EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 1, 2023, 363-372 ISSN 1307-5543 – ejpam.com Published by New York Business Global Strong Resolving Domination in the Lexicographic Product of Graphs Gerald B. Monsanto1,∗, Penelyn L. Acal2, Helen M. Rara3 1 College of Teacher Education, Arts and Sciences, Visayas State University Villaba, 6537 Villaba, Leyte, Philippines 2 Department of Mathematical Sciences, University of Science and Technology of Southern Philippines, 9023, Cagayan de Oro City, Philippines 3 Department of Mathematics and Statistics, College of Science and Mathematics, Center of Graph Theory, Algebra, and Analysis-Premier Research Institute of Science and Mathematics, Mindanao State University-Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. Let G be a connected graph. A subset S ⊆ V (G) is a strong resolving dominating set of G if S is a dominating set and for every pair of vertices u, v ∈ V (G), there exists a vertex w ∈ S such that u ∈ IG[v, w] or IG[u,w]. The smallest cardinality of a strong resolving dominating set of G is called the strong resolving domination number of G. In this paper, we characterize the strong resolving dominating sets in the lexicographic product of graphs and determine the corresponding strong resolving domination number. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Strong resolving dominating set, strong resolving domination number, lexicographic product 1. Introduction All graphs considered in this study are finite, simple, and undirected connected graphs, that is, without loops and multiple edges. For some basic concepts in Graph Theory, we refer readers to [7]. Let G = ( V (G), E(G) ) be a connected graph. The open neighborhood of v ∈ V (G) is NG(v) = {u ∈ V (G) : uv ∈ E(G)}. Any element u of NG(v) is called a neighbor of v. The closed neighborhood of v ∈ V (G) is NG[v] = NG(v) ∪ {v}. Thus, the degree of v ∈ V (G) is given by degG(v) = |NG(v)|. Customary, for S ⊆ V (G), NG(S) = ⋃ v∈S NG(v) and NG[S] = ⋃ v∈S NG[v]. DOI: https://doi.org/10.29020/nybg.ejpam.v16i1.4652 Email addresses: gerald.monsanto@vsu.edu.ph (G. Monsanto), penelyn.acal@g.msuiit.edu.ph (P. Acal), helen.rara@g.msuiit.edu.ph (H. Rara) https://www.ejpam.com 363 © 2023 EJPAM All rights reserved. G. Monsanto, P. Acal, H. Rara / Eur. J. Pure Appl. Math, 16 (1) (2023), 363-372 364 A nonempty set S ⊆ V (G) is a dominating set in graph G if NG[S] = V (G). Otherwise, we say S is a non-dominating set of G. The domination number of a graph G, denoted by γ(G), is given by γ(G) = min|S| : S is a dominating set of G. If S is a dominating set of G and if |S| = γ(G), then S is called a minimum dominating set or a γ-set of G. A vertex w ∈ S strongly resolves two different vertices u, v ∈ V (G) if v ∈ IG[u,w] or u ∈ IG[v, w]. A set W of vertices in G is a strong resolving set of G if every two vertices of G are strongly resolved by some vertices of W . The smallest cardinality of a strong resolving set of G is called the strong metric dimension of G and is denoted by sdim(G). A subset S ⊆ V (G) is a strong resolving dominating set of G if it is both strong resolving