Signed edge domination numbers of complete tripartite graphs: Part 2

TL;DR

This paper determines the signed edge domination number for complete tripartite graphs, completing the classification for all such graphs and advancing understanding of their domination properties.

Contribution

It provides the exact signed edge domination number for all complete tripartite graphs $K_{m,n,p}$ with specified parameters, filling a gap in the existing literature.

Findings

Exact signed edge domination numbers for $K_{m,n,p}$ with $p \\geq m+n$

Completes the classification of signed edge domination for all complete tripartite graphs

Advances understanding of domination functions in complex graph structures

Abstract

The closed neighborhood $N_G[e]$ of an edge $e$ in a graph $G$ is the set consisting of $e$ and of all edges having an end-vertex in common with $e$. Let $f$ be a function on $E(G)$, the edge set of $G$, into the set $\{-1,1\}$. If $\sum_{x\in{N[e]}}f(x)\geq 1$ for each edge $e \in E(G)$, then $f$ is called a signed edge dominating function of $G$. The signed edge domination number of $G$ is the minimum weight of a signed edge dominating function of $G$. In this paper, we find the signed edge domination number of the complete tripartite graph $K_{m,n,p}$, where $1\leq m\leq n$ and $p\geq m+n$. This completes the search for the signed edge domination numbers of the complete tripartite graphs.

Equations

$$f[uv]=\mathrm{deg}(u)+\mathrm{deg}(v)-2x-2y-f(uv)\ge 1.$$

$$\begin{array}[]{lclr}g[v_{1}v_{j}]&=&\deg(v_{1})+\deg(v_{j})-2y_{1}-2y_{j}-g(v_{1}v_{j})&\\ &=&\deg(v_{1})+\deg(v_{j})-2x_{1}-2-2x_{j}-f(v_{1}v_{j})&\\ &\geq&\deg(v_{2})+\deg(v_{j})-2x_{2}+2-2x_{j}-f(v_{1}v_{j})&\\ &\geq&\deg(v_{2})+\deg(v_{j})-2x_{2}-2x_{j}-f(v_{2}v_{j})&\\ &\geq&k&\hfill{\mbox{ by (\ref{eq1})}},\\ \end{array}$$

$$\begin{array}[]{lclr}g[v_{2}v_{j}]&=&\deg(v_{2})+\deg(v_{j})-2y_{2}-2y_{j}-g(v_{2}v_{j})&\\ &=&\deg(v_{2})+\deg(v_{j})-2x_{2}+2-2x_{j}-f(v_{2}v_{j})&\\ &\geq&k+2&\hfill{\mbox{ by (\ref{eq1})}}.\end{array}$$

$$\begin{array}[]{lclr}g[v_{1}v_{2}]&=&\deg(v_{1})+\deg(v_{2})-2y_{1}-2y_{2}-g(v_{1}v_{2})&\\ &=&\deg(v_{1})+\deg(v_{2})-2x_{1}-2x_{2}-f(v_{1}v_{2})&\\ &\geq&k&\hfill{\mbox{ by (\ref{eq1})}}.\end{array}$$

$$|W_3\mid \ge p-[(m+p-2k)/2+(n-m)/2]=(p-n+2k)/2\ge 1.$$

$$|W_3\mid \ge p-[(m+p-2k-1)/2+(n-m)/2]=(p-n+2k+1)/2\ge 1.$$

$$|W_3\mid \ge p-[(m+p-2k-1)/2+(n-m-1)/2]=(p-n+2k+2)/2\ge 2.$$

$$|W_3\mid \ge p-[(m+p-2k-2)/2+(n-m+1)/2]=(p-n+2k+1)/2\ge 1.$$

$$m(n+p-2)/2+n(m+p-2)/2-p(m+n)/2=mn-m-n$$

$$\begin{array}[]{l}m(n+p-2)/2+n(m+p-2)/2-(p-1)(m+n)/2\\ -(m+n-2)/2=mn-m-n+1,\end{array}$$

$$\begin{array}[]{l}(m-1)(n+p-3)/2+(n+p-5)/2+n(m+p-3)/2\\ -p(m+n)/2=mn-1-(3m+3n)/2\end{array}$$

$$m(p+n-3)/2+n(m+p-3)/2-p(m+n)/2=mn-(3m+3n)/2$$

$$\begin{array}[]{l}m(n+p-3)/2+n(m+p-3)/2-p(m+n)/2=mn-(3m+3n)/2\\ \end{array}$$

$$\begin{array}[]{l}(m-1)(n+p-3)/2+(n+p-5)/2+n(m+p-3)/2\\ -p(m+n)/2=mn-1-(3m+3n)/2\\ \end{array}$$

$$\begin{array}[]{l}m(n+p-2)/2+((n-m-1)/2)(m+p-1)/2\\ +((n+m+1)/2)(m+p-3)/2-(p/2)(m+n+1)/2\\ -(p/2)(m+n-1)/2=mn-n-(3m+1)/2\end{array}$$

$$\begin{array}[]{l}m(n+p-2)/2+((n-m+1)/2)(m+p-1)/2\\ +((n+m-1)/2)(m+p-3)/2-(p/2)(m+n+1)/2\\ -(p/2)(m+n-1)/2=mn-n-(3m-1)/2\end{array}$$

$$\begin{array}[]{l}m(n+p-1)/2+((n-m-1)/2)(m+p-2)/2\\ +((n+m+1)/2)(m+p-4)/2-(p/2)(m+n+1)/2\\ -(p/2)(m+n-1)/2=mn-m-(3n+1)/2\end{array}$$

$$\begin{array}[]{l}m(n+p-1)/2+((n-m-1)/2)(m+p-2)/2\\ +((n+m+1)/2)(m+p-4)/2-(p/2)(m+n+1)/2\\ -(p/2)(m+n-1)/2=mn-m-(3n-1)/2\end{array}$$

$$\begin{array}[]{l}m(n+p-1)/2+((n-m-1)/2)(m+p-2)/2\\ +((n+m+1)/2)(m+p-4)/2-((p+1)/2)(m+n+1)/2\\ -((p-1)/2)(m+n-1)/2=mn-m-(3n+2)/2\end{array}$$

$$\begin{array}[]{l}m(n+p-1)/2+((n-m+1)/2)(m+p-2)/2\\ +((n+m-1)/2)(m+p-4)/2-((p+1)/2)(m+n+1)/2\\ -((p-1)/2)(m+n-1)/2=mn-m-3n/2\end{array}$$

$$\begin{array}[]{l}m(n+p-2)/2+((n-m+1)/2)(m+p-1)/2\\ +((n+m-1)/2)(m+p-3)/2-((p-1)/2)(m+n+1)/2\\ -((p+1)/2)(m+n-1)/2=mn-n-(3m-2)/2\end{array}$$

$$\begin{array}[]{l}m(n+p-2)/2+((n-m+1)/2)(m+p-1)/2\\ +((n+m-1)/2)(m+p-3)/2-((p+1)/2)(m+n+1)/2\\ -((p-1)/2)(m+n-1)/2=mn-n-(3m)/2\end{array}$$

$$m(n+p-2)/2+n(m+p-2)/2-(p-1)(m+n)/2-(m+n-2)/2=0$$

$$\begin{array}[]{l}m(n+p-2)/2+((n-m+1)/2)(m+p-1)/2\\ +((n+m-1)/2)(m+p-3)/2-(p/2)(m+n+1)/2\\ -((p-2)/2)(m+n-1)/2-(m+n-3)/2=mn-n-(3m-3)/2=0\end{array}$$

$$\begin{array}[]{l}m(n+p-1)/2+((n-m+1)/2)(m+p-2)/2\\ +((n+m-1)/2)(m+p-4)/2-((p-n-1)/2)(m+n+1)/2\\ -((p+n+1)/2)(m+n-1)/2=mn-m-n+1=0\end{array}$$

$$\begin{array}[]{l}m(n+p-3)/2+n(m+p-3)/2-(p-(n+3)/2)(m+n)/2\\ -((n+3)/2)(m+n-2)/2=(2mn-3m-2n+3)/2=0\end{array}$$

