EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 1, Article Number 5658 ISSN 1307-5543 – ejpam.com Published by New York Business Global Another Look at Hop Independence in Graphs 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. A set S ⊆ V (G) is a hop independent set in an undirected graph G if dG(v, w) ̸= 2 for any two distinct vertices v, w ∈ S. The maximum cardinality among the hop independent sets in G, denoted by αh(G), is called the hop independence number of G. The hop independent sets in the shadow graph, complementary prism, edge corona and disjunctive products of two graphs are characterized. These characterizations are used to determine the exact or sharp bounds of the hop independence numbers of these graphs. Furthermore, we show that the hop independent set decision problem (HISP) is NP -complete. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Hop independence number, shadow graph, complementary prism, edge corona, disjunction, strong product 1. Introduction Hop domination is a domination-related concept introduced and studied by Natara- jan and Ayyaswamy in [7]. This parameter has been widely studied since its introduction and some variations of the concept have been defined and investigated (see for example [1], [2], [5], [8], [9], and [10]). Recently, Hassan et al. [4] introduced an independent-type parameter called hop inde- pendence. As mentioned in the paper, the motivation of such study is the ever increasing figure of research on hop-domination related topics. It’s worth noting that the hop in- dependence number provides a sharp upper bound for the hop domination number of a graph. The authors also showed that the absolute difference of the (ordinary) indepen- dence number and the hop independence number can be made arbitrarily large. Moreover, hop independent sets in the join, corona, lexicographic product, and Cartesian product of two graphs have been characterized. Subsequently, sharp bounds (exact values for others) DOI: https://doi.org/10.29020/nybg.ejpam.v18i1.5658 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 (1) (2025), 5658 2 of 15 of the hop independence numbers of these graphs have been obtained. In [3], the authors used the concept of hop independence to define a variation of hop domination. Karp in [6], as one of his many original problems, showed that the clique decision problem is NP -complete. We shall use this result to show that the hop independence decision problem is also NP -complete. 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. A vertex v of G is isolated if |NG(v)| = 0. 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 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 set S ⊆ V (G) is an independent set of G if no two pair of distinct vertices of S are adjacent. The maximum cardinality of an independent set of G, denoted by α(G), is called the independence number of G. Set S is a hop independent set of G if dG(v, w) ̸= 2 for any two distinct vertices v and w of S. The maximum cardinality of a hop independent set of G, denoted by αh(G), is called the hop independence number of G. Any independent (hop independent) set with cardinality α(G) (resp. αh(G)) is referred to as a maximum independent set or α-set (resp. maximum hop independent set or αh-set) of G. A set S is clique of a graph G if the graph ⟨S⟩ induced by S is a complete graph. The maximum size or cardinality of a clique of G, denoted by ω(G), is called the clique number of G. Any clique in G with cardinality ω(G) is called an ω-set in G. A matching of a graph G is a set M = {e1, e2, · · · , en} ⊆ E(G) such that each vertex v ∈ V (G) appears in at most one edge in M (i.e., the edges in M do not have a common vertex). A matching M is maximum if M ∪{e} is not a matching for any e ∈ E(G)\M . The matching number of a graph G, denoted by ν(G) is the size of a maximum matching in G. LetG andH be undirected graphs. The shadow graph D2(G) ofG is the graph obtained by taking two copies of