. Moreover, is -free as G is -free, and thus . Therefore, . This proves Proposition 8. □
Proof of Theorem 6. By Propositions 7 and 8, we may assume that G is -free. By Proposition 6, we have that . This proves Theorem 6. □