The f-Chromatic Index of Claw-free Graphs Whose f-Coreis 2-regular Key Words: f-coloring, f-Core, f-Class 1. 2010 Mathematics Subject Classification: 05C15, 05C38.

The -Chromatic Index of Claw-free Graphs Whose -Core
is -regular thanks: Key Words: -coloring, -Core, -Class . thanks: 2010 Mathematics Subject Classification: 05c15, 05c38.

S. Akbari,  M. Chavooshi,  M. Ghanbari,  R. Manaviyat
Department of Mathematics, Payame Noor University, Tehran, Iran
Department of Mathematical Sciences, Sharif University of Technology, Tehran, Iran
School of Mathematics, Institute for Research in Fundamental Sciences (IPM),
P.O. Box: 19395-5746, Tehran, Iran.
E-mail addresses: , , , .
Abstract

Let be a graph and be a function. An -coloring of a graph is an edge coloring such that each color appears at each vertex at most times. The minimum number of colors needed to -color is called the -chromatic index of and is denoted by . It was shown that for every graph , , where . A graph is said to be -Class if , and -Class , otherwise. Also, is the induced subgraph of on . In this paper, we show that if is a connected graph with and has an edge cut of size at most which is a matching or a star, then is -Class . Also, we prove that if is a connected graph and every connected component of is a unicyclic graph or a tree and is not -regular, then is -Class . Moreover, we show that except one graph, every connected claw-free graph whose -core is -regular with a vertex such that is -Class .

1 Introduction

All graphs considered in this paper are simple and finite. Let be a graph. The number of vertices of is called the order of and is denoted by . Also, and denote the vertex set and the edge set of , respectively. The degree of a vertex in is denoted by and denotes the set of all vertices adjacent to . For a subgraph of , . Also, let and denote the maximum degree and the minimum degree of , respectively. A star graph is a graph containing a vertex adjacent to all other vertices and with no extra edges. A matching in a graph is a set of pairwise non-adjacent edges. An edge cut is a set of edges whose removal produces a subgraph with more connected components than the original graph. Moreover, a graph is -edge connected if the minimum number of edges whose removal would disconnect the graph is at least . We mean , the induced subgraph on . For two subsets and of , where , denotes the number of edges with one end in and other end in . For a subset , we denote the induced subgraph of on by . A graph is called a unicyclic graph if it is connected and contains exactly one cycle.

A -edge coloring of a graph is a function , where and , for every two adjacent edges of . The minimum number of colors needed to color the edges of properly is called the chromatic index of and is denoted by . Vizing [6] proved that , for any graph . A graph is said to be Class if and Class if . A graph is called critical if is connected, Class and , for every edge . Also, is the induced subgraph on all vertices of degree .

For a function which assigns a positive integer to each vertex , an -coloring of is an edge coloring of such that each vertex has at most edges colored with the same color. The minimum number of colors needed to -color is called the -chromatic index of , and denoted by . For a graph , if for all , then the -coloring of is reduced to the proper edge coloring of . Let . A graph is said to be -Class if and -Class , otherwise. Also, we say that has a -coloring if is -Class . A vertex is called an -maximum vertex if . A graph is called -critical if is connected, -Class and , for every edge . The -core of a graph is the induced subgraph of on the -maximum vertices and denoted by . The following example introduces an -Class graph.

Example 1

. Let be a graph shown in the following figure such that and , for . It is easy to see that , and is -Class .

Figure  : An -Class graph

In [3], Hakimi and Kariv obtained the following three results.

Theorem 1

. Let be a graph. Then

Theorem 2

. Let be a bipartite graph. Then is -Class .

Theorem 3

. Let be a graph and be even, for all . Then is -Class .

Theorem 4

.[7] Let be a graph. If for all , then is -Class .

Following result due to Zhang, Wang and Liu gave a series of sufficient conditions for a graph to be -Class based on the -core of .

Theorem 5

.[9] Let be a graph. If is a forest, then is -Class .

In [5], some properties of -critical graphs are given. In the following, we review one of them.

Theorem 6

. For every vertex of an -critical graph , is adjacent to at least -maximum vertices and contains at least three -maximum vertices.

There are some theorems in proper edge coloring of graphs as follows:

Theorem 7

.[4] Let be a connected Class graph with . Then:

is critical;

;

, unless G is an odd cycle.

