EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 1, Article Number 5716 ISSN 1307-5543 – ejpam.com Published by New York Business Global k-Hop Domination Defect in a Graph Jesica M. Anoche 1, 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. In this paper, we introduce a new graph parameter called the hop domination defect and investigate it for some classes of graphs. The hop domination number γh(G) of a graph G is the minimum number of vertices required to hop dominate all the vertices of G. The minimality of γh(G) implies that if W ⊆ V (G) and |W | < γh(G), then there is at least one vertex in G that is not hop dominated by W . Given a positive integer k < γh(G), where γh(G) ≥ 2, the k-hop domination defect of G, denoted by ζhk (G), is the minimum number of vertices of G that is not hop dominated by any subset of vertices of G with cardinality γh(G)− k. We give some bounds on the k-hop domination defect of a graph in terms of its order and maximum hop degree. Furthermore, we determine the k-hop domination defects of the join of some graphs. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Hop domination, k-hop domination defect, hop degree of a graph, join 1. Introduction Natarajan and Ayyaswamy in [13] introduced and studied hop domination, which, in some sense, is related to the standard domination. According to the authors, the concept has its origin from the field of Inorganic Chemistry. This parameter has been widely studied since its appearance in the literature and a significant number of variants of the parameter have already been defined and investigated (see for example [1], [2], [4], [5], [6], [7], [8], [9], [14], [15]), and [16]). Recently, Das et al. [3] introduced a domination parameter called the domination defect. As mentioned in their paper, the motivation of this study was mainly on dealing with problems associated with guarding facilities or placing monitoring devices in networks when there is only fewer than the minimum number of guards or devices required. In their DOI: https://doi.org/10.29020/nybg.ejpam.v18i1.5716 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.R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5716 2 of 13 study, the authors were able to establish various bounds on the domination defect of a graph in terms of, among others, the domination number, order, degree sequence, graph homomorphisms, and efficient dominating set. Other studies on the topic (see [10–12]) focused on characterizing the k-domination defect sets and determining the k-domination defect in the join, corona, edge corona, and composition of two graphs. Since hop domination has similar applications (e.g. in facility location, protection strategy, management in social networks) as domination, it is also a bit interesting to study the effect of having fewer than the required minimum number of nodes in a hop dominating set. In this study, we define the parameter k-hop domination defect and study it for some known graphs. The study hopes to give bounds of the newly defined parameter in terms of the order, hop degree of the graph, and other parameters. In particular, the authors would like to find what specific conditions to impose so that these bounds and some results which seemingly run parallel to the ones found in [3], hold in the sense of hop domination. The k-hop domination defects of some join of graphs are also obtained. 