Maximum Independent Sets Partition of (n,k)-Star Graphs

TL;DR

This paper establishes the maximum independent sets partition of (n,k)-star graphs, enabling the precise calculation of their independent and chromatic numbers, which were previously unknown or incorrectly proven.

Contribution

It generalizes prior results and provides the correct maximum independent sets partition for (n,k)-star graphs, resolving previous inaccuracies.

Findings

Exact values of independent number and chromatic number derived

Maximum independent sets partition explicitly constructed

Corrected previous incorrect proof

Abstract

The (n,k)-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 did not know the values of this two parameters of the (n,k)-star graph since it was proposed. In [18], Wei et. al. declared that they determined the independent number of the (n,k)-star graph, unfortunately their proof is wrong. This paper generalize their result and present a maximum independent sets partition of (n,k)-star graph. From that we can immediately deduce the exact value of the independent number and chromatic number of (n,k)-star graph.

Equations

$$V(S_n)=\{u_1{u}_2\dots u_n:u_i\in [n],u_i\ne u_j\mathrm{for}i\ne j\}.$$

$$\begin{array}[]{ll}&I_{1}^{2}=\{21,32,43,\ldots,n(n-1),1n\},\\ &I_{2}^{2}=\{31,42,53,\ldots,1(n-1),2n\},\\ &\ldots\\ &I_{n-1}^{2}=\{n1,12,23,\ldots,(n-2)(n-1),(n-1)n\}.\end{array}$$

$$\begin{array}[]{ll}&p_{1}p_{2}\ldots p_{i-1}\leftrightarrow p_{1}(n-k+i)\ldots p_{i-1}p_{2},\\ &p_{2}p_{1}\ldots p_{i-1}\leftrightarrow(n-k+i)p_{1}\ldots p_{i-1}p_{2}.\end{array}$$

$$\begin{array}[]{ll}&p_{1}p_{2}\ldots p_{s}\ldots p_{i-1}\leftrightarrow p_{1}(n-k+i)\ldots p_{s}\ldots p_{i-1}p_{2},\\ &p_{s}p_{2}\ldots p_{1}\ldots p_{i-1}\leftrightarrow p_{s}(n-k+i)\ldots p_{1}\ldots p_{i-1}p_{2}.\end{array}$$

$$\begin{array}[]{ll}&p_{1}p_{2}\ldots p_{i-1}\leftrightarrow p_{1}(n-k+i)\ldots p_{i-1}p_{2},\\ &p_{i}p_{2}\ldots p_{i-1}\leftrightarrow p_{i}(n-k+i)\ldots p_{i-1}p_{2}.\end{array}$$

$$\begin{array}[]{ll}&(n-k+3)p_{1}p_{2}\leftrightarrow p_{2}p_{1},\\ &(n-k+3)p_{2}p_{1}\leftrightarrow p_{1}p_{2}.\end{array}$$

$$\begin{array}[]{ll}&pp_{1}p_{2}\leftrightarrow pp_{1},\\ &pp_{2}p_{1}\leftrightarrow pp_{2}.\end{array}$$

$$\begin{array}[]{ll}&(n-k+3)p_{1}p_{2}\leftrightarrow p_{2}p_{1},\\ &(n-k+3)p_{3}p_{2}\leftrightarrow p_{2}p_{3}.\end{array}$$

$$\begin{array}[]{ll}&pp_{1}p_{2}\leftrightarrow pp_{1},\\ &pp_{2}p_{1}\leftrightarrow pp_{2}.\end{array}$$

$$\begin{array}[]{ll}&(n-k+i)p_{2}p_{3}\ldots p_{i-1}p_{i}\leftrightarrow p_{i}p_{2}p_{3}\ldots p_{i-1},\\ &(n-k+i)p_{i}p_{3}\ldots p_{i-1}p_{2}\leftrightarrow p_{2}p_{i}p_{3}\ldots p_{i-1}.\end{array}$$

$$\begin{array}[]{ll}&pp_{2}p_{3}\ldots p_{i-1}p_{i}\leftrightarrow pp_{2}p_{3}\ldots p_{i-1},\\ &pp_{i}p_{3}\ldots p_{i-1}p_{2}\leftrightarrow pp_{i}p_{3}\ldots p_{i-1}.\end{array}$$

$$p_2{p}_1,p_2{p}_3\in I_j^2,\mathrm{if}x=n-k+3,$$

$$p_{1}x,p_{3}x\in I_j^2,\mathrm{if}{p}_2=n-k+3,$$

$$x{p}_2,p_3{p}_2\in I_j^2,\mathrm{if}{p}_1=n-k+3,$$

$$p_1{p}_2,x{p}_2\in I_j^2,\mathrm{if}{p}_3=n-k+3,$$

$$p_1{p}_2,p_3{p}_2\in I_j^2,\mathrm{otherwise}.$$

$$p_2{p}_1\dots p_s\dots p_{i-1},p_2{p}_s\dots p_1\dots p_{i-1}\in I_j^{i-1},\mathrm{if}s\ge 3,$$