Full text

Keywords: domination in graphs, edge domination, signed edge domination

MSC 2000: 05C69

1 Introduction

Let $G$ be a simple non-empty graph with vertex set $V(G)$ and edge set $E(G)$. We use [4] for terminology and notation not defined here. Two edges $e_{1}$, $e_{2}$ of $G$ are called adjacent if they are distinct and have a common end-vertex. The open neighborhood $N_{G}(e)$ of an edge $e\in E(G)$ is the set of all edges adjacent to $e$. Its closed neighborhood is $N_{G}[e]=N_{G}(e)\cup\{e\}$. For a function $f:E(G)\rightarrow\{-1,1\}$ and a subset $S$ of $E(G)$ we define $f(S)=\sum_{x\in S}f(x)$. If $S=N_{G}[e]$ for some $e\in E$, then we denote $f(S)$ by $f[e]$. The weight of vertex $v\in V(G)$ is defined by $f(v)=\sum_{e\in E(v)}f(e)$, where $E(v)$ is the set of all edges at vertex $v$. A function $f:E(G)\rightarrow\{-1,1\}$ is called a signed edge dominating function (SEDF) of $G$ if $f[e]\geq 1$ for each edge $e\in E(G)$. The SEDF of a graph was first defined in [5]. The weight of $f$, denoted $w(f)$, is defined to be $w(f)=\sum_{e\in E(G)}f(e)$. The signed edge domination number (SEDN) $\gamma^{\prime}_{s}(G)$ is defined as $\gamma^{\prime}_{s}(G)$ = min{$w(f)\mid f$ is an SEDF of $G$}. An SEDF $f$ is called a $\gamma^{\prime}_{s}(G)$-function if $\omega(f)=\gamma^{\prime}_{s}(G)$. In [5] it was conjectured that $\gamma_{s}^{\prime}(G)\leq|V(G)|-1$ for every graph $G$ of order at least 2.

The signed edge domination numbers of the complete graph $K_{n}$ and the complete bipartite graph $K_{m,n}$ were determined in [6] and [1], respectively. In [3], the signed edge domination number of $K_{m,n,p}$ was calculated when $1\leq m\leq n\leq p\leq m+n$. For completeness, we state the main theorem of [3].

Theorem 1**.**

Let $m,n$ and $p$ be positive integers and $m\leq n\leq p\leq m+n$. Let $(m,n,p)\not\in\{(1,1,1),$ $(2,3,5)\}$.

• A.

Let $m$, $n$ and $p$ be even.

If $m+n+p\equiv 0$ (mod 4), then $\gamma_{s}^{\prime}(K_{m,n,p})=(m+n+p)/2$. 2. 2.

If $m+n+p\equiv 2$ (mod 4), then $\gamma_{s}^{\prime}(K_{m,n,p})=(m+n+p+2)/2$.

• B.

Let $m,n$ and $p$ be odd.

If $m+n+p\equiv 1$ (mod 4), then $\gamma_{s}^{\prime}(K_{m,n,p})=(m+n+p+1)/2$. 2. 2.

If $m+n+p\equiv 3$ (mod 4), then $\gamma_{s}^{\prime}(K_{m,n,p})=(m+n+p+3)/2$.

• C.

Let $m,n$ be odd and $p$ be even or $m,n$ be even and $p$ be odd.

If $m+n\equiv 0$ (mod 4), then $\gamma_{s}^{\prime}(K_{m,n,p})=(m+n)/2+p+1$. 2. 2.

If $m+n\equiv 2$ (mod 4), then $\gamma_{s}^{\prime}(K_{m,n,p})=(m+n)/2+p$.

• D.

Let $m,p$ be odd and $n$ be even or $m,p$ be even and $n$ be odd.

If $m+p\equiv 0$ (mod 4), then $\gamma_{s}^{\prime}(K_{m,n,p})=(m+p)/2+n+1$. 2. 2.

If $m+p\equiv 2$ (mod 4), then $\gamma_{s}^{\prime}(K_{m,n,p})=(m+p)/2+n$.

• E.

Let $n,p$ be odd and $m$ be even or $n,p$ be even and $m$ be odd.

If $n+p\equiv 0$ (mod 4), then $\gamma_{s}^{\prime}(K_{m,n,p})=(n+p)/2+m+1$. 2. 2.

If $n+p\equiv 2$ (mod 4), then $\gamma_{s}^{\prime}(K_{m,n,p})=(n+p)/2+m$.

In addition, $\gamma_{s}^{\prime}(K_{1,1,1})=1$ and $\gamma_{s}^{\prime}(K_{2,3,5})=5$. **

In this paper, we find the signed edge domination number of the complete tripartite graph $K_{m,n,p}$, when $m,n\geq 1$ and $p\geq m+n$. In Section 2, we present some crucial results which will be employed in the rest of this paper. In Section 3, we prove that if $f$ is a $\gamma_{s}^{\prime}(K_{m,n,p})$-function, then $f(w)\geq-1$ for every vertex $w$ in the largest partite set of $K_{m,n,p}$. In Section 4, we present general constructions for the SEDFs of $K_{m,n,p}$ with minimum weight. In Section 5, we calculate the signed edge domination numbers of $K_{1,n,p}$ and $K{2,2,p}$, where $p$ is even. These cases do not follow the constructions given in Section 4. In addition, we notice that $\gamma_{s}^{\prime}(K_{1,n,n+3})=2n+3$ when $n$ is odd. So there is an infinite family of graphs which achieve the upper bound given in Xu’s conjecture (see [5]). The main theorem of this paper is presented in Section 6.

2 Preliminary results

Let $f$ be a SEDF of $G$. An edge $e\in E(G)$ is called a negative edge (positive edge) if $f(e)=-1$ ($f(e)=1$). Let $uv$ be an edge of $G$ and suppose that $x,y$ are the number of negative edges at vertices $u$ and $v$, respectively. Then

[TABLE]

Hence, if $e=uv$, then $f(u)+f(v)\geq 0$ for every edge $e\in E(G)$. In addition, if $f(u)+f(v)=0$ or $1$, then $f(e)=-1$.

The following result can be found in [2]. Since there are typos in the proof given in [2] we modify the proof and present it here.

Lemma 2**.**

Let $G$ be a graph, $u,v\in V(G)$ and $N(u)\setminus\{v\}=N(v)\setminus\{u\}$. Let $f$ be a SEDF of $G$. Then there exists a SEDF of $G$, say $g$, with $w(g)=w(f)$ such that the difference between the number of negative edges at $u$ and at $v$ is at most 1.**

Proof.

Let $V(G)=\{v_{1}=u,v_{2}=v,v_{3},\ldots,v_{n}\}$ and let $x_{i}$, $1\leq i\leq n$, be the number of edges $e$ at $v_{i}$ with $f(e)=-1$.

If $x_{1}\leq x_{2}-2$, then there exists a vertex $v_{\ell}$ such that $f(v_{1}v_{\ell})=1$ and $f(v_{2}v_{\ell})=-1$ for some $\ell\in\{3,4,\ldots,n\}$. Define $g:E(G)\rightarrow\{-1,1\}$ by $g(v_{1}v_{\ell})=-1$, $g(v_{2}v_{\ell})=1$ and $g(e)=f(e)$ for $e\in E(G)\setminus\{v_{1}v_{\ell},v_{2}v_{\ell}\}$. Let $y_{i}$, $1\leq i\leq n$, be the number of edges $e$ at $v_{i}$ with $g(e)=-1$. Obviously, $y_{1}=x_{1}+1$, $y_{2}=x_{2}-1$ and $y_{i}=x_{i}$ for $3\leq i\leq n$. In addition, $w(g)=w(f)$. We prove that $g$ is a SEDF of $G$.

If $v_{1}v_{j}\in E(G)$ and $j\not\in\{1,2,\ell\}$

[TABLE]

[TABLE]

Similarly, for $e\in E(G)\setminus\{v_{1}v_{j},v_{2}v_{j}\mid j\not\in\{1,2,\ell\}\}$, we obtain $g[e]=f[e]\geq k$. In addition, if $v_{1}v_{2}\in E(G)$, then

[TABLE]

Hence, $g$ is a SEDF of $G$. If $g$ satisfies the required condition, the proof is complete. Otherwise, by repeating this process we can obtain the required function. ∎

Corollary 3**.**

Let $G$ be a complete multipartite graph. There exists a $\gamma_{sk}^{\prime}(G)$-function $f$ such that the difference between the number of negative edges at every two vertices in the same partite set is at most 1.**

3 SEDFs of complete tripartite graphs with vertices of negative weight

Consider the complete tripartite graph $K_{m,n,p}$ with partite sets $U,V$ and $W$. Throughout this section we assume $|U|=m$, $|V|=n$ and $|W|=p$, where $1\leq m\leq n\leq p$. In this section, we study SEDFs $f$ of $K_{m,n,p}$ such that $f(w)<0$ for some $w\in W$.

