EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 5903 ISSN 1307-5543 – ejpam.com Published by New York Business Global k-Geodetic Hop Domination Defect in a Graph Jesica M. Anoche1,2,∗, 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 of Mathematical and Theoretical Physical Sciences-PRISM, MSU-Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. Let G = (V (G), E(G)) be a simple undirected graph. A set S ⊆ V (G) is a geodetic hop dominating set in G if for every v ∈ V (G)\S, there exist vertices x, y, z ∈ S such that dG(x, v) = 2 and v lies in a y-z geodesic, that is, v ∈ IG(y, z). The minimum cardinality of a geodetic hop dominating set of G, denoted by γhg(G), is called the geodetic hop domination number of G. The minimality of γhg(G) implies that if S ⊆ V (G) such that |S| < γhg(G), then there is at least one vertex of G that is not geodetically hop-dominated by S. The k-geodetic hop domination defect of G, denoted by ζhgk (G), is the minimum number of vertices of G that is not geodetically hop-dominated by any subset of vertices of G with cardinality γhg(G) − k. A set S ⊆ V (G) of cardinality γhg(G)−k for which |V (G)\Nhg G [S]| = ζhgk (G), where Nhg G [S] = N2 G[S]∩IG[S], is called a ζhgk -set of G. In this paper, we initiate the study of the concept of k-geodetic hop domination defect of a non-trivial graph G and investigate it for some known classes of graphs. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Geodetic set, hop domination, geodetic hop domination, k-geodetic hop domination defect 1. Introduction The domination number γ(G) of a graph G refers to the smallest number of vertices required to dominate all the vertices of G. Hence, if a set S of vertices of G has cardinality strictly less than γ(G), then definitely, there will be vertices ofG that will not be dominated by any of the vertices in S. Recently, Das et al. [1] introduced and studied the notion of k-domination defect of a graph, where k is a positive integer strictly less than the domination number of the graph. The authors in this study established various bounds on the k-domination defect of a graph in terms of the maximum degree and domination number of the graph, and other parameters. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5903 Email addresses: jesica.anoche@g.msuiit.edu.ph (J. Anoche), sergio.canoy@g.msuiit.edu.ph (S. Canoy, Jr.) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 2 of 17 For hop domination, a concept which is studied by many researchers (see for example, [2], [3], [4], [5], [6], and [7]), Anoche et al. [8] introduced and studied the notion of k-hop domination defect in a graph. The study obtained some bounds on the k-hop domination defect of a graph in terms of its order and maximum hop degree. Moreover, the k-hop domination defects of some classes of graphs have been determined. Some variants of domination and hop domination utilize the concept of geodetic set. The associated term geodetic number of a graph was introduced by Harary et al. [9]. Geodetic number and geodetic domination were considered in [10], [11], [12], [13], and [14]. Recently, Anoche et al. [15] introduced and studied the notion of k-geodetic domination defect of a graph. The authors in this study established some sharp bounds on the k- geodetic domination defect of a graph and computed its values for several well-known graphs. On the other hand, the notion of geodetic hop domination was introduced and investigated in [16], [17], [18] and [19]. Note that if G is a graph and S is a set of vertices of G with cardinality strictly less than the geodetic hop domination number γhg(G) of G, then there is at least one vertex outside S that is not geodetically hop-dominated, that is, has no hop neighbor in S or is not in any shortest path joining any two vertices in S. In this paper, we introduce the notion k-geodetic hop domination defect and study it for some