$$p_2{p}_1\dots p_{i-1},p_1{p}_2\dots p_{i-1}\in I_j^{i-1},\mathrm{if}s=2.$$

$$\begin{array}[]{rl}I_{1}^{3}&=\{124,234,314\}\cup\{241,321,431\}\cup\{412,342,132\}\cup\{213,423,143\}\\ &=I_{1}^{3}(4)\cup I_{1}^{3}(1)\cup I_{1}^{3}(2)\cup I_{1}^{3}(3),\end{array}$$

$$\begin{array}[]{rl}I_{2}^{3}&=\{134,214,324\}\cup\{341,421,231\}\cup\{312,142,432\}\cup\{413,123,243\}\\ &=I_{2}^{3}(4)\cup I_{2}^{3}(1)\cup I_{2}^{3}(2)\cup I_{2}^{3}(3).\end{array}$$

Full text

Maximum Independent Sets Partition of $(n,k)$-Star Graphs††thanks: The work was supported by NNSF

of China (Nos. 11401004, 11371028, 11471016).

Fu-Tao Hu111 Correspondence to: F.-T. Hu; e-mail: hufu@ahu.edu.cn

School of Mathematical Sciences, Anhui University, Hefei, 230601, P.R. China

Abstract: The $(n,k)$-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 did not know the values of this two parameters of the $(n,k)$-star graph since it was proposed. In [18], Wei et. al. declared that they determined the independent number of the $(n,k)$-star graph, unfortunately their proof is wrong. This paper generalize their result and present a maximum independent sets partition of $(n,k)$-star graph. From that we can immediately deduce the exact value of the independent number and chromatic number of $(n,k)$-star graph.

Keywords: $(n,k)$-star graph, independent set, independent number, chromatic number.

**AMS Subject Classification: ** 05C69

1 Introduction

For graph-theoretical notation and terminology not defined here we follow [20]. In particular, let $G=(V,E)$ be a simple undirected graph without loops and multi-edges, where $V=V(G)$ is the vertex set, and $E=E(G)$ is the edge set. If $xy\in E(G)$, we call two vertices $x$ and $y$ are adjacent. For a vertex $x$, all the vertices adjacent to it are the neighbors of $x$.

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

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

In 1989, Akers and Krishnamurthy [1] introduced the $n$-dimensional star graph $S_{n}$, which has superior degree and diameter compared to the hypercube as well as it is highly hierarchical and symmetrical [9]. However, the vertex cardinality of the $n$-dimensional star is $n!$. The gap between $n!$ and $(n+1)!$ is very large when $S_{n}$ is extended to $S_{n+1}$. Chiang and Chen [7] in 1995 generalized the star graph $S_{n}$ to the $(n,k)$-star graph, which preserves many good properties of the star graph and has smaller scale. Since he $(n,k)$-star graph was introduced, it has received great attention in the literature [2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 19, 21, 22] since then on.

The independent number and chromatic number of a graph are two important parameters in graph theory. In [18], Wei et. al. declared that they determined the independent number of the $(n,k)$-star graph, their result is right, unfortunately their proof is wrong. In Section 3, we will show a counterexample of their result. In Section 4, we will present a maximum independent sets partition and determine the exact value of the independent number and chromatic number of $(n,k)$-star graph.

2 Preliminary results

We use $[n]$ to denote the set $\{1,2,\ldots,n\}$, where $n$ is a positive integer. A permutation of $[n]$ is a sequence of $n$ distinct symbols of $u_{i}\in[n]$, $u_{1}u_{2}\ldots u_{n}$. The $n$-dimensional star network, denoted by $S_{n}$, is a graph with the vertex set

[TABLE]

The edges are specified as follows:

$E(S_{n})$: $u_{1}u_{2}\ldots u_{n}$ is adjacent to $v_{1}v_{2}\ldots v_{n}$ iff there exists $i$ with $2\leq i\leq n$ such that $v_{j}=u_{j}$ for $j\notin\{1,i\}$, $v_{1}=u_{i}$ and $v_{i}=u_{1}$;

The Star graphs are vertex-transitive $(n-1)$-regular of order $n!$.

Let $n$ and $k$ be two positive integers with $k\in[n-1]$, and let $\Gamma_{n,k}$ be the set of all $k$-permutations on $[n]$, that is, $\Gamma_{n,k}=\{p_{1}p_{2}\ldots p_{k}:\,p_{i}\in[n]~{}{\rm and}~{}p_{i}\neq p_{j}~{}{\rm for}~{}i\neq j\}$. In 1995, Chiang and Chen [7] generalized the Star graph to $(n,k)$-Star graph denoted by $S_{n,k}$ with vertex set $V(S_{n,k})=\Gamma_{n,k}$. The adjacency is defined as follows: $p_{1}p_{2}\ldots p_{i}\ldots p_{k}$ is adjacent to

(1) $p_{i}p_{2}\ldots p_{1}\ldots p_{k}$ where $2\leq i\leq k$;

(2) $xp_{2}\ldots p_{i}\ldots p_{k}$ where $x\in[n]\setminus\{p_{i}:\,i\in[k]\}$.

