EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 6074 ISSN 1307-5543 – ejpam.com Published by New York Business Global 2-Step Movability of Hop Dominating Sets in Graphs Roger L. Estrella1,∗, Gina M. Malacas1,2, Sergio R. Canoy, Jr.1,2 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- PRISM, MSU-Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. Let G be an undirected connected graph with vertex and edge sets V (G) and E(G), respectively. A hop dominating set S in G is 2-step movable hop dominating if for each v ∈ S, S \ {v} or [S \ {v}] ∪ {w} for some w ∈ [V (G) \ S] ∩ N2 G(v) is a hop dominating set in G. The minimum cardinality of a 2-step movable hop dominating set in G, denoted by γ2 mh(G), is called the 2-step movable hop domination number of G. In this paper, we characterize those graphs which admit a 2-step movable hop dominating set. We give bounds on the 2-step movable hop domination number and give necessary and sufficient conditions for those graphs that attain these bounds. We also characterize the 2-step movable hop dominating sets in the shadow graph and determine the 2-step movable hop domination numbers of the shadow graph and complementary prism. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: hop domination, 2-step movable hop dominating, 2-step movable hop domination number 1. Introduction Movability of dominating sets was introduced and studied by Blair et al. in [1]. Ap- parently, this is a variation on dominating sets in which vertices in a dominating set are either removed or replaced. A motivation of this study can be seen, for example, in a net- work with sensors located at some nodes or vertices to serve their purpose (e.g. monitor activities in the network). It may happen that malfunctioning of some of these sensors occurs due to loss of battery supply or destruction by natural calamities. When such a case happens, a new nearby location for a sensor can be chosen appropriately so as to preserve the desired activity, connectivity, or security these sensors are purposely designed ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.6074 Email addresses: roger.estrella@g.msuiit.edu.ph (R. Estrella) gina.malacas@g.msuiit.edu.ph (G. Malacas) sergio.canoy@g.msuiit.edu.ph (S. Canoy) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) R. Estrella, Gina M. Malacas , S. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6074 2 of 15 in the network. Movability of different types of dominating sets had been considered in [2], [3], [4], [5], [6], and [7]. Hop domination, a kind of domination introduced by Natarajan et al. in [8], has also gained popularity and interest among the researchers in the field. Through the years, a great number of variants of hop domination have already been considered and studied (see, for example, [9], [10], [11], [12], [13], [14], [15], [16], [17], [18], and [19]). Since hop domination and domination have similar applications in networks, it is also worthwhile to consider movability of hop dominating sets in graphs. In this paper, we introduce 2-step movability of hop dominating sets. Since these types of sets need not be present in some graphs, we characterize those graphs that admit such sets. We also give bounds on the 2-step movable hop domination number and determine the values of the parameter in some known graphs including the shadow graph and complementary prism. 2. Terminology and Notation Let G = V (G), E(G)) be an undirected graph. For any two vertices u and v of G, the distance dG(u, v) is the length of a shortest path joining u and v. Any u-v path of length dG(u, v) is called a u-v geodesic. The interval IG [u, v] consists of u, v, and all vertices lying on a u-v geodesic. The interval IG(u, v) = IG [u, v] \ {u, v}. Vertices u and v are adjacent (or neighbors) if uv ∈ E(G). The set of neighbors of a vertex u in G, denoted by NG(u), is called the open neighborhood of u. The closed neighborhood of u is the set NG[u] = NG(u) ∪ {u}. If X ⊆ V (G), the open neighborhood of X is the set NG(X) = ⋃ u∈X NG(u). The closed neighborhood of X is the set NG[X] = NG(X) ∪X. A set D ⊆ V (G) is a dominating set (resp. total dominating set) of G if for every v ∈ V (G) \ D (resp. v ∈ V (G)), there exists u ∈ D such that uv ∈ E(G), that is, NG[D] = V (G) (resp. NG(D) = V (G)). The domination number (resp. total domination number) of G, denoted by γ(G) (resp. γt(G)), is the minimum cardinality of a dominating (resp. total dominating) set in G. Any dominating (resp. total dominating) set in G with cardinality γ(G) (resp. γt(G)), is called a γ-set (resp. γt-set) in G. If γ(G) = 1 and {v} is a dominating set in G, then we call v a dominating vertex in G. A vertex v in G is a hop neighbor of vertex u in G if dG(u, v) = 2. The set N2 G(u) = {v ∈ V (G) : dG(v, u) = 2} is called the open hop neighborhood of u. The closed hop neighborhood of u is given by N2 G[u] = N2 G(u) ∪ {u}. The open hop neighborhood of X ⊆ V (G) is the set N2 G(X) = ⋃ u∈X N2 G(u). The closed hop neighborhood of X is the set N2 G[X] = N2 G(X) ∪X. A set S ⊆ V (G) is a hop dominating set in G if N2 G[S] = V (G), that is, for every v ∈ V (G)\S, there exists u ∈ S such that dG(u, v) = 2. The minimum cardinality among all hop dominating sets in 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 hop dominating set S is 2-step movable hop dominating if for each v ∈ S, S \ {v} is hop dominating or there exists w ∈ (V (G) \ S) ∩ N2 G(v) such that (S \ {v}) ∪ {w} is hop R. Estrella, Gina M. Malacas , S. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6074 3 of 15 dominating in G. The minimum cardinality among all 2-step movable hop dominating sets in G, denoted by γ2mh(G), is called the 2-step movable hop domination number of G. Any 2-step movable hop dominating set with cardinality equal to γ2mh(G) is called a γ2mh-set. The shadow graph D2(G) of graph G is constructed by taking two copies of G, say G1 and G2, and then joining each vertex u ∈ V (G1) to the neighbors of its corresponding vertex u′ ∈ V (G2). For a graph G, the complementary prism GG, is the graph formed from the disjoint union of G and its complement G by adding a perfect matching between corresponding vertices of G and G. For each v ∈ V (G), let v denote the vertex in G corresponding to v. In simple terms, the graph GG is form from G∪G by adding the edge vv for every vertex v ∈ V (G). 3. Results Our first result characterizes those connected graphs which admit a 2-step movable hop dominating set. Theorem 1. Let G be a connected graph. Then G admits a 2-step movable hop dominating set if and only if γ(G) ̸= 1. Proof. Suppose G admits a 2-step movable hop dominating set, say S. Suppose γ(G) = 1, say v is a dominating vertex in G. Then v ∈ S because S is a hop dominating set in G. Since every hop dominating set contains all dominating vertices of G where v is one of them, it follows that S \ {v} is not a hop dominating set. Also, since N2 G(v) = ∅, there exists no w ∈ [V (G) \ S] ∩N2 G(v) such that [S \ {v}] ∪ {w} is a hop dominating set in G. This implies that S is not a 2-step movable hop dominating set, contrary to our assumption. Thus, γ(G) ̸= 1. For the converse, suppose that γ(G) ̸= 1. Let x ∈ V (G). Since x is not a dominating vertex of G, there exists y ∈ N2 G(x). It follows that V (G) \ {x} is a hop dominating set in G. This implies that V (G) is a 2-step movable hop dominating set in G. Remark 1. Let G1, G2, . . . , Gk be the components of a graph G. Then S is a hop dom- inating set in G if and only if Sj = S ∩ V (Gj) is a hop dominating set in Gj for each j ∈ [k] = {1, 2, · · · , k}. Moreover, γh(G) = ∑k j=1 γh(Gj). Theorem 2. Let G1, G2, . . . , Gk be the components of G. Then G admits a 2-step movable hop dominating set if and only if γ(Gj) ̸= 1 for every j ∈ [k] = {1, 2, · · · , k}. In this case, γ2mh(G) = ∑k j=1 γ 2 mh(Gj). Proof. Suppose G admits a 2-step movable hop dominating set, say S. Let Sj = S ∩ V (Gj) for each j ∈ [k]. By Remark 1, each set Sj is a hop dominating set in Gj . For an arbitrary j ∈ [k], let x ∈ Sj . Then x ∈ S. Since S a 2-step movable hop dominating set in G, S \{x} or [S \{x}]∪{y} for some y ∈ [V (G)\S]∩N2 G(x), is a hop dominating set in R. Estrella, Gina M. Malacas , S. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6074 4 of 15 G. If S \ {x} is a hop dominating set in G, then Sj \ {x} is a hop dominating set in Gj by Remark 1. If (S \{x})∪{y} is a hop dominating set in G for some y ∈ (V (G)\S)∩N2 G(x), then y ∈ (V (Gj)\Sj)∩N2 Gj (x) and (Sj \{x})∪{y} is a hop dominating set in Gj . Hence, Sj is a 2-step movable hop dominating set in Gj . By Theorem 1, γ(Gj) ̸= 1. Note that if, in particular, S is a γ2mh-set in G, then γ2mh(G) = |S| = | ∪j∈[k] Sj | ≥ ∑ j∈[k] γ2mh(Gj). Conversely, suppose γ(Gj) ̸= 1 for each j ∈ [k]. Then each Gj admits a 2-step movable hop dominating set Dj by Theorem 1. Clearly, D = ∪j∈[k]Dj is a 2-step movable hop dominating set in G. Moreover, if Dj is a γ2mh-set in Gj for each j ∈ [k], then we have γ2mh(G) ≤ |D| = | ∪j∈[k] Dj | = ∑ j∈[k] γ2mh(Gj). Therefore, the assertion holds. Corollary 1. If G admits a 2-step movable hop dominating set, then |V (G)| ≥ 4. Proof. Suppose