Examining Captive and Inverse Captive Domination in Selected Graphs and Their Complements

Let $\mathrm{G}=(\mathrm{V}, \mathrm{E})$ be a finite simple undirected graph. Let $N(v)=\{u \in V, u v \in E\}$ be the open neighborhood of a vertex $v$ and $N[v]=N(v) \cup\{v\}$ is the closed neighborhood set. Also, let $G[D]$ be the subgraph of $G$ induced by vertices in $D$ [1]. A subset $D \subseteq V(G)$ is a dominating set of $G$ if $N(v) \cap$ $D \neq \emptyset ;$ for all $v \in V-D$.

A dominating set $D$ is minimal dominating if it does not have a proper subset dominating set. The domination number $\gamma(G)$ is the minimum cardinality of a dominating set $D$ of $G$. If $V-D$ contains a dominating set, then this set is called an inverse set of $D$ in $G$. The symbol $\gamma^{-1}(G)$ represents the minimum cardinality over all inverse dominating set of G [2]. The concept of domination is used to solve many problems in various fields of mathematics subjects such as topological graphs [2, 3], fuzzy graphs [4, 5], and [6, 7], number theory graph [8], general graphs [9-18], and others. The reader can be found all notions not mentioned [19-22]. The captive domination is initiated by Al-Harere et al. [1], and they get many properties and bounded. the main targeted applications are placing security cameras, or personnel on a site, minimizing the effect of losing a dominating vertex by keeping another vertex with the same capabilities and each dominating vertex is dominating other vertices outside the dominating set-in addition to dominating a neighbouring vertex in the dominating set. In this work, the new properties are discussed, also for some graphs and it is complementing the captive domination and the inverse captive domination are determined for some graphs as a ladder graph, corona graph of two paths, lollipop graph, barbell graph, corona graph of a cycle of order n and null graph of order p, and helm graph.

3.1 Ladder graph

Proposition 3.1.1 If G is ladder graph, then:

$\gamma_{\mathrm{ca}}(\mathrm{G})=\left\{\begin{array}{c}2, \text { if } \mathrm{n}=2 \\ 2\left\lceil\frac{\mathrm{n}}{3}\right\rceil, \text { if } \mathrm{n} \geq 3\end{array}\right\}$ and there is no $\mathrm{CDN}$ if $n=1$

Proof. In the ladder graph, there are two copies of $P_n$ as an induced subgraph. The vertex set of the first path is $\left\{v_1, v_2, \ldots \ldots, v_n\right\}$ and the vertex set of the second path is $\left\{v_{n+1}, v_{n+2}, \ldots \ldots, v_{2 n}\right\}$ such that for each vertex in the first path $v_i ; 1 \leq i \leq n$ there is a corresponding vertex $v_{i+n}$ in the second path and these corresponding vertices are adjacent as shown in the Figure 1.

1.png

Figure 1. P₂×P_n

Let $v_1, v_2, \ldots \ldots, v_n, v_{n+1}, v_{n+2}, \ldots \ldots, v_{2 n}$ be the vertices of graph $G$ and $D \subset V(G)$ such that:

$\mathrm{D}=\left\{\begin{array}{c}\left\{v_{2+3 i}, v_{(n+2)+3 i}, i=0,1, \ldots \ldots, \frac{n}{3}-1\right\} \text { if } n \equiv 0(\bmod 3) \\ \left\{v_{2+3 i}, v_{(n+2)+3 i}, i=0,1, \ldots \ldots,\left[\frac{n}{3}\right]-2\right\} \cup\left\{v_n, v_{2 n}\right\} \text { if } \\ n \equiv 1(\bmod 3) \\ n \equiv 2(\bmod 3)\end{array}\right.$

For each copy of the path, each vertex can be dominated three vertices as a maximum, so each two corresponding vertices can be CDS at most six vertices. Thus, for every consecutive three vertices, the middle vertex can be chosen.

Therefore, depending on n there are three classifications as below.

Case 1 . If $n=1$, then the graph $P_2 \times P_1 \equiv P_2$, so one can be concluded that there is no $C D S$ in this case.

Case 2. If $n=2$, then the graph $P_2 \times P_2 \equiv C_4$, and it is obvious that $\gamma_{c a}\left(P_2 \times P_2\right)=2$.

