EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 3, 2023, 1568-1579 ISSN 1307-5543 – ejpam.com Published by New York Business Global Another Look at Geodetic Hop Domination in a Graph Chrisley Jade C. Saromines1,∗, Sergio R. Canoy, Jr.1,2 1 Department of Mathematics and Statistics, College of Science and Mathematics, MSU- Iligan Institute of Technology, Philippines 2 Center for Mathematical and Theoretical Physical Sciences-PRISM, MSU-Iligan Institute of Technology, Philippines Abstract. Let G be an undirected graph with vertex and edge sets V (G) and E(G), respectively. A subset S of vertices of G is a geodetic hop dominating set if it is both a geodetic set and a hop dominating set. The geodetic hop domination number of G is the minimum cardinality among all geodetic hop dominating sets in G. In this paper, we characterize the geodetic hop dominating sets in the join of two graphs. These characterizations which use the concept of pointwise non-dominating 2-path closure absorbing set are, in turn, used to determine the geodetic hop domination number of the join of graphs. Moreover, a realization result involving the hop domination number and geodetic hop domination number is also obtained. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Geodetic hop domination, join 1. Introduction Over the years, a number of studies dealing with the topic on hop domination, a concept introduced and initially studied by Natarajan and S. K. Ayyaswamy [13], had been done. In particular, some variations of hop domination had been introduced and considered in many studies (see [2], [3], [4], [5], [6], [7], [10], [11], [12], [14], [15], and [16]). Henning and Rad [9] gave a probabilitics upper bound of the hop domination number of a graph and showed that the hop dominating set problem is NP-complete for planar bipartite graphs and planar chordal graphs. In a recent study, Henning et al. [8] presented a linear time algorithm for computing a minimum hop dominating set in bipartite permutation graphs. The idea of combining the concepts of hop domination and geodetic has led to the introduction of the notion of geodetic hop domination. This hop domination variant was first defined and examined by Anusha and Robin [1]. Motivated by the new concept, Saromines and Canoy [16] gave characterizations of the geodetic hop dominating sets in the corona and lexicographic product of two graphs. In this present paper, we revisit the concept of geodetic hop domination and give further results of this new parameter. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i3.4810 Email addresses: chrisleyjade.saromines@g.msuiit.edu.ph (C.J. Saromines), sergio.canoy@g.msuiit.edu.ph (S. Canoy, Jr.) https://www.ejpam.com 1568 © 2023 EJPAM All rights reserved. C.J. Saromines, S. Canoy, Jr., / Eur. J. Pure Appl. Math, 16 (3) (2023), 1568-1579 1569 2. Terminology and Notation For any two vertices u and v in an undirected connected graph 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 u, v and all vertices lying on a u-v geodesic. The interval IG(u, v) = IG [u, v] \ {u, v}. The open neighborhood of a vertex u is the set NG(u) consisting of all vertices v which are adjacent to u. The closed neighborhood of u is NG[u] = NG(u) ∪ {u}. For any A ⊆ V (G), NG(A) = ⋃ v∈A NG(v) is called the open neighborhood of A andNG[A] = NG(A)∪A is called the closed neighborhood of A. The open hop neighborhood of a vertex u is the set N2 G(u) = {v ∈ V (G) : dG(v, u) = 2}. The closed hop neighborhood of u is N2 G[u] = N2 G(u) ∪ {u}. For any A ⊆ V (G), N2 G(A) = ⋃ v∈A N2 G(v) is called the open hop neighborhood of A and N2 G[A] = N2 G(A)∪A is called the closed hop neighborhood of A. A set S ⊆ V (G) is a dominating set in G if NG[S] = V (G). The smallest cardinality of a dominating set in G, denoted by γ(G) is called the domination number of G. The geodetic closure of a set S ⊆ V (G), denoted by IG [S], is the union of the intervals IG[u, v], where u, v ∈ S. Set S is geodetic set in G if IG[S] = V (G). The smallest