EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 14, No. 3, 2021, 1015-1023 ISSN 1307-5543 – ejpam.com Published by New York Business Global On Resolving Hop Domination in Graphs Jerson S. Mohamad1,∗, Helen M. Rara2 1 Department of Mathematics and Statistics, College of Science and Mathematics, Western Mindanao State University, 7000 Zamboanga City, Philippines 2 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. A set S of vertices in a connected graph G is a resolving hop dominating set of G if S is a resolving set in G and for every vertex v ∈ V (G) \S there exists u ∈ S such that dG(u, v) = 2. The smallest cardinality of such a set S is called the resolving hop domination number of G. This paper presents the characterizations of the resolving hop dominating sets in the join, corona and lexicographic product of two graphs and determines the exact values of their corresponding resolving hop domination number. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Resolving hop dominating set, resolving hop domination number, join, corona, lexicographic product 1. Introduction Domination in graphs was first introduced by C. Berge in 1958 [3]. There are now many studies involving domination and its variations. Natarajan and Ayyaswamy [9] introduced and studied the concept of hop domination in graphs. Hop domination in graphs were also studied in [6, 10, 11]. Slater [12] introduced and studied the concept of resolving set. Resolving sets and resolving dominating sets were studied in [1, 2, 4, 7, 8]. This paper combines the idea of resolving and hop domination sets by introducing the concept of resolving hop domination in graphs. Resolving hop dominating sets in graphs can have real world applications. One possible application is in the minimization problem with specific conditions. For example, a company that makes electric cars with a smart navigation feature, wants to build the least number of charging stations in a given city, such that any car with a low charge, from any area, can reach a charging station before ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v14i3.4055 Email addresses: mohamadjerson@gmail.com (J. Mohamad), helen.rara@g.msuiit.edu.ph (H. Rara) http://www.ejpam.com 1015 c© 2021 EJPAM All rights reserved. J. Mohamad, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 1015-1023 1016 running out of either or both its remaining charge and auxiliary power. However, fully charging a car takes time, which at some point can overwhelm a station’s capacity. To reduce the chance of this from happening, the company may require, as much as possible, that no such two cars from different areas arrive at the same station at about the same time. Assuming that both the remaining low charge and auxiliary power can each cover the same travel distance d, the graph-theoretic model for this scenario could be that vertices represents the areas, and adjacency of vertices represent a connected route of distance d. Resolving hop domination in graphs can be used to determine the minimum number of charging stations and where to build them in such a manner that cars from different areas have relatively distinct distances from these stations. In this study, we only consider graphs that are finite, simple, undirected and connected. Readers are referred to [5] for elementary Graph Theory concepts. Let G = ( V (G), E(G) ) be a graph. NG(v) = {u ∈ V (G) : uv ∈ E(G)} is a neighbor- hood of v. An element u ∈ NG(v) is called a neighbor of v. NG[v] = NG(v) ∪ {v} is a closed neighborhood of v. The degree of v, denoted by degG(v), is equal to |NG(v)|. For S ⊆ V (G), NG(S) = ⋃ v∈S NG(v) and NG[S] = ⋃ v∈S NG[v]. A connected graph G is said to be point determining if distinct