Case 3. If $n \geq 3$, then two subcases that depend on $n$ are discussed as the following.

Subcase $1 . \quad$ If $n \equiv 0(\bmod 3)$, then let $D_1=$ $\left\{v_{2+3 i}, v_{(n+2)+3 i}, i=0,1, \ldots \ldots, \frac{n}{3}-1\right\}$, so it is obvious that the set $D_1$ is $C D S$ and it is the minimum, because each vertex dominates the maximum number as possible. Thus, $\gamma_{c a}(G)=$ $\frac{2 n}{3}$.

Subcase 2 . If $n \not \equiv 0(\bmod 3)$, then two subcases are discussed as the following.

I) If $n \equiv 1(\bmod 3)$, then $n-1 \equiv 0(\bmod 3)$. In the same manner in Subcaes 1, the set $D_2=\left\{v_{2+3 i}, v_{(n+2)+3 i}, i=\right.$ $\left.0,1, \ldots,\left\lceil\frac{n}{3}\right]-2\right\}$ is minimum $C D S$ for all the graph $G$ except the two vertices $v_n$ and $v_{2 n}$, so the set $D_3=D_2 \cup\left\{v_n, v_{2 n}\right\}$ is the minimum $C D S$ and $\gamma_{c a}(G)=2\left\lceil\frac{n}{3}\right\rceil$.

II) If $n \equiv 2(\bmod 3)$, then $n-2 \equiv 0(\bmod 3)$.

In the same manner in subcase 1 , the set $D_4=$ $\left\{v_{2+3 i}, v_{(n+2)+3 i}, i=0,1, \ldots,\left\lceil\frac{n}{3}\right\rceil-2\right\}$ is minimum $C D S$ for all the graph $G$ except the four vertices $v_{n-1}, v_n, v_{2 n-1}$, and $v_{2 n}$, so the set $D_5=D_4 \cup\left\{v_n, v_{2 n}\right\}$ is the minimum $C D S$ and $\gamma_{c a}(G)=2\left[\frac{n}{3}\right]$.

From each case above, the required is done.

Proposition 3.1.2 If G is ladder graph, then:

$\gamma_{c a}^{-1}(G)=\left\{\begin{array}{c}2, \text { if } n=2 \\ \frac{2 n}{3}+2, \text { if } n \equiv 0(\bmod 3) \\ 2\left[\frac{n}{3}\right], \text { if } n \equiv 1,2(\bmod 3)\end{array}\right\}$

Proof. Depending on n there are three classification as below.

Case 1. If $n=2$, then the graph $P_2 \times P_2 \equiv C_4$, and it is obvious that $\gamma^{-1}{ }_{c a}\left(P_2 \times P_2\right)=2$.

Case 2. If $n \equiv 0(\bmod 3)$, then let $D=\left\{v_{2+3 i}, v_{(n+2)+3 i}\right.$, $\left.i=0,1, \ldots ., \frac{n}{3}-1\right\}$, so it is obvious that the set $D$ is $C D S$ and it is the minimum, and let $S=\left\{v_{1+3 i}, v_{(n+1)+3 i}, i=\right.$ $\left.0,1, \ldots, \frac{n}{3}-1\right\} \cup\left\{v_n, v_{2 n}\right\}$ is inverse $C D S$ with respect to $D$ then $\gamma_{c a}^{-1}(G)=\frac{2 n}{3}+2$.

Case 3. If $n \equiv 1,2(\bmod 3)$, then let $D_1=$ $\left\{v_{2+3 i}, v_{(n+2)+3 i}, i=0,1, \ldots,\left[\frac{n}{3}\right]-2\right\} \cup\left\{v_n, v_{2 n}\right\}$, so it is $C D S$ and it is the minimum, and let $S_1=\left\{v_{1+3 i}, v_{(n+1)+3 i}, i=\right.$ $\left.0,1, \ldots,\left\lceil\frac{n}{3}\right\rceil-2\right\} \cup\left\{v_{n-1}, v_{2 n-1}\right\}$ is inverse $C D S$ with depending on the set $D_1$ then $\gamma_{c a}^{-1}(G)=2\left[\frac{n}{3}\right]$.

From each case above, the required is done.

