*Article* **On the Outer-Independent Roman Domination in Graphs**

#### **Abel Cabrera Martínez 1,\* , Suitberto Cabrera García <sup>2</sup> , Andrés Carrión García <sup>3</sup> and Angela María Grisales del Rio <sup>3</sup>**


Received: 21 October 2020; Accepted: 6 November 2020; Published: 9 November 2020

**Abstract:** Let *G* be a graph with no isolated vertex and *f* : *V*(*G*) → {0, 1, 2} a function. Let *V<sup>i</sup>* = {*v* ∈ *V*(*G*) : *f*(*v*) = *i*} for every *i* ∈ {0, 1, 2}. The function *f* is an outer-independent Roman dominating function on *G* if *V*<sup>0</sup> is an independent set and every vertex in *V*<sup>0</sup> is adjacent to at least one vertex in *V*2. The minimum weight *<sup>ω</sup>*(*f*) = <sup>∑</sup>*v*∈*V*(*G*) *f*(*v*) among all outer-independent Roman dominating functions *f* on *G* is the outer-independent Roman domination number of *G*. This paper is devoted to the study of the outer-independent Roman domination number of a graph, and it is a contribution to the special issue "Theoretical Computer Science and Discrete Mathematics" of *Symmetry*. In particular, we obtain new tight bounds for this parameter, and some of them improve some well-known results. We also provide closed formulas for the outer-independent Roman domination number of rooted product graphs.

**Keywords:** outer-independent Roman domination; Roman domination; vertex cover; rooted product graph

#### **1. Introduction**

Throughout this paper, we consider *G* = (*V*(*G*), *E*(*G*)) as a simple graph with no isolated vertex. Given a vertex *v* of *G*, *N*(*v*) and *N*[*v*] represent the open neighbourhood and the closed neighbourhood of *v*, respectively. We also denote by deg(*v*) = |*N*(*v*)| the degree of vertex *v*. For a set *D* ⊆ *V*(*G*), its open neighbourhood and closed neighbourhood are *N*(*D*) = ∪*v*∈*DN*(*v*) and *N*[*D*] = *N*(*D*) ∪ *D*, respectively. Moreover, the subgraph of *G* induced by *D* ⊆ *V*(*G*) will be denoted by *G*[*D*].

Domination theory is an interesting topic in the theory of graphs, as well as one of the most active topic of research in this area. A set *D* ⊆ *V*(*G*) is a dominating set of *G* if *N*[*D*] = *V*(*G*). The domination number of *G*, denoted by *γ*(*G*), is the minimum cardinality amongst all dominating sets of *G*. Numerous results on this issue obtained in the previous century are shown in [1,2]. We define a *γ*(*G*)-set as a dominating set of cardinality *γ*(*G*). The same terminology will be assumed for optimal parameters associated with other sets or functions defined in the paper.

Moreover, in the last two decades, the interest in the domination theory in graphs has increased. In that sense, a very high number of variants of domination parameters have been studied, many of which are combinations of two or more parameters. Next, we expose some of them.


All the previous parameters are, in one way or another, related to each other. Next, we show the most natural relationships that exist between them, which are easily deductible by definition.

**Remark 1.** *For any graph G of order n with no isolated vertex,*


For the graphs shown in Figure 1 we have the following.


**Figure 1.** The labels of (gray and black) coloured vertices describe the positive weights of a *γoiR*(*G<sup>i</sup>* )-function, for *i* ∈ {1, 2}.

In this paper, we continue the study of the outer-independent Roman domination number of graphs. For instance, in Section 2 we give some new relationships between this parameter and the others mentioned above. Several of these results improve other bounds previously given. Finally, in Section 3 we provide closed formulas for this parameter in rooted product graphs. In particular, we show that there are four possible expressions for the outer-independent Roman domination number of a rooted product graph, and we characterize the graphs reaching these expressions.

#### **2. Bounds and Relationships with Other Parameters**

Abdollahzadeh Ahangar et al. [8] in 2017, established the following result.

**Theorem 1** ([8])**.** *For any graph G with no isolated vertex,*

$$
\mathfrak{a}(G) + 1 \le \gamma\_{o\bar{i}R}(G) \le 2\mathfrak{a}(G).
$$

Observe that any graph *G* with no isolated vertex, order *n* and maximum degree ∆, satisfies that 1 ≤ l *n*−*α*(*G*) ∆ m . It is also well-know that *γ*(*G*) ≤ *α*(*G*), which implies *α*(*G*) + *γ*(*G*) ≤ 2*α*(*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* ∆*,*

$$
\mathfrak{a}(\mathcal{G}) + \left\lceil \frac{n - \mathfrak{a}(\mathcal{G})}{\Delta} \right\rceil \le \gamma\_{o i R}(\mathcal{G}) \le \mathfrak{a}(\mathcal{G}) + \gamma(\mathcal{G}).
$$

**Proof.** We first prove the upper bound. Let *D* be a *γ*(*G*)-set and *S* an *α*(*G*)-set. Let *g*(*W*0, *W*1, *W*2) be a function defined by *W*<sup>0</sup> = *V*(*G*) \ (*D* ∪ *S*), *W*<sup>1</sup> = (*D* ∪ *S*) \ (*D* ∩ *S*) and *W*<sup>2</sup> = *D* ∩ *S*. We claim that *g* is an OIRDF on *G*. Without loss of generality, we may assume that *W*<sup>0</sup> 6= ∅. Notice that *W*<sup>0</sup> = *V*(*G*) \ (*D* ∪ *S*) is an independent set of *G* as *S* is a vertex cover. Now, we prove that every vertex in *W*<sup>0</sup> has a neighbour in *W*2. Let *x* ∈ *W*<sup>0</sup> = *V*(*G*) \ (*D* ∪ *S*). Since *S* is a vertex cover and *D* is a dominating set, we deduce that *N*(*x*) ⊆ *S* and *N*(*x*) ∩ *D* 6= ∅, respectively. Hence *N*(*x*) ∩ *D* ∩ *S* 6= ∅, or equivalently, *N*(*x*) ∩ *W*<sup>2</sup> 6= ∅. Thus, *g* is an OIRDF on *G*, as required. Therefore, *γoiR*(*G*) ≤ *ω*(*g*) = |(*D* ∪ *S*) \ (*D* ∩ *S*)| + 2|*D* ∩ *S*| = *α*(*G*) + *γ*(*G*).

