EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 4, 2023, 2431-2449 ISSN 1307-5543 – ejpam.com Published by New York Business Global Hop Italian domination in graphs Sergio R. Canoy,Jr.1,2, Ferdinand P. Jamil1,2 and Sheila M. Menchavez1,2,∗ 1 Department of Mathematics and Statistics, College of Science and Mathematics, Mindanao State University-Iligan Institute of Technology, 9200 Iligan City, Philippines 2 Center for Graph Theory, Algebra and Analysis, Premier Research Institute of Science and Mathematics (PRISM), Mindanao State University - Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. Given a simple graph G = (V (G), E(G)), a function f : V (G) → {0, 1, 2} is a hop Italian dominating function if for every vertex v with f(v) = 0 there exists a vertex u with f(u) = 2 for which u and v are of distance 2 from each other or there exist two vertices w and z for which f(w) = 1 = f(z) and each of w and z is of distance 2 from v. The minimum weight∑ v∈V (G) f(v) of a hop Italian dominating function is the hop Italian domination number of G, and is denoted by γhI(G). In this paper, we initiate the study of the hop Italian domination. First, we establish some properties of the the hop Italian dominating function and characterize graphs G with smaller values for γhI(G). Next, we explore the relationships of the hop Italian domination number with closely related concepts, particularly the hop Roman domination number and the 2- hop domination number. Finally, we investigate the hop Italian domination in the complementary prism, join, corona and lexicographic product of graphs. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Hop Italian dominating function, Hop Italian domination number 1. Introduction The history of the Roman domination in graphs can be traced back to the military strategy adapted by Constantine the Great (Emperor of Rome) during the fourth century AD (see [23, 26]). In order to defend his cities Constantine issued a decree that any city without a legion stationed to secure it must neighbor another city having two stationed legions. If the first were attacked, then the second could deploy a legion to protect it without becoming vulnerable itself. It is called defense-in-depth strategy, which used only four Field Armies (FA) available for deployment to defend a total of eight regions. Roman domination as a mathematical concept was introduced by Cockayne, Dreyer, S.M. Hedetniemi and S.T. Hedetniemi [9] in 2004. Thereafter, it has become an active ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i4.4914 Email addresses: sergio.canoy@g.msuiit.edu.ph (S.R.Jr. Canoy), ferdinand.jamil@g.msuiit.edu.ph (F. Jamil), sheila.menchavez@g.msuiit.edu.ph (S. Menchavez) https://www.ejpam.com 2431 © 2023 EJPAM All rights reserved. S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2432 research area (see [1, 6, 12, 18, 19, 21, 22, 25, 27]). It models many facility location problems (see [7]), where f(v) is viewed as cost function. Units with cost 2 may be able to serve neighboring locations, while units with costs 1 can serve only their own location. In a communication network, f(v) = 2 is assigned to locations where we install wireless hubs which are more expensive but can serve neighboring locations, while f(v) = 1 is assigned to locations where we install wired hubs which function at low-range but are cheaper. In 2016, the Roman 2-domination was introduced by Chellali, Haynes, Hedetniemi and McRae [8]. It is also called Italian domination. A function f : V (G) → {0, 1, 2} is an Italian dominating function provided for every vertex v with f(v) = 0 we have∑ x∈NG(v) f(x) ≥ 2, where N(v) is the set of all vertices adjacent to v. Apparently, a Roman dominating function is an Italian dominating function. The Italian domination number is the minimum weight of an Italian dominating function. Excellent references for Italian domination include [8, 20]. In 2017, Shabani [24] introduced the hop Roman domination. A hop Roman domi- nating function on G is a function f : V (G) → {0, 1, 2} satisfying the property that for every vertex v of G with f(v) = 0 there is a vertex u with f(u) = 2 for which the distance dG(u, v) between u and v is 2. It was largely motivated by the concept of hop domination which is relatively well-known to have a wide range of applications in social network. Hop Roman domination in graphs was further studied in [21, 22]. This present paper intends to introduce and initiate the study of the hop Italian domination. We will establish some of its properties and make characterizations for some special graphs. We will explore its relationships with the hop Roman domination and other related hop domination concepts. Finally, we will investigate the hop Italian domination in graphs under some binary operations. All throughout this paper, we consider only graphs which are simple, finite and undi- rected. Given a graph G = (V (G), E(G)), we call V (G) the vertex set of G and E(G) its edge set. The cardinality |V (G)| of V (G) is the order of G. All terminologies used here which are not being defined are adapted from [3]. Let G and H be disjoint graphs. The complementary prism GG is formed from G and its complement G by adding a perfect matching between corresponding vertices of G and G. If for each v ∈ V (G), v is the vertex in G corresponding to v, then GG is formed by adding the edge vv for every v ∈ V (G). The corona of G and H is the graph G ◦H obtained by taking one copy of G and |V (G)| copies of H, and then joining the ith vertex of G to every vertex in the ith copy of H. In particular, we call G ◦K1 