Proposition 3.1.3 If $G$ be a graph denoted by $G \equiv$ $\overline{\mathrm{P}_2 \times \mathrm{P}_{\mathrm{n}}}$ for $\mathrm{n} \geq 3$ then $\gamma_{\mathrm{ca}}\left(\overline{\mathrm{P}_2 \times \mathrm{P}_{\mathrm{n}}}\right)=2$ and there is no captive domination when $n=1,2$.

Proof. Let G be a graph of order 2n, depending on n  there are three classifications as below.

Case 1. If $n=1$, then the graph $\overline{P_2 \times P_1} \equiv \overline{K_2}$, so one can be concluded that there is no $C D S$ in this case.

Case 2. If $n=2$, then the graph $\overline{P_2 \times P_2} \equiv \overline{C_4}$, so there is no $C D S$ in this case.

Case 3. If $n \geq 3$, suppose that the vertices $v_1$ and $v_{2 n} \in$ $P_2 \times P_n$ such that $v_1$ is adjacent to two vertices $v_2, v_{n+1}$ and the vertex $v_{2 \mathrm{n}}$ is adjacent to two vertices $v_{\mathrm{n}}, v_{2 \mathrm{n}-1}$ by proposition 3.1.1 (as shown in the Figure 1) since four vertices in $\overline{P_2 \times P_n}$ has regular degree $2 n-3$ and $2 n-4$ vertices in $\overline{P_2 \times P_n}$ has regular degree $2 n-4$, (as shown in the Figure 2). therefore, the vertex $v_1$ is adjacent to all vertices (except the two vertices $v_2, v_{n+1}$ ) and the vertex $v_{2 n}$ is adjacent to all vertices (except the two vertices $v_n, v_{2 n-1}$ ) in $\overline{P_2 \times P_n}$ the $C D N$ is 2.

From each case above, the required is done.

2.png

Figure 2. $\overline{P_2 \times P_n}$

Proposition 3.1.4 If $G$ be a graph denoted by $G \equiv$ $\overline{\mathrm{P}_2 \times \mathrm{P}_{\mathrm{n}}}$ for $\mathrm{n} \geq 3$ then $\gamma^{-1}{ }_{\text {ca }}\left(\overline{\left.\mathrm{P}_2 \times \mathrm{P}_{\mathrm{n}}\right)}=\gamma_{\mathrm{ca}}\left(\overline{\left.\mathrm{P}_2 \times \mathrm{P}_{\mathrm{n}}\right)}=2\right.\right.$ and there is no inverse captive domination when $\mathrm{n}=1,2$.

Proof. Depending on $n$ there are three classifications as below.

Case 1 . If $n=1$, then the graph $\overline{P_2 \times P_1} \equiv \overline{K_2}$, so one can be concluded that there is no inverse $C D S$ in this case.

Case 2. If $n=2$, then the graph $\overline{P_2 \times P_1} \equiv \overline{C_4}$, so there is no inverse $C D S$ in this case.

Case 3. If $n \geq 3$, by Proposition 2.1 since four vertices in $\overline{P_2 \times P_1}$ has (2n-3)-regular degree and $2 n-4$ vertices in $\overline{P_2 \times P_1}$ has (2n-4)-regular degree, then there exist $C D S$ in V-D this set is inverse $C D S$ respect with $D$ then $\gamma^{-1}{ }_{c a}\left(\overline{P_2 \times P_1}\right)=2$.

From each case above, the required is done.

3.2 The corona graph of two paths

Proposition 3.2.1 If $G$ be a corona graph denoted by $G \equiv$ $P_n \odot P_m$ then $\gamma_{c a}(G)=\left\{\begin{array}{l}2 \text { if } n=1, m \geq 2 \\ n \text { if } n>1, m \geq 1\end{array}\right\}$, the graph $G$ has no $C D S$ when $\mathrm{n}=1, \mathrm{~m}=1$.

Proof. Let $G$ be a graph of order $n+n m$ such that $P_n$ be a path of order $n$ and $P_m$ be a path of order $\mathrm{m}$ then depending on $n$ as shown in the Figure 3. There are three classifications as below.

Case 1. If $n=1, m=1$ then the graph $P_n \odot P_m \equiv K_2$, so one can be concluded that there is no $C D S$ in this case.