By definition, $S_{n,k}$ is $(n-1)$-regular vertex-transitive with $n!/(n-k)!$ vertices. Moreover, $S_{n,n-1}\cong S_{n}$ and $S_{n,1}\cong K_{n}$ where $K_{n}$ is the complete graph with order $n$.

Let $S_{n-1,k-1}^{i}$ denote a subgraph of $S_{n,k}$ induced by all the vertices with the same last symbol $i$, for each $i\in[n]$. See Figure 1 for instance.

Lemma 2.1** (Chiang and Chen [7], 1995)**

$S_{n,k}$* can be decomposed into $n$ subgraphs $S_{n,k}^{i}$, $i\in[n]$, and each subgraph $S_{n-1,k-1}^{i}$ is isomorphic to $S_{n-1,k-1}$.*

Lemma 2.2** (Li and Xu [13], 2014)**

For any $\alpha=p_{2}p_{3}\ldots p_{k}\in\Gamma_{n,k-1}~{}(k\geq 2)$, let $V_{\alpha}=\{p_{1}\alpha:\,p_{1}\in[n],~{}p_{1}\neq p_{i},~{}2\leq i\leq k\}$. Then the subgraph of $S_{n,k}$ induced by $V_{\alpha}$ is a complete graph of order $n-k+1$, denoted by $K_{n-k+1}^{\alpha}$.

3 Comments on Wei et. al.’s result

In [18], Wei et. al. declared that they determined the independent number of the $(n,k)$-star graph which was $\alpha(S_{n,k})=\frac{n!}{(n-k+1)!}$. We have checked their proof carefully. However, their proof is wrong. They constructed an independent set $I_{k}$ of $S_{n,k}$ with cardinality $\frac{n!}{(n-k+1)!}$ step by step. The following are their details.

Let $I_{k}^{(i,j)}$ be a vertex-set, which includes the vertices of $I_{k}$ if the vertices don’t include element $i$, and $I_{k}^{(i,j)}$ includes the vertices of $I_{k}$ if the vertices include element $i$ but swap $i$ with $j$.

Step 1: Clearly, $I_{2}=\{12,23,\ldots,(n-k+1)(n-k+2),(n-k+2)1\}$ is an independent set of $S_{n-k+2,2}$;

Step 2: In $S_{n-k+3,3}$, let $I_{3(n-k+3)}=\{\beta(n-k+3):\,\beta\in I_{2}\}$ and $I_{3x}=\{\beta x:\,x\in[n-k+2],\beta\in I_{2}^{(x,n-k+3)}\}$. Now, we let $I_{3}=\bigcup_{x\in[n-k+3]}I_{3x}$. Clearly, $|I_{3}|=(n-k+3)(n-k+2)$;

$\ldots\ldots\ldots$

Step k: In $S_{n-k+3,3}$, let $I_{k(n-k+k)}=\{\beta(n-k+3):\,\beta\in I_{k-1}\}$ and $I_{kx}=\{\beta x:\,x\in[n-1],\beta\in I_{k-1}^{(x,n-k+k)}\}$. Now, we let $I_{k}=\bigcup_{x\in[n]}I_{kx}$. Clearly, $|I_{k}|=n(n-1)\ldots(n-k+3)(n-k+2)=\frac{n!}{(n-k+1)!}$.

The above $I_{k}$ is constructed in Wei’s paper [18]. However, their construction is wrong and without detailed proof. Next we use their method to construct the independent set $I_{3}$ for $S_{4,3}$.

By Step 1, $I_{2}=\{12,23,31\}$ is a maximum independent set of $S_{3,2}$. By Step 2, $I_{34}=\{124,234,314\}$, $I_{31}=\{421,231,341\}$, $I_{32}=\{142,432,312\}$, $I_{33}=\{123,243,413\}$. Then $I_{4}=\bigcup_{x\in[4]}I_{3x}$. However, the vertices $124$ and $421$ are adjacent in $S_{4,3}$ (see Figure 4). So their construction is wrong.

In the next section, we will give a new method to construct a maximum independent sets partition of $S_{n,k}$ and show detailed proof for our result, Which generalized Wei et. al.’s result.

4 Maximum independent sets partition of $S_{n,k}$

Proposition 4.1

The independent number of $S_{n,k}$ is $\alpha(S_{n,k})\leq\frac{n!}{(n-k+1)!}$.