G, say G1 and G2, and then joining each vertex v ∈ V (G1) to the neighbors of v′ ∈ V (G2), where v′ is the vertex in V (G2) corresponding to v, i.e., v and v′ represent the same vertex in G. The complementary prism of graph G, denoted J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5658 3 of 15 by GG, is the graph obtained from the disjoint union of G and G by adding the edges vv, where v ∈ V (G) and v is the vertex of G corresponding to vertex v. The disjunction of graphs G and H, denoted by G ∨ H, is the graph with V (G ∨ H) = V (G) × V (H) and (x, p)(y, q) ∈ E(G ∨ H) if and only if xy ∈ E(G) or pq ∈ E(H). The edge corona G ⋄ H of G and H is the graph obtained by taking one copy of G and |E(G)| copies of H, and then joining two end-vertices of the i-th edge of G to every vertex in the i-th copy of H. The strong product G ⊠ H of graphs G and H is the graph with vertex set V (G) × V (H) and (u, v) is adjacent with (u′, v′) whenever [uu′ ∈ E(G) and v = v′] or [vv′ ∈ E(H) and u = u′] or [uu′ ∈ E(G) and vv′ ∈ E(H)]. 3. Results Proposition 1 ([4]). Let G be any graph on n vertices. If S is a maximun hop independent set of G, then S is a hop dominating set. In particular, γh(G) ≤ αh(G). Theorem 1 ([4]). Let G be any graph on n vertices. If S is a hop independent set of G, then every component of ⟨S⟩ is complete. Moreover, (i) αh(G) = n if and only if every component of G is complete; and (ii) for n ≥ 3, αh(G) = n − 1 if and only if all but a single component C of G are complete and C \ v is a complete graph for some vertex v ∈ V (C). Corollary 1 ([4]). Let G be a connected graph on n vertices. Then (i) αh(G) = n if and only if G = Kn; and (ii) for n ≥ 3, αh(G) = n− 1 if and only if G ̸= Kn and there exists v ∈ V (G) such that G \ v = Kn−1. Observation 1: Let G be a graph of order n. (i) ∅ and cliques in G are hop independent sets. (ii) If S is a hop independent set in G and I ⊆ I(G), where I(G) denotes the set consisting of the isolated vertices of G, then S ∪ I is also hop independent in G. Moreover, if S is an αh-set in G, then I(G) ⊆ S. Observation 2: Let G1, G2, · · · , Gk be the components of graph G, where k ≥ 1. Then S is hop independent in G if and only if Sj = S∩V (Gj) is hop independent in Gj for each j ∈ {1, 2, · · · , k}. Moreover, αh(G) = ∑k i=1 αh(Gj). J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5658 4 of 15 4. Shadow Graph If G1 and G2 are the copies of graph G in the definition of the shadow graph D2(G) and if SG1 ⊆ V (G1) and SG2 ⊆ V (G2), then the sets S′ G1 and S′ G2 are the sets given by S′ G1 = {a′ ∈ V (G2) : a ∈ SG1} and S′ G2 = {a ∈ V (G1) : a ′ ∈ SG2}. We denote by I(G) the set containing all the isolated vertices of G. Theorem 2. Let G be a graph. Then a subset S of V (D2(G)) is hop independent in D2(G) if and only if one of the following conditions holds: (i) S is a hop independent set in G1. (ii) S is a hop independent set in G2. (iii) S = SG1 ∪ SG2 ∪ I1 ∪ I2 and satisfies the following conditions: (a) I1 ⊆ I(G1) and I2 ⊆ I(G2). (b) SG1 and SG2 are hop independent sets in G1 and G2, respectively, not containing isolated vertices. (c) SG1 ∩ S′ G2 = ∅ and S′ G1 ∩ SG2 = ∅. (d) SG1 ∪ S′ G2 and S′ G1 ∪ SG2 are hop independent sets in G1 and G2, respectively. Proof. Suppose S is a hop independent set in D2(G). If S ∩ V (G2) = ∅, then S ⊆ V (G1). Since S is hop independent in D2(G), it follows that S is hop independent in G1. This shows that (i) holds. Similarly, (ii) holds if S ∩ V (G1) = ∅. Next, suppose that S ∩ V (G1) ̸= ∅ and S ∩ V (G2) ̸= ∅. Let I1 = I(G1) ∩ S, I2 = I(G2) ∩ S, SG1 = S ∩ (V (G1) \ I1), and SG2 = S ∩ (V (G2) \ I2). Then S = SG1 ∪ SG2 ∪ I1 ∪ I2. Clearly, (a) holds. Since S is a hop independent set in D2(G), property (b) also holds. Now, if SG1 = ∅ or SG2 = ∅, then (c) and (d) hold. So suppose SG1 ̸= ∅ and SG2 ̸= ∅. Suppose further that SG1 ∩ S′ G2 ̸= ∅, say v ∈ SG1 ∩ S′ G2 . Then v, v′ ∈ S. Since v /∈ I(G1), there exists w ∈ NG1(v). It follows that dD2(G)(w, v ′) = 1. Thus, dD2(G)(v, v ′) = 2, contradicting the assumption that S is hop independent in D2(G). Therefore, SG1 ∩ S′ G2 = ∅. Similarly, S′ G1 ∩ SG2 = ∅, showing that (c) holds. Finally, suppose that SG1 ∪ S′ G2 is not hop independent in G1. Since SG1 and SG2 are hop independent in G1 and G2, respectively, according to (b), there exist x ∈ SG1 and y ∈ S′ G2 such that dD2(G)(x, y) = 2. This, however, would imply that dD2(G)(x, y ′) = 2, contradicting the fact that y′ ∈ SG2 and S is a hop independent set in D2(G). Hence, SG1 ∪ S′ G2 is hop independent in G1. Similarly, S′ G1 ∪ SG2 is hop independent in G2. This shows that (d) holds. Conversely, if (i) or (ii) holds, then S is a hop independent set in D2(G). Suppose now that (iii) holds. Since (b) holds, SG1 ∪ I1 and SG2 ∪ I2 are hop independent sets in G1 and G2, respectively (and hence, in D2(G)). Let x ∈ SG1 and y′ ∈ SG2 . Then y ∈ S′ G2 and by (c), x ̸= y. By property (d), dD2(G)(x, y ′) = dD2(G)(x, y) ̸= 2. Therefore, S is a hop independent set in D2(G). J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5658 5 of 15 Corollary 2. Let G be a graph. Then αh(D2(G)) = αh(G) + |I(G)|. Proof. If G is an empty graph, then D2(G) is an empty graph. Hence, αh(G) = |I(D2(G))| = |I(G1)|+ |I(G2)| = αh(G) + |I(G)|. Suppose that G has an edge. Let S be an αh-set in G1. Then S∗ = S ∪ I(G2) is a hop independent set in D2(G). This implies that αh(G) ≥ |S∗| = αh(G) + |I(G)|. Next, suppose that Q is αh-set in D2(G). Then I(D2(G)) = I(G1)∪I(G2) ⊂ Q. Let QG1 = (Q∩ V (G1))\I(G1) and QG2 = (Q∩V (G2))\I(G2). Then Q = QG1∪QG2∪I(G1)∪I(G2). From property (d), QG1 ∪Q′ G2 is a hop independent set in G1. It follows that QG1 ∪Q′ G2 ∪I(G1) is a hop independent set in G1. Thus, αh(D2(G)) = |Q| = |QG1 ∪QG2 ∪ I(G1) ∪ I(G2)| = |QG1 ∪QG2 ∪ I(G1)|+ |I(G2)| = |QG1 ∪Q′ G2 ∪ I(G1)|+ |I(G2)| ≤ αh(G) + |I(G)|. This establishes the desired equality. 5. Edge Corona of Two Graphs For every edge e = uv of G, denote by He = Huv the copy of H where the vertices are joined to vertices u and v. Theorem 3. Let G be a non-trivial connected graph and let H be any graph. Then S is a hop independent set in G ⋄ H if and only if S = A ∪ (∪uv∈E(G)Suv) and satisfies the following conditions: (i) A is a hop independent set in G. (ii) If Suv ̸= ∅, then Suv is a clique in Huv. (iii) Suv = ∅ whenever any of the following holds: (a) {u, v} ∩NG(A)) ̸= ∅ (b) Suw ̸= ∅ for some w ∈ NG(u) (c) Svz ̸= ∅ for some z ∈ NG(v) Proof. Suppose S is a hop independent set in G ⋄ H and let A = S ∩ V (G) and Suv = S ∩ V (Huv) for each uv ∈ E(G). Then S = A ∪ (∪uv∈E(G)Suv). Since S is a hop independent set in G ⋄ H, A is a hop independent set in G. This shows that (i) holds. J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5658 6 of 15 Next, let uv ∈ E(G) and Suv ̸= ∅. If Suv is not a clique in Huv, then there exist p, q ∈ Suv such that dHuv(p, q) ̸= 1. It follows that dG⋄H(p, q) = 2, contrary to the assumption that S is hop independent in G ⋄ H. Thus, Suv is a clique in Huv, showing that (ii) holds. Finally, suppose that uv ∈ V (G). Since S is hop independent in G⋄H, Suv = ∅ whenever (a) or (b) or (c) holds. This shows that (iii) holds. For the converse, suppose that S has the given form and satisfies (i), (ii), and (iii). Let a, b ∈ S, where a ̸= b, and let uv, xy ∈ E(G) such that a ∈ V (⟨{u, v}⟩ + Huv) and b ∈ V (⟨{x, y}⟩+Hxy). If a, b ∈ A, then dG⋄H(a, b) ̸= 2 because of (i). Suppose that a or b is not in A. Consider the following cases: Case 1. uv ̸= xy. Suppose first that uv and xy have a common vertex, say x = v. By (iii), Suv and Sxy cannot be both nonempty. Assume that Sxy = ∅. Suppose b = y. Then Suv = ∅ by (iii). It follows that a ∈ {u, v}, contrary to our assumption that a or b is not in A. Hence, b = x, a ∈ Suv, and ab ∈ E(G ⋄H). So suppose uv and xy do not have a common vertex. Assume first that one of a and b is in A, say a = u ∈ A. Then b ∈ Sxy. By (iii), x, y /∈ NG(a). This implies that dG⋄H(a, b) ̸= 2. Next, suppose that a ∈ Suv and b ∈ Sxy. Since uv