We now proceed to prove the lower bound. Let *f*(*V*0, *V*1, *V*2) be a *γoiR*(*G*)-function. By definition, we have that *V*<sup>0</sup> is an independent set, and so, *V*<sup>1</sup> ∪ *V*<sup>2</sup> is a vertex cover. Moreover, we note that every vertex in *V*<sup>2</sup> has at most ∆ neighbours in *V*0. Hence, |*V*0| ≤ ∆|*V*2|. By inequality above, and the fact that *n* − *α*(*G*) = *β*(*G*) ≥ |*V*0|, we have

$$\begin{aligned} \Delta \gamma\_{o\bar{n}\bar{R}}(G) &= \Delta(|V\_1| + 2|V\_2|) \\ &= \Delta(|V\_1| + |V\_2|) + \Delta|V\_2| \\ &\ge \Delta(n - |V\_0|) + |V\_0| \\ &= \left| n\Delta - (\Delta - 1)|V\_0| \\ &\ge \left| n\Delta - (\Delta - 1)(n - \alpha(G)) \right| \\ &= \left| \Delta\alpha(G) + (n - \alpha(G)) \right|. \end{aligned}$$

Therefore, *<sup>γ</sup>oiR*(*G*) <sup>≥</sup> *<sup>α</sup>*(*G*) + <sup>l</sup> *n*−*α*(*G*) ∆ m , 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 [8] and satisfy the equality *γoiR*(*G*) = 2*α*(*G*). Since *γ*(*G*) ≤ *α*(*G*), we deduce that for

every vertex cover Roman graph *G* it follows that *γoiR*(*G*) = *α*(*G*) + *γ*(*G*). Note also that both bounds are achieved for the graph *<sup>G</sup>*<sup>1</sup> given in Figure 1, i.e., *<sup>α</sup>*(*G*1) + <sup>l</sup> |*V*(*G*1)|−*α*(*G*1) ∆(*G*1) m = *γoiR*(*G*1) = *α*(*G*1) + *γ*(*G*1).

The following result is an immediate consequence of Theorem 2.

**Corollary 1.** *If G is a graph such that γ*(*G*) = 1*, then*

$$
\gamma\_{o\ddot{o}\mathbb{R}}(G) = \mathfrak{a}(G) + 1.
$$

However, the graphs *G* with *γ*(*G*) = 1 are not the only ones that satisfy the equality *γoiR*(*G*) = *α*(*G*) + 1. For instance, the path *P*<sup>4</sup> satisfies that *γ*(*P*4) = 2 and *γoiR*(*P*4) = 3 = *α*(*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.*


**Proof.** We first suppose that (i) holds, i.e., *γoiR*(*G*) = *α*(*G*) + 1. Let *f*(*V*0, *V*1, *V*2) be a *γoiR*(*G*)-function such that |*V*2| is maximum. Hence, *V*<sup>2</sup> 6= ∅. Let *v* ∈ *V*2. Since *V*<sup>1</sup> ∪ *V*<sup>2</sup> is a vertex cover of *G*, it follows that *α*(*G*) + 1 ≤ (|*V*1| + |*V*2|) + |*V*2| = *γoiR*(*G*) = *α*(*G*) + 1. Hence, we have equalities in the previous inequality chain, which implies that *S* = *V*<sup>1</sup> ∪ *V*<sup>2</sup> is an *α*(*G*)-set and *V*<sup>2</sup> = {*v*}. So, *V*(*G*) \ *S* = *V*<sup>0</sup> ⊆ *N*(*V*2) = *N*(*v*). Therefore, (ii) follows.

On the other hand, suppose that (ii) holds, i.e., suppose there exist an *α*(*G*)-set *S* and *v* ∈ *S* such that *V*(*G*) \ *S* ⊆ *N*(*v*). Observe that the function *g*(*W*0, *W*1, *W*2), defined by *W*<sup>2</sup> = {*v*}, *W*<sup>1</sup> = *S* \ {*v*} and *W*<sup>0</sup> = *V*(*G*) \ *S*, is an OIRDF on *G*. Therefore, and using the lower bound given in the Theorem 1, we obtain that *α*(*G*) + 1 ≤ *γoiR*(*G*) ≤ *ω*(*g*) = |*S*| + 1 = *α*(*G*) + 1. Hence, *γoiR*(*G*) = *α*(*G*) + 1, which completes the proof.

A tree *T* is an acyclic connected graph. A leaf vertex of *T* is a vertex of degree one. The set of leaves is denoted by *L*(*T*). We say that a vertex *v* ∈ *V*(*T*) is a support vertex (strong support vertex) if |*N*(*v*) ∩ *L*(*T*)| ≥ 1 (|*N*(*v*) ∩ *L*(*T*)| ≥ 2). The set of support vertices and strong support vertices are denoted by *S*(*T*) and *Ss*(*T*), respectively.

With this notation in mind, we next characterize the trees *T* with *γoiR*(*T*) = *α*(*T*) + 1. Before we do this, we shall need to state the following useful lemma, in which *diam*(*T*) represents the diameter of *T*.

**Lemma 1.** *If T is a tree such that γoiR*(*T*) = *α*(*T*) + 1*, then the following statements hold.*

$$\begin{array}{ll}(i) & \operatorname{diam}(T) \le 4. \\ (ii) & V(T) = L(T) \cup S(T). \end{array}$$

**Proof.** We first proceed to prove (i). By Theorem 3 there exist an *α*(*T*)-set *S* and *v* ∈ *S* such that *V*(*T*) \ *S* ⊆ *<sup>N</sup>*(*v*). Now, we suppose that *<sup>k</sup>* = *diam*(*T*) ≥ 5. Let *<sup>P</sup>* = *<sup>v</sup>*0*v*<sup>1</sup> · · · *<sup>v</sup>k*−1*v<sup>k</sup>* be a diametrical path of *<sup>T</sup>*. Hence, <sup>∅</sup> 6= {*v*0, *<sup>v</sup>*1, *<sup>v</sup>k*−<sup>1</sup> , *vk*} ∩ (*V*(*T*) \ *S*) 6⊆ *N*(*v*), which is a contradiction. Therefore, *diam*(*T*) ≤ 4, as desired.

