EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 4, Article Number 6264 ISSN 1307-5543 – ejpam.com Published by New York Business Global Total Modern Roman Dominating Functions in Graphs Sherihatha R. Ahamad1,2, Jerry Boy G. Cariaga1,2, Sheila M. Menchavez1,2, Ferdinand P. Jamil1,2,∗ 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(u) = 3. In addition, if every subgraph induced by the set {v ∈ V (G) : f(v) > 0} is isolated-free, then we say that f is a total modern Roman dominating function of G. The minimum weight ωtmR G (f) = ∑ v∈V (G) f(v) of a total modern Roman dominating function f of G is called the total modern Roman domination number, γtmR(G), of G. In this paper, we initiate the study of total modern Roman domination. We characterize graphs with smaller total modern Roman domination number and obtain the γtmR(G) of some special graphs. Moreover, we investigate and characterize the total modern Roman domination in the join and corona of graphs. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Dominating set, domination number, modern Roman dominating function, modern Roman domination number, total modern Roman dominating function, total modern Roman domination number 1. Introduction The concept of Roman domination was introduced by Cockayne et al. [1] in 2004, inspired by the strategies for defending the Roman Empire presented in the work of ReVelle and Rosing[2] and Stewart [3]. Since then, it has become an active research field in graph theory, with numerous studies exploring this concept (see [4],[5],[6],[7],[8],[9],[10], ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6264 Email addresses: sherihatha.ahamad@g.msuiit.edu.ph (S. R. Ahamad), jerryboy.cariaga@g.msuiit.edu.ph (J. B. G. Cariaga), sheila.menchavez@g.msuiit.edu.ph (S. M. Menchavez), ferdinand.jamil@g.msuiit.edu.ph (F. P. Jamil) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 2 of 20 [11],[12],[13],[14]). A new model of graph domination is introduced in [5] based on the Roman domination and is called modern Roman domination. In this paper, we introduce the concept of total modern Roman domination domination in graphs. It focuses on providing the total modern Roman domination number of some special graphs and some characterizations for the total modern Roman domination 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 [15]. 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 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. The authors always refer to [16] 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. For k = 2, we have D as 2-dominating set with 2-domination number denoted by γ2(G) [8]. A Roman dominating function (RDF) on G is a function f : V (G) → {0, 1, 2} such that every vertex u ∈ V (G) with f(u) = 0 is adjacent to at least one vertex v with f(v) = 2. The weight of an RDF is the value ωG(f) = ∑ u∈V (G) f(u). The Roman domination num- ber γR(G) is the minimum weight of an RDF on G. An RDF with ωG(f) = γR(G) is referred to as a γR-function [8]. A modern Roman dominating function (MRDF) on 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 S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 3 of 20 (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 is called the modern Roman domi- nation number γmR(G) of G [5]. 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|. 3. Known Results We make use of the following known results from [7]. Proposition 1. 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). Proposition 2. 