and dominating. The smallest cardinality of a strong resolving dominating set of G is called the strong resolving domination number of G and is denoted by γsR(G). A strong resolving dominating set of cardinality γsR(G) is called a γsR-set of G. A clique in a graph G is a complete induced subgraph of G. A clique C in G is called a superclique if for every pair of distinct vertices u, v ∈ C, there exists w ∈ V (G) \ C such that w ∈ NG(u) \ NG(v) or w ∈ NG(v) \ NG(u). A superclique C in G is called a dominated superclique if for every u ∈ C, there exists v ∈ V (G)\C such that uv ∈ E(G).[5] A superclique (resp. dominated superclique) C is maximum in G if |C| ≥ |C∗| for all supercliques (resp. dominated supercliques) C∗ in G. The superclique (resp. dominated superclique) number, ωS(G) (resp. ωDS) of G is the cardinality of a maximum superclique (resp. maximum dominated superclique) in G. A vertex u of G is maximally distant from vertex v of G, u ̸= v, if for every vertex w ∈ NG(u), dG(v, w) ≤ dG(u, v). If u is maximally distant from v and v is maximally distant from u, then we say that u and v are mutually maximally distant, denoted by uMMDv. In recent years, the concept of domination in graphs has been studied extensively and several research papers have been published on this topic. The said concept was not formally defined mathematically until the publications of the books by Claude Berge [3] in 1958 and Oystein Ore in 1962. In 1977, a survey paper by Cockayne and Hedetniemi [4] began to study the concept of domination. On the other hand, the problem of uniquely recognizing the possible position of an in- truder such as fault in a computer network and spoiled device was the principal motivation in introducing the concept of metric dimension in graphs. Slater [10] brought in the notion of locating sets and its minimum cardinality as locat- ing number. The same concept was also introduced by Harary and Melter [7] but using the terms resolving sets and metric dimension to refer to locating sets and locating num- ber, respectively. However, in recent studies, locating sets and resolving sets are defined differently. In 2013, Canoy and Omega [8], defined a locating set as a subset S of V (G) in a connected graph G satisfying that NG(u) ∩ S ̸= NG(v) ∩ S for all u, v ∈ V (G) \ S. Meanwhile, Bailey et al. [2] defined a resolving set as a set of vertices S in a graph G such that for any two vertices u, v, there exists x ∈ S such that the distances d(u, x) ̸= d(v, x). In 2007, Oellerman and Peter-Fransen [9] introduced the strong resolving graph GSR of a connected graph G as a tool to study the strong metric dimension of G. This study aims to define and characterize the strong resolving dominating sets in G. Monsanto, P. Acal, H. Rara / Eur. J. Pure Appl. Math, 16 (1) (2023), 363-372 365 the lexicographic product of graphs and determine their corresponding strong resolving domination number. 