Finally, we proceed to prove (ii). By (i) we have that *diam*(*T*) ≤ 4. If *V*(*T*) \ (*L*(*T*) ∪ *S*(*T*)) 6= ∅, then for every *α*(*T*)-set *S* and *v* ∈ *S* it follows that *V*(*T*) \ *S* 6⊆ *N*(*v*), which is a contradiction with Theorem 3. Hence, *V*(*T*) = *L*(*T*) ∪ *S*(*T*), which completes the proof.

Let T be the family of trees *Tr*,*<sup>s</sup>* of order *r* + *s* + 1 with *r* ≥ 1 and *r* − 1 ≥ *s* ≥ 0, obtained from a star *K*1,*<sup>r</sup>* by subdividing *s* edges exactly once. In Figure 2 we show the tree *T*5,3.

**Figure 2.** The tree *T*5,3.

**Theorem 4.** *Let T be a nontrivial tree. Then γoiR*(*T*) = *α*(*T*) + 1 *if and only if T* ∈ T *.*

**Proof.** If *T* ∈ T , then it is easy to check that *γoiR*(*T*) = *α*(*T*) + 1. Now, we prove the converse. Let *T* be a nontrivial tree such that *γoiR*(*T*) = *α*(*T*) + 1. By Lemma 1-(i) we have that *diam*(*T*) ≤ 4. If *diam*(*T*) ≤ 2, then *T* ∼= *Tr*,0 ∈ T . If *diam*(*T*) = 3, then *T* ∼= *Tr*,1 ∈ T . We now suppose that *diam*(*T*) = 4. By Lemma 1-(ii) we have that *V*(*T*) = *L*(*T*) ∪ *S*(*T*). We claim that for any diametrical path *P* = *v*0*v*1*v*2*v*3*v*<sup>4</sup> of *T*, it follows that *v*1, *v*<sup>3</sup> ∈ *S*(*T*) \ *Ss*(*T*). First, we observe that *v*1, *v*<sup>3</sup> ∈ *S*(*T*). Without loss of generality, suppose that *v*<sup>1</sup> ∈ *Ss*(*T*). Hence, *v*<sup>1</sup> belongs to every *α*(*T*)-set. By Theorem 3 there exist an *α*(*T*)-set *S* and *v* ∈ *S* such that *V*(*T*) \ *S* ⊆ *N*(*v*). Since *v*<sup>0</sup> ∈ *V*(*T*) \ *S*, then *v* = *v*1. Notice also that ∅ 6= {*v*3, *v*4} ∩ (*V*(*T*) \ *S*) 6⊆ *N*(*v*1), which is a contradiction. Therefore, *v*1, *v*<sup>3</sup> ∈ *S*(*T*) \ *Ss*(*T*), as desired. From above, we deduce that *T* ∼= *Tr*,*<sup>s</sup>* ∈ T , where *r* ≥ 3 and *r* − 1 ≥ *s* ≥ 2. Therefore, the proof is complete.

The following result is another consequence of Theorem 2.

**Theorem 5.** *Let G be a graph with no isolated vertex. For any γR*(*G*)*-function f*(*V*0, *V*1, *V*2)*,*

$$
\gamma\_{oiR}(G) \le \gamma\_R(G) + \mathfrak{a}(G) - |V\_2|.
$$

**Proof.** Let *f*(*V*0, *V*1, *V*2) be a *γR*(*G*)-function. Since *V*<sup>1</sup> ∪ *V*<sup>2</sup> is a dominating set of *G*, it follows that *γ*(*G*) ≤ |*V*1| + |*V*2| = *γR*(*G*) − |*V*2|. Therefore, Theorem 2 leads to *γoiR*(*G*) ≤ *α*(*G*) + *γ*(*G*) ≤ *γR*(*G*) + *α*(*G*) − |*V*2|, which completes the proof.

The bound above is tight. For instance, in the corona graph *G N<sup>r</sup>* with *r* ≥ 3, the unique *γR*(*G Nr*)-function *f*(*V*0, *V*1, *V*2), defined by *V*<sup>2</sup> = *V*(*G*) and *V*<sup>1</sup> = ∅, is also a *γoiR*(*G Nr*)-function, and so, *γR*(*G Nr*) = *γoiR*(*G Nr*) = *γR*(*G Nr*) + *α*(*G Nr*) − |*V*2| = 2|*V*(*G*)|. The following result, which is a consequence of Remark 1 and Theorem 5, generalizes the previous example.

**Proposition 1.** *If there exists a γR*(*G*)*-function f*(*V*0, *V*1, *V*2) *such that* |*V*2| = *α*(*G*)*, then*

$$
\gamma\_{oiR}(G) = \gamma\_R(G).
$$

We now relate the outer-independent Roman domination number with other domination parameters of graphs. Before, we shall state the following proposition.

**Proposition 2.** *For any graph G with no isolated vertex, there exists a γoiR*(*G*)*-function f*(*V*0, *V*1, *V*2) *such that V*<sup>0</sup> *is an independent dominating set of G.*

**Proof.** Let *f*(*V*0, *V*1, *V*2) be a *γoiR*(*G*)-function such that |*V*2| is maximum. By definition we have that *V*<sup>0</sup> is an independent set. We next prove that *V*<sup>0</sup> is a dominating set of *G*. It is clear that *V*<sup>2</sup> ⊆ *N*(*V*0). Let *v* ∈ *V*1. If *N*(*v*) ⊆ *V*<sup>1</sup> ∪ *V*2, then the function *f* 0 (*V* 0 0 , *V* 0 1 , *V* 0 2 ), defined by *f* 0 (*v*) = 0, *f* 0 (*u*) = *f*(*u*) + 1 for some vertex *u* ∈ *N*(*v*) ∩ *V*<sup>1</sup> and *f* 0 (*x*) = *f*(*x*) whenever *x* ∈ *V*(*G*) \ {*v*, *u*}, is a *γoiR*(*G*)-function and