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 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 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 and NG[A] = NG(A) ∪ A is called the closed neighborhood of A. The degree of vertex v is degG(v) = |NG(v)|. A vertex v of G is isolated if |NG(v)| = 0. The maximum degree ∆(G) of G is given by ∆(G) = max{|NG(v)| : v ∈ V (G)} and the minimum degree δ(G) of G is given by δ(G) = min{|NG(v)| : v ∈ V (G)}. A vertex is called an endvertex if its degree is 1. A vertex is called a support vertex if it is adjacent to an end-vertex. The I(G) is the set containing all the isolated vertices of 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. The maximum hop degree and minimum hop degree of G, denoted by ∆h(G) and δh(G), respectively, is given by ∆h(G) = max{|N2 G(v)| : v ∈ V (G)} and δh(G) = min{|N2 G(v)| : v ∈ V (G)}. 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. Any hop dominating set with cardinality equal to γh(G) is called a γh-set. A set S ⊆ V (G) is a point-wise 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 point-wise non- dominating set of G is denoted by pnd(G). J. Anoche, S.R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5716 3 of 13 Let G be a non-trivial graph of order n and let 1 ≤ k < γh(G). Let S ⊆ V (G) with cardinality |S| = γh(G)− k. The set V (G) \N2 G[S] is called the k-hop defect set of S and the k-hop defect of S is ζhk (S) = |V (G) \N2 G[S]| = n− |N2 G[S]|. The minimum cardinality of a k-hop defect set in G, denoted by ζhk (G), is called the k-hop domination defect of G, i.e., ζhk (G) = min{ζhk (S) : S ⊆ V (G) with |S| = γh(G)− k}. A set S ⊆ V (G) of cardinality γh(G) − k for which |V (G) \ N2 G[S]| = ζhk (G) is called a ζhk -set of G. Thus, 〈 N2 G[S] 〉 is an induced subgraph of G with n− ζhk (G) vertices and hop domination number γh(G)− k. Let G and H be undirected graphs. The join G + H of G and 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)}. The graph G− v is the graph ⟨V (G) \ {v}⟩ induced by V (G) \ {v} and G− {u, v} = ⟨V (G) \ {u, v}⟩. 3. Results Theorem 1 ([16]). Let m and n be positive integers. Then each of the following holds. (i) For a complete graph Kn, γh(Kn) = n. (ii) For a complete bipartite graph Km,n, γh(Km,n) = 2. (iii) For a path Pn on n vertices, we have γh(Pn) =  2r if n = 6r 2r + 1 if n = 6r + 1 2r + 2 if n = 6r + s; 2 ≤ s ≤ 5. (iv) For a cycle Cn on n vertices, we have γh(Cn) =  2r if n = 6r 2r + 1 if n = 6r + 1 2r + 2 if n = 6r + s; 2 ≤ s ≤ 5. (v) γh(Wn) = 3 where Wn is a wheel with n spokes. (vi) γh(P ) = 2 where P denotes the Petersen graph. Theorem 2. [8] Let G and H be any two graphs of orders m and n, respectively. Then γh(G+H) = pnd(G) + pnd(H). In particular, (i) γh(G+H) = m+ n if G and H are complete; (ii) γh(G+H) = 2 if G and H have isolated vertices; J. Anoche, S.R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5716 4 of 13 (iii) γh(G+H) = 1 + pnd(H) if G = K1; (iv) γh(G+H) = 4 if G = Pm and H = Pn (m,n ≥ 2); and (v) γh(G+H) = 4 if G = Cm and H = Cn (mn ≥ 4). Theorem 3. Let G be a graph with I(G) ̸= ∅ and suppose |I(G)| = r. Then ζhj (G) = j for every j ∈ [r] = {1, 2, · · · , r} and ζhk (G) = r+ζhk−r(G ′) for every k ∈ {r+1, · · · , γh(G)−1}, where G′ = ⟨V (G) \ I(G)⟩. Proof. Let I(G) = {v1, v2, · · · , vr} and let S be a γh-set in G. Then I(G) ⊆ S. Let j ∈ [r]. Then D = S \ {v1, v2, · · · , vj} is a ζhj -set in G and |N2 G[D]| = |N2 G[S]| − |N2 G[{v1, v2, · · · , vj}]| = |V (G)| − j. Hence, ζhj (G) = |V (G)| − (|V (G)| − j) = j. Next, let k ∈ {r+1, · · · , γh(G)−1}. Then S0 = S\I(G) is γh-set inG′ = ⟨V (G) \ I(G)⟩. Hence, γh(G ′) = γh(G)−r. Since k ≤ γh(G)−1, k−r ≤ γh(G)−(r+1) < γh(G)−r. Let S′ be a ζhk−r-set in G′. Then |S′| = (γh(G)−r)−(k−r) = γh(G)−k and ζhk−r(G ′) = |V (G′)|− |N2 G′ [S′]| = (|V (G)| − r) − |N2 G[S ′]|. This implies that |V (G)| − |N2 G[S ′]| = r + ζhk−r(G ′). Therefore, since S′ is also a ζhk -set in G, ζhj (G) = r + ζhk−r(G ′). Remark 1. A ζhk -set of a graph G need not be contained in any γh-set of G. To see this, consider the graph in Figure 1 with γh(G) = 2. The set {1, 4} is the only γh-set and {7} is the only ζh1 -set in G and hence, ζh1 (G) = 5. Observe that {7} ⊈ {1, 4}. ..................................................................................................................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ... ................. ................ ................ ................ ................ ................ ................ ................ ................ ................ ................ ................ ................ ................ ................ ................. ................ ................ ................ ................ ................ ................ ................ ................ ................ ................ ................ ................ ................ ................ ..................................................................................................................................... ................................................................................................................................................. ................................................................................................................................... .............................................................................................................................................................................................................................................................. ..................................................................................................................................... ........... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ... ................................................................................................................................................................................................. ........................................................................................... ................................................................................................................................................. ........... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .... .......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... . ......... ........ ........ ........ ........ ........ ........ ........ ........ ... ................ ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ............... ... ........... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .... . ................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ........................................................................ .................................... .................................... 