vertices have distinct neighborhoods, that is, NG(a) 6= NG(b) whenever a, b ∈ V (G) and a 6= b. A connected graph G of order n ≥ 3 is totally point determining if for any two distinct vertices u and v of G, NG(u) 6= NG(v) and NG[u] 6= NG[v]. A vertex x of a graph G is said to resolve two vertices u and v of G if dG(x, u) 6= dG(x, v). For an ordered set W = {x1, ..., xk} ⊆ V (G) and a vertex v in G, the k-vector rG(v/W ) = (dG(v, x1), dG(v, x2), ..., dG(v, xk)) is called the representation of v with respect to W . The set W is a resolving set for G if and only if no two vertices of G have the same representation with respect to W . The metric dimension of G, denoted by dim(G), is the minimum cardinality over all resolving sets of G. A resolving set of cardinality dim(G) is called basis. A set S ⊆ V (G) of vertices of G is a dominating set if every u ∈ V (G) \ S is adjacent to at least one vertex v ∈ S. 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}. A set S ⊆ V (G) is a hop dominating set of G if for every v ∈ V (G)\S, there exists u ∈ S such that dG(u, v) = 2. The minimum cardinality of a hop dominating set of G, denoted by γh(G), is called the hop domination number of G. Any hop dominating set with cardinality equal to γh(G) is called a γh-set. A vertex v in G is a hop neighbor of vertex u in G if dG(u, v) = 2. The set NG(u, 2) = {v ∈ V (G) : dG(v, u) = 2} is called the open hop neighborhood of u. The closed hop neighborhood of u in G is given by NG[u, 2] = NG(u, 2)∪ {u}. The open hop neighborhood of X ⊆ V (G) is the set NG(X, 2) = ⋃ u∈X NG(u, 2). The closed hop neighborhood of X in G is the set NG[X, 2] = NG(X, 2) ∪X. J. Mohamad, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 1015-1023 1017 A set S ⊆ V (G) is a locating set of G if for every two distinct vertices u and v of V (G)\S, NG(u) ∩ S 6= NG(v) ∩ S. The locating number of G, denoted by ln(G), is the smallest cardinality of a locating set of G. A locating set of G of cardinality ln(G) is referred to as a ln-set of G. A set S ⊆ V (G) is a strictly locating set of G if it is a locating set of G and NG(u)∩S 6= S for all u ∈ V (G)\S. The strictly locating number of G, denoted by sln(G), is the smallest cardinality of a strictly locating set of G. A strictly locating set of G of cardinality sln(G) is referred to as a sln-set of G. A set S ⊆ V (G) is a resolving hop dominating set of G if S is both a resolving set and a hop dominating set. The minimum cardinality of a resolving hop dominating set of G, denoted by γRh(G), is called the resolving hop domination number of G. Any resolving hop dominating set with cardinality equal to γRh(G) is called a γRh-set. 2. Preliminary Results Remark 1. For any connected graph G of order n ≥ 2, 2 ≤ γRh(G) ≤ n. Moreover, γRh(P2) = 2 and γRh(Kn) = n. Proposition 1. For any connected graph G of order n ≥ 2. Then, γRh(G) = n if and only if G = Kn. Proof: If G = Kn, then γRh(G) = n. Suppose γRh(G) = n and G 6= Kn. Then there exists x, y ∈ V (G) such that d(x, y) = 2. Let S = V (G) \ {y}). Then S is a resolving hop dominating set of G. Hence, γRh(G) ≤ |S| = n− 1, a contradiction. Remark 2. Let G be a connected graph and S ⊆ V (G). Then for any two distinct vertices x, y ∈ V (G) \ S with NG(x, 2) ∩ S 6= NG(y, 2) ∩ S, we have rG(x/S) 6= rG(y/S). Remark 3. Every resolving hop dominating set of a connected graph G is a resolving set of G. Thus, dim(G) ≤ γRh(G). Proposition 2. Let G be a connected graph of order 4. Then γRh(G) = 2 if and only if G = C4 or G = P4. Proof: If G = C4 or P4, then γRh(G) = 2. Suppose that γRh(G) = 2. Let W = {x1, x2} be a γRh-set of G. Since W is a hop dominating set, possible representations of distinct vertices u, v ∈ V (G) \W are (1,2), (2,1) or (2,2). Clearly (2,2) cannot be a representation of vertex u or v since G is of order 4. Thus we consider the following cases: Case 1. rG(u/W ) = (1, 2) and rG(v/W ) = (2, 1) Case 2. rG(u/W ) = (2, 1) and rG(v/W ) = (1, 2) For case 1, ux1, vx2 ∈ E(G) and either x1x2 ∈ E(G) or uv ∈ E(G) or both x1x2, uv ∈ E(G). Hence, G = [u, x1, x2, v] or G = [x1, u, v, x2] or G = [u, x1, x2, v, u]. Thus, G is either a path P4 or a cycle C4. Similarly, if case 2 holds, then G = P4 or G = C4. Proposition 3. Let n be a positive number. (i) For a path Pn on n vertices, n > 1 J. Mohamad, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 1015-1023 1018 γRh(Pn) =  2 if n = 2, 3, 4, 5 2r if n = 6r 2r + 1 if n = 6r + 1 2r + 2 if n = 6r + s; 2 ≤ s ≤ 5. (ii) For a cycle Cn of length n, γRh(Cn) =  2 if n = 4, 5 2r if n = 6r 2r + 1 if n = 6r + 1 2r + 2 if n = 6r + s; 2 ≤ s ≤ 5. 3. On Resolving Hop Domination in the Join of Graphs The join of two graphs G and H is the graph G + H with vertex set V (G + H) = V (G) • ∪ V (H) and edge set E(G + H) = E(G) • ∪ E(H) ∪ {uv : u ∈ V (G), v ∈ V (H)}. Theorem 1. [7, 8] Let G and H be non-trivial connected graphs. A set W ⊆ V (G+H) is a resolving set of G+H if and only if W = WG∪WH where WG ⊆ V (G) and WH ⊆ V (H) are locating sets of G and H, respectively, where WG or WH is a strictly locating set. Theorem 2. Let G and H be non-trivial connected graphs. A set W ⊆ V (G + H) is a resolving hop dominating set of G+H if and only if W = WG ∪WH where WG and WH are strictly locating sets of G and H, respectively. Proof: Suppose that W is a resolving hop dominating set of G + H. Then W is a resolving set of G + H. By Theorem 1, W = WG ∪ WH where WG ⊆ V (G) and WH ⊆ V (H) are locating sets of G and H, respectively. Suppose WG or WH is not strictly locating set, say WG is not strictly locating. Then there exists v ∈ V (G) \WG such that NG(v) ∩WG = WG. Hence, v ∈ V (G+H) \W and dG+H(v, w) = 1 for all w ∈ W . This contradicts the assumption that W is a hop dominating set of G+H. Similarly, if WH is not strictly locating, then a contradiction follows. Hence, WG and WH are both strictly locating. For the converse, suppose that W = WG ∪WH where WG ⊆ V (G), WH ⊆ V (H) and both WG and WH are strictly locating sets of G and H, respectively. Since WG and WH are locating sets by Theorem 1, W is a resolving set of G+H. Let v ∈ V (G+H) \W . If v ∈ V (G), then v /∈WG. Since WG is strictly locating there exists u ∈WG \NG(v). Hence, dG+H(v, u) = 2. Similarly, if v ∈ V (H), then v /∈ WH and there exists w ∈ WG \NG(v). Thus, dG(v, w) = 2. Therefore W is a hop dominating set of G+H. Accordingly, W is a resolving hop dominating set of G+H. The next result follows immediately from Theorem 2. Corollary 1. Let G and H be non-trivial connected graphs. Then γRh(G+H) = sln(G) + sln(H). J. Mohamad, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 1015-1023 1019 4. On Resolving Hop Domination in the Corona of Graphs The corona of two graphs G and H, denoted by G◦H, is the graph obtained by taking one copy of G of order n and n copies of H, and then joining every vertex of the ith copy of H to the ith vertex of G. For v ∈ V (G), denote by Hv the copy of H whose vertices are attached one by one to the vertex v. Subsequently, denote by v+Hv the subgraph of the corona G ◦H corresponding to the join 〈{v}〉+Hv, v ∈ V (G). Theorem 3. [7, 8] Let G and H be non-trivial connected graphs. Then W ⊆ V (G◦H) is a resolving set of G ◦H if and only if W ∩ V (Hv) 6= ∅ for all v ∈ V (G) and W = A ∪B, where A ⊆ V (G), and B = ∪{Bv : v ∈ V (G) and Bv is a locating set of Hv}. Theorem 4. Let G and H be non-trivial connected graphs. Then W ⊆ V (G ◦ H) is a resolving hop dominating set of G ◦H if