Lemma 4**.**

Let $m,n,p$ be all even or all odd and $1\leq m\leq n\leq p$. If $f$ is a $\gamma_{s}^{\prime}(K_{m,n,p})$-function, then $f(w)\geq 0$ for every vertex of $w\in W$. **

Proof.

The proof is by contradiction. By Corollary 3, we may assume that the difference between the number of negative edges at every two vertices in the same partite set of $K_{m,n,p}$ is at most 1. Assume $f(w)=-2k$, where $1\leq k\leq(m+n)/2$, for some $w\in W$, and $f(w^{\prime})\geq-2k$ for all $w^{\prime}\in W$. Then there are $(m+n+2k)/2$ negative edges and $(m+n-2k)/2$ positive edges at $w$. Therefore the weight of $(m+n+2k)/2$ vertices in $U\cup V$ must be at least $2k$ and the weight of $(m+n-2k)/2$ vertices in $U\cup V$ must be at least $2k+2$. Since $m\leq n$ and $f$ is a $\gamma_{s}^{\prime}(K_{m,n,p})$-function, we can assume $f(u)=2k$ for every $u\in U$. So there are $(n+p-2k)/2$ negative edges at every vertex $u\in U$. Let $U\cup V_{1}$, where $V_{1}\subseteq V$, consist of vertices of weight $2k$ and let $V_{2}=V\setminus V_{1}$ consist of vertices of weight $2k+2$. Since $(m+n+2k)/2>m$ and $f$ is a $\gamma_{s}^{\prime}(K_{m,n,p})$-function, it follows that there is a vertex $v\in V_{1}$ of weight $2k$. Indeed, $|V_{1}|=(n-m+2k)/2$. Therefore there are $(m+p-2k)/2$ negative edges at $v$. Let $W_{1}\subseteq W$ consist of vertices of weight $-2k$. Since every vertex in $V_{1}$ must be joined to every vertex in $W_{1}$ with a negative edge by (1), it follows that $|W_{1}|\leq(m+p-2k)/2$. Hence, $|W\setminus W_{1}|\geq(p-m+2k)/2$ and every vertex in this set has weight $-2k+2$. Note that $(n+p-2k)/2-(m+p-2k)/2=(n-m)/2$. Let $W_{2}$ be a subset of $W\setminus W_{1}$ with $(n-m)/2$ vertices and the edges between $W_{2}$ and $U$ are all negative edges. Let $W_{3}=W\setminus(W_{1}\cup W_{2})$. Then

[TABLE]

Now let $w^{\prime}\in W_{3}$. Then the edges between $w^{\prime}$ and $U\cup V_{1}$ are all positive edges. Therefor the number of negative edges at $w^{\prime}$ is at most $(n+m-2k)/2$. On the other hand, for every vertex $w\in W_{1}$ there are $(n+m+2k)/2$ negative edges. Since $k\geq 1$, the difference between the number of negative edges at $w$ and at $w^{\prime}$ is $2k\geq 2$, which is a contradiction. ∎

The proof of the following result is similar to the proof of Lemma 4.

Lemma 5**.**

Let $m,n$ be even and $p$ be odd or $m,n$ be odd and $p$ be even, where $1\leq m\leq n<p$. If $f$ is a $\gamma_{s}^{\prime}(K_{m,n,p})$-function, then $f(w)\geq 0$ for every vertex of $w\in W$.

Proof.

The proof is by contradiction. Let $m,n$ be even and $p$ be odd. (The case $m,n$ odd and $p$ even is similar.) By Corollary 3, we may assume that the difference between the number of negative edges at every two vertices in the same partite set of $K_{m,n,p}$ is at most 1. Assume $f(w)=-2k$, where $1\leq k\leq(m+n)/2$, for some $w\in W$, and $f(w^{\prime})\geq-2k$ for all $w^{\prime}\in W$. Then there are $(m+n+2k)/2$ negative edges and $(m+n-2k)/2$ positive edges at $w$. Therefore the weight of $(m+n+2k)/2$ vertices in $U\cup V$ must be at least $2k+1$ and the weight of $(m+n-2k)/2$ vertices in $U\cup V$ must be at least $2k+3$. Since $m\leq n$ and $f$ is a $\gamma_{s}^{\prime}(K_{m,n,p})$-function, we can assume $f(u)=2k+1$ for every $u\in U$. So there are $(n+p-2k-1)/2$ negative edges at every vertex $u\in U$. Let $U\cup V_{1}$, where $V_{1}\subseteq V$, consist of vertices of weight $2k+1$ and let $V_{2}=V\setminus V_{1}$ consist of vertices of weight $2k+3$. Since $(m+n+2k)/2>m$ and $f$ is a $\gamma_{s}^{\prime}(K_{m,n,p})$-function, it follows that there is a vertex $v\in V_{1}$ of weight $2k+1$. Indeed, $|V_{1}|=(n-m+2k)/2$. Therefore there are $(m+p-2k-1)/2$ negative edges at $v$. Let $W_{1}\subseteq W$ consist of vertices of weight $-2k$. Since every vertex in $V_{1}$ must be joined to every vertex in $W_{1}$ with a negative edge by (1), it follows that $|W_{1}|\leq(m+p-2k-1)/2$. Hence, $|W\setminus W_{1}|\geq(p-m+2k+1)/2$ and every vertex in this set has weight $-2k+2$. Note that $(n+p-2k-1)/2-(m+p-2k-1)/2=(n-m)/2$. Let $W_{2}$ be a subset of $W\setminus W_{1}$ with $(n-m)/2$ vertices and the edges between $W_{2}$ and $U$ are all negative edges. Let $W_{3}=W\setminus(W_{1}\cup W_{2})$. Then

[TABLE]

Now let $w^{\prime}\in W_{3}$. Then the edges between $w^{\prime}$ and $U\cup V_{1}$ are all positive edges. Therefor the number of negative edges at $w^{\prime}$ is at most $(n+m-2k)/2$. On the other hand, for every vertex $w\in W_{1}$ there are $(n+m+2k)/2$ negative edges. Since $k\geq 1$, the difference between the number of negative edges at $w$ and at $w^{\prime}$ is $2k\geq 2$, which is a contradiction. ∎

Lemma 6**.**

Let $m$ be odd and $n,p$ be even or $m,p$ be even and $n$ be odd, where $1\leq m<n\leq p$. If $f$ is a $\gamma_{s}^{\prime}(K_{m,n,p})$-function, then $f(w)\geq-1$ for every vertex of $w\in W$. **

Proof.

The proof is by contradiction. Let $m$ be odd and $n,p$ be even. (The case $m,p$ even and $n$ odd is similar.) By Corollary 3, we may assume that the difference between the number of negative edges at every two vertices in the same partite set of $K_{m,n,p}$ is at most 1. Assume $f(w)=-2k-1$, where $1\leq k\leq(m+n-1)/2$, for some $w\in W$, and $f(w^{\prime})\geq-2k-1$ for all $w^{\prime}\in W$. Then there are $(m+n+2k+1)/2$ negative edges and $(m+n-2k-1)/2$ positive edges at $w$. Therefore the weight of $(m+n+2k+1)/2$ vertices in $U\cup V$ must be at least $2k+1$ and the weight of $(m+n-2k-1)/2$ vertices in $U\cup V$ must be at least $2k+3$. Since $m<n$ and $f$ is a $\gamma_{s}^{\prime}(K_{m,n,p})$-function, we can assume $f(u)=2k+2$ for every vertex $u\in U$. So there are $(n+p-2k-2)/2$ negative edges at every vertex $u\in U$. Let $V_{1}\subset V$ consist of vertices of weight $2k+1$ and let $V_{2}=V\setminus V_{1}$ consist of vertices of weight $2k+3$. Since $(m+n+2k+1)/2>m$ and $f$ is a $\gamma_{s}^{\prime}(K_{m,n,p})$-function, it follows that there is a vertex $v\in V_{1}$ of weight $2k+1$. Indeed, $|V_{1}|=(n-m+2k+1)/2$. Therefore there are $(m+p-2k-1)/2$ negative edges at $v$. Let $W_{1}\subseteq W$ consist of vertices of weight $-2k-1$. Since every vertex in $V_{1}$ must be joined to every vertex in $W_{1}$ with a negative edge by (1), it follows that $|W_{1}|\leq(m+p-2k-1)/2$. Hence, $|W\setminus W_{1}|\geq(p-m+2k+1)/2$ and every vertex in this set has weight $-2k+1$.