Proof. This conclusion is true for $k=1$ since $S_{n,1}\cong K_{n}$. Next, assume $k\geq 2$. Let $I$ be any maximum independent set of $S_{n,k}$. For any $\alpha=p_{2}p_{3}\ldots p_{k}\in\Gamma_{n,k-1}~{}(k\geq 2)$, let $V_{\alpha}=\{p_{1}\alpha:\,p_{1}\in[n],~{}p_{1}\neq p_{i},~{}2\leq i\leq k\}$. Then the subgraph of $S_{n,k}$ induced by $V_{\alpha}$ is a complete graph of order $n-k+1$, denoted by $K_{n-k+1}^{\alpha}$ by Lemma 2.2. Thus, $I$ contains at most one vertex in $K_{n-k+1}^{\alpha}$. By definition, there are exactly $\frac{n!}{(n-k+1)!}$ such $K_{n-k+1}^{\alpha}$. Therefore, $\alpha(S_{n,k})=|I|\leq\frac{n!}{(n-k+1)!}$.

Proposition 4.2

Let $I_{1}^{1}=\{1\},I_{2}^{1}=\{2\},\ldots,I_{n}^{1}=\{n\}$. Then $\{I_{1}^{1},I_{2}^{1},\ldots,I_{n}^{1}\}$ is a maximum independent sets partition of $S_{n,1}$.

Proposition 4.3

Let

[TABLE]

Then $\{I_{1}^{2},I_{2}^{2},\ldots,I_{n-1}^{2}\}$ is a maximum independent sets partition of $S_{n,2}$.

For each $j\in[n-k+1]$, we use $I_{j}^{2}$ of $S_{n-k+2,2}$ to generate a maximum independent set $I_{j}^{3}$ of $S_{n-k+3,3}$. Step by step, we generate a maximum independent set $I_{j}^{k}$ of $S_{n,k}$ in the following. For each $i\in[k]\setminus\{1,2\}$, $\pi\in I_{j}^{i-1}$ and $x\in[n-k+i-1]$, denote by $\pi(x,n-k+i)$ be such a permutation that replace $x$ by $n-k+i$ if $x\in\pi$ ($x\in\pi$ means $x$ is equal to some symbol in $\pi$), otherwise $\pi(x,n-k+i)=\pi$. Let $\pi=p_{1}p_{2}\ldots p_{i}$ be any vertex in $S_{n-k+i,i}$. Denote by $\pi^{\prime}=p_{2}p_{1}\ldots p_{i}$ be the vertex by exchanging the first two symbols in $\pi$.

Step 1: By Proposition 4.3, denote by $I_{j}^{2}=\{(j+1)1,(j+2)2,\ldots,(n-k+2)(n-k+2-j),1(n-k+3-j),2(n-k+4-j),\ldots,j(n-k+2)\}$ be a maximum independent set of $S_{n-k+2,2}$ for each $j\in[n-k+1]$.

Step 2: Let $I_{j}^{3}(n-k+3)=\{\pi^{{}^{\prime}}(n-k+3):\,\pi\in I_{j}^{2}\}$ and $I_{j}^{3}(x)=\{\pi(x,n-k+3)x:\,\pi\in I_{j}^{2}\}$ for each $x\in[n-k+2]$. Let $I_{j}^{3}=\bigcup_{x\in[n-k+3]}I_{j}^{3}(x)$.

$\ldots\ldots\ldots$

Step i-1: Let $I_{j}^{i}(n-k+i)=\{\pi^{{}^{\prime}}(n-k+i):\,\pi\in I_{j}^{i-1}\}$ and $I_{j}^{i}(x)=\{\pi(x,n-k+i)x:\,\pi\in I_{j}^{i-1}\}$ for each $x\in[n-k+i-1]$. Let $I_{j}^{i}=\bigcup_{x\in[n]}I_{j}^{i}(x)$.

$\ldots\ldots\ldots$

Step k-1: Let $I_{j}^{k}(n)=\{\pi^{{}^{\prime}}n:\,\pi\in I_{j}^{k-1}\}$ and $I_{j}^{k}(x)=\{\pi(x,n)x:\,\pi\in I_{j}^{k-1}\}$ for each $x\in[n-1]$. Let $I_{j}^{k}=\bigcup_{x\in[n]}I_{j}^{k}(x)$.

Let $j\in[n-k+1]$ and $i\in[k]\setminus\{1,2\}$. By the above construction, it is easy to see that $I_{j}^{i}(x)$ is one corresponding to one with $I_{j}^{i-1}$ for each $x\in[n-k+i]$ and $I_{j}^{i}(x)\cap I_{j}^{i}(x^{\prime})=\emptyset$ for every $x^{\prime}\in[n-k+i]\setminus\{x\}$. We can easily show the following conclusion by induction on $i$.

Proposition 4.4

For each $j\in[n-k+1]$ and $i\in[k]\setminus\{1\}$, $|I_{j}^{i}|=\frac{(n-k+i)!}{(n-k+1)!}$, $I_{j}^{i}\cap I_{j^{\prime}}^{i}=\emptyset$ for any $j^{\prime}\in[n-k+1]\setminus\{j\}$. Therefore $\{I_{1}^{i},I_{2}^{i},\ldots,I_{n-k+1}^{i}\}$ is a vertex sets partition of $S_{n-k+i,i}$.