2. Preliminary Results Remark 1. [1] Every strong resolving dominating set of a connected graph G is a domi- nating set of G. Hence, γ(G) ≤ γsR(G). Remark 2. [1] Every strong resolving dominating set of a connected graph G is a strong resolving set of G. Hence, sdim(G) ≤ γsR(G). Remark 3. For any connected graph G of order n ≥ 2, 1 ≤ γsR(G) ≤ n− 1. Example 1. γsR(P2) = 1 and γsR(Kn) = n− 1 for n ≥ 2. Remark 4. [1] Any superset of a strong resolving dominating set is a strong resolving dominating set. Proposition 1. Every strong resolving set of a connected graph G is a resolving set. Proof: Let S ⊆ V (G) is a strong resolving set of G. Then for any pair of distinct vertices u, v ∈ V (G), there exists w ∈ S such that u ∈ IG[v, w] or v ∈ IG[u,w]. If u ∈ IG[v, w], then dG(u,w) < dG(v, w). Thus, rG(u/S) ̸= rG(v/S), showing that w resolves u and v. Hence, S is a resolving set of G. Proposition 2. Every strong resolving dominating set of a connected graph G is a resolv- ing dominating set. Hence, γR(G) ≤ γsR(G). Proof: Follows from Proposition 1. Remark 5. The converse of Proposition 2 is not true. To see this, consider the graph in Figure 1, the set W = {v2, v5} is a resolving set since the representation of each vertex in G, with respect to W is unique. These representations are as follows: rG(v1/W ) = (1, 2), rG(v2/W ) = (0, 1), rG(v3/W ), rG(v4/W ) = (2, 1) and rG(v5/W ) = (1, 0). However, none among the vertices in W strongly resolves the vertices v1 and v3. Thus, W is a resolving set but it is not a strong resolving set of G. In the same graph, it is easy to verify that the set {v1, v3} is a strong resolving set of G, hence a resolving set as well. Figure 1. A strong resolving set {v1, v3} of G G. Monsanto, P. Acal, H. Rara / Eur. J. Pure Appl. Math, 16 (1) (2023), 363-372 366 Proposition 3. [1] Let G be a connected graph of order n ≥ 2. Then, (i) γsR(Pn) = ⌈ n+1 3 ⌉ (ii) γsR(Kn) = n− 1 (iii) γsR(Cn) =  2 , if n = 3 n− 2 , if n > 3 and n is odd n 2 , if n > 3 and n is even. Remark 6. Let G be a connected graph (i) A set {u} ⊂ V (G) induces a dominated superclique of G. (ii) A clique ⟨C⟩ of G is a dominated superclique of G if NG[u] ̸= NG[v] for every pair of distinct vertices u, v ∈ C and V (G) \ C is a dominating set of G. (iii) Every dominated superclique of G is a superclique of G. Example 2. Let n be a positive integer. (i) ωDS(Kn) = ωS(Kn) = 1. (ii) If Pn = [v1, v2, . . . , vn] for n ≥ 4, then the dominated supercliques of Pn are induced from the singletons {vj} for all j = 1, 2, . . . , n and {vi, vi+1} for i = 2, 3, . . . , n− 2. (iii) The dominated supercliques of a cycle Cn for n ≥ 4, are ⟨{uj}⟩ and induced from {ui, uj} ⊆ V (Cn) where uiuj ∈ E(Cn). (iv) The dominated supercliques of a complete bipartite graph Km,n are the singleton sets {v} ⊂ V (Km,n) and n ̸= 1, m ̸= 1. (v) The maximum dominated supercliques of a complete bipartite graph Km,n are in- duced from the sets {xi, xj} ⊆ V (Km,n) where xixj ∈ E(Km,n) and n ̸= 1, m ̸= 1. Theorem 1. Let G be a connected graph of order n. Then ωDS(G) = 1 if and only if γ(G) = Kn or G = K1,n−1. Proof: Suppose ωDS(G) = 1. If n = 1 or n = 2, then G = Kn. If n = 3, then G = K3 or G = K1,2. Suppose n ≥ 4 and G ̸= Kn. Then there exist distinct vertices a and b of G such that dG(a, b) = 2. Let v ∈ NG(a) ∩ NG(b). Since {a, v} is a superclique