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 V0 ∩ V (H). (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 V0 ∩ V (H). (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. S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 4 of 20 (b2) V2 ∩ V (G) is a dominating set of V0 ∩ V (G). (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 V0 ∩ V (G). (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 ∩ V (G). (b) V3 ∩ V (H) ̸= ∅ whenever NG(x) ∩ V3 = ∅ for some x ∈ V0 ∩ V (G). (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 ∩ V (H). (e) V3 ∩ V (G) ̸= ∅ whenever NH(x) ∩ V3 = ∅ for some x ∈ V0 ∩ V (H). (f) V2 ∩ V (G) ̸= ∅ or V3 ∩ V (G) ̸= ∅ whenever ∃x ∈ V1 with NH(x)∩ V2 = ∅ and NH(x) ∩ V3 = ∅ 4. Results Definition 1. If f = (V0, V1, V2, V3) is a modern Roman dominating function of a non-isolated graph G, then it is said to be a total modern Roman dominating function (TMRDF (G)) of G provided it satisfies the following additional property: (P3) the set {v ∈ V (G) : f(v) > 0} induces an isolated-free subgraph. The minimum weight ωtmR G (f) = ∑ v∈V (G) f(v) of a total modern Roman dominating function f of G is called the total modern Roman domination number γtmR(G) of G. A total modern Roman dominating function of G with weight ωG(f) = γtmR(G) is called a γtmR-function of G. Example 1. Consider the graph G in Figure 1. The function f : V (G) → {0, 1, 2, 3} given by f(a) = 3, f(g) = 2, f(b) = 1 and f(c) = f(d) = f(e) = f(h) = 0 is a total modern Roman dominating function of G. It can be verified that, γtmR(G) = 6. 0 d 0e 0c 0h 3a 2 g G : 1b S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 5 of 20 Figure 1: Graph G of order 7 with γtmR(G) = 6 Proposition 3. Let G be any isolated-free graph. Then γ(G) + γt(G) ≤ γtmR(G). Moreover, this bound is sharp whenever, G = •∪n i=1P2i, where i = 1, 2, · · ·n. Proof. Let G be a graph with no isolated vertex and f = (V0, V1, V2, V3) be a γtmR- function on G. Then V2 ∪ V3 is a dominating set of G. Moreover, V1 ∪ V2 ∪ V3 is a total dominating set of G. Thus, γtmR(G) =ωtmR G (f) =|V1|+ 2|V2|+ 3|V3| =(|V1|+ |V2|+ |V3|) + (V2|+ |V3|) + |V3| ≥γt(G) + γ(G). Hence, we obtained the desired lower bound. Moreover, the bound is sharp by considering the disjoint union of copies of K2. Proposition 4. Let G be any graph with no isolated vertex and f = (V0, V1, V2, V3) a γtmR-function of G. Then V0 = ∅ if and only if V3 = ∅ and V2 is a γ-set of G. Proof. The result follows immediately from Proposition 1. Proposition 5. Let G be any graph with no isolated vertex and f = (V0, V1, V2, V3) a γtmR-function of G. Then one of the following holds: (i) If |V0| = 0, then |V1| ̸= 0; (ii) If |V1| = 0, then |V0| ̸= 0. Proof. (i) Let |V0| = 0 and suppose that |V1| = 0. Then γtmR(G) = 2|V2| + 3|V3| and ⟨V2 ∪ V3⟩ is a total dominating set of G. Let D ⊆ V2 ∪ V3 be a dominating set of the induced subgraph ⟨V2 ∪ V3⟩. Define a function h = (V ′ 0 , V ′ 1 , V ′ 2 , V ′ 3), where V ′ 0 = ∅, V ′ 1 = (V2 ∪ V3) \D,V ′ 2 = D and V ′ 3 = ∅. Then h ∈ TMRDF (G) and thus, ωtmR G (h) = |V ′ 1 |+ 2|V ′ 2 | ≤ 2|V2|+ 3|V3| = ωtmR G (f), a contradiction. Hence, |V1| ̸= 0. (ii) Let |V1| = 0 with |V3| ̸= 0 and suppose that |V0| = 0. Then γtmR(G) = ωtmR G (f) = 2|V2| + 3|V3| and since f is a γtmR-function, ⟨V2 ∪ V3⟩ is a total dominating set in G. Consider the vertices x ∈ V (G) such that x ∈ NG(V2) ∩ NG(V3). Let V ′′ 1 = ∅, V ′′ 0 = NG(V2) ∩ NG(V3), V ′′ 2 = V2 \ (NG(V2) ∩ NG(V3)). Then the function defined by h = (V ′′ 0 , V ′′ 1 , V ′′ 2 , V ′′ 3 ) ∈ TMRDF (G). Thus, ωtmR G (h) = 2|V ′′ 2 | + 3|V ′′ 3 | ≤ 2|V2| + 3|V3| = ωtmR G (f) = γtmR(G), a contradiction. Hence, |V0| ̸= 0. S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 6 of 20 Remark 1. The converse of (ii) and (iii) need not be true. Proposition 6. Let G be a nontrivial connected graph and f = (V0, V1, V2, V3) a γtmR- function of G. Then (i) |V0| = 0 if and only if V2 is a γ-set