Note that $(n+p-2k-2)/2-(m+p-2k-1)/2=(n-m-1)/2$. Let $W_{2}$ be a subset of $W\setminus W_{1}$ with $(n-m-1)/2$ vertices and the edges between $W_{2}$ and $U$ are all negative edges. Let $W_{3}=W\setminus(W_{1}\cup W_{2})$. Then

[TABLE]

Now let $w^{\prime}\in W_{3}$. Then the edges between $w^{\prime}$ and $U\cup V_{1}$ are all positive edges. Therefor the number of negative edges at $w^{\prime}$ is at most $(n+m-2k-1)/2$. On the other hand, for every vertex $w\in W_{1}$ there are $(n+m+2k+1)/2$ negative edges. Since $k\geq 1$, the difference between the number of negative edges at $w$ and at $w^{\prime}$ is $2k+1\geq 3$, which is a contradiction. ∎

The proof of the following result is similar to the proof of Lemma 6.

Lemma 7**.**

Let $m,p$ be odd and $n$ be even or $m$ be even and $n,p$ be odd, where $1\leq m<n\leq p$. If $f$ is a $\gamma_{s}^{\prime}(K_{m,n,p})$-function, then $f(w)\geq-1$ for every vertex of $w\in W$.

Proof.

The proof is by contradiction. Let $m,p$ be odd and $n$ be even. (The case $m$ even and $n,p$ odd is similar.) By Corollary 3, we may assume that the difference between the number of negative edges at every two vertices in the same partite set of $K_{m,n,p}$ is at most 1. Assume $f(w)=-2k-1$, where $1\leq k\leq(m+n-1)/2$, for some $w\in W$, and $f(w^{\prime})\geq-2k-1$ for all $w^{\prime}\in W$. Then there are $(m+n+2k+1)/2$ negative edges and $(m+n-2k-1)/2$ positive edges at $w$. Therefore the weight of $(m+n+2k+1)/2$ vertices in $U\cup V$ must be at least $2k+1$ and the weight of $(m+n-2k-1)/2$ vertices in $U\cup V$ must be at least $2k+3$. Since $m<n$ and $f$ is a $\gamma_{s}^{\prime}(K_{m,n,p})$-function, we can assume $f(u)=2k+1$ for every vertex $u\in U$. So there are $(n+p-2k-1)/2$ negative edges at every vertex $u\in U$. Let $V_{1}\subset V$ consist of vertices of weight $2k+2$ and let $V_{2}=V\setminus V_{1}$ consist of vertices of weight $2k+4$. Since $(m+n+2k+1)/2>m$ and $f$ is a $\gamma_{s}^{\prime}(K_{m,n,p})$-function, it follows that there is a vertex $v\in V_{1}$ of weight $2k+2$. Indeed, $|V_{1}|=(n-m+2k+1)/2$. Therefore there are $(m+p-2k-2)/2$ negative edges at $v$. Let $W_{1}\subseteq W$ consist of vertices of weight $-2k-1$. Since every vertex in $V_{1}$ must be joined to every vertex in $W_{1}$ with a negative edge by (1), it follows that $|W_{1}|\leq(m+p-2k-2)/2$. Hence, $|W\setminus W_{1}|\geq(p-m+2k+2)/2$ and every vertex in this set has weight $-2k+1$.

Note that $(n+p-2k-1)/2-(m+p-2k-2)/2=(n-m+1)/2$. Let $W_{2}$ be a subset of $W\setminus W_{1}$ with $(n-m+1)/2$ vertices and the edges between $W_{2}$ and $U$ are all negative edges. Let $W_{3}=W\setminus(W_{1}\cup W_{2})$. Then

[TABLE]

Now let $w^{\prime}\in W_{3}$. Then the edges between $w^{\prime}$ and $U\cup V_{1}$ are all positive edges. Therefor the number of negative edges at $w^{\prime}$ is at most $(n+m-2k-1)/2$. On the other hand, for every vertex $w\in W_{1}$ there are $(n+m+2k+1)/2$ negative edges. Since $k\geq 1$, the difference between the number of negative edges at $w$ and at $w^{\prime}$ is $2k+1\geq 3$, which is a contradiction. ∎

4 The SEDN of $K_{m,n,p}$

Consider the complete tripartite graph $K_{m,n,p}$ whose partite sets are $U,V$ and $W$. Throughout this section we assume $|U|=m$, $|V|=n$ and $|W\mid=p$, where $m,n$ and $p$ are positive integers, $m\leq n$ and $p\geq m+n$ In this section we compute the signed edge domination number of $K_{m,n,p}$, where $m\geq 2$ and $(m,n)\neq(2,2)$ if $p$ is odd.

Theorem 8**.**

Let $m,n$ and $p$ be even and $p\geq m+n$. Then $\gamma_{s}^{\prime}(K_{m,n,p})=m+n$.**

Proof.

Consider the graph $K_{m,n,p}$ whose partite sets are $U,V$ and $W$. By assumption

[TABLE]

is even. First we label $(mn-m-n)/2$ edges between $U$ and $V$ with $-1$ in the following way. Label an edge $uv$, where $u\in U$ and $v\in V$, with $-1$ if

the total number of negative edges between $U$ and $V$ is less than $(mn-m-n)/2$, 2. 2.

the number of negative edges at $u$ is less than $(n+p-2)/2$, 3. 3.

the number of negative edges at $v$ is less than $(m+p-2)/2$, 4. 4.

the number of negative edges at $u$ is less than or equal to the number of negative edges at $u^{\prime}$ for every $u^{\prime}\in(U\setminus\{u\})$, and 5. 5.

the number of negative edges at $v$ is less than or equal to the number of negative edges at $v^{\prime}$ for every $v^{\prime}\in(V\setminus\{v\})$.

Then we label $p(m+n)/2$ edges between $U\cup V$ and $W$ with $-1$ in a similar fashion described above. An edge $rw$, where $r\in(U\cup V)$ and $w\in W$ is labelled by $-1$ if

the number of negative edges between $U\cup V$ and $W$ is less than $p(m+n)/2$, 2. 2.

the number of negative edges at $r$ is less than $(n+p-2)/2$ if $r\in U$, 3. 3.

the number of negative edges at $r$ is less than $(m+p-2)/2$ if $r\in V$, 4. 4.

the number of negative edges at $w$ is less than $(m+n)/2$, 5. 5.

the number of negative edges at $r$ is less than or equal to the number of negative edges at $u$ for every $u\in(U\setminus\{r\})$ if $r\in U$, 6. 6.

the number of negative edges at $r$ is less than or equal to the number of negative edges at $v$ for every $v\in(V\setminus\{r\})$ if $r\in V$, 7. 7.

the number of negative edges at $w$ is less than or equal to the number of negative edges at $w^{\prime}$ for every $w^{\prime}\in(W\setminus\{w\})$.

Then there are exactly $((p+n)/2)-1$, $((p+m)/2)-1$ and $(m+n)/2$ negative edges at every vertex in $U,V$ and $W$, respectively. Label the remaining edges of $K_{m,n,p}$ by $+1$. Then all vertices in $U\cup V$ have weight $2$ and all the vertices in $W$ have weight zero. Hence, this labeling defines a SEDF $f$ of $K_{m,n,p}$ by (1), and $\omega(f)=m+n$. Note that since the weight of every vertex of $W$ is zero, no vertices in $U\cup V$ can have weight zero by (1). Now by Lemma 4 and the facts that $f(W)=0$ and $f(r)=2$ for every $r\in U\cup V$, it follows that $\gamma_{s}^{\prime}(K_{m,n,p})=m+n$. ∎

Theorem 9**.**

Let $m,n$ and $p$ be odd, $m,n\geq 3$ and $p\geq m+n+1$. Then $\gamma_{s}^{\prime}(K_{m,n,p})=m+n+1$. **

Proof.

