Characterization of Some Claw-Free Graphs in Co-Secure Domination Number

Abstract

For a vertex subset S of a graph $G,$ if each vertex of G is either in S or adjacent to some vertex in $S,$ then S is a dominating set of $G.$ Let S be a dominating set of a graph $G.$ If each vertex v not in S has a neighbor u in S such that $(S\backslash \{u\left\}\right)\cup \left\{v\right\}$ is also a dominating set of $G,$ then S is a secure dominating set of $G.$ If each vertex u in S has a neighbor v not in S such that $(S\backslash \{u\left\}\right)\cup \left\{v\right\}$ is also a dominating set of G, then S is a co-secure dominating set of G. The minimum cardinality of a secure (resp. co-secure) dominating set of G is the secure (resp. co-secure) domination number of $G.$ Arumugam et al. proposed the questions to characterize a graph G such that the co-secure domination number of G equals the independence number and the secure domination number of G, respectively. Inspired by those questions, in this paper, we obtain two classes of claw-free graphs such that the co-secure domination number equal the independence number and the secure domination number. Our results provide some theoretical basis of claw-free graphs for networks.

1. Introduction

We consider only simple and undirected graphs in this paper. For notation and terminology not defined in this paper, readers are referred to [1]. For a graph G and $S\subseteq V\left(G\right),$ we use $G\left[S\right]$ to denote the subgraph of G induced by S. Let $N_S\left(v\right)$ denote the set of all the neighbors of v in $S,$ and $d_S\left(v\right)={\displaystyle \left|N_S\left(v\right)\right|};$ moreover, we set $N\left(v\right)=N_G\left(v\right),$ and $d\left(v\right)=d_G\left(v\right)$ for simplicity. Let $N\left[v\right]=N\left(v\right)\cup \left\{v\right\}.$ For a graph $H_1,$ if a subgraph $H_2$ of a graph G is isomorphic to $H_1,$ then let $H_1=H_2.$ A complete graph consisting of n vertices is denoted by $K_n,$ and a complete bipartite graph consisting of two vertex-disjoint independent $A,B$ is denoted by $K_{a,b},$ where $\left|A\right|=a$ and $\left|B\right|=b.$ We use $G-S$ to denote the vertex set $V\left(G\right)-S,$ where S is a vertex subset of G.

Given a graph $H,$ a graph G is H-free if G contains no induced subgraph isomorphic to $H.$ A claw is a graph isomorphic to $K_{1,3}.$ Let $G\left[\{v,v_1,v_2,v_3\}\right]=K_{1,3}$ such that $d_H\left(v\right)=3,d_H\left(v_i\right)=1,i\in \{1,2,3\},$ if a subgraph H consisting of $\{v,v_1,v_2,v_3\}$ in G is isomorphic to a claw. Let $K_4^{-}$ denote the graph obtained by deleting an edge of $K_4.$ For a positive integer $k,$ we use $C_k,$ and $P_k$ to denote a cycle, and a path of order k, respectively. A bull is a graph (see Figure 1) obtained by joining two isolated vertices to two distinct vertices of a triangle, respectively. We use $G\left[\{v_1,v_2,v_3,v_4,v_5\}\right]$ to denote a subgraph H of G isomorphic to a bull, which consists of $\{v_1,v_2,v_3,v_4,v_5\}$ such that $d_H\left(v_1\right)=2, d_H\left(v_2\right)=d_H\left(v_3\right)=3, d_H\left(v_4\right)=d_H\left(v_5\right)=1$ and $v_4{v}_2,v_3{v}_5\in E\left(G\right).$ A $Z_2$ is a graph (see Figure 1) obtained by identifying an end of a path of length 2 to a vertex of a triangle. In this paper, we use $G\left[\{u_1,u_2,u_3,u_4,u_5\}\right]$ to denote a subgraph H of G isomorphic to $Z_2$, which consists of $\{u_1,u_2,u_3,u_4,u_5\}$ such that $d_H\left(u_1\right)=3, d_H\left(u_2\right)=d_H\left(u_3\right)=d_H\left(u_4\right)=2, d_H\left(u_5\right)=1$ and $H\left[\{u_1,u_2,u_3\}\right]=K_3.$ For two graphs $H_1$ and $H_2,$ the join of $H_1$ and $H_2,$ denoted by $H_1\vee H_2,$ is a graph with vertex set $V\left(H_1\right)\cup V\left(H_1\right)$ and edge set $E\left(H_1\right)\cup E\left(H_2\right)\cup \{uv:u\in V\left(H_1\right),v\in V\left(H_2\right)\}.$ Let $\alpha \left(G\right)$ denote the independence number of a graph G.

If every vertex of a graph G is in S or adjacent to some vertex in $S,$ then S is called a dominating set of $G.$ Let S be a dominating set of a graph G. If for every vertex $u\in G-S,$ S contains a neighbor of $u,$ denoted by $v,$ such that $(S\backslash \{v\left\}\right)\cup \left\{u\right\}$ is also a dominating set of $G,$ then S is called a secure dominating set of $G.$ S is called a co-secure dominating set of G if for every vertex v in $S,$ there is a neighbor of v in $G-S,$ denoted by $u,$ such that $(S\backslash \{v\left\}\right)\cup \left\{u\right\}$ is also a dominating set of $G.$ The minimum cardinality of a co-secure dominating set (resp. secure dominating set) of G is the called the co-secure (resp. secure) domination number of $G.$ We use ${\gamma }_{cs}\left(G\right)$ and ${\gamma }_s\left(G\right)$ to denote the co-secure domination number and secure domination number of $G$, respectively.

Let S be a dominating set of a graph $G.$ For a vertex $u\in G-S,$ if u has an exactly one neighbor in S, denoted by $v,$ then u is called an external private neighbor of v with respect to $S.$ For a vertex $v\in S,$ let $epn(v,S)=\{u\in V(G)-S:u$ is an external private neighbor of v with respect to S$\},$ and $rpn(v,S)=\{w\in V(G)-S:wv\in E(G)$ and $(S\backslash \{v\left\}\right)\cup \left\{w\right\}$ is a dominating set of G$\}.$

In [2], Cockayne et al. proposed the concepts of a secure dominating set and secure domination number; moreover, they also determined the exact values of the secure domination number of paths, cycles, and complete multipartite graphs. Araki and Yumoto [3] proved that the secure domination number of a maximal outerplane graph of n vertices is at most $\lceil 3n/7\rceil .$ Degawa and Saito [4] gave that the secure domination number of every $C_5$-free graph is no larger than its independence number. The secure domination number of Cartesian products of small graphs with paths, cycles, and some other graph classes were given in [5]. For the bounds of the secure domination number of more classes of graphs, the readers are referred to [6,7,8].

Arumugan et al. introduced the concept of co-secure domination in [9]. They also obtained the co-secure domination number of paths and cycles, and the sharp upper and lower bounds of some graphs; moreover, they proposed that the problem to determine the co-secure domination number of a graph is NP-complete even if it is a bipartite, chordal or planar graph. Joseph and Sangeetha [10] obtained the co-secure domination number of Friendship graphs, Jahangir graphs, and Helm graphs; they also studied a family of trees such that the co-secure domination number equals the secure domination number. Pothuvath et al. [11] gave the bounds of domination number and co-secure domination number for jump graphs. We summarize the results on the parameter in the bound of co-secure domination number in

Table 1. Parameter in the bounds of the secure domination number of a graph G.