and xy do not have a common vertex, dG⋄H(a, b) ̸= 2. Case 2. uv = xy. If a ∈ {u, v}, then b ∈ Suv and dG⋄H(a, b) = 1. If a, b ∈ Suv, then dG⋄H(a, b) = 1 by (ii). Therefore, S is a hop independent set in G ⋄H. Lemma 1. Let G be a connected graph of order n ≥ 3. Then ν(G) = 1 if and only if G = K3 or G = K1,n−1. Furthermore, if ν(G) = 1 and αh(G) > 2, then G = K3. Proof. Suppose that ν(G) = 1. If n = 3, then G ∈ {K3, P3}. Suppose n ≥ 4 and let M = {uv} be a maximum matching in G. Let x ∈ V (G) \ {u, v}. Since ν(G) = 1, xu ∈ E(G) or xv ∈ E(G). Assume that xu ∈ E(G). Suppose further that xv ∈ E(G). Then ⟨{x, u, v}⟩ is (isomorphic to) K3. Next, let y ∈ V (G)\{x, u, v}. Since G is connected and n ≥ 4, pick any y ∈ NG({x, u, v}). We may assume that xy ∈ E(G). Then xy and uv do not have a common vertex. This implies thatM∪{xy} is a matching in G, contradicting the maximality of M . Thus, xv /∈ E(G). Now let z ∈ V (G) \ {u, x}. Since M = {uv} is a maximum matching in G, zu ∈ E(G) and xz /∈ E(G). Therefore, G = K1,n−1. The converse is clear. For the second part, suppose that ν(G) = 1 and αh(G) > 2. Since αh(K1,n−1) = 2 for all n ≥ 3, it follows from the first part that G = K3. Corollary 3. Let G be a non-trivial connected graph and let H be any graph. (i) If ν(G) = 1, then αh(G ⋄H) = 2 + ω(H). (ii) If ν(G) ≥ 2, then αh(G ⋄H) ≥ max{αh(G), ν(G)ω(H)}. J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5658 7 of 15 Proof. (i) Assume that ν(G) = 1, sayM ′ = {pq} is a maximum matching inG. Clearly, αh(G ⋄ H) = 2 + ω(H) if G = ⟨{p, q}⟩ = K2. Suppose G ̸= K2. Let S0 = {p, q} ∪ Spq, where Spq is a maximum clique in H. Then, by Theorem 3, S0 is a hop independent set in G ⋄H. This implies that αh(G ⋄H) ≥ |S0| = 2 + ω(H). On the other hand, if S∗ is an αh-set in G⋄H, then S∗ = A∪(∪uv∈E(G)Suv) and satisifes conditions (i), (ii), and (iii) of Theorem 3. Suppose there exists an st ∈ E(G) such that Sst ̸= ∅. Without loss of generality, we may assume that st = pq, i.e., Sst = Spq ̸= ∅. Then Spq is a clique in Hpq by (ii). Let kl ∈ E(G)\M ′. Since M ′ is a maximum matching in G (or since ν(G) = 1), pq and kl must have a common vertex. We may assume that k = p. Then Skl = ∅ by (iii). Also, since w ∈ NG({p, q}) for all w ∈ V (G) \ {p, q} and Spq ̸= ∅, it follows from (iii) that w /∈ A for all w ∈ V (G) \ {p, q}. This implies that A ⊆ {p, q}. Hence, αh(G ⋄H) = |A|+ |Spq| ≤ 2 + ω(H). Next, suppose that Suv = ∅ for all uv ∈ E(G). Then S∗ = A. Since G is a non-trivial connected graph, |A| ≥ 2. Clearly, αh(G ⋄H) = |A| ≤ 2 + ω(H) if |A| = 2. So suppose |A| > 2. Since A is a hop independent set in G and |A| > 2, it follows from Lemma 1 that G = K3. Hence, αh(G ⋄H) = |A| = 3 ≤ 2 + ω(H). Accordingly, αh(G ⋄H) = 2 + ω(H). (ii) Suppose now that ν(G) ≥ 2. Let S be an αh-set in G. Then S is a hop independent set in G ⋄ H by Theorem 3. It follows that αh(G ⋄ H) ≥ |S| = αh(G). Next, let M be a maximum matching in G. Let Suv be a maximum clique in Huv for each uv ∈ M and set Sxy = ∅ for each xy ∈ E(G) \ M . Then S = ∪uv∈E(G)Suv = ∪uv∈MSuv is a hop independent set in G ⋄H by Theorem 3. Thus, αh(G ⋄H) ≥ |S| = ν(G)ω(H). Therefore, αh(G ⋄H) ≥ max{αh(G), ν(G)ω(H)}. Remark 1. The lower bound given in Corollary 3 is tight. However, strict inequality is also attainable. To see this, consider K5 ⋄ P3, C4 ⋄ P3, and G ⋄ P3, where G is the graph in Figure 3. ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ....... .................................... .......................................................................................................................................................... .................................... .................................... ................... .................. .................. .................. .................. .................. ......... .................................... .................................... ................... .................. .................. .................. .................. .................. ......... .................................... .................................... ...................................................................................................................... .