In this paper, the graphs are undirected simple graphs and for other terminologies we follow [1]. Let $G=(V, E)$ be a graph with vertex set $V=V(G)$ and edge set $E=E(G)$. Every maximal complete subgraph $K$ of graph $G$ is called a clique of $G$, the order of a largest complete subgraph is called the clique number of $G$, denoted by $\omega(G)$. A clique $K$ is called non-trivial if $K\neq K_{1}$. Let $G_1$ and $G_2$ be any two disjoint graphs. Then $G_{1}\vee G_{2}$ denotes the join graphs of $G_{1}$ and $G_{2}$:

$ \begin{array}{*{20}{l}} {V({G_1} \vee {G_2}) = V({G_1})\bigcup V ({G_2}), }\\ {E({G_1} \vee {G_2}) = E({G_1})\bigcup E ({G_2})\bigcup {\left\{ {uv:u \in V({G_1}), v \in V({G_2})} \right\}} .} \end{array} $

Let $G=(V, E)$ be a graph. For a function $f:E\rightarrow \{+1, -1\}$ and a subset $S$ of $E(G)$, define $f(S)=\sum\limits_{e\in S}f(e)$. For convenience, for a given graph $G=(V, E)$, an edge $e\in E(G)$ is said to be a +1 edge of $G$ if $f(e)=+1$, analogously, an edge $e\in E(G)$ is said to be a -1 edge of $G$ if $f(e)=-1$. Write $E_{1}=\{e\in E(G)|f(e)=+1\}$, $E_{2}=\{e\in E(G)|f(e)=-1\}$.

Definition 1.1 [2]  Let $G=(V, E)$ be a simple graph. A function $f : E\rightarrow \{+1, -1\}$ is said to be a signed clique edge dominating function of $G$ if $\sum\limits_{e\in E(K)}f(e)\geq 1$ for every non-trivial clique $K$ in $G$. The signed clique edge domination number of $G$ is defined to be ${\gamma '_{scl}}(G)= \min\{\sum\limits_{e\in E(G)}$ $f(e): f$ is a signed clique edge dominating function of $G$$\}$. In particular, for empty graph $\overline{K_{n}}$, define ${\gamma '_{scl}}(\overline{K_{n}}) = 0$.

In recent years, domination number and its variations were studied extensively. The monographs [2] contain extensive reviews of topics. Signed edge domination was studied in [3, 4], signed clique edge domination was studied in [5], signed star domination in [6], signed cycle domination in [7], minus edge domination in [8], signed edge total domination in [9]. In this paper, we determine the signed clique edge domination numbers of graphs $K_{n}\vee P_{m}$ and $K_{n}\vee C_{m}$.

Theorem 2.1   For any positive integer $n\geq3$ and $m\geq3$,

$ {\gamma '_{scl}}({K_n} \vee {P_m}) = \left\{ \begin{array}{l} 2(6 - n)\left\lfloor {\frac{m}{2}} \right\rfloor - (n + 1)m + \frac{{n(n - 1)}}{2} + 1, \;\;\;{\rm{when}}\;\;n{\rm{ = 3}}, {\rm{4}}, {\rm{5}}, \\ - (n + 1)m + 2n + 3 + \frac{{{{( - 1)}^{\left\lfloor {\frac{n}{2}} \right\rfloor + 1}} + 1}}{2}, \;\;\;\;\;\;\;\;\;{\rm{when}}\;\;n \ge {\rm{6}}. \end{array} \right. $

Proof   Let $f$ be a signed clique edge dominating function of graph $G=K_{n}\vee P_{m}$ such that ${\gamma '_{scl}}(G)=f(E)=\sum\limits_{e\in E}f(e)$. The vertices of $K_{n}$ are $v_{1}, v_{2}, \cdots v_{n}$ in this order, and the vertices of $P_{m}$ are $u_{1}, u_{2}, \cdots u_{m}$ in this order. Then $|E(G)|$=$\frac{n(n-1)}{2}+(n+1)m-1$. Let $A=\{v_{i}u_{j}|i=1, 2, \cdots n, j=1, 2, \cdots m\}\bigcup\{u_{i}u_{i+1}|i=1, 2, \cdots, m-1\}$.