1 2 3 4 5 6 7 8 9 10 11 12 13 14 • • Figure 1: A graph G with γh(G) = 2. In view of Theorem 3, all graphs considered henceforth, unless specified, do not have isolated vertices. Furthermore, given a graph G, the positive integer k, when not specified, always satisfies the condition k ≤ γh(G)− 1. Theorem 4. If G is a graph of order n ≥ 2, then 1 ≤ ζhk (G) ≤ n− 1. Proof. Let S be a ζhk -set in G. Then |S| = γh(G) − 1. Hence, V (G) \ N2 G[S] ̸= ∅. This implies that ζhk (G) = |V (G)| − |N2 G[S]| ≥ 1. Moreover, since |N2 G[S]| ≥ 1, ζhk (G) = |V (G)| − |N2 G[S]| ≤ n− 1. This proves the assertion. J. Anoche, S.R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5716 5 of 13 Theorem 5. Let G be a non-trivial graph of order n. Then ζh1 (G) = 1 if and only if there exists v ∈ V (G) such that γh(G− v) = γh(G)− 1. Proof. Suppose ζh1 (G) = 1 and let S be a ζh1 -set in G. Then |S| = γh(G) − 1 and |V (G) \ N2 G[S]| = 1. Let v ∈ V (G) \ N2 G[S]. Then N2 G[S] = V (G) \ {v}. Therefore, γh(G− v) = γh( 〈 N2 G[S] 〉 ) = γh(G)− 1. Conversely, let v ∈ V (G) such that γh(G−v) = γh(G)−1. Then, there exists S ′ ⊆ V (G) with |S′ | = γh(G)−1 and N2 G[S ′ ] = V (G)\{v}. It follows that ζh1 (G) = |V (G)\N2 G[S ′ ]| = |{v}| = 1. Therefore, ζh1 (G) = 1. Theorem 6. If G is a graph on n vertices, then ζhk (G) ≤ k(1 + ∆h(G)). Proof. Let S be a γh-set in G. For each v ∈ V (G), we have |N2 G[v]| ≤ 1 + ∆h(G). Let S′ ⊂ S with |S′| = k and set S∗ = S \ S′. Then ζhk (S ∗) = |V (G)−N2 G[S ∗]| = |N2 G[S]| − |N2 G[S ∗]| ≤ |N2 G[S ′]|+ |N2 G[S ∗]| − |N2 G[S ∗]| = |N2 G[S ′]| = ∑ v∈S′ |N2 G[v]| ≤ k(1 + ∆h(G)). Therefore, ζhk (G) ≤ k(1 + ∆h(G)). Remark 2. Let G1, G2, · · · , Gr be the components of a graph G. Then each of the following holds: (i) γh(G) = ∑r j=1 γh(Gj). (ii) If Aj ⊆ V (Gi) for each j ∈ [r] = {1, 2, · · · , r} and A = ∪r j=1Aj, then N2 G[A] = ∪r j=1N 2 G[Aj ] (a disjoint union). Theorem 7. Let G1, G2, · · · , Gr be the components of graph G and let ζh1 (Gi) be the hop domination defect of Gi for each i ∈ [r] = {1, 2, · · · , r}. Then ζh1 (G) = min{ζh1 (Gi) : i ∈ [r]}. Proof. Let γh(Gi) and γh(G) be the hop domination numbers of Gi and G, respectively. By Remark 2(i), γh(G) = ∑r j=1 γh(Gj). For each i ∈ [r], let Di be a ζh1 -set of Gi. Then |Di| = γh(Gi) − 1 and ζh1 (Gi) = |V (Gi) − N2 G[Di]|. Let j ∈ [r] be such that ζh1 (Gj) = min{ζh1 (Gi) : i ∈ [r]}. Let Si be a γh-set in Gi for each i ∈ [r] and let S = (∪i∈[r]\{j}Si) ∪Dj . Then |S| = ∑ i∈[r]\{j} |Si|+ |Dj | = γh(G)− 1 J. Anoche, S.R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5716 6 of 13 and, by Remark 2(ii), |N2 G[S]| = |N2 Gj [Dj ]|+ ∑ i∈[r]\{j} |N2 Gi [Si]| = |V (Gj)| − ζh1 (Gj) + ∑ i∈[r]\{j} |V (Gi)| = r∑ i=1 |V (Gi)| − ζh1 (Gj). Thus, in G, ζh1 (S) = |V (G)| − |N2 G[S]| = ζh1 (Gj). It now remains to show that ζh1 (S) is the minimum among all subsets of V (G) with cardinality γh(G) − 1. So assume there exists S ′ ⊆ V (G) such that |S′ | = γh(G)−1 and ζh1 (S ′ ) < ζh1 (S). Let S ′ = S ′ 1∪S ′ 2∪· · ·∪S ′ r where S ′ i ⊆ V (Gi) for each i ∈ [r]. Since |S′ | = γh(G)−1, at least one S ′ l is not a hop dominating set of Gl by Remark 2(i). Thus, |S′ l| = γh(Gl)− 1 and ζh1 (S ′ l ) ≥ ζh1 (Gl) ≥ ζh1 (Gj). Hence, ζh1 (S ′ ) = |V (G)| − |N2 G[S ′ ]| = r∑ i=1 (|V (Gi)| − |N2 Gi [S ′ i ]|) ≥ |V (Gl)| − |N2 Gl [S ′ l ]| = ζh1 (S ′ l ) ≥ ζh1 (Gj) = ζh1 (S), contrary to the assumption that ζh1 (S ′ ) < ζh1 (S). Therefore, ζ h 1 (G) = ζh1 (S) = ζh1 (Gj). Theorem 8. ζhk (Kn) = k for all n ≥ 2 and 1 ≤ k < n. Proof. By Theorem 1(i), γh(Kn) = n. Let k be a positive integer such that 1 ≤ k ≤ n − 1 and let S be any subset of V (Kn) with |S| = γh(Kn) − k = n − k. Since ⟨S⟩ is a complete graph, N2 Kn [S] = S. This implies that ζhk (S) = |V (Kn) \ N2 Kn [S]| = n − |S| = n− (n− k) = k. Since S was arbitrarily chosen, it follows that ζhk (Kn) = k. Note that γh(P3) = γh(P4) = γh(P5) = 2. Therefore, k = 1. Consider