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ....... .................................... .................................... Figure 3 Clearly, αh(K5) = 5, αh(C4) = 2, αh(G) = 3, ν(K5) = ν(C4) = ν(G) = 2, and ω(P3) = 2. It can be verified easily that αh(K5 ⋄P3) = αh(K5) = 5 > 4 = ν(K5)ω(P3), αh(C4 ⋄P3) = ν(C4)ω(P3) = 4 > 2 = αh(C4), and αh(G ⋄ P3) = 6 > 4 = max{αh(G), ν(G)ω(P3)}. J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5658 8 of 15 6. Complementary Prism Theorem 4. Let G be a graph. Then S is a hop independent set in GG if and only if one of the following holds. (i) S is a hop independent set in G. (ii) S is a hop independent set in G. (iii) S = {v, v} for some v ∈ V (G). Proof. Suppose S is a hop independent set in GG. If S ⊆ V (G), then S is a hop independent set in G. If S ⊆ V (G), then S is a hop independent set in G. Hence, (i) or (ii) holds. Suppose S ∩ V (G) ̸= ∅ and S ∩ V (G) ̸= ∅. Let v ∈ S ∩ V (G) and let w ∈ S ∩ V (G). Suppose w ̸= v. Then v ̸= w. Since vw ∈ E(G) if and only if v w /∈ E(G), it follows that dGG(v, w) = 2, contrary to the assumption that S is a hop independent set in GG. Thus, w = v. Suppose there exists x ∈ (S ∩ V (G)) \ {v}. Then dGG(x, v) = 2 which is not possible. Thus, S ∩V (G) = {v}. Similarly, S ∩V (G) = {v}. This shows that (iii) holds. The converse is clear. The next result is immediate from Theorem 4. Corollary 4. Let G be a graph. Then αh(GG) = max{2, αh(G), αh(G}. 7. Disjunction of Two Graphs Theorem 5. Let G and H be non-trivial connected graphs. Then C = ∪x∈S({x} × Tx) is a hop independent set in G ∨H if and only if following conditions hold. (i) S is a hop independent set in G. (ii) Tx is a clique in H for every x ∈ S. (iii) If x, y ∈ S and dG(x, y) ≥ 3, then Tx ∩ Ty = ∅ and dH(Tx, Ty) ≥ 3. (iv) ∪x∈S0Tx is a hop independent set in H for every independet set S0 ⊆ S. Proof. Suppose C is a hop independent set in G∨H. Suppose S is not hop independent in G. Then there exist u, v ∈ S such that dG(u, v) = 2. Let w ∈ NG(u) ∩NG(v), a ∈ Tu and b ∈ Tv. If a = b, then [(u, a), (w, a), (v, a)] is a (u, a)-(v, b) geodesic in G∨H. If a ̸= b, then [(u, a), (w, b), (v, b)] is a (u, a)-(v, b) geodesic in G ∨ H. Both cases are contrary to the assumption that C is a hop independent set in G ∨H. Hence, S is hop independent in G, showing that (i) holds. Now, let x ∈ S and let p, q ∈ Tx where p ̸= q. Since C is a hop independent set in G ∨ H, dG∨H((x, p), (x, q)) ̸= 2. From the fact that 1 ≤ dG∨H((x, p), (x, q)) ≤ 2 and because (x, p) ̸= (x, q), we must have dG∨H((x, p), (x, q)) = 1. This implies that pq ∈ E(G). Therefore, Tx is a clique in H, showing that (ii) holds. J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5658 9 of 15 Next, let x, y ∈ S with dG(x, y) ≥ 3. Suppose Tx ∩ Ty ̸= ∅, say t ∈ Tx ∩ Ty. Let s ∈ NH(t). Then [(x, t), (y, s), (y, t)] is an (x, t)-(y, t) geodesic in G ∨H, contrary to the assumption that C is hop independent in G ∨ H. Thus, Tx ∩ Ty = ∅. Let a1 ∈ Tx and a2 ∈ Ty. Suppose dH(a1, a2) = 2 and let q ∈ NH(a1, a2). Then [(x, a1), (y, q), (y, a2)] is an (x, a1)-(y, a2) geodesic in G ∨ H, a contradiction. This implies that dH(a1, a2) ̸= 2. Since a1 and a2 were arbitrarily chosen, it follows that dH(Tx, Ty) ≥ 3, showing that (iii) holds. Finally, suppose that S0 is an independent subset of S. Suppose ∪x∈S0Tx is not hop independent in H. Then there exist c, d ∈ ∪x∈S0Tx with dH(c, d) = 2. By (ii), it follows that c ∈ Tz and d ∈ Tw, where z ̸= w and z, w ∈ S0. Let r ∈ NH(c)∩NH(d). Then [(z, c), (w, r), (w, d)] is a (z, c)-(w, d) geodesic in G ∨H which is not possible. Therefore, ∪x∈S0Tx is hop independent in H, showing that (iv) holds. For the converse, suppose that C satisfies properties (i), (ii), (iii), and (iv). Let (v, p), (w, q) ∈ C such that (v, p) ̸= (w, q). Consider the following cases: Case 1. v = w. Then p, q ∈ Tv and p ̸= q. By property (ii), dH(p, q) = 1. Hence, dG∨H((v, p), (w, q)) = 1. Case 2. v ̸= w. Suppose first that dG(v, w) = 1. Then dG∨H((v, p), (w, q)) = 1. Next, suppose that dG(v, w) > 1. By property (i), we must have dG(v, w) ≥ 3. Now, by property (iii), p ̸= q. If dH(p, q) = 1, then dG∨H((v, p), (w, q)) = 1. Suppose dH(p, q) ̸= 1. Then, by property (iv) (using the fact that S0 = {v, w} is an independent set), dH(p, q) ≥ 3. Therefore, dG∨H((v, p), (w, q)) ≥ 3. Accordingly, C is a hop independent set in G ∨H. A sequence of cliques ⟨Sq1 , Sq2 , · · · , Sqm⟩ in an undirected graph G, where each qj = |Sqj |, is a decreasing d3-sequence if dG(Sqi , Sqj ) ≥ 3 for every pair of distinct indices i, j ∈ {1, 2, · · · ,m} and q1 ≥ q2 ≥ . . . ≥ qm. It is a maximum decreasing d3-sequence if for every decreasing d3-sequence of cliques ⟨St1 , St2 , · · · , Sts⟩ in G, it holds that s ≤ m and tj ≤ qj for each j ∈ {1, 2, . . . , s}. Corollary 5. Let G and H be non-trivial connected graphs and let ⟨Sq1 , Sq2 , · · · , Sqm⟩ and ⟨Dp1 , Dp2 , · · · , Dpr⟩ be maximum decreasing d3-sequences of cliques in G and H, respec- tively. Then αh(G ∨H) = ρHG∑ k=1 qkpk, where ρHG = min{m, r}. Proof. Let ⟨Sq1 , Sq2 , · · · , Sqm⟩ and ⟨Dp1 , Dp2 , · · · , Dpr⟩ be maximum decreasing d3- sequences of cliques in G and H, respectively. Then S = ∪ρHG k=1Sqk is a hop dominating set inG. For each x ∈ S, set Tx = Dpj if x ∈ Sqj , where 1 ≤ j ≤ ρHG . Then C = ∪x∈S({x}×Tx) J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5658 10 of 15 is a hop independent set in G ∨H by Theorem 5. This implies that αh(G ∨H) ≥ |C| = ∑ x∈S |Tx| = ρHG∑ k=1 ∑ x∈Sqk |Tx| = ρHG∑ k=1 ∑ x∈Sqk |Dpk | = ρHG∑ k=1 qkpk. On the other hand, suppose C0 = ∪x∈S0({x}×Rx) is an αh-set in G∨H. Then, by (i) and (ii) of Theorem 5, S0 is a hop independent set in G and Rx is a clique in H for each x ∈ S0. By Theorem 1, the components of S0 are cliques in G. Let Gr1 , Gr2 , · · · , and Grn be the components of G, where r1 ≥ r2 · · · ≥ rn, where rj = |V (Grj )| for each j ∈ {1, 2, · · · , n}. By (iii) of Theorem 5, ⟨V (Gr1), V (Gr2), · · · , V (Grn)⟩ is a decreasing d3-sequence of cliques in G. Hence, n ≤ m and V (Grj ) ⊆ Sqj for each j ∈ {1, 2, · · · , n}. Since C0 is an αh-set in G ∨H, it follows that for each j with 1 ≤ j ≤ n and for each x ∈ V (Grj ), Rx = Qtj and tj ≥ tj+1 for j ∈ {1, 2, · · · , n − 1}. Thus, ⟨Qt1 , Qt2 , · · · , Qtn⟩ is a decreasing d3-sequence of cliques in H. Hence, n ≤ r and Qtj ⊆ Dpj for each j ∈ {1, 2, · · · , n}. It follows that n ≤ ρHG . Therefore, αh(G ∨H) = |C0| = ∑ x∈S0 |Rx| = n∑ k=1 rktk ≤ ρHG∑ k=1 qkpk. This proves the desired equality. Corollary 6. Let G and H be non-trivial connected graphs. If G or in H has a maximum decreasing d3-sequence consisting of a single clique as a term, then αh(G∨H) = ω(G)ω(H). Proof. Suppose, without loss of generality, that ⟨Sq⟩ is a maximum d3-sequence of a single clique in G. Then ρHG = 1 and Sq is a maximum clique in G. Let D be a maximum clique in H. Then αh(G ∨H) = |Sq||D| = ω(G)ω(H) by Corollary 5. Example 1. For any non-trivial connected graph H, αh(Kn ∨H) = nω(H) and αh(P3 ∨ H) = 2ω(H). 8. Strong Product of Two Graphs Theorem 6. Let G and H be non-trivial connected graphs. Then C = ∪x∈S({x} × Tx) where S ⊆ V (G) and Tx ⊆ V (H) for every x ∈ S, is a hop independent set in G ⊠H if and only if following conditions hold. (i) Tx is a hop independent set in H for every x ∈ S. (ii) Tx∪Ty is hop independent in H for every pair of vertices x, y ∈ S with dG(x, y) ≤ 2. (iii) Tx∩Ty = ∅ and dH(Tx, Ty) ≥ 3 for every pair of vertices x, y ∈ S with dG(x, y) = 2. J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5658 11 of 15 Proof. Suppose C is a hop independent set in G⊠H. Let x ∈ S and a, b ∈ Tx such that a ̸= b. If ab ∈ E(G), then (x, a)(x, b) ∈ E(G⊠H). Suppose ab /∈ E(G). Then (x, a)(x, b) /∈ E(G ⊠H). Since C is hop independent in G ⊠H, dG⊠H((x, a), (x, b)) ≥ 3. This implies