Diagonal Ramsey numbers in multipartite graphs related to stars

TL;DR

This paper determines the exact size Ramsey numbers for multipartite graphs with respect to stars, providing key insights into edge colorings that force the presence of star subgraphs.

Contribution

It establishes the exact values of size Ramsey numbers for complete balanced multipartite graphs concerning stars, a novel result in this area.

Findings

Exact values of $m_j(S_n,S_m)$ for $n,m geq 3$ and $j geq 3$

Characterization of edge colorings avoiding certain star subgraphs

Advancement in understanding Ramsey properties of multipartite graphs

Abstract

Let the star on $n$ vertices, namely $K_{1,n-1}$ be denoted by $S_n$. If every two coloring of the edges of a complete balanced multipartite graph $K_{j \times s}$ there is a copy of $S_n$ in the first color or a copy of $S_m$ in the second color, then we will say $K_{j \times s}\rightarrow (S_n,S_m)$. The size Ramsey multipartite number $m_j(S_n, S_m)$ is the smallest natural number $s$ such that $K_{j \times s}\rightarrow (S_n,S_m)$. In this paper, we obtain the exact values of the size Ramsey numbers $m_j(S_n,S_m)$ for $n,m \ge 3$ and $j \ge 3$.

Equations

$$a_1=\{\begin{array}{ll}i+c& \text{ if }i+c⩽w-1\\ & \\ i+c-w& \text{ if }i+c>w-1\end{array}$$

$$a_2=\{\begin{array}{ll}i-c& \text{ if }i-c⩾0\\ & \\ i-c+w& \text{ if }i-c<0\end{array}$$

$$d_B(v)⩾\lceil \frac{n+m-3}{j-1}\rceil (j-1)-(n-2)⩾m-1$$

$$d_B(v)⩾\lceil \frac{n+m-4}{j-1}\rceil (j-1)-(n-3)⩾m-1$$

$$m_j(S_n,S_m)=\{\begin{array}{ll}\lceil \frac{\max \{n,m\}-1}{j-1}\rceil & \text{ if }n=2\text{ or }m=2\\ & \\ \lceil \frac{n+m-4}{j-1}\rceil & \text{ if }n+m-4=(j-1)s;j,s,n\\ & \text{ are odd and }n,m⩾3\\ & \\ \lceil \frac{n+m-3}{j-1}\rceil & \text{ otherwise }\end{array}$$

Full text

1 Introduction

In this paper we concentrate on simple graphs without loops and multiple edges. Let the complete multipartite graph having $j$ uniform sets of size $s$ be denoted by $K_{j\times s}$ and the complete bipartite graph on $n+m$ vertices be denoted by $K_{n,m}$. Given, three graphs $K_{N}$, $G$ and $H$, we say that $K_{N}\rightarrow(G,H)$ if $K_{N}$ is colored by two colors red and blue and it contains a copy of G(in the first color) or a copy of H(in the second color). Using this notation we define the classical Ramsey number $r(n,m)$ as the smallest integer $N$ such that $K_{N}\rightarrow(K_{n},K_{m})$. Sadly to say, even in the case of diagonal classical Ramsey numbers $r(n,n)$ almost nothing significant is known beyond the case $n=5$ (see Radziszowski et al 2014 for a survey).

In the decades that followed there are several interesting variations that have originated from these classical Ramsey number. One obvious variation is the case of size Ramsey numbers build up mainly by Erdös, Faudree, Rousseau and Shelph (see Erdös et al 1978 and Faudree et al 1975). Another variation introduced recently by Buger et al 2004 and Syafrizal et al 2005, is the concept of balanced multipartite Ramsey numbers. This concept is based on exploring the two colorings of multipartite graphs $K_{j\times s}$ instead of the complete graph. Formally define size Ramsey multipartite number $m_{j}(G,H)$ as the smallest natural number $s$ such that $K_{j\times s}\rightarrow(G,H)$. However, sad to say currently there are very few known balanced multipartite Ramsey numbers between pairs of graphs and pairs of classes of graphs other than the ones introduced initially by Syafrizal et al 2005 and 2009).