the corona of G, and write cor(G) = G ◦ K1. The composition (or lexicographic product) of G and H is the graph G[H] with V (G[H]) = V (G) × V (H) and (u, v)(u′, v′) ∈ E(G[H]) if and only if either uu′ ∈ E(G) or u = u′ and vv′ ∈ E(H). In any of these graphs, G and H are referred to as their basic component graphs. For vertices u and v of a graph G, a u-v geodesic is any shortest path in G joining u and v. The length of a u-v geodesic is the distance between u and v, and is denoted by dG(u, v). S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2433 The eccentricity of v refers to the quantity e(v) = max{dG(u, v) : v ∈ V (G)}. Customarily, diam(G) = max{e(v) : v ∈ V (G)}. In this paper, we write e(G) = min{e(v) : v ∈ V (G)}. Vertices u and v of a graph G are neighbors if uv ∈ E(G). The open neighborhood of v refers to the set NG(v) consisting of all neighbors of v. The degree of v refers to the cardinality |NG(v)| of the open neighborhood of v, and δ(G) is the minimum degree of a vertex of G. The closed neighborhood of v is the set NG[v] = NG(v) ∪ {v}. Customarily, for S ⊆ V (G), NG(S) = ∪v∈SNG(v) and NG[S] = ∪v∈SNG[v]. A subset S ⊆ V (G) is a dominating set of G if NG[S] = V (G). The minimum cardinality γ(G) of a dominating set of G is the domination number of G. A dominating set of cardinality γ(G) is called a γ-set of G. The reader is referred to [2, 10, 11, 13, 16, 17] for the history, fundamental concepts and recent developments of domination in graphs as well as its various applications. A set S ⊆ V (G) is a pointwise nondominating set of G (or PND-set of G) if for each v ∈ V (G)\S, there exists u ∈ S such that uv /∈ E(G). The smallest cardinality of a point- wise nondominating set of G, denoted by pnd(G), is called the pointwise nondomination number of G. Any point-wise nondominating (resp. dominating pointwise nondominat- ing) set S of G of cardinality |S| = pnd(G) (resp. |S| = γpnd(G)), is called a pnd-set (resp. γpnd-set) of G. PND-sets are introduced and discussed in [5]. A subset S of V (G) is a hop dominating set of G if for each v ∈ V (G) \ S, there exists u ∈ S for which dG(u, v) = 2. The minimum cardinality of a hop dominating set is called the hop domination number of G, and is denoted by γh(G). Any hop dominating set of cardinality γh(G) is called γh-set of G. Good references on hop domination include [4, 5, 15]. At times we write S ∈ HD(G) to mean that S is a hop dominating set of G. For a vertex v of a connected graph G, NG(v, 2) = {u ∈ V (G) : dG(u, v) = 2}. Each element of NG(v, 2) is called a hop-neighbor of v. For S ⊆ V (G), NG(S, 2) = ∪v∈SNG(v, 2) and NG[S, 2] = NG(S, 2)∪S. Precisely, S is a hop dominating set if and only if NG[S, 2] = V (G). A subset S of V (G) is a 2-hop dominating set of G if for each v ∈ V (G)\S, there exist distinct vertices u,w ∈ S for which dG(u, v) = 2 = dG(w, v). The minimum cardinality of a 2-hop dominating set of G is the 2-hop domination number of G, denoted by γ2h(G). A comprehensive study on 2-hop domination is given in [14], where 2-hop domination is referred to as double hop domination. Here we also write S ∈ 2-HD(G) to mean that S is a 2-hop dominating set of G A set S ⊆ V (G) is a (1, 2)∗-dominating set of G (resp. (1, 2)∗-total dominating set) if it is both a dominating (resp. a total dominating) set and a hop dominating set of G. The smallest cardinality of a (1, 2)∗ -dominating (resp. (1, 2)∗-total dominating) set of G, denoted by γ∗1,2(G) (resp.γ∗t1,2(G)) is called the (1, 2)∗ -domination number (resp. (1, 2)∗- total domination number) of G. A (1, 2)∗ -dominating (resp. (1, 2)∗ -total dominating) set S with |S| = γ∗1,2(G) (resp. |S| = γ∗t1,2(G)) is called a γ∗1,2-set (resp. γ∗t1,2-set) of G. The concept of (1, 2)∗-domination (a variation of (1, 2)-domination) is introduced in [4]. A function f : V (G) → {0, 1, 2} is a hop Roman dominating function of G if for each v ∈ V (G) with f(v) = 0 there exists u ∈ V (G) for which dG(u, v) = 2 and f(u) = 2. The S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2434 sum ωG(f) = ∑ v∈V (G) f(v) is the weight of f in G. The minimum weight of a hop Roman dominating function of G is the hop Roman domination number of G, and is denoted by γhR(G). 1.1. Some known results Theorem 1.1. [24] For any graph G, γhR(G) ≤ 2γh(G). Observation 1.2. (i) γhR(Pn) = { 4k + r, if n = 6k + r; 0 ≤ r ≤ 3; k ≥ 0 4k + 4, if n = 6k + r; 4 ≤ r ≤ 5; k ≥ 0, and (ii) γhR(Cn) =  3, if n = 3; 4, if n = 4, 5; 4k + r, if n = 6k + r; 0 ≤ r ≤ 3; k ≥ 1 4k + 4, if n = 6k + r; 4 ≤ r ≤ 5; k ≥ 1. 2. Hop Italian domination A function f : V (G) → {0, 1, 2} is a hop Italian dominating function (or hID-function) of G if for each v ∈ V (G) with f(v) = 0, ∑ x∈NG(v;2) f(x) ≥ 2. More precisely, f is an hID-function of G if and only if at least one of the following holds for each v ∈ V (G) with f(v) = 0: (i) There exists u ∈ V (G) for which f(u) = 2 and dG(u, v) = 2; (ii) There exist distinct u,w ∈ V (G) for which f(u) = 1 = f(w) and dG(u, v) = 2 = dG(w, v). The minimum weight ∑ v∈V (G) f(v) of an hID-function of G is the hop Italian domination number of G, and is denoted by γhI(G). If hID(G) denotes the collection of all hID- functions of G, then γhI(G) = min{ωG(f) : f ∈ hID(G)}. A hop Italian dominating function f of G with ωG(f) = γhI(G) is called γhI-function of G. As usual, for f : V (G) → {0, 1, 2} we write f = (V0, V1, V2), where Vk = {v ∈ V (G) : f(v) = k} for each k ∈ {0, 1, 2}. Thus, f = (V0, V1, V2) ∈ hID(G) if and only if for each v ∈ V0, V2 ∩NG(v, 2) ̸= ∅ or |V1 ∩NG(v)| ≥ 2. If f = (V0, V1, V2) is a γhI -function of G, then V1 ∪ V2 is a hop dominating set