cardinality among all geodetic sets in G, denoted by g(G), is called the geodetic number of G. A geodetic set of cardinality g(G) is called a g-set of G. A set S ⊆ V (G) is a geodetic dominating set in G if it is both a dominating and a geodetic set. A set S ⊆ V (G) is a hop dominating set if N2 G[S] = V (G). The minimum cardinality of a hop dominating set of a graph G, denoted by γh(G), is called the hop domination number of G. A subset S of V (G) is a total hop dominating set of G if for every v ∈ V (G), there exists u ∈ S such that dG(u, v) = 2. The smallest cardinality of a total hop dominating set of G, denoted by γth(G) is called the total hop domination number of G. Any total hop dominating set of G with cardinality γth(G) is called a γth-set. A subset S of vertices of G is a geodetic hop dominating set if it is both a geodetic and a hop dominating set. The geodetic hop domination number γhg(G) of G is the minimum cardinality among all geodetic hop dominating sets in G. Any geodetic hop dominating set of G with cardinality γhg(G) is called a γhg-set. A set S ⊆ V (G) of a graph G is called a 2-path closure absorbing if for each x ∈ V (G)\S there exist u, v ∈ S such that dG(u, v) = 2 and x ∈ IG(u, v). The minimum cardinality of a 2-path closure absorbing set in G is denoted by ρ2(G). Any 2-path closure absorbing set of G with cardinality ρ2(G) is called a ρ2-set. A set D ⊆ V (G) is a pointwise non-dominating set of G if for each v ∈ V (G) \ S, there exists u ∈ S such that v /∈ NG(u). The smallest cardinality of a pointwise non-dominating set of G, denoted by pnd(G), is called the pointwise non-domination number of G. A pointwise non-dominating set S ⊆ V (G) of a graph G is called a 2-path closure absorbing pointwise non-dominating set if it is a 2-path closure absorbing set. The minimum cardinality of a 2-path closure absorbing pointwise non-dominating set in G is denoted by ρ2pnd(G). Any 2-path closure absorbing pointwise non-dominating set of G C.J. Saromines, S. Canoy, Jr., / Eur. J. Pure Appl. Math, 16 (3) (2023), 1568-1579 1570 with cardinality ρ2pnd(G) is called a ρ2pnd-set. Let G and H be two graphs. The join G+H is the graph with vertex set V (G+H) = V (G) ∪ V (H) and edge set E(G+H) = E(G) ∪ E(H) ∪ {uv : u ∈ V (G), v ∈ V (H)}. 3. Results Since every geodetic hop dominating set is a hop dominating set, we have the following remark. Remark 1. Let G be any connected graph on n vertices. Then γh(G) ≤ γhg(G). Remark 2. The bound given in Remark 1 is tight. Moreover, strict inequality can also be attained. To see this, consider G = K4 and H = K1,3. It can easily be verified that γh(G) = γhg(G) = 4 and γh(H) = 2 < 4 = γhg(H). Theorem 1. Let a and b be positive integers such that 2 ≤ a ≤ b. Then there exists a connected graph G such that γh(G) = a and γhg(G) = b. Proof. Consider the following cases: Case 1. a = b. Let G = Ka. Then γh(G) = a = γhg(G). Case 2. a < b. Consider the following subcases: Subcase 2.1. a is even. Suppose a = 2 and let m = b−a. Consider the graph G in Figure 1. Let S1 = {x1, x2} and S2 = {y1, y2, z1, z2, ..., zm}. Then S1 and S2 are, respectively, γh-set and γhg-set of G. Hence, γh(G) = a and γhg(G) = a+m = b. .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ................................................................................... ................................................................................... .................................... ........... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ...... .................................... ........... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ...... .................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... .................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... .................................... ................................................................................ .................................... .................................... ... y1 y2 x1 x2 z1 z2 zm G : Figure 1 Suppose a ≥ 4. Consider the graph G′ in Figure 2. Let S3 = {x1, x2, ..., xa−1, xa} and S4 = {y1, y2, ..., ya−1, ya, z1, z2, ..., zm}. Then S3 and S4 are, respectively, γh-set and γhg-set of G ′. Hence, γh(G ′) = a and γhg(G ′) = a+m = b. C.J. Saromines, S. Canoy, Jr., / Eur. J. Pure Appl. Math, 16 (3) (2023), 1568-1579 1571 .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .. .................................. ....................................................................................................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ................................................................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ................................................................................... ................................................................................... ................................................................................... ................................................................................... ................................................................................... ................................................................................... .................................... ................................................................................... ................................................................................... ................................................................................... .................................... ........... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ...... .................................... ........... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ...... .................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... .................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... .................................... ................................................................................ .................................... .................................... . . . . . . ... y1 y2 x1 x2 y3 y4 x3 x4 ya−1 ya xa−1 xa z1 z2 zm G′ : Figure 2 Subcase 2.2. a is odd. Suppose a = 3 and let m = b − a + 1. Consider the graph H in Figure 3. Let S5 = {x1, x2, x3} and S6 = {y1, y2, z1, z2, ..., zm}. Then S5 and S6 are, respectively, γh-set and γhg-set of H. Hence, γh(H) = a and γhg(H) = m+ a− 1 = b. .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ................................................................................... ................................................................................... .................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... .................................... .................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... .................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... .................................... ................................. ................................ ................................ ........... ........................................ ................................ ................................ ................................ ........ .................................... ...................................................................................................................... .......................................................................................................................................................... .................................... ............................................................................................................... .................................... ........................................................................................................... .................................... .................................... ... y1 y2 x1 x2 x3 z1 z2 zm−1 zm H : Figure 3 Suppose a ≥ 5 and let m = b − a + 1. Consider the graph H ′ in Figure 4. Let S7 = {x1, x2, ..., xa−1, xa} and S8 = {y1, y2, ..., ya−1, z1, z2, ..., zm}. Then S7 and S8 are, respectively, γh-set and γhg-set of H ′. Hence, γh(H ′) = a and γhg(H ′) = m+ a− 1 = b. .