Notation

Given a graph $G=(V,E)$ with the order of the graph is denoted by $|V(G)|$ and the size of the graph is denoted by $|E(G)|$. For a vertex $v$ of a graph $G$, the neighborhood of $v$ is denoted by $N(v)$ and is defined as the set of vertices adjacent to $v$. Further the cardinality of this set, denoted $d(v)$, is defined as the degree of $v$. We say that a graph $G$ is a $k$-regular graph if $d(v)=k$ for all $v\in V(G)$. Given a red-blue coloring of $K_{j\times s}=H_{R}\oplus H_{B}$. The red degree and blue degree of any vertex $v$ belonging to V(K_{j\times s})=\{$$v_{k,i}$$| $0\leqslant i\leqslant s-1,$ $0\leqslant k\leqslant j-1\}$ denoted by $d_{R}(v)$ and $d_{B}(v)$ respectively, are defined as the degree of vertex $v$ in $H_{R}$ and $H_{B}$ respectively.

Given $w\geqslant 2$, $0\leqslant i\leqslant w-1$ and $0<c\leqslant w-1$, define $\sigma_{c,w}(i)=\{a_{1}\}\cup\{a_{2}\}$, $\sigma^{+}_{c,w}(i)=\{a_{1}\}$, $\sigma^{-}_{c,w}(i)=\{a_{2}\}$ and $B_{0,w}(i)=\phi$ and if $k>0$, $B_{k,w}(i)=\cup_{c=1}^{k}\sigma_{c,w}(i)$ where

[TABLE]

[TABLE]

2 **Some Lemmas **

In all the following lemmas assume $d>0$ as the results are trivially true when $d=0$.

Lemma 1**.**

There exists a regular induced subgraph of degree $d$ of $K_{j\times s}$ on the vertex set $V(K_{j\times s})$ provided that $d$ is even, $j$ is odd and $s$ is odd.

Proof.

If $d=2k_{1}(j-1)+2k_{2}$ for some non negative integers $k_{1}$ and $k_{2}$ such that $2k_{1}\leqslant s-1$ and $0<2k_{2}\leqslant j-1$. Construct a graph by connecting the vertices $v_{i,l}$ and $v_{p,r}$ if one of the following situations hold

a) If $r\in B_{k_{1},s}(l)$ and $p\neq i$.       b) If $r=l$ and $p\in B_{k_{2},j}(i)$.

v_{2,2}$$v_{2,4}$$v_{2,3}$$v_{1,2}$$v_{0,2}$$v_{2,1}$$v_{1,4}$$v_{1,3}$$v_{1,1}$$v_{0,1}$$v_{2,0}$$v_{0,4}$$v_{0,3}$$v_{1,0}$$v_{0,0}

Figure 1: In the case when $d=6=2k_{1}(j-1)+2k_{2}=(2\times 1)(3-1)+2\times 1$

We know that $K_{j\times s}$ consists of $j$ partite sets of size $s$. Given $v_{i,l}$ $0\leqslant i\leqslant j-1$, $0\leqslant l\leqslant s-1$, the set $\{v_{p,r}|$ $p\neq i$ and $r\in B_{k_{1},s}(l)\}$ will represent the vertices not belonging to the $i^{th}$ partite set (denoted by $V_{i}$) that are at most $2k_{1}$ distance apart inside a partite set (with respect to the second coordinate), as illustrated in the following figure.

v_{i,l}$$v_{i,i-k}$$V_{i-1}$$v_{i,s-1}$$v_{i,0}$$v_{i,l+k}$$V_{i}$$V_{i+1}$$V_{j-1}$$V_{0}$$v_{0,l}$$v_{0,s-1}$$v_{0,0}$$v_{i-1,l}$$v_{i-1,s-1}$$v_{i-1,0}$$v_{i+1,l}$$v_{i+1,s-1}$$v_{i+1,0}$$v_{j-1,l}$$v_{j-1,s-1}$$v_{j-1,0}$$k_{1}i\>\>vertices$$(j-1-i)k_{1}\>\>vertices$$k_{1}i\>\>vertices$$(j-1-i)k_{1}\>\>vertices

Figure 2: The set consisting of $2k_{1}(j-1)$ vertices corresponding to part(a), namely $\{v_{p,r}|$ $p\neq i$ and $r\in B_{k_{1},s}(l)\}$