Case 2. If $n=1, m \geq 2$ then the graph $P_n \odot P_m \equiv F_n$, then $\gamma_{c a}(G)=2$.

Case 3. $n>1, m \geq 1$, let $v_1, v_2, \ldots \ldots, v_n$ the vertices of path $P_n$ and let $u_1, u_2, \ldots \ldots, u_m$ the vertices of path $P_m$, since every vertex in $P_n$ join with vertices of copy path $P_m$ from $u_1$ to $u_m$ and $D-v, v \in p_n$ not $C D S$ then the minimum dominating set $\mathrm{D}=\left\{v_1, v_2, \ldots \ldots, v_n\right\}$, and $\gamma_{c a}(G)=n$.

From each case above, the required is done.

3.png

Figure 3. $G \equiv P_n \odot P_m$

Proposition 3.2.2 If G be a corona graph denoted by $G \equiv P_n \odot P_m$ then: 

$\gamma_{\text {ca }}^{-1}(\mathrm{G})=\left\{\begin{array}{c}2\left\lceil\frac{\mathrm{m}}{4}\right\rceil \text { if } \mathrm{n}=1, \mathrm{~m} \neq 5,9 \\ 2\left\lceil\frac{\mathrm{m}}{4}\right\rceil-1 \text { if } \mathrm{n} =1, \mathrm{~m}=5,9 \\ \mathrm{n}\left(2\left\lceil\frac{\mathrm{m}}{4}\right\rceil\right) \text { if } \mathrm{n} \geq 2, \mathrm{~m} \geq 2\end{array}\right\}$

and the graph G has no inverse CDS when $\mathrm{n}=1, \mathrm{~m}=1,2$ and when n=2, m=1.

Proof. Let $G$ be a graph of order $n+n m$ such that $P_n$ be a path of order $n$ and $P_m$ be a path of order $m$ then depending on $n$ there are six classifications as below.

Case 1. If $n=1, m=1$ then the graph $P_1 \odot P_1 \equiv K_2$, so one can be concluded that there is no inverse $C D S$ in this case.

Case 2. If $n=1, m=2$ then the graph $P_1 \odot P_2 \equiv K_3$, so one can be concluded that there is no inverse $C D S$ in this case.

Case 3. If $n=2, m=1$ then the graph $P_2 \odot P_1 \equiv K_3$, so one can be concluded that there is no inverse $C D S$ in this case.

Case 4. If $n=1, m \neq 5,9$ and since $P_1 \odot P_m \equiv F_n$, the $C D S D$ contains one vertex say $v$ in $P_1$ of degree $n-1$ such that $D=\left\{v, u_1\right\}$, the vertex $u_1$ be in $P_m$, so these vertices must be out of inverse $C D S D^{-1}$, then $\gamma_{c a}^{-1}\left(P_1 \odot P_m\right) \equiv \gamma_{c a}\left(P_n\right)$ by theorem 2.12 [1].

Case 5. If n=1, m=5,9 then there exist two subcases as follows:

i) If $n=1, m=5$ then let $D=\left\{v_1, u_1\right\}$, so it is obvious that the set $D$ is $C D S$ and it is the minimum, so the $D^{-1}=\left\{u_2, u_3, u_4\right\}$ is inverse $C D S$ with respect to $D$ then $\gamma^{-1}{ }_{c a}(G)=2\left\lceil\frac{m}{4}\right\rceil-1$.

ii) If $n=1, m=9$ then let $D=\left\{v_1, u_1\right\}$, so it is obvious that the set $D$ is $C D S$ and it is the minimum, so the $D^{-1}=\left\{u_2, u_3\right\} \cup$ $\left\{u_{m-3}, u_{m-2}, u_{m-1}\right\}$ is inverse $C D S$ depending on the set $D$ then $\gamma_{c a}^{-1}(G)=2\left\lceil\frac{m}{4}\right\rceil-1$.

Case 6 . If $n \geq 2, m \geq 2$, now by Proposition 3.1 , then the minimum dominating set $\mathrm{D}=\left\{v_1, v_2, \ldots \ldots, v_n\right\}$, these vertices must be out of the minimum inverse dominating set $D^{-1}$, so $\gamma_{c a}^{-1}\left(P_1 \odot P_m\right) \equiv n\left(\gamma_{c a}\left(P_n\right)\right)$

From each case above, the required is done.