P3 = [a, b, c], P4 = [p, q, r, s], and P5 = [v, w, x, y, z] below. Then S1 = {a}, S2 = {p}, and S3 = {x} are ζh1 -sets in P3, P4, and P5, respectively. Since |N2 P3 [S1]| = 2, |N2 P4 [S2]| = 2, and |N2 P5 [S3]| = 3, it follows that ζh1 (P3) = 1, and ζh1 (P4) = ζh1 (P5) = 2. ................................................................................................................ .................................................................................................................................................... .................................... ................................................................................................................ ................................................................................................................ ................................................................................................................ .................................... ................................................................................................................ ................................................................................................................ ................................................................................................................ ................................................................................................................ .................................... a b c p q r s v w x y z • • • J. Anoche, S.R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5716 7 of 13 Theorem 9. If Pn is a path on n ≥ 6 vertices, then ζhk (Pn) =  3k if n = 6r 3k − 2 if n = 6r + 1 3k + s− 6 if n = 6r + s; 2 ≤ s ≤ 5 and 2 ≤ k ≤ γh(Pn)− 1 Proof. We denote the vertices of Pn as {1, 2, · · · , n}. Now, consider the following cases: Case 1: Suppose n = 6r. By Theorem 1 (iii), γh(Pn) = 2r. Choose a (2r−k)-element set S = {3, 6, · · · , 6r − 3k}. This implies that ζhk (S) = n− |N2 Pn [S]| = n− [(2r − k) + 2(2r − k)] = 6r − [6r − 3k] = 3k. Thus, it is the minimum value for any set S with 2r − k vertices. Hence, ζhk (Pn) = 3k. Case 2: Suppose n = 6r + 1. By Theorem 1 (iii), γh(Pn) = 2r + 1. Choose a (2r−k+1)-element set S = {3, 6, · · · , 6r−3k+3}. This implies that ζhk (S) = n−|N2 Pn [S]| = n− [(2r−k+1)+2(2r−k+1)] = 6r− [6r−3k+3]+1 = 3k−2. Thus, it is the minimum value for any set S with 2r − k + 1 vertices. Hence, ζhk (Pn) = 3k − 2. Case 3: Suppose n = 6r + s where 2 ≤ s ≤ 5. By Theorem 1 (iii), γh(Pn) = 2r + 2. Choose a (2r−k+2)-element set S = {3, 6, · · · , 6r− 3k+6} for 2 ≤ k ≤ γh(Pn)− 1. This implies that ζhk (S) = n−|N2 Pn [S]| = n−[(2r−k+2)+2(2r−k+2)] = 6r−[6r−3k+6]+s = 3k + s− 6. Thus, it is the minimum value for any set S with 2r − k + 2 vertices. Hence, ζhk (Pn) = 3k + s− 6. From Theorem 8, we have ζhk (C3) = k for k ∈ {1, 2}. Now, since γh(C4) = γh(C5) = 2, k = 1. It can easily be verified that |N2 C4 [S]| = 2 and |N2 C5 [S′]| = 3 for any singleton subsets S and S′ of V (C4) and V (C5), respectively. Hence, ζ h k (C4) = ζhk (C5) = 2. The proof of the next result uses Theorem 1(iv) and follows along the same lines as that of Theorem 9. Theorem 10. If Cn is a cycle on n ≥ 6 vertices, then ζhk (Cn) =  3k if n = 6r 3k − 2 if n = 6r + 1 3k + s− 6 if n = 6r + s; 2 ≤ s ≤ 5 and 2 ≤ k ≤ γh(Cn)− 1. Lemma 1. Let G be a nontrivial connected graph with γh(G) ≥ 2 and let k = γh(G)− 1. Then S ⊆ V (G) is a ζhk -set of G if and only if S = {x} for some x ∈ V (G) with N2 G(x) = △h(G). Proof. Let k = γh(G) − 1 and let S ⊆ V (G) be a ζhk -set of