Theorem 8

.[1] Let be a connected graph and . Suppose that has an edge cut of size at most which is a matching or a star. Then is Class .

Theorem 9

.[1] Let be a connected graph. If every connected component of is a unicyclic graph or a tree and is not -regular, then is Class .

In [2], Theorem 7 and Theorem 8 in a case that the edge cut is a matching were generalized to -colorings.

Theorem 10

.[2] Let be a connected -Class graph with . Then the followings hold:

is -critical;

is -regular;

, for every .

Theorem 11

.[2] Let be a connected graph and . Suppose that has an edge cut of size at most which is a matching. Then is -Class and has a -coloring in which the edges of the edge cut have different colors.

In this paper, we prove Theorems 8 and 9 in -coloring of graphs. Moreover, we show that except one graph, every connected claw-free graph whose -core is -regular with a vertex such that is -Class .

2 Results

In this section, we generalize Theorems 8, 9 and we obtain some results in -coloring of claw-free graphs whose -core is -regular. To see this, first we want to prove that if a connected graph with has an edge cut of size at most which is a matching or a star, then is -Class . To do this, first we need a lemma which is proved in [2].

Lemma 1

.[2] Let be a connected graph with . Suppose that , , is an edge cut of and . Then is -Class .

Theorem 12

. Let be a connected graph and . Suppose that has an edge cut of size at most which is a matching or a star. Then is -Class .

Proof.

Clearly, we can assume that . By Theorem 11, we can assume that the edge cut is a star. Let , , is an edge cut and , and every edge of has one end point in and other end point in . Let and be the induced subgraphs on and , respectively. With no loss of generality, assume that and , for . By Lemma 1 and Theorem 11, we can assume that . To the contrary assume that is -Class . Since by Theorem 10, is -critical and noting that , by Theorem 6, . Thus by Theorem 10, . Also, note that since is -critical, by Theorem 6, , for . Let . Note that since is -Class and , by Theorem 11, has no cut edge. Thus there exists a component of such that . With no loss of generality, let . Add three new vertices , and to . Then join and to , and , respectively. Let and . Let be a function defined by

Note that and are connected. Moreover, , because

and since , for , .

We claim that both and are -Class . Note that if is -Class , then by Theorem 10, , but , a contradiction. So, there exists an -coloring of by colors . Similarly, there is an -coloring of by colors and the claim is proved.

By a suitable permutation of colors, one may assume that are distinct. Now, define an -coloring as follows:

This implies that is -Class which is a contradiction and the proof is complete.

Now, we want to prove another result in -coloring of graphs which classify some of -Class graphs. To do this, we need the following lemma.

Lemma 2

.[8] Let denote the set of colors available to color the edges of a simple graph . Suppose that is an uncolored edge in , and graph is -colored with the colors in . If for every neighbor of either or , there exists a color which appears at most times, then there exists an -coloring of using colors of .

Theorem 13

. Let be a connected graph. If every connected component of is a unicyclic graph or a tree and is not -regular, then is -Class .

Proof.

First suppose that . Toward a contradiction, assume that is -Class . By Theorem 10, is -regular, which is a contradiction. So, one may suppose that . Now, the proof is by induction on . Since , we have . First assume that . Then . Consider , where and . Consider the graph with function . We want to show that is -Class . If is connected, then is the union of two isolated vertices. Then by Theorem 10, is -Class . Now, assume that is not connected. Let and be two connected components of such that and . Note that . Since , by Theorem 10, is -Class . Now, if , then by Theorem 1, has an -coloring with colors . Also, if , then clearly and by Theorem 4, is -Class . Also, Now, since for every , we have , there exists a color which appears at most times in and so by Lemma 2, is -Class and we are done.

Let be a graph and . Assume that the assertion holds for all graphs with . Consider , where is one of edges of such that and . Consider the graph with function . We would like to show that is -Class . Two cases may occur.

First assume that is connected. If , then by the induction hypothesis we are done. If and is not -regular, then by Theorem 10, is -Class . Thus assume that is -regular. Then it is not hard to see that is a disjoint union of some cycles and the graph shown in the following figure:

Figure 2: A part of

Now, by Theorem 10, is -critical and so by Theorem 6, should have at least two neighbors in , a contradiction.

