EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 1, Article Number 5749 ISSN 1307-5543 – ejpam.com Published by New York Business Global Semitotal Roman Domination in Graphs Brayan F. Bullang1, Imelda S. Aniversario1,2, Alkajim A. Aradais2,3, 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 Center for Mathematical and Theoretical Physical Sciences, Premier Research Institute of Science and Mathematics, MSU-Iligan Institute of Technology, 9200 Iligan City, Philippines 3 Integrated Laboratory School, College of Education, Mindanao State University - Tawi-Tawi College of Technology and Oceanography, Tawi-Tawi, Philippines Abstract. Let G be a nontrivial graph without isolated vertices. A function f : V (G) → {0, 1, 2} is a semitotal Roman dominating function of G if for each v ∈ V (G) with f(v) = 0, there exists u ∈ V (G) for which f(u) = 2 and uv ∈ E(G) and for each v ∈ V (G) with f(v) ̸= 0, there exists u ∈ V (G) for which f(u) ̸= 0 and dG(u, v) ≤ 2. The minimum weight ωG(f) = ∑ u∈V (G) f(u) of a semitotal Roman dominating function f of G is the semitotal Roman domination number of G, denoted by γt2R(G). In this paper, we initiate the study of semitotal Roman domination. We characterize graphs G with small values of γt2R(G) and solve some realization problems with other existing related concepts. We also investigate the semitotal Roman domination in the join, corona and complimentary prism of graphs. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Semitotal Roman dominating function, semitotal Roman domina- tion number, Roman dominating function, Roman domination number, total Roman dominating function, and total Roman domination number 1. Introduction Semitotal domination and Roman domination are two among intriguing concepts in domination in graphs. Both extend the classical notion of domination by incorporating additional structural or functional constraints, offering alternative perspective and appli- cations. DOI: https://doi.org/10.29020/nybg.ejpam.v18i1.5749 Email addresses: brayan.bullang@g.msuiit.edu.ph (B. Bullang), imelda.aniversario@g.msuiit.edu.ph (I. Aniversario), alkajimaradais@msutawi-tawi.edu.ph (A. Aradais), ferdinand.jamil@g.msuiit.edu.ph (F. Jamil) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) B. F. Bullang et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5749 2 of 15 The semitotal domination in graph was introduced by W. Goddard et al. in [9], and it bridges the gap between total domination and traditional domination. A semitotal dominating set S requires that S is a dominating set, but unlike the total domination, only requires that each vertex in S is of distance 1 or 2 from at least one vertex in S. It has practical applications in network resilience and communication. Roman domination, which draws inspiration from the Roman empire’s defense strate- gies, was introduced in 2004 by E.J. Cockayne et al. [8]. In this context, a Roman dominating function assigns weights (0, 1, or 2) to the vertices of a graph such that every vertex with weight 0 has an adjacent vertex with weight 2. It reflects a scenario where resources (represented by weights) are allocated to ensure the protection of vulnerable nodes, offering insights into resource optimization and strategic planning in networks. Since its introduction, Roman domination has become a very active area of research (see for example [1], [6], [8], [11], [16], [17], [18]). Part of its development was the introduction of the total Roman domination in 2016 by A. Ahangar et al. [1], some of the follow-up studies of which can be found in [15] and [16]. In this paper, we introduce and initiate the study of the semitotal Roman domination. We explore graphsG with values of γt2R(G) equal to 2, 3 or 4, and we solve some realization problems involving the concept with other existing related concepts. We also investigate the semitotal Roman domination in the join, corona and complimentary prism of graphs. All throughout this