/ / Article

Research Article | Open Access

Volume 2019 |Article ID 8247419 | https://doi.org/10.1155/2019/8247419

Fu-Tao Hu, "Maximum Independent Sets Partition of ()-Star Graphs", Complexity, vol. 2019, Article ID 8247419, 6 pages, 2019. https://doi.org/10.1155/2019/8247419

# Maximum Independent Sets Partition of ()-Star Graphs

Academic Editor: Alejandro F. Villaverde
Received19 Apr 2019
Accepted24 Jun 2019
Published04 Jul 2019

#### Abstract

The -star graph is a very important computer modelling. The independent number and chromatic number of a graph are two important parameters in graph theory. However, we have not known the values of these two parameters of the -star graph since it was proposed. In this paper, we show a maximum independent sets partition of -star graph. From that, we can immediately deduce the exact value of the independent number and chromatic number of -star graph.

#### 1. Introduction

For graph-theoretical notation and terminology not defined here, we follow . In particular, let be a simple undirected graph without loops and multiedges, where is the vertex set and is the edge set. If , we call that two vertices and are adjacent. For a vertex , all the vertices adjacent to it are the neighbors of .

A subset of is said to be an independent set if no two of vertices are adjacent in of a graph . The cardinality of a maximum independent set in a graph is called the independent number of and is denoted by . Let be a set of colours. A -vertex-colouring (simply a -colouring) is a mapping such that any two adjacent vertices are assigned the different colours of graph . A graph is -colourable if it has a -colouring. The chromatic number, which is denoted by , is the minimum , for which graph is -colourable.

As we know, the interconnection networks take an important part in the parallel computing/communication systems. An interconnection network can be modeled by a graph, where the processors are the vertices and the edges are the communication links.

In 1989, Akers and Krishnamurthy  introduced the -dimensional star graph , which has superior degree and diameter compared to the hypercube and it is highly hierarchical and symmetrical . However, the vertex cardinality of the -dimensional star is . The gap between and is very large when is extended to . Chiang and Chen  in 1995 generalized the star graph to the -star graph, which preserves many good properties of the star graph and has smaller scale. Since the -star graph was introduced, it has received great attention in the literature .

The independent number and chromatic number of a graph are two important parameters in graph theory. However, we did not know the values of these two parameters of the -star graph as far as we know. In this paper, we show a maximum independent sets partition and determine the exact value of the independent number and chromatic number of -star graph.

#### 2. Preliminary Results

We use to denote the set , where is a positive integer. A permutation of is a sequence of distinct symbols of , . The -dimensional star network, denoted by , is a graph with the vertex set The edges are specified as follows:

: is adjacent to if there exists with such that for , , and .

The star graphs are vertex-transitive -regular of order .

Let and be two positive integers with , and let be the set of all -permutations on ; that is, . In 1995, Chiang and Chen  generalized the star graph to -star graph denoted by with vertex set . The adjacency is defined as follows: is adjacent to

(1) , where ;

(2) , where .

By definition, is -regular vertex-transitive with vertices. Moreover, and , where is the complete graph with order .

Let denote a subgraph of induced by all the vertices with the same last symbol , for each . See Figure 1 for instance.

Lemma 1 (Chiang and Chen , 1995). can be decomposed into subgraphs , , and each subgraph is isomorphic to .

Lemma 2 (Li and Xu , 2014). For any , let . Then the subgraph of induced by is a complete graph of order , denoted by .

#### 3. Maximum Independent Sets Partition of

Proposition 3. The independent number of is .

Proof. This conclusion is true for since . Next, assume that . Let be any maximum independent set of . For any , let . Then the subgraph of induced by is a complete graph of order , denoted by by Lemma 2. Thus, contains at most one vertex in . By definition, there are exactly such . Therefore, .

Proposition 4. Let . Then is a maximum independent sets partition of .

Proposition 5. Let Then is a maximum independent sets partition of .

For each , we use of to generate a maximum independent set of . Step by step, we generate a maximum independent set of in the following. For each , , and , denote by a permutation that replaces by if ( means is equal to some symbol in ); otherwise . Let be any vertex in . Denote by the vertex by exchanging the first two symbols in .

Step 1. By Proposition 5, denoted by , , and is a maximum independent set of for each .

Step 2. Let and for each . Let .

Step i-1. Let and for each . Let .

Step k-1. Let and for each . Let .

Let and . By the above construction, it is easy to see that is one corresponding to one with for each and for every . We can easily show the following conclusion by induction on .

Proposition 6. For each and , and for any . Therefore is a vertex sets partition of .

In the following, we show that is an independent set of for each , , and .

Lemma 7. Let , , and . If is an independent set of , then is an independent set of .