G. Then |S| = 1, say, J. Anoche, S.R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5716 8 of 13 S = {x} and ζhk (S) = n − |N2 G[x]| = ζhk (G). This implies that |N2 G[x]| is maximum in G. Thus, N2 G(x) = △h(G). Conversely, suppose S = {x} with N2 G(x) = △h(G). Since n−|N2 G[S]| = n−|N2 G[x]| = n−△h(G) is the minimum possible value for any singleton subset of V (G), it follows that S is a ζhk -set of G. Theorem 11. If G = Km1,m2,··· ,mt is a complete multipartite graph with 1 ≤ m1 ≤ m2 ≤ · · · ≤ mt, then γh(G) = t. Proof. Let Q1, Q2, · · · , Qt be the partite sets of G and let S be a γh-set of G. Suppose there exists j ∈ [t] = {1, 2, · · · , t} such that S ∩Qj = ∅. Then Qj ⊆ NG(S). Hence, the vertices in Qj are not hop dominated by any element of S. This implies that S is not a hop dominating set in G, a contradiction. Therefore, S ∩Qj ̸= ∅ for every j ∈ [t]. Since S is a γh-set of G, |S ∩Qj | = 1 for every j ∈ [t]. Accordingly, γh(G) = |S| = t. Theorem 12. For a complete multipartite graph G = Km1,m2,··· ,mt where m1 ≤ m2 ≤ · · · ≤ mt, ζ h k (G) = ∑k j=1mj . Proof. Let Q1, Q2, · · · , Qt be the partite sets of G with cardinalities m1 ≤ m2 ≤ · · · ≤ mt. By Theorem 11, γh(G) = t. Choose a (t − k)-element set S = {qk+1, qk+2, · · · , qt} where qj ∈ Qj for each j ∈ {k+1, k+2, · · · , t}. Then N2 G[S] = ∪t j=k+1N 2 G[qj ] = ∪t j=k+1Qj . It follows that |N2 G[S]| = ∑t j=k+1 |Qj | = ∑t j=k+1mj . This is the maximum value that can be obtained for any set S with t− k vertices because m1 ≤ m2 ≤ · · · ≤ mt. Thus, ζhk (G) = ζhk (S) = |V (G)| − |N2 G[S]| = t∑ j=1 mj − t∑ j=k+1 mj = k∑ j=1 mj . The next result is a consequence of Theorem 12. Corollary 1. For a complete bipartite graph Km,n where 2 ≤ m ≤ n, ζhk (Km,n) = m. Theorem 13. For a Petersen graph P , ζhk (P ) = 3. Proof. By Theorem 1(vi), γh(P ) = 2. It follows that k = 1. Let S be a ζhk -set of P . Then |S| = 1, say S = {v}. Since |N2 P [x]| = 7 for every x ∈ V (P ), it follows that |N2 P [S]| = 7. Therefore, ζhk (P ) = ζhk (S) = |V (P )| − |N2 P [S]| = 10− 7 = 3. Theorem 14. Let G be a graph with diam(G) ≥ 3 and let G′ be a graph obtained by adding any number of edges xy to E(G) with dG(x, y) ≥ 3 such that γh(G) = γh(G ′). Then ζhk (G ′) ≤ ζhk (G) where 1 ≤ k ≤ γh(G)− 1. J. Anoche, S.R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5716 9 of 13 Proof. Let 1 ≤ k ≤ γh(G) − 1 and let S be a ζhk -set of G. Since γh(G) = γh(G ′), |S| = γh(G)−k = γh(G ′)−k. Let x ∈ N2 G[S]. If x ∈ S, then x ∈ N2 G′ [S]. If x ∈ N2 G(S)\S, then there exists z ∈ S such that dG(x, z) = 2. It follows that xz /∈ E(G′). Hence, dG′(x, z) = 2. This implies that x ∈ N2 G′(S). Thus, N2 G[S] ⊆ N2 G′ [S]. Therefore, ζhk (G ′) ≤ n− |N2 G′ [S]| ≤ n− |N2 G[S]| = ζhk (G). Theorem 15. Let G be a graph of order n and let v be a vertex with the property that for every pair of vertices x and y with dG(x, y) = 2 and v ∈ NG(x) ∩NG(y), it holds that |NG(x) ∩ NG(y)| ≥ 2. If G′ = ⟨V (G) \ {v}⟩ and γh(G) > γh(G ′) ≥ 2, then ζhk+1(G) ≤ ζhk (G ′) + 1 where 1 ≤ k ≤ γh(G)− 2. Proof. Let S be a γh-set in G′. Then v /∈ S. Let S′ = S ∪ {v} and let x ∈ V (G) \ S′. Since x /∈ S′, x /∈ S and x ̸= v. Hence, x ∈ V (G′)\S. Since S is a hop dominating set in G′, there exists z ∈ S such that dG′(z, x) = 2. This implies that there exists z ∈ S′ such that dG(z, x) = 2. Therefore, S′ is a hop dominating set in G and γh(G) ≤ |S′| = γh(G ′) + 1. Since γh(G ′) < γh(G), γh(G ′) + 1 ≤ γh(G). Thus, γh(G) = γh(G ′) + 1. Now let D be a ζhk -set of G′ where 1 ≤ k ≤ γh(G) − 2. Then |D| = γh(G ′) − k and ζhk (G ′) = (n − 1) − |N2 G′ [D]|. This implies that |N2 G′ [D]| = (n − 1) − ζhk (G ′). Since |D| = γh(G ′)− k = γh(G)− (k + 1), it follows that ζhk+1(G) ≤ n− |N2 G[D]|. Consider the following cases: Case 1: |N2 G[D]| = |N2 G′ [D]|. Then (n − 1) − ζhk (G ′) = |N2 G′ [D]| = |N2 G[D]| ≤ n − ζhk+1(G). Thus, we have ζhk+1(G) ≤ ζhk (G ′) + 1. Case 2: |N2 G[D]| ≠ |N2 G′ [D]|. Since N2 G′ [D] ⊆ N2 G[D], the assumption implies that there exists w ∈ N2 G(D) \ N2 G′(D). Suppose w ̸= v. Since w ∈ N2 G(D), there exists u ∈ D such that dG(w, u) = 2. Since u ∈ D, u ̸= v. The assumption that w /∈ N2 G′(D) would imply that v ∈ NG(w)∩NG(u) and the path [w, v, u] is the only w-u geodesic in G, contradicting the property of v. Hence, w = v. Therefore, |N2 G[D]| = |N2 G′ [D]|+ 1 and (n− 1)− ζhk (G ′) = |N2 G′ [D]| = |N2 G[D]| − 1 ≤ n− ζhk+1(G)− 1. Thus, ζhk+1(G) ≤ ζhk (G ′) ≤ ζhk (G ′) + 1. Therefore, the assertion holds. Remark 3. Equality of the two expressions (defects) given in Theorem 15 is attainable. J. Anoche, S.R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5716 10 of 13 Consider G = W4 = ⟨{v}⟩+ [a, b, c, d, a]. Let G′ = G− v = C4 = [a, b, c, d, a]. Observe that v satisfies the property given in Theorem 15, and γh(G) = 3 > 2 = γh(G ′). Then k = γh(G)− 2 = 1. If D is ζhk -set in G′, then |D| = γh(G ′)− 1 = 1. In this case, we may take any of the four vertices of C4 as the element of D, say D = {a}. Then N2 G[a] = {a, c} and N2 G′ [a] = {a, c}. It follows that ζhk+1(G) = |V (G)| − |N2 G[a]| = 3 = (|V (G′)| − |N2 G′ [a]|) + 1 = ζhk (G ′) + 1. Theorem 16. Let G be a graph of order n such that γh(G − v) = γh(G) for every v ∈ V (G). If there exist u, v ∈ V (G) such that uv /∈ E(G) and γh(G − {u, v}) < γh(G), then ζh1 (G) = 2. Proof. Let u, v ∈ V (G) be such that uv /∈ E(G) and γh(G − {u, v}) < γh(G). For convenience, let H = G−{u, v}. Then γh(H) + 1 ≤ γh(G). Let S be a γh-set of H. Then u /∈ S. Set S′ = S ∪ {u} and let H ′ = G − v. Let x ∈ H ′ \ S′. Then x /∈ {u, v}. Hence, x ∈ H \S. Since S is a hop dominating set of H, there exists y ∈ S such that dH(x, y) = 2. It follows that y ∈ S′ and dH′(x, y) = 2. This shows that S′ is a hop dominating set in H ′. Since γh(H)+1 ≤ γh(G) = γh(H ′) ≤ |S′| = γh(H)+1, γh(G) = γh(H ′) = |S′| = γh(H)+1, that is, γh(H) = γh(G) − 1. Hence, |S| = γh(G) − 1. Clearly, V (H) ⊆ N2 G[S]. Suppose N2 G[S] ̸= V (H). Then u ∈ N2 G(S) or v ∈ N2 G(S), say v ∈ N2 G(S). Then there exists w ∈ S such that dG(v, w) = 2. Let [v, p, w] be a v-w geodesic in G. Since uv /∈ E(G), p ̸= u. This implies that [v, p, w] is a v-w geodesic in H∗ = G \ u. Hence, dH∗(v, w) = 2. It follows that S is a hop dominating set H∗. This, however, is not possible because γh(H) = |S| < γh(G) = γh(H ∗) by assumption. Therefore, N2 G[S] = V (H). It follows that ζh1 (S) = n − |V (H)| = n − (n − 2) = 2. Since γh(G − {v}) = γh(G) for every v ∈ V (G), ζh1 (G) ̸= 1 by Theorem 5. Therefore, ζh1 (G) = ζh1 (S) = 2. Example 1. Let G be a graph obtained from C7 = [v1, v2, · · · , v7, v1] by adding the pendant edge pv1. Then γh(G) = 7 (S = {v1, v2, v5} is a γh-set in G). One can easily verify that γh(G \ v) = γh(G) for every v ∈ V (G). Consider the non-adjacent vertices p and v2 of G. Then G \ {p, v2} = P6. Hence, γh(G \ {p, v2}) = 2 < γh(G). By Theorem 16, ζh1 (G) = 2. It should be noted that the converse of Theorem 16 is not true. To see this, consider G = C4. Then γh(G) = 2 and γh(G \ v) = γh(P3) = 2 = γh(G) for every v ∈ V (C4). Moreover, ζh1 (G) = 2. However, one cannot