of G so that γh(G) ≤ |V1 ∪ V2| ≤ ωG(f) = γhI(G). Now observe that if S ⊆ V (G) is a γ2h-set of G, then f = (V (G) \ S, S,∅) ∈ hID(G). Thus, γhI(G) ≤ |S| = γ2h(G). Moreover, since a hop Roman dominating function is a hop Italian dominating function, γhI(G) ≤ min{γhR(G), γ2h(G)}. (1) S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2435 Let G be the graph in Figure 1 obtained by joining two copies of P5, say [x1, x2, x3, x4, x5] and [y1, y2, y3, y4, y5], using the edges x1y1, x3y3 and x5, y5. Then γhI(G) = γhR(G) = .................................................................................................... .................................................................................................... .................................................................................................... .................................................................................................... .................................... .................................................................................................... .................................................................................................... .................................................................................................... .................................................................................................... .................................... • • • • • • • • • • ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ..... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ..... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ..... x1 x2 x3 x4 x5 y1 y2 y3 y4 y5 Figure 1: A graph G with γhI(G) = γhR(G) 4 < 6 = γ2h(G). In this case, γhR(G) = γhI(G) is determined by the function f = (V (G) \ {x3, y3},∅, {x3, y3}). On the other hand, if G = C5, then γhI(G) = γ2h(G) = 3 < 4 = γhR(G). However, if G = Kp (the complete graph on p vertices), then γhI(G) = γhR(G) = γ2h(G) = p. Observation 2.1. Let G be any graph. Then (i) γhI(G) = γhR(G) if and only if G has a γhI-function that is a hop Roman dominating function of G; (ii) γhI(G) = γ2h(G) if and only if G has a γhI-function (V0, V1, V0) for which V2 = ∅. Observation 2.2. On paths, cycles and complete bipartite graphs: (i) γhI(Pn) =  2, if n = 2; 3, if n = 3; 2k + 2, if n = 4k + r with 0 ≤ r ≤ 2; k ≥ 1; 2k + 3, if n = 4k + 3; k ≥ 1. (ii) γhI(Cn) =  3, if n = 3, 5; 4, if n = 4; 2k + 2, if n = 4k + 2 + r with 0 ≤ r ≤ 2; k ≥ 1; 2k + 3, if n = 4k + 5; k ≥ 1. (iii) γhI(Km,n) =  2, if m = n = 1; 3, if m = 1 (resp. n = 1) and n ≥ 2 (resp m ≥ 2); 4, if m ≥ 2 and n ≥ 2 2.1. Some properties and graphs with small values of γhI Let f = (V0, V1, V2) be a γhI -function of G. A vertex w ∈ V0 is an Italian private hop-neighbor of v ∈ V1 ∪ V2 under f provided ∑ u∈NG(w,2)\{v} f(u) < 2. If no confusion arises, instead of saying Itaian private hop-neighbor of v under f , we simply say Italian private hop-neighbor of v. S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2436 Observe that the function given by f(x) = 1 for all x ∈ V (G) is a γhI -function of G = Kp. In this case, V2 = ∅, and such is a particular case of the following proposition. Proposition 2.3. For every graph G, there exists a γhI-function f = (V0, V1, V2) such that either V2 = ∅ or V2 ̸= ∅ and v has at least three Italian private hop-neighbors for each v ∈ V2. Proof : Let f = (V0, V1, V2) be a γhI -function of G with a minimum |V2|. If V2 = ∅, then the proposition holds. Suppose that V2 ̸= ∅, and let v ∈ V2. We claim that v has at least three Italian private hop-neighbors. First, note that if v has no Italian private hop-neighbor in V0, then g = (V0, V1 ∪ {v}, V2 \ {v}) ∈ hID(G) with ωG(g) < ωG(f), a contradiction. Next, suppose that v has exactly one Italian private hop-neighbor w ∈ V0. If ∑ u∈NG(w,2)\{v} f(u) = 0, then g = (V ∗ 0 , V ∗ 1 , V ∗ 2 ) ∈ hID(G) with ωG(g) = ωG(f), where V ∗ 0 = V0 \ {w}, V ∗ 1 = V1 ∪ {w, v} and V ∗ 2 = V2 \ {v}. Since |V ∗ 2 | < |V2|, this is a contradiction to the choice of f . On the other hand, if ∑ u∈NG(w,2)\{v} f(u) = 1, then g = (V0, V1 ∪ {v}, V2 \ {v}) ∈ hID(G) with ωG(g) < ωG(f), a contradiction. Finally, suppose that v has exactly two neighbors w and z in V0. Exactly one of the following holds: (a) ∑ u∈NG(w,2)\{v} f(u) = 0 and ∑ u∈NG(z,2)\{v} f(u) = 0; (b) ∑ u∈NG(w,2)\{v} f(u) = 1 and ∑ u∈NG(z,2)\{v} f(u) = 1; (c) ∑ u∈NG(w,2)\{v} f(u) = 0 and ∑ u∈NG(z,2)\{v} f(u) = 1; and (d) ∑ u∈NG(w,2)\{v} f(u) = 1 and ∑ u∈NG(z,2)\{v} f(u) = 0. Suppose that (a) holds for f . Put V ∗ 0 = {v} ∪ (V0 \ {w, z}), V ∗ 1 = V1 ∪ {w, z} and V ∗ 2 = V2 \ {v}. Then g = (V ∗ 0 , V ∗ 1 , V ∗ 2 ) ∈ hID(G) with wG(g) = wG(f). Since |V ∗ 2 | < |V2|, this is a contradiction to the assumption of f . Next, suppose that (b) holds for f . In this case, define V ∗ 0 = V0, V ∗ 1 = V1 ∪ {v} and V ∗ 2 = V2 \ {v}. Then g = (V ∗ 0 , V ∗ 1 , V ∗ 2 ) ∈ hID(G) with wG(g) < wG(f), a contradiction. Next, suppose that (c) holds for f . Define V ∗ 0 = V0 \ {w}, V ∗ 1 = V1 ∪ {w, v} and V ∗ 2 = V2 \ {v}. Then g = (V ∗ 0 , V ∗ 1 , V ∗ 2 ) ∈ hID(G) with wG(g) = wG(f). Since |V ∗ 2 | < |V2|, this is a contradiction. Similar contradiction is attained if (d) holds for f . The above contradictions imply that v has at least three Italian private hop-neighbors . ■ Proposition 2.4. Let G be a connected graph of order n. Then (i) γhI(G) = 1 if and only if G = K1; (ii) γhI(G) = 2 if and only if G = K2; (iii) γhI(G) = 3 if and only if γ2h(G) = 3 or G = K1 + (K1 ∪H) for some graph H of order ≥ 3. S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2437 Proof : For (i): If G = K1, then γhI(G) = 1. Conversely, if