EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 1, Article Number 5252 ISSN 1307-5543 – ejpam.com Published by New York Business Global Modern Roman Dominating Functions in Graphs Sherihatha R. Ahamad1,2,∗, Jerry Boy G. Cariaga1,2, Sheila M. Menchavez1,2 1 Department of Mathematics and Statistics, College of Science and Mathematics, MSU-Iligan Institute of Technology, 9200 Iligan City, Philippines 2 CMTPS, Premier Research Institute of Science and Mathematics, MSU-Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. Let G = (V (G), E(G)) be any connected graph. A function f : V (G) → {0, 1, 2, 3} is a modern Roman dominating function of G if for each v ∈ V (G) with f(v) = 0, there exist u,w ∈ NG(v) such that f(u) = 2 and f(w) = 3; and for each v ∈ V (G) with f(v) = 1, there exists u ∈ NG(v) such that f(u) = 2 or f(w) = 3. The weight of a modern Roman dominating function f of G is the sum ωmR G (f) = ∑ v∈V (G) f(v) and the minimum weight among all of the modern dominating functions on G is called the modern Roman domination number γmR(G) of G. In this paper, we characterize graphs with smaller modern Roman domination number and obtain the γmR(G) of some special graphs. Moreover, we investigate and characterize the modern Roman domination of the join and corona of graphs. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Dominating set, Domination number, Modern Roman dominating function, and Modern Roman domination number. 1. Introduction The concept of Roman domination is introduced in 2004 [6]. It is inspired by the strategies for defending the Roman Empire presented in the work of ReVelle and Rosing in [13] and Stewart, Cockayne, et al. in [15]. Since then, it has emerged as an active research field in graph theory (see [10],[8],[1],[4],[3],[12],[9],[7],[14],[11]). A new model of graph domination based on Roman domination is introduced in [8], called modern Roman domination. Studies and exploration on this variant can be found in [1, 11, 14]. Explicity, a function f : V (G) → {0, 1, 2, 3} is a modern Roman dominating function (MRDF ) of G if for each v ∈ V (G) with f(v) = 0, there exist u,w ∈ NG(v) such that f(u) = 2 and f(w) = 3; and for each v ∈ V (G) with f(v) = 1, there exists u ∈ NG(v) such that f(u) = 2 or f(u) = 3. The minimum weight among all of the MRDF is called the modern Roman ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i1.5252 Email addresses: sherihatha.ahamad@g.msuiit.edu.ph (S. Ahamad), jerryboy.cariaga@g.msuiit.edu.ph (J. Cariaga), sheila.menchavez@g.msuiit.edu.ph (S. Menchavez) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 2 of 18 domination number and is denoted by γmR(G). In this model, the label of a vertex under the function f represents a type of weapon in a war zone. The four defensive weapon types are represented by the set of weights {0, 1, 2, 3} under the function f . Weapon types are given ascending weights: light, medium, heavy, and air force. Light weapons are for pedestrians; heavy weapons can be tanks and rockets. The defense strategy of modern Roman domination relies on a support system of heavy weapons and air forces to back up the light and medium weapons. [1]. This study explores further the concept of modern Roman domination in graphs. It focuses on providing the modern Roman domination number of some specials graphs and some characterizations for the modern Roman dominating functions of the join and corona of graphs. 2. Terminology and Notation The symbols V (G) and E(G) denote the vertex set and edge set, respectively, of a graph G. For S ⊆ V (G), |S| is the cardinality of S. In particular, |V (G)| and |E(G)| are the order and size, respectively, of G. All graph terminologies that are not introduced but are being used here are adapted from [2]. The set of neighbors of a vertex u in G, denoted by NG(u), is called the open neigh- borhood of u in G. The closed neighborhood of u in G is the set NG[u] = NG(u) ∪ {u}. If S ⊆ V (G), the open neighborhood of S in G is the set NG(S) = ⋃ u∈S NG(u). The closed neighborhood of S in G is the set NG[S] = NG(S)∪S. For S ⊆ V (G) of a connected graph G, NG(S) = ⋃ v∈S NG(v) and NG[S] = S∪NG(S). A graph whose edge-set is empty is called an empty graph (also called null graph or totally disconnected graph). An empty graph of order n is denoted by Kn. A set S ⊆ V (G) is a dominating set in G if NG[S] = V (G). Thus, S is a dominating set in G if and only if for each v ∈ V (G) \ S, there exists u ∈ S such that uv ∈ E(G). The minimum cardinality of a dominating set in G, denoted by γ(G), is the domination number of G. A dominating set S of G with |S| = γ(G) is called a γ - set of G. Readers may refer to [5] for the introduction and more comprehensive discussion of the development of the concept of domination in graphs. For a positive integer k, a set D ⊆ V (G) is called a k-dominating set if each x ∈ V (G) \D is adjacent to at least k vertices in D. The k-domination number γk(G) is then defined to be the smallest cardinality of a k-dominating set of G. A Roman dominating function (RDF) on G is a function f : V (G) → {0, 1, 2} such that every vertex u ∈ V (G) for which f(u) = 0 is adjacent to at least one vertex v for which f(v) = 2. The weight of an RDF is the value ωG(f) = ∑ u∈V (G) f(u). The Roman domination number γR(G) is the minimum weight among all of the RDF on G. An RDF with ωG(f) = γR(G) is referred to as a γR-function [3]. S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 3 of 18 A function f : V (G) → {0, 1, 2, 3} is a double Roman dominating function of G, written f ∈ DRD(G), if each of the following holds: (1) for each v ∈ V (G) with f(v) = 0 at least one of the following holds: (a) v has two adjacent vertices u and w for which f(u) = f(w) = 2; or (b) v has an adjacent vertex u for which f(u) = 3, and (2) for each v ∈ V (G) with f(v) = 1, v is adjacent to a vertex u for which either f(u) = 2 or f(u) = 3. The double Roman domination number of G denoted by γdR(G), is the minimum weight ωG(f) = ∑ v∈V (G) f(v) of all the double Roman dominating functions f of G. Any f ∈ DRD(G) of weight equal to γdR(G) is referred to as γdR-function of G [3]. A modern Roman dominating function (MRDF ) of G is a function f : V (G) → {0, 1, 2, 3} if (P1) for each v ∈ V (G) with f(v) = 0, there exist u,w ∈ NG(v) such that f(u) = 2 and f(w) = 3; and (P2) for each v ∈ V (G) with f(v) = 1, there exists u ∈ NG(v) such that f(u) = 2 or f(u) = 3. The weight of a modern Roman dominating function f of G is the sum ωmR G (f) =∑ v∈V (G) f(v) and its minimum weight among all of the modern Roman dominating func- tion is called the modern Roman domination number γmR(G) of G. A modern Roman dominating function of G with weight ωmR G (f) = γmR(G) is called a γmR-function of G [8]. For a function f : V (G) → {0, 1, 2, 3} on a graph G, let (V0, V1, V2, V3) be the ordered partition induced by f , where Vi = {v ∈ V (G) : f(v) = i} for i ∈ {0, 1, 2, 3}. Then we can write f = (V0, V1, V2, V3). The weight of f is defined by ωG(f) = |V1|+ 2|V2|+ 3|V3|. Example 1. Consider the given graph G with V (G) = {a, b, c, d, e, g, h} in Figure 1. The function f : V (G) → {0, 1, 2, 3} given by f(v) =  3, if v = a. 2, if v = g. 0, otherwise. is a modern Roman dominating function of G. It can be verified that the γmR(G) = 5. S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 4 of 18 0b 0 d 0e 0c 0h 3a 2 g G : Figure 1: Graph G of order 7 with γmR(G) = 5. 3. Known Results We make use of the following known results from [8]. Proposition 1. Let G be a graph of order n and let f = (V0, V1, V2, V3) be a γmR-function on G. Then each of the following statements holds: (i) If n ≥ 4, then 5 ≤ γmR(G) ≤ 2n. (ii) If there are two vertices that are adjacent to all other vertices in G, then γmR = 5. (iii) If G is empty graph, then 2γ(G) = γmR(G). (iv) V2 ̸= ∅ (v) V2 ∪ V3 is a dominating set of G. Moreover, it is a 2-dominating set of G [V0] (vi) If v is a pendant vertex, then f(v) ̸= 0. (vii) If v is an isolated vertex, then f(v) = 2. Proposition 2. For path Pn, n ≥ 1, γmR(Pn) = n+ ⌈n 3 ⌉ Proposition 3. For cycle Cn, n ≥ 3, γmR(Cn) = { 5, if n = 4 n+ ⌈ n 3 ⌉ , if n ̸= 4 4. Main Results This section begins with the general and useful properties of modern Roman domi- nating functions. It also presents the characterizations of some graphs G with γmR(G) ∈ {2, 3, 4, 5} and the modern Roman domination number of the n-barbell graph Bn, wind- mill graph Wd(k, n), friendship graph Gn 3 , butterfly graph G2 3, complete bipartite graph S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 5 of 18 Km,n, star graph Sn and fan graph Fn. For simplicity, we denote by MRDF (G) the set of all modern Roman dominating functions on a graph G. Remark 1. If f = (V0, V1, V2, V3) is a γmR-function of G and v ∈ V1, then v need not be in NG(V2) ∩NG(V3). Proposition 4. Let G be any graph with no isolated vertex. If f = (V0, V1, V2, V3) a