G admits a 2-step movable hop dominating set. Let G′ be a component of G. Then γ(G′) ̸= 1 by Theorem 2. It follows that G′ /∈ {K1,K2,K3, P3}. Thus, 4 ≤ |V (G′)| ≤ |V (G)|. Throughout this section, unless specified, it is assumed that every component of a graph does not have a dominating vertex, i.e., every graph admits a 2-step movable hop dominating set. Remark 2. Let G be any graph. Then γh(G) ≤ γ2mh(G). Moreover, for each positive integer n, there exists a connected graph G such that γ2mh(G)−γh(G) = n. In other words, the difference γ2mh(G)− γh(G) can be made arbitrarily large. Note that for a graph that admits a 2-step movable hop dominating set, every 2-step movable hop dominating set is hop dominating. Thus, γh(G) ≤ γ2mh(G). To see that the second part of Remark 2 holds, let n be a positive integer and con- sider the graph G in Figure 1 obtained from Kn+2 by adding the edges ab and bx1, where V (Kn+1) = {x1, x2, · · · , xn+2}. Clearly, {a, b} is a γh-set inG. Hence, γh(G) = 2. Let S be a γ2mh-set in G. Suppose b /∈ S. If x1 ∈ S, then S = {x1, x2, · · · , xn+2} is a γ2mh-set in G. If x1 /∈ S, then S = {a, x2, · · · , xn+2}. Suppose b ∈ S. Suppose |S∩{x2, · · · , xn+2}| ≤ n−1. We may assume x2, x3 /∈ S. Since for each j ∈ {2, 3} the set (S \ {b}) ∪ {xj} is not hop dominating, it follows that S is not 2-step hop dominating, a contradiction. R. Estrella, Gina M. Malacas , S. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6074 5 of 15 ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... .................................... ............................................................................................................................................ ............................................................................................................... .................................... ................................................................................................................................................... .................................... .................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... .................................... .................................... ..................... .................... .................... .................... .................... .................... .................... .................... .................... .................... ........ .................................... .................................... ......................................................................................................................................................................................................................................................................................................................... ..................................................................................................................................................................................................................................................... .................................... .................................... ............................................................................................................................................ ..................................................................................................................................................................................................................................................... .................................... .................................... ..................................................................................................................................................................................................... .................................... ................................................................................................................................................................................ .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ........... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ...... .................................... .................................... ..................... .................... .................... .................... .................... .................... .................... .................... .................... .................... ........ .