**Theorem 6.** *For any graph G with no isolated vertex, order n, minimum degree δ and maximum degree* ∆*,*

$$
\left\lceil \frac{i(G)\delta}{\Delta} \right\rceil + 1 \le \gamma\_{o\bar{o}R}(G) \le n - i(G) + \gamma(G).
$$

**Proof.** The upper bound follows by Theorem 2 and the fact that *α*(*G*) = *n* − *β*(*G*) ≤ *n* − *i*(*G*). Now, we proceed to prove the lower bound. Let *f*(*V*0, *V*1, *V*2) be a *γoiR*(*G*)-function which satisfies Proposition 2. Since every vertex in *V*<sup>1</sup> ∪ *V*<sup>2</sup> has at most ∆ neighbours in *V*<sup>0</sup> and *V*<sup>0</sup> is an independent dominating set, it follows that *δ*|*V*0| ≤ ∆(|*V*1| + |*V*2|) and |*V*0| ≥ *i*(*G*). Hence,

$$\begin{array}{rcl}\gamma\_{o\bar{i}R}(G) &=& (|V\_1| + |V\_2|) + |V\_2| \\ &\geq& \frac{|V\_0|\delta}{\Delta} + |V\_2| \\ &\geq& \frac{i(G)\delta}{\Delta} + 1. \end{array}$$

Therefore, the proof is complete.

The bounds above are tight. For example, the lower bound is achieved for the complete bipartite graphs *Kr*,*<sup>r</sup>* , where *γoiR*(*Kr*,*r*) = *r* + 1 = l *r* 2 *r* m + 1 = l *i*(*Kr*,*r*)*δ*(*Kr*,*r*) ∆(*Kr*,*r*) m + 1. In addition, the upper bound is achieved for the case of complete graphs, and in connection with this fact, we pose the following question.

**Open question:** Is it the case that *γoiR*(*G*) = *n* − *i*(*G*) + *γ*(*G*) if and only if *G* is a complete graph?

Next, we give new bounds for the outer-independent Roman domination number of triangle-free graphs. Recall that in these graphs, no pair of adjacent vertices can have a common neighbor. For this purpose, we shall need to introduce the following definitions.

A set *S* ⊆ *V*(*G*) is a 3-packing if the distance between *u* and *v* is greater than three for every pair of different vertices *u*, *v* ∈ *S*. The 3-packing number of *G*, denoted by *ρ*3(*G*), is the maximum cardinality among all 3-packings of *G*. We also define

$$\mathcal{P}\_3(G) = \{ S \subseteq V(G) : S \text{ is a 3-packing of } G \}.$$

**Theorem 7.** *For any triangle-free graph G of order n,*

$$\gamma\_{o\bar{o}\mathcal{R}}(G) \le n - \max\_{S \in \mathcal{P}\_{\mathcal{I}}(G)} \left\{ \sum\_{v \in S} (\deg(v) - 1) \right\}.$$

**Proof.** Let *S* ∈ P3(*G*). As *G* is triangle-free, it follows that *N*(*v*) is an independent set of *G* for every *v* ∈ *V*(*G*). Hence, *N*(*S*) is an independent set of *G*, which implies that the function *f*(*V*0, *V*1, *V*2), defined by *V*<sup>2</sup> = *S*, *V*<sup>0</sup> = *N*(*S*) and *V*<sup>1</sup> = *V*(*G*) \ *N*[*S*], is an OIRDF on *G*. Thus, *γoiR*(*G*) ≤ 2|*V*2| + |*V*1| = 2|*S*| + (*n* − |*N*[*S*]|) = *<sup>n</sup>* − <sup>∑</sup>*v*∈*<sup>S</sup>* (deg(*v*) − 1). Since the inequality holds for every *S* ∈ P3(*G*), the result follows.

**Corollary 2.** *For any triangle-free graph G of order n and minimum degree δ,*

$$
\gamma\_{oiR}(G) \le n - \rho\_3(G)(\delta - 1).
$$

In [8], the bound *γoiR*(*G*) ≤ *n* − ∆(*G*) + 1 was given for the case of triangle-free graph. Next, we state a result which improve the bound above for the triangle-free graphs *G* that satisfy the condition *diam*(*G*)(*δ*(*G*) − 1) ≥ 4(∆(*G*) − 1).

**Proposition 3.** *Let G be a connected triangle-free graph of order n, minimum degree δ and maximum degree* ∆*. If diam*(*G*) ≥ 4*, then*

$$
\gamma\_{o\dot{o}R}(G) \le n - \left\lceil \frac{diam(G)}{4} \right\rceil (\delta - 1).
$$

**Proof.** Assume that *diam*(*G*) ≥ 4. Let *P* = *v*0*v*<sup>1</sup> · · · *v<sup>k</sup>* be a diametrical path of *G* (notice that *k* = *diam*(*G*)), and *<sup>S</sup>* = {*v*0, *<sup>v</sup>*4, . . . , *<sup>v</sup>*4b*k*/4c}. It is easy to see that *<sup>S</sup>* ∈ P3(*G*), and so, by Theorem <sup>7</sup> we deduce that *<sup>γ</sup>oiR*(*G*) ≤ *<sup>n</sup>* − <sup>∑</sup>*v*∈*<sup>S</sup>* (deg(*v*) − 1) ≤ *n* − l *diam*(*G*) 4 m (*δ* − 1) which completes the proof.

The bounds given in Corollary 2 and Proposition 3 are tight. For instance, they are achieved for the cycle *C*10.