Consider the graph $K_{m,n,p}$ whose partite sets are $U,V$ and $W$. By assumption,

[TABLE]

is odd. In addition, $(n+p-2)/2$, $(m+p-2)/2$ and $(m+n)/2$ are all odd, or two are even and one is odd. Hence, there is no graph whose $m$ vertices have degree $(n+p-2)/2$, $n$ vertices have degree $(n+p-2)/2$ and $p$ vertices have degree $(m+n)/2$.

On the other hand,

[TABLE]

is an even number. We label $(mn-m-n+1)/2$ edges between $U$ and $V$ and $(p-1)(m+n)/2+(m+n-2)/2$ edges between $U\cup V$ and $W$ with $-1$ in a similar fashion described in Theorem 8. Then there are $(n+p-2)/2$ negative edges at each vertex of $U$, $(m+p-2)/2$ negative edges at each vertex of $V$ and $(m+n)/2$ negative edges at each vertex of $W$ except one vertex which is incident with $(m+n-2)/2$ negative edges. We label the remaining edges of $K_{m,n,p}$ with $+1$. Then the weight of vertices in $U\cup V$ are all $2$ and the weight of vertices in $W$ are all zero except one vertex of $W$ whose weight is $2$. Hence, this labeling defines a SEDF $f$ of $K_{m,n,p}$ by (1), and $\omega(f)=m+n+1$.

Note that since the weight of every vertex of $W$ is zero, no vertices in $U\cup V$ can have weight zero by (1). Now by Lemma 4 and the facts that $f(W)=2$ and $f(r)=2$ for every $r\in U\cup V$, it follows that $\gamma_{s}^{\prime}(K_{m,n,p})=m+n+1$. ∎

Theorem 10**.**

Let $m,n\geq 3$ be odd, $p$ be even and $p\geq m+n$. Then

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+3n+2}{2}$ if $m+n\equiv 0\pmod{4}$, 2. 2.

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+3n}{2}$ if $m+n\equiv 2\pmod{4}$.

Proof.

Consider the graph $K_{m,n,p}$ whose partite sets are $U,V$ and $W$.

Case 1. $m+n\equiv 0\pmod{4}$.

By assumption,

[TABLE]

is even. Label $(1/2)[mn-1-(3m+3n)/2]$ edges between $U$ and $V$ and $p(m+n)/2$ edges between $U\cup V$ and $W$ with $-1$ in a similar fashion described in the proof of Theorem 8. Then every vertex in $U$ is incident with $(n+p-3)/2$ negative edges except one vertex which is incident with $(n+p-5)/2$ negative edges. Every vertices in $V$ is incident with $(m+p-3)/2$ negative edges and every vertex in $W$ is incident with $(m+n)/2$ negative edges. Label the remaining edges of $K_{n,m,p}$ by $+1$. Then the weight of vertices in $U\cup V$ are all $3$ except one vertex of $U$ whose weight is $5$, and the weight of vertices in $W$ are all zero. Hence, this labeling defines a SEDF $f$ with $w(f)=(3m+3n+2)/2$.

Note that since the weight of every vertex of $W$ is zero, no vertices in $U\cup V$ can have weight one by (1). Now by Lemma 5 and the facts that $f(W)=0$ and $f(r)=3$ for every $r\in U\cup V$ except one vertex which has weight 5, it follows that $\gamma_{s}^{\prime}(K_{m,n,p})=(3m+3n+2)/2$.

Case 2. $m+n\equiv 2\pmod{4}$.

By assumption,

[TABLE]

is even. Label $(1/2)[mn-(3m+3n)/2]$ edges between $U$ and $V$ and $p(m+n)/2$ edges between $U\cup V$ and $W$ with $-1$ in a similar fashion described in the proof of Theorem 8. Then every vertex in $U$ is incident with $(n+p-3)/2$ negative edges, every vertex in $V$ is incident with $(m+p-3)/2$ negative edges and every vertex in $W$ is incident with $(m+n)/2$ negative edges. Label the remaining edges of $K_{n,m,p}$ with $+1$. Then the weight of vertices in $U\cup V$ are all $3$ and the weight of vertices in $W$ are all zero. Hence, this labeling defines a SEDF $f$ with $w(f)=(3m+3n)/2$.

Note that since the weight of every vertex of $W$ is zero, no vertices in $U\cup V$ can have weight one by (1). Now by Lemma 5 and the facts that $f(W)=0$ and $f(r)=3$ for every $r\in U\cup V$, it follows that $\gamma_{s}^{\prime}(K_{m,n,p})=(3m+3n)/2$. ∎

Theorem 11**.**

Let $m$ and $n$ be even, $(m,n)\neq(2,2)$, $p$ be odd and $p\geq m+n+1$. Then

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+3n}{2}$ if $m+n\equiv 0\pmod{4}$, 2. 2.

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+3n+2}{2}$ if $m+n\equiv 2\pmod{4}$.

Proof.

Consider the graph $K_{m,n,p}$ whose partite sets are $U,V$ and $W$.

Case 1. $m+n\equiv 0\pmod{4}$.

By assumption,

[TABLE]

is even. Label $(1/2)[mn-3(m+n)/2]$ edges between $U$ and $V$ and $p(m+n)/2$ edges between $U\cup V$ and $W$ with $-1$ as described in the proof of Theorem 8. Then every vertex in $U$ is incident with $(n+p-3)/2$ negative edges, every vertex in $V$ is incident with $(m+p-3)/2$ negative edges and every vertex in $W$ is incident with $(m+n)/2$ negative edges. Label the remaining edges of $K_{n,m,p}$ by $+1$. Then the weight of vertices in $U\cup V$ are all $3$ and the weight of vertices in $W$ are all zero. Hence, this labeling defines a SEDF $f$ with $w(f)=(3m+3n)/2$.

Note that since the weight of every vertex of $W$ is zero, no vertices in $U\cup V$ can have weight one by (1). Now by Lemma 5 and the facts that $f(W)=0$ and $f(r)=3$ for every $r\in U\cup V$, it follows that $\gamma_{s}^{\prime}(K_{m,n,p})=(3m+3n)/2$.

Case 2. $m+n\equiv 2\pmod{4}$.

By assumption,

[TABLE]

is even. Label $(1/2)[mn-1-3(m+n)/2]$ edges between $U$ and $V$ and $p(m+n)/2$ edges between $U\cup V$ and $W$ with $-1$ as described in the proof of Theorem 8. Then every vertex in $U$ is incident with $(n+p-3)/2$ negative edges except one vertex which is incident with $(n+p-5)/2$ negative edges, every vertex in $V$ is incident with $(m+p-3)/2$ negative edges, and every vertex in $W$ is incident with $(m+n)/2$ negative edges. Label the remaining edges of $K_{n,m,p}$ with $+1$. Then the weight of vertices in $U\cup V$ are all $3$ except one vertex whose weight is 5 and the weight of vertices in $W$ are all zero. Hence, this labeling defines a SEDF $f$ with $w(f)=(3m+3n+2)/2$.

Note that since the weight of every vertex of $W$ is zero, no vertices in $U\cup V$ can have weight one by (1). Now by Lemma 5 and the facts that $f(W)=0$, $f(u)=3$ for every vertex $u$ in $U$ except one vertex whose weight is 5, and $f(v)=3$ for every $v\in V$, it follows that $\gamma_{s}^{\prime}(K_{m,n,p})=(3m+3n+2)/2$. ∎

Theorem 12**.**

Let $m$ be odd, $n,p$ be even, $3\leq m<n$ and $p\geq m+n+1$. Then

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+2n+1}{2}$ if $m\equiv 1\pmod{4}$, 2. 2.

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+2n-1}{2}$ if $m\equiv 3\pmod{4}$

Proof.

Consider the graph $K_{m,n,p}$ whose partite sets are $U,V$ and $W$.

Case 1. $m\equiv 1\pmod{4}$.