γmR-function of G, then the following holds: (i) V0 = ∅ if and only if V3 = ∅ and V2 is a γ-set of G. Moreover, γmR(G) = |V (G)|+ γ(G). (ii) V1 = ∅ if and only if V2 ∪ V3 is a 2-dominating set of G. Moreover, if V1 = ∅, ⟨V2 ∪ V3⟩ is connected and V3 is a γ-set of G, then γmR(G) ≥ γ(G) + 2γ2(G). Proof. Clearly, V0 = ∅ if and only if V3 = ∅. Suppose V0 = ∅. Since V2 ∪ V3 is a dominating set of G and V3 = ∅, it follows that V2 is a dominating set of G. Suppose V2 is not a γ-set of G. Let S be a γ-set of G and define g = (V ′ 0 , V ′ 1 , V ′ 2 , V ′ 3) where V ′ 0 = V ′ 3 = ∅, V ′ 1 = V (G)\S, and V ′ 2 = S. Then there exists V ∗ 2 ⊆ V (G) such that V ∗ 2 is a γ-set of G. Let V ′ 2 = V ∗ 2 , V ′ 0 = V ′ 3 = ∅ and V ′ 1 = V (G)\V ∗ 2 . Thus, g = (V ′ 0 , V ′ 1 , V ′ 2 , V ′ 3) ∈ MRDF (G), and so, ωmR G (g) < ωmR G (f), a contradiction. Hence, V2 is a γ-set of G. Furthermore, γmR(G) = |V1|+2|V2| = |V (G)\V2|+2|V2| = |V (G)\V2|+2γ(G) = |V (G)|−γ(G)+2γ(G) = |V (G)|+ γ(G). This proves (i). Now we prove (ii). Suppose V1 = ∅. Then by Proposition 1, V2 ∪V3 is a 2-dominating set of G. Conversely, suppose that V1 ̸= ∅ and take {v} ∈ V1. Then by Remark 1, v need not be in NG(V2) ∩ NG(V3), which is a contradiction. Hence, the assertion follows. Moreover, assume that ⟨V2 ∪ V3⟩ is connected and let V3 be a γ-set of G. Since V1 = ∅, γmR(G) = 2|V2|+ 3|V3| = 2|V2 ∪ V3|+ |V3| ≥ 2γ2(G) + γ(G). Proposition 5. Let G be a connected graph. Then (i) γmR(G) = 2 if and only if G = K1. (ii) γmR(G) = 3 if and only if G = K2. (iii) γmR(G) = 4 if and only if G ∈ {K3, P3}. (iv) γmR(G) = 5 if and only if |V (G)| = 4 and γ(G) = 1 or γ2(G) = 2 and |V (G)| ≥ 4. Proof. (i) Suppose γmR(G) = 2, say f = (V0, V1, V2, V3) is a γmR-function on G. By Proposition 1(iv), V2 = {v}. Hence, V0 = V1 = V3 = ∅. The converse is clear. (ii) Suppose γmR(G) = 3, say f = (V0, V1, V2, V3) is a γmR-function on G. By (i), |V2| ≥ 2. By Proposition 1(iv), and the assumption that γmR(G) = 3, |V2| = 1, |V1| = 1 and V0 = V3 = ∅. Therefore, |V (G)| = 2. Since G is connected, G = K2. Conversely, suppose that G = K2, say V (G) = {x, y}. Then g = {∅, {x}, {y},∅} ∈ MRDF (G) and S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 6 of 18 ωmR G (g) = 3. Since γmR(G) ≥ 2, it follows that γmR(G) = 3. (iii) Note that if γmR(G) = 4, then 1 ≤ |V2| ≤ 2 by Proposition 1(iv). Hence, there are only two cases to consider, namely, |V2| = 1 and |V2| = 2. If |V2| = 2, then |V1| = 0. By (P2), this cases is not possible. So if |V2| = 1, we have |V1| = 2. By (P2), ⟨V1 ∪ V2⟩ must be connected. Thus, the result follows. The converse follows directly from Propositions 2 and 3. (iv) If γmR(G) = 5, then |V3| ≤ 1 and 1 ≤ |V2| ≤ 2. Also, by (iii), |V (G)| ≥ 4. Now, if |V3| = 0, then |V0| = 0. Hence, there are only two cases to consider, namely, |V2| = 1 and |V2| ≤ 2. If |V2| = 2, then |V1| = 1. Therefore, |V (G)| = 3 which is not possible by (iii). If |V2| = 1, then |V1| = 3. By (P2), ⟨V1 ∪ V2⟩ must be connected and V2 is a dominating set in G, it follows that V2 is a γ-set in G. Therefore, |V (G)| = 4 and γ(G) = 1. Now, suppose that |V3| = 1. If |V2| = 0, then |V1| = 2. Consequently, |V (G)| = 3. Thus, G ∈ {K3, P3}, a contradiction by (iii). If |V2| = 1, then |V1| = 0. Since V2 ∪ V3 is a 2-dominating set in G and |V2 ∪ V3| = 2, it follows that V2 ∪ V3 is a γ2-set in G. Hence, |V (G)| ≥ 4 and γ2(G) = 2. Conversely, suppose |V (G)| = 4 and γ(G) = 1. By (iii), γ(G) ≥ 5. Let v be a dominating vertex of G and define a function f = (V0, V1, V2, V3) on V (G) such that V0 = ∅ = V3, V2 = {v}, V1 = V (G) \ {v}. Then f ∈ MRDF (G) and ωmR G (f) = 5. This implies that γmR(G) = 5. Next, suppose that γ2(G) = 2 and |V (G)| ≥ 4. Let D = {u, v} be the γ2-set of G. Define a function g = (V0, V1, V2, V3) such that V1 = ∅ and g(x) =  3, if x = u. 2, if x = v 0, if x ∈ V (G) \D. Then g ∈ MRDF (G) and ωmR G (g) = 5. Since G ̸∈ {K3, P3}, we must have ωmR G (g) = 5. Hence, γmR(G) = 5. Corollary 1. For a connected graph G of order 4, γmR(G) = 5 if and only if G ∈ {K1 + (K1 ∪K2),K1 +K3,K1 +K3,K1 + P3}. Proof. The proof follows directly from Proposition 5 (iv). Remark 2. Let G be a graph, then every γmR-function of G is a γdR-function of G if V0 = ∅. Proposition 6. For a complete graph Kn, γmR(Kn) = 5 for all n ≥ 4. Proof. Pick any x, y ∈ V (Kn) with x ̸= y. Clearly, g = (V (Kn) \ {x, y},∅, {x}, {y}) ∈ MRDF (Kn). It follows that γmR(Kn) ≤ 5. On the other hand, suppose that f = (V0, V1, V2, V3) is