#### **3. Rooted Product Graphs**

Let *G* be a graph of order *n* with vertex set {*u*1, . . . , *un*} and *H* a graph with root *v* ∈ *V*(*H*). The rooted product graph *G* ◦*<sup>v</sup> H* is defined as the graph obtained from *G* and *n* copies of *H*, by identifying the vertex *u<sup>i</sup>* of *G* with the root *v* in the *i th*-copy of *<sup>H</sup>*, where *<sup>i</sup>* ∈ {1, . . . , *<sup>n</sup>*} [12]. If *<sup>H</sup>* or *<sup>G</sup>* is a trivial graph, then *<sup>G</sup>* ◦*<sup>v</sup> <sup>H</sup>* is equal to *G* or *H*, respectively. In this sense, to obtain the rooted product *G* ◦*<sup>v</sup> 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* ∈ *V*(*G*), *H<sup>x</sup>* will denote the copy of *H* in *G* ◦*<sup>v</sup> H* containing *x*. The restriction of any *γoiR*(*G* ◦*<sup>v</sup> H*)-function *f* to *V*(*Hx*) will be denoted by *f<sup>x</sup>* and the restriction to *V*(*Hx*) \ {*x*} will be denoted by *f* − *x* .

**Figure 3.** The rooted product graph *G* ◦*<sup>v</sup> H*.

If *v* is a vertex of a graph *H*, then the subgraph *H* − *v* is the subgraph of *H* induced by *V*(*H*) \ {*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* ∈ *V*(*H*)*,*

$$
\gamma\_{o\dot{o}R}(H - v) \ge \gamma\_{o\dot{o}R}(H) - 1.
$$

**Proof.** Let *g* <sup>0</sup> be a *γoiR*(*H* − *v*)-function. Notice that the function *g*, defined by *g*(*v*) = 1 and *g*(*u*) = *g* 0 (*u*) whenever *u* ∈ *V*(*H*) \ {*v*}, is an OIRDF on *H*. Hence, *γoiR*(*H*) − 1 ≤ *ω*(*g*) − 1 = *ω*(*g* 0 ) = *γoiR*(*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* ∈ *V*(*H*)*, then the following statements hold.*


**Proof.** From any *γoiR*(*H*)-function *g* such that *g*(*v*) = 0 and any *α*(*G*)-set, we can construct an OIRDF on *G* ◦*<sup>v</sup> H* of weight *α*(*G*) + *nγoiR*(*H*). Thus, *γoiR*(*G* ◦*<sup>v</sup> H*) ≤ *α*(*G*) + *nγoiR*(*H*) and (i) follows.

Now, if there exists a *γoiR*(*H*)-function *g* such that *g*(*v*) > 0, then from *g* we can construct an OIRDF on *G* ◦*<sup>v</sup> H* of weight *nω*(*g*). Thus, *γoiR*(*G* ◦*<sup>v</sup> H*) ≤ *nω*(*g*) = *nγoiR*(*H*), and (ii) follows.

Finally, if there exists a *γoiR*(*H* − *v*)-function *g* such that *g*(*x*) > 0 for every *x* ∈ *N*(*v*), then from *g* and any *γoiR*(*G*)-function we can construct an OIRDF on *G* ◦*<sup>v</sup> H* of weight *γoiR*(*G*) + *nγoiR*(*H* − *v*), which completes the proof.

**Lemma 4.** *Let f*(*V*0, *V*1, *V*2) *be a γoiR*(*G* ◦*<sup>v</sup> H*)*-function. The following statements hold for any vertex x* ∈ *V*(*G*)*.*

*(i) ω*(*fx*) ≥ *γoiR*(*H*) − 1*. (ii) If ω*(*fx*) = *γoiR*(*H*) − 1*, then x* ∈ *V*<sup>0</sup> *and N*(*x*) ∩ *V*(*Hx*) ⊆ *V*1*.*

**Proof.** Let *x* ∈ *V*(*G*). Observe that *V*<sup>0</sup> ∩ *V*(*Hx*) is an independent set of *H<sup>x</sup>* and also, every vertex in *V*<sup>0</sup> ∩ (*V*(*Hx*) \ {*x*}) has a neighbour in *V*<sup>2</sup> ∩ *V*(*Hx*). 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* ∈ *V*(*Hx*) \ {*x*}, is an OIRDF on *Hx*. Hence, *γoiR*(*H*) − 1 = *γoiR*(*Hx*) − 1 ≤ *ω*(*g*) − 1 ≤ *ω*(*fx*), which completes the proof of (i).

Now, we suppose that *ω*(*fx*) = *γoiR*(*H*) − 1. If *x* ∈ *V*<sup>1</sup> ∪ *V*<sup>2</sup> or *x* ∈ *V*<sup>0</sup> and *N*(*x*) ∩ *V*(*Hx*) ∩ *V*<sup>2</sup> 6= ∅, then *f<sup>x</sup>* is an OIRDF on *Hx*, which is a contradiction. Hence, *x* ∈ *V*<sup>0</sup> and as *V*<sup>0</sup> ∩ *V*(*Hx*) is an independent set, we deduce that *N*(*x*) ∩ *V*(*Hx*) ⊆ *V*1, which completes the proof.

From Lemma 4 (i) we deduce that any *γoiR*(*G* ◦*<sup>v</sup> H*)-function *f* induces three subsets A*<sup>f</sup>* , B*<sup>f</sup>* and C*<sup>f</sup>* of *V*(*G*) as follows.

$$\begin{aligned} \mathcal{A}\_f &= \{ \mathbf{x} \in V(G) : \omega(f\_\mathbf{x}) > \gamma\_{o\dot{o}R}(H) \}, \\ \mathcal{B}\_f &= \{ \mathbf{x} \in V(G) : \omega(f\_\mathbf{x}) = \gamma\_{o\dot{o}R}(H) \}, \\ \mathcal{C}\_f &= \{ \mathbf{x} \in V(G) : \omega(f\_\mathbf{x}) = \gamma\_{o\dot{o}R}(H) - 1 \}. \end{aligned}$$

Next, we state the four possible values of *γoiR*(*G* ◦*<sup>v</sup> H*).