γhI(G) = 1, then γh(G) = 1 and so G = K1. For (ii): If G = K2, then γhI(G) = 2. Assume that γhI(G) = 2. By Proposition 2.3, G has a γhI -function f = (V0, V1, V2) for which either V2 = ∅ or V2 ̸= ∅ and each v ∈ V2 has at least 3 private hop-neighbors in V0. Suppose that V2 ̸= ∅. Let v ∈ V2 and let u ∈ V0 be a private hop-neighbor of v. Then there exists a u-v geodesic [u,w, v] in G. If w ∈ V1∪V2, then wG(f) ≥ f(v) + f(w) ≥ 3. If w ∈ V0 and a ∈ V2 \ {v} for which dG(a,w) = 2, then wG(f) ≥ f(v) + f(a) = 3. Either case is a contradiction. Thus, V2 = ∅ and |V1| = 2. It follows that V0 = ∅ and |V (G)| = |V1| = 2. Therefore, G = K2. For (iii): If G = K1 + (K1 ∪H) for some graph H of order ≥ 3, then clearly γhI(G) = 3. Suppose that γ2h(G) = 3. Then G /∈ {K1,K2}. By (ii) and Equation 1, γhI(G) = 3. Conversely, suppose that γhI(G) = 3, and let f = (V0, V1, V2) be a γhI -function of G such that either V2 = ∅ or V2 ̸= ∅ and each v ∈ V2 has at least 3 private hop-neighbors in V0. If V2 = ∅, then by γ2h(G) = γhI(G) = 3 by Observation 2.1(ii). Suppose that V2 ̸= ∅. Then |V2| = 1 = |V1|, say V2 = {v} and V1 = {u}. By Proposition 2.3, v has at least 3 Italian private hop-neighbors in V0. Thus, G = ⟨{u}⟩ + (⟨{v}⟩ ∪H) = K1 + (K1 ∪H), where H = ⟨V0⟩ of order ≥ 3. ■ It is worth noting that the family of graphs G for which γ2h(G) = γhI(G) = 3 includes P3, K3, C5, K1-gluing of C5 and K2, K2-gluing of C5 and K2; graph G containing H = K3 such that each v ∈ V (G)\V (H) is adjacent to (exactly) one vertex of H (may be viewed as one generate by a triangle K3; the graph G1 in Figure 2 (may be viewed as one generated by mutually nonadjacent x, y and z); and the graph G2 in Figure 2 (may be viewed as one generated by path [x, y, z]). • • • • • • • • • ............................................................................................................................ .................................... ............... .............. .............. .............. .............. .............. .............. .............. ........... .................................... ......... ........ ........ ........ ........ ........ ........ ....... .................................... ............................................................................................................................ .................................... ............... .............. .............. .............. .............. .............. .............. .............. ........... .................................... ......... ........ ........ ........ ........ ........ ........ ....... .................................... .................................... ............................................................................................................ .................................... .......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ........ .................................... .................................... ........................................................................................................................................... .................................... ............ ........... ........... ........... ........... ........... ........... ........... ...... .................................... .................................... ........................................................................................................................................... .................................... ............................................................................................... .................................... .................................... x y z G1 • • • • • • • ............................................................................................................................ .................................... ............... .............. .............. .............. .............. .............. .............. .............. ........... .................................... ......... ........ ........ ........ ........ ........ ........ ....... .................................... ............................................................................................................................ .................................... ............... .............. .............. .............. .............. .............. .............. .............. ........... .................................... ......... ........ ........ ........ ........ ........ ........ ....... .................................... ...................................... ......................... ......................... ......................... ......................... ...... .................................... .......................... ......................... ......................... ......................... ....... .................................... .................................... x y z G2 .......................................................................................................................................... .................................... ......... ........ ........ ........ ........ ........ ........ ....... .................................... ......... ........ ........ ........ ........ ........ ........ ....... .................................... ............... .............. .............. .............. .............. .............. .............. .............. ........... .................................... .......................................................................................................................................... .................................... • • • • • .................................................................................................................................................................................................................................. ............... .............. .............. .............. .............. .............. .............. .............. ........... 21 0 0 0 G3 Figure 2: Examples of graphs G described in Proposition 2.4(iii) with γhI(G) = 3 Graph G3 in Figure 2 shows an example of a graph G with γ2h(G) ̸= 3 = γhI(G). Proposition 2.5. (i) For every nonnegative integer k, there exists a connected graph G for which γhR(G) = γhI(G) + k. (ii) For every pair of positive integers a and b with 4 ≤ a ≤ b, there exists a connected graph G for which γhI(G) = a and γ2h(G) = b. Consequently, for each nonnegative integer k, there exists a connected graph G with γ2h(G) = γhI(G) + k. S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2438 Proof : For (i): If k = 0, then we take a complete graph G. Suppose that k ≥ 1. First, suppose that k is even, say k = 2j for some integer j ≥ 1. Choose G to be the path Pn, where n = 12j + 3. Writing n = 6(2j) + 3, Observation 1.2 yields γhR(G) = γhR(Pn) = 4(2j) + 3 = 8j + 3. Similarly, by Observation 2.2, γhI(G) = 2(3j) + 3. Thus, γhR(G) = (6j + 3) + 2j = γhI(G) + k. Next, suppose that k = 2j + 1 for some integer j ≥ 0. If j = 0, then we take G = C7. Assume that j ≥ 1. Consider the graph G = Cn, a cycle on n vertices, where n = 12j+3. By Observation 1.2 and Observation 2.2, γhR(G) = 4(2j) + 3 = (6j + 2) + (2j + 1) = [(2(3j) + 2] + (2j + 1) = γhI(G) + k. For (ii): If a = b, then we take G = Ka, the complete graph on a vertices. Suppose that b = a+ k with k ≥ 1. We consider the following cases: Case 1: Suppose that a = 2n+2 for some n ≥ 1. Let t = 4n, and put Pt = [x1, x2, . . . , xt], a path on t vertices. If n = 1, then we take G = G1, where G1 is the graph in Figure 3 obtained from P4 by adding k + 1 distinct paths [x3, yj , zj ], j = 1, 2, . . . , k + 1. Define ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................ .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ...................................................................... .................................... ..................................................................................................... .................................... .................................... ...................................................................... .................................... ..................................................................... .................................... .................................... .......... ......... ......... ......... ......... ......... ......... ...... .................................... .......... ......... ......... ......... ......... ......... ......... ..... .................................... .................................... .................... ................... ................... ............ .................................... .................. ................. ................. ................. ................. ............... .................................... .................................... • • • • • • • • • • • • · · · · · · x1 x2 x3 x4 y1 y2 yk yk+1 z1 z2 zk zk+1 G1 ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... .................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................ .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ...................................................................... .................................... ..................................................................................................... .................................... .................................... ...................................................................... .................................... ..................................................................... .................................... .................................... .......... ......... ......... ......... ......... ......... ......... ...... .................................... .......... ......... ......... ......... ......... ......... ......... ..... .................................... .................................... .................... ................... ................... ............ .................................... .................. ................. ................. ................. ................. ............... .................................... .................................... • • • • • • • • • • • • • • • • • • · · · · · · · · · x1 x2 x3 x4 x5 x6 xt−3 xt−2 xt−1 xt y1 y2 yk−1 ykz1 z2 zk−1 zk ....................................... ....................................... G2 Figure 3: Examples of graphs G with γhI(G) = a and γ2h(G) = b V2 = {x3, x4}, V1 = ∅ and V0 = V (G)\{x3, x4}. Then f = (V0, V1, V2) is a γhI -function of G. Thus, γhI(G) = 4 = a. On the other hand, the set {x1, x2, x4}∪{zj : j = 1, 2, . . . , k+1} is a γ2h-set of G, implying that γ2h(G) = 3+k+1 = 4+k = b. Suppose that n ≥ 2. Obtain G as the graph G2 in Figure 3 from Pt by adding k distinct paths [x3, yj , zj ], j = 1, 2, . . . , k. Define V2 = {x3, x4}, V1 = {x7, x8, x11, x12, . . . , xt−1, xt} and V0 = V (G)\ (V1 ∪ V2). Then f = (V0, V1, V2) is a γhI -function of G, implying that γhI(G) = γhI(Pt) = 2n+ 2 = a. On the other hand, necessarily, S = {zj : j = 1, 2, . . . , k} is contained in any 2-hop dominating set of G. Observe that S ∪ {x1, x2, x4, x5, x8, x9, . . . , xt−4, xt−3, xt−1, xt} is a γ2h-set of G. Thus, γ2h(G) = (2n+ 2) + k = a+ k = b. S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2439 Case 2: Suppose that a = 2n+3 for some n ≥ 1. Put t = 4n and let Pt = [x1, x2, . . . , xt]. If n = 1, then obtain G as the graph G1 in Figure 4by adding to P6 k geodesics, namely ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................ .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ...................................................................... .................................... ..................................................................................................... .................................... .................................... ...................................................................... .................................... ..................................................................... .................................... .................................... .......... ......... ......... ......... ......... ......... ......... ...... .................................... .......... ......... ......... ......... ......... ......... ......... ..... .................................... .................................... .................... ................... ................... ............ .................................... .................. ................. ................. ................. ................. ............... .................................... .................................... • • • • • • • • • • • • • •• • • • · · · · · · x1 x2 x3 x4 x5 x6 y1 y2 yk−1 ykz1 z2 zk−1 zk G1 ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... .................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................ .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ...................................................................... .................................... ..................................................................................................... .................................... .................................... ...................................................................... .................................... ..................................................................... .................................... .................................... .......... ......... ......... ......... ......... ......... ......... ...... .................................... .......... ......... ......... ......... ......... ......... ......... ..... .................................... .................................... .................... ................... ................... ............ .................................... .................. ................. ................. ................. ................. ............... .................................... .................................... • • • • • • • • • • • • •• • • • • • • • · · · · · · · · · x1 x2 x3 x4 x5 xt−3 xt−2 xt−1 xt y1 y2 yk+1 yk+2z1 z2 zk+1 zk+2 ....................................... ....................................... G2 Figure 4: Examples of graphs G with γhI(G) = a and γ2h(G) = b [x2, yj , zj ], j = 1, 2, . . . , k. Then γhI(G) = 5 = a, which is determined by the γhI -function f = (V0, V1, V2) with V1 = {x6}, V2 = {x2, x3} and V0 = V (G) \ {x2, x3, x6}. On the other hand, S = {zj : j = 1, 2, . . . , k} is always contained in a 2-hop dominating set of G so that S ∪ (V (P6) \ {x2}) is a γ2h-set of G. Thus, γ2h(G) = 5 + k = b. Now, suppose that n ≥ 2. Obtain G as the graph G2 in Figure 4 from Pt by adding k distinct paths [x3, yj , zj ], j = 1, 2, . . . , k + 2. Define V2 = {x2}, V1 = {x1, x3, x4, x7, x8, . . . , xt−1, xt} and V0 = V (G) \ (V1 ∪ V2). Then f = (V0, V1, V2) is a γhI -function of G, implying that γhI(G) = γhI(Pt) = 2n + 3 = a. On the other hand, if S = {zj : j = 1, 2, . . . , k}, then S ∪ {x1, x3, x4, x7, x8, x11, x12, . . . , xt−1, xt} is a γ2h-set of G. Thus, γ2h(G) = (2n + 1) + (k + 2) = a+ k = b. ■ 2.2. PNDI-functions A function f = (V0, V1, V2) on V (G) is a PNDI-function of G if for each v ∈ V0 one of the following holds: (i) there exists u ∈ V2 for which v /∈ NG(u); (ii) there exist vertices u and w in V1 for which v /∈ NG(u) ∪NG(w). The minimum weight of an PNDI -function of G is the PNDI number of G, denoted by pndI(G). Any PNDI -function of G with weight equal to pndI(G) is a pndI-function. S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2440 Example 2.6. (1) pndI(Pn) =  1, if n = 1; 2, if n = 2; 3, if n ≥ 3. (2) pndI(Cn) = { 4, if n = 4; 3, otherwise . (3) pndI(Kp) = p for p ≥ 1 and pndI(Km,n) = 4 for m,n ≥ 2. If f = (V0, V1, V2) is a PNDI -function of G, then V1 ∪ V2 is a PND-set of G. Thus, pnd(G) ≤ |V1| + |V2| ≤ ωG(f) for all PNDI -functions f = (V0, V1, V2) of G. On the other hand, if S ⊆ V (G) is a PND-set of G, then f = (V (G) \ S,∅, S) is a PNDI - function of G. Also, every hop Italian dominating function is a PNDI -function. Thus, pnd(G) ≤ pndI(G) ≤ min{2 pnd(G), γhI(G)}. Observation 2.7. Let G be any graph. Then (i) pndI(G) = 1 if and only if G = K1; (ii) pndI(G) = 2 if and only if either G = K2 or G is a nontrivial graph with an isolated vertex; (iii) pndI(G) = 3 if and only if one of the following holds: (a) G has an endvertex; (b) G has a set of vertices S = {x, y, z} for which every v ∈ V (G) \ S is adjacent to at most one vertex in S. Lemma 2.8. Let G be a noncomplete graph. Then G admits a pndI-function f = (V0, V1, V2) for which V2 ̸= ∅. Proof : Let f = (V0, V1, V2) be a pndI -function of