By assumption,

[TABLE]

is even. Partition $V$ into $V_{1}$ and $V_{2}$ with $|V_{1}\mid=(n-m-1)/2$ and $|V_{2}\mid=(n+m+1)/2$. Also partition $W$ into $W_{1}$ and $W_{2}$ with $|W_{1}\mid=|W_{2}\mid$. In a similar fashion described in the proof of Theorem 8, label $(1/2)[mn-n-(3m+1)/2]$ edges between $U$ and $V$ and $(p/2)(m+n+1)/2+(p/2)(m+n-1)/2$ edges between $U\cup V$ and $W$ with $-1$ such that the edges between $V_{1}$ and $W_{1}$ are all negative edges. In addition, every vertex in $U$ is incident with $(n+p-2)/2$ negative edges, every vertex in $V_{1},V_{2}$ is incident with $(m+p-1)/2$ and $(m+p-3)/2$ negative edges, respectively, and every vertex in $W_{1},W_{2}$ is incident with $(m+n+1)/2$ and $(m+n-1)/2$ negative edges, respectively. Label the remaining edges of $K_{n,m,p}$ by $+1$. Then the weight of vertices in $U$ are all $2$, the weight of vertices in $V_{1},V_{2}$ are all $1,3$, respectively, and the weight of vertices in $W_{1}$ are all $-1$ and in $W_{2}$ are $+1$. Hence, this labeling defines a SEDF $f$ with $w(f)=(3m+2n+1)/2$.

Note that since the weight of some vertices in $W$ is $-1$, no vertices in $U\cup V$ can have weight zero or $-1$ by (1). Now by Lemma 6 and the facts that $f(W)=0$, $f(u)=2$ for every $u\in U$, $f(v)=1$ for every $v\in V_{1}$ and $f(v)=3$ for every $v\in V_{2}$, it follows that $\gamma_{s}^{\prime}(K_{m,n,p})=(3m+2n+1)/2$.

Case 2. $m\equiv 3\pmod{4}$.

By assumption,

[TABLE]

is even. Partition $V$ into $V_{1}$ and $V_{2}$ with $|V_{1}\mid=(n-m+1)/2$ and $|V_{2}\mid=(n+m-1)/2$. Also partition, $W$ into $W_{1}$ and $W_{2}$ with $|W_{1}\mid=|W_{2}\mid$. In a similar fashion described in the proof of Theorem 8, label $(1/2)[mn-n-(3m-1)/2]$ edges between $U$ and $V$ and $(p/2)(m+n+1)/2+(p/2)(m+n-1)/2$ edges between $U\cup V$ and $W$ with $-1$ such that the edges between $V_{1}$ and $W_{1}$ are all negative edges. In addition, every vertex in $U$ is incident with $(n+p-2)/2$ negative edges, every vertex in $V_{1},V_{2}$ is incident with $(m+p-1)/2$ and $(m+p-3)/2$ negative edges, respectively, and every vertex in $W_{1},W_{2}$ is incident with $(m+n+1)/2$ and $(m+n-1)/2$ negative edges, respectively. Label the remaining edges of $K_{n,m,p}$ by $+1$. Then the weight of vertices in $U$ are all $2$, the weight of vertices in $V_{1},V_{2}$ are all $1,3$, respectively, and the weight of vertices in $W_{1}$ are all $-1$ and in $W_{2}$ are $+1$. Hence, this labeling defines a SEDF $f$ with $w(f)=(3m+2n-1)/2$.

Note that since the weight of some vertices in $W$ is $-1$, no vertices in $U\cup V$ can have weight zero or $-1$ by (1). Now by Lemma 6 and the facts that $f(W)=0$, $f(u)=2$ for every $u\in U$, $f(v)=1$ for every $v\in V_{1}$ and $f(v)=3$ for every $v\in V_{2}$, it follows that $\gamma_{s}^{\prime}(K_{m,n,p})=(3m+2n-1)/2$. ∎

Theorem 13**.**

Let $m,p$ be even, $n$ be odd, $m<n$ and $p\geq m+n+1$. Then

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{2m+3n+1}{2}$ if $n\equiv 1\pmod{4}$, 2. 2.

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{2m+3n-1}{2}$ if $n\equiv 3\pmod{4}$.

Proof.

Consider the graph $K_{m,n,p}$ whose partite sets are $U,V$ and $W$.

Case 1. $n\equiv 1\pmod{4}$.

By assumption,

[TABLE]

is even. Partition $V$ into $V_{1}$ and $V_{2}$ with $|V_{1}\mid=(n-m-1)/2$ and $|V_{2}\mid=(n+m+1)/2$. Also partition, $W$ into $W_{1}$ and $W_{2}$ with $|W_{1}\mid=|W_{2}\mid$. In a similar fashion described in the proof of Theorem 8, label $(1/2)[mn-n-(3n+1)/2]$ edges between $U$ and $V$ and $(p/2)(m+n+1)/2+(p/2)(m+n-1)/2$ edges between $U\cup V$ and $W$ with $-1$ such that the edges between $V_{1}$ and $W_{1}$ are all negative edges. In addition, every vertex in $U$ is incident with $(n+p-1)/2$ negative edges, every vertex in $V_{1},V_{2}$ is incident with $(m+p-2)/2$ and $(m+p-4)/2$ negative edges, respectively, and every vertex in $W_{1},W_{2}$ is incident with $(m+n+1)/2$ and $(m+n-1)/2$ negative edges, respectively. Label the remaining edges of $K_{n,m,p}$ by $+1$. Then the weight of vertices in $U$ are all $1$, the weight of vertices in $V_{1},V_{2}$ are all $2,4$, respectively, the weight of vertices in $W_{1}$ are all $-1$ and in $W_{2}$ are $+1$. Hence, this labeling defines a SEDF $f$ with $w(f)=(2m+3n+1)/2$.

Note that since the weight of some vertices in $W$ is $-1$, no vertices in $U\cup V$ can have weight zero or $-1$ by (1). Now by Lemma 6 and the facts that $f(W)=0$, $f(u)=1$ for every $u\in U$, $f(v)=2$ for every $v\in V_{1}$ and $f(v)=4$ for every $v\in V_{2}$, it follows that $\gamma_{s}^{\prime}(K_{m,n,p})=(2m+3n+1)/2$.

Case 2. $n\equiv 3\pmod{4}$.

By assumption,

[TABLE]

is even. Partition $V$ into $V_{1}$ and $V_{2}$ with $|V_{1}\mid=(n-m+1)/2$ and $|V_{2}\mid=(n+m-1)/2$. Also partition, $W$ into $W_{1}$ and $W_{2}$ with $|W_{1}\mid=|W_{2}\mid$. In a similar fashion described in the proof of Theorem 8, label $(1/2)[mn-n-(3n+1)/2]$ edges between $U$ and $V$ and $(p/2)(m+n+1)/2+(p/2)(m+n-1)/2$ edges between $U\cup V$ and $W$ with $-1$ such that the edges between $V_{1}$ and $W_{1}$ are all negative edges. In addition, every vertex in $U$ is incident with $(n+p-1)/2$ negative edges, every vertex in $V_{1},V_{2}$ is incident with $(m+p-2)/2$ and $(m+p-4)/2$ negative edges, respectively, and every vertex in $W_{1},W_{2}$ is incident with $(m+n+1)/2$ and $(m+n-1)/2$ negative edges, respectively. Label the remaining edges of $K_{n,m,p}$ by $+1$. Then the weight of vertices in $U$ are all $1$, the weight of vertices in $V_{1},V_{2}$ are all $2,4$, respectively, and the weight of vertices in $W_{1}$ are all $-1$ and in $W_{2}$ are $+1$. Hence, this labeling defines a SEDF $f$ with $w(f)=(2m+3n-1)/2$.

Note that since the weight of some vertices in $W$ is $-1$, no vertices in $U\cup V$ can have weight zero or $-1$ by (1). Now by Lemma 6 and the facts that $f(W)=0$, $f(u)=1$ for every $u\in U$, $f(v)=2$ for every $v\in V_{1}$ and $f(v)=4$ for every $v\in V_{2}$, it follows that $\gamma_{s}^{\prime}(K_{m,n,p})=(2m+3n-1)/2$. ∎

Theorem 14**.**

Let $m,p$ be odd, $n$ be even, $3\leq m<n$, and $p\geq m+n$. Then

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{2m+3n}{2}$ if $n\equiv 0\pmod{4}$, 2. 2.

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{2m+3n-2}{2}$ if $n\equiv 2\pmod{4}$,

Proof.

Consider the graph $K_{m,n,p}$ whose partite sets are $U,V$ and $W$.

Case 1. $n\equiv 0\pmod{4}$.