Proposition 3.2.3 If $G$ be a graph denoted by $G \equiv \overline{P_n \odot P_m}$ then $\gamma_{\mathrm{ca}}(\mathrm{G})=2$, is no captive domination when $\mathrm{n}=1, \mathrm{~m} \geq$ 1.

Proof. Let $G$ be a graph of order $n+n m$ such that $P_n$ be a path of order $n$ and $P_m$ be a path of order $m$ then depending on $n$ there are two classification as below.

Case 1. If $n=1, m \geq 1$ then the graph $\overline{P_1 \odot P_m}$ has isolated vertex, so one can be concluded that there is no $C D S$ in this case.

Case 2. If $n \geq 2, m \geq 1$ let $v_1, v_2, \ldots \ldots, v_n$ the vertices of path $P_n$ and let $u_1, u_2, \ldots \ldots, u_m$ the vertices of path $P_m$ since the vertex $u_1$ in copy path $P_m$ adjacent to every vertex in $P_n$ and copy path $P_m$ in the graph $\overline{P_1 \odot P_m}$ except two vertices the vertex $v_1$ and $u_2$ whose adjacent with it in $P_n \odot P_m$ and since there exist another vertex in the graph $\overline{P_1 \odot P_m}$ adjacent with $u_1$ and dominates on these vertices $v_1$ and $u_2$ then the $C D N$ is 2 ( as shown in Figure 4).

Depending to two cases above, the required is done.

4.png

Figure 4. $G \equiv \overline{P_n \odot P_m}$

Proposition 3.2.4 If $G$ be a graph denoted by $\mathrm{G} \equiv \overline{\mathrm{p}_{\mathrm{n}} \odot \mathrm{p}_{\mathrm{m}}}$ then $\gamma_{\mathrm{ca}}^{-1}(\mathrm{G})=2$, the graph $G$ has no inverse CDS when $n=$ $1, m \geq 1$ and $n=2, m=1$.

Proof. Depending on n there are three classifications as below.

Case 1. If $n=1, m \geq 1$ then the graph $\overline{P_1 \odot P_m}$ has isolate vertex, so one can be concluded that there is no $C D S$ in this case and there is no inverse $C D S$.

Case 2. If $n=2, m=1$ then the graph $\overline{P_2 \odot P_1} \equiv P_4$, so one can be concluded that there is no inverse $C D S$ in this case by Note 3.3 [1].

Case 3. If $n \geq 2, m>1$ then by above proposition there exist another dominating set in $V$ - $D$ such that this set is inverse $C D S$ and minimum, so $\gamma_{c a}^{-1}(G)=2$.

From all cases above, the required is done.

3.3 The lollipop graph

Proposition 3.3.1 If $G$ is a lollipop graph denoted by $G \equiv$ $\mathrm{L}_{\mathrm{m}, \mathrm{n}}$ then:

$\gamma_{\mathrm{ca}}(\mathrm{G})=\left\{\begin{array}{c}2\left\lceil\frac{\mathrm{n}}{4}\right\rceil, \text { if } \mathrm{n} \equiv 1,2(\bmod 4) \\ 2\left\lceil\frac{\mathrm{n}}{4}\right\rceil+2, \text { if } \mathrm{n} \equiv 0,3(\bmod 4)\end{array}\right.$

Proof. Let $v_1=u_0$, then depending on $n$ there are three classifications as below.

Case 1. If $n \equiv 2(\bmod 4)$, then let $D_1=\left\{u_{4 i}, u_{1+4 i}, i=\right.$ $\left.0, \ldots,\left[\frac{n}{4}\right]-1\right\}$. The vertex $u_0=v_1$, so this vertex dominates all vertices in the induced subgraph which is isomorphic to the complete graph and the vertex $u_1$ is adjacent to the vertex $u_0$ and dominates the vertex $u_2$ as shown in Figure 5. Thus, the two vertices $u_0$ and $u_1$ are taken in a $C D S$, after this leave two vertices $\left(u_2\right.$ and $\left.u_3\right)$ and take two vertices $\left(u_4\right.$ and $\left.u_5\right)$. The vertex $u_4$ dominates the vertex $u_3$ and in this way we keep the totality condition of $C D S$ and so on. Thus, the set $D_1$ is a minimum $C D S$.