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ................................................................................................................ ................................................................................................................ .................................... . . . x1 x2x3 x4 x5 xn+2 b a Figure 1: Graph G with γ2 mh(G)− γh(G) = n Thus, |S ∩ {x2, · · · , xn+2}| = n. Moreover, |S ∩ {a, x1}| = 1. Therefore, in any case, γ2mh(G) = n+ 2. Accordingly, γ2mh(G)− γh(G) = n. Theorem 3. Let G be any graph of order n ≥ 4 and let S be a hop dominating set in G. Then S is a 2-step movable hop dominating set in G if and only if for each v ∈ S such that S \ {v} is not hop dominating, it holds that there exists w ∈ (V (G) \S)∩N2 G(v) such that ephn(v;S) ⊆ N2 G[w]. Proof. Suppose S is a 2-step movable hop dominating set in G. Let v ∈ S such that S \ {v} is not hop dominating. Since S is 2-step movable hop dominating, there exists w ∈ (V (G) \ S) ∩ N2 G(v) such that Sw = (S \ {v}) ∪ {w} is a hop dominating set in G. Now let z ∈ ephn(v;S). Then N2 G(z) ∩ S = {v}. Since Sw is a hop dominating set in G, it follows that z ∈ N2 G[w]. Thus, ephn(v;S) ⊆ N2 G[w]. For the converse, suppose that the given property holds. Let v ∈ S such that S \ {v} is not hop dominating. Then by assumption, there exists w ∈ (V (G) \ S) ∩ N2 G(v) such that ephn(v;S) ⊆ N2 G[w]. Let Sw = (S \ {v})∪ {w} and let x ∈ V (G) \Sw. If x = v, then x ∈ N2 G(w). Suppose x ̸= v. If x /∈ ephn(v;S), then there exists u ∈ (S\{v})∩N2 G(x) since S is a hop dominating set in G. Hence, x ∈ N2 G(Sw). Next, suppose that x ∈ ephn(v;S). Then x ∈ N2 G(w) because ephn(v;S) ⊆ N2 G[w] and x ̸= w. Therefore, Sw is a hop dominating set in G. Since this is true for every v ∈ S such that S \ {v} is not hop dominating, it follows that S is a 2-step movable hop dominating set in G. Corollary 2. Let G be a non-trivial graph and let S be a hop dominating set in G. If each v ∈ S satisfies the property that |ephn(v;S)| ≤ 1 or |ephn(v;S)| ≥ 2 such that there exists q ∈ ephn(v;S) with the property that dG(q, w) = 2 for every w ∈ ephn(v;S) \ {q}, then S is 2-step movable hop dominating in G. Proof. Suppose S satisfies the given property. Let v ∈ S and suppose S \ {v} is not hop dominating. Since γ(H) ̸= 1 for every component H of G, |N2 G(v)| ̸= 0. Suppose N2 G(v)∩(V (G)\S) = ∅, i.e., N2 G(v) ⊆ S \{v}. This and the fact that S is hop dominating imply that S \ {v} is hop dominating, a contradiction. Thus, N2 G(v) ∩ (V (G) \ S) ̸= ∅. If |ephn(v;S)| = 0, then ephn(v;S) ⊆ N2 G[u] for each u ∈ N2 G(v) ∩ (V (G) \ S). Suppose |ephn(v;S)| = 1, say xv ∈ ephn(v;S). Then xv ∈ (V (G) \ S) ∩ N2 G(v) and ephn(v;S) = {xv} ⊆ N2 G[xv]. Finally, suppose |ephn(v;S)| ≥ 2 such that there exists q ∈ ephn(v : S) such that dG(q, w) = 2 for every w ∈ ephn(v;S) \ {q}. Then q ∈ (V (G) \ S) ∩ N2 G(v). R. Estrella, Gina M. Malacas , S. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6074 6 of 15 Moreover, since dG(q, w) = 2 for every w ∈ ephn(v;S) \ {q}, it follows that ephn(v;S) ⊆ N2 G[q]. Therefore, S is a 2-step movable hop dominating set in G by Theorem 3. Theorem 4. Let G be a graph of order n ≥ 4. Then 2 ≤ γ2mh(G) ≤ n − 2. Moreover, each of the following holds: (i) γ2mh(G) = 2 if and only if there exist distinct vertices p, q, vp, vq ∈ V (G) satisfying the following conditions: (i1) N2 G[{p, q}] = V (G), i.e., {p, q} is hop dominating in G. (i2) dG(p, vp) = dG(q, vq) = 2, V (G) \ (N2 G[q]∪ {p}) ⊆ N2 G[vp], and V (G) \ (N2 G[p]∪ {q}) ⊆ N2 G[vq]. (ii) γ2mh(G) = n− 2 if and only if G satisfies the following conditions: (j1) there exist distinct vertices v and w of G such that dG(v, w) = 2 whenever |ephn(x;V (G) \ {v, w})| = 2 for some x /∈ {v, w}; and (j2) for each hop dominating set S in G with |S| < n − 2, |ephn(p;S)| ≥ 2 for some p ∈ S and there exists no q ∈ ephn(p;S) such that dG(q, s) = 2 for all s ∈ ephn(p;S) \ {q}. Proof. Since γh(G) ≥ 2, it follows from Remark 2 that γ2mh(G) ≥ 2. Next, let v ∈ V (G). Since γ(G) ̸= 1, |N2 G(v)| ≥ 1. Let z ∈ N2 G(v) and let w ∈ NG(v) ∩NG(w). Set S = V (G)\{v, w}. Then S is a hop dominating set in G. Let x ∈ S. If x /∈ N2 G(v)∪N2 G(w), then let S1 = S \ {x} = V (G) \ {x, v, w}. Clearly, S1 is hop dominating in G. Suppose x ∈ N2 G(v) ∪ N2 G(w). If x ∈ N2 G(w) \ N2 G(v) or x ∈ N2 G(v) ∩ N2 G(w), then x ̸= z because zw ∈ E(G). Then (S \ {x}) ∪ {w} = V (G) \ {x, v} is hop dominating in G because x ∈ N2 G(w) and v ∈ N2 G(z). If x ∈ N2 G(v) \N2 G(w), then (S \ {x}) ∪ {v} = V (G) \ {x,w} is a hop dominating in G. This implies that S is a 2-step movable hop dominating set in G. Therefore, γ2mh(G) ≤ |S| = n− 2. (i) Suppose γ2mh(G) = 2, say S = {p, q} is a γ2mh-set of G. Then S is a hop dominating set and γh(G) = 