By assumption,

[TABLE]

is even. Partition $V$ into $V_{1}$ and $V_{2}$ with $|V_{1}\mid=(n-m-1)/2$ and $|V_{2}\mid=(n+m+1)/2$. Also partition $W$ into $W_{1}$ and $W_{2}$ with $|W_{1}\mid=|W_{2}\mid+1$. In a similar fashion described in the proof of Theorem 8, label $(1/2)[mn-m-(3n+2)/2]$ edges between $U$ and $V$ and $((p+1)/2)(m+n+1)/2+((p-1)/2)(m+n-1)/2$ edges between $U\cup V$ and $W$ with $-1$ such that the edges between $V_{1}$ and $W_{1}$ are all negative edges. In addition, every vertex in $U$ is incident with $(n+p-1)/2$ negative edges, every vertex in $V_{1},V_{2}$ is incident with $(m+p-2)/2$ and $(m+p-4)/2$ negative edges, respectively, and every vertex in $W_{1},W_{2}$ is incident with $(m+n+1)/2$ and $(m+n-1)/2$ negative edges, respectively. Label the remaining edges of $K_{n,m,p}$ by $+1$. Then the weight of vertices in $U$ are all $1$, the weight of vertices in $V_{1},V_{2}$ are all $2,4$, respectively, and the weight of vertices in $W_{1}$ are $-1$ and in $W_{2}$ are $+1$. Hence, this labeling defines a SEDF $f$ with $w(f)=(2m+3n)/2$.

Note that since the weight of some vertices in $W$ is $-1$, no vertices in $U\cup V$ can have weight zero or $-1$ by (1). Now by Lemma 7 and the facts that $f(W)=-1$, $f(u)=1$ for every $u\in U$, $f(v)=2$ for every $v\in V_{1}$ and $f(v)=4$ for every $v\in V_{2}$, it follows that $\gamma_{s}^{\prime}(K_{m,n,p})=(2m+3n)/2$.

Case 2. $n\equiv 2\pmod{4}$.

By assumption,

[TABLE]

is even. Partition $V$ into $V_{1}$ and $V_{2}$ with $|V_{1}\mid=(n-m+1)/2$ and $|V_{2}\mid=(n+m-1)/2$. Also partition $W$ into $W_{1}$ and $W_{2}$ with $|W_{1}\mid=|W_{2}\mid+1$. In a similar fashion described in the proof of Theorem 8, label $(1/2)[mn-m-3n/2]$ edges between $U$ and $V$ and $((p+1)/2)(m+n+1)/2+((p-1)/2)(m+n-1)/2$ edges between $U\cup V$ and $W$ with $-1$ such that the edges between $V_{1}$ and $W_{1}$ are all negative edges. In addition, every vertex in $U$ is incident with $(n+p-1)/2$ negative edges, every vertex in $V_{1},V_{2}$ is incident with $(m+p-2)/2$ and $(m+p-4)/2$ negative edges, respectively, and every vertex in $W_{1},W_{2}$ is incident with $(m+n+1)/2$ and $(m+n-1)/2$ negative edges, respectively. Label the remaining edges of $K_{n,m,p}$ by $+1$. Then the weight of vertices in $U$ are all $1$, the weight of vertices in $V_{1},V_{2}$ are all $2,4$, respectively, and the weight of vertices in $W_{1}$ are all $-1$ and in $W_{2}$ are $+1$. Hence, this labeling defines a SEDF $f$ with $w(f)=(2m+3n-2)/2$.

Note that since the weight of some vertices in $W$ is $-1$, no vertices in $U\cup V$ can have weight zero or $-1$ by (1). Now by Lemma 7 and the facts that $f(W)=-1$, $f(u)=1$ for every $u\in U$, $f(v)=2$ for every $v\in V_{1}$ and $f(v)=4$ for every $v\in V_{2}$, it follows that $\gamma_{s}^{\prime}(K_{m,n,p})=(2m+3n-2)/2$. ∎

Theorem 15**.**

Let $m$ be even, $n,p$ be odd, $m<n$, and $p\geq m+n$. Then there is an SEDF $f$ of $K_{m,n,p}$ with

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+2n}{2}$ if $m\equiv 0\pmod{4}$, 2. 2.

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+2n-2}{2}$ if $m\equiv 2\pmod{4}$.

Proof.

Consider the graph $K_{m,n,p}$ whose partite sets are $U,V$ and $W$.

Case 1. $m\equiv 0\pmod{4}$.

By assumption,

[TABLE]

is even. Partition $V$ into $V_{1}$ and $V_{2}$ with $|V_{1}\mid=(n-m+1)/2$ and $|V_{2}\mid=(n+m-1)/2$. Also partition $W$ into $W_{1}$ and $W_{2}$ with $|W_{1}\mid=|W_{2}\mid-1$. In a similar fashion described in the proof of Theorem 8, label $(1/2)[mn-n-(3m-2)/2]$ edges between $U$ and $V$ and $((p-1)/2)(m+n+1)/2+((p+1)/2)(m+n-1)/2$ edges between $U\cup V$ and $W$ with $-1$ such that the edges between $V_{1}$ and $W_{1}$ are all negative edges. In addition, every vertex in $U$ is incident with $(n+p-2)/2$ negative edges, every vertex in $V_{1},V_{2}$ is incident with $(m+p-1)/2$ and $(m+p-3)/2$ negative edges, respectively, and every vertex in $W_{1},W_{2}$ is incident with $(m+n+1)/2$ and $(m+n-1)/2$ negative edges, respectively. Label the remaining edges of $K_{n,m,p}$ by $+1$. Then the weight of vertices in $U$ are all $2$, the weight of vertices in $V_{1},V_{2}$ are all $1,3$, respectively, and the weight of vertices in $W_{1}$ are all $-1$ and in $W_{2}$ are $+1$. Hence, this labeling defines a SEDF $f$ with $w(f)=(3m+2n)/2$.

Note that since the weight of some vertices in $W$ is $-1$, no vertices in $U\cup V$ can have weight zero or $-1$ by (1). Now by Lemma 7 and the facts that $f(W)=1$, $f(u)=2$ for every $u\in U$, $f(v)=1$ for every $v\in V_{1}$ and $f(v)=3$ for every $v\in V_{2}$, it follows that $\gamma_{s}^{\prime}(K_{m,n,p})=(3m+2n)/2$.

Case 2. $m\equiv 2\pmod{4}$.

By assumption,

[TABLE]

is even. Partition $V$ into $V_{1}$ and $V_{2}$ with $|V_{1}\mid=(n-m+1)/2$ and $|V_{2}\mid=(n+m-1)/2$. Also partition $W$ into $W_{1}$ and $W_{2}$ with $|W_{1}\mid=|W_{2}\mid+1$. In a similar fashion described in the proof of Theorem 8, label $(1/2)[mn-n-3m/2]$ edges between $U$ and $V$ and $((p+1)/2)(m+n+1)/2+((p-1)/2)(m+n-1)/2$ edges between $U\cup V$ and $W$ with $-1$ such that the edges between $V_{1}$ and $W_{1}$ are all negative edges. In addition, every vertex in $U$ is incident with $(n+p-2)/2$ negative edges, every vertex in $V_{1},V_{2}$ is incident with $(m+p-1)/2$ and $(m+p-3)/2$ negative edges, respectively, and every vertex in $W_{1},W_{2}$ is incident with $(m+n+1)/2$ and $(m+n-1)/2$ negative edges, respectively. Label the remaining edges of $K_{n,m,p}$ by $+1$. Then the weight of vertices in $U$ are all $2$, the weight of vertices in $V_{1},V_{2}$ are all $1,3$, respectively, and the weight of vertices in $W_{1}$ are all $-1$ and in $W_{2}$ are $+1$. Hence, this labeling defines a SEDF $f$ with $w(f)=(3m+2n-2)/2$.

Note that since the weight of some vertices in $W$ is $-1$, no vertices in $U\cup V$ can have weight zero or $-1$ by (1). Now by Lemma 7 and the facts that $f(W)=-1$, $f(u)=2$ for every $u\in U$, $f(v)=1$ for every $v\in V_{1}$ and $f(v)=3$ for every $v\in V_{2}$, it follows that $\gamma_{s}^{\prime}(K_{m,n,p})=(3m+2n-2)/2$. ∎

5 The SEDNs of $K_{1,n,p}$ and $K_{2,2,p}$

The constructions given in Section 4 work if the sum of the desired negative edges at vertices in $U$ and at vertices in $V$ is not less than the desired negative edges at vertices in $W$. In this section we calculate the signed edge domination numbers of $K_{1,n,p}$ and $K_{2,2,p}$ which are not covered by constructions given in Section 4.