and ωDS(G) = 1, |NG(a)| = 1. Similarly, |NG(b)| = 1. Suppose there exists y ∈ V (G)\NG(v). We may assume that dG(y, v) = 2. Let z ∈ NG(y) ∩ NG(v). Then {z, v} is a dominated superclique of G, contrary to the assumption that ωDS(G) = 1. Hence, x ∈ NG(v) for all x ∈ V (G) \ {v}. Also, for any distinct vertices x, y ∈ V (G) \ {v}, xy /∈ E(G). Therefore, G = ⟨v⟩+ ⋃ x∈V (G)\{v} ⟨x⟩ = K1,n−1. For the converse, suppose G = Kn or G = K1,n−1. Then, clearly, ωDS(G) = 1. G. Monsanto, P. Acal, H. Rara / Eur. J. Pure Appl. Math, 16 (1) (2023), 363-372 367 3. Strong Resolving Domination in the Lexicographic Product of Graphs Lemma 1. Let G = Kn for n > 1 and H a non-trivial connected graph with γ(H) ̸= 1. Then A×C ⊆ V (G[H]) is a superclique in G[H] if and only if A is a nonempty subset of V (G) and C is a superclique in H. Proof: Let G = Kn for n > 1 and H a non-trivial connected graph. Suppose A × C ⊆ V (G[H]) is a superclique in G[H]. Then A ⊆ V (G), C ⊆ V (H), A ̸= ∅ and C ̸= ∅. If |C| = 1, then we are done. Suppose |C| ≥ 2. Let x, y ∈ C, x ̸= y. Since A × C is a superclique in G[H], (v, x)(v, y) ∈ E(G[H]) for all v ∈ A and there exists (w, z) ∈ V (G[H]) \ (A × C) such that (w, z) ∈ NG[H]((v, x)) \ NG[H]((v, y)) or (w, z) ∈ NG[H]((v, y)) \ NG[H]((v, x)). Hence, xy ∈ E(H). Since G is complete, v = w and z ∈ NH(x) \ NH(y) or z ∈ NH(y) \ NH(x), where z ∈ V (H) \ C. Thus, C is a superclique in H. Conversely, suppose A ⊆ V (G), A ̸= ∅ and C is a superclique in H. Then A×C ⊆ V (G[H]) and A×C ̸= ∅. If |A×C| = 1, then we are done. Suppose |A×C| ≥ 2. Let (u, x), (v, y) ∈ A× C, (u, x) ̸= (v, y). Consider the following cases: Case 1. u = v Then x ̸= y. Since x, y ∈ C and C is a superclique in H, xy ∈ E(H) and there exists z ∈ V (H) \ C such that z ∈ NH(x) \ NH(y) or z ∈ NH(y) \ NH(x). Thus, (u, x)(v, y) ∈ E(G[H]), (u, z) /∈ A×C and (u, z) ∈ NG[H]((u, x))\NG[H]((v, y)) or (u, z) ∈ NG[H]((v, y))\ NG[H]((u, x)). Case 2. u ̸= v Subcase 2.1 x = y Since γ(H) ̸= 1, there exists z ∈ V (H) such that z /∈ NH(x). Also, since x ∈ C and z /∈ C, we have (u, z) /∈ A × C and (u, z) ∈ NG[H]((v, x)) \ NG[H]((u, x)). Since G is complete and u ̸= v, (u, x)(v, x) ∈ E(G[H]). Subcase 2.2 x ̸= y Since x, y ∈ C and C is a superclique in H, xy ∈ E(H) and there exists z /∈ C such that z ∈ NH(x)\NH(y) or z ∈ NH(y)\NH(x). Hence, (v, z) ∈ NG[H]((u, x))\NG[H]((v, y)) or (v, z) ∈ NG[H]((v, y)) \NG[H]((u, x)) for some (v, z) /∈ A× C. In any case, A× C is a superclique in G[H]. Theorem 2. [6] Let G and H be connected graphs. Then C ⊆ V (G+H) is a dominating set in G+H if and only if at least one of the following is true: (i) C ∩ V (G) is a dominating set in G. (ii) C ∩ V (H) is a dominating set in H. (iii) C ∩ V (G) ̸= ∅ and C ∩ V (H) ̸= ∅. Theorem 3. [6] Let G and H be connected graphs. Then C ⊆ V (G+H) is a dominating set in G[H] if and only if C = ⋃ x∈S ( {x} × Tx ) and either (i) S is a total dominating set in G or G. Monsanto, P. Acal, H. Rara / Eur. J. Pure Appl. Math, 16 (1) (2023), 363-372 368 (ii) S is a dominating set in G and Tx is a dominating set in H for every x ∈ S \NG(S). Lemma 2. Let G = Kn for n > 1 and H a non-trivial