More precisely, it will consist of the vertices

$v_{0,l-k_{1}},...,v_{i-1,l-k_{1}},v_{i+1,l-k_{1}},...,v_{j-1,l-k_{1}}$

$v_{0,l-k_{1}+1},...,v_{i-1,l-k_{1}+1},v_{i+1,l-k_{1}+1},...,v_{j-1,l-k_{1}+1}$

…

$v_{0,l-1},...,v_{i-1,l-1},v_{i+1,l-1},...,v_{j-1,l-1}$

$v_{0,l+1},...,v_{i-1,l+1},v_{i+1,l+1},...,v_{j-1,l+1}$

…

$v_{0,l+k_{1}-1},...,v_{i-1,l+k_{1}-1},v_{i+1,l+k_{1}-1},...,v_{j-1,l+k_{1}-1}$

$v_{0,l+k_{1}},...,v_{i-1,l+k_{1}},v_{i+1,l+k_{1}},...,v_{j-1,l+k_{1}}$

That is such a set consists of $2k_{1}(j-1)$ vertices.

Similarly, given $v_{i,l}$ where $0\leqslant i\leqslant j-1$ and $0\leqslant l\leqslant s-1$ the set $\{v_{p,r}|$ $r\neq l$ and $p\in B_{k_{2},j}(i)\}$ will represent the vertices not belonging to the $i^{th}$ partite set(denoted by $V_{i}$) that are at most $2k_{2}$ distance apart between partite sets (with respect to the first coordinate), as illustrated in the following figure. More precisely, it will consist of the vertices

$v_{i-k_{2},l},...,v_{i-1,l},v_{i+1,l},...,v_{i+k_{2},l}$

That is such a set consists of $2k_{2}$ vertices.

v_{i,l}$$v_{i,i-k_{1}}$$V_{i-1}$$v_{i,s-1}$$v_{i,0}$$v_{i,l+k_{1}}$$V_{i}$$V_{i+1}$$V_{j-1}$$V_{0}$$v_{0,l}$$v_{0,i-k_{1}}$$v_{0,s-1}$$v_{0,0}$$v_{0,l+k_{1}}$$v_{i-1,i-k_{1}}$$v_{i-1,s-1}$$v_{i-1,0}$$v_{i-1,l+k_{1}}$$v_{i+1,i-k_{1}}$$v_{i+1,s-1}$$v_{i+1,0}$$v_{i+1,l+k_{1}}$$v_{j-1,l}$$v_{j-1,i-k_{1}}$$v_{j-1,s-1}$$v_{j-1,0}$$v_{j-1,l+k_{1}}$$k_{2}\>\>vertices$$k_{2}\>\>vertices

Figure 3: The set consisting of $2k_{2}$ vertices corresponding to part(b), namely $\{v_{p,r}|$ $r=l$ and $p\in B_{k_{2},j}(i)\}$

Thus, by the above definition, part $(a)$ will represent $2k_{1}(j-1)$ vertices adjacent to $v_{i,l}$ belonging to $V_{0},V_{1},V_{2},...,V_{i-1},V_{i+1},...V_{j-1}$ and part $(b)$ will represent another $2k_{2}$ vertices adjacent to $v_{i,l}$ belonging to $V_{i-k_{2}},$ $V_{i-k_{2}+1},...,$ $V_{i-1},V_{i+1},...,V_{i+k_{2}-1},$ $V_{i+k_{2}}$.

Therefore, the degree of $v_{i,l}$ will be equal to $2k_{1}(j-1)+2k_{2}$. Also by the remark $l\in B_{k,w}(i)$ if and only if $i\in B_{k,w}(l)$, before the beginning of this section, we get that the graph is well defined.

Next, if $d=(2k_{1}+1)(j-1)+2k_{2}$ for some non negative integers $k_{1}$ and $k_{2}$ such that $2k_{1}\leqslant s-3$ and $0<2k_{2}\leqslant j-1$. Construct a graph by connecting the vertices $v_{i,l}$ and $v_{p,r}$ if one of the following situations hold

a) If $r\in B_{k_{1},s}(l)$ and $p\neq i$.

b) If there exists $w$ such that $4w=(j-1+2k_{2})$ and $r\in\sigma_{k_{1}+1,s}(l)$ and $p\in B_{w,j}(i)$.

c) If there exists $w$ such that $(j-1+2k_{2})-4w=2$ and $r\in\sigma_{k_{1}+1,s}(l)$ and $p\in B_{w,j}(i)$ or else $r=l$ and $p\in B_{1,j}(i)$.

v_{2,2}$$v_{1,2}$$v_{0,2}$$v_{2,1}$$v_{1,1}$$v_{0,1}$$v_{2,0}$$v_{1,0}$$v_{0,0}

Figure 4: In the case when $d=4=(2k_{1}+1)(j-1)+2k_{2}=(2\times 0+1)\times(3-1)+2\times 1$

It should be noted that the vertex sets of part (b) and part (c) are disjoint and that $j-1+2k_{2}$ is even as $j$ is odd. Therefore, given $v_{i,l}$, it will be either adjacent the vertices corresponding to part (a) and part (b) or else adjacent the vertices corresponding to part (a) and part (c) according to whether $4w=(j-1+2k_{2})$ or else $(j-1+2k_{2})-4w=2$, respectively.