Lemma 16**.**

Let $n\geq 1$ and $p\geq n+1$.

If $n,p$ are odd, then $\gamma_{s}^{\prime}(K_{1,n,p})=n+2$. 2. 2.

If $n,p$ are even, then $\gamma_{s}^{\prime}(K_{1,n,p})=n+2$. 3. 3.

If $n$ is even and $p$ is odd, then $\gamma_{s}^{\prime}(K_{1,n,p})=2n+1$. 4. 4.

If $n$ is odd and $p$ is even, then $\gamma_{s}^{\prime}(K_{1,n,p})=2n+3$.

Proof.

Consider the graph $K_{1,n,p}$ whose partite sets are $U,V$ and $W$.

Case 1. $n$ and $p$ are odd.

Since $mn-m-n=n-1-n=-1$ when $m=1$, the construction given in Theorem 9 does not work. On the other hand,

[TABLE]

when $m=1$. So we can label $(n+p-2)/2+n(1+p-2)/2$ edges between $U\cup V$ and $W$ such that the vertex in $U$ is incident with $(n+p-2)/2$ negative edges, each vertex in $V$ is incident with $(p-1)/2$ negative edges and all the vertices in $W$ are incident with $(n+1)/2$ negative edges except one vertex which is incident with $(n-1)/2$ negative edges. We label the remaining edges with $+1$. This yields a signed edge dominating function of weight $n+2$. It is easy to see that this is the minimum weight of a SEDF of $K_{1,n,p}$ when $n,p$ are odd.

Case 2. $n$ and $p$ are even.

Since $mn-n-(3m+1)/2=-2$ when $m=1$, the construction given in Theorem 12 does not work. On the other hand,

[TABLE]

when $m=1$. So we can label $m(n+p-2)/2+((n-m+1)/2)(m+p-1)/2\\ +((n+m-1)/2)(m+p-3)/2$ edges between $U\cup V$ and $W$ such that the vertex in $U$ is incident with $(n+p-2)/2$ negative edges, half of the vertices in $V$ are incident with $p/2$ negative edges and the other half are incident with $(p-2)/2$ negative edges, $p/2$ vertices in $W$ are incident with $(n+2)/2$, $(p-2)/2$ vertices are incident with $n/2$ and one vertex is incident with $(n-2)/2$ negative edges. We label the remaining edges with $+1$. This yields a signed edge dominating function of weight $n+2$. It is easy to see that this is the minimum weight of a SEDF of $K_{1,n,p}$ when $n,p$ are even.

Case 3. $n$ is even and $p$ is odd.

If $m=1$, then $mn-m-3n/2<0$, so the constructions given in Theorem 14 do not work. On the other hand,

[TABLE]

when $m=1$. Place $m(n+p-1)/2+((n-m+1)/2)(m+p-2)/2+((n+m-1)/2)(m+p-4)/2$ negative edges between $U\cup V$ and $W$ such that the vertex in $U$ is incident with $(n+p-1)/2$ negative edges, $(n-m+1)/2$ vertices in $V$ are incident with $(m+p-2)/2$ negative edges, $(n+m-1)$/2 vertices are incident with $(m+p-4)/2$ negative edges, $(p-n-1)/2$ vertices of $W$ are incident with (m+n+1)/2 negative edges and $(p+n+1)/2$ vertices of $W$ are incident with (m+n-1)/2 negative edges. Label the remaining edges with $+1$. This yields a signed edge dominating function of weight $2n+m$ which is $2n+1$ when $m=1$. It is an easy to see that $\gamma(K_{1,n,p})=2n+1$.

Case 4. $n$ is odd and $p$ is even.

If $m=1$, then $mn-(3m+3n)/2<0$, so the constructions given in Theorem 10 do not work. On the other hand,

[TABLE]

when $m=1$. So we can find a signed edge dominating function of weight $(1/2)[3+3n+2(n+3)/2]=2n+3$. It is easy to verify that $\gamma_{s}^{\prime}(K_{1,n,p})=2n+3$. ∎

In [5] it was conjectured that $\gamma_{s}^{\prime}(G)\leq|V(G)|-1$ for every graph $G$ of order at least 2. Note that if $n$ is odd, then $\gamma_{s}^{\prime}(K_{1,n,n+3})=2n+3$ by Lemma 16, Part 4. Hence, the graph $K_{1,n,n+3}$ achieves the upper bound in this conjecture.

Lemma 17**.**

Let $p\geq 5$ be odd. Then $\gamma_{s}^{\prime}(K_{2,2,p})=8$. **

Proof.

Consider the graph $K_{m,n,p}$ with partite sets $U$, $V$ and $W$. Partition $W$ into $W_{1},W_{2}$ and $W_{3}$ such that $|W_{1}\mid=|W_{2}\mid$ and $|W_{3}|=1$. Label the edges between $U$ and $W_{1}$ and between $V$ and $W_{2}$ with $-1$ and the remaining edges with $+1$. Then the weight of vertices in $W_{1}\cup W_{2}$ are zero and the weight of the vertex in $W_{3}$ is $4$. The weight of the vertices in $U\cup V$ are all 3. This leads to $\gamma_{s}^{\prime}(K_{2,2,p})=8$. ∎

6 Main Theorem

Let $m,n,p$ be positive integers, $m\leq n$ and $p\geq m+n$. In this section we state the main theorem of this paper. This result together with the main result of [3] provide the signed edge domination number of $K_{m,n,p}$ for all positive integers $m,n$ and $p$.

Main Theorem Let $m,n$ and $p$ be positive integers, $m\leq n$ and $p\geq m+n$. Let $m\geq 2$ and if $p$ is odd, $(m,n)\neq(2,2)$.

• A.

If $m,n$ and $p$ are even, then $\gamma_{s}^{\prime}(K_{m,n,p})=m+n$.

• B.

If $m,n$ and $p$ are odd and $m,n\geq 3$, then $\gamma_{s}^{\prime}(K_{m,n,p})=m+n+1$.

• C.

If $m,n\geq 3$ are odd and $p$ is even, then

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+3n+2}{2}$ if $m+n\equiv 0\pmod{4}$, 2. 2.

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+3n}{2}$ if $m+n\equiv 2\pmod{4}$.

• D.

If $m,n$ are even, $p$ is odd and $(m,n)\neq(2,2)$, then

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+3n}{2}$ if $m+n\equiv 0\pmod{4}$, 2. 2.

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+3n+2}{2}$ if $m+n\equiv 2\pmod{4}$.

• E.

If $m$ is odd, $n,p$ are even and $3\leq m<n$, then

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+2n+1}{2}$ if $m\equiv 1\pmod{4}$, 2. 2.

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+2n-1}{2}$ if $m\equiv 3\pmod{4}$.

• F.

If $m,p$ are even, $n$ is odd and $m<n$, then

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{2m+3n+1}{2}$ if $n\equiv 1\pmod{4}$, 2. 2.

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{2m+3n-1}{2}$ if $n\equiv 3\pmod{4}$.

• G.

If $m,p$ are odd, $n$ is even and $3\leq m<n$, then

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{2m+3n}{2}$ if $n\equiv 0\pmod{4}$, 2. 2.

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{2m+3n-2}{2}$ if $n\equiv 2\pmod{4}$.

• H.

If $m$ is even, $n,p$ are odd and $m<n$, then

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+2n}{2}$ if $m\equiv 0\pmod{4}$, 2. 2.

$\gamma_{s}^{\prime}(K_{m,n,p})=\dfrac{3m+2n-2}{2}$ if $m\equiv 2\pmod{4}$.

In addition, $\gamma_{s}^{\prime}(K_{1,n,p})=n+2$ if $n,p$ are both odd or both even, $\gamma_{s}^{\prime}(K_{1,n,p})=2n+1$ if $n$ is even and $p$ is odd, $\gamma_{s}^{\prime}(K_{1,n,p})=2n+3$ if $n$ is odd and $p$ is even, and $\gamma_{s}^{\prime}(K_{2,2,p})=8$ if $p$ is odd.