connected graph with γ(H) ̸= 1. Then A×C ⊆ V ( G[H] ) induces a dominated superclique in G[H] if and only if one of the following hold: (i) A ⊆ V (G) with 1 ≤ |A| ≤ n− 2 and ⟨C⟩ is a superclique of H. (ii) A ⊆ V (G) with |A| = n− 1 and ⟨V (H) \ C⟩ is a dominated superclique of H. Proof: Let G = Kn for n > 1 and H a nontrivial connected graph with γ(H) ̸= 1. Suppose A × C ⊆ V ( G[H] ) induces a dominated superclique of G[H]. By Remark 6(ii), V ( G[H] ) \ (A × C) = (V (G) \ A) × ( V (H) \ C ) is a dominating set of G[H]. Thus, by Theorem 2, V (G) \A is a total dominating set of G or V (G) \A is a dominating set of G and V (H) \ C is a dominating set of H. Since G = Kn, 1 ≤ |A| ≤ n − 2 or |A| = n − 1 and V (H) \ C is a dominating set of H. By Lemma 1, ⟨C⟩ is a superclique of H. Thus, (i) and (ii) hold. The converse follows immediately from Theorem 3 and Lemma 1. Proposition 4. [1] Let G be a non-trivial connected graph with diam(G) ≤ 2. Then W ⊆ V (G) \C is a strong resolving set of G if and only if C = ∅ or C is a superclique in G. In particular, sdim(G) = |V (G)| − ωS(G). Theorem 4. Let G = Kn for n > 1 and H a non-trivial connected graph with γ(H) ̸= 1. A subset S of V ( G[H] ) is a strong resolving dominating set of G[H] if and only if S = V (G[H]) \ (A× C) satisying either of the following: (i) A ⊆ V (G) with 1 ≤ |A| ≤ n− 2 and ⟨C⟩ is a superclique of H. (ii) A ⊆ V (G) with |A| = n− 1 and ⟨C⟩ is a dominated superclique of G[H]. Proof: Let S be a strong resolving dominating set of G[H]. Since diam(G[H]) = 2, by Proposition 4, S = V (G[H]) \ C0, where C0 = ∅ or ⟨C⟩ is a superclique of G[H]. By Lemma 2, (i) and (ii) follow. For the converse, suppose S = V (G[H]) \ (A× C) satisfying condition (i) or (ii). By Lemma 2, ⟨A× C⟩ is a dominated superclique of G[H]. Since diam(G[H]) = 2, S is a strong resolving dominating set of G[H] by Proposition 4. Lemma 3. Let G = Kn for n > 1 and H a non-trivial connected graph with γ(H) = 1. Then A×C ⊆ V (G[H]) is a superclique in G[H] if and only if A is a nonempty subset of V (G) and C is a superclique in H such that |A| = 1 whenever C ∩C∗ ̸= ∅ for some γ-set C∗ of H. Proof: Suppose A × C ⊆ V (G[H]) is a superclique in G[H]. Then A ⊆ V (G), C ⊆ V (H), A ̸= ∅ and C ̸= ∅. If |C| = 1, then C is a superclique in H. Suppose |C| ≥ 2 and x, y ∈ C where x ̸= y. Then (u, x)(u, y) ∈ A×C for all u ∈ A. Since (u, x) ̸= (u, y) and A× C is a superclique inG[H], (u, x)(u, y) ∈ E(G[H]) and (w, z) ∈ NG[H]((u, x))\NG[H]((u, y)) G. Monsanto, P. Acal, H. Rara / Eur. J. Pure Appl. Math, 16 (1) (2023), 363-372 369 or (w, z) ∈ NG[H]((u, y)) \ NG[H]((u, x)) for some (w, z) ∈ V (G[H]) \ (A × C). Thus, xy ∈ E(H), w = u and z ∈ NH(x)\NH(y) or z ∈ NH(y)\NH(x), where z /∈ C. Hence, C is a superclique in H. Let a ∈ C where {a} is a γ-set of H. Suppose |A| ≥ 2 and let u, v ∈ A, where u ̸= v. Then (u, a), (v, a) ∈ A × C, (u, a) ̸= (v, a). Since A × C is a superclique, there exists (w, b) ∈ V (G[H]) \ (A×C) such that (w, b) ∈ NG[H]((v, a)) \NG[H]((u, a)) or (w, b) ∈ NG[H]((u, a)) \NG[H]((v, a)). Since G is complete, w = u w = v and b /∈ NH(a). This is a contrdiction since {a} is a