classes of graphs. For a motivation of the study, consider an establishment with a large number of em- ployees which needs to make an annual evaluation of their workers. The manager chooses some workers to form a team of assessors to evaluate the performance of their co-workers. To be cost effective or to minimize costs, the manager ensures that this team will con- sist of the smallest number of members that can do the task. Moreover, to avoid bias in the assessment, an inspector should be non-biased, that is, neither be close friends nor enemies with any of the workers he or she is assigned to assess. This situation can be modeled by constructing a graph where each vertex represents a worker and an edge between two workers represents possible bias, that is, if the two workers are either close friends or enemies. Here, every worker who is not in the team will be evaluated by a non-biased inspector who is at a distance two from him/her. Moreover, for the purpose of visibility and monitoring that the policy is strictly followed, it is imposed that every worker must be in a shortest path connecting two members of the team. However, due to possible budgetary constraints, the required minimum number of evaluators may not always be attained or sometimes, it may happen that during the course of the evaluation process an evaluator may be absent and, subsequently, unable to perform his or her task. Consequently, some workers may not be evaluated accordingly. Finding the number of unevaluated workers with respect to a given team of evaluators not reaching the required minimum number of membership may be of help to the management. Situations such as this led us to introduce the concept of geodetic hop domination defect in a graph. 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 J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 3 of 17 called a u-v geodesic. The distance between two subsets A and B of V (G) is given by dG(A,B) = min{dG(a, b) : a ∈ A and b ∈ B}. The open neighborhood of a point u is the set NG(u) consisting of all points v which are adjacent to u. The closed neighbor- hood 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 and NG[A] = NG(A) ∪ A is called the closed neighborhood of A. A vertex v of G is isolated if |NG(v)| = 0. The set containing all the isolated vertices of G is denoted by I(G). The open hop neighborhood of a point 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 of G if NG[S] = V (G). The smallest cardinality of a dominating set of G, denoted by γ(G), is called the domination number of G. A dominating set S of G with |S| = γ(G), is called a γ-set 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. The set S is a 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 of G if for each x ∈ V (G) \ S, there exists z ∈ S such that dG(x, z) = 2. The smallest cardinality of a hop dominating set of G, denoted by γh(G), is called the hop domination number of G. A hop dominating set S of G with |S| = γh(G) is called a γh-set of G. A set S ⊆ V (G) is geodetic hop dominating if it is both a geodetic and a hop domi- nating 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. Let G be a non-trivial graph of order n and let 1 ≤ k < γhg(G). Let S ⊆ V (G) with cardinality |S| = γhg(G)− k and let Nhg G [S] = N2 G[S] ∩ IG[S], the set of geodetically hop- dominated set of vertices of G. The set V (G) \ Nhg G [S] is called the k-geodetic hop domination defect set of S and the k-geodetic hop domination defect of S is ζhgk (S) = |V (G) \ Nhg G [S]| = n − |Nhg G [S]|. The minimum cardinality of a k-geodetic hop domination defect set in G, denoted by ζhgk (G), is called the k-geodetic hop domination defect of G, i.e., ζhgk (G) = min{ζhgk (S) : S ⊆ V (G) with |S| = γhg(G)− k}. A set S ⊆ V (G) of cardinality γhg(G)− k for which |V (G) \Nhg G [S]| = ζhgk (G) is called a ζhgk -set of G. Thus, 〈 Nhg G [S] 〉 is an induced subgraph of G with