**Theorem 8.** *Let G and H be two graphs with no isolated vertex and* |*V*(*G*)| = *n. If v* ∈ *V*(*H*)*, then*

$$\gamma\_{\dot{o}\mathcal{R}}(G\circ\_{\mathcal{O}}H)\in\{\mathfrak{a}(G)+\mathfrak{n}\gamma\_{\dot{o}\mathcal{R}}(H),\ \mathfrak{n}\gamma\_{\dot{o}\mathcal{R}}(H),\ \gamma\_{\dot{o}\mathcal{R}}(G)+\mathfrak{n}(\gamma\_{\dot{o}\mathcal{R}}(H)-1),\ \mathfrak{a}(G)+\mathfrak{n}(\gamma\_{\dot{o}\mathcal{R}}(H)-1)\}.$$

**Proof.** Let *f*(*V*0, *V*1, *V*2) be a *γoiR*(*G* ◦*<sup>v</sup> H*)-function. By Lemma 3 (i) and (ii) we deduce the upper bound *γoiR*(*G* ◦*<sup>v</sup> H*) ≤ *α*(*G*) + *nγoiR*(*H*). Now, we consider the subsets A*<sup>f</sup>* , B*<sup>f</sup>* , C*<sup>f</sup>* ⊆ *V*(*G*) associated to *f* and distinguish the following cases.

Case 1. C*<sup>f</sup>* = ∅. In this case, for any *x* ∈ *V*(*G*) we have that *ω*(*fx*) ≥ *γoiR*(*H*) and, as a consequence, *γoiR*(*G* ◦*<sup>v</sup> H*) = *ω*(*f*) ≥ *nγoiR*(*H*). If A*<sup>f</sup>* = ∅, then *γoiR*(*G* ◦*<sup>v</sup> H*) = *nγoiR*(*H*). Hence, assume that A*<sup>f</sup>* 6= ∅. This implies that *ω*(*f*) > *nγoiR*(*H*). Moreover, we note that B*<sup>f</sup>* 6= ∅ because *α*(*G*) < *n* and

*ω*(*f*) ≤ *α*(*G*) + *nγoiR*(*H*). Thus, by Lemma 3 (ii) we obtain that B*<sup>f</sup>* ⊆ *V*0, and as *V*<sup>0</sup> is an independent set, we have that A*<sup>f</sup>* is a vertex cover of *G*. Therefore,

$$\begin{split} \gamma\_{o\bar{o}R}(G\circ\_{\mathbb{P}}H) &= \sum\_{\boldsymbol{x}\in\mathcal{A}\_{f}}\omega(f\_{\boldsymbol{x}}) + \sum\_{\boldsymbol{x}\in\mathcal{B}\_{f}}\omega(f\_{\boldsymbol{x}}) \\ &\geq \sum\_{\boldsymbol{x}\in\mathcal{A}\_{f}}(\gamma\_{o\bar{o}R}(H)+1) + \sum\_{\boldsymbol{x}\in\mathcal{B}\_{f}}\gamma\_{o\bar{o}R}(H) \\ &= |\mathcal{A}\_{f}| + \sum\_{\boldsymbol{x}\in V(\mathcal{G})}\gamma\_{o\bar{o}R}(H) \\ &\geq \mathfrak{a}(\mathcal{G}) + n\gamma\_{o\bar{o}R}(H). \end{split}$$

Hence, *γoiR*(*G* ◦*<sup>v</sup> H*) = *α*(*G*) + *nγoiR*(*H*).

Case 2. C*<sup>f</sup>* 6= ∅. Let *z* ∈ C*<sup>f</sup>* . By Lemma 4 (ii) we obtain that *z* ∈ *V*<sup>0</sup> and *N*(*z*) ∩ *V*(*Hz*) ⊆ *V*1. Hence, *f* − *z* is an OIRDF on *H<sup>z</sup>* − *z*, and so *γoiR*(*H* − *v*) = *γoiR*(*H<sup>z</sup>* − *z*) ≤ *ω*(*f* − *z* ) = *γoiR*(*H*) − 1. Thus, Lemma 2 leads to *γoiR*(*H<sup>z</sup>* − *z*) = *γoiR*(*H*) − 1. This implies that *f* − *z* is a *γoiR*(*H<sup>z</sup>* − *z*)-function which satisfies Lemma 3 (iii). Therefore, *γoiR*(*G* ◦*<sup>v</sup> H*) ≤ *γoiR*(*G*) + *n*(*γoiR*(*H*) − 1).

Now, observe the following inequality chain.

$$\gamma\_{o\bar{o}R}(G\circ\_{\upsilon}H) = \sum\_{\mathbf{x}\in\mathcal{A}\_{f}\cup\mathcal{B}\_{f}}\omega(f\_{\mathbf{x}}) + \sum\_{\mathbf{x}\in\mathcal{C}\_{f}}\omega(f\_{\mathbf{x}}) \geq (2|\mathcal{A}\_{f}| + |\mathcal{B}\_{f}|) + n(\gamma\_{o\bar{o}R}(H) - 1). \tag{1}$$

By Lemma 4 (ii) we have that C*<sup>f</sup>* ⊆ *V*0, which implies that A*<sup>f</sup>* ∪ B*<sup>f</sup>* is a vertex cover of *G*. Thus, Inequality chain (1) leads to *γoiR*(*G* ◦*<sup>v</sup> H*) = *ω*(*f*) ≥ *α*(*G*) + *n*(*γoiR*(*H*) − 1). Next, we consider the following two subcases.

Subcase 1. There exists a *γoiR*(*H*)-function *g* such that *g*(*v*) = 2. Let *D* be an *α*(*G*)-set. From *D*, *g* and *f<sup>z</sup>* , we define a function *h* on *G* ◦*<sup>v</sup> H* as follows. For every *x* ∈ *D*, the restriction of *h* to *V*(*Hx*) is induced from *g*. Moreover, if *x* ∈ *V*(*G*) \ *D*, then the restriction of *h* to *V*(*Hx*) is induced from *fz*. By the construction of *g* and *fz*, it is straightforward to see that *h* is an OIRDF on *G* ◦*<sup>v</sup> H*. Thus,

$$\begin{split} \gamma\_{o\bar{o}R}(G\circ\_{v}H) &\leq \sum\_{\boldsymbol{x}\in D} \omega(h\_{\boldsymbol{x}}) + \sum\_{\boldsymbol{x}\in V(G)\backslash D} \omega(h\_{\boldsymbol{x}}) \\ &= \sum\_{\boldsymbol{x}\in D} \omega(g) + \sum\_{\boldsymbol{x}\in V(G)\backslash D} \omega(f\_{\boldsymbol{x}}) \\ &= \sum\_{\boldsymbol{x}\in D} \gamma\_{o\bar{o}R}(H) + \sum\_{\boldsymbol{x}\in V(G)\backslash D} (\gamma\_{o\bar{o}R}(H) - 1) \\ &= |D| + \sum\_{\boldsymbol{x}\in V(G)} (\gamma\_{o\bar{o}R}(H) - 1) \\ &= \mathfrak{a}(G) + \mathfrak{n}(\gamma\_{o\bar{o}R}(H) - 1). \end{split}$$

Therefore, *γoiR*(*G* ◦*<sup>v</sup> H*) = *α*(*G*) + *n*(*γoiR*(*H*) − 1).