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .. .................................. ....................................................................................................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ................................................................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ................................................................................... ................................................................................... ................................................................................... ................................................................................... ................................................................................... ................................................................................... .................................... ................................................................................... ................................................................................... ................................................................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ . .................................... .................................... ................................................................................... ................................................................................... .................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... .................................... .................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... .................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... .................................... ................................. ................................ ................................ ........... ........................................ ................................ ................................ ................................ ........ .................................... ...................................................................................................................... .......................................................................................................................................................... .................................... ............................................................................................................... .................................... ........................................................................................................... .................................... .................................... ... . . . . . . y1 y2 x1 x2 y3 y4 x3 x4 ya−2 ya−1 xa−2 xa−1 xa z1 z2 zm−1 zm H ′ : Figure 4 This proves the assertion. Corollary 1. Let n be a positive integer. Then there exists a connected graph such that γhg(G)−γh(G) = n. In other words, the difference γhg(G)−γh(G) can be made arbitrarily large. C.J. Saromines, S. Canoy, Jr., / Eur. J. Pure Appl. Math, 16 (3) (2023), 1568-1579 1572 The next few results deal with the concept of pointwise non-dominating 2-path closure absorbing sets. Remark 3. Every pointwise non-dominating 2-path closure absorbing is both a pointwise non-dominating set and a 2-path closure absorbing set in G. Hence, ρ2pnd(G) ≥ max {pnd(G), ρ2(G)} . Theorem 2. Let G be a graph on n ≥ 3 vertices. Then 3 ≤ ρ2pnd(G) ≤ n. Moreover, (i) ρ2pnd(G) = 3 if and only if n = 3 or n > 3 and there exists S ⊆ V (G) with |S| = 3 such that for each v ∈ V (G) \ S, |NG(v) ∩ S| = 2 and dG(a, b) = 2 for a, b ∈ NG(v) ∩ S. (ii) ρ2pnd(G) = n if and only if one of the following holds: (a) G is connected and for every pair of vertices x, y with dG(x, y) = 2, the set NG(x) ∩NG(y) contains dominating vertices of G only or (b) G is disconnected such that every component H of G is complete. Proof. Let S be a ρ2pnd-set of G. Suppose |S| ≤ 2 and let v ∈ V (G) \ S. Since S is a 2-path closure