find non-adjacent vertices p, q ∈ V (G) such that γh(G \ {p, q} = 1. Lemma 2. Let G and H be any two graphs of orders m and n, respectively. If x ∈ V (G+H) and |N2 G+H(x)| = ∆h(G+H), then |N2 G+H [x]| = max{m− δ(G), n− δ(H)}. Proof. Let x ∈ V (G). Since V (H) ⊆ NG(x), it follows that N2 G+H [x] ∩ V (H) = ∅. Hence, N2 G+H [x] ⊆ V (G). Now p ∈ N2 G+H [x] if and only if p = x or dG+H(x, p) = 2. This J. Anoche, S.R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5716 11 of 13 implies that p ∈ N2 G+H [x] if and only if p ∈ V (G)\NG(x). Thus, N 2 G+H [x] = V (G)\NG(x). Consequently, |N2 G+H [x]| = m− |NG(x)| = m− degG(x). Clearly, max{|N2 G+H [x]| : x ∈ V (G)} = max{m− degG(x) : x ∈ V (G)} = m− δ(G). Similarly, max{|N2 G+H [x]| : x ∈ V (H)} = max{n− degH(x) : x ∈ V (H)} = n− δ(H). Therefore, if x ∈ V (G+H) such that ∆h(G+H) = |N2 G+H(x)|, then |N2 G+H [x]| = max{m− δ(G), n− δ(H)}. This proves the assertion. Theorem 17. Let G and H be graphs of orders m and n, respectively. Then each of the following holds: (i) ζhk (G+H) = min{m+ δ(H), n+ δ(G)} where k = γh(G+H)− 1. (ii) ζh1 (G + H) = 1 if and only if there exists v ∈ V (G + H) such that v ∈ V (G) and pnd(G− v) = pnd(G)− 1 or v ∈ V (H) and pnd(H − v) = pnd(H)− 1. (iii) If I(G) ̸= ∅ and I(H) ̸= ∅, then ζh1 (G+H) = min{m,n}. (iv) If I(G) ̸= ∅ and H = Kn, then ζhk (G+H) = k, where 1 ≤ k ≤ n. Proof. (i) Let S be a ζhk -set in G + H. By Lemma 1, S = {x} where |N2 G+H(x)| = ∆h(G + H). By Lemma 2, we may assume without loss of generality that |N2 G+H [x] = m− δ(G) ≥ n− δ(H) for x ∈ V (G). Then ζhk (G+H) = (m+ n)− (m− δ(G)) = n+ δ(G) ≤ m+ δ(H) = (m+ n)− (n− δ(H)). (ii) By Theorem 5 and Theorem 2, ζh1 (G + H) = 1 if and only if there exists a vertex v ∈ V (G + H) such that γh((G + H) − v) = γh((G + H)) − 1 = pnd(G) + pnd(H) − 1. If v ∈ V (G), then (G +H) − v = (G − v) +H. Otherwise, (G +H) − v = G + (H − v). By Theorem 2, γh((G+H)− v) = pnd(G− v) + pnd(H) or γh((G+H)− v) = pnd(G) + pnd(H − v). Therefore, ζh1 (G+H) = 1 if and only if there exists a vertex v ∈ V (G) such that pnd(G− v) = pnd(G)− 1 or v ∈ V (H) with pnd(H − v) = pnd(H)− 1. (iii) If I(G) ̸= ∅ and I(H) ̸= ∅, then γh(G + H) = 2, by Theorem 2. Thus, k = 1. Let p ∈ I(G) and q ∈ I(H). Since N2 G(p) = V (G) \ {p} and N2 H(q) = V (H) \ {q}, it follows that ∆h(G +H) = max{|N2 G(p)|, |N2 H(q)|}. Therefore, since δ(G) = |NG(p)| = 0 and δ(H) = |NH(q)| = 0, it follows from (i) that ζh1 (G+H) = min{m,n}. (iv) Since pnd(G) = 1 and pnd(Kn) = n, γh(G + Kn) = n + 1 by Theorem 2. Let k be such that 1 ≤ k ≤ n and let S = SG ∪ Sn be a ζhk -set in G +Kn, where SG ⊆ V (G) and J. Anoche, S.R. Canoy Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5716 12 of 13 Sn ⊆ V (Kn). Then |S| = |SG|+ |Sn| = (n+ 1)− k and |N2 G+Kn [S]| = |N2 G+Kn [SG]|+ |N2 G+Kn [Sn]| = |N2 G+Kn [SG]|+ |Sn|. Since N2 G+Kn [t] = V (G) for any t ∈ I(G) and the value |N2 G+Kn [S]| is maximum, it follows that SG = {w} for some w ∈ I(G). This implies that |Sn| = (n + 1) − k − 1 = n − k. Therefore, ζhk (G+H) = (m+ n)− |N2 G+Kn [S]| = (m+ n)− (m+ n− k) = k. 4. Conclusion In this paper, we introduced and studied a new graph invariant called the k-hop domi- nation defect of a graph. We obtained the k-hop domination defects of some known graphs including the join of some graphs. Also, we provided some bounds on the k-hop domina- tion defect of a graph G in terms of its order and maximum hop degree and characterized the graphs that yield a k-domination defect equal to 1. 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