a γmR-function of Kn. If V0 = ∅, then V3 = ∅. Since f is a γmR- function of Kn, |V2| = 1 and |V1| = n − 1. Hence, γmR(Kn) = ωmR Kn (f) = n + 1 ≥ 5. If S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 7 of 18 V0 ̸= ∅, then |V2| ≥ 1 and |V3| ≥ 1. It follows that γmR(Kn) = ωmR Kn (f) = 2|V2|+3|V3| ≥ 5. Therefore, γmR(Kn) = 5. In what follows, we denote by f |G the restriction of f on the subgraph G of the graph H. Proposition 7. Let G be a disconnected graph with nontrivial components G1, G2, · · · , Gn. Then γmR(G) = ∑n i=1 γmR(Gi). Proof. Let G1, G2, · · · , Gn be the components of G. Let f1, f2, · · · , fn be γmR-functions of G1, G2, · · · , Gn respectively. Define a function f : V (G) −→ {0, 1, 2, 3} given by f(x) =  f1(x), if x ∈ V (G1). f2(x), if x ∈ V (G2). ... fn(x), if x ∈ V (Gn). Then f is a γmR-function ofG. Thus γmR(G) ≤ ∑n i=1 γmR(Gi). Conversely, let f be a γmR- function of G. Then the restriction f |Gi of f to Gi, where i = 1, 2, · · · , n is a γmR-function of Gi. Thus, γmR(Gi) ≤ ωmR G (f |Gi) for all i = 1, 2, · · · , n. Hence, ∑n i=1 γmR(Gi) ≤ γmR(G). Hence, the assertion follows by combining the results. Corollary 2. Let G be a graph of order n. Then γmR(G) = 2n if and only if G = Kn. The n-barbell graph is the simple graph obtained by joining two copies of complete graph Kn≥3 by a bridge and is denoted by Bn. Figure 2 shows the n-barbell graphs B3 and B5, respectively. 1 1 1 1 0 0 0 0 0 0 B3 : B5 : 2 2 2 3 3 2 Figure 2: The graphs B3 and B5 with γmR(B3) = 8 and γmR(B5) = 10, respectively. Proposition 8. For any n-barbell graph Bn where n ≥ 3, γmR(Bn) = { 8, if n = 3. 10, if n ≥ 4. S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 8 of 18 Proof. Let Bn be an n-barbell graph and uv ∈ E(Bn) be the bridge that joins the two copies of Kn. If n = 3, define a function f = (V0, V1, V2, V3) given by f(x) = { 2, x ∈ {u, v}. 1, otherwise. Then f ∈ MRDF of B3. It follows that γmR(B3) ≤ 8. Now, suppose that f ′ = (V ′ 0 , V ′ 1 , V ′ 2 , V ′ 3) is a γmR-function of B3. If V ′ 0 = ∅, then V ′ 3 = ∅. Since f ′ is a γmR- function of B3, |V ′ 2 | = 2 and |V ′ 1 | = V (B3) \ |V ′ 2 |. Hence, γmR(B3) = ωmR B3 (f ′) ≥ 8. If |V ′ 0 | ̸= 0, then |V ′ 2 | ≥ 2 and |V ′ 3 | ≥ 1. It follows that γmR(B3) = ωmR B3 (g) ≥ 8. There- fore, γmR(B3) = 8. If n ≥ 4. Pick any v′, u′ ∈ V (Bn) such that v′ ̸= u, u′ ̸= v, and v′v, u′u ∈ E(Bn). Now, define a function f = (V0, V1, V2, V3) given by f(x) =  0, x ∈ V (Bn) \ {u, v, u′, v′}. 3, x ∈ {u, v}. 2, x ∈ {u′, v′}. Then f ∈ MRDF of Bn, n ≥ 4. It follows that γmR(Bn) ≤ 10. Now, suppose that g = (W0,W1,W2,W3) is a γmR-function of Bn. If W0 = ∅, then W3 = ∅. Since g is a γmR-function of Bn, |W2| = 2 and |W1| = V (Bn) \ |W2|. Hence, γmR(Bn) = ωmR Bn (g) ≥ 10. If |W0| ̸= 0, then |W2| ≥ 2 and |W3| ≥ 2. It follows that γmR(Bn) = ωmR Bn (g) = 2|W2| + 3|W3| ≥ 10. Therefore, γmR(B3) = 10. The windmill graph Wd(k, n)= G = K1 + nKk−1 is constructed for k ≥ 2 and n ≥ 2 by joining n copies of the complete graph Kk at a shared vertex. It has n(k − 1) + 1 vertices and 1 2nk(k − 1) edges. The case k = 3 corresponds to the dutch windmill graph (also called friendship graph) Gn 3 = K1 + nK2 and the case n = 2 corresponds to the butterfly graph G2 3 = K1+2K2. The graphs in Figures 3, 4, and 5 are the windmill graph Wd(4, 2), friendship graph G4 3 and butterfly graphs G2 3, respectively. 3 22 0 00 0 Figure 3: A windmill graph Wd(4, 2) with γmR(Wd(4, 2)) = 7 S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 9 of 18 2 1 1 1 1 11 1 1 Figure 4: A friendship graph G4 3 with γmR(G 4 3) = 10 2 1 11 1 Figure 5: A butterfly graph G2 3 with γmR(G 2 3) = 6 Proposition 9. For any windmill graph G = K1 + nKk−1, where k ≥ 4 and n ≥ 2, γmR(G) = 2n+ 3. Proof. Let G = K1 + nKk−1, where k ≥ 4 and n ≥ 2. Suppose V (K1) = {u} be the central vertex in G, then pick a vertex v in each n copies of the complete graph Kk−1 and define a function f = (V0, V1, V2, V3) given by f(x) =  3, x = u. 2, x = v. 0, otherwise. Then f ∈ MRDF (G). It follows that γmR(G) ≤ 2n + 3. Now, suppose that g = (W0,W1,W2,W3) is a γmR-function of G. If W0 = ∅, then W3 = ∅. Since g is a γmR- function of G, |W2| = {u} = 1 and |W1| = V (G)\{u}. Hence, γmR(G) = ωmR G (g) ≥ 2n+3. If |W0| ̸= 0, then |W2| ≥ 2 and |W3| ≥ 1. It follows that γmR(G) = ωmR G (g) = 2|W2|+ 3|W3| ≥ 2n+ 3. Therefore, γmR(G) = 2n+ 3. Proposition 10. For any friendship graph G, γmR(G) = 2n+ 2. Proof. Let G = K1 + nK2, n ≥ 2. Let