G with V2 = ∅. Then V (G) = V1. Since G is noncomplete, there exist u, v ∈ V (G) such thatdG(u, v) = 2. Observe that g = ({u}, V1 \ {u, v}, {v}) is a PNDI -function of G with ωG(g) = ωG(f). ■ 3. Graphs under binary operations In view of Proposition 2.4, γhI(GG) ≥ 2 for any graph G. Proposition 3.1. (complementary prism of graphs) Let G be any graph. Then (i) γhI(GG) = 2 if and only if G = K1. (ii) γhI(GG) = 4 for all nontrivial graphs G. S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2441 Proof : If G = K1, then γhI(GG) = γhI(P2) = 2. Conversely, if γhI(GG) = 2, then GG = K2 by Proposition 2.4(ii). This means that G = K1. Suppose that G ̸= K1. First, we claim that γhI(GG) ≥ 4. By (i), γhI(GG) ≥ 3. Suppose that γhI(GG) = 3. In view of Proposition 2.4(iii), γ(GG) = 1. This is possible only when GG = K2 so that γhI(GG) = 2, a contradiction. Thus, γhI(GG) ≥ 4. Let v ∈ V (G) and define V2 = {v, v}, V1 = ∅ and V0 = V (GG) \ {v, v}. Let z ∈ V0. Assume that z ∈ V (G) (the case where z ∈ V (G) is done similarly). If zv ∈ E(G), then [z, v, v] is a geodesic in GG so that v ∈ V2 ∩NGG(z, 2). On the other hand, if zv /∈ E(G), then [z, z, v] is a geodesic in GG so that v ∈ V2∩NGG(z, 2). This shows that f = (V0, V1, V2) ∈ hI(GG), and consequently, γhI(GG) ≤ ωGG(f) = 4. ■ From Proposition 3.1(ii), for n ≥ 5, γhI(KnKn) = 4 while max{γhI(Kn), γhI(Kn)} = n. Thus, contrary to the case of Italian domination in complementary prisms, it is not always true that γhI(GG) ≥ max{γhI(G), γhI(G)}. Corollary 3.2. For all graphs G, γhI(GG) ≤ γhI(G) + γhI(G), and this bound is sharp. Proof : If G = K1, then by Proposition 3.1(i) and Proposition 2.4(i), γhI(GG) = 2 = γhI(G) + γhI(G). Suppose that G ̸= K1. Since 2 ≤ γhI(G) and 2 ≤ γhI(G), 4 ≤ γhI(G) + γhI(G). The conclusion follows immediately from Proposition 3.1. To show sharpness of the bound, consider G = P2. By Observation 2.1(ii), γhI(GG) = γhI(P4) = 4 = γhI(G) + γhI(G). ■ Strict inequality can be obtained in Theorem 3.2. Consider G = K1 ∪K3. The graph GG is as shown in Figure 5. For this graph, γhI(GG) = 4, γhI(G) = 4 and γhI(G) = 3. ............................................................................................................................................................................ ............ ........... ........... ........... ........... ........... ........... ........... .... .. ............................................................................................................................... .................................... ........................................................................ .................................... .................................... .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ....... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ....... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ....... .......................... ......................... ......................... ......................... ......................... .................. ............................................................................................................................................................................................................................. .......................................................................................................................................................................................................................................................................................... Figure 5: The graph of GG where G = K1 ∪K3 Theorem 3.3. (join of graphs) Let G and H be any graphs, and f = (V0, V1, V0) be a function on V (G + H). Then f ∈ hID(G + H) if and only if f |G ∈ PNDI(G) and f |H ∈ PNDI(H), where f |G and f |H are the restrictions of f to G and H, respectively. S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2442 Proof : Let f = (V0, V1, V2) ∈ hID(G+H). Let v ∈ V0∩V (G). Then |V2∩NG+H(v, 2)| ≥ 1 or |V1∩NG+H(v, 2)| ≥ 2. Suppose that |V2∩NG+H(v, 2)| ≥ 1, and let u ∈ V2∩NG+H(v, 2). Since dG+H(u, v) = 2, u ∈ V2 ∩ V (G) and u /∈ NG(v). Suppose, on the other hand, that |V1 ∩ NG+H(v, 2)| ≥ 2, say u,w ∈ V1 ∩ NG+H(v, 2). Then u,w ∈ V1 ∩ V (G) and v /∈ NG(u)∪NG(w). This shows that f |G = (V0∩V (G), V1∩V (G), V2∩V (G)) ∈ PNDI(G). Similarly, f |H ∈ PNDI(H). Conversely, let v ∈ V0. Suppose that v ∈ V (G). If f |G ∈ PNDI(G), then there exists u ∈ V2 ∩ V (G) for which v /∈ NG(u) or there exist w, z ∈ V1 ∩ V (G) for which v /∈ NG(w) ∪ NG(z). The former implies that u ∈ V2 ∩ NG+H(v, 2), while the latter implies that w, z ∈ V1 ∩ NG+H(v, 2). Similarly, if v ∈ V (H) and f |H ∈ PNDI(H), then |V2 ∩NG+H(v, 2)| ≥ 1 or |V1 ∩NG+H(v, 2)| ≥ 2. Therefore, f ∈ hID(G+H). ■ Corollary 3.4. Let G and H be any graphs of orders m and n, respectively. Then γhI(G+H) = pndI(G) + pndI(H). In particular, (i) γhI(G+H) = m+ n if G and H are complete graphs; (ii) γhI(G+H) = 4 if both G and H have isolated vertices; (iii) γhI(G+H) = 1 + pndI(H) if G = K1; Proposition 3.5. Let G be a graph with no isolated vertices. Then γhI(G◦H) ≤ γ∗t1,2(G). Proof : Let S ⊆ V (G) be a γ∗t1,2-set of G, and define f = (V0, V1, V2), where V0 = V (G ◦ H) \ S, V1 = ∅ and V2 = S. Let v ∈ V0 ∩ V (G). Since V2 is a hop dominating set of G, there exists u ∈ V2 for which dG(u, v) = 2. Let v ∈ V0∩V (Hu), where u ∈ V (G). Since V2 is a total dominating set of G, there exists w ∈ V2 ∩NG(u). Then dG◦H(u,w) = 2. Thus, f ∈ hID(G ◦H). Consequently, γhI(G ◦H) ≤ ωG◦H(f) = 2|S| = 2γ∗t1,2(G). ■ Theorem 3.6. (corona of graphs) Let G be a nontrivial connected graph and H any graph, and let f = (V0, V1, V2) be a