paper, by a graph G we mean simple and undirected. We refer to [5] for all graph terminologies we used but are not defined here. As usual, the symbols V (G) and E(G) denote the vertex set and edge set, respectively, of 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. Given two graphs G and H with disjoint vertex sets, the join G+H of G and H is the graph with vertex set V (G)∪V (H) and edge set E(G) ∪ E(H)∪{uv : u ∈ V (G) and v ∈ V (H)}. The corona G◦H of graph G and H is obtained by taking one copy of G of order n and n copies of H, and then joining the ith vertex of G to every vertex of the ith copy of H. The complementary prism, denoted GG, is formed from the disjoint union of G and its complement G by adding a perfect matching between corresponding vertices of G and G. For u ∈ V (G), the open neighborhood of u is the set NG(u) of all vertices adjacent to u. The closed neighborhood of u in G is the set NG[u] = NG(u) ∪ {u}. For S ⊆ V (G),the open neighborhood of S is the set NG(S) = ∪u∈SNG(u). The closed neighborhood of S is the set NG[S] = NG(S) ∪ S. A set S ⊆ V (G) is a dominating set in G if NG[S] = V (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 [4] for the introduction and more comprehensive discussion of the development of the concept of domination in graphs. Provided that G has no isolated vertices, a set S ⊆ V (G) is a total dominating set in G if for every v ∈ V (G), there exists u ∈ S such that uv ∈ E(G). The minimum cardinality B. F. Bullang et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5749 3 of 15 of a total dominating set in G, denoted by γt(G), is the total domination number of G. A total dominating set S in G with |S| = γt(G) is called a γt-set of G. M. Henning and A. Yeo provides in [12] a very comprehensive discussion of total domination, including its history and the succeeding developments. A set S ⊆ V (G) is a semitotal dominating set in G if S is a dominating set in G such that for every x ∈ S, there exists y ∈ S\{x} for which dG(x, y) ≤ 2. The smallest cardinality of a semitotal dominating set in G, denoted by γt2(G), is called a semitotal domination number in G. A semitotal dominating set S in G with cardinality γt2(G) is called a γt2-set of G. It was introduced in [9] and further studied in [1, 10, 13, 14, 16]. A Roman dominating function (RDF) of 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. Provided G has no isolated vertices, a total Roman dominating function of G is a dominating Roman function f : V (G) → {0, 1, 2} such that for every v ∈ V (G) with f(v) ̸= 0, there exists u ∈ V (G) with f(u) ̸= 0 and uv ∈ V (G). The Roman domination number (resp. total Roman domination number) of G is the minimum weight ωG(f) = ∑ v∈V (G) f(v) of an RDF (resp. TRDF) of G. We write f ∈ RDF (G) and f ∈ TRDF (G) to mean that f is an RDF and TRDF, respectively, of G. An RDF (resp. TRDF) with ωG(f) = γR(G) is referred to as a γR-function (resp. γtR-function ) of G. 2. The semitotal Roman domination We start by introducing a semitotal Roman dominating function. A function f : V (G) → {0, 1, 2} is a semitotal Roman dominating function of a graph G, and write f ∈ SRDF (G), if each of the following holds: (i) For each v ∈ V (G) with f(v) = 0, there exists u ∈ V (G) with f(u) = 2 and uv ∈ E(G); (ii) For each v ∈ V (G) with f(v) ̸= 0, there exists u ∈ V (G) for which f(u) ̸= 0 and dG(u, v) ≤ 2. The minimum weight ωG(f) = ∑ u∈V (G) f(u) of a semitotal Roman dominating function f of G is the semitotal Roman domination number of G, denoted by γt2R(G). As usual, we write f = (V0, V1, V2) for a function f : V (G) → {0, 1, 2} of a graph G where Vi = {v ∈ V (G) : f(v) = i for each i ∈ {0, 1, 2}}. More precisely, f ∈ SRDF (G) if and only if V2 ∩NG(v) ̸= ∅ for each v ∈ V0 and V1 ∪ V2 is a semitotal dominating set of G. In this case, ωG(f) = |V1|+ 2|V2|. Proposition 1. Let G be any graph. Then G admits a semitotal Roman dominating function if and only if G has no isolated