Case 2 . If $n \equiv 1(\bmod 4)$, then depending on $n$ there are two classifications as below.

I) If $n=1$, then $D=\left\{v_1, v_2\right\}$ it is clear that set $D$ is the minimum $C D S$.

II) Let $D_2=\left\{v_1, v_2\right\} \cup\left\{u_{3+4 i}, u_{4+4 i}, i=0, \ldots,\left\lceil\frac{n}{4}\right\rceil-2\right\}$. two vertices $v_1, v_2$ so these vertices dominate all vertices in the induced subgraph which is isomorphic to the complete graph and vertex $u_0=v_1$ is adjacent to the vertex $u_1$. The two vertices $v_1$ and $v_2$ are taken in a $C D S$, after this leave two vertices $\left(u_1\right.$ and $\left.u_2\right)$ and take two vertices $\left(u_3\right.$ and $\left.u_4\right)$. The vertex $u_3$ dominates the vertex $u_2$ and in this way, we keep the totality condition of $C D S$ and so on. Thus, the set $D_2$ is a minimum $C D S$.

Case 3. If $n \equiv 0(\bmod 4)$, then let $D_3=\left\{v_1, v_2\right\} \cup$ $\left\{u_{2+4 i}, u_{3+4 i}, i=0, \ldots,\left[\frac{n}{4}\right]-1\right\}$. As the same technique in Case 1, Subcase 2, the set $D_3$ is a $C D S$ of all vertices in the graph.

Case 4. If $n \equiv 3(\bmod 4)$, then let $D_4=\left\{v_2, v_3\right\} \cup$ $\left\{u_{1+4 i}, u_{2+4 i}, i=0, \ldots,\left[\frac{n}{4}\right]-1\right\}$. As the same technique in Case 1 , Subcase 2 , the set $D_4$ is a $C D S$ of all vertices in the graph.

From all cases above, the required is done.

5.png

Figure 5. $G \equiv L_{m, n}$

Proposition 3.3.2 If $G$ is a lollipop graph denoted by $G \equiv$ $L_{m, n}$ then $G$ has no $C D N \gamma_{c a}^{-1}$.

Proof: The graph $G$ has path induced subgraph and according to Observation 2.2, the $G$ has no inverse $C D N \gamma_{c a}^{-1}$.

Proposition 3.3.3 If $G \equiv \overline{L_{m, n}}$ be a graph then $\gamma_{c a}(G)=2$, if $n>1$, and has no $C D S$ if $n=1$.

Proof: Depending on $n$ there are three classifications as below.

I) If n=1, then graph G has an isolated vertex, then there is no CDS.

II) Since every vertex in the induced subgraph which is isomorphic to the complete graph dominates on all vertices in the induced subgraph which is isomorphic to the path graph except the vertex $v_1$ this vertex is adjacent to n-1 vertices in induced subgraph which is isomorphic to the path graph, then $\gamma_{c a}(G)=2$.

From all cases above, the required is done.

Proposition 3.3.4 If $G \equiv \overline{L_{m, n}}, n>1$ be a graph then $\gamma_{c a}^{-1}(G)=2$.

Proof: By the Proposition above since every vertex in the induced subgraph which is isomorphic to the complete graph dominates on all vertices in induced subgraph which is isomorphic to the path graph except the vertex $v_1$  this vertex adjacent to n-1 vertices in the induced subgraph which is isomorphic to the path graph then there exist another set in V-D such that this set is minimum and inverse CDN.

3.4 The barbell graph

Proposition 3.4.1 If $G \equiv B_{n, n}$ be a barbell graph of order $2 n$ then $\gamma_{\mathrm{ca}}(\mathrm{G})=2$.

Proof: The vertex set of $B_{n, n}$ is $\left\{v_i: 1 \leq i \leq 2 n\right\}$, the vertex set of the first complete graph $K_n$ is $\left\{v_1, v_2, \ldots \ldots, v_n\right\}$ and the vertex set of the second complete graph $K_n$ is $\left\{v_{n+1}, v_{n+2}, \ldots \ldots, v_{2 n}\right\}$ as shown in Figure 6

From the definition of a barbell graph, there are two copies of a complete graph each of which can dominate by one vertex and to keep the totality of a dominating set, two adjacent vertices are taken. Thus, the set $D=\left\{v_n, v_{n+1}\right\}$ is a minimum $C D S$, and $\gamma_{c a}(G)=2$.