Proof. By Proposition 5, is an independent set of for each . Next assume that . Suppose to the contrary that is not an independent set of . Firstly, assume that and are two adjacent vertices in of . If , then and belong to by the construction of ; the one to one correspondence is However, and are adjacent in , a contradiction. If , then and belong to by the construction of ; the one to one correspondence is However, and are adjacent in , a contradiction.
Secondly, assume that and are two adjacent vertices in of . By the construction of , and belong to ; the one to one correspondence is However, and are adjacent in , a contradiction.

Lemma 8. Let ,  , and . If is an independent set of , then the two vertices and can not both belong to of , where and are adjacent in .

Proof. By Proposition 5, , , and is an independent set of .
We consider the case for . Suppose to the contrary that there exist two vertices and in but and are adjacent in .
Assume that and . Then and . If , then and are in by the construction of ; the one to one correspondence is However, and are adjacent in , a contradiction with being an independent set of . If , then and are two vertices in by the construction of ; the one to one correspondence is a contradiction with the construction of .
Now, assume that and . Then and . If , then and are in by the construction of ; the one to one correspondence is a contradiction with the construction of . If , then and are two vertices in by the construction of ; the one to one correspondence is a contradiction with the construction of .
Therefore, the conclusion is true for .
We prove this Lemma by induction on . Assume that the induction hypothesis is true for with . We prove the case for . Assume that is an independent set of . Then is an independent set of by Lemma 7. Suppose to the contrary that there exist two vertices and in but and are adjacent in .
Firstly, assume that and . Suppose that . If , then and are in by the construction of ; the one to one correspondence is However, and are adjacent in , a contradiction.
Now assume that ; then and are two vertices in by the construction of ; the one to one correspondence is However, and are adjacent in , a contradiction with the induction hypothesis.
Next, suppose that . If , then and are two vertices in by the construction of . Now assume that . Then and are two vertices in by the construction of . However, and are two adjacent vertices in , a contradiction with the induction hypothesis.
Secondly, assume that and . If , then and are two vertices in by the construction of . Now suppose that . Then and are two vertices in by the construction of . However, and are two adjacent vertices in , a contradiction with the induction hypothesis.
By the principle of induction, this Lemma completes.

Lemma 9. Let , , and . If is an independent set of , then is an independent set of for each .

Proof. Assume that . Suppose to the contrary that is not an independent set of . Assume that and are two adjacent vertices in . By the construction of , if and if (the case for is similar), a contradiction with the construction of in Proposition 5. Assume that and are two adjacent vertices in . By the construction of , we have Any case above makes a contradiction with the construction of in Proposition 5.
We proceed by induction on . Assume that the induction hypothesis is true for with . Next we prove that is an independent set of for each . Suppose to the contrary that is not an independent set of .
Firstly, assume that and are two adjacent vertices in . On one hand, suppose that . By the construction of , we have The first case makes a contradiction with the result in Lemma 8, since and are adjacent in . The second case makes a contradiction with being an independent set. On the other hand, suppose that . By the construction of , we have if for each (if , for some , we just replace by ). However, and are adjacent in , a contradiction with being an independent set.
Secondly, assume that and are two adjacent vertices in . If , then and are two vertices in by the construction of . However, and are two adjacent vertices in , a contradiction with the result in Lemma 8. Assume that . By the construction of , we have if for each (if , for some , we just replace by ). However, and are adjacent in , a contradiction with being an independent set.
By the principle of induction, this Lemma completes.

Theorem 10. The vertex set is an independent set of for each and .

Proof. We proceed by induction on . By Proposition 5, is an independent set of . Assume that the induction hypothesis is true for with .
Assume that is an independent set of . By Lemma 9, is an independent set of for each . Suppose to the contrary that is not independent. Assume that and are two adjacent vertices in (this is the only possible case since is an independent set of for each ). If , then and should be in by the construction of , but they are two adjacent vertices in , a contradiction (the case for is similar). If , for some with , then (replace by ) and (replace by ) should be in by the construction of , but they are two adjacent vertices in , a contradiction. Now assume that for each . Then and should be in by the construction of , but they are two adjacent vertices in , a contradiction. This Theorem completes.

Theorem 11. The constructed is a maximum independent set of for each and . Moreover, is a maximum independent sets partition of .

Proof. By Proposition 3, . By Proposition 6, . By Theorem 10, is an independent set of . Therefore, for each is a maximum independent set of and . By Proposition 6, is a vertex sets partition of , so is a maximum independent sets partition of .

Since is a maximum independent sets partition of , we immediately obtain the chromatic number of .

Corollary 12. The chromatic number of is .

Next, we show the maximum independent sets partition of by our construction.

Example 13 (see Figure 2). By Proposition 5, and are two maximum independent sets of . The constructed two maximum independent sets of are

#### Data Availability

The data used to support the findings of this study are included within the article.

#### Conflicts of Interest

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

#### Acknowledgments

The work was supported by NSFC (no. 11401004) and Anhui Provincial Natural Science Foundation (no. 1708085MA01).

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