In the following, we show that $I_{j}^{i}$ is an independent set of $S_{n-k+i,i}$ for each $i\in\{3,4,\ldots,k\}$, $k\geq 3$ and $j\in[n-k+1]$.

Lemma 4.1

Let $i\in\{3,4,\ldots,k\}$, $k\geq 3$ and $j\in[n-k+1]$. If $I_{j}^{i}$ is an independent set of $S_{n-k+i,i}$, then $I_{j}^{i-1}$ is an independent set of $S_{n-k+i-1,i-1}$.

Proof. By Proposition 4.3, $I_{j}^{2}$ is an independent set of $S_{n-k+i-1,i-1}$ for each $j\in[n-k+1]$. Next assume $i\geq 4$. Suppose to the contrary that $I_{j}^{i-1}$ is not an independent set of $S_{n-k+i-1,i-1}$. Firstly, assume $p_{1}p_{2}\ldots p_{s}\ldots p_{i-1}$ and $p_{s}p_{2}\ldots p_{1}\ldots p_{i-1}$ be two adjacent vertices in $I_{j}^{i-1}$ of $S_{n-k+i-1,i-1}$. If $s=2$, then $p_{1}(n-k+i)\ldots p_{i-1}p_{2}$ and $(n-k+i)p_{1}\ldots p_{i-1}p_{2}$ are belong to $I_{j}^{i}(p_{2})\subseteq I_{j}^{i}$ by the construction of $I_{j}^{i}$, the one to one correspondence is

[TABLE]

However $p_{1}(n-k+i)\ldots p_{i-1}p_{2}$ and $(n-k+i)p_{1}\ldots\ldots p_{i-1}p_{2}$ are adjacent in $S_{n-k+i,i}$, a contradiction. If $s>2$, then $p_{1}(n-k+i)\ldots p_{s}\ldots p_{i-1}p_{2}$ and $p_{s}(n-k+i)\ldots p_{1}\ldots p_{i-1}p_{2}$ are belong to $I_{j}^{i}(p_{2})\subseteq I_{j}^{i}$ by the construction of $I_{j}^{i}$, the one to one correspondence is

[TABLE]

However $p_{1}(n-k+i)\ldots p_{s}\ldots p_{i-1}p_{2}$ and $p_{s}(n-k+i)\ldots p_{1}\ldots p_{i-1}p_{2}$ are adjacent in $S_{n-k+i,i}$, a contradiction.

Secondly, assume $p_{1}p_{2}\ldots p_{i-1}$ and $p_{i}p_{2}\ldots p_{i-1}$ be two adjacent vertices in $I_{j}^{i-1}$ of $S_{n-k+i-1,i-1}$. By the construction of $I_{j}^{i}$, $p_{1}(n-k+i)\ldots p_{i-1}p_{2}$ and $p_{i}(n-k+i)\ldots p_{i-1}p_{2}$ are belong to $I_{j}^{i}(p_{2})\subseteq I_{j}^{i}$, the one to one correspondence is

[TABLE]

However $p_{1}(n-k+i)\ldots p_{s}\ldots p_{i-1}p_{2}$ and $p_{i}(n-k+i)\ldots p_{s}\ldots p_{i-1}p_{2}$ are adjacent in $S_{n-k+i,i}$, a contradiction.

Lemma 4.2

Let $i\in\{3,\ldots,k\}$, $k\geq 3$ and $j\in[n-k+1]$. If $I_{j}^{i-1}$ is an independent set of $S_{n-k+i-1,i-1}$, then the two vertices $p\pi_{1}$ and $p\pi_{2}$ can not both belong to $I_{j}^{i}$ of $S_{n-k+i,i}$ where $\pi_{1}$ and $\pi_{2}$ are adjacent in $S_{n-k+i-1,i-1}$.

Proof. By Proposition 4.3, $I_{j}^{2}=\{(j+1)1,(j+2)2,\ldots,(n-k+2)(n-k+2-j),1(n-k+3-j),2(n-k+4-j),\ldots,j(n-k+2)\}$ is an independent set of $S_{n-k+2,2}$.

We consider the case for $i=3$. Suppose to the contrary that there exist two vertices $p\pi_{1}$ and $p\pi_{2}$ in $I_{j}^{3}$ but $\pi_{1}$ and $\pi_{2}$ are adjacent in $S_{n-k+2,2}$.