V (K1) = {u} be the central vertex in G. Define a function f = (∅, V (G)\{u}, {u},∅). Then for all vi ∈ V1, 1 ≤ i ≤ n, f(NG[vi]) = 2n + 2. Thus, f ∈ MRDF (G). It follows that γmR(G) ≤ 2n + 2. Now, suppose that f ′ = (W0,W1,W2,W3) is a γmR-function of G. If W0 = ∅, then W3 = ∅. Since f ′ is a γmR-function of G, by Proposition 4 (i), γmR(G) = 2n+ 2. Corollary 3. For a butterfly graph G, γmR(G) = 6. S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 10 of 18 Proof. The result follows from Proposition 10. A graph G is called bipartite if the vertex set V (G) of G can be partitioned into two subsets V1 and V2 such that every edge in G joins a vertex in V1 with a vertex in V2. If G is bipartite such that G contains every edge incident with any pair of vertices in V1 and V2, then G is a complete bipartite graph; in this case, G = Km,n if |V1| = m and |V2| = n. Figure 6 shows the complete bipartite graph K7,5. .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ......................................................................................................................................................................................................................... ............................................................................................................................................................................................................................................... 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K7,5: Figure 6: A complete bipartite G = K7,5 Proposition 11. For a complete bipartite graph Km,n, let p = min{m,n},m, n ≥ 2. Then γmR(Km,n) =  5, if p = 2. 7, if p = 3 9, if p = 4 10, if p ≥ 5. Proof. Let G be a complete bipartite graph Km,n and X and Y be partite sets of Km,n, where |X| = m and |Y | = n. Let p = min{m,n}. For p = 2, since |V5| = 5 and γ2(G) = 2, it follows from 5(iv) that γmR(G) = 5. For p = 3, let X and Y be partite sets of G and assume that |X| = 3, say X = {x1, x2, x3}. Let V0 = Y, V1 = ∅, V2 = {x1, x2} and V3 = {x3}. Then f = (V0, V1, V2, V3) ∈ MRDF (G) and its weight is 7. Hence, γmR(G) ≤ 7. Now, suppose that g = (W0,W1,W2,W3) is a γmR-function. If W0 = ∅ then W3 = ∅. Since g is a γmR-function, then γmR(G) ≥ 7. If W0 ̸= ∅, let W0 = Y,W1 = ∅,W2 = {x1, x2} and W3 = {x3}. Since g is a γmR-function, then γmR(G) ≥ 7. Thus, γmR(G) = 7. If p = 4 and WLOG, let X = {x1, x2, x3, x4} such that f(x1) = 2, f(x2) = 3 and f(x3) = 2 = f(x4). Then f(yi) = 0, i = 1, 2, · · · , n, and for every yi ∈ Y , yi ∈ NG(X) by (P2). Define f = (V0, V1, V2, V3) such that V0 ̸= ∅, V1 = ∅, V2 = {x1, x3, x4}, V3 = {x2}. Then f ∈ MRDF (G). Thus, γmR(G) ≤ 9. Suppose to the contrary that γmR(G) < 9 and p = 4. So, γmR(G) = 8. Now, if X = {x1, x2, x3, x4} then {x1, x2, x3} is a γ3-set of G \ {x4}. Clearly, 7 = γmR(G \ {x4}) ≤ γmR(G) = 8. This means that f(x4) = 1, a contradiction. Therefore, γmR(G) = 9. If p ≥ 5, let X = Km and Y = Kn. WLOG, let S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 11 of 18 {u1, u2} ∈ V (Km), {v1, v2} ∈ V (Kn) and define a function f = (V0, V1, V2, V3) given by f(z) =  3, z ∈ {u1, v1}. 2, z ∈ {u2, v2}. 0, otherwise. Then f ∈ MRDF (G). Since |V1| = 0 and uivi ∈ E(G), for all i = 1, 2, {u1, v1} = V3 is a dominating set of G. Thus, γ(G) = 2. Since ⟨V2 ∪ V3⟩ is connected, by Proposition 4 (ii), γmR(G) = 10. The fan Fn of order n+ 1 is the graph Pn +K1 and the star Sn of order n+ 1 is the graph Kn +K1. The graphs in Figures 7 and 8 are the star graph S6 and fan graph F6, respectively. 2 1 1 1 1 1 2 Figure 7: A star graph S6 with γmR(S6) = 7 1 1 1 1 1 2 Figure 8: A fan graph F6 with γmR(F6) = 7 Proposition 12. If G ∈ {Fn, Sn}, n ≥ 1, then γmR(G) = n+ 2. Proof. WLOG, let G = Fn where V (G) = V (K1 + Pn) and V (K1) = {u} is a central vertex of G. Now, define a function f = (V0, V1, V2, V3) given by f(x) = { 2, x = {u}. 