vertices. Proposition 2. For all graphs G without isolated vertices, γt2(G) ≤ γR(G) ≤ γt2R(G) ≤ min{γtR(G), 2γt2(G)}. (1) B. F. Bullang et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5749 4 of 15 Proof. Clearly, a total Roman dominating function is a semitotal Roman dominating function and a semitotal Roman dominating function ofG is a Roman dominating function. Thus, γR(G) ≤ γt2R(G) ≤ γtR(G). Also, if S ⊆ V (G) is a semitotal dominating set of G, then f = (V (G) \ S,∅, S) ∈ SRDF (G), showing that γt2R(G) ≤ 2γt2(G). Now, let f = (V0, V1, V2) ∈ RDF (G). Then S = V1 ∪ V2 is a total dominating set of G. We claim that V1 ⊆ NG(S, 2). Let x ∈ V1. Suppose that dG(x, y) ≥ 3 for all y ∈ S \ {x}. Since x is not an isolated vertex, there exists z ∈ NG(x) ∩ V0. Then there exists w ∈ V2 for which zw ∈ E(G), implying that dG(x,w) ≤ 2, a contradiction. This establishes the claim, which implies that S \NG(S, 2) ⊆ V2. For each u ∈ S \NG(S, 2), pick one vu ∈ V (G) for which dG(u, vu) ≤ 2. If S∗ = {vu : u ∈ S \NG(S, 2)}, then S ∪ S∗ is a semitotal dominating set of G with |S∗| ≤ |V2|. Thus, γt2(G) ≤ |S|+ |S∗| ≤ |V1|+ 2|V2| = γR(G). The following examples show that the inequalities in Equation 1 are sharp. For all n ≥ 1, γR(K1,n) < γt2R(K1,n) = 3 = γtR(K1,n) < 2γt2(K1,n). On the other hand, γR(P5) = γt2R(P5) = 2γt2(P5) = 4 < γtR(P5). Proposition 3. Let G be any graph without isolated vertices, and let f = (V0, V1, V2) be a γt2R-function in G. Then each of the following holds: (i) V0 = ∅ if and only if V2 = ∅; and (ii) V1 = ∅ if and only if V2 is a γt2-set in G. Proof. Clearly, if V2 = ∅, then V0 = ∅. Conversely, if V0 = ∅ and V2 ̸= ∅, then g = (V0, V1 ∪ V2,∅) ∈ SRDF (G) with ωG(g) = |V1|+ |V2| < ωG(f), a contradiction. This proves (i). If V1 = ∅, then γt2R(G) = 2|V2| ≥ 2γt2(G). Then Inequality 1 completes the equation γt2R(G) = 2γt2(G). Consequently, |V2| = γt2(G) and V2 is a γt2-set of G. Conversely, suppose that V2 is a γt2-set of G. Then Inequality 1 yields |V1| + 2γt2(G) = γt2R(G) ≤ 2γt2(G), showing that V1 = ∅. This completes (ii). Proposition 4. Let G be a graph without isolated vertices. Then (i) γt2R(G) = γt2(G) if and only if G has a γt2R-function f = (V0, V1, V2) for which V0 = ∅; and (ii) γt2R(G) = 2γt2(G) if and only if G has a γt2R-function f = (V0, V1, V2) for which V2 is a γt2-set of G. Proof. For (i), suppose that γt2R(G) = γt2(G), and let f = (V0, V1, V2) be a γt2R- function of G. Then |V1 + |V2| = |V1|+ 2|V2|, showing that V2 = ∅. The converse follows from Proposition 3. B. F. Bullang et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5749 5 of 15 To prove (ii), if γt2R(G) = 2γt2(G) and f = (V0, V1, V2) is a γt2R- function of G, then |V1| + 2|V2| = 2|V1| + 2|V2|. Necessarily, V1 = ∅. By Proposition 3, V2 is a γt2-set of G. Conversely, by Proposition 3, V1 = ∅ and γt2R(G) = 2|V2| = 2γt2(G). Proposition 5. Let G be a connected graph of order n ≥ 2. Then γt2R(G) = 2 if and only if G = K2. Proof. Let G = K2, then γt2R(G) = 2. Conversely, assume that γt2R(G) = 2 and let f = (V0, V1, V2) be a γt2R-function of G. Since V1 ∪ V2 is a semitotal dominating set, 2 ≤ |V1| + |V2| ≤ |V1| + 2|V2| = 2. Necessarily, V2 = ∅ and, by Proposition 3, V0 = ∅. Therefore, V (G) = V1, that is, G = K2. Proposition 6. Let G be a connected graph of order n ≥ 3. Then γt2R(G) = 3 if and only if either γ(G) = 1 or there exist u and v for which NG[v] = V (G) \ {u} and dG(u, v) = 2. Proof. Suppose that γt2R(G) = 3, and let f = (V0, V1, V2) be a γt2R-function of G. If |V1| = 3, then since G is connected, G = K3 or G = P3. In any case, γ(G) = 1. Suppose that |V2| = 1 = |V1|, say V2 = {v} and V1 = {u}. Then V0 ̸= ∅ and V0 = V (G) \ {u} ⊆ NG[v]. If uv ∈ E(G), then NG[v] = V (G) and γ(G) = 1. Otherwise, dG(u, v) = 2. Conversely, suppose that either there exists v ∈ V (G) for which NG[v] = V (G) or there exist u, v ∈ V (G) such that V (G) \ {u} = NG[v] and dG(u, v) = 2. In any case, let V0 = V (G)\{u, v}, V1 = {u} and V2 = {v}. Then f ∈ SRDF (G) so