# On the Outer-Independent Roman Domination in Graphs
## 2 Bounds and Relationships with Other Parameters  
[PERSON] et al. (2017) in 2017, established the following result. **Theorem 1** (([PERSON] et al., 2017)). : _For any graph \(G\) with no isolated vertex,_  
\[\alpha(G)+1\leq\gamma_{oiR}(G)\leq 2\alpha(G).\]  
Observe that any graph \(G\) with no isolated vertex, order \(n\) and maximum degree \(\Delta\), satisfies that \(1\leq\left\lceil\frac{n-\alpha(G)}{\Delta}\right\rceil\). It is also well-know that \(\gamma(G)\leq\alpha(G)\), which implies \(\alpha(G)+\gamma(G)\leq 2\alpha(G)\). With the above inequalities in mind, we state the following theorem, which improves the bounds given in Theorem 1. **Theorem 2**. : _For any graph \(G\) with no isolated vertex, order \(n\) and maximum degree \(\Delta\),_  
\[\alpha(G)+\left\lceil\frac{n-\alpha(G)}{\Delta}\right\rceil\leq\gamma_{oiR}( G)\leq\alpha(G)+\gamma(G).\]  
Proof. : We first prove the upper bound. Let \(D\) be a \(\gamma(G)\)-set and \(S\) an \(\alpha(G)\)-set. Let \(g(W_{0},W_{1},W_{2})\) be a function defined by \(W_{0}=V(G)\setminus(D\cup S)\), \(W_{1}=(D\cup S)\setminus(D\cap S)\) and \(W_{2}=D\cap S\). We claim that \(g\) is an OIRDF on \(G\). Without loss of generality, we may assume that \(W_{0}\
eq\emptyset\). Notice that \(W_{0}=V(G)\setminus(D\cup S)\) is an independent set of \(G\) as \(S\) is a vertex cover. Now, we prove that every vertex in \(W_{0}\) has a neighbour in \(W_{2}\). Let \(x\in W_{0}=V(G)\setminus(D\cup S)\). Since \(S\) is a vertex cover and \(D\) is a dominating set, we deduce that \(N(x)\subseteq S\) and \(N(x)\cap D\
eq\emptyset\), respectively. Hence \(N(x)\cap D\cap S\
eq\emptyset\), or equivalently, \(N(x)\cap W_{2}\
eq\emptyset\). Thus, \(g\) is an OIRDF on \(G\), as required. Therefore, \(\gamma_{oiR}(G)\leq\omega(g)=|(D\cup S)\setminus(D\cap S)|+2|D\cap S|=\alpha(G )+\gamma(G)\). We now proceed to prove the lower bound. Let \(f(V_{0},V_{1},V_{2})\) be a \(\gamma_{oiR}(G)\)-function. By definition, we have that \(V_{0}\) is an independent set, and so, \(V_{1}\cup V_{2}\) is a vertex cover. Moreover, we note that every vertex in \(V_{2}\) has at most \(\Delta\) neighbours in \(V_{0}\). Hence, \(|V_{0}|\leq\Delta|V_{2}|\). By inequality above, and the fact that \(n-\alpha(G)=\beta(G)\geq|V_{0}|\), we have  
\[\Delta\gamma_{oiR}(G) = \Delta(|V_{1}|+2|V_{2}|)\] \[= \Delta(|V_{1}|+|V_{2}|)+\Delta|V_{2}|\] \[\geq \Delta(n-|V_{0}|)+|V_{0}|\] \[= n\Delta-(\Delta-1)|V_{0}|\] \[\geq n\Delta-(\Delta-1)(n-\alpha(G))\] \[= \Delta\alpha(G)+(n-\alpha(G)).\]  
Therefore, \(\gamma_{oiR}(G)\geq\alpha(G)+\left\lceil\frac{n-\alpha(G)}{\Delta}\right\rceil\), which completes the proof. The bounds above are tight. To see this, let us consider the vertex cover Roman graphs \(G\). These graphs were defined in [PERSON] et al. (2017) and satisfy the equality \(\gamma_{oiR}(G)=2\alpha(G)\). Since \(\gamma(G)\leq\alpha(G)\), we deduce that for every vertex cover Roman graph \(G\) it follows that \(\gamma_{oiR}(G)=\alpha(G)+\gamma(G)\). Note also that both bounds are achieved for the graph \(G_{1}\) given in Figure 1, i.e., \(\alpha(G_{1})+\left\lceil\frac{|V(G_{1})|-\alpha(G_{1})}{\Delta(G_{1})}\right\rceil =\gamma_{oiR}(G_{1})=\alpha(G_{1})+\gamma(G_{1})\). The following result is an immediate consequence of Theorem 2. Corollary 1: _If \(G\) is a graph such that \(\gamma(G)=1\), then_  
\[\gamma_{oiR}(G)=\alpha(G)+1.\]  
However, the graphs \(G\) with \(\gamma(G)=1\) are not the only ones that satisfy the equality \(\gamma_{oiR}(G)=\alpha(G)+1\). For instance, the path \(P_{4}\) satisfies that \(\gamma(P_{4})=2\) and \(\gamma_{oiR}(P_{4})=3=\alpha(P_{4})+1\). In such a sense, we next give a theoretical characterization of the graphs that satisfy this equality above. Theorem 3: _If \(G\) is a graph with no isolated vertex, then the following statements are equivalent._  
1. \(\gamma_{oiR}(G)=\alpha(G)+1\)_._
2.