# On the Outer-Independent Roman Domination in Graphs
## 3 Rooted Product Graphs
Thus,  
\[\mathcal{A}_{f} =\{x\in V(G):\omega(f_{x})>\gamma_{oiR}(H)\},\]0 \[\mathcal{A}_{f} =\{x\in V(G):\omega(f_{x})>\gamma_{oiR}(H)\},\]1 \[\mathcal{A}_{f} =\{x\in V(G):\omega(f_{x})>\gamma_{oiR}(H)\},\]2 \[\mathcal{A}_{f} =\{x\in V(G):\omega(f_{x})>\gamma_{oiR}(H)\},\]3 \[\mathcal{A}_{f} =\{x\in V(G):\omega(f_{x})>\gamma_{oiR}(H)\},\]4  
Therefore, \(\gamma_{oiR}(G\circ_{v}H)=\alpha(G)+n(\gamma_{oiR}(H)-1)\). Subcase 2. \(g(v)\leq 1\) for every \(\gamma_{oiR}(H)\)-function \(g\). This condition implies that \(V_{2}\cap\mathcal{B}_{f}=\emptyset\). Since every vertex \(x\in\mathcal{C}_{f}\) has a neighbour in \(V_{2}\), and as Lemma 4 (ii) leads to \(N(x)\cap V(H_{x})\subseteq V_{1}\), then we deduce that \(N(x)\cap V_{2}\cap\mathcal{A}_{f}\
eq\emptyset\). Hence, and as \(\mathcal{C}_{f}\subseteq V_{0}\), the function \(f^{\prime}(V_{0}^{\prime},V_{1}^{\prime},V_{2}^{\prime})\), defined by \(V_{2}^{\prime}=\mathcal{A}_{f}\), \(V_{1}^{\prime}=\mathcal{B}_{f}\) and \(V_{0}^{\prime}=\mathcal{C}_{f}\), is an OIRDF on \(G\). So \(\gamma_{oiR}(G)\leq\omega(f^{\prime})=2|\mathcal{A}_{f}|+|\mathcal{B}_{f}|\). Therefore, Inequality chain (1) leads to \(\gamma_{oiR}(G\circ_{v}H)\geq\gamma_{oiR}(G)+n(\gamma_{oiR}(H)-1)\), which implies that \(\gamma_{oiR}(G\circ_{v}H)=\gamma_{oiR}(G)+n(\gamma_{oiR}(H)-1)\). Therefore, the proof is complete. In order to see that the four possible values of \(\gamma_{oiR}(G\circ_{v}H)\) described in Theorem 8 are realizable, we consider the following example. **Example 1**. : _Let \(G\) be a graph with no isolated vertex. If \(H\) is the graph shown in Figure 4, then the resulting values of \(\gamma_{oiR}(G\circ_{v}H)\) for some specific roots \(x\in V(H)\) are described below._  
* \(\gamma_{oiR}(G\circ_{v}H)=\alpha(G)+n\gamma_{oiR}(H)\)_._
* \(\gamma_{oiR}(G\circ_{w}H)=n\gamma_{oiR}(H)\)_._
* \(\gamma_{oiR}(G\circ_{v^{\prime}}H)=\gamma_{oiR}(G)+n(\gamma_{oiR}(H)-1)\)_._
* \(\gamma_{oiR}(G\circ_{w^{\prime}}H)=\alpha(G)+n(\gamma_{oiR}(H)-1)\)_._  
Now, we characterize the graphs with \(\gamma_{oiR}(G\circ_{v}H)=\alpha(G)+n\gamma_{oiR}(H)\). **Theorem 9**. : _Let \(G\) and \(H\) be two graphs with no isolated vertex, let \(|V(G)|=n\) and \(v\in V(H)\). The following statements are equivalent._  
* \(\gamma_{oiR}(G\circ_{v}H)=\alpha(G)+n\gamma_{oiR}(H)\)_._
* \(g(v)=0\) _for every_ \(\gamma_{oiR}(H)\)_-function_ \(g\)_._  
Proof. : We first assume that (i) holds, i.e., \(\gamma_{oiR}(G\circ_{v}H)=\alpha(G)+n\gamma_{oiR}(H)\). If there exists a \(\gamma_{oiR}(H)\)-function \(g\) such that \(g(v)>0\), then by Lemma 3 (ii) it follows that \(\gamma_{oiR}(G\circ_{v}H)\leq n\gamma_{oiR}(H)\), which is a contradiction. Therefore, (ii) holds. On the other hand, we assume that (ii) holds, i.e., \(g(v)=0\) for every \(\gamma_{oiR}(H)\)-function \(g\). Let \(f(V_{0},V_{1},V_{2})\) be a \(\gamma_{oiR}(G\circ_{v}H)\)-function. If \(\mathcal{C}_{f}\
eq\emptyset\), then by Lemma 4 (ii) we can obtain a \(\gamma_{oiR}(H)\)-function \(g\) such that \(g(v)=1\), which is a contradiction. Hence, \(\mathcal{C}_{f}=\emptyset\), and so, by Theorem 8 we deduce that \(\gamma_{oiR}(G\circ_{v}H)\in\{\alpha(G)+n\gamma_{oiR}(H)\), \(n\gamma_{oiR}(H)\}\). Now, suppose that \(\gamma_{oiR}(G\circ_{v}H)=n\gamma_{oiR}(H)\). Since \(\mathcal{C}_{f}=\emptyset\), it follows that \(\mathcal{B}_{f}=V(G)\) and as \(V_{0}\) is an independent set, there exists \(x\in\mathcal{B}_{f}\setminus V_{0}\). This implies that \(f_{x}\) is a \(\gamma_{oiR}(H_{x})\)-function such that \(f_{x}(x)>0\), which is a contradiction. Therefore, \(\gamma_{oiR}(G\circ_{v}H)=\alpha(G)+n\gamma_{oiR}(H)\), which completes the proof. Next, we characterize the graphs with \(\gamma_{oiR}(G\circ_{v}H)=\alpha(G)+n(\gamma_{oiR}(H)-1)\). **Theorem 10**. : _Let \(G\) and \(H\) be two graphs with no isolated vertex, let \(|V(G)|=n\) and \(v\in V(H)\).