As illustrated in figure 2, the set generated by part (a) namely, $\{v_{p,r}|$ $p\neq i$ and $r\in B_{k_{1},s}(l)\}$ will consist of $2k_{1}(j-1)$ vertices.

Similarly, given $v_{i,l}$ the set generated by part (b), will represent the vertices belonging to $V_{i-w},...,V_{i-1},V_{i+1},...V_{i+w}$ sets, that are at most $2(k_{1}+1)$ distance apart between partite sets (with respect to the first coordinate), as illustrated in the following figure. More precisely, it will consist of the vertices

$v_{i-w,l-(k_{1}+1)},...,v_{i-1,l-(k_{1}+1)},v_{i+1,l-(k_{1}+1)},...v_{i+w,l-(k_{1}+1)}$

$v_{i-w,l+(k_{1}+1)},...,v_{i-1,l+(k_{1}+1)},v_{i+1,l+(k_{1}+1)},...v_{i+w,l+(k_{1}+1)}$

Such a set consists of $4w$ vertices. That is, the set consists of $(j-1)+2k_{2}$ vertices.

v_{i,l}$$v_{i,i-k-1}$$V_{i-w}$$v_{i,s-1}$$v_{i,0}$$v_{i,l+k+1}$$V_{i}$$V_{i+w}$$V_{j-1}$$V_{0}$$v_{0,l}$$v_{0,s-1}$$v_{0,0}$$v_{i-w,l}$$v_{i-w,s-1}$$v_{i-w,0}$$v_{i+w,l}$$v_{i+w,s-1}$$v_{i+w,0}$$v_{j-1,l}$$v_{j-1,s-1}$$v_{j-1,0}$$w\>\>vertices$$w\>\>vertices$$w\>\>vertices$$w\>\>vertices

Figure 5: The set consisting of $4w$ vertices corresponding to part (b), namely $\{v_{p,r}|$ $r\in\sigma_{k_{1}+1,s}(l)$ and $p\in B_{w,j}(i)\}$

Similarly, given $v_{i,l}$ the set generated by the later part (c), will represent the two vertices belonging to $V_{i-1},V_{i+1}$ sets, namely $v_{i-1,l-1}$, $v_{i+1,l+1}$. More precisely the set generated by part (c), will consist of the vertices

$v_{i-w,l-(k_{1}-1)},...,v_{i-1,l-(k_{1}-1)},v_{i+1,l-(k_{1}-1)},...v_{i+w,l-(k_{1}-1)}$

$v_{i-1,l-1},v_{i+1,l+1}$

$v_{i-w,l-(k_{1}+1)},...,v_{i-1,l-(k_{1}+1)},v_{i+1,l-(k_{1}+1)},...v_{i+w,l-(k_{1}+1)}$

Such a set consists of $4w+2$ vertices. That is, the set consists of $(j-1)+2k_{2}$ vertices.

v_{i,l}$$V_{i-1}$$v_{i,s-1}$$v_{i,0}$$V_{i}$$V_{i+1}$$V_{j-1}$$V_{0}$$v_{0,l}$$v_{0,s-1}$$v_{0,0}$$v_{i-1,l}$$v_{i-1,s-1}$$v_{i-1,0}$$v_{i+1,l}$$v_{i+1,s-1}$$v_{i+1,0}$$v_{j-1,l}$$v_{j-1,s-1}$$v_{j-1,0}$$1\>vertex$$1\>vertex

Figure 6: The set consisting of $2$ vertices corresponding to later section of part (c), namely $\{v_{p,r}|$ $p\in B_{w,j}(i)$ or else $r=l$ and $p\in B_{1,j}(i)\}$

Therefore, the degree of $v_{i,l}$ will be equal to $2k_{1}(j-1)+2k_{2}$ when part (a)+(b) situation arrises or when part (a)+(c) situation arrises. ∎

Lemma 2**.**

There exist some uniform graphs of degree $d$ on $V(K_{j\times s})$ if $j$ is even or $s$ is even.