2. It follows that S \ {p} and S \ {q} are not hop dominating sets. By Theorem 3, there exist vertices vp ∈ (V (G) \ S) ∩ N2 G(p) and vq ∈ (V (G) \ S) ∩ N2 G(q) such that ephn(p;S) ⊆ N2 G[vp] and ephn(q;S) ⊆ N2 G[vq]. Now let a ∈ ephn(p;S). Then a ∈ V (G) \ {p, q} and N2 G(a)∩ {p, q} = {p}. This shows that a ∈ V (G) \ (N2 G[q]∪ {p}). It follows that ephn(p;S) ⊆ V (G) \ (N2 G[q]∪ {p}). Next, let b ∈ V (G) \ (N2 G[q]∪ {p}). Then b ∈ V (G) \N2 G(q). Since S is hop dominating and b /∈ S, it follows that b ∈ N2 G(p). Thus, b ∈ ephn(p;S), showing that ephn(p;S) = V (G) \ (N2 G[q] ∪ {p}). Similarly, ephn(q;S) = V (G) \ (N2 G[p] ∪ {q}). It remains to show that vp ̸= vq. To this end, suppose vp = vq. Let [p, s, vp] and [q, t, vq] be p-vp and q-vq geodesics, respectively. Since S is hop dominating, t ̸= s. It follows that t ∈ ephn(p;S) \N2 G[vp], a contradiction. Therefore, p, q, vp, and vq are distinct vertices of G satisfying conditions (i1) and (i2). For the converse, suppose there exist distinct vertices p, q, vp, vq ∈ V (G) satisfying conditions (i1) and (i2). Set D = {p, q} and let Dp = (D \ {p}) ∪ {vp} = {vp, q}. Then D R. Estrella, Gina M. Malacas , S. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6074 7 of 15 is a hop dominating set by condition (i1). Let v ∈ V (G) \Dp. If v = p, then dG(v, vp) = 2 by assumption. So suppose v ̸= p. If v ∈ N2 G[q] ∪ {p}, then v ∈ N2 G(q) because v ̸= q. If v ∈ V (G)\(N2 G[q]∪{p}), then v ∈ N2 G(vp) by (i2). This shows that Dp is a hop dominating set in G. Similarly, Dq = {vq, p} is a hop dominating set in G. Therefore, D is a 2-step movable hop dominating set in G. By Remark 1, γ2mh(G) = 2. (ii) Suppose γ2mh(G) = n − 2. Let D = V (G) \ {v, w} be a γ2mh-set in G. Suppose there exists z ∈ D such that ephn(z;D) = {v, w}. Suppose further that dG(v, w) ̸= 2. Then D\{z}, (D\{z})∪{v}, and (D\{z})∪{w} are not hop dominating sets. This implies that D is not a 2-step movable hop dominating set, contrary to our assumption. Therefore, (j1) holds. Let S be a hop dominating set in G with |S| < n − 2. Since γ1h(G) = n − 2, S is not 2-step movable hop dominating in G. Hence, there exists p ∈ S such that none of S \ {p} and the sets (S \ {p}) ∪ {s} for s ∈ (V (G) \ S) ∩ N2 G(p) is a hop dominating set in G. By the contrapositive of Corollary 2, this implies that ephn(p;S) ≥ 2 and there exists no q ∈ ephn(p;S) such that dG(q, t) = 2 for all t ∈ ephn(p;S) \ {q}. This shows that (j2) holds. For the converse, suppose G satisfies (j1) and (j2). Let Q = V (G) \ {v, w}. Then Q is a 2-step hop dominating set in G by (j1) and Corollary 2. By (j2), it follows that Q is a γ2mh-set in G. Thus, γ2mh(G) = |Q| = n− 2. Theorem 5. Let n be any positive integer such that n ≥ 4. Then γ2mh(Pn) =  ⌊n 3 ⌋ + 2, if n ∈ {5, 6, 7, 9, 10} 2t, if n = 4t (t ≥ 1) or n = 4t+ 1 (t ≥ 3) or n = 4t+ 2 (t ≥ 3) 2t+ 1, if n = 4t+ 3 (t ≥ 2). Proof. Let Pn = [v1, v2, . . . , vn] and consider the following cases: Case 1. n = 4t. Clearly, if n = 4, then γ2mh(P4) = 2. If n = 8, then R = {v1, v4, v7, v8} is a γ2mh- set of P8. Hence, γ2mh(P8) = 4. Next, let t ≥ 3. For each j ∈ {1, 2, . . . , ⌊ t−1 2 ⌋ , set Sj = {v8j−1, v8j , v8j+1, v8j+2}. Let S = {v1, v2} ∪ (⋃⌊ t−1 2 ⌋ j=1 Sj ) . Since S is hop dominat- ing and |ephn(v;S)| ≤ 1 for each v ∈ S, S is a 2-step movable hop dominating set in Pn by Corollary 2. Moreover, |S| = 2 + ∑⌊ t−1 2 ⌋ j=1 |Sj | = 2 + 4( t−1 2 ) = 2t. It can be shown that if S′ is a hop dominating set in Pn with |S′| < |S|, then there exists a vertex w ∈ S′ with |ephn(w;S′)| = 2. This implies that S′ is not a 2-step movable hop dominating set. R. Estrella, Gina M. Malacas , S. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6074 8 of 15 Therefore, γ2mh(Pn) = |S| = 2t. Case 2. n = 4t+ 1 or n = 4t+ 2. If n = 5, then S = {v1, v2, v5} is a γ2mh-set in P5. Hence, γ2mh(P5) = 3. If n = 6, then S = {v1, v2, v5, v6} is a γ2mh-set in P6. Thus, γ2mh(P6) = 4. If n = 9, then S = {v1, v2, v5, v6, v7} is a γ2mh-set in P9 and if n = 10, then S = {v1, v4, v7, v8, v9} is a γ2mh- set in P10. Hence, γ2mh(P9) = γ2mh(P10) = 5. Next, let t ≥ 3. Let Sj = {v4j+3, v4j+4} for each j ∈ {1, . . . , t − 1}. Let S = {v1, v4} ∪ ⋃t−1 j=1 Sj . Since S is hop dominating and |ephn(v;S)| ≤ 1 for each v ∈ S, S is a 2-step movable hop dominating set in Pn by Corollary 2. Again, it can be verified that every hop dominating set S′ in Pn with |S′| < |S| has a vertex w ∈ S′ with |ephn(w;S′)| = 2 and so cannot be a 2-step movable hop dominating set in Pn. Therefore, γ2mh(Pn) = |S| = 2 + t−1∑ j=1 |Sj | = 2 + 2(t− 1) = 2t. Case 3: n = 4t+ 3. If n = 7, then