absorbing set, there exist p, q ∈ S such that dG(p, q) = 2 and v ∈ IG(p, q). Hence, S cannot be a pointwise non-dominating set, contradicting our assumption that S is a ρ2pnd-set of G. Therefore, 3 ≤ ρ2pnd(G). (i) Suppose ρ2pnd(G) = 3. Suppose further that n > 3 and let S be a ρ2pnd-set of G. Then |S| = 3. Let v ∈ V (G) \ S. Then there exist vertices a, b ∈ S such that dG(a, b) = 2 and v ∈ IG(a, b) because S is a 2-path closure absorbing set. Also, since S is a pointwise non-dominating set, there exists z ∈ S \ {a, b} such that v /∈ NG(z). Therefore, |NG(v) ∩ S| = 2. For the converse, suppose that n > 3 and there exists S ⊆ V (G) with |S| = 3 that satisfies the given conditions. Let v ∈ V (G) \ S. Then, by assumption, there exist a, b ∈ S with dG(a, b) = 2 and v ∈ IG(a, b). This implies that S is a 2-path closure absorbing set of G. Since |NG(v) ∩ S| = 2, S is also a pointwise non-dominating set of G. Finally, suppose that n = 3. Then S = V (G) is both a pointwise non-dominating and 2-path closure absorbing set. Since 3 ≤ ρ2pnd(G), it follows that ρ2pnd(G) = 3. (ii) Suppose ρ2pnd(G) = n. Consider the following cases: Case 1. G is connected. C.J. Saromines, S. Canoy, Jr., / Eur. J. Pure Appl. Math, 16 (3) (2023), 1568-1579 1573 Suppose there exist p, q ∈ V (G) with dG(p, q) = 2 such that NG(p)∩NG(q) contains a non-dominating vertex, say z. Then there is a vertex w ∈ V (G)\NG(z). This implies that V (G) \ {z} is a pointwise non-dominating and 2-path closure absorbing set of G, contrary to the assumption that ρ2pnd(G) = n. Thus, NG(x) ∩ NG(y) contains dominating vertices of G only for every pair of vertices x, y with dG(x, y) = 2, showing that (a) holds. Case 2. G is disconnected. Suppose there exists a component H of G that is not complete. Then there exist v, w ∈ V (H) such that dG(v, w) = 2. Let u ∈ NG(v) ∩ NG(w). Then V (G) \ {u} is a 2-path closure absorbing set of G. Let H ′ be a component of G with H ′ ̸= H and pick u ′ ∈ V (H). Then u ′ ∈ V (G) \ {u} and uu ′ ∈ E(G). Hence, V (G) \ {u} is also a pointwise non-dominating set of G. This gives a contradiction. Thus, every component of G is complete. For the converse, suppose first that (a) holds. Let S be a ρ2pnd-set of G. Suppose S ̸= V (G), say v ∈ V (G) \ S. Since S is a 2-path closure absorbing set of G, there exist x, y ∈ S such that dG(x, y) = 2 and v ∈ IG(x, y). By assumption, v is a dominating vertex of G. Therefore, S is not a pointwise non-dominating set, a contradiction. Hence, S = V (G) and ρ2pnd(G) = n. Next, suppose that (b) holds. Then the only 2-path closure absorbing set of G is V (G). Therefore, V (G) is the only pointwise non-dominating and 2-path closure absorbing set of G. Accordingly, ρ2pnd(G) = n. The next result follows from Theorem 2. Corollary 2. Let n be a positive integer and n ≥ 2. Then ρ2pnd(Kn) = ρ2pnd(Kn) = ρ2pnd(K1,n−1) = n. Proposition 1. Let m and n be positive integers with m,n ≥ 2. Then ρ2pnd(Km,n) = { 3 if m = 2 or n = 2 4 if m ≥ 3 and n ≥ 2 . Proof. Suppose m = 2 or n = 2, say m = 2. Choose any w ∈ V (Kn). Then S = V (Km) ∪ {w} is a pointwise non-dominating and 2-path closure absorbing set of Km,n. By Theorem 2, ρ2pnd(Km,n) = |S| = 3. Next, suppose that m ≥ 3 and n ≥ 3. Pick any x, y ∈ V (Km) and p, q ∈ V (Kn). Then {x, y, p, q} is a pointwise non-dominating and 2-path closure absorbing set of Km,n. This implies that ρ2pnd(Km,n) ≤ 4. Let S◦ be a ρ2pnd-set of Km,n. Suppose further that |S◦| = 3. Since S◦ is a pointwise non-dominating set, S1 = S◦ ∩ V (Km) ̸= ∅ and S2 = S◦∩V (Kn) ̸= ∅. We may assume that |S1| = 1. Then |S2| = 2. Let z ∈ V (Kn)\S◦. Then z /∈ IKm,n(u, v) for all u, v ∈ S◦, a contradiction. Therefore, |S◦| ≥ 4. Accordingly, ρ2pnd(Km,n) = 4. C.J. Saromines, S. Canoy, Jr., / Eur. J. Pure Appl. Math, 16 (3) (2023), 1568-1579 1574 Proposition 2. For each positive integer n ≥ 2, (i) ρ2pnd(Pn) =  2 if n = 2 3 if n = 3, 4 ⌈n+1 2 ⌉ if n ≥ 5 (ii) ρ2pnd(Cn) = { 3 if n = 3, 4 ⌈n2 ⌉ if n ≥ 5 Proof. (i) Clearly, ρ2pnd(P2) = 2 and ρ2pnd(P3) = ρ2pnd(P4) = 3. Let n ≥ 5. If