1, otherwise. Then f ∈ MRDF (G). It follows that γmR(G) ≤ n + 2. Now, suppose that g = (W0,W1,W2,W3) is a γmR-function of G. If W0 = ∅, then W3 = ∅. Since g is a γmR-function of G, |W2| = |V (K1)| = 1 and |W1| = |V (Pn)| = n. Hence, γmR(G) = ωmR G (g) ≥ n + 2. If |W0| ≠ 0, then |W2| ≥ 1 and |W3| ≥ 1. It follows that γmR(G) = ωmR G (g) = 2|W2|+ 3|W3| ≥ n+ 2. Therefore, γmR(G) = n+ 2. S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 12 of 18 5. On the join of graphs Given two graphs G and H with disjoint vertex sets, the join G+H of graphs G and H, is the graph with vertex-set V (G+H) = V (G)∪V (H) and edge-set E(G + H) = E(G) ∪ E(H) ∪ {uv : u ∈ V (G) and v ∈ V (H)} [3]. In this section, the following proposition characterizes allMRDF on the join of graphs. Proposition 13. Let G and H be any graphs and let f ∈ (V0, V1, V2, V3) be a function on V (G+H) with V2 ̸= ∅ and V3 ̸= ∅. Then f ∈ MRDF (G+H) if and only if one of the following holds: (i) f |G ∈ MRDF (G) and one of the following holds: (a) |V2 ∩ V (G)| ≥ 1 and |V3 ∩ V (G)| ≥ 1 (b) V2 ∩ V (G) = ∅ and each of the following holds: (b1) V3 is a dominating set of G. (b2) V2 ∩ V (H) is a dominating set of H[V0]. (c) V3 ∩ V (G) = ∅ and each of the following holds: (c1) V2 is a dominating set of G. (c2) V3 ∩ V (H) is a dominating set of H[V0]. (ii) f |H ∈ MRDF (H) and one of the following holds: (a) |V2 ∩ V (H)| ≥ 1 and |V3 ∩ V (H)| ≥ 1 (b) V2 ∩ V (H) = ∅ and each of the following holds: (b1) V3 is a dominating set of H. (b2) V2 ∩ V (G) is a dominating set of G[V0]. (c) V3 ∩ V (H) = ∅ and each of the following holds: (c1) V2 is a dominating set of H. (c2) V3 ∩ V (G) is a dominating set of G[V0]. (iii) f |G ̸∈ MRDF (G), f |H ̸∈ MRDF (H) and each of the following holds: (a) V2 ∩ V (H) ̸= ∅ whenever NG(x) ∩ V2 = ∅ for some x ∈ V0. (b) V3 ∩ V (H) ̸= ∅ whenever NG(x) ∩ V3 = ∅ for some x ∈ V0. (c) V2 ∩V (H) ̸= ∅ or V3 ∩V (H) ̸= ∅ whenever ∃x ∈ V1 with NG(x)∩V2 = ∅ and NG(x) ∩ V3 = ∅ (d) V2 ∩ V (G) ̸= ∅ whenever NH(x) ∩ V2 = ∅ for some x ∈ V0. (e) V3 ∩ V (G) ̸= ∅ whenever NH(x) ∩ V3 = ∅ for some x ∈ V0. S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 13 of 18 (f) V2 ∩ V (G) ̸= ∅ or V3 ∩ V (G) ̸= ∅ whenever ∃x ∈ V1 with NH(x)∩ V2 = ∅ and NH(x) ∩ V3 = ∅ Proof. Suppose f |G ∈ MRDF (G). Assume that (i)(a) holds. Let v ∈ V0. If v ∈ V (G), then there exist u,w ∈ V (G) such that {u,w} ⊆ NG(v) and f(u) = 2 and f(w) = 3, by (P1). This implies that {u,w} ⊆ NG+H(v). Now, assume that v ∈ V (H). Note that |V2 ∩ V (G)| ≥ 1 and |V3 ∩ V (G)| ≥ 1. Now, take u ∈ V2 ∩ V (G) and w ∈ V3 ∩ V (G) such that vu, vw ∈ E(G +H). Thus, {u, v} ⊆ NG+H(v). Moreover, let v ∈ V1. Assume v ∈ V (G). Then there exists z ∈ V2 ∩ V (G) or z ∈ V3 ∩ V (G) such that z ∈ NG(v) by (P2). This means that z ∈ NG+H(v). Now, assume v ∈ V (H). Since |V2 ∩ V (G)| ≥ 1 and |V3 ∩ V (G)| ≥ 1, there exists z ∈ V2 ∩ V (G) or z ∈ V3 ∩ V (G) such that z ∈ NG+H(v). Thus, f ∈ MRDF (G + H). Similarly, if f |H ∈ MRDF (H) with |V2 ∩ V (H)| ≥ 1 and |V3 ∩ V (H)| ≥ 1, then f ∈ MRDF (G+H). Assume (i)(b) holds. Since V0 ∩ V (G) = ∅, V0 ⊆ V (H). Let v ∈ V0. By (b2), there exists u ∈ V2 ∩ V (H) such that uv ∈ E(H) ⊆ E(G+H). Also, since V3 is a dominating set of G, V3 ∩ V (G) ̸= ∅. Pick u ∈ V3 ∩ V (G). Then uv ∈ E(G+H). Let v ∈ V1 ∩ V (G). By (b1), there exists u ∈ V3 ∩ V (G) such that uv ∈ E(G) ⊆ E(G +H). Now, let v ∈ V1 ∩ V (H). By (b1), there exists u ∈ V3 ∩ V (G). Then uv ∈ E(G + H). Therefore, f ∈ MDRF (G + H). Similarly, if (ii)(b) holds, then f ∈ MRDF (G +H). Assume (i)(c) holds. Since V0 ∩ V (G) = ∅, then V0 ⊆ V (H). Let v ∈ V0. By (c2), there exists u ∈ V3∩V (H) such that uv ∈ E(H) ⊆ E(G+H). Also, since V2 is a dominating set of G, V2 ∩ V (G) ̸= ∅. Pick u ∈ V2 ∩ V (G). Then uv ∈ E(G+H). Let v ∈ V1∩V (G). By (c1), there exists u ∈ V2∩V (G) such that uv ∈ E(G) ⊆ E(G+H). Now, let v ∈ V1 ∩ V (H). By (c1), there exists u ∈ V2 ∩ V (G). Then uv ∈ E(G + H). Therefore, f ∈ MDRF (G+H). Similarly, if (ii)(c) holds, then f ∈ MRDF (G+H). Suppose (iii) holds, that is f |G ̸∈ MRDF (G) and f |H ̸∈ MRDF (G). Let v ∈ V0 ∩ V (G). If NG(v) ∩ V2 = ∅ and NG(v) ∩ V3 ̸= ∅. Take u ∈ V3 ∩ V (G) such that uv ∈ E(G) ⊆ E(G + H). Since NG(v) ∩ V2 = ∅, by assumption there exists w ∈ V2 ∩ V (H) such that vw ∈ E(G + H). If NG(v) ∩ V2 ̸= ∅ and NG(v) ∩ V3 = ∅. Pick u ∈ V2 ∩ V (G) such that uv ∈ E(G) ⊆ E(G + H). Since NG(v) ∩ V3 = ∅, by assumption there exists w ∈ V3 ∩ V (H) such that vw ∈ E(G+H). If NG(v)∩ V2 = ∅ and NG(v)∩ V3 = ∅. Then by assumption, V2∩V (H) ̸= ∅ and V3∩V (H) ̸= ∅ and so, there exist u ∈ V2∩V (H) and w ∈ V3 ∩ V (H) such that vu, vw ∈ E(G+H). Now, suppose f(v) = 1. If NG(v)∩ V2 = ∅ and NG(v) ∩ V3 = ∅. Then by assumption, there exist z ∈ V2 ∩ V (H) or z ∈ V3 ∩ V (H) such that vz ∈ E(G+H) satisfying (P2). Therefore, f ∈ MRDF (G+H). Similarly, for v ∈ V (H) such that f(v) ∈ {0, 1}, f ∈ MRDF (G+H). Conversely, suppose f ∈ MRDF (G+H). Consider the following cases: Case 1: Suppose f |G ∈ MRDF (G). If (i)(a) holds, we are done. Suppose (i)(a) does not hold. Thus, either V2∩V (G) = ∅ or V3∩V (G) = ∅. Suppose V2∩V (G) = ∅. Necessarily, V0 ∩ V (G) = ∅. Let v ∈ V1 ∩ V (G). Since f |G ∈ MRDF (G), there exists u ∈ V3 such that uv ∈ E(G). Thus, V3 is a dominating set of G, and so, (b1) holds. Also, since V2∩V (G) = ∅, we have V2 ⊆ V (H). This means that V2∩V (H) ̸= ∅, say w ∈ V2∩V (H). Suppose v ∈ V0 ∩ V (H). Then since f ∈ MRDF (G+H), vw ∈ E(H) ⊆ E(G+H). And so, (b2) holds. Also, since V3 is a dominating set of G, there exists u ∈ V3 ∩ V (G) where S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 14 of 18 uv ∈ E(G+H). Suppose V3 ∩ V (G) = ∅, then similarly, (i)(c1) and (i)(c2) hold. Case 2: Suppose f |H ∈ MRDF (H). This case can be proven similarly with Case 1. Case 3: Suppose f |G /∈ MRDF (G) and f |H /∈ MRDF (H). If f |G /∈ MRDF (G), then there exists x ∈ V0 ∩V (G) such that NG(x)∩V2 = ∅ or NG(x)∩V3 = ∅. Moreover, there exists y ∈ V1∩V (G) such that NG(y)∩V2 = ∅ and NG(y)∩V3 = ∅. If NG(x)∩V2 = ∅ and NG(x)∩V3 ̸= ∅. Note that f ∈ MRDF (G+H). Then, there exists u ∈ (V2 ∩V (G+H)) such that u ∈ NG+H(x) for some x ∈ V0 ∩ V (G). Since NG(x) ∩ V2 = ∅, u ∈ NH(x). Consequently, u ∈ V2 ∩ V (H) for some x ∈ V0. Thus, (iii)(a) holds. If NG(x) ∩ V3 = ∅ and NG(x) ∩ V2 ̸= ∅. Since f ∈ MRDF (G+H), then there exists w ∈ (V3 ∩ V (G+H)) such that w ∈ NG+H(x) for some x ∈ V0 ∩ V (G). Consequently, by assumption, (iii)(b) holds. If NG(x) ∩ V2 = ∅ and NG(x) ∩ V3 = ∅. Since f ∈ MRDF (G + H), then there exist u ∈ (V2 ∩ V (G+H)) and w ∈ (V3 ∩ V (G+H)) such that u,w ∈ NG+H(x) for some x ∈ V0∩V (G). Thus, by assumption, u,w ∈ NH(x) and consequently, V2∩V (H) ̸= ∅ and V3 ∩ V (H) ̸= ∅. Hence, (iii)(a) and (iii)(b) hold. Furthermore, suppose NG(y) ∩ V2 = ∅ and NG(y) ∩ V3 = ∅ for some y ∈ V1 ∩ V (G). Since f ∈ MRDF (G + H), there exist u ∈ V2 ∩ V (G + H) or w ∈ V3 ∩ V (G + H) such that u,w ∈ NG+H(y). By assumption, u,w ∈ NH(y) and consequently, (iii)(c) holds. Similarly, if f |H /∈ MRDF (H), then (iii)(d), (iii)(e), and (iii)(f) hold. Proposition 14. Let G and H be any graphs. Then 3 ≤ γmR(G+H) ≤ 10 Proof. Suppose G and H are trivial graphs. Then by Proposition 5 (ii), γmR(G+H) = γmR(K2) = 3. Suppose G and H are not trivial graphs, then γmR(G + H) > 2. That is, γmR(G + H) ≥ 3. On the other hand, let V (G) = {v1, v2, · · · , vn} and V (H) = {u1, u2, · · · , un}. Now, define a function f = (V0, V1, V2, V3) on V (G+H) given by f(x) =  2, if x ∈ {v1, u1}. 3, if x ∈ {v2, u2}. 0, if x ∈ V (G+H) \ {v1, v2, u1, u2}. for every x ∈ V (G+H). Then f ∈ MRDF (G+H). Thus, γmR(G+H) ≤ ωmR G+H(f) = 10. Hence, 3 ≤ γmR(G+H) ≤ 10. Proposition 15. Let G and H be any graphs. Then (i) γmR(G+H) = 3 if and only if G = K1 and H = K1 (ii) γmR(G+H) = 4 if and only if G = K1 and H ∈ { K2,K2 } (iii) γmR(G+H) = 5 if and only if one of the following holds: (a) G = K1 and H ∈ {P3,K3,K3,K1 ∪K2} or H = K1 and G ∈ {P3,K3,K3,K1 ∪ K2}. S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 15 of 18 (b) If |V (G+H)| ≥ 4, then γ2(G) = 2 or γ2(H) = 2. (c) If |V (G+H)| ≥ 4, then γ(G) = 1 and γ(H) = 1. Proof. The proof follows immediately from Proposition 5. Corollary 4. Let m and n be positive integers. (i) If G = Kn and H = Km with n,m ≥ 2, γmR(G+H) = 5. (ii) If G = Kn and H = Km with n,m ≥ 5, γmR(G+H) = 10. 6. On the corona of graphs Let G and H be graphs with disjoint vertex sets. The corona of G and H is the graph G ◦H obtained by taking one copy of G and |V (G)| copies of H, and then joining the ith vertex of G to every vertex of the ith copy of H. For convenience, we adapt the notation Hv + v used in [3] to denote the subgraph of G ◦ H corresponding to the join Hv+⟨{v}⟩, v ∈ V (G). Moreover, for convenience, we define for i = 0, 1, 2, 3 and v ∈ V (G), V v i = {u ∈ V (Hv)|f(u) = i}. Proposition 16. Let G be any nontrivial connected graph and H be any graph. Let f = (V0, V1, V2, V3) be any function on V (G ◦ H). Then f ∈ MRDF (G) if and only if each of the following holds: (i) For every v ∈ (V0 ∪ V1) ∩ V (G), f |Hv ∈ MRDF (Hv). Moreover, if v ∈ V0 ∩ V (G), then the following holds: (a) If |V v 2 | ≠ 0 and |V v 3 | = 0, then |NG(v) ∩ V3| ≥ 1; and (b) If |V v 2 | = 0 and |V v 3 | ≠ 0, then |NG(v) ∩ V2| ≥ 1. (ii) For every v ∈ V2 ∩ V (G), V v 3 dominates V v 0 . (iii) For every v ∈ V3 ∩ V (G), V v 2 dominates V v 0 . Proof. Suppose f ∈ MRDF (G ◦ H) and let v ∈ (V0 ∪ V1) ∩ V (G). Let u ∈ V v 0 . Then u ∈ V0 and by definition, there exist w, z ∈ NG◦H(u) such that w ∈ V2 and z ∈ V3. But NG◦H(u) = {v} ∪ NHv(u) and v ∈ V0 ∪ V1, and thus, w, z ∈ NHv(u). Moreover, let u ∈ V v 1 . Then u ∈ V1 and by definition, there exists x ∈ V2 ∪ V3 such that x ∈ NG◦H(u), so that x ∈ (V v 2 ∪ V v 3 ) and so, x ∈ NHv(u). Hence, f |Hv ∈ MRDF (G ◦ H). Now, let v ∈ V0. Suppose that |V v 2 | ≠ 0 and |V v 3 | = 0. Then if u ∈ NG◦H(v) and u ∈ V3, we have u ∈ NG(v) ∩ V3. Thus, |NG(v) ∩ V3| ≥ 1. Moreover, suppose that |V v 3 | ̸= 0 and |V v 2 | = 0. Similarly, if w ∈ NG◦H(v) and w ∈ V2, then u ∈ NG(v)∩V2. Thus, |NG(v)∩V2| ≥ 1. This proves (i). Suppose that v ∈ V2 ∩ V (G) and let u ∈ V v 0 . By definition, there exists {x, y} ⊆ NG◦H(u) = {v} + NHv(u) such that x ∈ V2 and y ∈ V3. If v ∈ V2 and take x = v, then S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 16 of 18 y ∈ V v 3 and y ∈ NHv(u). Thus, V v 3 dominates V v 0 . This proves (ii). Similarly, if v ∈ V3 and taking y = v, then x ∈ V v 2 and x ∈ NHv(u). This means that V v 2 dominates V v 0 . This proves (iii). Conversely, let u ∈ V0 and let v ∈ V (G) for which u ∈ V (Hv + v). If u = v, by (i), f |Hv ∈ MRDF (G ◦H). Thus, there exist w ∈ V v 2 and z ∈ V v 3 such that w, z ∈ NHv(u) and so, w, z ∈ NG◦H(u). Now, if |V v 3 | = 0, by (i) |NG(v) ∩ V3| ≥ 1. Thus, there exists w ∈ V v 2 and z ∈ NG(v) ∩ V3 such that w, z ∈ NG◦H(u). Similarly, if |V v 2 | = 0, by (i), |NG(v)∩V2| ≥ 1. Thus, there exist w ∈ NG(v)∩V2 and z ∈ V v 3 such that w, z ∈ NG◦H(u). If u ̸= v, then u ∈ V v 0 . Suppose v ∈ V1∩V (G). By (i), f |Hv ∈ MRDF (G◦H). Thus, there exist x, y ∈ NHv(u) such that x ∈ V v 2 and y ∈ V v 3 . If v ∈ V2∩V (G). By (ii), V v 3 dominates V v 0 . Thus, there exist w, v ∈ NHv(u) such that w ∈ V v 3 . Also, if v ∈ V3 ∩ V (G). By (iii), V v 2 dominates V v 0 . Thus, there exist w, v ∈ NHv(u) such that w ∈ V v 2 . Now, let u ∈ V1. If u ∈ V (G), then there exist x ∈ V u 2 ∪V u 3 such that x ∈ NHu(u) since f |Hu ∈ MRDF (Hu). This implies that x ∈ V2 ∪ V3 and x ∈ NG◦H(u). Suppose u ∈ V (Hv) for some v ∈ V (G). If v ∈ (V2 ∪ V3) ∩ V (G), then v ∈ (V2 ∪ V3) ∩ NG◦H(u). If v ∈ V0 ∪ V1, then there exist w ∈ V v 2 ∪ V v 3 such that w ∈ NHv(u) since f |Hv ∈ MRDF (Hv) by (i). It implies that w ∈ V2 ∪ V3 and w ∈ NG◦H(u). Therefore, f ∈ MRDF (G ◦H). Proposition 17. Let G and H be any graph with |V (G)| = n and |V (H)| = m and let f = (V0, V1, V2, V3) be a γmR-function of G ◦H. Then 3n ≤ ∑ a∈V (v+Hv) f(a) ≤ 2n+ nm, for each v ∈ V (G). Proof. Let v ∈ V (G). If v ∈ V2 ∪ V3, then 3n ≤ ∑ p∈V (Hv) f(p) ≤ ∑ a∈V (v+Hv) f(a) ≤ 2n + nm. Suppose that v ∈ V0. By Proposition 16, f |Hv ∈ MRDF (Hv). Thus, 3n ≤∑ a∈V (v+Hv) f(a) ≤ 2n + nm. If v ∈ V1, then by Proposition 16, f |Hv ∈ MRDF (Hv). Thus, 3n ≤ ∑ p∈V (Hv) f(p) ≤ ∑ a∈V (v+Hv) f(a) ≤ 2n + nm. Moreover, the bounds are sharp if H = K1 and G ∈ {Pn, Cn,Kn}. Proposition 18. Let G be a connected graph of order n ≥ 1 and Km be the complete graph of order m ≥ 2, then γmR(G ◦Km) = { 4n, if m = 2. 5n, if m ≥ 3. Proof. If n = 1, then G ◦Km = Km+1. Hence, if m = 2, γmR(G ◦K2) = γmR(K3) = 4 by Proposition 5 (iii). If m ≥ 4, then γmR(Km+1) = 5 by Proposition 6. Now, If n > 1, then for m = 2, let V (K2) = {x, y} and V (G) = {v1, v2, · · · , vn}. Define a function f = (V0, V1, V2, V3) on V (G ◦K2) where V0 = ∅ = V3, V1 = ⋃ v∈V (G) V (Hv), V2 = V (G). Then f ∈ MRDF (G ◦ K2). It follows that γmR(G ◦ K2) ≤ 4n. Now, suppose that g = (W0,W1,W2,W3) is a γmR-function of G ◦K2. If W0 = ∅, then W3 = ∅. Since g is a γmR-function of G ◦K2, by Proposition 4, γmR(G ◦K2) = 4n. S. Ahamad, J. Cariaga, S. Menchavez / Eur. J. Pure Appl. Math, 18 (1) (2025), 5252 17 of 18 For m ≥ 3, let V (G) = {v1, v2, · · · , vn} and WLOG, pick a vertex u ∈ V (Km). Define a function f = (V0, V1, V2, V3) on V (G ◦Km) by f(x) =  3, if x ∈ V (G). 2, if x ∈ ⋃ v∈V (G) V (uv). 0, if x ∈ ⋃ v∈V (G) V ((H \ u)v). Then f ∈ MRDF (G ◦ Km). It follows that γmR(G ◦ Km) ≤ 5n. Now, suppose that g = (W0,W1,W2,W3) is a γmR-function of G ◦ Km. If W0 = ∅, then W3 = ∅. Since g is a γmR-function of G ◦Km, |W2| = V (G) and |W1| = V (Hv). Hence, γmR(G ◦Km) = ωmR G◦Km (g) ≥ 5n. If |W0| ≠ 0, then |W2| ≥ 1 and |W3| ≥ 1. It follows that γmR(G ◦Km) = ωmR G◦Km (g) = 2|W2|+ 3|W3| ≥ 5n. Therefore, γmR(G ◦Km) = 5n. Corollary 5. If Kn is a complete graph of order n ≥ 1, then (i) γmR(K1 ◦Kn) = n+ 2. (ii) γmR(Kn ◦K1) = 3n. Proof. Statement (i) follows from the fact that K1 ◦Kn = Sn and by Proposition 12 (ii), γmR(K1 ◦Kn) = γmR(Sn) = n+2. For (ii), note that Kn ◦K1 is the disjoint union n copies of K2. 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