We first prove lower bound.

Case 1   $n=3, 4, 5$, then

$ {\gamma '_{scl}}(G) \ge 2(6 - n)\left\lfloor {\frac{m}{2}} \right\rfloor - (n + 1)m + \frac{{n(n - 1)}}{2} + 1. $

(2.1)

Let $s$ (respectively $t$) be the number of +1 (respectively -1) edges of $G$, thus $\frac{n(n-1)}{2}+(n+1)m-1=s+t$, ${\gamma '_{scl}}(G)=s-t$.

Suppose that (2.1) does not hold. Then ${\gamma '_{scl}}(G)< 2(6-n)\lfloor \frac{m}{2}\rfloor-(n+1)m+\frac{n(n-1)}{2}+1$, Hence $t>(n+1)m-(6-n)\lfloor \frac{m}{2}\rfloor-1$. Let the number of -1 edges in $A$ be $r$.

Case 1.1   $m\equiv 0 \pmod 2$.

Suppose $3(n-2)\frac{m}{2} +(m-1)<r\leq (n+1)m-1$. By the pigeonhole principle, there exists a clique $K_{n+2}\in G $, that the number of -1 edges is at least $3n-4$, such that $\sum\limits_{e\in E(K_{n+2})}f(e)\leq 0$. This is a contradiction.

If

$ (n + 1)m - (6 - n)\frac{m}{2} - \frac{{n(n - 1)}}{2} \le r \le 3(n - 2)\frac{m}{2} + (m - 1), $

then the number of -1 edges in $E(G)\setminus A $ is at least 1. By the pigeonhole principle, there exists a $K_{n+2}\in G $, such that $\sum_{e\in E(K_{n+2})}f(e)\leq0$. This is a contradiction.

Case 1.2   $m\equiv 1 \pmod 2$.

Suppose $n\lceil\frac{m}{2}\rceil+2(n-3)\lfloor\frac{m}{2}\rfloor +(m-1)<r\leq (n+1)m-1$. By the pigeonhole principle, there exists a clique $K_{n+2}\in G $, that the number of -1 edges is at least $3n-4$, such that $\sum\limits_{e\in E(K_{n+2})}f(e)\leq0$. This is a contradiction.

If

$ (n + 1)m - (6 - n)\left\lfloor {\frac{m}{2}} \right\rfloor - \frac{{n(n - 1)}}{2} \le r \le 3(n - 2)\left\lfloor {\frac{m}{2}} \right\rfloor + n + (m - 1), $

then the number of -1 edges in $E(G)\setminus A $ is at least 1. By the pigeonhole principle, there exists a $K_{n+2}\in G $, such that $\sum\limits_{e\in E(K_{n+2})}f(e)\leq 0$. This is a contradiction.

Hence ${\gamma '_{scl}}(G)\geq 2(6-n)\lfloor \frac{m}{2} \rfloor -(n+1)m+\frac{n(n-1)}{2}+1$.

Case 2   $n\geq 6$. Then

$ {\gamma '_{scl}}(G) \ge - (n + 1)m + 2n + 3 + \frac{{{{( - 1)}^{\left\lfloor {\frac{n}{2}} \right\rfloor + 1}} + 1}}{2}. $

Let $f$ be a signed clique edge dominating function of G such that ${\gamma '_{scl}}(G)=f(G)$, and $s$ the number of +1 edges of $G$. Then ${\gamma '_{scl}}(G)=2s-|E(G)|$. And $\sum\limits_{e\in E(K_{n+2})}f(e)\geq1$ for every non-trivial clique $K_{n+2}$ in $G$. Hence $s\geq s_{0}=|\{e\in E(K_{n+2})\mid f(e)=1\}|$.