n is odd, then S1 = {v1, v3, ..., vn−2, vn} is the only ρ2pnd-set of Pn. Hence, ρ2pnd(Pn) = n+1 2 . If n is even, then S2 = {v1, v3, ..., vn−3, vn−1, vn} and S3 = {v1, v3, v4..., vn−2, vn} are the only ρ2pnd-sets of Pn. It follows that ρ2pnd(Pn) = n+2 2 . (ii) By Theorem 2(i), ρ2pnd(C3) = ρ2pnd(C4) = 3. Let n ≥ 5. If n is odd, then {v1, v3, v5, ..., vn−2, vn} is a ρ2pnd-set of Cn. If n is even, then {v1, v3, v5, ..., vn−1} is a ρ2pnd-set of Cn. Therefore, ρ2pnd(Cn) = ⌈n2 ⌉. From the proof of Proposition 2, the next result follows. Corollary 3. Let n be a positive integer. Then (i) ρ2pnd(Pn) = ρ2(Pn) for all n ̸= 3 and (ii) ρ2pnd(Cn) = ρ2(Cn) for all n ≥ 3. The next result is found in [11]. Theorem 3. Let G and H be any two graphs. A set S ⊆ V (G +H) is hop dominating set of G+H if and only if S = SG ∪SH , where SG and SH are pointwise non-dominating sets of G and H, respectively. Theorem 4. Let G and H be any two graphs. A set S ⊆ V (G + H) is geodetic hop dominating set of G + H if and only if S = SG ∪ SH , where SG and SH are pointwise non-dominating sets of G and H, respectively, such that (i) SG is a 2-path closure absorbing set in G whenever ⟨SH⟩ is a complete subgraph of H and (ii) SH is a 2-path closure absorbing set in H whenever ⟨SG⟩ is a complete subgraph of G. Proof. Suppose that S is a geodetic hop dominating set of G+H. Let SG = S ∩V (G) and SH = S ∩ V (H). Since S is a hop dominating set, by Theorem 3, SG and SH are pointwise non-dominating sets of G and H, respectively. Next, suppose that ⟨SH⟩ is a C.J. Saromines, S. Canoy, Jr., / Eur. J. Pure Appl. Math, 16 (3) (2023), 1568-1579 1575 complete subgraph of H. If SG = V (G), then we are done. Suppose SG ̸= V (G). Let x ∈ V (G) \ SG. Since S is a geodetic set of G + H, there exist y, z ∈ S such that x ∈ IG+H(y, z). Since ⟨SH⟩ is complete, y, z ∈ SG. Thus, dG(y, z) = 2. Hence, SG is 2-path closure absorbing set in G. Similarly, (ii) holds. Conversely, suppose S satisfies the given conditions. By Theorem 3, S is a hop dom- inating set of G + H. Let u ∈ V (G + H) \ S. Suppose u ∈ V (H) \ SH . If ⟨SG⟩ is non-complete, then there exist v, w ∈ SG ⊆ S such that dG+H(v, w) = 2. Hence, u ∈ IG+H(v, w). If ⟨SG⟩ is complete, then there exist s, t ∈ SH ⊆ S such that dH(s, t) = 2 and u ∈ IG(s, t) = IG+H(s, t) by (ii). Similarly, there exist p, q ∈ S such that [p, u, q] is a geodesic in G +H if u ∈ V (G) \ SG. Therefore, S is a geodetic hop dominating set of G+H. Lemma 1. Let G be a non-complete graph. Then the following hold: (i) If D is a pnd-set of G such that ⟨D⟩ is complete, then D ∪ {v} is a pointwise non- dominating set and ⟨D ∪ {v}⟩ is non-complete for every v ∈ V (G) \D. (ii) If E is a ρ2pnd-set of G, then ⟨E⟩ is non-complete. Proof. (i) Let v ∈ V (G)\D. Since D is a pointwise non-dominating set, D∪{v} is a pointwise non-dominating set and there exists w ∈ D \ NG(v). Therefore, ⟨D ∪ {v}⟩ is non- complete. (ii) If E = V (G), then we are done. Suppose E ̸= V (G). Let x ∈ V (G) \ E. Since E is a 2-path closure absorbing, there exist p, q ∈ E such that dG(p, q) = 2 and x ∈ IG(p, q). Therefore, ⟨E⟩ is non-complete. Before proceeding to the next result, we denote the family C of graphs by C = {G : G has a pnd-set which induces a non-complete graph} . Lemma 2. Let G be a non-complete graph. If G /∈ C, then pnd(G) < ρ2pnd(G), that is, pnd(G) + 1 ≤ ρ2pnd(G). Proof. Let S be a pnd-set of G. Then ⟨S⟩ is complete because G /∈ C. Therefore, pnd(G) < ρ2pnd(G) by Remark 3 and Lemma 1(ii). Corollary 4. Let G and H be any two non-complete graphs of orders m and n, respectively. Then γhg(G) =  pnd(G) + pnd(H), if G,H ∈ C min {ρ2pnd(G) + pnd(H), pnd(G) + pnd(H) + 1} if G ∈ C and H /∈ C min {pnd(G) + ρ2pnd(H), pnd(G) + pnd(H) + 1} if G /∈ C, H ∈ C min{ρ2pnd(G) + pnd(H), pnd(G) + ρ2pnd(H) pnd(G) + pnd(H) + 2} if G,H /∈ C. C.J. Saromines, S. Canoy, Jr., / Eur. J. Pure Appl. Math, 16 (3) (2023), 1568-1579 1576 Proof. Let S be a γhg-set of G + H. Then SG = S ∩ V (G) and SH = S ∩ V (H) are pointwise non-dominating sets of G and