n− ζhgk (G) vertices and geodetic hop domination number γhg − k. Consider the graph G in Figure 1. Then D = {a, b, x, y} is a γhg-set of G, i.e., J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 4 of 17 γhg(G) = 4. If k = 1, then S1 = {a, b, x} is a ζhg1 -set ofG. SinceNhg G [S1] = {a, b, c, d, u, v, x}, it follows that ζhg1 (G) = ζhg1 (S) = |V (G)| − |Nhg G [S1]| = 8− 7 = 1. Clearly, the set {a, b, d} is not a ζhg1 -set of G because Nhg G [{a, b, d}] = {a, b, c, d, u, v}, i.e., ζhg1 ({a, b, d}) = 8 − 6 = 2. If k = 2, then S2 = {a, v} is a ζhg2 -set of G and Nhg G [S2] = {a, c, d, u, v}. Hence, ζhg2 (G) = 8 − 5 = 3. Observe that the sets {a, x} and {b, x} are not ζhg2 -sets of G. Finally, if k = 3, then any 1-element subset S3 of V (G) is a ζhg3 -set of G. Since Nhg G [S3] = S3, it follows that ζ hg 3 (G) = |V (G)|− |Nhg G [S3]| = 8− 1 = 7. ................................................................................................................ ................................................................................................................ ................................................................................................................ .................................... ................................................................................................................ ................................................................................................................ ................................................................................................................ .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ... .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ... .................................... .................................... a b x y c u d v Figure 1: Graph G with γhg(G) = 4 3. Results Theorem 1 ([19]). Let n be positive integer. Then each of the following holds. (i) For a path Pn on n vertices, we have γhg(Pn) =  n if n = 1, 2 n+6 3 if n ≡ 0(mod 3) n+2 3 if n ≡ 1(mod 3) n+4 3 if n ≡ 2(mod 3). (ii) For a cycle Cn on n vertices, we have γhg(Cn) =  3 if n = 3, 4, 5 n 3 if n ≡ 0(mod 3) n+2 3 if n ≡ 1(mod 3) n+4 3 if n ≡ 2(mod 3). (iii) For a complete graph Kn, we have γhg(Kn) = n. Corollary 1 ([18]). 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; J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 5 of 17 (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. Remark 1. Let G1, G2, · · · , Gr be the components of a graph G. Then each of the following holds: (i) γhg(G) = ∑r j=1 γhg(Gj). (ii) If Aj ⊆ V (Gj) for each j ∈ [r] = {1, 2, · · · , r} and A = ∪r j=1Aj, then Nhg G [A] = ∪r j=1N hg G [Aj ] (a disjoint union). Theorem 2. Let G1, G2, · · · , Gr be the components of graph G and let ζhg1 (Gi) be the 1-geodetic hop domination defect of Gi for each i ∈ [r] = {1, 2, · · · , r}. Then ζhg1 (G) = min{ζhg1 (Gi) : i ∈ [r]}. Proof. Let γhg(Gi) and γhg(G) be the geodetic hop domination numbers of Gi and G, respectively. By Remark 1(i), γhg(G) = ∑r i=1 γhg(Gi). For each i ∈ [r], let Di be a ζhg1 -set of Gi. Then |Di| = γhg(Gi) − 1 and ζhg1 (Gi) = |V (Gi) − Nhg G [Di]|. Let j ∈ [r] be such that ζhg1 (Gj) = min{ζhg1 (Gi) : i ∈ [r]}. Let Si be a γhg-set in Gi for each i ∈ [r] and let S = (∪i∈[r]\{j}Si) ∪Dj . Then |S| = ∑ i∈[r]\{j} |Si|+ |Dj | = γhg(G)− 1 and, by Remark 1(ii), |Nhg G [S]| = |Nhg Gj [Dj ]|+ ∑ i∈[r]\{j} |Nhg Gi [Si]| = |V (Gj)| − ζhg1 (Gj) + ∑ i∈[r]\{j} |V (Gi)| = r∑ i=1 |V (Gi)| − ζhg1 (Gj). Thus, in G, ζhg1 (S) = |V (G)| − |Nhg G [S]| = ζhg1 (Gj). We claim that ζhg1 (S) is the minimum among all subsets of V (G) with cardinality γhg(G)− 1. To this end, suppose there exists Q ⊆ V (G) such that |Q| = γhg(G)− 1 and ζhg1 (Q) < ζhg1 (S). Let Q = Q1 ∪Q2 ∪ · · · ∪Qr where Qi ⊆ V (Gi) for each i ∈ [r]. Since |Q| = γhg(G) − 1, at least one Qt is not J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 6 of 17 a geodetic hop dominating set of Gt