Note that

$ |E(G)| = \frac{{n(n - 1)}}{2} + (n + 1)m - 1, |E({K_{n + 2}})| = \frac{{(n + 2)(n + 1)}}{2}. $

Since $f(K_{n+2})\geq1$, $s_{0}\geq \lfloor \frac{(n+2)(n+1)}{4}\rfloor+1$. Then $s\geq \lfloor \frac{(n+2)(n+1)}{4}\rfloor+1$. Hence

$ {\gamma '_{scl}}(G) = 2s - |E(G)| \ge - (n + 1)m + 2n + 3 + \frac{{{{( - 1)}^{\left\lfloor {\frac{n}{2}} \right\rfloor + 1}} + 1}}{2}. $

Next consider the upper bound.

We define the signed clique edge dominating function $f$ of graph $G$ as follows:

For $n=3$, let

$ f(e) = \left\{ {\begin{array}{*{20}{l}} { + 1,\;\;\;\;{\rm{when}}\;e \in {{\rm{K}}_{\rm{3}}}\bigcup {\{ {{v}_{i}}{{u}_{j}}{\rm{|}}{i}{\rm{ = 1}},{\rm{2}},{\rm{3}},j \equiv {\rm{0}}\;\;\;\left( {\,\bmod \,{\rm{2}}} \right)\} } ;}\\ { - 1,\;\;\;\;{\rm{otherwise}}.} \end{array}} \right. $

For $n=4$, let

$ f(e) = \left\{ {\begin{array}{*{20}{l}} { + 1,\;\;\;\;{\rm{when}}\;e \in {{\rm{K}}_4}\bigcup {\{ {{v}_{i}}{{u}_{j}}{\rm{|}}{i}{\rm{ = 3}},{\rm{4}},{j} \equiv {\rm{0}}\;\;\;\left( {\,\bmod \,{\rm{2}}} \right)\} } ;}\\ { - 1,\;\;\;\;{\rm{otherwise}}.} \end{array}} \right. $

For $n=5$, let

$ f(e) = \left\{ {\begin{array}{*{20}{l}} { + 1,\;\;\;\;{\rm{when}}\;e \in {{\rm{K}}_{\rm{5}}}\bigcup {\{ {{v}_{i}}{{u}_{j}}{\rm{|}}{i}{\rm{ = 5}},{j} \equiv {\rm{0}}\;\;\;\left( {\,\bmod \,{\rm{2}}} \right)\} } ;}\\ { - 1,\;\;\;\;{\rm{otherwise}}.} \end{array}} \right. $

For $n=3, 4, 5$, every non-trivial clique $K_{n+2}$ in $G$, we have

$ \begin{array}{l} \sum\limits_{e \in E({K_{n + 2}})} f (e) = \sum\limits_{e \in {E_1}\bigcap E ({K_{n + 2}})} f (e) - \sum\limits_{e \in {E_2}\bigcap E ({K_{n + 2}})} f (e)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \frac{{n(n - 1)}}{2} + (6 - n) - (3n - 5)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \frac{{n(n - 1)}}{2} - 4n + 11 \ge 1. \end{array} $

Hence ${\gamma '_{scl}}(G)\leq \sum\limits_{e\in E(G)}f(e)=2(6-n)\lfloor \frac{m}{2} \rfloor -(n+1)m+\frac{n(n-1)}{2}+1.$

For $n\geq6$, let the number of +1 edges in $K_{n}$ is $\lfloor \frac{(n+2)(n+1)}{4}\rfloor+1$. All other edges are assigned -1. For every non-trivial clique $K_{n+2}$ in $G$, we have