H, respectively, by Theorem 4. Hence, if G and H are in C, then SG and SH are pnd-sets of G and H, respectively. Therefore, γhg(G+H) = pnd(G) + pnd(H) if G,H ∈ C. Next, suppose that G ∈ C and H /∈ C. Let DG be a ρ2pnd-set of G and let DH be a pnd- set of H. Then ⟨DG⟩ is non-complete by Lemma 1(ii). Since H /∈ C, ⟨DH⟩ is complete. Hence, DH ̸= V (H) because H is non-complete. Let w ∈ V (H) \ DH . By Lemma 1(i), D ′ H = DH ∪ {w} is a pointwise non-dominating set of H and 〈 D ′ H 〉 is non-complete. Let D ′ G be a pnd-set of G such that 〈 D ′ G 〉 is non-complete. Then S1 = DG ∪ DH and S2 = D ′ G ∪ D ′ H are geodetic hop dominating sets of G + H by Theorem 4. Thus, γhg(G+H) ≤ |S1| = ρ2pnd(G) + pnd(H) and γhg(G+H) ≤ |S2| = pnd(G) + pnd(H) + 1. Consequently, γhg(G+H) ≤ min {ρ2pnd(G) + pnd(H), pnd(G) + pnd(H) + 1} . Now, suppose that S∗ = S∗ G ∪ S∗ H is a γhg-set of G + H. Then S∗ G and S∗ H satisfy the conditions in Theorem 4. Suppose ρ2pnd(G) + pnd(H) ≤ pnd(G) + pnd(H) + 1. If ⟨S∗ H⟩ is complete, then S∗ G is pointwise non-dominating 2-path closure absorbing set of G by Theorem 4. It follows that γhg(G + H) = |S∗ G| + |S∗ H | ≥ ρ2pnd(G) + pnd(H). Suppose ⟨S∗ H⟩ is non-complete. Since H /∈ C, H is non-complete, and S∗ is γhg-set of G + H, |S∗ H | ≥ pnd(H) + 1 (see Lemma 1(i)). It follows that γhg(G+H) = |S∗| = |S∗ G|+ |S∗ H | ≥ pnd(G) + pnd(H) + 1 ≥ ρ2pnd(G) + pnd(H). Similar arguments may be used to show that γhg(G + H) ≥ pnd(G) + pnd(H) + 1 if pnd(G) + pnd(H) + 1 ≤ ρ2pnd(G) + pnd(H). Therefore, γhg(G+H) = min {ρ2pnd(G) + pnd(H), pnd(G) + pnd(H) + 1} if G ∈ C and H /∈ C. Similarly, γhg(G + H) = min {ρ2pnd(H) + pnd(G), pnd(G) + pnd(H) + 1} if G /∈ C and H ∈ C. Suppose G,H /∈ C. Let R = min{ρ2pnd(G) + pnd(H), pnd(G) + ρ2pnd(H), pnd(G) + pnd(H) + 2}. Clearly, γhg(G+H) ≤ min {ρ2pnd(G) + pnd(H), ρ2pnd(H) + pnd(G))} . Let S1 and S2 be pnd-sets of G and H, respectively. Let S ′ 1 = S1 ∪{p} and S ′ 2 = S2 ∪{q}, where p ∈ V (G)\S1 and q ∈ V (H)\S2. Then S ′ 1 and S ′ 2 are pointwise non-dominating sets of G and H, respectively, and 〈 S ′ 1 〉 and 〈 S ′ 2 〉 are non-complete by Lemma 1(i). Hence, C.J. Saromines, S. Canoy, Jr., / Eur. J. Pure Appl. Math, 16 (3) (2023), 1568-1579 1577 S ′ = S ′ 1 ∪ S ′ 2 is a geodetic hop dominating set of G+H by Theorem 5. It follows that γhg(G+H) ≤ ∣∣∣S′ ∣∣∣ = pnd(G) + pnd(H) + 2. Therefore, γhg(G+H) ≤ R. Let S◦ = S◦ G ∪ S◦ H be a γhg-set of G +H. Then S◦ H and S◦ H satisfy the conditions in Theorem 5. Consider the following cases: Case 1. ⟨S◦ H⟩ is complete. By Theorem 4, S◦ G is a pointwise non-dominating 2-path closure absorbing set of G and |S◦ G| ≥ ρ2pnd(G). Hence, γhg(G+H) = |S◦| = |S◦ G|+ |S◦ H | ≥ ρ2pnd(G) + pnd(H) ≥ R. Case 2. ⟨S◦ G⟩ is complete. Then S◦ H is a pointwise non-dominating and 2-path closure absorbing set of H. Hence, γhg(G+H) = |S◦| = |S◦ G|+ |S◦ H | ≥ ρ2pnd(H) + pnd(G) ≥ R. Case 3. ⟨S◦ G⟩ and ⟨S◦ H⟩ are non-complete. Then |S◦ G| ≥ pnd(G) + 1 and |S◦ H | ≥ pnd(H) + 1 by Lemma 1(i). Hence, γhg(G+H) = |S◦| = |S◦ G|+ |S◦ H | ≥ pnd(G) + pnd(H) + 2 ≥ R. Accordingly, γhg(G+H) = R. Corollary 5. Let G be a non-complete graph and n a positive integer. Then S ⊆ V (Kn+ G) is a geodetic hop dominating set of Kn + G if and only if S = V (Kn) ∪ SG, where SG is a pointwise non-dominating and 2-path closure absorbing set in G. In particular, γhg(Kn +G) = n+ ρ2pnd(G). Corollary 6. Let G and H be any two graphs of orders m and n respectively. Then (i) γhg(G+H) = m+ n if G and H are complete; (ii) γhg(K1,n−1) = γhg(K1 +Kn−1) = n for n ≥ 2; (iii) γhg(Fn) = 1 + ρ2pnd(Pn); (iv) γhg(Wn) = 1 + ρ2pnd(Cn); and (v) γhg(Km,n)= { 3 if m = 2 or n = 2. 4 otherwise. REFERENCES 1578 4. Conclusion A realization result involving the hop domination number and the geodetic hop domi- nation number was obtained. This result shows that the difference of these two parameters can be made arbitrarily large. The concept of pointwise non-dominating 2-path closure absorbing set was defined and studied for some graphs. 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