that γt2R(G) ≤ ωG(f) = 3. Since n ≥ 3, Proposition 5 implies that γt2R(G) = 3. Proposition 7. Let G be a connected graph. Then γt2R(G) = 4 if and only if γ(G) ≥ 2 and either (i) γt2(G) = 2 and for all γt2-sets {u, v} of G, NG(u)\NG(v) ̸= ∅ and NG(v)\NG(u) ̸= ∅; or (ii) γt2(G) = 3 and G has distinct vertices u, v and w for which V (G) \ {u,w} = NG[v] and dG(u, v) = 2 and dG(w, v) = 2. Proof. Suppose that γt2R(G) = 4. By Proposition 5 and Proposition 6, γ(G) ≥ 2. Let f = (V0, V1, V2) is a γt2R-function of G. In view of Proposition 3, if V2 = ∅, then |V (G)| = 4 and γt2R(G) ≤ 3, a contradiction. Thus, either |V2| = 2 and V1 = ∅ or |V2| = 1 and |V1| = 2. By Proposition 3, the former implies that γt2(G) = |V2| = 2 and, by Proposition 6, (i) holds. Now, suppose that V2 = {v} and V1 = {u,w}. Since V1 ∪ V2 is a semitotal dominating set of G, γt2(G) ≤ 3. If γt2(G) = 2, then (i) holds. Suppose that γt2(G) = 3. Since f ∈ SRDF (G), V (G) \ {u,w} ⊆ NG[v]. Suppose that uv ∈ E(G). If dG(u,w) = 1, then f∗ = (V0∪{u}, {w}, {v}) ∈ SRDF (G) with ωG(f ∗) = 3, a contradiction. If dG(u,w) ≥ 2 and P = [u = x1, x2, . . . , xk = w] is a u-w geodesic in G, then xj ∈ V0 for all j ∈ {2, 3, . . . , k − 1}. Since xk−1v ∈ E(G), dG(w, v) = 2. Thus, f∗∗ = (V0 ∪ {u}, {w}, {v}) ∈ SRDF (G) with ωG(f ∗∗) = 3, a contradiction. Thus, B. F. Bullang et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5749 6 of 15 uv /∈ E(G). Similarly, wv /∈ E(G). Therefore, NG[v] = V (G) \ {u,w}. Suppose that dG(u, v) ≥ 3, and let P be a v-u geodesic in G. Since NG[v] = V (G)\{u,w}, P has length 3 and w lies on P so that dG(u, v) = 2 and uw ∈ E(G). Then {u, v} is a γt2-set of G, a contradiction. Therefore, dG(u, v) = 2 = dG(w, v). Thus, (ii) holds. Conversely, assume that γ(G) ≥ 2. Suppose that (i) holds. In view of Proposition 6, γt2R(G) ≥ 4. Now let S = {u, v} be a γt2-set of G. Since f = (V (G) \ {u, v},∅, {u, v}) ∈ SRDF (G), γt2R(G) ≤ ωG(f) ≤ 4. Now, suppose that (ii) holds. Since γt2(G) = 3, γt2R(G) ≥ 4 by Proposition 6. Moreover, since f = (V (G) \ {u,w}, {u,w}, {v}) ∈ SRDF (G), γt2R(G) ≤ wG(f) = 4. In any case, γt2R(G) = 4, Proposition 6 asserts that if γt2R(G) = 3, then γt2(G) = 2. Proposition 7 also provides that the converse need not be true. In particular, for G2 in Figure 1, γt2(G2) = 2 but γt2R(G) = 4 by Proposition 7. The graph G1 in Figure 1 provides a graph which perfectly satisfies condition (ii) of Proposition 7. vu w G1 : v u G2 : Figure 1: Graphs satisfying the conditions of Proposition 7 Proposition 8. For each positive integer n and integer 0 ≤ k ≤ n, there exists a connected graph G for which γR(G) = 4n and γt2R(G) = 4n + k. Consequently, the difference γt2R(·)− γR(·) can be made arbitrarily large. Proof. We consider the following cases: Case 1: Suppose that k = 0. Consider the corona G = P2n ◦ K4 provided in Figure 2. Then γR(G) = 4n = γt2R(G). x1 x2 x2n−1 x2n · · · Figure 2: The graph P2n ◦K4 Case 2: Suppose that k = n. In this case take the graph G as in Figure 3 obtained from P6n−2 = [x1, x2, . . . , x6n−2] by joining K4 with x3j−2 for all j = 1, 2, . . . , 2n. Then γR(G) = 4n and γt2R(G) = 4n+ k. Case 3: Finally, suppose that 1 ≤ k ≤ n − 1. Consider the graph G obtained from P2n+4k = [x1, x2, . . . , 2n+4k] by joining K4 with xj for all j ∈ {1, 2, . . . , 2n− 2k}∪ {2n− 2k + 3l : l = 1, 2, . . . , 2k} (see Figure 4 below). Then γR(G) = 2(2n− 2k) + 4k = 4n and γt2R(G) = 2(2n− 2k) + (4k + k) = 4n+ k. B. F. Bullang et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5749 7 of 15 x1 x2 x3 x4 x5 x6 x7 x6n−5 x6n−4 x6n−3 x6n−2 · · · Figure 3: A graph G with γR(G) = 4n and γt2R(G) = 4n+ n x1 x2 x2n−2k−1 x2n−2k x2n−2k+1 x2n−2k+2 x2n−2k+3 x2n−2k+4 x2n−2k+5 x2n−2k+6 x2n−2k+7 x2n−2k+8 x2n−2k+9 x2n+4k−3 x2n+4k−2 x2n+4k−1 x2n+4k · · · · · · Figure 4: A graph G with γR(G) = 4n and γt2R(G) = 4n+ k, 1 ≤ k ≤ n− 1 Proposition 9. For each positive integer n and integer 0 ≤ k ≤ n, there exists a connected graph G for which γt2R(G) = 4n and γtR(G) = 4n + k. Consequently, the difference γtR(·)− γt2R(·) can be made arbitrarily large. Proof. For k = 0, we take the graph G = P2n ◦ K4 in Figure 2. For this graph, γt2R(G) = γtR(G) = 4n. Suppose that k = n. Consider the graph G given in Figure 5 obtained from P4n−1 = [x1, x2, . . . , x4n−1] by joining K4 with x2j−1 for all j = 1, 2, . . . , 2n. Then γt2R(G) = 2(2n) = 4n and γtR(G) = 2(2n) + n = 4n+ k. x1 x2 x3 x4 x7 x4n−3 x4n−2 x4n−1 · · · Figure 5: A graph G with γR(G) = 4n and γt2R(G) = 4n+ n Now, suppose that 1 ≤ k ≤ n−1. Obtain G as the graph in Figure 6 from P2n+2k+1 = [x1, x2, . . . , x2n+2k+1] by joiningK4 with xj for all j ∈ {1, 2, . . . , 2n−2k}∪{2n−2k+2l+1 : l = 1, 2, . . . , 2k}. Then γt2R(G) = 2(2n− 2k) + 2(2k) = 4n and γtR(G) = 4n+ k. Proposition 10. Let G be a disconnected graph without isolated vertices, and let C1, C2, ..., Ck be the components of G. Then γt2R(G) = ∑k i=1 γt2R(Ci). B. F. Bullang et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5749 8 of 15 x1 x2 x2n−2k−1 x2n−2k x2n−2k+1 x2n−2k+2 x2n−2k+3 x2n−2k+4 x2n−2k+5 x2n−2k+6 x2n−2k+7 x2n+2k−1 x2n+2k x2n+2k+1 · · · · · · Figure 6: A graph G with γt2R(G) = 4n and γtR(G) = 4n+ k, 1 ≤ k ≤ n− 1 Proof. Let f1, f2, ..., fk be γt2R-functions of C1, C2, ..., Ck, respectively. Then the func- tion f : V (G) → {0, 1, 2} given by f(x) =  f1(x), if x ∈ V (C1), f2(x), if x ∈ V (C2), ... fk(x), if x ∈ V (Ck), is a SRDF of G. Thus, γt2R(G) ≤ ωG(f) = ∑k i=1 γt2R(Ci). Conversely, let f be a γt2R- function of G. Then, for each i ∈ {1, 2, . . . , k}, the restriction f |Ci of f to Ci is a SRDF of C1, showing γt2R(Ci) ≤ ωCi(f |Ci). Thus ∑k i=1 γt2R(Ci) ≤ ∑k i=k ωCi(f) = ωG(f) = γt2R(G). Proposition 11. For paths, cycles and complete multipartite graphs, we have the follow- ing: (i) For n ≥ 2, γt2R(Pn) =  n, n = 2, 3, 3k + r, n = 4k + r and 0 ≤ r ≤ 2, 3k + 2, n = 4k + 3. (ii) For n ≥ 3, γt2R(Cn) =  3, n = 3, 3k + r, n = 4k + r and 0 ≤ r ≤ 2, 3k + 2, n = 4k + 3. (iii) For the complete k-partite G = Kn1,n2,...,nk with 2 ≤ n1 ≤ n2 ≤ · · · ≤ nk, γt2R(G) = { 3, if n1 = 2; 4, otherwise. Proof. For (i), if n = 2, 3, then the result follows immediately. For n ≥ 4, write n = 4k + r where 0 ≤ r ≤ 3. We define f = (V0, V1, V2) as follows: Case 1: Suppose that r = 0. Put V2 = {x4l−2 : l = 1, 2, . . . , k}, V1 = {x4l : l = 1, 2, . . . , k} and V0 = V (Pn) \ (V1 ∪ V2). B. F. Bullang et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5749 9 of 15 Case 2: Suppose that r = 1. Put V2 = {x4k, x4l−2 : l = 1, 2, . . . , k}, V1 = {x4l : l = 1, 2, . . . , k − 1} and V0 = V (Pn) \ (V1 ∪ V2). Case 3: Suppose that r = 2. Define V2 = {x4l−2 : l = 1, 2, . . . , k + 1}, V1 = {x4l : l = 1, 2, . . . , k} and V0 = V (Pn) \ (V1 ∪ V2). Case 4: Finally, suppose that r = 3. Define V2 = {v4l−2 : l = 1, 2, . . . , k + 1}, V1 = {v4l : l = 1, 2, . . . , k} and V0 = V (Pn) \ (V1 ∪ V2). In any case, f = (V0, V1, V2) is a γt2R-function of Pn with ωPn(f) = 3k+r for 0 ≤ r ≤ 2 and ωPn(f) = 3k + 2 when r = 3. Similar arguments prove statement (ii). Whereas, statement (iii) follows from Propo- sition 6 and Proposition 7. 3. Semitotal Roman domination in graphs under some operations Here we investigate the semitotal Roman domination in the join, corona and comple- mentary prism of graphs. 3.1. In the join of graphs In what follows, by f |G we mean the restriction of f to G. Theorem 1. Let G and H be any graphs and f = (V0, V1, V2) : V (G+H) → {0, 1, 2} be a function with V0 ̸= ∅. Then f ∈ SRDF (G+H) if and only if one of the following holds: (i) V2 ⊆ V (G), f |G ∈ RDF (G) and one of the following holds: (a) V1 ∩ V (H) ̸= ∅; (b) |(V1 ∪ V2) ∩ V (G)| ≥ 2; (ii) V2 ⊆ V (H), f |H ∈ RDF (H), and one of the following holds: (a) V1 ∩ V (G) ̸= ∅; (b) |(V1 ∪ V2) ∩ V (H)| ≥ 2; (iii) V2 ∩ V (G) ̸= ∅ and V2 ∩ V (H) ̸= ∅. Proof. Suppose that f ∈ SRDF (G + H). Since V0 ̸= ∅, V2 ̸= ∅. Suppose that V2 ⊆ V (G), and let v ∈ V0 ∩ V (G). Then there exists u ∈ V2 for which uv ∈ E(G +H). Since V2 ⊆ V (G), uv ∈ E(G). Accordingly, f |G ∈ RDF (G). If V1 ∩ V (H) ̸= ∅, then (i)(a) holds. Suppose that V1 ∩ V (H) = ∅. Pick v ∈ V2. Since f ∈ SRDF (G+H), there exists u ∈ V1 ∪ V2 for which 1 ≤ dG+H(v, w) ≤ 2. The assumptions imply that w ∈ V (G). Thus, (i)(b) holds. Similarly, if V2 ⊆ V (H), then (ii) holds. Clearly, if both (i) and (ii) do not hold, then (iii) holds. B. F. Bullang et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5749 10 of 15 Conversely, assume that (i) holds. Let v ∈ V0. If v ∈ V (H), then pick any u ∈ V2. If v ∈ V (G), then since f |G ∈ RDF (G), there exists u ∈ V2 for which uv ∈ E(G). In any case, uv ∈ E(G+H). Let v ∈ V1 ∪ V2. If v ∈ V (H), then v ∈ V1 and uv ∈ E(G+H) for all u ∈ V2 ⊆ V (G). Suppose that v ∈ V (G). If (i)(a) holds, then dG+H(u, v) = 1 for all u ∈ V1 ∩ V (H). If (i)(b) holds, then dG+H(u, v) ≤ 2 for all u ∈ [(V1 ∪ V2) ∩ V (G)] \ {v}. Accordingly, f ∈ SRDF (G +H). Similarly, if (ii) holds, then f ∈ SRDF (G +H). It is also clear that if (iii) holds, then f ∈ SRDF (G+H). Corollary 1. Let G and H be any graphs. Then 2 ≤ γtR(G+H) ≤ 4. (2) Proof. Since G + H is at least P2, γt2R(G + H) ≥ 2 by Proposition 5. Now let u ∈ V (G) and v ∈ V (H). Define V0 = V (G+H)∖ {u, v}, V1 = ∅ and V2 = {u, v}. Then V2 ∩ V (G) ̸= ∅ and V2 ∩ V (H) ̸= ∅. By Theorem 1, f = (V0, V1, V2) ∈ TRD(G + H). Thus, γt2R(G+H) ≤ wG+H(f) = 4. Proposition 12. Let G and H be any graphs of orders n and m, respectively. Then (i) γt2R(G+H) = 2 if and only if n = m = 1; (ii) γt2R(G+H) = 3 if and only if n ̸= 1 or m ̸= 1 and one of the following holds: (a) γ(G) = 1 or γ(H) = 1; (b) ∆(G) = n− 2 or ∆(H) = m− 2. (iii) γt2R(G+H) = 4 if and only if ∆(G) ≤ n− 3 and ∆(H) ≤ m− 3. Proof. By Proposition 5, γt2R(G+H) = 2 if and only if G+H = P2. Thus, (i) holds. Suppose that γt2R(G + H) = 3. By (i), G ̸= K1 or H ̸= K1. If γ(G + H) = 1, then γ(G) = 1 or γ(H) = 1 and (ii)(a) holds. Otherwise, by Proposition 6, there exists u, v ∈ V (G+H) for which NG+H [v] = V (G+H) \ {u} and dG+H(u, v) = 2. Necessarily, either u, v ∈ V (G) or u, v ∈ V (H). This means that if γ(G) ̸= 1 and γ(H) ̸= 1, then (ii)(b) holds. Conversely, if G ̸= K1 or H ̸= K1, then γt2R(G +H) ≥ 3 by (i). If (ii)(a) holds, then γ(G+H) = 1 and γt2R(G+H) = 3 (by Proposition 6). If (ii)(b) holds, then ∆(G) = m+n−2. Thus, there exists u, v ∈ V (G+H) such that NG+H [v] = V (G+H)\{u} and dG+H(u, v) = 2. By Proposition 6, γt2R(G+H) = 3. This proves (ii). Statement (iii) follows immediately from statements (i) and (ii) and the Inequality 2. The join G + H, where G = K2, is a good example of Proposition 12(ii)(b), where ∆(G) = n − 2 = 0. Proposition 12(iii) also implies that γt2R(Km,n) = 4 for all integers m,n ≥ 3. B. F. Bullang et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5749 11 of 15 3.2. In the corona of graphs Let G and H be connected graphs. We adapt the notation Hv to denote the copy of H whose vertices is joined to v ∈ V (G). Theorem 2. Let G be nontrivial connected graph and H be any graph, and let f = (V0, V1, V2) : V (G ◦ H) → {0, 1, 2}. Then f ∈ SRDF (G ◦ H) if and only if each of the following holds: (i) For each v ∈ V0 ∩ V (G), f |Hv ∈ RDF (Hv) and each of the following holds: (a) |(V1 ∪ V2) ∩ V (Hv)| ≥ 2 whenever NG(v) ⊆ V0; (b) V2 ∩ V (Hv) ̸= ∅ whenever NG(v) ∩ V2 = ∅. (ii) For each v ∈ V1 ∩ V (G), f |Hv ∈ RDF (Hv). (iii) For each v ∈ V2 ∩ V (G), (V1 ∪ V2) ∩NG(v, 2) ̸= ∅ whenever V (Hv) ⊆ V0. Proof. If f ∈ SRDF (G◦H), then the statements (i)-(iii) are clear. Conversely, assume that all conditions (i), (ii) and (iii) hold for f . First, let x ∈ V0 and let v ∈ V (G) for which x ∈ V (v +Hv). Suppose that x = v. If NG(v) ∩ V2 ̸= ∅, say u ∈ NG(v) ∩ V2, then u ∈ V2∩NG◦H(x). If, on the other hand, NG(v)∩V2 = ∅, then by (i)(b), V2∩V (Hv) ̸= ∅. Pick u ∈ V2 ∩ V (Hv). Then u ∈ V2 ∩NG◦H(x). Suppose that x ̸= v. If v ∈ V0 ∪ V1, then since f |Hv ∈ RDF (Hv), there exists u ∈ V2∩NHv(x). This means that u ∈ V2∩NG◦H(x). And, if v ∈ V2, then v ∈ V2 ∩NG◦H(x). Next, let x ∈ V1 ∪ V2 and let v ∈ V (G) for