$ \begin{array}{*{20}{l}} {\sum\limits_{e \in E({K_{n + 2}})} f (e) = \sum\limits_{e \in {E_1}\bigcap E ({K_{n + 2}})} f (e) - \sum\limits_{e \in {E_2}\bigcap E ({K_{n + 2}})} f (e)}\\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 2\left\lfloor {\frac{{(n + 2)(n + 1)}}{4}} \right\rfloor + 2 - \frac{{(n + 2)(n + 1)}}{2}}\\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \left\{ {\begin{array}{*{20}{l}} {1,\;\;\;{n} \equiv {\rm{0}},{\rm{1}}\;\;\;\left( {\,\bmod \,{\rm{4}}} \right);}\\ {2,\;\;\;{n} \equiv {\rm{2}},{\rm{3}}\;\;\;\left( {\,\bmod \,{\rm{4}}} \right).} \end{array}} \right.} \end{array} $

Hence

$ {\gamma '_{scl}}(G) \le \sum\limits_{e \in E(G)} f (e) = - (n + 1)m + 2n + 3 + \frac{{{{( - 1)}^{\left\lfloor {\frac{n}{2}} \right\rfloor + 1}} + 1}}{2}. $

Theorem 2.2   For any positive integer $n\geq3$ and $m\geq3$,

$ {\gamma '_{scl}}({K_n} \vee {C_m}) = \left\{ \begin{array}{l} 2(6 - n)\left[{\frac{m}{2}} \right] - (n + 1)m + \frac{{n(n - 1)}}{2}, \;\;\;\;\;{\rm{when}}\;{n}{\rm{ = 3}}, {\rm{4}}, {\rm{5}}\\ - (n + 1)m + 2n + 2 + \frac{{{{( - 1)}^{\left[{\frac{n}{2}} \right] + 1}} + 1}}{2}, \;\;\;\;\;\;\;{\rm{when}}\;{n} \ge {\rm{6}}. \end{array} \right. $

Proof   Let $f$ be a signed clique edge dominating function of graph $G=K_{n}\vee C_{m}$ such that ${\gamma '_{scl}}(G)=f(E)=\sum\limits_{e\in E}f(e)$. The vertices of $K_{n}$ are $v_{1}, v_{2}, \cdots v_{n}$ in this order, and the vertices of $C_{m}$ are $u_{1}, u_{2}, \cdots u_{m}$ in this order. Then $|E(G)|$=$\frac{n(n-1)}{2}+(n+1)m$. Write

$ A = \{ {v_i}{u_j}|i = 1, 2, \cdots n, j = 1, 2, \cdots m\} \bigcup {\{ {u_i}{u_{i + 1}}|i = 1, 2, \cdots, m - 1\} } \bigcup {\{ {u_1}{u_n}\} } . $

$n=3, 4, 5$, we first prove lower bound.

$ {\gamma '_{scl}}(G) \ge 2(6 - n)\left\lfloor {\frac{m}{2}} \right\rfloor - (n + 1)m + \frac{{n(n - 1)}}{2}. $

(2.2)

Let $s$ (respectively$t$) be the number of +1 (respectively -1) edges of $G$. Thus

$ \frac{{n(n - 1)}}{2} + (n + 1)m = s + t, {\gamma '_{scl}}(G) = s - t. $

Suppose that (2.2) does not hold. Then ${\gamma '_{scl}}(G)< 2(6-n)\lceil \frac{m}{2} \rceil -(n+1)m+\frac{n(n-1)}{2}$. Hence $t>(n+1)m-(6-n)\lceil \frac{m}{2} \rceil$. Let the number of -1 edges in $A$ be $r$.

Case 1   $m\equiv0\pmod 2$.

Suppose $3(n-2)\frac{m}{2}+m<r\leq (n+1)m$. By the pigeonhole principle, there exists a clique $K_{n+2}\in G $, That the number of -1 edges is at least $3n-4$, such that $\sum\limits_{e\in E(K_{n+2})}f(e)\leq 0$. This is a contradiction.

If

$ (n + 1)m - (6 - n)\frac{m}{2} - \frac{{n(n - 1)}}{2} + 1 \le r \le 3(n - 2)\frac{m}{2} + m. $

Then the number of -1 edges in $E(G)\setminus A $ is at least 1. By the pigeonhole principle, there exists a $K_{n+2}\in G $, such that $\sum\limits_{e\in E(K_{n+2})}f(e)\leq 0$. This is a contradiction.