Graph Class	The Parameter in the ${\mathit{\gamma }}_{\mathit{cs}}\left(\mathit{G}\right)$-Bound	Reference
Path, cycle, wheel G	The order of G	[9]
Tree G	The order and the number of leaves of G	[9]
Connected graph G	$\alpha \left(G\right)$	[9]
General graph G	The order of G and $\gamma \left(G\right)$	[9]
General graph G with $\delta \left(G\right)\ge 2$	${\gamma }_s\left(G\right)$	[9]
Friendship graph G	The order of G	[10]
Jahangir graph G	The order of G	[10]
Helm graph G	The order of G	[10]
Jump graph G	Specific number	[11]
Claw-free graph G	$\alpha \left(G\right)$, ${\gamma }_s\left(G\right)$	[12]

.

Finding the minimum dominating set (MDS) and co-secure dominating set (MCSDS) are NP-complete. However, MCSDS is practically harder due to the co-secure constraint, which introduces combinatorial challenges. The co-secure condition (the “swap” for every $u\in S$ in a dominating set S) forces solutions to be larger and more structured. Specifically, algorithms must consider larger sets S, and the number of candidate sets grows exponentially with size (e.g., from $O\left(2^n\right)$ for MDS to $O\left(2^{cn}\right)$ for larger c in MCSDS).

Arumugan et al. [9] gave the bounds of the co-secure domination number by independence number, and secure domination number, as follows.

Theorem 1

([9]). If G is a non-trivial graph, then ${\gamma }_{cs}\left(G\right)\le \alpha \left(G\right).$

Theorem 2

([9]). If G is a graph and $\delta \left(G\right)\ge 2,$ then ${\gamma }_{cs}\left(G\right)\le {\gamma }_s\left(G\right).$

Moreover, in [9], Arumugan et al. proposed the following two problems.

Problem 1

([9]). Characterize graphs G with ${\gamma }_{cs}\left(G\right)=\alpha \left(G\right).$

Problem 2

([9]). Characterize graphs G with ${\gamma }_{cs}\left(G\right)={\gamma }_s\left(G\right).$

There is very little research on Problems 1 and 2. We think the dense graphs may satisfy the characterization by the structural analysis. Claw-free graphs can be utilized to design deadlock-free routing algorithms in the context of supercomputer networks and distributed systems, because their structure can avoid complex dependencies that may lead to deadlocks. $P_5$-free graphs are useful for creating low-diameter and fault-tolerant network topologies, as their restricted path structure helps minimize communication latency and improve robustness. Inspired by the significant application value of $\{claw,P_5\}$-free graphs and the above open problems, in this paper, we characterize the co-secure domination number of $\{claw,P_5\}$-free graphs.

Notice that if a graph G contains an isolate vertex v, then v is in every dominating set of $G.$ By the definition, for a co-secure dominating set S of a graph $G,$ each vertex in S has a neighbor in $G-S,$ which implies that G contains no isolated vertex. Thus, $\delta \left(G\right)\ge 1$ if G contains a co-secure dominating set.

2. Characterization of Claw-Free Graphs $\mathit{G}$ with ${\mathit{\gamma }}_{\mathit{cs}}\left(\mathit{G}\right)=\mathit{\alpha }\left(\mathit{G}\right)$

In this section, we characterize the claw-free graphs G with ${\gamma }_{cs}\left(G\right)=\alpha \left(G\right).$ Firstly, we obtain the following preliminaries.

Lemma 1.

Let G be a non-trivial claw-free graph such that G contains exactly one vertex of degree $\left|V\right(G\left)\right|-1.$ Then ${\gamma }_{cs}\left(G\right)=\alpha \left(G\right).$

Proof. 

Let v be the only one vertex of G of degree $\left|V\right(G\left)\right|-1.$ Then, G is not a complete graph, and hence, $\alpha \left(G\right)\ge 2.$ Moreover, each vertex of G is in $N\left[v\right]$, which implies $\alpha \left(G\right)=2$ since G is claw-free. Suppose ${\gamma }_{cs}\left(G\right)=1$ and $\left\{x\right\}$ is a minimum co-secure dominating set of $G.$ Then, there is a vertex y in $G-\left\{x\right\}$ such that $\left\{y\right\}$ is dominating set of $G,$ which implies $d\left(x\right)=d\left(y\right)=\left|V\right(G\left)\right|-1,$ a contradiction. Thus, ${\gamma }_{cs}\left(G\right)\ge 2.$ Assume $v_1,v_2$ are two nonadjacent vertices of G. We claim that $S=\{v_1,v_2\}$ is a co-secure dominating set of G, and then, S is a minimum co-secure dominating set of G by ${\gamma }_{cs}\left(G\right)\ge 2.$ Since $\alpha \left(G\right)=2,$S is a maximum independent set of $G,$ and hence, S is a dominating set of $G.$ Clearly, $v\notin S$ since each vertex of G is adjacent to v and $v_1{v}_2\notin E\left(G\right).$ Thus, $S_i=(S\backslash \left\{v_i\right\})\cup \left\{v\right\}$ is a dominating set of G for each $i\in \{1,2\}.$ It follows that S is a minimum co-secure dominating set of G and then ${\gamma }_{cs}\left(G\right)=\alpha \left(G\right).$ □

Let $\mathcal{F}$ be the family of claw-free graphs G of such that G is not complete, and G contains at least two vertices of degree $\left|V\right(G\left)\right|-1$ (see Figure 2). It is easy to verify that ${\gamma }_{cs}\left(G\right)=1$ and $\alpha \left(G\right)=2.$ Thus, by Lemma 1, ${\gamma }_{cs}\left(G\right)=\alpha \left(G\right)$ if G contains a vertex of degree $\left|V\right(G\left)\right|-1$ and $G\notin \mathcal{F}.$ In Theorem 4 of [12], we prove that every $\{claw,bull,Z_2\}$-free graph G with $\delta \left(G\right)\ge 3$ and $\alpha \left(G\right)\ne 2$ contains a minimum co-secure dominating set that is a maximum independent set of $G,$ i.e., ${\gamma }_{cs}\left(G\right)=\alpha \left(G\right)$. However, there is a $\{claw,bull,Z_2\}$-free graph G with $\delta \left(G\right)\ge 3$ such that ${\gamma }_{cs}\left(G\right)=\alpha \left(G\right)=2$. In this paper, we give the following result, which is more precise.

Theorem 3.

Let G be a $\{claw,bull,Z_2\}$-free graph with $\delta \left(G\right)\ge 3$ and $G\notin \mathcal{F}.$ Then, G contains a minimum co-secure dominating set that is a maximum independent set of $G.$

Assume that $G_1=K_m,m\ge 1,$ and $G_2=v_1{v}_2{v}_3{v}_4{v}_5{v}_1$ is a 5-cycle. Let G be the graph (see Figure 3) obtained by joining each vertex in $G_1$ to every vertex in $\{v_1,v_2,v_5\}$ of $G_2.$ Clearly, G is $\{claw,P_5,bull\}$-free, but not $Z_2$-free. Thus, there is a $\{claw,P_5,bull\}$-free graph that is not $\{claw,bull,Z_2\}$-free. Conversely, there is a $\{claw,bull,Z_2\}$-free graph that is not $\{claw,P_5,bull\}$-free; for example, $P_k,k\ge 5,$ is a $\{claw,bull,Z_2\}$-free graph, but not $P_5$-free.