γ-set of H. Thus, |A| ≤ 1. Since A ̸= ∅, |A| = 1. Coversely, suppose A ⊆ V (G), A ̸= ∅ and C is a superclique in H such that |A| = 1 whenever C ∩ C∗ ̸= ∅ for some γ-set C∗ of H. Then A× C ⊆ V (G[H]) and A× C ̸= ∅. If |A × C| = 1, then we are done. Suppose |A × C| ≥ 2. Let (u, x), (v, y) ∈ A × C, (u, x) ̸= (v, y). Consider the following cases: Case 1. u = v Then x ̸= y. Since x, y ∈ C and C is a superclique in H, xy ∈ E(H) and there exists z ∈ NH(x) \ NH(y) or z ∈ NH(y) \ NH(x) for some z ∈ V (H) \ C. Hence, (u, x)(v, y) ∈ E(G[H]) and (u, z) ∈ NG[H]((u, x))\NG[H]((v, y)) or (u, z) ∈ NG[H]((v, y))\NG[H]((u, x)), where (u, z) ∈ V (G[H]) \ C. Case 2. u ̸= v Subcase 2.1. x = y Since u, v ∈ A, u ̸= v, |A| ≥ 2. By assumption, if x ∈ C, then {x} is not a γ-set of H. Thus, there exists z ∈ V (H) \ NH(x). Hence, (u, z) ∈ V (G[H]) \ (A × C) and (u, z) ∈ NG[H]((v, y)) \NG[H]((u, x)). Since G is complete, (u, x)(v, y) ∈ E(G[H]). Subcase 2.2. x ̸= y Since x, y ∈ C and C is a superclique in H, xy ∈ E(H) and z ∈ NH(x) \ NH(y) or z ∈ NH(y) \ NH(x) for some z ∈ V (H) \ C. Thus, (u, z) ∈ NG[H]((v, y)) \ NG[H]((u, x)) where (u, z) ∈ V (G[H]) \ (A×C). Since G is complete and u ̸= v, (u, x)(v, y) ∈ E(G[H]). In any case, A× C is a superclique in G[H]. The next result follows immediately from Lemma 3 and Theorem 2. Lemma 4. Let G = Kn for n > 1 and H a non-trivial connected graph with γ(H) = 1. Then A×C ⊆ V (G[H]) induces a dominated superclique of G[H] if and only if the following hold: (i) A ⊆ V (G) with 1 ≤ |A| ≤ n − 2 and ⟨C⟩ is a superclique of H such that |A| = 1 whenever C ∩ C∗ ̸= ∅ for some γ-set C∗ of H. (ii) A ⊆ V (G) with |A| = n− 1 and ⟨C⟩ is a dominated superclique of G[H]. Theorem 5. Let G = Kn for n > 1 and H a non-trivial connected graph with γ(H) = 1. A subset S of V ( G[H] ) is a strong resolving dominating set of G[H] if and only if the following hold: (i) A ⊆ V (G) with 1 ≤ |A| ≤ n − 2 and ⟨C⟩ is a superclique of H such that |A| = 1 whenever C ∩ C∗ ̸= ∅ for some γ-set C∗ of H. (ii) A ⊆ V (G) with |A| = n− 1 and ⟨C⟩ is a dominated superclique of G[H]. G. Monsanto, P. Acal, H. Rara / Eur. J. Pure Appl. Math, 16 (1) (2023), 363-372 370 Proof: The proof is similar to the proof of Theorem 4 and by using Lemma 4 Corollary 1. Let G = Kn for n > 1 and H a non-trivial connected graph of order m. Then γsR ( G[H] ) = nm− ωS ( G[H] ) . Proof: Let S be a strong metric basis of G[H]. Then S is a strong resolving set of G[H]. By Lemma 1, S = V (G[H]) \ (A × C), where A × C is a superclique of G[H]. Since S is a strong resolving dominating set, A×C is a maximum dominated superclique. Hence, γSR(G[H]) = |S| = |V (G[H])| − |A× C| = nm− ωDS(G[H]). The next result follows immediately from Lemma 1 and Corollary 1. Corollary 2. Let G = Kn for n > 1 and H a non-trivial connected graph of order m. Then γsR ( G[H] ) = nm− ωDS ( G[H] ) . Theorem 6. Let G be a non-trivial connected graph andH be non-trivial complete graph. A subset C = (⋃ x∈S { {x} × Tx })⋃ ⋃ x∈WG { {x} × (V (H) \ Tx) } of V ( G[H] ) where WG ⊆ S and Tx ⊆ V (H), ∀x ∈ S, is a strong resolving dominating set of G[H] if and only if (i) S = V (G). (ii) V (H) \ Tx is a superclique of H. (iii) WG is a strong resolving set of G. Proof: Let C be a strong resolving dominating set of G[H]. Suppose there exists x ∈ V (G) \ S. Let p, q ∈ V (H) with dH(p, q) = diam(H). Then (x, p) MMD (x, p) implying that (x, p) and (x, q) cannot be resolved by C since (x, p)(x, q) /∈ C. Hence, S = V (G) and (i) holds. Let u, v ∈ V (H) \ Tx, u ̸= v. Then (x, u), (x, v) /∈ C. Since C is a strong resolving dominating set of G[H], (x, u) and (x, v) can be strongly resolved by (y, w) ∈ C. If (x, u) ∈ IG[H][(x, v), (y, w)], then dG[H]((x, v), (x, u)) + dG[H]((x, u), (y, w)) = dG[H]((x, v), (y, w)) implying that x = y and w ∈ NH(u) \ NH(v). Similarly, if (x, v) ∈ IG[H][(x, u), (y, v)], then x = y and w ∈ NH(v) \NH(u). Hence, V (H) \ Tx is a superclique of H. Thus, (ii) holds. G. Monsanto, P. Acal, H. Rara / Eur. J. Pure Appl. Math, 16 (1) (2023), 363-372 371 Let p, q ∈ V (G) \ WG. Then (p, r), (q, r) /∈ C, where r /∈ Tp, Tq. Hence, there exists (s, t) ∈ C that resolves (p, r) and (q, r). If (p, r) ∈ IG[H][(q, r), (s, t)], then dG[H]((q, r), (p, r)) + dG[H]((p, r), (s, t)) = dG[H]((q, r), (s, t)) implying that r = t and p ∈ IG[q, s]. Similarly, if (q, r) ∈ IG[H][(p, r), (s, t)], then r = t and q ∈ IG[p, s]. Hence, s strongly resolves p and q. Therefore, WG is a strongly resolving dominating set of G and (iii) holds. For the converse, suppose C satisfies the given property. Let x = (x1, x2), y = (y1, y2) /∈ C, x ̸= y. Then consider the following cases: Case 1. x1 ∈ V (G) \WG and y1 ∈ V (G) \WG, x1 ̸= y1. By (iii), there exists z1 ∈ V (G) ∩WG that resolves x1 and y1. If x1 ∈ IG[y1, z1], then dG(y1, x1) + dG(x1, z1) = dG(y1, z1).(1) Choose z2 ∈ V (H) \ Tz1 . Clearly, (z1, z2) ∈ C. We claim that (z1, z2) strongly resolves x = (x1, x2) and y = (y1, y2) in G[H]. Using equation 1, we have dG[H]((y1, y2), (x1, x2)) + dG[H]((x1, x2), (z1, z2)) = dG[H]((y1, y2), (z1, z2)) implying that (x1, x2) ∈ IG[H][(y1, y2), (z1, z2)]. Similarly, if y1 ∈ IG[x1, z1], then (z1, z2) strongly resolves x = (x1, x2) and y = (y1, y2). Case 2. x1 = y1 Then x2 ̸= y2 since x ̸= y. It follows that Tx1 = Ty1 . Note that x = (x1, x2) and y = (y1, y2) /∈ C, then x2 /∈ Tx1 and y2 /∈ Ty1 . By (ii), there exists z ∈ Tx1 such that z ∈ NG(y2) \ NG(x2) or z ∈ NG(x2) \ NG(y2). Clearly, (x1, z) ∈ C and either (y1, y2) ∈ IG[H][(x1, x2), (x1, z)]. Thus, (x1, z) strongly resolves (x1, x2) and (y1, y2). The next result follows immediately from Theorem 6. Theorem 7. Let G be a non-trivial connected graph andH be non-trivial complete graph. A subset C = (⋃ x∈S { {x} × Tx })⋃ ⋃ x∈WG { {x} × (V (H) \ Tx) } of V ( G[H] ) where WG ⊆ S and Tx ⊆ V (H), ∀x ∈ S, is a strong resolving dominating set of G[H] if and only if (i) S = V (G). (ii) V (H) \ Tx is a dominated superclique of H. (iii) WG is a strong resolving dominating set of G. Acknowledgements This research is funded by the Commission on Higher Education (CHED) and Min- danao State University-Iligan Institute of Technology, Philippines. REFERENCES 372 References [1] P.L. Acal, G.B. Monsanto, and H.M. Rara. 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