Proof.

We approach this problem by considering the following three cases.

Case 1) If $j$ is even and $s$ is odd.

If $d=2k_{1}(j-1)+2k_{2}$ for some non negative integers $k_{1}$ and $k_{2}$ such that $2k_{1}\leqslant s-1$ and $0<2k_{2}\leqslant j-2$. Construct a graph by connecting the vertices $v_{i,l}$ and $v_{p,r}$ if one of the following situations hold

a) If $r\in B_{k_{1},s}(l)$ and $p\neq i$.      b) If $r=l$ and $p\in B_{k_{2},j}(i)$.

The vertex $v_{i,l}$ will be either adjacent the vertices corresponding to part (a) or part (b) and they are respectively equal to $2k_{1}(j-1)$ and $2k_{2}$. Therefore, we get that the degree of $v_{i,l}$ is equal to $2k_{1}(j-1)+2k_{2}$ as required.

If $d=2k_{1}(j-1)+2k_{2}+1$ for some non negative integers $k_{1}$ and $k_{2}$ such that $2k_{1}\leqslant s-1$ and $2k_{2}\leqslant j-2$. Construct a graph by connecting the vertices $v_{i,l}$ and $v_{p,r}$ if one of the following situations hold

a) If $r\in B_{k_{1},s}(l)$ and $p\neq i$.       b) If $r=l$ and $p\in B_{k_{2},j}(i)$.

c) If $r=l$ and $p\in\sigma_{\frac{j}{2},j}(i)$.

v_{0,0}$$v_{1,0}$$v_{2,0}$$v_{0,2}$$v_{1,2}$$v_{2,2}$$v_{0,1}$$v_{2,1}$$v_{3,0}$$v_{3,1}$$v_{3,2}$$Example:When d=3=2\times 0\times(4-1)+2\times 1+1$$v_{1,1}

Figure 7: In the case(1) when $d=3=2k_{1}(j-1)+2k_{2}+1=2\times 0\times(4-1)+2\times 1+1$

The vertex $v_{i,l}$ will be either adjacent the vertices corresponding to part (a), part (b) or part (c) and they are respectively equal to $2k_{1}(j-1)$, $2k_{2}$ and one. Therefore, we get that the degree of $v_{i,l}$ is equal to $2k_{1}(j-1)+2k_{2}+1$ as required.

If $d=(2k_{1}+1)(j-1)+m$ for some non negative integers $k_{1}$, $k_{2}$ and $m$ such that $2k_{1}\leqslant s-3$ and $0<m\leqslant j-1$ where $m=2k_{2}$ or $m=2k_{2}+1$. Construct a graph by connecting the vertices $v_{i,l}$ and $v_{p,r}$ if one of the following situations hold

a) If $r\in B_{k_{1},s}(l)$ and $p\neq i$.

b) If there exists $w$ such that $w=(j-1+m)$ div $4$, $r\in\sigma_{k_{1}+1,s}(l)$ and $p\in B_{w,j}(i)$.

c) If there exists $w$ such that $(j-1+m)-4w=1$, $r=l$ and $p\in\sigma_{\frac{j}{2},j}(i)$.

d) If there exists $w$ such that $(j-1+m)-4w=2$, $r=l$ and $p\in\sigma_{1,j}(i)$.

e) If there exists $w$ such that $(j-1+m)-4w=3$, $r=l$, $p\in\sigma_{1,j}(i)$ and $p\in\sigma_{\frac{j}{2},j}(i)$ (as $j$ is even).

v_{2,2}$$v_{1,2}$$v_{0,2}$$v_{2,1}$$v_{1,1}$$v_{0,1}$$v_{2,0}$$v_{1,0}$$v_{0,0}$$v_{3,0}$$v_{3,1}$$v_{3,2}$$v_{4,0}$$v_{4,1}$$v_{4,2}$$v_{5,0}$$v_{5,1}$$v_{5,2}

Figure 8: In the case(1) when $d=6=(2k_{1}+1)(j-1)+2k_{2}+1=(2\times 0+1)(6-1)+2\times 0+1$

It should be noted that the vertex sets of part (b), (c), (d) and part (e) are disjoint. Therefore, $v_{i,l}$ will be either adjacent the vertices corresponding to part (a) and part (b) or (a) and part (c) or (a) and part (d) or else part (a) and part (e) according to whether $4w=j-1+2k_{2}$, $(j-1+2k_{2})-4w=1$, $(j-1+2k_{2})-4w=2$ or $(j-1+2k_{2})-4w=3$ respectively. As done before, in all these scenarios we get $d=(2k_{1}+1)(j-1)+m$ as required.