and only if W ∩ V (Hv) 6= ∅ for every v ∈ V (G) and W = A ∪B ∪D where A ⊆ V (G), B = ∪ {Bv : v ∈ V (G) ∩NG(A) and Bv is a locating set of Hv} and D = ∪ {Du : u ∈ V (G) \NG(A) and Du is a strictly locating set of Hu} . Proof: Suppose W is a resolving hop dominating set of G ◦H. Then by Theorem 3, W ∩ V (Hv) 6= ∅ for every v ∈ V (G). Let A = W ∩ V (G), Bv =W ∩ V (Hv) for each v ∈ V (G) ∩NG(A) and Du =W ∩ V (Hu) for each u ∈ V (G) \NG(A). Set B = ⋃ Bv and D = ⋃ Du. Then W = A ∪ B ∪ D where A ⊆ V (G). By Theorem 3, Bv and Du are locating sets of Hv and Hu, respectively. Let x ∈ V (Hu) \Du. Then x ∈ V (G ◦H) \W . Since W is a hop dominating set of G ◦H, there exists y ∈ W such that dG◦H(x, y) = 2. Since u ∈ V (G) \NG(A), y ∈ V (Hu)∩Du. Hence, y ∈ Du \NHu(x). Thus, NHu(x) ∩Du 6= Du, showing that Du is strictly locating. For the converse, suppose that W ∩V (Hv) 6= ∅ for every v ∈ V (G) and W = A∪B∪D where A, B and D satisfy the given conditions. Let x ∈ V (G ◦H) \W and let v ∈ V (G) such that x ∈ V (v + Hv). Suppose x = v. Then v /∈ A. Let u ∈ V (G) ∩ NG(v). Since W ∩ V (Hu) 6= ∅, there exists y ∈ W ∩ V (Hu) and dG◦H(x, y) = 2. Suppose x 6= v. If v ∈ NG(A), then there exists z ∈ A ∩ NG(v). Hence, z ∈ W and dG◦H(x, z) = 2. Suppose v /∈ NG(A). Then x ∈ V (Hv) \ Dv. Since Dv is strictly locating there exists y ∈ Dv \ NHv(x). Thus, y ∈ W and dG◦H(x, y) = 2. This shows that W is a hop dominating set of G ◦H. Since Bv or Dv is a locating set for each v ∈ V (G), by Theorem 3, W is a resolving set of G ◦H. Accordingly, W is a resolving hop dominating set of G ◦H. Corollary 2. Let G be a non-trivial graph of order m and H be any graph. Then the following statements hold. (i) γRh(G ◦H) ≤ m(1 + ln(H)). J. Mohamad, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 1015-1023 1020 (ii) If sln(H) = ln(H), then γRh(G ◦H) = m(sln(H)). Proof: (i) Set A1 = V (G) and let Bv be an ln-set of H for each v ∈ V (G). Then W1 = A1 ∪ (⋃ v∈V (G)Bv ) is a resolving hop dominating set of G ◦ H by Theorem 4, Hence, γRh(G ◦H) ≤ |W1| = |V (G)|+ |V (G)||Bv| = m(1 + ln(H)). (ii) Suppose that sln(H) = ln(H). Set A2 = ∅ and let Du be an sln-set of H for each u ∈ V (G). Then W2 = A2 ∪ (⋃ u∈V (G)Du ) is a resolving hop dominating set of G ◦H by Theorem 4. Thus, γRh(G ◦H) ≤ |W2| = |A2|+ |V (G)||Du| = m(sln(H)). Now, let W0 = A0 ∪ (⋃ u∈V (G)\S0 Bv ) ∪ (⋃ u∈S0 Du ) be a γRh-set of G ◦H. By Theorem 4, A0 ⊆ V (G), S0 = {x ∈ V (G) : x /∈ NG(A0)}, Bv is a locating set of Hv for each v ∈ V (G) \ S0 and Du is a strict locating set of Hu for each u ∈ S0. Thus, γRh(G ◦H) = |W0| = |A0|+ |V (G) \ S0| |Bv|+ |S0| |Du| ≥ |V (G) \ S0| ln(H) + |S0| sln(H) = (|V (G)| − |S0|) sln(H) + |S0| sln(H) = m(sln(H)). Therefore, γRh(G ◦H) = m(sln(H)). 5. On Resolving Hop Domination in the Lexicographic Product of Graphs The lexicographic product of two graphs G and H, denoted by G[H], is the graph with vertex-set V (G[H]) = V (G) × V (H) such that (u1, u2)(v1, v2) ∈ E(G[H]) if either u1v1 ∈ E(G) or u1 = v1 and u2v2 ∈ E(H). Theorem 5. [7, 8] Let G and H be non-trivial connected graphs with4(H) ≤ |V (H)|−2. Then W = ⋃ x∈S [{x}×Tx], where S ⊆ V (G) and Tx ⊆ V (H) for each x ∈ S, is a resolving set of G[H] if and only if (i) S = V (G); (ii) Tx is a locating set for every x ∈ V (G); (iii) Tx or Ty is a strictly locating set of H whenever x and y are adjacent vertices of G with NG[x] = NG[y]; and (iv) Tx or Ty is a (locating) dominating set of H whenever x and y are nonadjacent vertices of G with NG(x) = NG(y). J. Mohamad, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 1015-1023 1021 Theorem 6. Let G and H be non-trivial connected graphs with 4(H) ≤ |V (H)| − 2. Then W = ⋃ x∈S [{x}×Tx], where S ⊆ V (G) and Tx ⊆ V (H) for each x ∈ S, is a resolving hop dominating set of G[H] if and only if (i) S = V (G); (ii) Tx is a locating set for every x ∈ V (G); (iii) Tx or Ty is a strictly locating set of H whenever x and y are adjacent vertices of G with NG[x] = NG[y]; (iv) Tx or Ty is a (locating) dominating set of H whenever x and y are nonadjacent vertices of G with NG(x) = NG(y); and (v) Tx is a strictly locating set of H for each x ∈ S \NG(S, 2). Proof: Suppose W is a resolving hop dominating set of G[H]. Then W is a resolving set. By Theorem 5, (i) to (iv) hold. Let x ∈ S \ NG(S, 2). If Tx = V (H), then Tx is a strictly locating set of H. So suppose that Tx 6= V (H) and let a ∈ V (H) \ Tx. Since W is hop dominating and (x, a) /∈W , there exists (y, b) ∈W such that dG[H]((x, a), (y, b)) = 2. The condition x ∈ S \ N(S, 2) would imply that y = x and b ∈ (V (H) \ NH(a)) ∩ Tx. Hence, Tx is a strictly locating set of H. Conversely, supose that W satisfies (i) to (v). By Theorem 5, W is a resolving set. Let (x, a) ∈ V (G[H]) \W . Since S = V (G), a ∈ V (H) \ Tx. If x ∈ NG(S, 2), then there exists z ∈ NG(x, 2). Let b ∈ Tz. Then (z, b) ∈ W ∩NG[H]((x, a), 2). Suppose x ∈ S \NG(S, 2). By (v), Tx is a strictly locating set of H. Hence, there exists p ∈ [V (H) \ NH(a)] ∩ Tx. This implies that (x, p) ∈ W ∩NG[H]((x, a), 2). Therefore, W is a hop dominating set of G[H]. Accordingly, W is a resolving hop dominating set of G[H]. Corollary 3. Let G and H be non-trivial connected graphs. Then γRh(G[H]) ≤ |V (G)|sln(H). If G is totally point determining graph and γ(G) 6= 1, then γRh(G[H]) = |V (G)|ln(H). Proof: Let S = V (G) and let Tx be an sln-set of H. By Theorem 6, W = ⋃ x∈S [{x} × Tx] is a resolving hop dominating set of G[H]. It follows that γRh(G[H]) ≤ |W | = |V (G)||Tx| = |V (G)|sln(H). Next, supppose that G is totally point determining graph and γ(G) 6= 1. Let S = V (G) and let Rx be an ln-set of H for each x ∈ S. Since γ(G) 6= 1, x ∈ NG(S, 2) for each x ∈ S. By Theorem 6, W = ⋃ x∈S [{x}×Rx] is a resolving hop dominating set of G[H]. It follows that γRh(G[H]) ≤ |W | = |V (G)||Rx| = |V (G)|ln(H). J. Mohamad, H. Rara / Eur. J. Pure Appl. Math, 14 (3) (2021), 1015-1023 1022 Now, if W0 = ⋃ x∈S0 [{x} × Tx] is a γRh-set of G[H], then S0 = V (G) and Tx is a locating set of H for each x ∈ V (G) by Theorem 6. Hence, γRh(G[H]) = |W0| = |V (G)||Tx| ≥ |V (G)|ln(H). Therefore, γRh(G[H]) = |V (G)|ln(H). Corollary 4. Let G and H be non-trivial connected graphs. If G is totally point deter- mining and γ(G) = 1, then γRh(G[H]) = sln(H) + (|V (G)| − 1)ln(H). Proof: Let DG = {v ∈ V (G) : {v} is a dominating set of G}. Since G is totally point determining, it follows that |DG| = 1. Set S = V (G). Let Tv be an sln-set of H for v ∈ DG and let Tx be an ln-set of H for each x ∈ V (G) \ {v}. Then by Theorem 6, W = ⋃ x∈S\{v}[{x} × Tx] ∪ ({v} × Tv) is a resolving hop dominating set of G[H]. Hence, γRh(G[H]) ≤ |W | = (|V (G)| − 1)ln(H) + sln(H). Supose now that W ∗ = ⋃ x∈S∗ [{x} × Rx] is a γRh-set of G[H]. Then there exists a unique vertex v such that {v} is a dominating set of G. By Theorem 6, S∗ = V (G), Rv is a strictly locating set of H and Rx is a locating set of H for each x ∈ V (G) \ {v}. Thus, γRh(G[H]) =|W ∗| =|Rv|+ ∑ x∈S∗\{v} |Rx| ≥sln(H) + (|V (G)| − 1)ln(H). Therefore, γRh(G[H]) = sln(H) + (|V (G)| − 1)ln(H). Corollary 5. Let H be a non-trivial connected graph and let n ≥ 2 be an integer. Then γRh(Kn[H]) = n(sln(H)). Proof: Let G = Kn. Then v is a dominating vertex of G for each v ∈ V (G). Thus, if W0 = ⋃ x∈S0 ({x} × Tx) is a γRh-set of G[H], then S0 = V (G) and Tx is an sln-set of H for each x ∈ S0, by Theorem 6. Hence, γRh(Kn[H]) = |W0| = |V (Kn)|sln(H) = n(sln(H)). 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