Assume $\pi_{1}=p_{1}p_{2}$ and $\pi_{2}=p_{2}p_{1}$. Then $pp_{1}p_{2}\in I_{j}^{3}(p_{2})$ and $pp_{2}p_{1}\in I_{j}^{3}(p_{1})$. If $p=n-k+3$, then $p_{2}p_{1}$ and $p_{1}p_{2}$ are in $I_{j}^{2}$ by the construction of $I_{j}^{3}$, the one to one correspondence is

[TABLE]

However $p_{2}p_{1}$ and $p_{1}p_{2}$ are adjacent in $S_{n-k+2,2}$, a contradiction with $I_{j}^{2}$ is an independent set of $S_{n-k+2,2}$. If $p<n-k+3$, then $pp_{1}$ and $pp_{2}$ are two vertices in $I_{j}^{2}$ by the construction of $I_{j}^{3}$, the one to one correspondence is

[TABLE]

A contradiction with the construction of $I_{j}^{2}$.

Now, assume $\pi_{1}=p_{1}p_{2}$ and $\pi_{2}=p_{3}p_{2}$. Then $pp_{1}p_{2}\in I_{j}^{3}(p_{2})$ and $pp_{3}p_{2}\in I_{j}^{3}(p_{2})$. If $p=n-k+3$, then $p_{2}p_{1}$ and $p_{2}p_{3}$ are in $I_{j}^{2}$ by the construction of $I_{j}^{3}$, the one to one correspondence is

[TABLE]

A contradiction with the construction of $I_{j}^{2}$. If $p<n-k+3$, then $pp_{1}$ and $pp_{3}$ are two vertices in $I_{j}^{2}$ by the construction of $I_{j}^{3}$, the one to one correspondence is

[TABLE]

A contradiction with the construction of $I_{j}^{2}$.

Therefore, the conclusion is true for $i=3$.

We prove this Lemma by induction on $i$. Assume that the induction hypothesis is true for $i-1$ with $i\geq 4$. We prove the case for $i\geq 4$. Assume $I_{j}^{i-1}$ is an independent set of $S_{n-k+i-1,i-1}$. Then $I_{j}^{i-2}$ is an independent set of $S_{n-k+i-2,i-2}$ by Lemma 4.1. Suppose to the contrary that there exist two vertices $p\pi_{1}$ and $p\pi_{2}$ in $I_{j}^{i}$ but $\pi_{1}$ and $\pi_{2}$ are adjacent in $S_{n-k+i-1,i-1}$.

Firstly, assume $\pi_{1}=p_{2}p_{3}\ldots p_{s}\ldots p_{i-1}p_{i}$ and $\pi_{2}=p_{s}p_{3}\ldots p_{2}\ldots p_{i-1}p_{i}$. Suppose $s=i$. If $p=n-k+i$, then $p_{i}p_{2}p_{3}\ldots p_{i-1}$ and $p_{2}p_{i}p_{3}\ldots p_{i-1}$ are in $I_{j}^{i-1}$ by the construction of $I_{j}^{i}$, the one to one correspondence is

[TABLE]

However $p_{i}p_{2}p_{3}\ldots p_{i-1}$ and $p_{2}p_{i}p_{3}\ldots p_{i-1}$ are adjacent in $S_{n-k+i-1,i-1}$, a contradiction.

Now assume $p<n-k+i$, then $pp_{2}p_{3}\ldots p_{i-1}$ and $pp_{i}p_{3}\ldots p_{i-1}$ are two vertices in $I_{j}^{i-1}$ by the construction of $I_{j}^{i}$, the one to one correspondence is

[TABLE]

However $p_{2}p_{3}\ldots p_{i-1}$ and $p_{i}p_{3}\ldots p_{i-1}$ are adjacent in $S_{n-k+i-2,i-2}$, a contradiction with the induction hypothesis.

Next, suppose $s\neq i$. If $p=n-k+i$, then $p_{i}p_{2}p_{3}\ldots p_{s}\ldots p_{i-1}$ and $p_{i}p_{s}p_{3}\ldots p_{2}\ldots p_{i-1}$ are two vertices in $I_{j}^{i-1}$ by the construction of $I_{j}^{i}$. Now assume $p<n-k+i$. Then $pp_{2}p_{3}\ldots p_{s}\ldots p_{i-1}$ and $pp_{s}p_{3}\ldots p_{2}\ldots p_{i-1}$ are two vertices in $I_{j}^{i-1}$ by the construction of $I_{j}^{i}$. However, $p_{2}p_{3}\ldots p_{s}\ldots p_{i-1}$ and $p_{s}p_{3}\ldots p_{2}\ldots p_{i-1}$ are two adjacent vertices in $S_{n-k+i-2,i-2}$, a contradiction with the induction hypothesis.

Secondly, assume $\pi_{1}=p_{2}p_{3}\ldots p_{i-1}p_{i}$ and $\pi_{2}=p_{i+1}p_{3}\ldots p_{i-1}p_{i}$. If $p=n-k+i$, then $p_{i}p_{2}p_{3}\ldots p_{i-1}$ and $p_{i}p_{i+1}p_{3}\ldots p_{i-1}$ are two vertices in $I_{j}^{i-1}$ by the construction of $I_{j}^{i}$. Now suppose $p<n-k+i$. Then $pp_{2}p_{3}\ldots p_{i-1}$ and $pp_{i+1}p_{3}\ldots p_{i-1}$ are two vertices in $I_{j}^{i-1}$ by the construction of $I_{j}^{i}$, However $p_{2}p_{3}\ldots p_{i-1}$ and $p_{i+1}p_{3}\ldots p_{i-1}$ are two adjacent vertices in $S_{n-k+i-2,i-2}$, a contradiction with the induction hypothesis.

