# The -Chromatic Index of Claw-free Graphs Whose -Core

is -regular
^{†}^{†}thanks: Key Words: -coloring, -Core, -Class .
^{†}^{†}thanks: 2010 Mathematics Subject Classification: 05c15,
05c38.

###### 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.

## 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 |