of G and V (G) is a total dominating set of G. (ii) |V1| = 0 if and only if V2 ∪ V3 is both a γ2-set and γt-set of G. Proof. (i) Suppose |V0| = 0, then |V3| = 0 by Proposition 4, V2 is a γ-set of G. It remains to show that V (G) is a γt-set of G. Note that V (G) = V1 ∪ V2 and ⟨V1 ∪ V2⟩ is isolated-free. It follows that V (G) is a total dominating set of G. Conversely, since V (G) is a total dominating set of G, f(v) ̸= 0 for all v ∈ V (G) by definition. Hence, |V0| = 0. (ii) Suppose |V1| = 0. Then |V0| ̸= 0 by Proposition 5 (ii). By (P1), for every v ∈ V0, |V2∩NG(v)| ≥ 1 and |V3∩NG(v)| ≥ 1. This implies that V2∪V3 is a 2-dominating set of G. Suppose V2 ∪ V3 is not a γ2-set of G. Then there exists D ⊆ V2 ∪ V3 such that D is a γ2-set of G. Define a function g = (V ∗ 0 , V ∗ 1 , V ∗ 2 , V ∗ 3 ), such that V ∗ 0 = ∅, V ∗ 1 = (V2 ∪ V3) \D,V ∗ 2 = D ∩ V2 and V ∗ 3 = D ∩ V3. Then g ∈ TMRDF (G) and so, ωtmR G (g) = |V ∗ 1 |+2|V ∗ 2 |+3|V ∗ 3 | = |(V2∪V3)\D|+2|D∩V2|+3|D∩V3| ≤ 2|V2|+3|V3|−|(V2∪V3)\D| = ωtmR G (f)−|(V2∪V3)∩D| ≤ ωtmR G (f). This is a contradiction. Thus, V2∪V3 is a γ2-set of G. Now, by (P3) V2 ∪ V3 is a total dominating set of G. Suppose V2 ∪ V3 is not a γt-set of G. Then there exists D ⊆ V2∪V3 such that D is a γt−set of G. Let V ” 0 = (V2∪V3)\D,V ” 1 = ∅, V ” 2 = D ∩ V2 and V ” 3 = D ∩ V3. Then g = (V ” 0 ,∅, V ” 2 , V ” 3 ) ∈ TMRDF (G) and ωtmR G (g) = 2|V ” 2 | + 3|V ” 3 | ≤ 2|V2| + 3|V3| = ωtmR G (f), a contradiction. Hence, V2 ∪ V3 is a γt-set of G. Conversely, let V2 ∪ V3 be a γ2-set and γt-set of G. Suppose |V1| ̸= 0. Then there exists y ∈ V1 such that y ∈ NG(x) ∩ NG(z) where x ∈ V2 and z ∈ V3. Define a function g = (V ∗ 0 , V ∗ 1 , V ∗ 2 , V ∗ 3 ) for which V0 = V ∗ 0 , V1 = V ∗ 1 ̸= ∅, V2 = V ∗ 2 and V3 = V ∗ 3 . Then g ∈ TMRDF (G). Thus, ωtmR G (g) = 3|V ∗ 3 |+2|V ∗ 2 |+ |V ∗ 1 | ≤ 3|V3|+2|V2| = ωtmR G (f). This contradicts the fact that f is a γtmR-function of G. Hence, |V1| = 0. Proposition 7. Let G be a connected graph of order n ≥ 2. Then (i) γtmR(G) = 3 if and only if G ∈ {P2,K2}. (ii) γtmR(G) = 4 if and only if G ∈ {P3,K3}. Proof. (i) WLOG, Suppose G = K2. Let V (G) = {a, b}, then f = (∅, {a}, {b},∅) ∈ TMRDF (G+H). Thus, γtmR(G) ≤ ωtmR G = 3. Since G cannot be trivial, γtmR(G) ≥ 3. Hence, γtmR(G + H) = 3. Conversely, suppose γtmR(G) = 3 and f = (V0, V1, V2, V3) a γtmR-function of G. Then V (G) = V1 ∪ V2 with |V1| = |V2| = 1 and so, |V (G +H)| = 2. Moreover, ⟨V1 ∪ V2⟩ is connected. Thus, G = K2. (ii) WLOG, we assume that G = K3. Let V (G) = {x, y, z}. Define a function f = (V0, V1, V2, V3) where V0 = ∅ = V3, V1 = {x, z} and V2 = {y}. Then f ∈ TMRDF (G) and thus, γtmR(G) ≤ ωtmR G+H(f) = |V1| + 2|V2| = 4. Since, G ̸= K2, γtmR(G) ≥ 4 by (i). Hence, γtmR(G) = 4. Conversely, suppose that γtmR(G) = 4 and f = (V0, V1, V2, V3) is a S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 7 of 20 γtmR-function of G. Then γtmR(G) = |V1| + |V2| + |V3| = 4. If |V3| = 1, then |V1| = 1. This implies that |V (G)| = 2 and so, G = K2 since ⟨V1 ∪ V3⟩ must be connected. But γtmR(K2) = 3. Thus, |V3| = 0. It follows that |V0| = 0. Similarly, |V2| ̸= 2 and so, |V2| ≤ 1. Moreover, if V2 = ∅, that is V (G) = V1, then (P2) is not satisfied. Thus, |V2| = 1 and hence, |V1| = 2. Moreover, the induced subgraph ⟨V1 ∪ V2⟩ is isolated-free. This means that since G is connected, |V (G)| = |V1 ∪ V2| = 3. Therefore, G = K3. Proposition 8. For a connected graph G, γtmR(G) = 5 if and only if |V (G)| = 4 and γ(G) = 1 or γt2(G) = 2 and |V (G)| ≥ 4. Proof. If γtmR(G) = 5, then |V3| ≤ 1 and 1 ≤ |V2| ≤ 2. Also, by Proposition 7 (ii), |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 Proposition 7 (ii). 