In the following, we give the relation between the co-secure domination number and independence number of $\{claw,P_5,bull\}$-free graphs.

Theorem 4.

Let G be a $\{claw,P_5,bull\}$-free graph such that $G\notin \mathcal{F}$ and $\delta \left(G\right)\ge 1.$ Then, G contains a minimum co-secure dominating set that is a maximum independent set of G.

Proof. 

Since for a disconnected graph, its co-secure domination number is the sum of the co-secure domination number of all the components, and hence, it suffices to consider G is connected in the following proof. If G is complete, then we are done. Thus, we assume that G is a connected and non-complete graph. Then, $\alpha \left(G\right)\ge 2,$ which implies that ${\gamma }_{cs}\left(G\right)\ge 2$ by $G\notin \mathcal{F}$. We take a minimum co-secure dominating set S of G such that $G\left[S\right]$ contains the minimum number of edges. We have $\left|S\right|\ge 2$ by ${\gamma }_{cs}\left(G\right)\ge 2$. In the following proof, firstly, we prove that S is an independent set, and then, we claim that S is a maximum independent set of G. □

We need the following result to prove S is an independent set of G.

Claim 1.

$epn(v,S)=\varnothing$ if $v\in S$ and $d_S\left(v\right)\ge 1.$

Proof. 

Suppose to the contrary that there is a vertex $v\in S$ such that $d_S\left(v\right)\ge 1$ and $epn(v,S)\ne \varnothing .$ Let $u\in S$ such that $uv\in E\left(G\right).$ We claim that $G\left[epn\right(v,S\left)\right]$ is a clique. Otherwise, $epn(v,S)$ contains two non-adjacent vertices. Let $v_1,v_2\in epn(v,S)$ such that $v_1{v}_2\notin E\left(G\right).$ Since $v_1,v_2\in epn(v,S)$ and $u\in S,$ we have $v_{1}u,v_{2}u\notin E\left(G\right).$ It follows that $G\left[\{v,v_1,v_2,u\}\right]=K_{1,3},$ a contradiction. Thus, $G\left[epn\right(v,S\left)\right]$ is a clique.

Let $v_1\in epn(v,S)$ and $S^{\prime }=(S\backslash \left\{v\right\})\cup \left\{v_1\right\}$. Since $G\left[epn\right(v,S\left)\right]$ is a clique, we have $v_1\in rpn(v,S),$ and hence, $S^{\prime }$ is a dominating set of G with $|S^{\prime }| = |S|.$ By $v_1\in epn(v,S),$$G\left[S^{\prime }\right]$ contains fewer edges than $G\left[S\right].$ Thus, $S^{\prime }$ is not a co-secure dominating set of G by the choice of $S.$ Clearly, $v\in rpn(v_1,S^{\prime }).$ Then, there is vertex $w\in S^{\prime }\backslash \left\{v_1\right\}$ such that $rpn(w,S^{\prime })=\varnothing$ and $epn(w,S^{\prime })$ contains two nonadjacent vertices $w_1,w_2$ with $\{w_1,w_2\}\cap N\left(v\right)\ne \varnothing .$ We have $w_1{v}_1,w_2{v}_1\notin E\left(G\right)$ by $w_1,w_2\in epn(w,S^{\prime }).$

Suppose $\{w_1,w_2\}\subseteq N\left(v\right).$ Then, $w_1{w}_2\in E\left(G\right)$ by $G[v,w_1,w_2,v_1]\ne K_{1,3}$ and $w_1{v}_1,$$w_2{v}_1\notin E\left(G\right),$ a contradiction. Thus, $|\{w_1,w_2\}\cap N\left(v\right)|=1.$ Without loss of generality, assume that $w_{1}v\in E\left(G\right)$ and $w_{2}v\notin E\left(G\right).$ Then, $w_2\in epn(w,S).$ Recall that $u\in N_S\left(v\right).$ Suppose $w\ne u.$ Then, $w_{1}u\notin E\left(G\right)$ by $w_1\in epn(w,S^{\prime }),$ which implies $G\left[\{v,v_1,u,w_1\}\right]=K_{1,3},$ a contradiction. Thus, $w=u,$ and then, $G\left[\{w_1,v,u,v_1,w_2\}\right]=bull,$ a contradiction. The claim is true. □

Next, we prove S is an independent set.

Claim 2.

S is an independent set.

Proof. 

Suppose to the contrary that S is not an independent set. Let $u,v$ be in S such that $uv\in E\left(G\right),$ and set $S^{\prime }=S\backslash \left\{v\right\}.$ By Claim 1, $epn(v,S)=\varnothing ,$ and hence, $S^{\prime }$ is a dominating set of $G.$ Since $|S^{\prime }| < |S|$ and S is a minimum co-secure dominating set of $G,$$S^{\prime }$ is not a co-secure dominating set of $G.$ Thus, there is a vertex $w\in S^{\prime }$ such that $rpn(w,S^{\prime })=\varnothing$ and $epn(w,S^{\prime })$ contains two nonadjacent vertices $w_1,w_2$ with $\{w_1,w_2\}\cap N\left(v\right)\ne \varnothing .$

Suppose $\{w_1,w_2\}\subseteq N\left(v\right).$ Then, we have $w_{1}u\in E\left(G\right)$ or $w_{2}u\in E\left(G\right)$ by $G[v,w_1,w_2,u]$$\ne K_{1,3}$ and $w_1{w}_2\notin E\left(G\right),$ which implies $u=w$ by $w_i\in epn(w,S^{\prime })$ for each $i\in \{1,2\}.$ We have $epn(u,S)=\varnothing$ by Claim 1, and hence, each vertex in $epn(u,S^{\prime })$ is adjacent to $v.$ Then, $v\in rpn(u,S^{\prime }),$ a contradiction to $rpn(w,S^{\prime })=\varnothing .$ Thus, $|\{w_1,w_2\}\cap N\left(v\right)|=1.$

Without loss of generality, assume $w_{1}v\in E\left(G\right)$ and $w_{2}v\notin E\left(G\right).$ Then, $w_2\in epn(w,S),$ which implies $w\ne u$ since $epn(u,S)=\varnothing$ by Claim 1. Thus, $w_{1}u,w_{2}u\notin E\left(G\right)$ by $w_1,w_2\in epn(w,S^{\prime }).$ It follows that $G\left[\{u,v,w_1,w,w_2\}\right]=P_5,$ a contradiction. Thus, S is an independent set of G. □

If there is a minimum co-secure dominating set of G that is a maximum independent set of G, then we are done. Suppose on the contrary, that every minimum co-secure dominating set of G is not a maximum independent set of G. Let $S^{\ast }$ be a maximum independent set of G. Subject to the original assumption that S is a minimum co-secure dominating set of G such that $G\left[S\right]$ contains the minimum number of edges; we choose S such that $|S\cap S^{\ast }|$ is maximum. S is an independent set of G by Claim 2, but it is not a maximum independent set of G. Then, $\left|S\right| < |S^{\ast }|$ and $|S^{\ast }\backslash S|>|S\backslash S^{\ast }|.$

Claim 3.