Subcase 2. *g*(*v*) ≤ 1 for every *γoiR*(*H*)-function *g*. This condition implies that *V*<sup>2</sup> ∩ B*<sup>f</sup>* = ∅. Since every vertex *x* ∈ C*<sup>f</sup>* has a neighbour in *V*2, and as Lemma 4 (ii) leads to *N*(*x*)∩ *V*(*Hx*) ⊆ *V*1, then we deduce that *N*(*x*) ∩ *V*<sup>2</sup> ∩ A*<sup>f</sup>* 6= ∅. Hence, and as C*<sup>f</sup>* ⊆ *V*0, the function *f* 0 (*V* 0 0 , *V* 0 1 , *V* 0 2 ), defined by *V* 0 <sup>2</sup> = A*<sup>f</sup>* , *V* 0 <sup>1</sup> = B*<sup>f</sup>* and *V* 0 <sup>0</sup> = C*<sup>f</sup>* , is an OIRDF on *G*. So *γoiR*(*G*) ≤ *ω*(*f* 0 ) = 2|A*<sup>f</sup>* | + |B*<sup>f</sup>* |. Therefore, Inequality chain (1) leads to *γoiR*(*G* ◦*<sup>v</sup> H*) ≥ *γoiR*(*G*) + *n*(*γoiR*(*H*) − 1), which implies that *γoiR*(*G* ◦*<sup>v</sup> H*) = *γoiR*(*G*) + *n*(*γoiR*(*H*) − 1).

Therefore, the proof is complete.

In order to see that the four possible values of *γoiR*(*G* ◦*<sup>v</sup> 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 γoiR*(*G* ◦*<sup>x</sup> H*) *for some specific roots x* ∈ *V*(*H*) *are described below.*


Now, we characterize the graphs with *γoiR*(*G* ◦*<sup>v</sup> H*) = *α*(*G*) + *nγoiR*(*H*).

**Figure 4.** The labels of (gray and black) coloured vertices describe the positive weights of a *γoiR*(*H*)-function.

**Theorem 9.** *Let G and H be two graphs with no isolated vertex, let* |*V*(*G*)| = *n and v* ∈ *V*(*H*)*. The following statements are equivalent.*

*(i) γoiR*(*G* ◦*<sup>v</sup> H*) = *α*(*G*) + *nγoiR*(*H*)*. (ii) g*(*v*) = 0 *for every γoiR*(*H*)*-function g.*

**Proof.** We first assume that (i) holds, i.e., *γoiR*(*G* ◦*<sup>v</sup> H*) = *α*(*G*) + *nγoiR*(*H*). If there exists a *γoiR*(*H*)-function *g* such that *g*(*v*) > 0, then by Lemma 3 (ii) it follows that *γoiR*(*G* ◦*<sup>v</sup> H*) ≤ *nγoiR*(*H*), which is a contradiction. Therefore, (ii) holds.

On the other hand, we assume that (ii) holds, i.e., *g*(*v*) = 0 for every *γoiR*(*H*)-function *g*. Let *f*(*V*0, *V*1, *V*2) be a *γoiR*(*G* ◦*<sup>v</sup> H*)-function. If C*<sup>f</sup>* 6= ∅, then by Lemma 4 (ii) we can obtain a *γoiR*(*H*)-function *g* such that *g*(*v*) = 1, which is a contradiction. Hence, C*<sup>f</sup>* = ∅, and so, by Theorem 8 we deduce that *γoiR*(*G* ◦*<sup>v</sup> H*) ∈ {*α*(*G*) + *nγoiR*(*H*), *nγoiR*(*H*)}. Now, suppose that *γoiR*(*G* ◦*<sup>v</sup> H*) = *nγoiR*(*H*). Since C*<sup>f</sup>* = ∅, it follows that B*<sup>f</sup>* = *V*(*G*) and as *V*<sup>0</sup> is an independent set, there exists *x* ∈ B*<sup>f</sup>* \ *V*0. This implies that *f<sup>x</sup>* is a *γoiR*(*Hx*)-function such that *fx*(*x*) > 0, which is a contradiction. Therefore, *γoiR*(*G* ◦*<sup>v</sup> H*) = *α*(*G*) + *nγoiR*(*H*), which completes the proof.

Next, we characterize the graphs with *γoiR*(*G* ◦*<sup>v</sup> H*) = *α*(*G*) + *n*(*γoiR*(*H*) − 1).