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 Proposition 7 (ii). If |V2| = 1, then |V1| = 0. It follows that |V2 ∪ V3| = 2. By Proposition 6, V2 ∪ V3 is a γt2-set in G. Hence, |V (G)| ≥ 4 and γt2(G) = 2. Conversely, suppose |V (G)| = 4 and γ(G) = 1. By Proposition 7 (ii), 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 ∈ TMRDF (G) and ωtmR G (f) = 5. This implies that γtmR(G) = 5. Next, suppose that γt2(G) = 2 and |V (G)| ≥ 4. Let D = {u, v} be the γt2-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 ∈ TMRDF (G) and ωtmR G (g) = 5. Since G ̸∈ {K3, P3}, we must have ωtmR G (g) = 5. Hence, γtmR(G) = 5. Corollary 1. For a connected graph G of order 4, γtmR(G) = 5 if and only if G ∈ {K1 + (K1 ∪K2),K1 +K3,K1 +K3,K1 + P3}. Proof. The proof follows directly from Proposition 8. Proposition 9. Let G be a disconnected graph with nontrivial components G1, G2, · · · , Gn. Then γtmR(G) = ∑n i=1 γtmR(Gi). Proof. Let G1, G2, · · · , Gn be the components of G. Let f1, f2, · · · , fn be γtmR- functions of G1, G2, · · · , Gn respectively. Define a function f : V (G) −→ {0, 1, 2, 3} given by S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 8 of 20 f(x) =  f1(x), if x ∈ V (G1). f2(x), if x ∈ V (G2). ... fn(x), if x ∈ V (Gn). Then f is a γtmR-function of G. Thus γtmR(G) ≤ ∑n i=1 γtmR(Gi). Conversely, let f be a γtmR-function of G. Then the restriction f |Gi of f to Gi, where i = 1, 2, · · · , n is a γtmR-function of Gi. Thus, γtmR(Gi) ≤ ωtmR G (f |Gi) for all i = 1, 2, · · · , n. Hence,∑n i=1 γtmR(Gi) ≤ γtmR(G). Hence, combining the results, ∑n i=1 γtmR(Gi) ≤ γtmR(G) ≤∑n i=1 γtmR(Gi). Therefore, γtmR(G) = ∑n i=1 γtmR(Gi). Proposition 10. If G ∈ {Pn, Cn}, then γtmR(G) = n+ ⌈n 3 ⌉ . Proof. Let G ∈ {Pn, Cn}. Suppose G = [v1, v2, · · · , vn]. If n ≡ 0(mod 3). Put V2 = {v2, v5, · · · , vn−7, vn−4, vn−1}, V1 = V (G) \ V2, |V0| = 0 = |V3|. Then f = (V0, V1, V2, V3) is a TMRDF on G. If n ≡ 1(mod 3). Put V2 = {v2, · · · , vn−8, vn−5, vn−2, vn}, V1 = V (G) \ V2, |V0| = 0 = |V3|. Then f = (V0, V1, V2, V3) is a TMRDF on G. If n ≡ 2(mod 3). Put V2 = {v2, v5, · · · , vn−9, vn−6, vn−3, vn}, V1 = V (G) \ V2, |V0| = 0 = |V3|. Then f = (V0, V1, V2, V3) is a TMRDF on G. In any case, γtmR(G) ≤ ωtmR G (f) = |V1|+ |V2| = n + ⌈n 3 ⌉ . Since |V0| = 0, by Proposition 6, V2 is a γ-set of G and V (G) is a γt-set of G. By Proposition 3, γ(G) + γt(G) = |V2| + |V (G)| = n + ⌈n 3 ⌉ ≤ γtmR(G). Thus, γtmR(G) = n+ ⌈n 3 ⌉ . The n-barbell graph is a 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 γtmR(B3) = 8 and γtmR(B5) = 10, respectively. Proposition 11. For any n-barbell graph Bn where n ≥ 3, γtmR(Bn) = { 8, if n = 3. 10, if n ≥ 4. S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 9 of 20 Proof. Let Bn be any 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 is a TMRDF of B3. Thus, γtmR(B3) ≤ ωtmR B3 (f) = 8. Clearly γ(B3) = 2 = γt(B3). Thus, by Proposition 3, γtmR(B3) ≥ 8. Therefore, γtmR(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 is a TMRDF of Bn. Thus, γtmR(Bn) ≤ ωtmR Bn (f) = 10. Clearly γ(Bn) = 2 = γt(Bn). Thus, by Proposition 3, γtmR(Bn) ≥ 10. Therefore, for all n ≥ 4, γtmR(Bn) = 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 γtmR(Wd(4, 2)) = 7 2 1 1 1 1 11 1 1 Figure 4: A Friendship graph G4 3 with γtmR(G 4 3) = 10 S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 10 of 20 2 1 11 1 Figure 5: A butterfly graph G2 3 with γtmR(G 2 3) = 6 Proposition 12. For any windmill graph G = K1 + nKk−1, where k ≥ 4 and n ≥ 2, γtmR(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 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 ∈ TMRDF (G). Thus, γtmR(G) ≤ ωtmR G (f) = 2n + 3. Now, since γ(G) = 1 and γt(G) = 2, by Proposition 3, γtmR(G) ≥ 2n+ 3. Hence, γtmR(G) = 2n+ 3. Proposition 13. For any friendship graph G, γtmR(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 ∈ TMRDF (G) and γtmR(G) ≤ ωtmR G (f) = 2n + 2. Clearly, {u} is a γ-set of G, and so γ(G) = 1. Moreover, γt(G) = 2 and so, by Proposition 3, γtmR(G) ≥ 2n + 2. Hence, γtmR(G) = 2n+ 2. Corollary 2. For a butterfly graph G, γtmR(G) = 6. Proof. The result follows from Proposition 13. 