By the principle of induction, this Lemma completes.

Lemma 4.3

Let $i\in\{3,\ldots,k\}$, $k\geq 3$ and $j\in[n-k+1]$. If $I_{j}^{i-1}$ is an independent set of $S_{n-k+i,i}$, then $I_{j}^{i}(x)$ is an independent set of $S_{n-k+i,i}$ for each $x\in[n-k+i]$.

Proof. Assume $i=3$. Suppose to the contrary that $I_{j}^{3}(x)$ is not an independent set of $S_{n-k+3,3}$. Assume $p_{1}p_{2}x$ and $p_{2}p_{1}x$ be two adjacent vertices in $I_{j}^{3}(x)$. By the construction of $I_{j}^{3}(x)$, $p_{1}p_{2},p_{2}p_{1}\in I_{j}^{2}$ if $n-k+3\notin\{p_{1},p_{2}\}$ and $xp_{2},p_{2}x\in I_{j}^{2}$ if $p_{1}=n-k+3$ (the case for $p_{2}=n-k+3$ is similar). A contradiction with the construction of $I_{j}^{2}$ in Proposition 4.3. Assume $p_{1}p_{2}x$ and $p_{3}p_{2}x$ be two adjacent vertices in $I_{j}^{3}(x)$. By the construction of $I_{j}^{3}(x)$, we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Any case above makes a contradiction with the construction of $I_{j}^{2}$ in Proposition 4.3.

We proceed by induction on $i\geq 3$. Assume that the induction hypothesis is true for $i-1$ with $i\geq 4$. Next we prove $I_{j}^{i}(x)$ is an independent set of $S_{n-k+i,i}$ for each $x\in[n-k+i]$. Suppose to the contrary that $I_{j}^{i}(x)$ is not an independent set of $S_{n-k+i,i}$.

Firstly, assume $p_{1}p_{2}\ldots p_{s}\ldots p_{i-1}x$ and $p_{s}p_{2}\ldots p_{1}\ldots p_{i-1}x$ be two adjacent vertices in $I_{j}^{i}(x)$. On one hand, suppose $x=n-k+i$. By the construction of $I_{j}^{i}(n-k+i)$, we have

[TABLE]

[TABLE]

The first case makes a contradiction with the result in Lemma 4.2, since $p_{1}p_{3}\ldots p_{s}\ldots p_{i-1}$ and $p_{s}p_{3}\ldots p_{1}\ldots p_{i-1}$ are adjacent in $S_{n-k+i-2,i-2}$. The second case makes a contradiction with $I_{j}^{i-1}$ is an independent set. On the other hand, suppose $x<n-k+i$. By the construction of $I_{j}^{i}(x)$, we have $p_{1}p_{2}\ldots p_{s}\ldots p_{i-1},p_{s}p_{2}\ldots p_{1}\ldots p_{i-1}\in I_{j}^{i-1}$ if $p_{t}\neq n-k+i$ for each $t=1,2,\ldots,i-1$ (if $p_{t}=n-k+i$ for some $t\in\{1,2,\ldots,i\}$, we just replace $p_{t}$ by $x$). However, $p_{1}p_{2}\ldots p_{s}\ldots p_{i-1}$ and $p_{s}p_{2}\ldots p_{1}\ldots p_{i-1}$ are adjacent in $S_{n-k+i-1,i-1}$, a contradiction with $I_{j}^{i-1}$ is an independent set.

Secondly, assume $p_{1}p_{2}\ldots p_{i-1}x$ and $p_{i}p_{2}\ldots p_{i-1}x$ be two adjacent vertices in $I_{j}^{i}(x)$. If $x=n-k+i$, then $p_{2}p_{1}\ldots p_{i-1}$ and $p_{2}p_{i}\ldots p_{i-1}$ are two vertices in $I_{j}^{i-1}$ by the construction of $I_{j}^{i}(n-k+i)$. However $p_{1}\ldots p_{i-1}$ and $p_{i}\ldots p_{i-1}$ are two adjacent vertices in $I_{j}^{i-2}$, a contradiction with the result in Lemma 4.2. Assume $x<n-k+i$. By the construction of $I_{j}^{i}(x)$, we have $p_{1}p_{2}\ldots p_{i-1},p_{i}p_{2}\ldots p_{i-1}\in I_{j}^{i-1}$ if $p_{t}\neq n-k+i$ for each $t=1,2,\ldots,i-1,i$ (if $p_{t}=n-k+i$ for some $t\in\{1,2,\ldots,i\}$, we just replace $p_{t}$ by $x$). However, $p_{1}p_{2}\ldots p_{i-1}$ and $p_{i}p_{2}\ldots p_{i-1}$ are adjacent in $S_{n-k+i-1,i-1}$, a contradiction with $I_{j}^{i-1}$ is an independent set.