which x ∈ V (v + Hv). If x = v ∈ V1, then since f |Hv ∈ RDF (Hv), (V1 ∪ V2) ∩ V (Hv) ̸= ∅. Then 1 ≤ dG◦H(x, u) ≤ 2 for all u ∈ (V1 ∪ V2) ∩ V (Hv). Suppose x = v ∈ V2. If V (Hv) \ V0 ̸= ∅, then 1 ≤ dG◦H(u, x) ≤ 2 for all u ∈ V (Hv) \V0. If V (Hv) ⊆ V0, then by (iii), there exists u ∈ (V1 ∪V2)∩NG(x, 2). Finally, suppose that x ̸= v. If v ∈ V1 ∪ V2, then we are done. Suppose that v ∈ V0. If NG(v) \ V0 ̸= ∅, then dG◦H(x, u) ≤ 2 for all u ∈ NG(v) \ V0 ⊆ (V1 ∪ V2). If NG(v) ⊆ V0, then by (i)(a), there exists u ∈ (V1 ∪ V2) ∩ V (Hv) with u ̸= x. Here, dG◦H(x, u) ≤ 2. The above results imply that f ∈ SRDF (G ◦H). It is worth noting that Theorem 2(i)(a) does not happen if H = K1. Corollary 2. Let G be a nontrivial connected graph of order n. Then (i) γt2R(G ◦K1) = n+ γ(G); (ii) γt2R(G ◦H) = 2n for all nontrivial graphs H. Proof. In the case where H = K1 = {u}, we define for each v ∈ V (G), V (Hv) = {uv}. Let S ⊆ V (G) be a γ-set of G. Define V0 = (V (G) \S)∪ (∪v∈S{uv}), V1 = ∪v∈V (G)\S{uv} and V2 = S. It is easy to see that f = (V0, V1, V2) ∈ SRDF (G ◦H). Thus, γt2R(G ◦H) ≤ ωG◦H(f) = 2|S| + (n − |S|) = n + γ(G). Now, let f = (V0, V1, V2) be a γt2R-function of B. F. Bullang et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5749 12 of 15 G ◦H. Since f is a γt2R-function, f(u v) = 1 for all v ∈ V1 ∩ V (G) and f(uv) = 0 for all v ∈ V2 ∩ V (G). Let X = {v ∈ V0 ∩ V (G) : f(uv) = 2}, Y = {uv : v ∈ X} and W = {uv : v ∈ V1 ∩ V (G)}, and define V ∗ 0 = (V0 \X) ∪ Y ∪W, V ∗ 1 = V1 \ [(V1 ∩ V (G)) ∪W ] and V ∗ 2 = ((V1 ∪ V2) ∩ V (G)) ∪X. First, we claim that g = (V ∗ 0 , V ∗ 1 , V ∗ 2 ) ∈ SRDF (G ◦ H). Let x ∈ V ∗ 0 . We consider the following cases: Case 1: Suppose x ∈ V0 \ X. If x = uv for some v, then v ∈ V2 ∩ V (G). Hence, v ∈ V ∗ 2 ∩NG◦H(x). If x ∈ V (G), then f(ux) ̸= 2 and thus, there exists y ∈ V2∩V (G) such that xy ∈ E(G ◦H). This means that y ∈ V ∗ 2 ∩NG◦H(x). Case 2: Suppose x ∈ Y ∪W . Then x = uv where v ∈ X or v ∈ V1 ∩ V (G). In any case, v ∈ V ∗ 2 ∩NG◦H(x). Now, let x ∈ V ∗ 1 . Then x = uv ∈ V1 with v /∈ V1 ∩ V (G). Thus, v ∈ V0 ∩ V (G). Since v /∈ X, there exists y ∈ V2 ∩NG(v). This means that dG◦H(x, y) = 2. Finally, let x ∈ V ∗ 2 . If x ∈ X, then x ∈ V0 ∩ V (G) for which ux ∈ V2, and there exists y ∈ V2 ∩ V (G) so that dG◦H(ux, y) = 2. This means that dG◦H(x, y) = 1. Suppose x ∈ (V1 ∪ V2) ∩ V (G). Take y ∈ V (G) such that xy ∈ E(G). If y ∈ V0, then either y ∈ X ⊆ V ∗ 2 or uy ∈ V1 \ W ⊆ V ∗ 1 . In any case, there exists u ∈ V ∗ 1 ∪ V ∗ 2 such that 1 ≤ dG◦H(x, u) ≤ 2. then since ux ∈ V0, there exists y ∈ V2 such that 1 ≤ dG◦H(x, y) ≤ 2. On the other hand, if y ∈ V1 ∪ V2, then y ∈ V ∗ 2 and dG◦H(x, y) = 1. The above arguments show that g ∈ SRDF (G ◦H). And since V (G) ⊆ V ∗ 0 , V ∗ 2 is a dominating set of G. Moreover, since X = V2 \ (V2 ∩ V (G) and |W | = |V1 ∩ V (G)|, ωG◦H(g) = |V ∗ 1 |+ 2|V ∗ 2 | = |V1| − 2|V1 ∩ V (G)|+ 2[|V2|+ |V1 ∩ V (G)|] = |V1|+ 2|V2| = γt2R(G ◦H). Thus, n+ γ(G) ≤ n+ |V ∗ 2 | = |V ∗ 1 |+ 2|V ∗ 2 | = γt2R(G ◦H). This proves (i). To prove (ii), since f = (∪v∈V (G)V (Hv),∅, V (G)) ∈ SRDF (G◦H), γt2R(G◦H) ≤ 2n. To get the other inequality, let f = (V0, V1, V2) ∈ SRDF (G ◦H). For each v ∈ V2 ∩V (G), clearly f(V (Hv + v) ≥ 2. For each v ∈ (V0 ∪ V1) ∩ V (G), since f |Hv ∈ RDF (Hv), f(V (Hv + v)) ≥ 2. Thus, ωG◦H(f) ≥ ∑ v∈V (G) f(V (Hv + v)) ≥ 2n. 3.3. In the complementary prism of graphs Proposition 13. Let G be a graph of order n. Then B. F. Bullang et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5749 13 of 15 (i) γt2R(GG) = 2 if and only if G = K1. (ii) γt2R(GG) = 3 if and only if G ∈ {K2,K2}. (iii) γt2R(GG) = 4 if and only if either (a) G ∈ {K3,K3}; or (b) G ̸= K2 and G (resp. G ̸= K2 and G) has vertices u and v for which NG[v] = V (G) and NG[u] = {u, v} (resp. NG[v] = V (G) and NG[u] = {u, v}). Proof. By Proposition 5, γt2R(GG) = 2 if and only if GG = K2. Thus, (i) holds. Suppose that γt2R(GG) = 3. Since GG ̸= K2, γ(GG) ≥ 2 by (i). By Proposition 6, GG has vertices u and v for which