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. Proposition 14. For a complete bipartite graph Km,n, let p = min{m,n},m, n ≥ 2. Then γtmR(Km,n) =  6, if p = 2. 8, if p = 3 9, if p = 4 10, if p ≥ 5. S. Ahamad et al. / Eur. J. Pure Appl. 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K7,5: Figure 6: A complete bipartite G = K7,5 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, m,n ≥ 2. Let p = min{m,n}. If p = 2 and WLOG, we let X = {x1, x2} and pick a vertex y1 ∈ Y and define a function f = (V0, V1, V2, V3) given by f(z) =  3, z = {x1}. 2, z = {x2}. 1, z = {y1}. 0, Otherwise. Then f is a TMRDF of G. Thus, γtmR(G) ≤ ωtmR G (f) = 6. By Proposition 3, γtmR(G) ≥ 6. Thus, γtmR(G) = 6. If p = 3 and WLOG, we let X = {x1, x2, x3} and pick a vertex y1 ∈ Y and define a function f = (V0, V1, V2, V3) given by f(z) =  3, z = {x1}. 2, z ∈ {x2, x3}. 1, z = {y1}. 0, Otherwise. Then f is a TMRDF of G. Thus, γtmR(G) ≤ ωtmR G (f) = 8. By Proposition 3, γtmR(G) ≥ 8. Thus, γtmR(G) = 8. If p = 4 and WLOG, we let X = {x1, x2, x3, x4} and pick a vertex y1 ∈ Y and define a function f = (V0, V1, V2, V3) given by f(z) =  3, z = {x1}. 2, z ∈ {x2, y1, }. 1, z ∈ {x3, x4}. 0, Otherwise. Then f is a TMRDF of G. Thus, γtmR(G) ≤ ωtmR G (f) = 9. By Proposition 3, γtmR(G) ≥ 9. Thus, γtmR(G) = 9. If p ≥ 5, let X = Km and Y = Kn. WLOG, let S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 12 of 20 {x1, x2} ∈ V (Km), {y1, y2} ∈ V (Kn) and define a function f = (V0, V1, V2, V3) given by f(z) =  3, z ∈ {x1, y1}. 2, z ∈ {x2, y2}. 0, Otherwise. Then f is a TMRDF of G. Thus, γtmR(G) ≤ ωtmR G (f) = 10. By Proposition 3, γtmR(G) ≥ 10. Thus, γtmR(G) = 10. The star Sn of order n+ 1 is the graph Kn +K1. The graph in Figures 7 is the star graph S6. 2 1 1 1 1 1 2 Figure 7: A star graph S6 with γtmR(S6) = 7 Proposition 15. If G = Sn, n ≥ 1, then γtmR(G) = n+ 2. Proof. Let G = Sn where V (G) = V (K1 +Kn) 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 ∈ TMRDF (G). It follows that γtmR(G) ≤ n + 2. Now, suppose that g = (W0,W1,W2,W3) is a γtmR-function of G. If W0 = ∅, then W3 = ∅. Since g is a γtmR-function of G, |W2| = |V (K1)| = 1 and |W1| = |V (Kn)| = n. Hence, γtmR(G) = ωtmR G (g) ≥ n + 2. If |W0| ̸= 0, then |W2| ≥ 1 and |W3| ≥ 1. It follows that γtmR(G) = ωtmR G (g) = 2|W2|+ 3|W3| ≥ n+ 2. Therefore, γtmR(G) = n+ 2. Proposition 16. For a complete graph Kn, γtmR(Kn) = 5 for all n ≥ 4. Proof. Pick any x, y ∈ V (Kn) with x ̸= y. Clearly, g = (V (Kn) \ {x, y},∅, {x}, {y}) ∈ TMRDF (Kn). It follows that γtmR(Kn) ≤ 5. On the other hand, suppose that f = (V0, V1, V2, V3) is a γtmR-function of Kn. If V0 = ∅, then V3 = ∅. Since f is a γtmR-function of Kn, |V2| = 1 and |V1| = n − 1. Hence, γtmR(Kn) = ωtmR Kn (f) = n + 1 ≥ 5. If V0 ̸= ∅, then |V2| ≥ 1 and |V3| ≥ 1. It follows that γtmR(Kn) = ωtmR Kn (f) = 2|V2| + 3|V3| ≥ 5. Therefore, γtmR(Kn) = 5. S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 13 of 20 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)} [8]. In this section, the following proposition characterizes all TMRDF on the join of graphs. Proposition 17. 