by Remark 1(i). Thus, |Qt| = γhg(Gt) − 1 and ζhg1 (Qt) ≥ ζhg1 (Gt) ≥ ζhg1 (Gj). Hence, ζhg1 (Q) = |V (G)| − |Nhg G [Q]| = r∑ i=1 (|V (Gi)| − |Nhg Gi [Qi]|) ≥ |V (Gt)| − |Nhg Gt [Qt]| = ζhg1 (Qt) ≥ ζhg1 (Gj) = ζhg1 (S), contrary to the assumption that ζhg1 (Q) < ζhg1 (S). Therefore, ζhg1 (G) = ζhg1 (S) = ζhg1 (Gj). Theorem 3. Let G be a graph with I(G) ̸= ∅ and suppose |I(G)| = r. Then ζhgj (G) = j for every j ∈ [r] = {1, 2, · · · , r} and ζhgk (G) = r + ζhgk−r(G ′) for every k ∈ {r + 1, · · · , γhg(G)− 1}, where G′ = ⟨V (G) \ I(G)⟩. Proof. Let I(G) = {v1, v2, · · · , vr} and let S be a γhg-set in G. Then I(G) ⊆ S. Let j ∈ [r]. Then D = S \ {v1, v2, · · · , vj} is a ζhgj -set of G and |Nhg G [D]| = |Nhg G [S]| − |Nhg G [{v1, v2, · · · , vj}]| = |V (G)| − j. Hence, ζhgj (G) = |V (G)| − (|V (G)| − j) = j. Next, let k ∈ {r + 1, · · · , γhg(G) − 1}. Then S0 = S \ I(G) is γhg-set in G′ = ⟨V (G) \ I(G)⟩. Hence, γhg(G ′) = γhg(G) − r. Since k ≤ γhg(G) − 1, k − r ≤ γhg(G) − (r + 1) < γhg(G) − r. Let S′ be a ζhgk−r-set of G′. Then |S′| = (γhg(G) − r) − (k − r) = γhg(G) − k and ζhgk−r(G ′) = |V (G′)| − |Nhg G′ [S′]| = (|V (G)| − r) − |Nhg G [S′]|. This implies that |V (G)| − |Nhg G [S′]| = r + ζhgk−r(G ′). There- fore, since S′ is also a ζhgk -set of G, ζhgk (G) = r + ζhgk−r(G ′). Theorem 4. Let G be a non-trivial graph of order n and let k be a positive integer with k ≤ γhg(G)− 1. Then ζhgk (G) ≤ n− γhg(G) + k. Proof. Let k be a positive integer with k ≤ γhg(G) − 1 and let S be a ζhgk -set of G. Then |S| = γhg(G)− k and ζhgk (G) = n− |Nhg G [S]|. Since S ⊆ Nhg G [S], it follows that ζhgk (G) = n− |Nhg G [S]| ≤ n− |S| = n− (γhg(G)− k) = n− γhg(G) + k. This proves the assertion. Remark 2. The bound in Theorem 4 is sharp. Strict inequality is also possible. To see this, consider the graph G = C5 in Figure 2. By Theorem 1(ii), γhg(G) = 3. Let k = 1. Then any set S1 ⊆ V (G) with |S1| = 2 is a ζhg1 -set of G and Nhg G [S1] = S1. J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 7 of 17 Thus, ζhg1 (G) = |V (G)| − γhg(G) + k = 5− 3 + 1 = 3. If k = 2, then any 1-element subset S2 is a ζhg2 -set of G and Nhg G [S2] = S2. Thus, ζhg2 (G) = 5 − 1 = 4. Since |V (G)| − γhg(G) + k = 5 − 3 + k = 4, the equality ζhgk (C5) = n− γhg(C5) + k holds. .............. ............. ............. ............. ............. ............. ............. ............. ............. ............. .. .................................... ..................................................................................................................................... .................................... ........... .......... .......... .......... .......... .......... .......... .......... .......... .. .. .................................. ........................................................................................................ .................................... ............................................................................................. .................................... .................................... a b c de Figure 2: G = C5 and γhg(G) = 3 For strict inequality, consider H = P10 in Figure 3. By Theorem 1(i), γhg(H) = 4. Let P10 = [v1, v2, · · · , v10] and let k = 1. It can easily be verified that S1 = {v1, v4, v7} is a ζhg1 -set of H and Nhg H [S1] = {v1, v2, v3, · · · , v7}. Hence, ζhg1 (H) = 10− 7 = 3 < 7 = |V (H)| − γhg(H) + k. If k = 2, then S2 = {v1, v4} is a ζhg2 -set of H and Nhg H [S2] = {v1, v2, v3, v4}. This implies that ζhg2 (H) = 6 < 8 = |V (H)| − γhg(H) + k. ................................................................................................................ ................................................................................................................ ................................................................................................................ ................................................................................................................ ................................................................................................................ ................................................................................................................ ................................................................................................................ ................................................................................................................ ................................................................................................................ .................................... v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 Figure 3: H = P10 and γhg(H) = 4 Theorem 5. Let G be a graph of order n ≥ 2 and k ≤ γhg(G)−1, then 1 ≤ ζhgk (G) ≤ n−1. Proof. Let S be a ζhgk -set of G. Then |S| = γhg(G)− k. Hence, V (G) \Nhg G [S] ̸= ∅. It follows that ζhgk (G) = |V (G)| − |Nhg G [S]| ≥ 1. Also, since |Nhg G [S]| ≥ 1, ζhgk (G) = |V (G)| − |Nhg G [S]| ≤ n− 1. This proves the assertion. Theorem 6. Let G be a non-trivial graph of order n. Then ζhg1 (G) = 1 if and only if there exists v ∈ V (G) such that γhg(G− v) = γhg(G)− 1. Proof. Suppose ζhg1 (G) = 1 and let S be a ζhg1 -set of G. Then |S| = γhg(G) − 1 and ζhgk (G) = |V (G) \ Nhg G [S]| = 1. Let v ∈ V (G) \ Nhg G [S]. Then Nhg G [S] = V (G) \ {v}. Therefore, γhg(G− v) = γhg( 〈 Nhg G [S] 〉 ) = γhg(G)− 1. J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 8 of 17 Conversely, let v ∈ V (G) such that γhg(G − v) = γhg(G) − 1. Then there ex- ists D ⊆ V (G) with |D| = γhg(G) − 1 and Nhg G [D] = V (G) \ {v}. This implies that ζhg1 (G) = |V (G) \Nhg G [D]| = |{v}| = 1. Therefore, ζhg1 (G) = 1. Lemma 1. Let G be a non-trivial graph of order n. If k = γhg(G)−1, then ζhgk (G) = n−1. Proof. Let k = γhg(G) − 1 and let S be a ζhgk -set of G. Then |S| = 1, say, S = {x} and ζhgk (G) = ζhgk (S) = n − |Nhg G [S]|. Since x ∈ N2 G[S] and IG[S] = S, it follows that Nhg G [S] = S. It follows that ζhgk (G) = n− |Nhg G [S]| = n− 1. Lemma 2. Let G be a connected graph of order n. If S is a clique in G, then Nhg G [S] = S. Proof. Let S is be clique in G. Then N2 G[S] = IG[S] = S. Thus, Nhg G [S] = N2 G[S] ∩ IG[S] = S. The next result shows that the bound given in Theorem 4 is sharp. Theorem 7. If Kn is a complete graph on n vertices, where n ≥ 2, and 1 ≤ k ≤ n − 1, then ζhgk (Kn) = k. Proof. Let k be a positive integer with k ≤ γhg(Kn)−1. Since γhg(Kn) = n, k ≤ n−1. Let S be a ζhgk -set of Kn. Then S is a clique, |S| = n − k and ζhgk (G) = n − |Nhg G [S]|. Therefore, by Lemma 2, ζhgk (G) = n− (n− k) = k. Theorem 8. For a path Pn with n vertices, ζhgk (Pn) =  1 if n = 2 or n = 3r and k = 1 3k − 4 if n = 3r and k ≥ 2 3k if n = 3r + 1 and k ≤ r 3k − 2 if n = 3r + 2 and k ≤ r + 1. Proof. Let Pn = [v1, v2, · · · , vn] and let r ≥ 1. Consider the following cases: Case 1: n = 3r. By Theorem 1 (i), γhg(Pn) = 3r+6 3 . Let k = 1 and let S1 = {v1, v4, · · · , v3r−2} ∪ {v3r−1}. Then Nhg G [S1] = V (Pn) \ {v3r}. Hence, ζhgk (S1) = n − |Nhg G [S1]| = 3r − (3r − 1) = 1. Thus, ζhgk (Pn) = 1. Next, let k ≥ 2. Then 3r+6 3 − k = r − k + 2. Choose an (r − k + 2)-element set S2 = {v1, v4, · · · , v3r−3k+4}. Then S2 is a ζhgk -set of Pn and Nhg G [S2] = {v1, v2, v3, v4, · · · , v3r−3k+4}. This implies that ζhgk (Pn) = ζhgk (S2) = n− |Nhg G [S2]| = 3r − (3r − 3k + 4) = 3k − 4. Case 2: n = 3r + 1. By Theorem 1 (i), γhg(Pn) = 3r+3 3 . Then k ≤ r and 3r+3 3 − k = r − k + 1. Choose J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 9 of 17 an (r − k + 1)-element set S3 = {v1, v4, · · · , v3r−3k+1}. Then S3 is a ζhgk -set of Pn and Nhg G [S3] = {v1, v2, v3, v4, · · · , v3r−3k+1}. Thus, ζhgk (Pn) = ζhgk (S3) = n− |Nhg G [S3]| = (3r + 1)− (3r − 3k + 1) = 3k. Case 3: n = 3r + 2. By Theorem 1 (i), γhg(Pn) = 3r+6 3 = r + 2. Here, k ≤ r + 1 and γhg(Pn)− k = r − k + 