Case 2) If $j$ is even and $s$ is even.

If $d=2k_{1}(j-1)+2k_{2}$ for some non negative integers $k_{1}$ and $k_{2}$ such that $2k_{1}\leqslant s-2$ and $0<k_{2}\leqslant j-1$. Construct a graph by connecting the vertices $v_{i,l}$ and $v_{p,r}$ if one of the following situations hold

a) If $r\in B_{k_{1},s}(l)$ and $p\neq i$.

b) If $k_{2}<\frac{j}{2}$, $r=l$ and $p\in B_{k_{2},j}(i)$.

c) If $\frac{j}{2}\leqslant k_{2}<j-1$ and (($r=l$ and $p\in B_{\frac{j-2}{2},j}(i)$) or ($r\in\sigma_{\frac{s}{2},s}(l)$ and $p\in B_{\frac{2k_{2}-(j-2)}{2},j}(i)$)).

d) If $k_{2}=j-1$ and (($r=l$ and $p\neq i$) or ($r\in\sigma_{\frac{s}{2},s}(l)$ and $p\neq i$)).

It should be noted that the vertex sets of part (b), (c) and part (d) are disjoint. Therefore, $v_{i,l}$ will be either adjacent the vertices corresponding to part (a) and part (b) or (a) and part (c) or else part (a) and part (d) according to whether $k_{2}<\frac{j}{2}$, $\frac{j}{2}\leqslant k_{2}<j-1$ or $k_{2}=j-1$ respectively. As done before, in all these scenarios we get $d=2k_{1}(j-1)+2k_{2}$ as required.

If $d=2k_{1}(j-1)+2k_{2}+1$ for some non negative integers $k_{1}$ and $k_{2}$ where such that $2k_{1}\leqslant s-2$ and $k_{2}<j-1$. Construct a graph by connecting the vertices $v_{i,l}$ and $v_{p,r}$ if one of the following situations hold

a) If $r\in B_{k_{1},s}(l)$ and $p\neq i$.

b) If $k_{2}\leqslant\frac{j-2}{2}$, $r=l$ and ($p\in B_{k_{2},j}(i)$ or $p\in\sigma_{\frac{j}{2},j}(i)$).

c) If $k_{2}\geqslant\frac{j}{2}$ and (($r=l$ and $p\neq i$) or ($r\in\sigma_{\frac{s}{2},s}(l)$ and $p\in B_{\frac{2k_{2}-(j-2)}{2},j}(i)$)).

v_{0,0}$$v_{1,0}$$v_{2,0}$$v_{0,2}$$v_{1,2}$$v_{2,2}$$v_{0,1}$$v_{3,1}$$v_{2,1}$$v_{0,3}$$v_{1,3}$$v_{2,3}$$v_{3,0}$$v_{1,1}$$v_{3,2}$$v_{3,3}

Figure 9: In the case(2) when $d=5=2k_{1}(j-1)+2k_{2}+1=2\times 0\times(4-1)+2\times 2+1$

It should be noted that the vertex sets of part (b) and part (c) are disjoint. Therefore, $v_{i,l}$ will be either adjacent the vertices corresponding to part (a) and part (b) or (a) and part (c) according to whether $k_{2}\leqslant\frac{j-2}{2}$ or $k_{2}\geqslant\frac{j}{2}$ respectively. As done before, in all these scenarios we get $d=2k_{1}(j-1)+2k_{2}+1$ as required.

Case 3) If $j$ is odd and $s$ is even.

If $d=2k_{1}(j-1)+2k_{2}$ for some non negative integers $k_{1}$ and $k_{2}$ such that $2k_{1}\leqslant s-2$ and $0<k_{2}\leqslant j-1$. Construct a graph by connecting the vertices $v_{i,l}$ and $v_{p,r}$ if one of the following situations hold

a) If $r\in B_{k_{1},s}(l)$ and $p\neq i$.

b) If $k_{2}<\frac{j}{2}$, $r=l$ and $p\in B_{k_{2},j}(i)$

c) If $k_{2}\geqslant\frac{j}{2}$ and (($r=l$ and $p\neq i$) or ($r\in\sigma_{\frac{s}{2},s}(l)$ and $p\in B_{\frac{2k_{2}-(j-1)}{2},j}(i)$)).