that dH(a, b) ≥ 3, showing that Tx is hop independent in H. This shows that (i) holds. Next, let x, y ∈ S with dG(x, y) ≤ 2 and suppose that Tx ∪ Ty is not hop independent in H. By (i), it follows that there exist p ∈ Tx and q ∈ Ty such that dH(p, q) = 2. Let t ∈ NH(p) ∩NH(q). Suppose first that dG(x, y) = 1. Then [(x, p)(y, t), (y, q)] is an (x, p)- (y, q) geodesic. Suppose dG(x, y) = 2 and let z ∈ NG(x)∩NG(y). Then [(x, p)(z, t), (y, q)] is an (x, p)-(y, q) geodesic. In both cases, we have dG⊠H((x, p), (y, q)) = 2, contrary to our assumption that C is hop independent. Thus, Tx ∪ Ty is hop independent in H, showing that (ii) holds. Finally, let x, y ∈ S such that dG(x, y) = 2. Let z ∈ NG(x) ∩ NG(y). Suppose there exists t ∈ Tx ∩ Ty. Then dG⊠H((x, t), (y, t)) = 2 which is not possible. Hence, Tx ∩ Ty = ∅. Suppose dH(Tx, Ty) = 1. Then this would imply that there exist c ∈ Tx and d ∈ Ty with dH(c, d) = 1. Consequently, [(x, c), (z, d), (y, d)] is an (x, c)-(y, d) geodesic in G⊠H. If dH(Tx, Ty) = 2, then there exist g ∈ Tx and h ∈ Ty with dH(g, h) = 2. Let l ∈ NH(g)∩NH(h). Then [(x, g), (z, l), (y, h)] is an (x, c)-(y, d) geodesic in G⊠H. In any case, we get a contradiction. Therefore, (iii) holds. For the converse, suppose that C satisfies properties (i), (ii), (iii), and (iv). Let (v, p), (w, q) ∈ C such that (v, p) ̸= (w, q). Consider the following cases: Case 1. v = w. Then p, q ∈ Tv and p ̸= q. By condition (i), dH(p, q) ̸= 2. Hence, dG⊠H((v, p), (w, q)) ̸= 2. Case 2. v ̸= w. Suppose first that vw ∈ E(G). If p = q, then dG⊠H((v, p), (w, q)) = 1. Suppose p ̸= q. If pq ∈ E(H), then dG⊠H((v, p), (w, q)) = 1. Suppose pq /∈ E(H). By (ii), Tv ∪ Tw is a hop independent set. It follows that dH(p, q) ≥ 3. Thus, dG⊠H((v, p), (w, q)) ≥ 3. Next, suppose that vw /∈ E(G). If dG(v, w) ≥ 3, then dG⊠H((v, p), (w, q)) ≥ 3. If dG(v, w) = 2, then Tv ∩ Tw = ∅ by (iii). Hence, p ̸= q and dH(p, q) ≥ 3 by (ii). Therefore, dG⊠H((v, p), (w, q)) ≥ 3. Accordingly, C is a hop independent set in G⊠H. Corollary 7. Let G and H be non-trivial connected graphs and let ⟨Sq1 , Sq2 , · · · , Sqm⟩ and ⟨Dp1 , Dp2 , · · · , Dpr⟩ be maximum decreasing d3-sequences of cliques in G and H, respec- tively. Then αh(G⊠H) ≥ m∑ i=1 r∑ j=1 qipj . Proof. Let ⟨Sq1 , Sq2 , · · · , Sqm⟩ and ⟨Dp1 , Dp2 , · · · , Dpr⟩ be maximum decreasing d3- sequences of cliques in G and H, respectively. Let S = ∪m i=1Sqi and for each x ∈ S, set Rx = ∪r j=1Dpj . Then each Rx is a hop independent set in H. Clearly, condition (ii) of Theorem 6 is satisfied. Moreover, because S is a hop independent set, condition (iii) of J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5658 12 of 15 Theorem 6 also holds. Thus, C = ∪x∈S({x} × Tx) is a hop independent set in G⊠H and αh(G⊠H) ≥ |C| = ∑ x∈S |Rx| = m∑ i=1 ∑ x∈Sqi | ∪r j=1 Dpj | = m∑ i=1 qi r∑ j=1 pj = m∑ i=1 r∑ j=1 qipj . This proves the assertion. Remark 2. The bound given in Corollary 7 is tight. Consider graphs G = P6 = [v1, v2, v3, v4, v5, v6], H = P6 = [p1, p2, p3, p4, p5, p6], and the strong product P6 ⊠ P6 in Figure 4. The sequences ⟨{v1, v2}, {v5, v6}⟩ and ⟨{p1, p2}, {p5, p6}⟩ are maximum (decreasing) d3-sequences of cliques in G and H, respectively. Clearly, αh(G⊠H) = |{v1, v2}||{p1, p2}|+ |{v1, v2}||{p5, p6}|+ |{v5, v6}||{p1, p2}|+ |{v5, v6}||{p5, p6}| = 16. J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5658 13 of 15 ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ....... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ....... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ....... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ....... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ....... .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ....... .................................... 