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 ∈ TMRDF (G+H) if and only if one of the following holds: (i) f |G ∈ TMRDF (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 V0 ∩ V (H). (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 V0 ∩ V (H). (ii) f |H ∈ TMRDF (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 V0 ∩ V (G). (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 V0 ∩ V (G). (iii) f |G ̸∈ TMRDF (G), f |H ̸∈ TMRDF (H) and each of the following holds: (a) V2 ∩ V (H) ̸= ∅ whenever NG(x) ∩ V2 = ∅ for some x ∈ V0 ∩ V (G). (b) V3 ∩ V (H) ̸= ∅ whenever NG(x) ∩ V3 = ∅ for some x ∈ V0 ∩ V (G). (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 ∩ V (H). (e) V3 ∩ V (G) ̸= ∅ whenever NH(x) ∩ V3 = ∅ for some x ∈ V0 ∩ V (H). S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 14 of 20 (f) V2 ∩ V (G) ̸= ∅ or V3 ∩ V (G) ̸= ∅ whenever ∃x ∈ V1 with NH(x)∩ V2 = ∅ and NH(x) ∩ V3 = ∅ (g) For every x ∈ V1 ∪ V2 ∪ V3, (V1 ∪ V2 ∪ V3)∩ V (H) ̸= ∅ whenever x ∈ V (G) and NG(x) ⊆ V0. (h) For every x ∈ V1 ∪ V2 ∪ V3, (V1 ∪ V2 ∪ V3)∩ V (G) ≠ ∅ whenever x ∈ V (H) and NH(x) ⊆ V0. Proof. Suppose f |G ∈ TMRDF (G). Assume that (i)(a) holds. Let v ∈ V0 ∩ V (G). By assumption, there exist u ∈ V2 ∩ V (G) and w ∈ V3 ∩ V (G) such that uv,wv ∈ E(G) ⊆ E(G+H). Let v ∈ V0 ∩ V (H). Note that |V2 ∩ V (G)| ≥ 1 and |V3 ∩ V (G)| ≥ 1. Take u ∈ V2∩V (G) and w ∈ V3∩V (G). Then vu, vw ∈ E(G+H). Also, since f |G ∈ TMRDF (G), uw ∈ E(G) ⊆ E(G + H). Let v ∈ V1 ∩ V (G). Since f |G ∈ TMRDF (G), there exists u ∈ (V2∪V3)∩V (G) such that uv ∈ E(G) ⊆ E(G+H). Now, assume v ∈ V1∩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 ∈ TMRDF (G +H). Similarly, if f |H ∈ TMRDF (H) with |V2 ∩ V (H)| ≥ 1 and |V3 ∩ V (H)| ≥ 1, then f ∈ TMRDF (G+H). Assume (i)(b) holds. Since f |G ∈ TMRDF (G), V0 ∩ V (G) = ∅. Thus, 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 w ∈ V3 ∩ V (G). Then vw ∈ E(G + H). Moreover, uw ∈ E(G+H). Let v ∈ V1∩V (G). By assumption, 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 ∈ TMDRF (G + H). Assume (i)(c) holds. Since f |G ∈ TMRDF (G), V0 ∩ V (G) = ∅. Thus, 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 w ∈ V2∩V (G). Then wv ∈ E(G+H). Moreover, uw ∈ 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 ∈ TMDRF (G + H). Similarly, if (ii) holds, then f ∈ TMRDF (G + H). Suppose (iii) holds, that is f |G ̸∈ TMRDF (G) and f |H ̸∈ TMRDF (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). Moreover, uw ∈ 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). Moreover, uw ∈ 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). Also, by assumption, there exists x ∈ V (G) \ V0 such that ux,wx ∈ 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 ∈ TMRDF (G + H). Similarly, for v ∈ V (H) such that f(v) ∈ {0, 1}, f ∈ TMRDF (G + H). Now, let v ∈ V1 ∪ V2 ∪ V3. If v ∈ V (G) with NG(v) ⊆ V0, then by (iii)(g), (V1 ∪ V2 ∪ V3) ∩ V (H) ̸= ∅. This means that NG+H(v) ∩ (V1 ∪ V2 ∪ V3) ̸= ∅. Similarly, NG+H(v) ∩ (V1 ∪ V2 ∪ V3) ̸= ∅, for each v ∈ V (H) with NH(v) ⊆ V0. Thus, S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 15 of 20 ⟨V1 ∪ V2 ∪ V3⟩ has no isolated vertex. Therefore, f ∈ TMRDF (G+H). Conversely, suppose f ∈ TMRDF (G+H). Consider the following cases: Case 1: Suppose f |G ∈ TMRDF (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 ∈ TMRDF (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) = ∅, then V2 ⊆ V (H). Hence, V2 ∩ V (H) ̸= ∅. Pick w ∈ V2 and let v ∈ V0 ∩ V (H). Since f ∈ TMRDF (G+H), vw ∈ E(H) ⊆ E(G+H). Thus, (b2) holds. Also, since V3 is a dominating set of G, there exists u ∈ V3 ∩ V (G) where uv ∈ E(G+H). Suppose V3 ∩ V (G) = ∅, then similarly, (i)(c1) and (i)(c2) hold. Furthermore, the case where f |H ∈ TMRDF (H) is similar. Case 2: Suppose f |H ∈ TMRDF (H). This can be proven similarly with Case 1. Case 3: Assume f |G /∈ TMRDF (G) and f |H /∈ TMRDF (H). Suppose NG(x) ∩ V2 = ∅ for some x ∈ V0. Since f ∈ TMRDF (G + H), there exists y ∈ V2 ∩ NG+H(x). Since y /∈ V (G), y ∈ V (H). Hence, y ∈ V2 ∩ V (H). Thus, V2 ∩ V (H) ̸= ∅ satisfying (iii)(a). Following similar argument, if NG(x) ∩ V3 = ∅ for some x ∈ V0, V3 ∩ V (H) ̸= ∅. Thus, (iii)(b) holds. Suppose NG(x) ∩ V2 = ∅ and NG(x) ∩ V3 = ∅ for some x ∈ V1. Since f ∈ TMRDF (G + H), there exists y ∈ V2 or y ∈ V3 such that xy ∈ E(G + H). Since y /∈ V (G), y ∈ V (H). Necessarily y ∈ V2 ∩ V (H) or y ∈ V3 ∩ V (H). Thus, V2 ∩ V (H) ̸= ∅ or V3 ∩ V (H) ̸= ∅ satisfying (iii)(c). Moreover, assume that ⟨(V1 ∪ V2 ∪ V3) ∩ V (G)⟩ has an isolated vertex. Let v ∈ [(V1 ∪ V2 ∪ V3) ∩ V (G)] such that NG(v) ⊆ V0. Since f ∈ TMRDF (G+H), ⟨V1∪V2∪V3⟩ is isolated-free vertex. Thus, (V1∪V2∪V3)∩V (H) ̸= ∅. Hence, (iii)(g) holds. Similarly, (iii)(d), (iii)(e), (iii)(f), and (iii)(h) follows. Corollary 3. Let G and H be any nontrivial graphs. Then 3 ≤ γtmR(G+H) ≤ 10. Proof. Since G+H is not a trivial graphs, γtmR(G+H) > 2. That is, γtmR(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 ∈ TMRDF (G+H). Thus, γtmR(G+H) ≤ ωtmR G+H(f) = 10. Hence, 3 ≤ γtmR(G+H) ≤ 10. Corollary 4. Let G be any graph of order n, then γtmR(K1+G) ≤ min{2+n, 3+2γ(G)}, and this upper bound is sharp. Proof. Define a function g = (V0, V1, V2, V3) for which V1 = V (G), V2 = V (K1) and S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 16 of 20 V0 = V3 = ∅. Then, g ∈ TMRDF (K1 +G). Thus, γtmR(K1 +G) ≤ ωtmR G (g) = 2|V2|+ |V1| = 2(1) + n = 2 + n. Now, let V (K1) = {x} and D a γ-set of G. Define a function h = (V ′ 0 , V ′ 1 , V ′ 2 , V ′ 3) for which V ′ 0 = V (G) \D,V ′ 1 = ∅, V ′ 2 = D and V ′ 3 = {x}. Then, h ∈ TMRDF (K1 +G). Thus, γtmR(K1 +G) ≤ ωtmR G (h) = 3|V ′ 3 |+ 2|V ′ 2 | = 3(1) + 2|D| = 3 + 2γ(G). Moreover, for the sharpness, consider the graphs K1+G,G ∈ {Kn,Kn}. Strict inequality may also be attained such as the graph K1 + P7. Example 2. Consider the graphs K1 + P7 and K1 + K5 in Figure 8 and Figure 9, respectively. 2b 0c 1 d 0 e 2 f 0a 0 g 3 u K1 + P7 : Figure 8: The graph K1 + P7 with γtmR(K1 + P7) = 8 In figure 8, the function f : V (K1 + P7) → {0, 1, 2, 3} given by f(u) = 3, f(b) = f(f) = 2 , f(a) = f(c) = f(g) = f(e) = 0, and f(d) = 1. Observe that {b, d, f} is a dominating set of P7. Thus, the γ(P7) = 3 and observe further that f is a total modern Roman dominating function of G with weight 8. It can be verified that 8 = γtmR(K1 + P7) < min{2 + n, 3 + 2(γ(P7))} = 9. 3 0 0 2 0 0K1 +K5 : Figure 9: The graph K1 +K5 with γtmR(K1 +K5) = 5. S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 17 of 20 In figure 9, Observe that f is a total modern Roman dominating function of K1+K5 with weight 5. It can be verified that γtmR(K1 +K5) = min{2 + n, 3 + 2(γ(K1 +K5))} = 5. Corollary 5. Let G and H be any graphs. Then (i) γtmR(G+H) = 3 if and only if G = K1 and H = K1. (ii) γtmR(G+H) = 4 if and only if G = K1 and H ∈ {K2,K2} or vise versa. (iii) γtmR(G+H) = 5 if and only if one of the following holds: (a) G = K1 and H ∈ {P3,K3,K3,K1 ∪K2} and vise versa. (b) γ(G) = 1 and γ(H) = 1. (c) γ2(G) = 2 or γ2(H) = 2. Proof. The proof follows immediately from Proposition 17. 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 [8] to denote the subgraph of G ◦ H corresponding to the join Hv + ⟨{v}⟩, v ∈ V (G). Proposition 18. Let G and H be nontrivial graphs. Let f = (V0, V1, V2, V3) be any function on V (G ◦ H). Then f ∈ TMRDF (G ◦ H) if and only if each of the following holds: (i) For every v ∈ (V0 ∪ V1)∩ V (G), f |Hv ∈ TMRDF (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), V3 ∩ V (Hv) is a dominating set of ⟨V0 ∩ V (Hv)⟩. (iii) For every v ∈ V3 ∩ V (G), V2 ∩ V (Hv) is a dominating set of ⟨V0 ∩ V (Hv)⟩. Proof. (i) Suppose f ∈ TMRDF (G ◦ H) and let v ∈ (V0 ∪ V1) ∩ V (G). Let u ∈ V v 0 . Then u ∈ V0 and 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. Thus, w, z ∈ NHv(u). Moreover, let u ∈ V v 1 . Then u ∈ V1 and there exists x ∈ V2 ∪ V3 such that x ∈ NG◦H(u), x ∈ (V v 2 ∪ V v 3 ) and so x ∈ NHv(u). Now, let u ∈ V v 2 ∪ V v 3 . By (P3), there exists z′ ∈ V (Hv) \ V v 0 such that z ∈ NG◦H(u). Thus, f |Hv ∈ TMRDF (Hv). Now, let v ∈ V0 ∩ V (G). Suppose that S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 18 of 20 |V v 2 | ̸= 0 and |V v 3 | = 0. Then u ∈ NG◦H(v) and u ∈ V3. Hence, u ∈ NG(v) ∩ V3. Thus, |NG(v)∩ V3| ≥ 1. Following similar argument, if |V v 3 | ̸= 0 and |V v 2 | = 0, |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 y ∈ V v 3 and y ∈ NHv(u). Thus, V v 3 dominates V v 0 . Hence, V3 ∩ V (Hv) is a dominating set of ⟨V0 ∩ V (Hv)⟩. This proves (ii). Similarly, (iii) follows. Conversely, suppose (i), (ii) and (iii) hold for f . Let u ∈ V0 and v ∈ V (G) for which u ∈ V (Hv + v). Consider the following cases: Case 1: Suppose u = v. By (i), f |Hv ∈ TMRDF (Hv). Thus, there exist w, z ∈ NHv(u) such that w ∈ V v 2 and z ∈ V v 3 . Hence, w ∈ V2 and z ∈ V3 and w, z ∈ NG◦H(u). Now, if |V v 3 | ̸= 0 and |V v 2 | = 0, then by (i)(a), |NG(u) ∩ V2| ≥ 1. Hence, there exist z ∈ V v 3 and w ∈ NG(u) ∩ V2 and w, z ∈ NG◦H(u). Similarly, if |V v 2 | ̸= 0 and |V v 3 | = 0, then by (i)(b), |NG(u) ∩ V3| ≥ 1. So, there exist w ∈ V v 2 and z ∈ NG(v) ∩ V3 such that w, z ∈ NG◦H(u). Case 2: Suppose u ̸= v. Then u ∈ V v 0 . Let v ∈ V1∩V (G). By (i), f |Hv ∈ TMRDF (G◦H). Thus, there exist w, z ∈ NHv(u) such that w ∈ V v 2 and z ∈ V v 3 . If v ∈ V2 ∩ V (G). By (ii), V v 3 dominates V v 0 . Hence, 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 . Case 1 and Case 2 shows that there exist w ∈ V2 and z ∈ V3 such that w, z ∈ NG◦H(u). 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 ∈ TMRDF (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 ∈ TMRDF (Hv) by (i). It implies that w ∈ V2∪V3 and w ∈ NG◦H(u). Therefore, f ∈ TMRDF (G◦H). Corollary 6. Let G and H be any graph with |V (G)| = n and |V (H)| = m and let f = (V0, V1, V2, V3) be a γtmR-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 18, f |Hv ∈ TMRDF (Hv). Thus, 3n ≤∑ a∈V (v+Hv) f(a) ≤ 2n + nm. If v ∈ V1, then by Proposition 18, f |Hv ∈ TMRDF (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}. Corollary 7. Let G be a connected graph of order n ≥ 1 and Km be the complete graph, then γtmR(G ◦Km) = { 4n, if m = 2. 5n, if m ≥ 3. Proof. If n = 1, then G◦Km = Km+1. Hence, if m = 2, γtmR(G◦K2) = γtmR(K3) = 4 by Proposition 7 (ii). If m ≥ 4, then γtmR(Km+1) = 5 by Proposition 8. Now, If n > 1, S. Ahamad et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6264 19 of 20 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 ∈ TMRDF (G ◦ K2). It follows that γtmR(G ◦ K2) ≤ 4n. Now, suppose that g = (W0,W1,W2,W3) is a γtmR-function of G ◦K2. If W0 = ∅, then W3 = ∅. Since g is a γtmR-function of G ◦ K2, γtmR(G ◦ K2) = ωtmR G◦K2 (g) ≥ 4n. If |W0| ̸= 0, then |W2| ≥ 1 and |W3| ≥ 1. It follows that γtmR(G ◦K2) = ωtmR G◦K2 (g) = 2|W2|+3|W3| ≥ 4n. Therefore, γtmR(G ◦ K2) = 4n. 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 ∈ TMRDF (G ◦ Km). It follows that γtmR(G ◦ Km) ≤ 5n. Now, suppose that g = (W0,W1,W2,W3) is a γtmR-function of G◦Km. If W0 = ∅, then W3 = ∅. Since g is a γtmR-function of G◦Km, γtmR(G◦Km) = ωtmR G◦Km (g) ≥ 5n. If |W0| ̸= 0, then |W2| ≥ 1 and |W3| ≥ 1. It follows that γtmR(G ◦ Km) = ωtmR G◦Km (g) = 2|W2| + 3|W3| ≥ 5n. Therefore, γtmR(G ◦Km) = 5n. Corollary 8. If Kn is a complete graph of order n ≥ 1, then (i) γtmR(K1 ◦Kn) = n+ 2. (ii) γtmR(Kn ◦K1) = 3n. Proof. Statement (i) follows from the fact that K1 ◦Kn = Sn and by Proposition 15, γtmR(K1 ◦Kn) = γtmR(Sn) = n + 2. 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