It should be noted that the vertex sets of part (b) and part (c) are disjoint. Therefore, $v_{i,l}$ will be either adjacent the vertices corresponding to part (a) and part (b) or (a) and part (c) according to whether $k_{2}<\frac{j}{2}$ or $k_{2}\geqslant\frac{j}{2}$ respectively. As done before, in all these scenarios we get $d=2k_{1}(j-1)+2k_{2}$ as required.

Case 4) If $j$ is odd and $s$ is even.

If $d=2k_{1}(j-1)+2k_{2}+1$ for some non negative integers $k_{1}$ and $k_{2}$ such that $2k_{1}\leqslant s-2$ and $0\leqslant k_{2}<j-1$. Construct a graph by connecting the vertices $v_{i,l}$ and $v_{p,r}$ if one of the following situations hold

a) If $r\in B_{k_{1},s}(l)$ and $p\neq i$.

b) If $k_{2}<\frac{j-1}{2}$ and (($r=l$ and $p\in B_{k_{2},j}(i)$) or ($r\in\sigma_{\frac{s}{2},s}(l)$ and $r>l$ and $p=\sigma^{+}_{k_{2}+1,j}(i)$) or ($r\in\sigma_{\frac{s}{2},s}(l)$ and $r<l$ and $p=\sigma^{-}_{k_{2}+1,j}(i)$)).

c) If $k_{2}=\frac{j-1}{2}$and (($r=l$ and $p\neq i$) or ($r\in\sigma_{\frac{s}{2},s}(l)$ and $r>l$ and $p=\sigma^{+}_{k_{2}+1,j}(i)$) or ($r\in\sigma_{\frac{s}{2},s}(l)$ and $r<l$ and $p=\sigma^{-}_{k_{2}+1,j}(i)$)).

d) If $k_{2}>\frac{j-1}{2}$ and (($r=l$ and $p\neq i$) or ($r\in\sigma_{\frac{s}{2},s}(l)$ and $p\in B_{\frac{2k_{2}-(j-1)}{2},j}(i)$) or ($r\in\sigma_{\frac{s}{2},s}(l)$ and $p=\sigma^{+}_{\frac{2k_{2}-(j-1)}{2}+1,j}(i)$ with $r>l$ ) or ($r\in\sigma_{\frac{s}{2},s}(l)$ and $p=\sigma^{-}_{\frac{2k_{2}-(j-1)}{2}+1,j}(i)$ with $r<l$)).

v_{2,2}$$v_{1,2}$$v_{0,2}$$v_{2,1}$$v_{1,1}$$v_{0,1}$$v_{2,0}$$v_{1,0}$$v_{0,0}$$v_{2,3}$$v_{0,3}$$v_{1,3}

Figure 10: In the case(3) when $d=5=2k_{1}(j-1)+2k_{2}+1=2\times 1\times(3-1)+2\times 0+1$

It should be noted that the vertex sets of part (b), (c) and part (d) are disjoint. Therefore, $v_{i,l}$ will be either adjacent the vertices corresponding to part (a) and part (b) or (a) and part (c) or else (a) and part (d) according to whether $k_{2}<\frac{j-1}{2}$, $k_{2}=\frac{j-1}{2}$ or $k_{2}>\frac{j-1}{2}$ respectively. As done before, in all these scenarios we get $d=2k_{1}(j-1)+2k_{2}+1$ as required. ∎

Lemma 3**.**

$m_{j}(S_{n},S_{m})\leqslant\left\lceil\dfrac{n+m-3}{j-1}\right\rceil$* where $j,n,m\geqslant 3$.*

Proof.

Consider any red/blue coloring given by $K_{j\times s}=H_{R}\oplus H_{B}$, where $s=\left\lceil\dfrac{n+m-3}{j-1}\right\rceil$, such that $H_{R}$ contains no red $S_{n}$. Let $v$ be any vertex of $K_{j\times s}$. Then $v$ is incident to at most $n-2$ red edge. Hence,

[TABLE]

Therefore, $H_{B}$ will contain a blue $S_{m}$. Hence the result. ∎

Lemma 4**.**

$m_{j}(S_{n},S_{m})\geqslant\left\lceil\dfrac{n+m-4}{j-1}\right\rceil$* where $j,n,m\geqslant 3$.*

Proof.

