# On the Outer-Independent Roman Domination in Graphs
## 3 Rooted Product Graphs  
Let \(G\) be a graph of order \(n\) with vertex set \(\{u_{1},\ldots,u_{n}\}\) and \(H\) a graph with root \(v\in V(H)\). The rooted product graph \(G\circ_{v}H\) is defined as the graph obtained from \(G\) and \(n\) copies of \(H\), by identifying the vertex \(u_{i}\) of \(G\) with the root \(v\) in the \(i^{th}\)-copy of \(H\), where \(i\in\{1,\ldots,n\}\)[12]. If \(H\) or \(G\) is a trivial graph, then \(G\circ_{v}H\) is equal to \(G\) or \(H\), respectively. In this sense, to obtain the rooted product \(G\circ_{v}H\), hereafter we will only consider graphs \(G\) and \(H\) of orders at least two. Figure 3 shows an example of a rooted product graph. For every \(x\in V(G)\), \(H_{x}\) will denote the copy of \(H\) in \(G\circ_{v}H\) containing \(x\). The restriction of any \(\gamma_{oiR}(G\circ_{v}H)\)-function \(f\) to \(V(H_{x})\) will be denoted by \(f_{x}\) and the restriction to \(V(H_{x})\setminus\{x\}\) will be denoted by \(f_{x}^{-}\). If \(v\) is a vertex of a graph \(H\), then the subgraph \(H-v\) is the subgraph of \(H\) induced by \(V(H)\setminus\{v\}\). The following three results will be the main tools to deduce our results. Lemma 2: _Let \(H\) be a graph without isolated vertices. For any \(v\in V(H)\),_  
\[\gamma_{oiR}(H-v)\geq\gamma_{oiR}(H)-1.\]  
Proof: Let \(g^{\prime}\) be a \(\gamma_{oiR}(H-v)\)-function. Notice that the function \(g\), defined by \(g(v)=1\) and \(g(u)=g^{\prime}(u)\) whenever \(u\in V(H)\setminus\{v\}\), is an OIRDF on \(H\). Hence, \(\gamma_{oiR}(H)-1\leq\omega(g)-1=\omega(g^{\prime})=\gamma_{oiR}(H-v)\), which completes the proof. **Lemma 3**. : _Let \(G\) and \(H\) be two graphs without isolated vertices. If \(G\) has order \(n\) and \(v\in V(H)\), then the following statements hold._  
* _If_ \(g(v)=0\) _for some_ \(\gamma_{oiR}(H)\)_-function_ \(g\)_, then_ \(\gamma_{oiR}(G\circ_{v}H)\leq\alpha(G)+n\gamma_{oiR}(H)\)_._
* _If_ \(g(v)>0\) _for some_ \(\gamma_{oiR}(H)\)_-function_ \(g\)_, then_ \(\gamma_{oiR}(G\circ_{v}H)\leq n\gamma_{oiR}(H)\)_._
* _If there exists a_ \(\gamma_{oiR}(H-v)\)_-function_ \(g\) _such that_ \(g(x)>0\) _for every_ \(x\in N(v)\)_, then_ \(\gamma_{oiR}(G\circ_{v}H)\leq\gamma_{oiR}(G)+n\gamma_{oiR}(H-v)\)_._  
Proof. : From any \(\gamma_{oiR}(H)\)-function \(g\) such that \(g(v)=0\) and any \(\alpha(G)\)-set, we can construct an OIRDF on \(G\circ_{v}H\) of weight \(\alpha(G)+n\gamma_{oiR}(H)\). Thus, \(\gamma_{oiR}(G\circ_{v}H)\leq\alpha(G)+n\gamma_{oiR}(H)\) and (i) follows. Now, if there exists a \(\gamma_{oiR}(H)\)-function \(g\) such that \(g(v)>0\), then from \(g\) we can construct an OIRDF on \(G\circ_{v}H\) of weight \(n\omega(g)\). Thus, \(\gamma_{oiR}(G\circ_{v}H)\leq n\omega(g)=n\gamma_{oiR}(H)\), and (ii) follows. Finally, if there exists a \(\gamma_{oiR}(H-v)\)-function \(g\) such that \(g(x)>0\) for every \(x\in N(v)\), then from \(g\) and any \(\gamma_{oiR}(G)\)-function we can construct an OIRDF on \(G\circ_{v}H\) of weight \(\gamma_{oiR}(G)+n\gamma_{oiR}(H-v)\), which completes the proof. **Lemma 4**. : _Let \(f(V_{0},V_{1},V_{2})\) be a \(\gamma_{oiR}(G\circ_{v}H)\)-function. The following statements hold for any vertex \(x\in V(G)\)._  
* \(\omega(f_{x})\geq\gamma_{oiR}(H)-1\)_._
* _If_ \(\omega(f_{x})=\gamma_{oiR}(H)-1\)_, then_ \(x\in V_{0}\) _and_ \(N(x)\cap V(H_{x})\subseteq V_{1}\)_._  
Proof. : Let \(x\in V(G)\). Observe that \(V_{0}\cap V(H_{x})\) is an independent set of \(H_{x}\) and also, every vertex in \(V_{0}\cap(V(H_{x})\setminus\{x\})\) has a neighbour in \(V_{2}\cap V(H_{x})\). So, it is easy to see that the function \(g\), defined by \(g(x)=\max\{1,f(x)\}\) and \(g(u)=f(u)\) whenever \(u\in V(H_{x})\setminus\{x\}\), is an OIRDF on \(H_{x}\). Hence, \(\gamma_{oiR}(H)-1=\gamma_{oiR}(H_{x})-1\leq\omega(g)-1\leq\omega(f_{x})\), which completes the proof of (i). Now, we suppose that \(\omega(f_{x})=\gamma_{oiR}(H)-1\). If \(x\in V_{1}\cup V_{2}\) or \(x\in V_{0}\) and \(N(x)\cap V(H_{x})\cap V_{2}\
eq\emptyset\), then \(f_{x}\) is an OIRDF on \(H_{x}\), which is a contradiction.