S = {v1, v2, v5, v6} is a γ2mh-set in P7. Thus, γ2mh(P7) = 4. Next, let t ≥ 2 andDj = {v4j+3, v4j+4} for each j ∈ {1, 2, . . . , t−1}. Let D = {v1, v4, v4t+3}∪ ⋃t−1 j=1 Sj . By Corollary 2, D is a 2-step movable hop dominating set in Pn because it is hop dominating and |ephn(v;D| ≤ 1 for each v ∈ D. Moreover, D is a γ2mh-set; hence, γ2mh(Pn) = |S| = 2 + t−1∑ j=1 |Sj | = 3 + 2(t− 1) = 2t+ 1. This proves the assertion. Theorem 6. Let n be any positive integer such that n ≥ 4. Then γ2mh(Cn) =  2t, if n = 4t, t ≥ 1 or n = 4t+ 1, t ≤ 3 or n = 4t+ 2, t ≥ 1 2t− 1, if n = 4t+ 1, t ≥ 4 2t+ 1, if n = 4t+ 3, t ≥ 1. Proof. Let Cn = [v1, v2, . . . , vn, v1], where n ≥ 4. Consider the following cases: Case 1. n = 4t, where t ≥ 1. Consider the following subcases: R. Estrella, Gina M. Malacas , S. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6074 9 of 15 Subcase 1: t is odd. If n = 4, then S = {v1, v2} is a γ2mh-set in C4. Hence, γ2mh(C4) = 2. Next, suppose that t ≥ 3. Let Sj = {v8j−1, v8j , v8j+1, v8j+2} for j = 1, . . . , t−1 2 . Let S = {v1, v2}∪ (⋃ t−1 2 j=1 Sj ) . Then S is a 2-step movable hop dominating set in Cn. If S′ is a hop dominating set with |S′| ≤ |S|, then ∃ a vertex v ∈ S′ with |ephn(v;S′)| = 2. Hence, S′ is not a 2-step movable hop dominating set. Therefore, γ2mh(Cn) = |S| = 2+ ∑ t−1 2 j=1 |Sj | = 2+4( t−1 2 ) = 2t. Subcase 2: t is even. Let Sj = {v8j−7, v8j−6, v8j−5, v8j−4} for j = 1, . . . , t 2 . Let S = ∪ (⋃ t 2 j=1 Sj ) . Then S is a γ2mh-set in Cn and γ2mh(Cn) = |S| = ( ∑ t 2 j=1 |Sj |) = 4( t2) = 2t. Case 2. n = 4t+ 1, where t ≤ 3. Let Sj = {v4j−3, v4j−2} for j = 1, . . . , t. Let S = (⋃t j=1 Sj ) . Then S is a γ2mh-set in Cn. Therefore, γ2mh(Cn) = |S| = ∑t j=1 |Sj | = 2t. Case 3. n = 4t+ 1, where t ≥ 4. Let Sj = {v4j+4, v4j+5} for j = 1, . . . , t − 3. Let S = {v1, v2, v5, vn−5, vn−2} ∪ (⋃t−3 j=1 Sj ) . Then S is a γ2mh-set in Cn and γ2mh(Cn) = |S| = 4 + ∑t−3 j=1 |Sj | = 5 + 2(t− 3) = 2t− 1. Case 4. n = 4t+ 2. If n = 6 and n = 10, then {v1, v4} and {v1, v2, v6, v7} are γ2mh-sets in C4 and C10, re- spetively. Hence, γ2mh(C6) = 2 and γ2mh(C10) = 4. Suppose t ≥ 3. Let Sj = {v4j+4, v4j+5} for j = 1, . . . , t− 2. Let S = {v1, v2, v5, vn−2}∪ (⋃t−2 j=1 Sj ) . Then S is a γ2mh-set in Cn and γ2mh(Cn) = |S| = 4 + ∑t−2 j=1 |Sj | = 4 + 2(t− 2) = 2t. Case 5. n = 4t+ 3. Let Sj = {v4j+4, v4j+5} for j = 1, . . . , t − 1. Let S = {v1, v2, v5} ∪ (⋃t−1 j=1 Sj ) . Then S is a γ2mh-set in Cn and γ2mh(Cn) = |S| = 4 + ∑t−1 j=1 |Sj | = 3 + 2(t− 1) = 2t+ 1. If G1 and G2 are the copies of graph G in the definition of the shadow graph D2(G) and if SG1 ⊆ V (G1) and SG2 ⊆ V (G2), then the sets S′ G1 and S′ G2 are the sets given by S′ G1 = {a′ ∈ V (G2) : a ∈ SG1} and S′ G2 = {a ∈ V (G1) : a ′ ∈ SG2}. R. Estrella, Gina M. Malacas , S. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6074 10 of 15 The next result is found in [20]. Theorem 7. Let G be a non-trivial connected graph. Then S is a hop dominating set in D2(G) if and only if one of the following conditions holds: (i) S is a hop dominating set in G1. (ii) S is a hop dominating set in G2. (iii) S = SG1 ∪ SG2 such that SG1 ∪ S′ G2 and S′ G1 ∪ SG2 are hop dominating sets in G1 and G2, respectively. Theorem 8. Let G be a non-trivial connected graph. Then D2(G) admits a 2-step movable hop dominating set. Moreover, a set S ⊆ V (D2(G)) is 2-step movable hop dominating in D2(G) if and only if one of the following conditions holds: (i) S is a 2-step movable hop dominating set in G1. (ii) S is a 2-step movable hop dominating set in G2. (iii) S = SG1∪SG2 such that SG1∪S′ G2 and S′ G1 ∪SG2 are 2-step movable hop dominating sets in G1 and G2, respectively. Proof. Since G is non-trivial and connected, it follows that D2(G) is connected and γ(D2(G)) ̸= 1. Thus, D2(G) admits a 2-step movable hop dominating set by Theorem 2. Let S be a 2-step movable hop dominating set in D2(G). Set SG1 = S ∩ V (G1) and SG2 = S ∩ V (G2). If SG2 = ∅, then S = SG1 is a hop dominating set in G1 by Theorem 7(i). Let v ∈ SG1 . Suppose SG1 \ {v} is not hop dominating in D2(G). Since S is a 2-step movable hop dominating set in D2(G), there exists w ∈ V (D2(G)) \ S) ∩N2 D2(G)(v) such that (S \ {v}) ∪ {w} is hop dominating in D2(G). If w ∈ V (G1), then (S \ {v}) ∪ {w} = (SG1 \ {v}) ∪ {w} is hop dominating in G1 by Theorem 7(i). Suppose w = u′ ∈ V (G2). Then u ∈ V (G1) and dD2(G)(v, u ′) = dD2(G)(v, u) = dG1(v, u) = 2. Since SG1 \ {v} is not hop dominating in D2(G), u ∈ (V (G1) \ {v}) ∩N2 G1 (v). Also, since (SG1 \ {v}) ∪ {u′} is hop dominating in D2(G), (SG1 \ {v}) ∪ {u} is hop dominating in G1 by