If $x\in S^{\ast }\backslash S$ and $x\in epn(y,S)$ for some vertex $y\in S,$ then $x\notin rpn(y,S).$

Proof. 

Let $x\in S^{\ast }\backslash S$ and $y\in S$ such that $x\in epn(y,S).$ Suppose to the contrary that $x\in rpn(y,S).$ Let $S^{\prime }=(S\backslash \left\{y\right\})\cup \left\{x\right\}.$ Then, we obtain that $epn(y,S)\subseteq N\left[x\right]$ by $x\in rpn(y,S).$ Thus, $S^{\prime }$ is a dominating set of $G,$ and $S^{\prime }$ is an independent set of G by $x\in epn(y,S).$ We have $y\notin S^{\ast }$ by $xy\in E\left(G\right)$ and $x\in S^{\ast },$ and hence, $|S^{\prime }\cap S^{\ast }| > |S\cap S^{\ast }|.$ Then, by the choice of $S,$$S^{\prime }$ is not a co-secure dominating set of G. Clearly, $y\in rpn(x,S^{\prime }).$ It follows that there is a vertex $z\in S^{\prime }\backslash \left\{x\right\}$ such that $rpn(z,S^{\prime })=\varnothing$ and $epn(z,S^{\prime })$ contains two non-adjacent vertices $z_1,z_2$ with $\{z_1,z_2\}\cap N\left(y\right)\ne \varnothing$.

By $z_1,z_2\in epn(z,S^{\prime })$ and $z\ne x,$ we have $z_{1}x,z_{2}x\notin E\left(G\right).$ We claim that $|\{z_1,z_2\}\cap N\left(y\right)|=1.$ For otherwise, $\{z_1,z_2\}\subseteq N\left(y\right),$ and then, $G\left[\{y,x,z_1,z_2\}\right]=K_{1,3}$, a contradiction. Without loss of generality, assume $z_{1}y\in E\left(G\right)$ and $z_{2}y\notin E\left(G\right)$. Then, $N\left(z_1\right)\cap S=\{z,y\}$ and $z_2\in epn(z,S).$ Clearly, $zx\notin E\left(G\right)$ by $x\in epn(y,S).$ Thus, $G\left[\{x,y,z_1,z,z_2\}\right]=P_5,$ a contradiction. It follows that $x\notin rpn(y,S).$ □

Claim 4.

If $x\in S^{\ast }\backslash S,$ then $x\notin epn(y,S)$ for any vertex y in $N_S\left(x\right).$

Proof. 

Let $x\in S^{\ast }\backslash S$ and $y\in N_S\left(x\right)$. Suppose to the contrary that $x\in epn(y,S).$ Then, by Claim 3, $x\notin rpn(y,S).$ Thus, there is a vertex $z\in epn(y,S)\backslash \left\{x\right\}$ such that $zx\notin E\left(G\right).$ Since S is a co-secure dominating set of $G,$$rpn(y,S)\ne \varnothing .$ Let $u\in rpn(y,S)$. Then, $zu,xu\in E\left(G\right).$ We claim that $u\in epn(y,S).$ Suppose on the contrary, $uw\in E\left(G\right)$ for some vertex $w\in S\backslash \left\{y\right\}.$ Then, $wx,wz\notin E\left(G\right)$ by $x,z\in epn(y,S),$ and hence, $G\left[\{u,x,z,w\}\right]=K_{1,3},$ a contradiction. Thus, $u\in epn(y,S).$

We claim that $N\left(y\right)=epn(y,S).$ Note that $N\left(y\right)\subseteq G-S$ since S is an independent set of G. Suppose to the contrary that there is a vertex $v\in N\left(y\right)\backslash epn(y,S).$ We have $d_S\left(v\right)=2$ since G is claw-free and S is an independent set. Let $N_S\left(v\right)=\{y,w\}.$ Clearly, $v\notin \{x,z,u\},$ and $xw,zw,uw\notin E\left(G\right)$ by $x,z,u\in epn(y,S).$ We have $vx\in E\left(G\right)$ or $vz\in E\left(G\right)$ by $G\left[\{y,v,x,z\}\right]\ne K_{1,3}$ and $zx\notin E\left(G\right).$ We claim that $\left|N\right(v)\cap \{z,x\left\}\right|=1$, for otherwise, $G\left[\{v,x,z,w\}\right]=K_{1,3},$ a contradiction. Without loss of generality, assume $vx\in E\left(G\right).$ It follows that $G\left[\right\{x,y,v,z,w\left\}\right]=bull,$ a contradiction. Thus, $N\left(y\right)=epn(y,S).$

Since $\left|S\right|\ge 2$ and G is connected, $N_{G-S}\left(y\right)$ contains a vertex adjacent to some vertex in $G-S$ that has a neighbor in $S\backslash \left\{y\right\}$ by $N\left(y\right)=epn(y,S).$ Let $x_1\in N\left(y\right),$$w_1\in G-S,$$w\in S\backslash \left\{y\right\}$ such that $x_1{w}_1,w_{1}w\in E\left(G\right).$ By $N\left(y\right)=epn(y,S),$ we have that $x_{1}w,w_{1}y\notin E\left(G\right).$ We claim that $w_{1}x,w_{1}z\notin E\left(G\right).$ Suppose on the contrary that $w_{1}x\in E\left(G\right).$ Then, $z{w}_1\in E\left(G\right)$ by $G\left[\{z,y,x,w_1,w\}\right]\ne P_5$ and $zx,w_{1}y,zw,yw,xw\notin E\left(G\right).$ It follows that $G\left[\{w_1,x,z,w\}\right]=K_{1,3},$ a contradiction. Thus, $w_{1}x\notin E\left(G\right),$ which implies $x\ne x_1.$ Similarly, $w_{1}z\notin E\left(G\right)$ and $z\ne x_1.$ We have $x_1\ne u,$ for otherwise, $G\left[\{u,w_1,x,z\}\right]=K_{1,3},$ a contradiction. We obtain that $x_{1}x\in E\left(G\right)$ or $x_{1}z\in E\left(G\right)$ by $G\left[\{y,x_1,x,z\}\right]\ne K_{1,3}$ and $xz\notin E\left(G\right).$ Moreover, we have $|N\left(x_1\right)\cap \{x,z\}|=1,$ for otherwise, $G\left[\{x_1,w_1,x,z\}\right]=K_{1,3},$ a contradiction. Without loss of generality, assume that $x_{1}z\in E\left(G\right).$ Then, $x_{1}x\notin E\left(G\right)$, and hence, $G\left[\{z,y,x_1,x,w_1\}\right]=bull,$ a contradiction. Thus, $x\notin epn(y,S)$ for any vertex $y\in S.$ □

Claim 5.

There are two vertices $x,y$ in $S^{\ast }\backslash S$ such that $d_S\left(x\right)=d_S\left(y\right)=2;$ moreover, $N_S\left(x\right)=N_S\left(y\right).$

Proof. 