2. Consider an (r− k+2)-element set S4 = {v1, v4, · · · , v3r−3k+4}. The set S4 is a ζhgk -set of Pn and Nhg G [S4] = {v1, v2, v3, v4, · · · , v3r−3k+4}. Therefore, ζhgk (Pn) = ζhgk (S4) = n− |Nhg G [S4]| = 3r + 2− (3r − 3k + 4) = 3k − 2. Theorem 9. If Cn is the cycle with n vertices and k ≤ γhg(Cn)− 1, then ζhgk (Cn) =  n+ k − 3 if n = 3, 4, 5 3k + 2 if n = 3r, r ≥ 2 and k = γhg(Cn)− 1 or r is odd, r ≥ 3, and k = γhg(Cn)− 2 3k if n = 3r, r ≥ 4 and, 1 ≤ k ≤ γhg(Cn)− 3 or n = 3r, r is even, r ≥ 4, and k = γhg(Cn)− 2 or n = 3r + 1 and k = γhg(Cn)− 1 or n = 3r + 1, r is even, and k = γhg(Cn)− 2 1 if n = 3r + 2, r ≥ 2, and k = 1 2 if n = 3r + 1, r ≥ 3 and k = 1 3k − 2 if n = 3r + 1, r ≥ 4, and 2 ≤ k ≤ γhg(Cn)− 3 or n = 3r + 1, r is odd, r ≥ 3, and k = γhg(Cn)− 2 or n = 3r + 2 and k = γhg(Cn)− 1 or n = 3r + 2, r is odd, r ≥ 3, and k = γhg(Cn)− 2 or n = 3r + 2, r = 2, and k = γhg(Cn)− 2 3k − 4 if n = 3r + 2, r ≥ 3 and 2 ≤ k ≤ γhg(Cn)− 3 or n = 3r + 2, r is even, r ≥ 4, and k = γhg(Cn)− 2 Proof. Let Cn = [v1, v2, · · · , vn, v1] and let k ≤ γhg(Cn) − 1. Consider the following cases: Case 1: n = 3 or 4, or 5. By Theorem 1(ii), γhg(Cn) = 3. Choose a (3 − k)-element set D1 = {v1, v3−k}. This J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 10 of 17 set is a ζhgk -set of Cn. Since D1 is a clique, Nhg G [D1] = D1. Hence, ζhgk (Cn) = ζhgk (D1) = n− |Nhg G [D1]| = n− (3− k) = n+ k − 3. Case 2: n = 3r where r ≥ 2. By Theorem 1(ii), γhg(Cn) = 3r 3 = r. Thus, k ≤ r−1. Consider the following subcases: Subcase 1: k = γhg(Cn)− 1. By Lemma 1, ζhgk (Cn) = n− 1 = 3r − 1 = 3k + 2. Subcase 2: r is odd and k = γhg(Cn)− 2. Let D2 = {v1, v4}. Then D2 is a ζhgk -set of Cn and Nhg G [D2] = {v1, v2, v3, v4}. Thus, ζhgk (Cn) = ζhgk (D2) = 3r − |Nhg G [D2]| = 3r − 4 = 3k + 2. Subcase 3: r ≥ 4 is even and k = γhg(Cn)− 2. Let D3 = {v1, v 3r+2 2 }. Then D3 is a ζhgk -set of Cn and Nhg G [D3] = {v1, v3, v 3r−2 2 , v 3r+2 2 , v 3r+6 2 , v3r−1}. Hence, ζhgk (Cn) = ζhgk (D3) = 3r + 1− |Nhg G [D3]| = 3r − 6 = 3k. Subcase 4: r ≥ 4 and 1 ≤ k ≤ γhg(Cn)− 3. Let k ≤ ⌊γhg(Cn)−2 2 ⌋ and let D4 = {v1, v4, · · · , v3r−3k−2}. Then D4 is a ζhgk -set of Cn and Nhg G [D4] = {v1, v2, v3, v4, · · · , v3r−3k−2, v3r−3k, v3r−1}. Thus, ζhgk (Cn) = ζhgk (D4) = n − |Nhg G [D4]| = 3r − [(3r − 3k − 2) + 2] = 3k. Next, let ⌊γhg(Cn)−2 2 ⌋ < k ≤ γhg(Cn) − 3. Choose an (r − k)-element set D5 = {v⌈ 3r+2 2 ⌉, v1, v4, · · · , v3r−3k−5}. Then D5 is a ζhgk -set of Cn and Nhg G [D5] = {v1, v2, v3, v4, · · · , v3r−3k−5, v3r−3k−3, v⌈ 3r+2 2 ⌉−2, v⌈ 3r+2 2 ⌉, v⌈ 3r+2 2 ⌉+2, v3r−1}. This implies that ζhgk (Cn) = ζhgk (D5) = n− |Nhg G [D5]| = 3r − [(3r − 3k − 5) + 5] = 3k. Case 3: n = 3r + 1. By Theorem 1(ii), γhg(Cn) = 3r+1+2 3 = r + 1. Thus, k ≤ r. Consider the following subcases: J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 11 of 17 Subcase 1: k = γhg(Cn)− 1. By Lemma 1, ζhgk (Cn) = n− 1 = 3r + 1− 1 = 3r = 3k. Subcase 2: r is even and k = γhg(Cn)− 2. Let D6 = {v1, v4}. Then D6 is a ζhgk -set of Cn and Nhg G [D6] = {v1, v2, v3, v4}. Hence, ζhgk (Cn) = ζhgk (D6) = 3r + 1− |Nhg G [D6]| = 3r + 1− 4 = 3k. Subcase 3: r ≥ 3 and k = 1. Let D7 = {v1, v4, v7, · · · , v3r−2} be an r-element set. Then Nhg G [D7] = {v1, v2, v3, v4, · · · , v3r−2, v3r}. Thus, D7 is a ζhg1 -set of Cn. Hence, ζhg1 (Cn) = ζhg1 (D7) = n− |Nhg G [D7]| = 3r + 1− (3r − 1) = 2. Subcase 4: r ≥ 4 and 2 ≤ k ≤ γhg(Cn)− 3 . Let k ≤ ⌊γhg(Cn)−2 2 ⌋ and D8 = {v1, v4, · · · , v3r−3k+1}. Then D8 is a ζhgk -set of Cn and Nhg G [D8] = {v1, v2, v3, v4, · · · , v3r−3k+1, v3r−3k+3, v3r}. This implies that ζhgk (Cn) = ζhgk (D8) = n−|Nhg G [D8]| = 3r+1− [(3r−3k+1)+2] = 3k−2. Hence, ζhgk (Cn) = 3k− 2. Next, let ⌊γhg(Cn)−2 2 ⌋ < k ≤ γhg(Cn)− 3. Choose an (r− k+1)- element set D9 = {v⌈ 3r+3 2 ⌉, v1, v4, · · · , v3r−3k−2}. Then D9 is a ζhgk -set of Cn and Nhg G [D9] = {v1, v2, v3, v4, · · · , v3r−3k−2, v3r−3k, v⌈ 3r+3 2 ⌉−2, v⌈ 3r+3 2 ⌉, v⌈ 3r+3 2 ⌉+2, v3r}. This implies that