6.png

Figure 6. $G \equiv B_{n, n}$

Proposition 3.4.2 If $G \equiv B_{n, n}, n>2$ be barbell graph of order $2 \mathrm{n}$ then $\gamma_{c a}{ }^{-1}(G)=4$.

Proof: By proposition 3.4.1, above let $=\left\{v_n, v_{n+1}\right\}$, so it is obvious that the set $D$ is captive dominating and it is the minimum, and since there exists another dominating set in $V-$ $D$ and it is minimum, let $S=\left\{v_1, v_2\right\} \cup\left\{v_{n+2}, v_{n+3}\right\}$ this set is inverse $C D N$ then $\gamma_{c a}{ }^{-1}(G)=4$.

Proposition 3.4.3 If $G \equiv \overline{B_{n, n}}, n>1$ be a graph of order $2 \mathrm{n}$ then $\gamma_{c a}(G)=2$.

Proof: From the definition of a barbell graph, there are two subgraphs that are isomorphic to a complete graph of order one $K_n \quad$ is $K_1=\left\{v_1, v_2, \ldots \ldots, v_n\right\}$ and $K_2=$ $\left\{v_{n+1}, v_{n+2}, \ldots \ldots, v_{2 n}\right\}$. In the graph $\overline{B_{n, n}}$ each vertex in the induced subgraph $K_1$ except the vertex $v_n$ dominates all vertices in the induced subgraph $K_2$. Also, each vertex in the induced subgraph $K_2$ except the vertex $v_n$ dominates all vertices in the induced subgraph $K_1$. Therefore, let $D_1=$ $\left\{v_{n-1}, v_{2 n-1}\right\}$ it is obvious that the set $D_1$ is a captive dominating set and it is minimum. Thus, $\gamma_{c a}(G)=2$.

Proposition 3.4.4 If $G \equiv \overline{B_{n, n}}, n>2$ be a graph of order $2 \mathrm{n}$ then $\gamma_{c a}{ }^{-1}(G)=2$.

Proof: By Proposition 3.4.3, since the set $S=\left\{v_1, v_{n+2}\right\}$ is captive dominating and it is the minimum, then there exists another set in $V-S$ say $W=\left\{v_2, v_{n+3}\right\}$, so it is inverse $C D N$ and minimum.

3.5 Corona graph of a cycle of order n and null graph of order p

Proposition 3.5.1 If $G \equiv C_n \odot \overline{K_p}$, where $C_n$ be cycle of order $n$ and $\overline{K_p}$ is null graph of order $p$, then $\gamma_{c a}(G)=n$.

Proof: Each vertex belongs to the induced subgraph which isomorphic to $\overline{K_p}$ not belong to each $C D S$ according to Observation 2.2. Moreover, each vertex in the induced subgraph which isomorphic to $C_n$ belongs to each $C D S$. These vertices are totally dominating set and each vertex of them dominates $p$ vertices that represent the set of vertices of a copy of $\overline{K_p}$ as shown in the Figure 7. Thus, $\gamma_{c a}(G)=n$.

Proposition 3.5.2 If $G \equiv C_n \odot \overline{K_p}$ then the graph $G$ has no inverse captive domination number.

Proof: The proof is straightforward according to Observation 1.2.

Proposition 3.5.3 If $G$ be a graph denoted by $G \equiv \overline{K_p} \odot C_n$ then $\gamma_{\mathrm{ca}}(\mathrm{G})=2 \mathrm{p}$, where $\mathrm{p}$ order of $\overline{\mathrm{K}_{\mathrm{p}}}$.

Proof: let $\overline{K_p}$ of order $p$ corona with $C_n$ be a cycle of order $n$, the graph $G$ denoted by $G \equiv \overline{K_p} \odot C_n$ of order $p(n+1)$. The vertex set of $\overline{K_p}$ is $\left\{v_1, v_2, \ldots \ldots, v_p\right\}$ and the vertex set of $C_n$ is $\left\{u_1, u_2, \ldots \ldots, u_n\right\}$ as shown in Figure 8.