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• • • • • • Figure 4: P6 ⊠ P6 v1 v2 v3 v4 v5 v6 p5 p6p4p3p2p1 We now show that the hop independent set decision problem (HISP) is NP -complete. To this end, consider the following hop independent set decision problem (HISP): Instance: Given a graph G = (V (G), E(G)) and a positive integer k ≤ |V (G)| Question: Does G contain a hop independent set of size k? On the other hand, the clique decision problem (CP) is stated as follows: Instance: Given a graph G = (V (G), E(G)) and a positive integer k ≤ |V (G)| Question: Does G contain a clique of size k? Theorem 7 ([6]). The clique problem is NP -complete. Theorem 8. The hop independent set problem is NP -complete. Proof. Given a subset S of vertices of G, one can check in polynomial time if S is a hop independent set. Hence, the hop independent set problem is NP . We now use the clique problem (CP) to show that HISP is NP -Hard. To this end, let G = (V (G), E(G) be a graph with V (G) = {v1, v2, ..., vn} and let k be a positive integer with k ≤ |V (G)|. Let H = G+K1, where V (K1) = {v}. If S is a clique in G and |S| = k, then S′ = S ∪{v} is a hop independent set in H and |S′| = k + 1. Conversely, suppose S∗ is a hop independent set in H with |S∗| = k+1. Suppose S∗ is not a clique in H. Then, by Theorem 1, ⟨S∗⟩ has at least two complete components, say H1 and H2. Pick x ∈ V (H1) and y ∈ V (H2). Then J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5658 14 of 15 dH(x, y) = 2, a contradiction. Thus, S∗ is a clique in H. If S∗ ⊆ V (G), then S∗ \ {x}, where x ∈ S∗, is a clique in G with size k. Suppose v ∈ S∗. Then S∗ \ {v} is a clique in G with size k. Therefore, G has a clique of size k if and only if H has a hop independent set of size k + 1. Accordingly, the hop independent set problem is NP -complete. 9. Conclusion The hop independence parameter has been explored for the shadow graph, complemen- tary prism, edge corona, disjunction, and strong product of two graphs. It is conjectured that the hop independence number of the edge corona of two graphs may take only three possible values, namely; αh(G), ν(G)ω(G), and ν(G)ω(G) + 2. It is also conjectured that the lower bound in Corollary 7 is the exact value of the parameter. Using the fact that the clique decision problem (CP) is NP -complete, it was shown that the hop independent set problem (HISP) is also NP -complete. Acknowledgements The authors would like to thank the referees for the comments and suggestions they offered us which led to the improvement of the paper. Also, the authors would like to thank the Department of Science and Technology - Accelerated Science and Technology Human Resource Development Program (DOST-ASTHRDP)-Philippines, and MSU-Iligan Institute of Technology for funding this research. References [1] S. Arriola and Jr. S. Canoy. (1, 2)∗-domination in graphs. Advances and Applications in Discrete Mathematics., 18(2):179–190, 2017. [2] S. Ayyaswamy, B. Krishnakumari, B. Natarjan, and Y. Venkatakrishnan. Bounds on the hop domination number of a tree. Proceedings-Mathematical Sciences, 125(4):449– 455, 2015. [3] J. Hassan1 and Jr. S. Canoy. Hop independent hop domination in graphs. Eur. J. Pure Appl. Math., 15(4):1783–1796, 2022. [4] J. Hassan1, Jr. S. Canoy, and A. Aradais. Hop independent sets in graphs. Eur. J. Pure Appl. Math., 15(2):467–477, 2022. [5] M. Henning and N. Rad. On 2-step and hop dominating sets in graphs. Graphs and Combinatorics., 33(4):913–927, 2017. [6] R. Karp. Reducibility among combinatorial problems. Complexity of Computer Com- putations (PDF), New York: Plenum, pages 85–103, 1972. [7] C. Natarajan and S. Ayyaswamy. Hop domination in graphs ii. Versita, 23(2):187– 199, 2015. [8] Jr. S. Canoy, R. Mollejon, and J. G. Canoy. Hop dominating sets in graphs under binary operations. Eur. J. Pure Appl. Math., 12(4):1455–1463, 2019. J. Anoche, S. Canoy, Jr. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5658 15 of 15 [9] Jr. S. Canoy and G. Salasalan. Locating-hop domination in graphs. Kyungpook Mathematical Journal, 62:193–204, 2022. [10] G. Salasalan and Jr. S. Canoy. Global hop domination numbers of graphs. Eur. J. Pure Appl. Math., 14(1):112–125, 2021.