Since S is a dominating set of G and $S^{\ast }$ is an independent set of G, each vertex in $S^{\ast }\backslash S$ is adjacent to some vertex in $S\backslash S^{\ast }.$ Moreover, there are at least two vertices in $S^{\ast }\backslash S$ that have a common neighbor in S by $|S^{\ast }\backslash S|>|S\backslash S^{\ast }|.$ Let $x,y\in S^{\ast }\backslash S,$ and $z\in N_S\left(x\right)\cap N_S\left(y\right).$ Since $S^{\ast }$ is an independent set with $x,y\in S^{\ast },$ we have $xy\notin E\left(G\right).$ By Claim 4, $x,y\notin epn(z,S),$ and hence, $min\{d_S\left(x\right),d_S\left(y\right)\}\ge 2.$ Moreover, $d_S\left(x\right)=d_S\left(y\right)=2$ since G is claw-free and S is an independent set of G.

Suppose on the contrary that $N_S\left(x\right)\ne N_S\left(y\right)$. Let $N_S\left(x\right)=\{z,u\}$ and $N_S\left(y\right)=\{z,v\}.$ Clearly, $v\ne u,$ and $zu,zv,uv,xv,xy,yu\notin E\left(G\right).$ Thus, $G\left[\{u,x,z,y,v\}\right]=P_5,$ a contradiction. It follows that x and y have two common neighbors in $S.$ □

Now, let us complete the proof of Theorem 4. By Claim 5, let $v_1,v_2$ be in $S^{\ast }\backslash S$ and $u,v\in S$ such that $N_S\left(v_1\right)=N_S\left(v_2\right)=\{u,v\}.$ Since $v_1,v_2$ are in $S^{\ast }$ and $S^{\ast }$ is an independent set, we have $v_1{v}_2\notin E\left(G\right).$ Let $S^{\prime }=(S\backslash \{u,v\})\cup \{v_1,v_2\}.$ We have $S^{\prime }$ is an independent set of G since S is an independent set of $G,$$N_S\left(v_1\right)=N_S\left(v_2\right)=\{u,v\}$ and $v_1{v}_2\notin E\left(G\right).$ For each vertex $x\in N\left(u\right)\cup N\left(v\right),$ we have $x{v}_1\in E\left(G\right)$ or $x{v}_2\in E\left(G\right)$ by $G\left[\{u\left(v\right),x,v_1,v_2\}\right]\ne K_{1,3}$ and $v_1{v}_2\notin E\left(G\right).$ Thus, $S^{\prime }$ is a dominating set with $|S^{\prime }| = |S|.$ By $v_i\in S^{\ast }$ and $u,v\in N\left(v_i\right),i\in \{1,2\},$ we have $u,v\notin S^{\ast }.$ Thus, $|S^{\prime }\cap S^{\ast }| > |S\cap S^{\ast }|.$ Then, by the choice of $S,$ we have $S^{\prime }$ is not a co-secure dominating set of G. Thus, there is a vertex x in $S^{\prime }$ such that $rpn(x,S^{\prime })=\varnothing .$ We claim that $x\in \{v_1,v_2\}.$

Suppose on the contrary that $x\notin \{v_1,v_2\}$ and $rpn(x,S^{\prime })=\varnothing .$ There are two non-adjacent vertices $x_1,x_2$ in $epn(x,S^{\prime })$ such that $\{x_1,x_2\}\cap (N\left(u\right)\cup N\left(v\right))\ne \varnothing$ by $rpn(x,S^{\prime })=\varnothing .$ Without loss of generality, assume $x_1\in N\left(u\right)\cup N\left(v\right).$ Then, $x_1{v}_1\in E\left(G\right)$ or $x_1{v}_2\in E\left(G\right)$ since each vertex in $N\left(u\right)\cup N\left(v\right)$ is adjacent to $v_1$ or $v_2.$ Thus, $d_{S^{\prime }}\left(x_1\right)\ge 2$ by $x\notin \{v_1,v_2\},$ a contradiction to $x_1\in epn(x,S^{\prime }).$ It follows that $x\in \{v_1,v_2\}.$ Without loss of generality, assume that $x=v_1.$ Then, $x_1{v}_2,x_2{v}_2\notin E\left(G\right)$ by $x_1,x_2\in epn(v_1,S^{\prime }).$ By $G\left[\{v_1,x_1,x_2,u\}\right]\ne K_{1,3}$ and $x_1{x}_2\notin E\left(G\right),$ we have $x_{1}u\in E\left(G\right)$ or $x_{2}u\in E\left(G\right).$ Without loss of generality, assume that $x_{1}u\in E\left(G\right).$ Then, we have $x_{2}u\notin E\left(G\right),$ for otherwise, $G\left[\{u,v_2,x_1,x_2\}\right]=K_{1,3},$ a contradiction. Thus, $G\left[\{x_1,v_1,u,x_2,v_2\}\right]=bull$ by $x_1{x}_2,v_2{x}_1,v_2{x}_2,x_{2}u,v_2{v}_1\notin E\left(G\right),$ a contradiction. Thus, Theorem 4 is true.

By Theorem 4, we obtain the following result, which characterizes a $\{claw,P_5,bull\}$-free graph G with ${\gamma }_{cs}\left(G\right)=\alpha \left(G\right).$

Corollary 1.

Let G be a $\{claw,P_5,bull\}$-free graph such that $\delta \left(G\right)\ge 1$ and $G\notin \mathcal{F}.$ Then, ${\gamma }_{cs}\left(G\right)=\alpha \left(G\right).$

A cograph is a graph that contains no induced subgraph isomorphic to $P_4.$ Obviously, a cograph is $P_5$-free. Then, by Theorem 4, we can obtain the following result.

Corollary 2.

Let G be a $\{claw,bull\}$-free cograph with $\delta \left(G\right)\ge 1$ and $G\notin \mathcal{F}.$ Then ${\gamma }_{cs}\left(G\right)=\alpha \left(G\right).$

3. Characterization of Claw-Free Graphs G with ${\mathit{\gamma }}_{\mathit{cs}}\left(\mathit{G}\right)={\mathit{\gamma }}_{\mathit{s}}\left(\mathit{G}\right)$

According to Problem 2 in Section 1, we characterize the claw-free graph G with ${\gamma }_{cs}\left(G\right)={\gamma }_s\left(G\right)$ in this section. By Theorem 2, for a graph G, if the minimum degree of G is at least 2, then ${\gamma }_{cs}\left(G\right)\le {\gamma }_s\left(G\right).$ For a claw-free graph, we can obtain the following result, in which the minimum degree can be decreased to 1. We will apply this result to the proof of the main result in this section.

Theorem 5.

If G is a claw-free graph and $\delta \left(G\right)\ge 1$, then ${\gamma }_{cs}\left(G\right)\le {\gamma }_s\left(G\right).$

Proof. 

It suffices to prove that there is a minimum secure dominating set of G such that it is also a co-secure domination set of G. If G is disconnected, then the minimum secure (resp. co-secure) domination number is the sum of the minimum secure (resp. co-secure) domination number of each component in G. Thus, without loss of generality, we assume that G is a connected claw-free graph. If G is complete, then we are done. Thus, we assume G is not complete. We prove the theorem by reduction to absurdity. Suppose to the contrary that each minimum secure dominating set of G is not a co-secure domination set of G. Firstly, we have the following result. □

Claim 6.