Graph $G$ consists of $p$ components and each component is isomorphic to a complete graph of order $n+l$, then the $C D S D=\left\{v_p, u_1\right\}$, and the $C D N$ is 2 . Thus, $\gamma_{c a}(G)=2 p$.

Proposition 3.5.4 If $G$ be a graph denoted by $G \equiv \overline{K_p} \odot C_n$ then $\gamma_{\mathrm{ca}}{ }^{-1}(\mathrm{G})=\left\{\begin{array}{c}2 \mathrm{p} \text { if } n=3 \\ 2 p\left[\frac{n}{4}\right] \text { if } n>3\end{array}\right\}$, where $\mathrm{p}$ order of $\overline{\mathrm{K}_{\mathrm{p}}}$.

Proof. Depending on the order of the cycle there are two cases as the following.

Case 1. If n=3, then each component is isomorphic to a complete graph of order 4, so the order of each of these components is equal to four. Two of these vertices are chosen in the set D, so the other two vertices of each component can be chosen to create the other a CDS which is disjoint from the set $D$. This set is $D^{-1}$ and it is obvious that is minimum cardinality since $\left|D^{-1}\right|=|D|=2$. Thus, $\gamma_{c a}{ }^{-1}(G)=2 p$.

Case 2. $n>3$, then each component is isomorphic to the wheel graph, the center of these components is used in the set $\mathrm{D}$, so we cannot use in another dominating set. Thus, the vertices of the set $D^{-1}$ lie in the induced subgrph isomorphic to the cycle graph. Thus, by using proposition 2.14 [1], $\gamma_{c a}{ }^{-1}(G)=2 p\left\lceil\frac{n}{4}\right\rceil$

From the two cases above, the required is done.

7.png

Figure 7. $G \equiv C_n \odot \overline{K_p}$

8.png

Figure 8. $G \equiv \overline{K_p} \odot C_n$

3.6 The helm graph

Proposition 3.6.1 If $G \equiv H_n$ be helm graph of order $(2 n-1)$ vertices, then $\gamma_{c a}(G)=n-1$.

Proof. Let $G$ be helm graph of order $(2 n-1)$, then the number of vertices in induced subgraph isomorphic to cycle is $n-1$ as shown in Figure 9. All these vertices are support vertices by definition must be in $C D S D$ then $\gamma_{c a}(G)=n-1$.

9.png

Figure 9. $G \equiv H_n$

Remark 3.6.2 If $G \equiv H_n$ be helm graph of order $(2 n-1)$ vertices, then $G$ has no inverse $C D S$ according to Observation 1.2 .

Proposition 3.6.3 If $G \equiv \overline{H_n}$ be a graph of order $(2 n-1)$ vertices, then $\gamma_{c a}(G)=2$.

Proof. Each pendent vertex in the helm graph $H_n$ will adjacent to every vertex in $G \equiv \overline{H_n}$ except the support vertex in $H_n$. Thus, $\gamma_{c a}(G) \geq 2$, now let $D \subseteq \overline{H_n}$ be set contains two pendants' vertices of $H_n$. It is obvious the set $D$ is dominating and the two vertices are adjacent in $\overline{H_n}$, moreover each one of them is adjacent to at least one vertex of the set $V-D$ in $\overline{H_n}$. Therefore, $\gamma_{c a}(G)=2$.

Proposition 3.6.4 If $G \equiv \overline{H_n}$ be a graph of order $(2 n-1)$ vertices, then $\gamma_{c a}{ }^{-1}(G)=2$, if $n \geq 5$ and has no inverse if $n=4$.

Proof. Depending on $n$ there are two classifications as below.

Case 1. If $n=4$, then by the previous proposition two pendants in $H_n$ make a dominating set in the graph $\overline{H_n}$ (D). In the graph $\overline{H_n}$, the remained pendant vertex in the graph $H_n$ not dominates the support vertex which is adjacent to it in the graph $H_n$ and there is no any vertex in $V-D$ dominates this vertex. Thus, the graph $\overline{H_4}$ has no inverse.

Case 2. If $n>4$, then there are at least two pendant vertices not belong to the set $D$. These two pendants' vertices make another dominating set which keep the all conditions of $C D S$ and disjoint from the set $D$. Thus, $\gamma_{c a}{ }^{-1}(G)=2$.

From the two cases above, the required is done.