ζhgk (Cn) = ζhgk (D9) = n−|Nhg G [D9]| = 3r+1− [(3r−3k−2)+5] = 3k−2. Hence, ζhgk (Cn) = 3k − 2. Subcase 5: r ≥ 3 is odd and k = γhg(Cn)− 2. Let D10 = {v1, v 3r+3 2 }. Then D10 is a ζhgk -set of Cn and Nhg G [D10] = {v1, v3, v 3r−1 2 , v 3r+3 2 , v 3r+7 2 , v3r}. Hence, ζhgk (Cn) = ζhgk (D10) = 3r + 1− |Nhg G [D10]| = 3r + 1− 6 = 3k − 2. Case 4: n = 3r + 2 where r ≥ 2. By Theorem 1(ii), γhg(Cn) = 3r+6 3 = r + 2. Thus, k ≤ r + 1. Consider the following subcases: J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 12 of 17 Subcase 1: k = 1. Let D11 = {v1, v4, · · · , v3r−2, v3r+1} be an (r + 1)-element set. Then Nhg G [D11] = V (Cn) \ {v3r+2}. Thus, D11 is a ζhg1 -set of Cn. Hence, ζhg1 (Cn) = ζhg1 (D11) = n − |Nhg G [D11]| = 3r + 2− (3r + 1) = 1. Subcase 2: k = γhg(Cn)− 1. By Lemma 1, ζhgk (Cn) = n− 1 = 3r + 2− 1 = 3r + 1 = 3k − 2. Subcase 3: r is odd and k = γhg(Cn)− 2. Let D12 = {v1, v4}. Then D12 is a ζhgk -set of Cn and Nhg G [D12] = {v1, v2, v3, v4}. Hence, ζhgk (Cn) = ζhgk (D12) = 3r + 2− |Nhg G [D12]| = 3r + 2− 4 = 3k − 2. Subcase 4: r is even and k = γhg(Cn)− 2. Let r = 2. Then D13 = {v1, v4} is a ζhgk -set of Cn and Nhg G [D13] = {v1, v2, v3, v4}. Hence, ζhgk (Cn) = ζhgk (D13) = 3r + 2− |Nhg G [D13]| = 3r + 2− 4 = 3k − 2. Suppose r ≥ 4. Then D14 = {v1, v 3r+4 2 } is a ζhgk -set of Cn and Nhg G [D14] = {v1, v3, v 3r+3 2 −2, v 3r+3 2 , v 3r+3 2 +2, v3r+1}. Hence, ζhgk (Cn) = ζhgk (D14) = 3r + 1− |Nhg G [D14]| = 3r + 2− 6 = 3k − 4. Subcase 5: r ≥ 3 and 2 ≤ k ≤ γhg(Cn)− 3 . Let k ≤ ⌊γhg(Cn)−2 2 ⌋ and let D15 = {v1, v4, · · · , v3r−3k+4}. Then D15 is a ζhgk -set of Cn and Nhg G [D15] = {v1, v2, v3, v4, · · · , v3r−3k+4, v3r−3k+6, v3r+1}. Thus, ζhgk (Cn) = ζhgk (D15) = n− |Nhg G [D15]| = 3r+2− [(3r− 3k+4)+2] = 3k− 4. Next, let ⌊γhg(Cn)−2 2 ⌋ < k ≤ γhg(Cn)− 3. Choose an (r − k + 2)-element set D16 = {v⌈ 3r+5 2 ⌉, v1, v4, · · · , v3r−3k+1}. Then D16 is a ζhgk -set of Cn and Nhg G [D16] = {v1, v2, v3, v4, · · · , v3r−3k+1, v3r−3k+3, v⌈ 3r+5 2 ⌉−2, v⌈ 3r+5 2 ⌉, v⌈ 3r+5 2 ⌉+2, v3r+1}. This implies that ζhgk (Cn) = ζhgk (D16) = n− |Nhg G [D16]| = 3r + 2− [(3r − 3k + 1) + 5] = 3k − 4. This proves the assertion. J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 13 of 17 Theorem 10. If G = Km,n is a complete bipartite graph with 1 ≤ m ≤ n and k is a postive integer with k ≤ γhg(Km, n)− 1, then ζhgk (G) =  k if m = 1 and k ≤ n n+ k − 1 if m = 2 and k ≤ 2 m− 2 if m ≥ 3 and k = 1 m+ n+ k − 4 if m ≥ 3 and k = 2, 3. Proof. Let k ≤ γhg(G)− 1. Consider the following cases: Case 1: m = 1. If n = 1, then k = 1 and ζhg1 (G) = 1. Suppose n ≥ 2. By Corollary 1(ii), γhg(G) = n+ 1. Let k ≤ n and let S ⊆ V (K1,n) with |S| = n− k + 1. Let v0 be the central vertex of K1,n and V (K1,n) = {v0, w1, w2, · · · , wn}. Suppose wj ∈ V (G) \ S for some 1 ≤ j ≤ n. Since wj /∈ IG[S], wj /∈ Nhg G [S]. Suppose v0 /∈ S. Since v0 /∈ N2 G[S], v0 /∈ Nhg G [S]. Therefore, Nhg G [S] ∩ (V (G) \ S) = ∅, i.e., Nhg G [S] = S. This implies that ζhgk (G) = ζhgk (S) = |V (G)| − |Nhg G [S]| = (n+ 1)− (n− k + 1) = k. Case 2: m = 2. By Corollary 1(v), γhg(G) = 3. Hence, k ≤ 2 and any set S ⊆ V (G) with |S| = 3− k is a ζhgk -set of G. Clearly, Nhg G [S] = S. Hence, ζhgk (G) = (n+ 2)− (3− k) = n+ k − 1. Case 3: m ≥ 3. By Corollary 1(v), γhg(G) = 4. This implies that k ≤ 3. If k = 3, then ζhg3 (G) = m+n− 1 = m+n+k− 4 by Lemma 1. If k = 2, then any set S ⊆ V (G) with |S| = 2 is a ζhg2 -set of G and Nhg G [S] = S. Hence, ζhg2 (G) = m+ n− |S| = m+ n− 2 = m+ n+ k− 4. Suppose k = 1. Let A and B be the partite sets of G with |A| = m and |B| = n, re- spectively, and let S be a set of vertices of G with |S| = 3. Consider the following subcases: Subcase 1: S ⊂ A or S ⊂ B. Then Nhg G [S] = S and ζhg1 (S) = m+ n− 3 . Subcase 2: |S ∩A| = 1 and |S ∩B| = 2. Then Nhg G [S] = A∪(S∩B). Thus, Nhg G [S] = m+2. Hence, ζhg1 (S) = m+n−(m+2) = J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 14 of 17 n− 2. Subcase 3: |S ∩A| = 2 and |S ∩B| = 1. Then Nhg G [S] = (S∩A)∪B. Thus, Nhg G [S] = 2+n. Hence, ζhg1 (S) = m+n− (n+2) = m− 2. Therefore, S is a ζhg1 -set ofG if |S∩A| = 2 and |S∩B| = 1 becausem−2 ≤ n−2 ≤ m+n−3. Accordingly, ζhg1 (G) = m− 2. Theorem 11. If G = Km1,m2,··· ,mr is a complete multipartite graph with 2 ≤ m1 ≤ m2 ≤ · · · ≤ mr, where r ≥ 3, then γhg(G) = { r + 1 if m1 = 2 r + 2 if m1 ≥ 3. Proof. Let Q1, Q2, · · · , Qr be the partite sets of G with |Qj | = mj for each j ∈ [r]. Let S be a geodetic hop dominating set of G. Suppose there exists j ∈ [r] = {1, 2, · · · , r} such that S ∩Qj = ∅. Then Qj ⊆ NG(S). This implies that Qj ∩N2 G(S) = ∅, contrary to the assumption that S is a hop dominating set. Therefore, S ∩Qj ̸= ∅ for each j ∈ [r]. Suppose |S ∩Qj | = 1 for each j ∈ [r]. Then [Q1 \ (S ∩Q1)] ∩ IG[S] = ∅, contrary to the assumption that S is a geodetic set. Thus, there exists t ∈ [r] such that |S ∩ Qt| ≥ 2. This implies that γhg(G) ≥ r + 1. If m1 = 2, then choose S1 such that |S1 ∩Q1| = 2 and |S1 ∩ Qj | = 1 for each j ∈ [r] \ {1}. Then S1 is a γhg-set of G. Hence, γhg(G) = r + 1. Supposem1 ≥ 3. Let S2 be a γhg-set of G. Suppose there exists exactly a single set Qt with |S2 ∩Qt| = 2. Then [Q1 \ (S2 ∩Q1)]∩ IG[S2] = ∅, a contradiction. This would imply that γhg(G) = |S2| ≥ r + 2. Consider a set D with the property that |D ∩Q1| = |D ∩Q2| = 2 and |D ∩Qj | = 1 for each j ∈ [r] \ {1, 2}. Then D is a geodetic hop dominating set of G and |D| = r+2. Since S2 is a γhg-set of G, it follows that γhg(G) = |S2| = |D| = r+2. Theorem 12. For a complete multipartite graph G = Km1,m2,··· ,mr where r ≥ 3 and 2 ≤ m1 ≤ m2 ≤ · · · ≤ mr, we have ζhgk (G) =  n− 1 if k = γhg(G)− 1 n− 2 if k = γhg(G)− 2∑k+1 j=2 mj if m1 = 2 and 1 ≤ k ≤ r − 2∑k j=1mj − 2 if m1 ≥ 3 and 1 ≤ k ≤ r − 2, where n = ∑r j=1mj. Proof. Let Q1, Q2, · · · , Qr be the partite sets of G with |Qj | = mj for each j ∈ [r]. Consider the following cases: J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 15 of 17 Case 1: k = γhg(G)− 1. By Lemma 1, ζhgk (G) = n− 1 = ∑r j=1mj − 1. Case 2: k = γhg(G)− 2. Then any 2-element subset S of V (G) is a ζhgk -set of G and Nhg G [S] = S. Hence, ζhgk (G) = ζhgk (S) = n− |Nhg G [S]| = n− 2 = ∑r j=1mj − 2. Case 3: m1 = 2 and 1 ≤ k ≤ r − 2. By Theorem 11, γhg(G) = r + 1. Consider the set D = {q11, q21, q1k+2, q 1 k+3, · · · , q1r} where qtj ∈ Qj for each j ∈ {1, k + 2, k + 3, · · · , r} and t ∈ {1, 2}. Then Nhg G [D] = ∪r j=k+2Qj ∪ {q11, q21}. It follows that ζhgk (G) ≤ ζhgk (D) = |V (G)| − |Nhg G [D]| = r∑ j=1 mj − [ r∑ j=k+2 mj + 2] = [2 + r∑ j=2 mj ]− [ r∑ j=k+2 mj + 2] = k+1∑ j=2 mj . Next, let S be a ζhgk -set of G. Then |S| = r − k + 1 and ζhgk (G) = ζhgk (S). Suppose S ⊆ Qj for some j ∈ [r]. Then Nhg G [S] = S and ζhgk (S) = n − r + k − 1. Suppose |S ∩ Qi| = 1 for each i ∈ R = {t ∈ [r] : S ∩ Qt ̸= ∅}. Then |R| = |S|, Nhg G [S] = S and ζhgk (S) = n− r + k − 1. For both cases, we have ζhgk (G) = ζhgk (S) = n− r+ k− 1 = [ k+1∑ j=2 mj +2+mk+2 + · · ·+mr]− (r− k+1) ≥ k+1∑ j=2 mj . Since only two vertices, say x and y, from a partite set are needed for the elements of the other partite sets to be in the interval IG(x, y), it can be shown that S always yields a k-geodetic hop domination defect greater than or equal to ∑k+1 j=2 mj . Therefore, ζhgk (G) = ζhgk (S) = ∑k+1 j=2 mj . Case 4: m1 ≥ 3 and 1 ≤ k ≤ r − 2. J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5903 16 of 17 By Theorem 11, γhg(G) = r + 2. Consider the set D′ = {q11, q21, q1k+1, q 1 k+2, · · · , q1r} where qtj ∈ Qj for each j ∈ {1, k + 1, k + 2, · · · , r} and t ∈ {1, 2}. Then Nhg G [D′] = ∪r j=k+1Qj ∪ {q11, q21}. It follows that ζhgk (G) ≤ ζhgk (D′) = |V (G)| − |Nhg G [D′]| = r∑ j=1 mj − [ r∑ j=k+1 mj + 2] = k∑ j=1 mj − 2. By following the arguments of the preceding case, it can be shown that if S′ is a ζhgk -set of G, then ζhgk (G) = ζhgk (S′) ≥ ∑k j=1mj − 2. 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