There is a vertex v in S such that all the neighbors of v are in S for every minimum secure dominating set S of G.

Proof. 

Suppose to the contrary that there is a minimum secure dominating set S of G such that each vertex in S has at least one neighbor in $G-S.$ Since S is not a co-secure dominating set of G, there is a vertex $x\in S$ such that $rpn(x,S)=\varnothing .$

If $epn(x,S)=\varnothing ,$ then each vertex in $N_{G-S}\left(x\right)$ is in $rpn(x,S),$ a contradiction. Thus, $epn(x,S)\ne \varnothing .$ Since S is a secure dominating set, $G\left[epn\right(x,S\left)\right]$ is a clique, which implies $epn(x,S)\subseteq rpn(x,S),$ a contradiction. Thus, the claim is true. □

In the following proof, let S be a minimum secure dominating set of G such that $G\left[S\right]$ contains the minimum number of edges among all the minimum secure dominating sets of $G.$ By Claim 6, let $v\in S$ such that $N\left[v\right]\subseteq S.$ Since $\delta \left(G\right)\ge 1,$$G\left[S\right]$ contains at least one edge. Then, $\left|S\right|\ge 2.$ Let $S^{\prime }=S\backslash \left\{v\right\}.$ Then, $S^{\prime }$ is a dominating set of G by $N\left[v\right]\subseteq S.$ Clearly, $|S^{\prime }| < |S|.$ By the minimality of $\left|S\right|,$$S^{\prime }$ is not a secure dominating set of G. Thus, there is a vertex x in $G-S^{\prime }$ such that $x\notin rpn(y,S^{\prime })$ for each neighbor y of x in $S^{\prime }.$ By $N\left[v\right]\subseteq S,$ it is easy to verify that v is the exactly one vertex in $G-S^{\prime }$ such that $v\notin rpn(y,S^{\prime })$ for every neighbor y of v in $S.$ Thus, there is a vertex in $epn(y,S^{\prime })$ that is non-adjacent to v for each neighbor y of v in $S.$

Let $y\in N\left(v\right)$ and $y_1\in epn(y,S^{\prime })$ such that $v{y}_1\notin E\left(G\right).$ Clearly, $y_1\in epn(y,S).$ We claim that $G\left[N_{G-S}\left(y\right)\right]$ is a clique. For two distinct vertices $u_1,u_2\in N_{G-S}\left(y\right),$ we have $u_{1}v,u_{2}v\notin E\left(G\right)$ by $N\left[v\right]\subseteq S,$ and hence, $u_1{u}_2\in E\left(G\right)$ by $G\left[\{y,u_1,u_2,v\}\right]\ne K_{1,3}.$ Thus, $G\left[N_{G-S}\left(y\right)\right]$ is a clique.

Let $S_1=(S\backslash \left\{y\right\})\cup \left\{y_1\right\}.$ Since $G\left[N_{G-S}\left(y\right)\right]$ is a clique, $S_1$ is a dominating set of G with $|S_1|=|S|.$ Moreover, $N_{S_1}\left(x\right)\subseteq N_S\left(x\right)\cup \left\{y_1\right\}$ for each vertex x in $G-S$ with $xy\notin E\left(G\right),$ and $N_{S_1}\left(z\right)=(N_S\left(z\right)\backslash \left\{y\right\})\cup \left\{y_1\right\}$ for each vertex in $z\in N_{G-S}\left(y\right).$ Thus, for each vertex $x\in G-S,$$x\in rpn(x^{\prime },S_1)$ for some neighbor $x^{\prime }$ of x in $S_1\backslash \{v,y_1\}$ if $\left|S\right|\ge 3$. Furthermore, it is easy to verify that $y\in rpn(v,S_1)$ and $y\in rpn(y_1,S_1).$ It follows that $S_1$ is a secure dominating set of $G.$ Clearly, $G\left[S_1\right]$ contains fewer edges than $G\left[S\right],$ a contradiction. Thus, G contains a minimum secure dominating set that is also a co-secure dominating set of $G,$ i.e., ${\gamma }_s\left(G\right)\ge {\gamma }_{cs}\left(G\right).$

Theorem 6.

If G is a $\{claw,bull,P_5\}$-free graph with $\delta \left(G\right)\ge 1$ and $\alpha \left(G\right)\ne 2,$ then ${\gamma }_{cs}\left(G\right)={\gamma }_s\left(G\right).$

Proof. 

By Theorem 5, it suffices to prove that ${\gamma }_{cs}\left(G\right)\ge {\gamma }_s\left(G\right).$ If G is complete, then we are done. We assume that G is not complete. Then, $\alpha \left(G\right)\ge 3$ by $\alpha \left(G\right)\ne 2.$ Without loss of generality, assume that G is connected. In the following, we will prove that G contains a minimum co-secure dominating set that is also a secure dominating set, and hence, ${\gamma }_{cs}\left(G\right)\ge {\gamma }_s\left(G\right).$ □

Note that $G\notin \mathcal{F}$ by $\alpha \left(G\right)\ne 2.$ Thus, by Theorem 4, let S be a minimum co-secure dominating set of G such that S is a maximum independent set of G. If S is also a secure dominating set of $G,$ then we are done. Thus, we assume that S is not a secure dominating set of $G.$ Notice that $\left|S\right|=\alpha \left(G\right)\ge 3$. Firstly, we have the following result.

Claim 7.

$G\left[epn\right(v,S\left)\right]$ is a clique if $epn(v,S)\ne \varnothing$ for a vertex $v\in S.$

Proof. 

Suppose to the contrary that $v\in S$ and $epn(v,S)$ contains two distinct non-adjacent vertices $v_1,v_2$. Clearly, $v_i$ is non-adjacent to any vertex in $S\backslash \left\{v\right\}$ by $v_i\in epn(v,S),$ for each $i\in \{1,2\}$. Let $S^{\prime }=(S\backslash \left\{v\right\})\cup \{v_1,v_2\}.$ Then, $S^{\prime }$ is an independent set of G with $|S^{\prime }| > |S|,$ a contradiction to the choice that S is a maximum independent set of G. Thus, the claim is true. □

Since S is not a secure dominating set of $G,$ in the following proof, let $x\in G-S$ such that $x\notin rpn(y,S)$ for each neighbor y of x in $S.$ By Claim 7, if $x\in epn(y,S)$ for some vertex $y\in S,$ then $x\in rpn(y,S),$ a contradiction. Thus, $d_S\left(x\right)\ge 2.$

Let $y_1,y_2\in N_S\left(x\right)$. Then, $epn(y_i,S)\ne \varnothing$ and $epn(y_i,S)$ contains at least one vertex that is non-adjacent to x by $x\notin rpn(y_i,S)$ for each $i\in \{1,2\}.$ Let $z_i\in epn(y_i,S)$ such that $z_{i}x\notin E\left(G\right)$ for each $i\in \{1,2\}.$ By $z_1\in epn(y_1,S)$ and $z_2\in epn(y_2,S),$ we have $z_1{y}_2,z_2{y}_1\notin E\left(G\right).$ Since S is an independent set of G, we have $y_1{y}_2\notin E\left(G\right)$. Thus, $z_1{z}_2\in E\left(G\right)$ by $G\left[\{z_1,y_1,x,y_2,z_2\}\right]\ne P_5.$