**Theorem 10.** *Let G and H be two graphs with no isolated vertex, let* |*V*(*G*)| = *n and v* ∈ *V*(*H*)*. The following statements are equivalent.*

*(i) γoiR*(*G* ◦*<sup>v</sup> H*) = *α*(*G*) + *n*(*γoiR*(*H*) − 1)*.*

*(ii) There exist two γoiR*(*H*)*-functions g*<sup>1</sup> *and g*<sup>2</sup> *such that g*1(*x*) = 1 *for every x* ∈ *N*[*v*] *and g*2(*v*) = 2*.*

**Proof.** We first assume that (i) holds, i.e., *γoiR*(*G* ◦*<sup>v</sup> H*) = *α*(*G*) + *n*(*γoiR*(*H*) − 1). Let *f*(*V*0, *V*1, *V*2) be a *γoiR*(*G* ◦*<sup>v</sup> H*)-function. As *α*(*G*) < *n*, it follows that C*<sup>f</sup>* 6= ∅, and so, by Lemma 4 (ii) we can obtain a *γoiR*(*H*)-function *g*<sup>1</sup> such that *g*1(*x*) = 1 for every *x* ∈ *N*[*v*]. Moreover, if *g*(*v*) ≤ 1 for every *γoiR*(*H*)-function *g*, then, by proceeding analogously to Subcase 2 in the proof of Theorem 8 we deduce

that *γoiR*(*G* ◦*<sup>v</sup> H*) ≥ *γoiR*(*G*) + *n*(*γoiR*(*H*) − 1), which is a contradiction as *γoiR*(*G*) > *α*(*G*). Therefore, there exists a *γoiR*(*H*)-function *g*<sup>2</sup> such that *g*2(*v*) = 2, and (ii) follows.

On the other hand, we assume that there exist two *γoiR*(*H*)-functions *g*<sup>1</sup> and *g*<sup>2</sup> such that *g*1(*x*) = 1 for every *x* ∈ *N*[*v*] and *g*2(*v*) = 2. Let *D* be an *α*(*G*)-set and let *g* 0 1 be a function on *H* such that *g* 0 1 (*v*) = 0 and *g* 0 1 (*x*) = *g*1(*x*) whenever *x* ∈ *V*(*H*) \ {*v*}. From *D*, *g* 0 1 and *g*2, we define a function *h* on *G* ◦*<sup>v</sup> H* as follows. For every *x* ∈ *D*, the restriction of *h* to *V*(*Hx*) is induced from *g*2. Moreover, if *x* ∈ *V*(*G*) \ *D*, then the restriction of *h* to *V*(*Hx*) is induced from *g* 0 1 . Notice that *h* is an OIRDF on *G* ◦*<sup>v</sup> H*, and so *γoiR*(*G* ◦*<sup>v</sup> H*) ≤ *ω*(*h*) = |*D*|*γoiR*(*H*) + |*V*(*G*) \ *D*|(*γoiR*(*H*) − 1) = *α*(*G*) + *n*(*γoiR*(*H*) − 1). Therefore, Theorem 8 leads to *γoiR*(*G* ◦*<sup>v</sup> H*) = *α*(*G*) + *n*(*γoiR*(*H*) − 1), which completes the proof.

Next we proceed to characterize the graphs with *γoiR*(*G* ◦*<sup>v</sup> H*) = *γoiR*(*G*) + *n*(*γoiR*(*H*) − 1). Notice that it is excluded the case *γoiR*(*G*) = *n*, since then *γoiR*(*G* ◦*<sup>v</sup> H*) = *nγoiR*(*H*).

**Theorem 11.** *Let G be a graph of order n with no isolated vertex such that γoiR*(*G*) < *n and let H be a graph with no isolated vertex and v* ∈ *V*(*H*)*. The following statements are equivalent.*


**Proof.** We first assume that (i) holds, i.e., *γoiR*(*G* ◦*<sup>v</sup> H*) = *γoiR*(*G*) + *n*(*γoiR*(*H*) − 1). Let *f*(*V*0, *V*1, *V*2) be a *γoiR*(*G* ◦*<sup>v</sup> H*)-function. Since *γoiR*(*G*) < *n*, it follows that C*<sup>f</sup>* 6= ∅, and so, by Lemma 4 (ii) we can obtain a *γoiR*(*H*)-function *g*<sup>1</sup> such that *g*1(*x*) = 1 for every *x* ∈ *N*[*v*]. Moreover, if there exists a *γoiR*(*H*)-function *g*<sup>2</sup> such that *g*2(*v*) = 2, then by Theorem 10 we deduce that *γoiR*(*G* ◦*<sup>v</sup> H*) = *α*(*G*) + *n*(*γoiR*(*H*) − 1), which is a contradiction as *γoiR*(*G*) > *α*(*G*). Therefore, *g*(*v*) ≤ 1 for every *γoiR*(*H*)-function *g*, which implies that (ii) follows.

On the other side, we assume that *g*(*v*) ≤ 1 for every *γoiR*(*H*)-function *g* and also, that there exists a *γoiR*(*H*)-function *g*<sup>1</sup> such that *g*1(*x*) = 1 for every *x* ∈ *N*[*v*]. Under these assumptions, observe that the function *g*<sup>1</sup> restricted to *V*(*H*) \ {*v*}, namely *g* 0 1 , is an OIRDF on *H* − *v*. Hence, *γoiR*(*H* − *v*) ≤ *ω*(*g* 0 1 ) = *ω*(*g*1) − 1 = *γoiR*(*H*) − 1 and by Lemma 2 we deduce that *γoiR*(*H* − *v*) = *γoiR*(*H*) − 1. Hence, *g* 0 1 is a *γoiR*(*H* − *v*)-function which satisfies Lemma 3 (iii). Therefore, Lemma 3 and Theorem 8 lead to *γoiR*(*G* ◦*<sup>v</sup> H*) ∈ {*γoiR*(*G*) + *n*(*γoiR*(*H*) − 1), *α*(*G*) + *n*(*γoiR*(*H*) − 1)}. Finally, as *g*(*v*) ≤ 1 for every *γoiR*(*H*)-function *g*, by Theorem 10 we deduce that *γoiR*(*G* ◦*<sup>v</sup> H*) = *γoiR*(*G*) + *n*(*γoiR*(*H*) − 1), which completes the proof.

From Theorem 8 we have that there are four possible expressions for *γoiR*(*G* ◦*<sup>v</sup> H*). Theorems 9–11 characterize three of these expressions. In the case of the expression *γoiR*(*G* ◦*<sup>v</sup> H*) = *nγoiR*(*H*), the corresponding characterization can be derived by elimination from the previous results.

**Author Contributions:** All authors contributed equally to this work. Investigation, A.C.M., S.C.G., A.C.G. and A.M.G.d.R; writing—review and editing, A.C.M., S.C.G., A.C.G. and A.M.G.d.R. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding.

**Conflicts of Interest:** The authors declare no conflict of interest.

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