Next assume that is not connected. Let and be two connected components of such that and . Clearly, . If , then by the induction hypothesis, is -Class . If and is not -regular, then by Theorem 10, is -Class . Thus assume that is -regular. Then it is not hard to see that is the disjoint union of some unicycles, trees and the graph shown in the Figure . Now, by Theorem 10, is -critical and so by Theorem 6, should have at least two neighbors in , a contradiction and is -Class . Now, if , then by Theorem 1, has an -coloring with colors . So, assume that . Now, if , then by Theorem 4, is -Class . Otherwise, similar to the argument about , is -Class . Now, since for every , we have , there exists a color which appears at most times in and so by Lemma 2, is -Class and we are done.

Theorem 14

. Let be a connected claw-free graph with . If there exists a vertex such that and , where is the graph shown in the following figure, then is -Class .

Figure  : The graph (The value of each vertex denotes )
Proof.

To the contrary assume that is -Class . Then by Theorem 10, is -critical and is -regular. Now, by Theorem 6, , for every and so by the definition we have

, for every . (1)

Note that, if , then since is connected and is -regular, and there is no vertex with . Thus we can assume that

(2)

Let . Now, if , for some , then clearly there exists an independent set of size in which implies that has a claw, a contradiction. Thus we have

, for every . (3)

Now, to prove the theorem, first we need the following claim:

, for every .

To see this by the contrary, assume that there exists a vertex such that . Clearly, . Now, by (3) and Theorem 6, we conclude that and . Then by Theorem 10, and so . Now, we want to show that for every , . Because otherwise, there are at least vertices, say , such that , for . Now, since is -regular, with no loss of generality, we can assume that . Then is a claw, a contradiction. Thus, we conclude that for every , . This implies that

which yields that , a contradiction and the claim is proved.

Now, by the assumption of theorem and Claim , we can assume that there exists a vertex such that . Then, by Theorem 10, . Now, by Theorem 6 and using (3), we have . Thus, three cases may occur:

.
Let and . Since is -regular, with no loss of generality, there are two non-adjacent vertices . Since is claw-free, or , for . Thus,

a contradiction.

.
Let and . First note that since is claw-free, does not contain an independent set of size and so it can be easily checked that is one of two following graphs:

Figure  : when

Three subcases may occur:

.
We have . Now, if , then is the graph shown in Figure , a contradiction. Thus, assume that is the graph shown in Figure . By Theorem 10, since is -regular, there exists and . Now, we divide the proof of this subcase into two parts:

. Let . Now, add a new vertex to and join to and . Call the resultant graph . Let be a function defined by

Clearly, is a connected graph with . Note that since , we have and so . Now, since and is not -regular, by Theorem 10, has an -coloring call , with colors . With no loss of generality, assume that and . Now, define an -coloring as follows.
Define , for every and

. Since , has a neighbor , where . Clearly, is a cut edge for and by Theorem 12, is -class , a contradiction.

.
Clearly, and . Now, we divide the proof of this subcase into two parts:

is the graph shown in Figure .
Since is claw-free, noting that , we have or . With no loss of generality assume that . Moreover, since is not a claw and , we have . Similarly, since is not a claw and , we conclude that . Also, since is not a claw and , we obtain that . Moreover, since is not a claw and , . Now, clearly is a claw which is a contradiction.

is the graph shown in Figure .
Similar to the previous argument, we can assume that . Now, since , . Now, by Claim we conclude that and so by Theorem 10, . Assume that . Now, by Theorem 6, we have . If , then by Case , we are done. So, we can assume that . Thus we have

(4)

Also, since is not a claw, for and , we obtain that

(5)

Now, we show that

(6)

To the contrary assume that . Then and since , we have . Let , where . Now, since and are not claws, we conclude that is a in and so . Then is a claw, a contradiction and (6) holds.

Now, consider . Add a new vertex to and join to . Call the resultant graph . Define be a function defined by

Clearly by (5), is connected and . Now, if , then clearly and and by Theorem 10, is -Class . So, assume that . Clearly is not -regular and each of the components is a unicyclic graph or a tree. Now, by Theorem 13, has an -coloring call , with colors . With no loss of generality, assume that , , and . Now, define an -coloring as follows.

Let , for every , and , , where and are the colors missed in coloring in and , respectively. Let . Now, by a suitable -coloring of , we extend the -coloring of to an -coloring of . With no loss of generality, one of the following cases may occur:

Figure  : -edge coloring of