We claim that $N_S\left(u\right)\subseteq \{y_1,y_2\}$ for each vertex u in $N\left(y_1\right)\cup N\left(y_2\right).$ Note that $N\left(y_1\right)\cup N\left(y_2\right)\subseteq G-S$ since S is an independent set of G. Without loss of generality, suppose to the contrary that $u_1\in N\left(y_1\right),y_3\in S\backslash \{y_1,y_2\}$ such that $u_1{y}_3\in E\left(G\right).$ Since S is an independent set of G and G is claw-free, we have $N_S\left(u_1\right)=\{y_1,y_3\}$ and $N_S\left(x\right)=\{y_1,y_2\},$ which implies $u_1\ne x.$ Thus, $u_1{y}_2,x{y}_3\notin E\left(G\right).$ We have $u_{1}x\in E\left(G\right)$ by $G\left[\{y_3,u_1,y_1,x,y_2\}\right]\ne P_5,$ which implies $G\left[\{y_1,u_1,x,y_3,y_2\}\right]=bull,$ a contradiction. It follows that $N_S\left(u\right)\subseteq \{y_1,y_2\}$ for each vertex u in $N\left(y_1\right)\cup N\left(y_2\right).$

Since $\left|S\right|\ge 3$ and G is connected, there is a vertex in $N\left(y_1\right)\cup N\left(y_2\right)$ adjacent to some vertex in $G-S$ that has a neighbor in $S\backslash \{y_1,y_2\}$ by $N_S\left(u\right)\subseteq \{y_1,y_2\}$ for each vertex u in $N\left(y_1\right)\cup N\left(y_2\right).$

Without loss of generality, assume that $u_1\in N_{G-S}\left(y_1\right),$$v_1\in G-S,v_2\in S\backslash \{y_1,y_2\}$ such that $u_1{v}_1,v_1{v}_2\in E\left(G\right).$ Then, $x{v}_2,v_1{y}_1,v_1{y}_2,u_1{v}_2\notin E\left(G\right),$ and $x\ne v_1$ by $N_S\left(u\right)\subseteq \{y_1,y_2\}$ for each vertex u in $N\left(y_1\right)\cup N\left(y_2\right).$ We have $v_{1}x\notin E\left(G\right)$ for otherwise, $G[x,v_1,y_1,y_2]$$=K_{1,3}$ by $v_1{y}_1,v_1{y}_2,y_1{y}_2\notin E\left(G\right),$ a contradiction. It follows that $u_1\ne x.$ We have $u_{1}x\in E\left(G\right)$ by $G\left[\{v_2,v_1,u_1,y_1,x\}\right]=P_5$ and $v_1{y}_1,u_1{v}_2,y_1{v}_2,x{v}_2,v_{1}x\notin E\left(G\right).$ Moreover, we have $u_1{y}_2\in E\left(G\right)$ by $G\left[\{y_1,u_1,x,v_1,y_2\}\right]\ne bull$ and $v_{1}x,v_1{y}_1,v_1{y}_2,y_1{y}_2\notin E\left(G\right).$ It follows that $G\left[\{u_1,v_1,y_1,y_2\}\right]=K_{1,3},$ a contradiction. Thus, Theorem 6 is true.

Note that $\alpha \left(G\right)\ne 2$ is necessary in Theorem 6. Let $G_1,G_2$ and $G_3$ be three vertex-disjoint graphs such that $G_1=K_{m_1}, G_2=K_{m_2},$ and $G_3=v_1{v}_2{v}_3{v}_4{v}_5{v}_1$ is a 5-cycle, where $m_1,m_2\ge 1$. Let G be a graph (see Figure 4) obtained by joining every vertex in $G_1$ to each vertex $\{v_1,v_4,v_5\}$ of $G_3,$ and joining every vertex in $G_2$ to each vertex in $\{v_1,v_2,v_3\}$ of $G_3.$ Clearly, G is a $\{claw,bull,P_5\}$-free graph with $\delta \left(G\right)\ge 1.$ It is easy to verify that $\alpha \left(G\right)=2,$${\gamma }_{cs}\left(G\right)=2,$ and ${\gamma }_s\left(G\right)=3.$

It is obvious that $\alpha \left(G\right)=2$ for each graph G in $\mathcal{F}.$ Thus, by Corollary 1 and Theorem 6, we can obtain the following result.

Corollary 3.

Let G be a $\{claw,bull,P_5\}$-free graph with $\delta \left(G\right)\ge 1$ and $\alpha \left(G\right)\ne 2.$ Then, ${\gamma }_{cs}\left(G\right)=\alpha \left(G\right)={\gamma }_s\left(G\right).$

In the following, we characterize the relationship between the co-secure domination number and secure domination number of $\{claw,bull,Z_2\}$-free graphs.

Theorem 7.

If G is a $\{claw,bull,Z_2\}$-free graph with $\delta \left(G\right)\ge 3$ and $\alpha \left(G\right)\ne 2,$ then ${\gamma }_{cs}\left(G\right)={\gamma }_s\left(G\right).$

Proof. 

If G is complete, then we are done. Thus, we assume G is not complete, and then, $\alpha \left(G\right)\ge 3$ by $\alpha \left(G\right)\ne 2.$ By Theorem 2, it suffices to prove that ${\gamma }_{cs}\left(G\right)\ge {\gamma }_s\left(G\right).$ As in the proof of Theorem 6, assume G is connected, and in the following proof, we show that G contains a minimum co-secure dominating set S such that S is also a secure dominating set of G.

By Theorem 3, let S be a minimum co-secure dominating set of G such that S is a maximum independent set of G. Then $\left|S\right|\ge 3$ by $\alpha \left(G\right)\ge 3$. If S is a secure dominating set of $G,$ then we are done. Suppose to the contrary that S is not a secure dominating set of $G.$ Let x be in $G-S$ such that $x\notin rpn(y,S)$ for each neighbor y of x in $S.$ As in the proof of Claim 7 of Theorem 6, we can also obtain that $G\left[epn\right(v,S\left)\right]$ is a clique, and then $epn(v,S)\subseteq rpn(v,S)$ if $epn(v,S)\ne \varnothing$ for a vertex $v\in S.$ Thus, $d_S\left(x\right)\ge 2$ and $epn(y,S)$ contains a vertex that is non-adjacent to x for each neighbor y of x in $S.$

Let $y_1,y_2\in N_S\left(x\right)$ and $z_i\in epn(y_i,S)$ such that $z_{i}x\notin E\left(G\right)$ for each $i\in \{1,2\}.$