Case 2   $m\equiv1\pmod 2$.

Suppose $n\lfloor \frac{m}{2}\rfloor+2(n-3)\lceil \frac{m}{2}\rceil+m<r\leq (n+1)m$. By the pigeonhole principle, there exists a clique $K_{n+2}\in G $, that the number of -1 edges is at least $3n-4$, such that $\sum\limits_{e\in E(K_{n+2})}f(e)\leq 0$. This is a contradiction.

If

$ (n + 1)m - (6 - n)\left\lceil {\frac{m}{2}} \right\rceil - \frac{{n(n - 1)}}{2} + 1 \le r \le n\left\lfloor {\frac{m}{2}} \right\rfloor + 2(n - 3)\left\lceil {\frac{m}{2}} \right\rceil + m, $

then the number of -1 edges in $E(G)\setminus A $ is at least 1. By the pigeonhole principle, there exists a $K_{n+2}\in G $, such that $\sum\limits_{e\in E(K_{n+2})}f(e)\leq 0$. This is a contradiction.

In summary,

$ {\gamma '_{scl}}(G) \ge 2(6 - n)\left\lceil {\frac{m}{2}} \right\rceil - (n + 1)m + \frac{{n(n - 1)}}{2}. $

Next we consider the upper bound. The upper bound is obtained by specifying a signed clique edge dominating function. We define the signed clique edge dominating function $f$ of $G$ as follows:

For $n=3$, let

$ f(e) = \left\{ \begin{array}{l} + 1,\;\;\;{\rm{when}}\;e \in {{\rm{K}}_{\rm{3}}}\bigcup {\{ {v_i}{u_j}{\rm{|}}i{\rm{ = 1}},{\rm{2}},{\rm{3}},j \equiv {\rm{1}}\;\;\left( {\,\bmod \,{\rm{2}}} \right)\} } ;\\ - 1,\;\;\;{\rm{otherwise}}. \end{array} \right. $

For $n=4$, let

$ f(e) = \left\{ \begin{array}{l} + 1, \;\;\;{\rm{when}}\;{e} \in {{\rm{K}}_{\rm{4}}}\bigcup {\{ {{v}_{i}}{{u}_{j}}{\rm{|}}{i}{\rm{ = 3}}, {\rm{4}}, {j} \equiv {\rm{1}}\;\;\left( {\bmod {\rm{2}}} \right)\} } ;\\ - 1, \;\;\;{\rm{otherwise}}. \end{array} \right. $

For $n=5$, let

$ f(e) = \left\{ \begin{array}{l} + 1, \;\;\;{\rm{when}}\;{e} \in {{\rm{K}}_5}\bigcup {\{ {{v}_{i}}{{u}_{j}}{\rm{|}}{i}{\rm{ = }}5, {j} \equiv {\rm{1}}\;\;\left( {\bmod {\rm{2}}} \right)\} } ;\\ - 1, \;\;\;{\rm{otherwise}}. \end{array} \right. $

For $n=3, 4, 5$, $m\equiv 1\pmod 2$, consider clique $K_{n+2}$ of include edge $u_{1}u_{m}$,

$ \begin{array}{l} \sum\limits_{e \in E({K_{n + 2}})} f (e) = \sum\limits_{e \in {E_1}} f (e) - \sum\limits_{e \in {E_2}} f (e)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \frac{{n(n - 1)}}{2} + 2(6 - n) - (4n - 11)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \frac{{n(n - 1)}}{2} - 6n + 23 \ge 1. \end{array} $

We have

$ {{\gamma '}_{scl}}(G) \le \sum\limits_{e \in E(G)} f (e) = 2(6 - n)\left\lceil {\frac{m}{2}} \right\rceil - (n + 1)m + \frac{{n(n - 1)}}{2}. $

For other cases the proof is similar to Theorem 2.1.