By the principle of induction, this Lemma completes.

Theorem 4.1

The vertex set $I_{j}^{i}$ is an independent set of $S_{n-k+i,i}$ for each $j\in[n-k+1]$ and $i\in\{2,3,\ldots,k\}$.

Proof. We proceed by induction on $i\geq 2$. By Proposition 4.3, $I_{j}^{2}$ is an independent set of $S_{n-k+2,2}$. Assume that the induction hypothesis is true for $i-1$ with $i\geq 3$.

Assume $I_{j}^{i-1}$ is an independent set of $S_{n-k+i-1,i-1}$. By Lemma 4.3, $I_{j}^{i}(x)$ is an independent set of $S_{n-k+i,i}$ for each $x\in[n-k+i]$. Suppose to the contrary that $I_{j}^{i}$ is not independent. Assume $p_{1}p_{2}\ldots p_{i-1}p_{i}$ and $p_{i}p_{2}\ldots p_{i-1}p_{1}$ be two adjacent vertices in $I_{j}^{i}$ (This is the only possible case since $I_{j}^{i}(x)$ is an independent set of $S_{n-k+i,i}$ for each $x\in[n-k+i]$). If $p_{1}=n-k+i$, then $p_{i}p_{2}\ldots p_{i-1}$ and $p_{2}p_{i}\ldots p_{i-1}$ should be in $I_{j}^{i-1}$ by the construction of $I_{j}^{i}$, but they are two adjacent vertices in $I_{j}^{i-1}$, a contradiction (the case for $p_{i}=n-k+i$ is similar). If $p_{s}=n-k+i$ for some $s$ with $2\leq s\leq i-1$, then $p_{1}p_{2}\ldots p_{i}\ldots p_{i-1}$ (replace $n-k+i$ by $p_{i}$) and $p_{i}p_{2}\ldots p_{1}\ldots p_{i-1}$ (replace $n-k+i$ by $p_{1}$) should be in $I_{j}^{i-1}$ by the construction of $I_{j}^{i}$, but they are two adjacent vertices in $I_{j}^{i-1}$, a contradiction. Now assume $p_{s}\neq n-k+i$ for each $s\in[i]$. Then $p_{1}p_{2}\ldots p_{i-1}$ and $p_{i}p_{2}\ldots p_{i-1}$ should be in $I_{j}^{i-1}$ by the construction of $I_{j}^{i}$, but they are two adjacent vertices in $I_{j}^{i-1}$, a contradiction. This Theorem completes.

Theorem 4.2

The constructed $I_{j}^{k}$ is a maximum independent set of $S_{n,k}$ for each $j\in[n-k+1]$ and $\alpha(S_{n,k})=\frac{n!}{(n-k+1)!}$. Moreover, $\{I_{1}^{k},I_{2}^{k},\ldots,I_{n-k+1}^{k}\}$ is a maximum independent sets partition of $S_{n,k}$.

Proof. By Proposition 4.1, $\alpha(S_{n,k})\leq\frac{n!}{(n-k+1)!}$. By Proposition 4.4, $I_{j}^{k}=\frac{n!}{(n-k+1)!}$. By Theorem 4.1, $I_{j}^{k}$ is an independent set of $S_{n,k}$. Therefore, $I_{j}^{k}$ for each $j\in[n-k+1]$ is a maximum independent set of $S_{n,k}$ and $\alpha(S_{n,k})=\frac{n!}{(n-k+1)!}$. By Proposition 4.4, $\{I_{1}^{k},I_{2}^{k},\ldots,I_{n-k+1}^{k}\}$ is a vertex sets partition of $S_{n,k}$, so $\{I_{1}^{k},I_{2}^{k},\ldots,I_{n-k+1}^{k}\}$ is a maximum independent sets partition of $S_{n,k}$.

Since $\{I_{1}^{k},I_{2}^{k},\ldots,I_{n-k+1}^{k}\}$ is a maximum independent sets partition of $S_{n,k}$, we immediately obtain the chromatic number of $S_{n,k}$.

Corollary 4.1

The chromatic number of $S_{n,k}$ is $\chi(S_{n,k})=n-k+1$.

Next, we show the maximum independent sets partition of $S_{4,3}$ by our construction.

Example 4.1** (See Figure 4)**

*By Proposition 4.3, $I_{1}^{2}=\{21,32,13\}$ and $I_{2}^{2}=\{31,12,23\}$ are two maximum independent sets of $S_{3,2}$. The constructed two maximum independent sets of $S_{4,3}$ are:

[TABLE]

[TABLE]