We claim that $N_S\left(u\right)\subseteq \{y_1,y_2\}$ for each vertex u in $N\left(y_1\right)\cup N\left(y_2\right)$. Without loss of generality, suppose on the contrary that $u_1\in N_S\left(y_1\right),$$v_1\in S\backslash \{y_1,y_2\}$ such that $u_1{v}_1\in E\left(G\right)$ by $\alpha \left(G\right)\ge 3$. Since G is claw-free and S is an independent set, we have $N_S\left(u_1\right)=\{y_1,v_1\}$ and $N_S\left(x\right)=\{y_1,y_2\}$. Thus, $u_1\ne x,$ and $x{v}_1,u_1{y}_2\notin E\left(G\right).$ Moreover, $u_1\notin \{z_1,z_2\}$ by $z_1\in epn(y_1,S)$ and $z_2\in epn(y_2,S)$. By $G\left[\{y_1,u_1,z_1,x\}\right]\ne K_{1,3}$ and $x{z}_1\notin E\left(G\right),$ we have $u_{1}x\in E\left(G\right)$ or $z_1{u}_1\in E\left(G\right).$ Suppose $u_{1}x\in E\left(G\right)$. Then, $G\left[\{y_1,x,u_1,y_2,v_1\}\right]=bull$ by $y_1{y}_2,u_1{y}_2,x{v}_1,v_1{y}_1,v_1{y}_2,x{v}_1\notin E\left(G\right),$ a contradiction. Thus, $u_{1}x\notin E\left(G\right),$ and then, $z_1{u}_1\in E\left(G\right).$ It follows that $G\left[\{z_1,y_1,u_1,x,v_1\}\right]=bull,$ a contradiction. Thus, $N_S\left(u\right)\subseteq \{y_1,y_2\}$ for each vertex u in $N\left(y_1\right)\cup N\left(y_2\right)$.

Since G is connected and $\left|S\right|\ge 3,$ we have $N\left(y_1\right)\cup N\left(y_2\right)$ contains a vertex adjacent to some vertex in $G-S$ that has a neighbor in $S\backslash \{y_1,y_2\}$ since G is connected. Without loss of generality, let $u_1\in N_{G-S}\left(y_1\right),$$u_2\in G-S,$$v\in S\backslash \{y_1,y_2\}$ such that $u_1{u}_2,v{u}_2\in E\left(G\right).$ Then, $u_{1}v,u_2{y}_1,u_2{y}_2\notin E\left(G\right)$ by $N_S\left(u\right)\subseteq \{y_1,y_2\}$ for each vertex u in $N\left(y_1\right)\cup N\left(y_2\right)$. We have $u_{2}x\notin E\left(G\right),$ for otherwise, $G\left[\{x,u_2,y_1,y_2\}\right]=K_{1,3},$ a contradiction. Thus, $u_1\ne x_1.$

We have that $u_{1}x\notin E\left(G\right),$ for otherwise $G\left[\{u_1,x,y_1,u_2,v\}\right]=Z_2,$ a contradiction. Moreover, we have $u_1{y}_2\notin E\left(G\right),$ for otherwise $G\left[\{u_1,y_1,y_2,u_2\}\right]=K_{1,3},$ a contradiction.

Since $\delta \left(G\right)\ge 3$ and $u_1\ne x,$ let $z\in N_{G-S}\left(y_1\right)\backslash \{u_1,x\}.$ Then, $zx\in E\left(G\right)$ or $z{u}_1\in E\left(G\right)$ by $G\left[\{y_1,u_1,x,z\}\right]\ne K_{1,3}$ and $u_{1}x\notin E\left(G\right).$ We have $zv\notin E\left(G\right)$ by $N_S\left(u\right)\subseteq \{y_1,y_2\}$ for each vertex u in $N\left(y_1\right)\cup N\left(y_2\right)$.

We claim that $z{u}_1\notin E\left(G\right).$ Otherwise, $z{u}_2\in E\left(G\right)$ by $G\left[\{u_1,z,y_1,u_2,v\}\right]\ne Z_2$ and $u_2{y}_1,v{y}_1,u_{1}v,zv\notin E\left(G\right)$. Then, we have $z{y}_2\notin E\left(G\right),$ for otherwise $G\left[\{z,u_2,y_1,y_2\}\right]=K_{1,3},$ a contradiction. Thus, $zx\in E\left(G\right)$ by $G\left[\{y_1,z,u_1,x,y_2\}\right]\ne Z_2$ and $u_1{y}_2,y_1{y}_2,z{y}_2,u_{1}x\notin E\left(G\right).$ It follows that $G\left[\{z,y_1,x,u_2,v\}\right]=Z_2,$ a contradiction. Thus, $z{u}_1\notin E\left(G\right),$ and then, $zx\in E\left(G\right).$

We have $z{u}_2\in E\left(G\right)$ by $G\left[\{y_1,x,z,u_1,u_2\}\right]\ne Z_2$ and $z{u}_1,u_{1}x,u_{2}x,u_2{y}_1\notin E\left(G\right).$ Thus, $G\left[\{u_2,u_1,z,v\}\right]=K_{1,3}$, a contradiction. Thus, the theorem is true. □

In the following, we construct a $\{claw,bull,Z_2\}$-free G with $\delta \left(G\right)\ge 3$ such that $\alpha \left(G\right)=2,$${\gamma }_{cs}\left(G\right)=2$ and ${\gamma }_s\left(G\right)=3,$ which shows that $\alpha \left(G\right)\ne 2$ is necessary in Theorem 7. Let $H_1$ and $H_2$ be two complete graphs of order $m, m\ge 2,$ respectively, such that $V\left(H_1\right)=\{u_1,\cdots ,u_m\}$ and $V\left(H_2\right)=\{v_1,\cdots ,v_m\}.$ Assume that $H_3=K_{m^2}$ such that $V\left(H_3\right)=\{w_{ij}:1\le i, j\le m\}.$ Let $G_1=H_1\vee H_2.$ Suppose that H is a graph obtained by joining every vertex $w_{ij}$ of $H_3$ to each vertex in $G_1$ except $u_i$ of $H_1$ and $v_j$ of $H_2,$$1\le i,j\le m.$ Let G be the graph (see Figure 5) with $V\left(G\right)=\{u,v\}\cup V\left(H\right)$ such that $vu\notin E\left(G\right),$u is adjacent to each vertex in $V\left(H_1\right)\cup V\left(H_3\right)$ of $H,$ and v is adjacent to each vertex in $V\left(H_2\right)\cup V\left(H_3\right)$ of $H.$

By Theorem 3 and Theorem 7, we can obtain the following results.

Corollary 4.

Let G be a $\{claw,bull,Z_2\}$-free graph with $\delta \left(G\right)\ge 3$ and $\alpha \left(G\right)\ne 2.$ Then, ${\gamma }_{cs}\left(G\right)=\alpha \left(G\right)={\gamma }_s\left(G\right).$

Corollary 5.

Let G be a 3-connected $\{claw,bull,Z_2\}$-free graph and $\alpha \left(G\right)\ne 2.$ Then, ${\gamma }_{cs}\left(G\right)=\alpha \left(G\right)={\gamma }_s\left(G\right).$

4. Conclusions

Co-secure dominating sets can be applied to wireless sensor networks, network security, combinatorial optimization problems, and other fields. Claw-free graphs are beneficial for designing deadlock-free routing algorithms and simplifying communication protocols in supercomputer networks. Moreover, inspired by Arumugam et al.’s [9] questions on the co-secure domination number, in this paper, we characterize two classes of claw-free graphs such that their co-secure domination numbers equal the independence number and the secure domination number. The research on degree conditions of graphs are always the focus problems in graph theory, and the minimum degree affects the bound of co-secure domination number. Thus, we propose the question: What is the minimum degree $\delta \left(G\right)\ge k$ such that if G is claw-free, then ${\gamma }_{cs}\left(G\right)=\alpha \left(G\right)={\gamma }_s\left(G\right)$ ?