NGG[v] = V (GG) \ {u} and dGG(u, v) = 2. This is possible only if GG = P4 or equivalently, G ∈ {K2,K2}. The converse of (ii) is immediate. Suppose that γt2R(GG) = 4. In view of Proposition 7, we consider two cases: Case 1: Suppose that γt2(GG) = 2, and let {u, v} be a γt2-set of GG. Since GG ̸= K2, if dGG(u, v) = 1, then GG = P4, a contradiction. Thus, dGG(u, v) = 2. Assume v ∈ V (G). Necessarily, u ∈ V (G) and u ̸= v. Since {u, v} is a dominating set of GG, NG[v] = V (G) and NG[u] = V (G) \ {v} or equivalently, NG[u] = {u, v}. Case 2: Suppose that γt2(GG) = 3 and u, v and w are vertices of GG for which the following hold: (a) V (GG) \ {u,w} = NGG[v]; (b) dGG(u, v) = 2 and dGG(w, v) = 2. WLOG assume that v ∈ V (G). If u ∈ V (G), then by (b), dG(u, v) = 2, say [u, z, v] is a u-v geodesic in G. Then by (a), u ∈ NGG[v] or z ∈ NGG[v], which is impossible. Thus u ∈ V (G). Similarly, w ∈ V (G). Moreover, (a) implies that V (G) = {v, u, w}. Thus, either G = P3 or G = K3. But if G = P3, then G is among the graphs described in Case 1. Thus, G = K3. The converse of (iii) is also immediate. Corollary 3. If G is a graph with isolated vertex v such that γ(G − v) = 1, then γt2R(GG) = 4. Proposition 14. Suppose that both G and G contain no isolated vertices. Then max{γt2R(G), γt2R(G)} ≤ γt2R(GG) ≤ γt2R(G) + γt2R(G). Proof. Let f = (V f 0 , V f 1 , V f 2 ) and g = (V g 0 , V g 1 , V g 2 ) be a γt2R-functions of G and G, respectively. It is straightforward to show that F = (V f 0 ∪ V g 0 , V f 1 ∪ V g 1 , V f 2 ∪ V g 2 ) ∈ SRDF (GG). Therefore, γt2R(GG) ≤ ωGG(F ) = γt2R(G) + γt2R(G). To get the other inequality, let p = (V0, V1, V2) be a γt2R-function of GG. Let A = {v ∈ V0 ∩ V (G) | V2 ∩NGG(v) = {v} and NG(v) ⊆ V0}, B. F. Bullang et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5749 14 of 15 B = {v ∈ V0 ∩ V (G) | V2 ∩NGG(v) = {v} and NG(v) ∩ V1 ̸= ∅}, and C = {v ∈ V (G) ∩ (V1 ∪ V2) | (V1 ∪ V2) ∩NGG(v, 2) ⊆ V (G)}. For each v ∈ A ∪ C, choose exactly one uv ∈ NG(v) and let D = {uv | v ∈ A} and E = {uv | v ∈ C}. Define q = (V ∗ 0 , V ∗ 1 , V ∗ 2 ) by V ∗ 0 = (V0 ∩ V (G))∖ (A ∪B ∪D ∪ E), V ∗ 1 = (V1 ∩ V (G)) ∪A ∪B ∪D ∪ E, and V ∗ 2 = V2 ∩ V (G). We claim that q ∈ SRDF (G). Let u ∈ V ∗ 0 . Since u ∈ V0, there exists v ∈ V2 for which uv ∈ E(GG). Beacause u /∈ (A ∪ B), v ∈ V (G). Thus, v ∈ V ∗ 2 and uv ∈ E(G). Let x ∈ V ∗ 1 ∪ V ∗ 2 . We consider the following cases: Case 1: x ∈ V ∗ 1 . Suppose that x ∈ V1 ∩ V (G). Then there exists y ∈ V1 ∪ V2 for which dGG(x, y) ≤ 2. If x /∈ C, then y ∈ (V1∪V2)∩V (G) ⊆ V ∗ 1 ∪V ∗ 2 . If x ∈ C, then there exists ux ∈ NG(x)∩E ⊆ V ∗ 1 . If x ∈ A, then there exists ux ∈ NG(x) ∩ D ⊆ V ∗ 1 . If x ∈ D, then x = uv for some v ∈ A ⊆ V ∗ 1 and v ∈ NG(x). Suppose that x ∈ B. Since NG(x) ∩ V1 ̸= ∅, say y ∈ NG(x) ∩ V1, we have y ∈ V ∗ 1 and xy ∈ E(G). Finally, suppose x ∈ E. Then x = uv for some v ∈ C. This means that v ∈ V ∗ 1 ∪ V ∗ 2 and xv ∈ E(G). Case 2: x ∈ V ∗ 2 . Then there exists y ∈ V1 ∪ V2 for which dGG(x, y) ≤ 2. If x /∈ C, then y ∈ V (G) so that y ∈ V ∗ 1 ∪ V ∗ 2 and dG(x, y) ≤ 2. Suppose that x ∈ C. Then ux ∈ E and xux ∈ E(G). Accordingly, q = (V ∗ 0 , V ∗ 1 , V ∗ 2 ) ∈ SRDF (G), showing that γt2R(GG) = ωGG(p) ≥ ωG(q) ≥ γt2R(G). Similarly, γt2R(GG) ≥ γt2R(G). Acknowledgment: The authors would like to thank the Department of Science and Technology - Accelerated Science and Technology Human Resource Development Program (DOST-ASTHRDP)-Philippines, and MSU-Iligan Institute of Technology for funding this research. B. F. Bullang et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5749 15 of 15 References [1] A.H. Ahangar, M. Henning, V. Samodivkin and I. Yero, Total Roman domination in graphs. Applicable Analysis and Discrete Mathematics , Vol. 10, No. 2 (October 2016), pp. 501-517. [2] I. Aniversario, S. Canoy Jr., and F.P. Jamil. On semitotal domination in graphs. Articles in European Journal of Pure and Applied Mathematics, October 2019. 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