Consider the red and blue coloring of $K_{j\times s}$ given by $K_{j\times s}=H_{R}\oplus H_{B}$, where $s=\left\lceil\dfrac{n+m-4}{j-1}\right\rceil-1$, where all the vertices will have uniform red degree of $n-2$ or $n-3$ (this is possible by lemma 2). Then clearly $H_{B}$ does not contain a red $S_{n}$. Let $v$ be any vertex of $K_{j\times s}$. Then,

$d_{B}(v)=\left(\left\lceil\dfrac{n+m-4}{j-1}\right\rceil-1\right)(j-1)-(n-3)$

$=\left\lceil\dfrac{n+m-4}{j-1}\right\rceil(j-1)-j+1-n+3$

$\geqslant n+m-4-j-n+4\geqslant m-j$

Therefore, $H_{B}$ will not contain a blue $S_{m}$. Hence the result. ∎

Lemma 5**.**

$m_{j}(S_{n},S_{m})=\left\lceil\dfrac{n+m-4}{j-1}\right\rceil$* if $(n+m-4)\neq 0$ mod $(j-1)$ where $j,n,m\geqslant 3$.*

Proof.

We know that if $(n+m-4)\neq 0$ mod $(j-1)$ then $\left\lceil\dfrac{n+m-4}{j-1}\right\rceil=\left\lceil\dfrac{n+m-3}{j-1}\right\rceil$. Hence the result follows by lemma 3 and lemma 4.

∎

Lemma 6**.**

Suppose that $j,n,m\geqslant 3$. Then, $m_{j}(S_{n},S_{m})\leqslant\left\lceil\dfrac{n+m-4}{j-1}\right\rceil$ provided that $(n+m-4)=0$ mod $(j-1)$ with $j$ is odd, $n$ is odd and $s=\dfrac{n+m-4}{j-1}$ is odd.

Proof.

Consider any red/blue coloring given by $K_{j\times s}=H_{R}\oplus H_{B}$, where $s=\left\lceil\dfrac{n+m-4}{j-1}\right\rceil$, such that $H_{R}$ contains no red $S_{n}$. Since, $j\times s\times(n-2)$ is odd, there will exist at least one vertex $v\in K_{j\times s}$ such it is not incident to $n-2$ red edges, as otherwise by handshake lemma $j\times s\times(n-2)=2|E(H_{R})|$, a contradiction. Hence,

[TABLE]

Therefore, $H_{B}$ will contain a blue $S_{m}$. Hence the result. ∎

Lemma 7**.**

Suppose that $j,n,m\geqslant 3$. Then, $m_{j}(S_{n},S_{m})\geqslant\left\lceil\dfrac{n+m-3}{j-1}\right\rceil$ provided that $(n+m-4)=0$ mod $(j-1)$ with $j$ is even or $s=\dfrac{n+m-4}{j-1}$ even or $n$ is even.

Proof.

By lemma 3 and lemma 4, $K_{j\times s}$ where $s=\left\lceil\dfrac{n+m-3}{j-1}\right\rceil-1=\left\lceil\dfrac{n+m-4}{j-1}\right\rceil$, will have a $n-2$ regular subgraph on $K_{j\times s}$. Using this subgraph generate a red/blue coloring given by $K_{j\times s}=H_{R}\oplus H_{B}$, where all the edges of this subgraph are colored red and all other edges colored blue. Then clearly $H_{R}$ is $S_{n}$ - free. Furthermore, for any vertex $v\in K_{j\times s}$, $d_{B}(v)=\left(\dfrac{n+m-4}{j-1}\right)(j-1)-(n-2)=m-2$. Therefore, $H_{B}$ will not contain a blue $S_{m}$. Hence the result. ∎

Theorem 8**.**

If $j\geqslant 3$ and $n,m\geqslant 2$ then,

[TABLE]

Proof.

The theorem clearly follows from lemmas 5, 6 and 7 as $m_{j}(S_{2},S_{m})=\left\lceil\dfrac{m-1}{j-1}\right\rceil$ (see Syafrizal et al 2005). ∎