Theorem 7(iii). Therefore, S = SG1 is 2-step movable hop dominating in G1. Similarly, S = SG2 is 2-step movable hop dominating in G2 whenever SG1 = ∅. Finally, suppose SG1 ̸= ∅ and SG2 ̸= ∅. By Theorem 7(iii), Q = SG1 ∪ S′ G2 and R = S′ G1 ∪ SG2 are hop dominating sets in G1 and G2, respectively. Let p ∈ Q such that Q\{p} = (SG1\{p})∪S′ G2 is not hop dominating in G1. Then S \ {p} is not hop dominating in D2(G) by Theorem 7. Since S is 2-step movable hop dominating in D2(G), there exists q ∈ (V (D2(G) \ S) ∩N2 D2(G)(p) such that (S \ {p})∪{q} is hop dominating in D2(G). If q ∈ V (G1), then q ∈ (V ((G1) \Q)∩N2 G1 (p) and (S\{p})∪{q} = [SG1\{p})∪{q}]∪SG2 . Hence, [SG1\{p})∪{q}]∪S′ G2 = (Q\{p})∪{q} is hop dominating in G1 by Theorem 7(iii). Suppose q = s′ ∈ V (G2). By assumption, s ∈ (V (G1) \ Q) ∩ N2 G1 (p) and (S \ {p}) ∪ {q} = (SG1 \ {p}) ∪ (SG2 ∪ {s′}. It follows from Theorem 7(iii) that (Q \ {p}) ∪ {s} = [SG1 \ {p}) ∪ {s}] ∪ S′ G2 is hop dominating in R. Estrella, Gina M. Malacas , S. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6074 11 of 15 G1. Thus, Q is 2-step movable hop dominating in G1. Similarly, R is 2-step movable hop dominating in G2. Therefore, (i) or (ii) or (iii) holds. For the converse, suppose (i) holds. Then S is hop dominating in D2(G) by Theorem 7. Let x ∈ S such that S \ {x} is not hop dominating in D2(G). By assumption, there exists y ∈ (V (G1) \S)∩N2 G1 (x) such that (S \ {x})∪{y} is hop dominating in G1. By Theorem 7(i), (S\{x})∪{y} is hop dominating inD2(G). Note that since dG1(x, y) = dD2(G)(x, y) = 2, y ∈ (V (D2(G)) \ S) ∩N2 D2(G)(x). Thus, S is 2-step movable hop dominating in D2(G). The same conclusion holds if (ii) holds. Finally, suppose S satisfies (iii). Let v ∈ S such that S \ {v} is not hop dominating in D2(G). Suppose, without loss of generality, that v ∈ SG1 . Then v ∈ (SG1 ∪ S′ G2 ). Suppose Q = (SG1 ∪ S′ G2 ) \ {v} = (SG1 \ {v}) ∪ S′ G2 is hop dominating in G1. Let p /∈ [(SG1 \ {v}) ∪ SG2 ]. Suppose p ∈ V (G1). If p′ ∈ SG2 , then p′ ∈ [(SG1 \ {v}) ∪ SG2 ] and dD2(G)(p, p ′) = 2. If p′ /∈ SG2 , then p /∈ Q. Since Q is hop dominating in G1, there exists q ∈ Q such that dG1(p, q) = dD2(G)(p, q) = 2. If q ∈ SG1 \ {v}, then q ∈ [(SG1 \ {v})∪SG2 ]. Suppose q /∈ SG1 \ {v}. Then q ∈ S′ G2 . Hence, q′ ∈ SG2 ⊆ [(SG1 \ {v}) ∪ SG2 ] and dD2(G)(p, q ′) = dD2(G)(p, q) = 2. This implies that S \ {v} = [(SG1 \ {v}) ∪ SG2 ] is hop dominating, a contradiction. Now, since SG1 ∪ S′ G2 is 2-step movable hop dominating in G1, there exists t ∈ [V (G1) \ (SG1 ∪ S′ G2 )] ∩N2 G1 (v) such that [(SG1 ∪ S′ G2 ) \ {v}] ∪ {t} is hop dominating in G1. By Theorem 7, it follows that [(SG1 \ {v}) ∪ {t}] ∪ S′ G2 is hop dominating in D2(G). Again, this will imply that [(SG1 \ {v}) ∪ {t}] ∪ SG2 is hop dominating in D2(G), showing that S is a 2-step movable hop dominating set in D2(G). The next result is a direct consequence of Theorem 8. Corollary 3. Let G be a non-trivial connected graph. Then γ2mh(D2(G)) = γ2mh(G). Proof. Let S be a γ2mh-set in D2(G). If S ⊆ V (G1) or S ⊆ V (G2), then S is a 2-step movable hop dominating set in G1 or in G2, respectively, by (i) and (ii) of Theorem 8. Hence, γ2mh(D2(G)) = |S| ≥ γ2mh(G). If S = SG1 ∪SG2 , then SG1 ∪S′ G2 and S′ G1 ∪SG2 are 2-step movable hop dominating sets in G1 and G2, respectively, by Theorem 8(iii). Thus, γ2mh(D2(G)) = |SG1 ∪ SG2 | = |SG1 ∪ S′ G2 | ≥ γ2mh(G). Next, suppose Q is a γ2mh-set in G = G1. Then Q is 2-step movable hop dominating set in D2(G) by Theorem 8. Hence, γ2mh(D2(G)) ≤ |Q| = γ2mh(G). This establishes the desired equality. Theorem 9. Let G be a non-trivial connected graph. Then GG admits a 2-step movable hop dominating set and 2 ≤ γ2mh(GG) ≤ 4. Moreover, each of the following statements hold: (i) γ2mh(GG) = 2 if and only if γh(G) = 2 and γh(G) = 2. (ii) γ2mh(GG) = 3 if and only if one of the following conditions holds. (a) γh(G) = 2 and γh(G) ≥ 3 or γh(G) = 2 and γh(G) ≥ 3. (b) γh(G) = 3 and γh(G) ≥ 3 (or γh(G) ≥ 3 and γh(G) = 3). R. Estrella, Gina M. Malacas , S. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6074 12 of 15 (c) There exist vertices x, y, z, w ∈ V (G) such z, w ∈ N2 G[x] ∪ N2 G[y[, z ∈ NG[w], and V (G) \ [NG(z) ∪NG(w)] ̸= ∅. (d) There exist vertices p, q, t, s ∈ V (G) such that s ∈ N2 G(t) and t, s ∈ N2 G [{p, q}]. (iii) γ2mh(GG) = 4 if and only if G does not satisfy any of the properties in (i) and (ii). Proof. Since G is non-trivial, γ(GG) ̸= 1. Hence, GG admits a 2-step movable hop dominating set by Theorem 1. Note that {v, v} is a hop dominating set in GG for each v ∈ V (G). Thus, {v, w, v, w} is a 2-step movable hop dominating set in GG for each pair of distinct vertices v and w of G. Therefore, 2 = γh(GG) ≤ γ2mh(GG) ≤ 4. (i) Suppose γ2mh(GG) = 2, say S = {p, q} is a γ2mh-set in GG. If p, q ∈ V (G), then S is a hop dominating set of G. If p, q ∈ V (G), then S is a hop dominating set of G. Thus, γh(G) = 2 or γh(G) = 2. Suppose p ∈ V (G) and q = t ∈ V (G). Suppose p ̸= t. If pt ∈ E(G), then p t /∈ E(G). It follows that t ∈ NGG(p) ∩ NGG(q). If pt /∈ E(G), then p t ∈ E(G). This implies that p ∈ NGG(p) ∩ NGG(q). Thus, S is not a hop dominating set in D2(G), a contradiction. Therefore, p = t, i.e., S = {p, p}. Now, since S is 2-step movable hop dominating in D2(GG), there exists s ∈ [V (GG) \ S] ∩ N2 GG (p) such that (S \{p})∪{s} = {p, s} is hop dominating in GG. If s ∈ V (G), then s = p, a contradiction. Thus, s ∈ V (G) and {p, s} is a hop dominating set in G. This implies that γh(G) = 2. Similarly, γh(G) = 2. For the converse, suppose γh(G) = 2 and γh(G) = 2. Let D = {x, y} be a γh-set in G. Note that whether xy ∈ E(G) or xy /∈ E(G), we find that (D \ {x}) ∪ {y} = {y, y} and (D \ {y})∪{x} = {x, x} are hop dominating sets. This implies that D is a 2-step movable hop dominating set in GG. Therefore, γ2mh(GG) = 2. (ii) Suppose γ2mh(GG) = 3. If γh(G) = 2 (γh(G) = 2), then γh(G) ≥ 3 (resp. γh(G) ≥ 3) by (i). Hence, (a) holds. Next, suppose γh(G) ≥ 3 and γh(G) ≥ 3. If γh(G) = 3 or γh(G) = 3, then (b) holds. Suppose γh(G) > 3 and γh(G) > 3. Let S = {x, y, t} be a γ2mh-set in GG. By assumption, S ∩ V (G) ̸= ∅ and S ∩ V (G) ̸= ∅. Suppose x, y ∈ V (G) and t = z ∈ V (G). Since z ∈ NGG(z) and S is hop dominating in GG, it follows that z ∈ N2 G(x) ∪ N2 G(y). Also, since γh(G) > 3 and S is 2-step movable hop dominating in GG, it follows that (S \ {z}) ∪ {w} = {x, y, w} is hop dominating in GG for some w ∈ [V (G) \ {z}] ∩ N2 G (z). This implies that w ∈ N2 G(x) ∪ N2 G(y), z ∈ NG(w), and V (G) \ [NG(z) ∪NG(w)] ̸= ∅. This shows that (c) holds. Suppose x, y ∈ V (G) and t ∈ V (G). Let x = p and x = q. Since tt ∈ E(GG), t ∈ N2 G [{p, q}] because S is hop dominating in GG. Since γh(G) > 3 and S is 2-step movable hop dominating in GG, there exists s ∈ N2 G(t) such that (S \ {t})∪ {s} = {p, q, s} is hop dominating in GG. This implies that s ∈ N2 G [{p, q}]. Thus, (d) holds. For the converse, suppose (a) holds. Let Q = {c, d} be a γh-set of G and let Q∗ = {c, d, d}. Since Q∗ \ {d} = {c, d}, Q∗ \ {c} = {d, d}, and (Q∗ \ {d}) ∪ {c} = {c, c} are hop dominating sets in GG, it follows that Q∗ = {c, d, d} is a γ2mh-set in GG. Hence, γ2mh(GG) = 3. The same conclusion holds if γh(G) ≥ 3 and γh(G) = 2. Suppose R. Estrella, Gina M. Malacas , S. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6074 13 of 15 (b) holds, i.e., γh(G) = 3 and γh(G) ≥ 3. Let S = {x, y, z} be a γh-set in G. Since (S \ {x}) ∪ {y} = {y, y, z}, (S \ {y}) ∪ {x} = {x, x, z}, and (S \ {z}) ∪ {y} = {x, y, y} are hop dominating sets inGG, it follows thatD is a 2-step movable hop dominating set inGG. Therefore, γ2mh(GG) = 3. This conclusion also holds if γh(G) ≥ 3 and γh(G) = 3. Next, suppose (c) holds. Let D = {x, y, z}. Since V (G) ∪ {z} ⊆ N2 GG ({x, y}) and v ∈ N2 GG (z) for all v ∈ V (G) \ {z, x, y} it follows that D is a hop dominating set in GG. If z ∈ {x, y}, say z = y, then D \ {x} = {y, z} is hop dominating in GG. Suppose z /∈ {x, y}. Then (D \{x})∪{y} = {y, y, z}, (D \{y})∪{x} = {x, x, z}, and (D \{z})∪{w} = {x, y, w} are hop dominating in GG. Therefore, D is a γ2mh-set in GG. Hence, γ2mh(GG) = 3. Finally, suppose (d) holds. Let Q = {p, q, t}. Cleary, Q is a hop dominating set in GG. Suppose t ∈ {p, q}, say t = q. Then Q \ {p} = {t, q is hop dominating in GG. Suppose t /∈ {p, q}. Then (Q\{p})∪{q} = {q, q, t}, (Q\{q})∪{p} = {p, p, t}, and (Q\{t})∪{s} = {p, q, s} are hop dominating sets in GG. This shows that Q is a γ2mh-set in GG. Thus, γ2mh(GG) = 3. (iii) This follows from (i) and (ii). The next result follows from Theorem 9. Corollary 4. Let n be a positive integer and n ≥ 2. Then γ2mh(KnKn) =  2, if n = 2 3, if n = 3 4, if n ≥ 4. 4. Conclusion The concepts of 2-step movability of hop dominating sets as well as the parameter 2-step movable hop domination number have been introduced in this paper. Graphs that admit a 2-step movable hop dominating set were characterized. Bounds on the 2-step movable hop domination number were given and graphs that attained these bounds were characterized. It was shown that the difference of the 2-step movable hop domination number and the hop domination can be made arbitrarily large. 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