function on V (G ◦H). Then f ∈ hID(G ◦H) if and only if each of the following holds: (i) One of the following holds for each v ∈ V0 ∩ V (G): (a) |V2 ∩NG(v, 2)| ≥ 1 or |V1 ∩NG(v, 2)| ≥ 2; (b) There exists w ∈ NG(v) for which |V2 ∩ V (Hw)| ≥ 1; c) There exists w ∈ NG(v) for which |V1 ∩ V (Hw)| ≥ 2; (d) There exist u,w ∈ NG(v) for which |V1 ∩ V (Hw)| = 1 = |V1 ∩ V (Hu)|; (e) |V1 ∩NG(v, 2)| = 1 and there exists w ∈ NG(v) for which |V1 ∩ V (Hw)| = 1. (ii) Each of the following holds for every v ∈ V (G) with V2 ∩NG(v) = ∅: (a) f |Hv is a PNDI-function of Hv if NG(v) ⊆ V0; S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2443 (b) V (Hv) \ V0 is a PND-set of Hv if |V1 ∩NG(v)| = 1. Proof : Suppose that f ∈ hID(G ◦H). Then (i) is clear. Let v ∈ V (G) with V2 ∩NG(v) = ∅. Suppose that NG(v) ⊆ V0, and let u ∈ V0 ∩ V (Hv). Then |V2 ∩ NG◦H(u, 2)| ≥ 1 of |V1 ∩ NG◦H(u, 2)| ≥ 2. Since NG(v) ⊆ V0, the preceding statement implies that |V2∩NHv(u, 2)| ≥ 1 or |V1∩NHv(u, 2)| ≥ 2. It means that there exists w ∈ V2∩V (Hv) for which u /∈ NHv(w) or there exist w and z in V1 ∩ V (Hv) for which u /∈ NHv(w)∪NHv(z). Thus, f |Hv = (V0∩V (Hv), V1∩V (Hv), V2∩V (Hv)) is a PNDI -function of Hv and (ii)(a) holds. Suppose that |V1 ∩NG(v)| = 1. Let u ∈ V0 ∩ V (Hv). Following similar argument, since |V1 ∩NG(v)| = 1, we have |V2 ∩NHv(u, 2)| ≥ 1 or |V1 ∩NHv(u, 2)| ≥ 1. In any case, there exists w ∈ V (Hv)\V0 such that u /∈ NHv(w), showing that V (Hv)\V0 is a PND-set of Hv. This proves (ii)(b). Conversely, suppose that (i) and (ii) hold for f . Let v ∈ V0. If v ∈ V (G), then (i) implies the existence of u ∈ V2 such that dG◦H(u, v) = 2 or of vertices u and w in V1 such that dG◦H(u, v) = 2 = dG◦H(w, v). Now, suppose that v ∈ V (Hu) for some u ∈ V (G). If V2 ∩NG(u) ̸= ∅, and w ∈ V2 ∩NG(u), then w is the desired vertex for which w ∈ V2 and dG◦H(v, w) = 2. Suppose that V2 ∩NG(u) = ∅. We consider two cases: Case 1: If NG(u) ⊆ V0, then by condition (ii)(a), there exists there exists w ∈ V2∩V (Hu) for which v /∈ NHu(w) or there exist vertices z and w in V1 ∩ V (Hu) for which v /∈ NHu(w) ∪ NHv(z). The former implies that dG◦H(w, v) = 2, while latter implies that dG◦H(w, v) = 2 = dG◦H(z, v). Case 2: Suppose that NG(u) ∩ V1 ̸= ∅. If |NG(u) ∩ V1| ≥ 2, say w, z ∈ NG(u) ∩ V1, then dG◦H(w, v) = 2 = dG◦H(z, v). Suppose that |NG(u) ∩ V1| = 1, say x ∈ NG(u) ∩ V1. By (ii)(b), V (Hu) \ V0 is a PND-set of Hu so that there exists w ∈ V (Hu) \ V0 such that v /∈ NHu(w). We either have w ∈ V2 and dG◦H(w, v) = 2 or w ∈ V1 and dG◦H(w, v) = 2 = dG◦H(x, v). Accordingly, f ∈ hID(G ◦H). ■ Corollary 3.7. Let G be a connected graph of order n and H be any graph. (i) If γ(G) = 1, then 4 ≤ γhI(G ◦H) ≤ 6. More precisely, (a) γhI(G ◦H) = 4 if γh(G) = 2 or H = K2 or H has an isolated vertex; (b) γhI(G ◦H) = 5 if pndI(H) = 3; and (c) γhI(G ◦H) = 6 if pndI(H) ≥ 4. (ii) In general, 4 ≤ γhI(G ◦H) ≤ ρH(G), where ρH(G) = min{2|S|+(n− |NG(S)|) pndI(H) : S ∈ HD(G)}, and this bound is tight. S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2444 Proof : In any case γhI(G ◦ H) ≥ 4 by Proposition 2.4. Suppose that γ(G) = 1, and let u ∈ V (G) for which NG[u] = V (G). Pick vu ∈ V (Hu) and w ∈ V (G) \ {u}. Put S = {u, vu, w}, and define V2 = S, V1 = ∅ and V0 = V (G ◦ H) \ S. By Theorem 3.6, f = (V0, V1, V2) ∈ hID(G ◦H). Thus, γhI(G ◦H) ≤ ωG◦H(f) = 6. Suppose that γh(G) = 2, and let S be a γh-set of G. Necessarily, u ∈ S. Put S = {u, v}, where dG(x, v) = 2 for all x ∈ V (G) \ {u}. Define V2 = S, V1 = ∅ and V0 = (V (G) \ S) ∪( ∪x∈V (G)V (Hx) ) . Since V (Hu) ∪ (V (G) \ S) ⊆ NG◦H(v, 2) and ∪x∈V (G)\{u}V (Hx) ⊆ NG◦H(u, 2), f ∈ hID(G ◦H). Thus, γhI(G ◦H) ≤ ωG◦H(f) = 4. Suppose that H has an isolated vertex v. Put S = {u, vu}, where vu is the copy of vertex v inHu. Define V2 = S, V1 = ∅ and V0 = (V (G) \ {u})∪ (( ∪x∈V (G)V (Hx) ) \ {vu} ) . Then (V (G) \ {u})∪(V (Hu) \ {vu}) ⊆ NG◦H(vu, 2) and ∪x∈V (G)\{u}V (Hx) ⊆ NG◦H(u, 2). Thus, f ∈ hID(G ◦H) so that γhI(G ◦H) ≤ ωG◦H(f) = 4. Suppose that γh(G) ̸= 2 and pndI(H) ≥ 2. Let fu = (V u 0 , V u 1 , V u 2 ) be a pndI -function of Hu. Define V2 = {u}∪V u 2 , V1 = V u 1 and V0 = (V (G) \ {u})∪ ( ∪x∈V (G)\{u}V (Hx) ) ∪V u 0 . Then f = (V0, V1, V2) ∈ hID(G) with ωG◦H(f) = 2 + (|V u 1 |+ 2|V u 2 |) = 2 + pndI(H). If pndI(H) = 2 (i.e., H = K2), then γhI(G ◦ H) = 4. If pndI(H) = 3, then the preceding result implies that γhI(G ◦H) = 5; and by a similar reason, if pndI(H) ≥ 4, then γhI(G ◦ H) = 6. To prove (ii), let S ⊆ V (G) be a hop dominating set of G. For each v ∈ V (G)\NG(S), let fv = (V v 0 , V v 1 , V v 2 ) be a pndI -function of H = Hv. Define the following • V0 = [V (G) \ S] ∪ [ ∪v∈V (G)∩NG(S)V (Hv) ] ∪ [ ∪v∈V (G)\NG(S)V v 0 ] ; • V1 = ∪v∈V (G)\NG(S)V v 1 ; • V2 = S ∪ [ ∪v∈V (G)\NG(S)V v 2 ] . Put f = (V0, V1, V2). Let v ∈ V0 ∩ V (G). Since S is a hop dominating set of G and v ∈ V (G) \ S, there exists u ∈ S ⊆ V2 ∩ V (G) for which dG(u, v) = 2, showing that condition (i)(b) of Theorem 3.6 is satisfied. Let v ∈ V (G) for which V2 ∩ NG(v) = ∅. Since V1 ∩ V (G) = ∅, v ∈ V (G) \ NG(S). Then f |Hv = fv, and therefore f |Hv is a PNDI -function of Hv. By Theorem 3.6, f ∈ hID(G ◦H). Moreover, γhI(G ◦H) ≤ ωG◦H(f) = |V1|+ 2|V2| = ∑ v∈V (G)\NG(S) |V v 1 |+ 2|S|+ ∑ v∈V (G)\NG(S) |V v 2 | = 2|S|+ ∑ v∈V (G)\NG(S) (|V v 1 |+ 2|V v 2 |) = 2|S|+ [n− |NG(S)|] pndI(H). Since S is arbitrary, γhI(G ◦H) ≤ ρH(G). S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2445 Consider G = P4. For any graph H, γhI(G ◦ H) = 4 = ρH(G). This proves the tightness of the bound. ■ It is also worth noting that for graphs G with no isolated vertices, Corollary 3.7 is an improvement of Proposition 3.5 as ρH(G) ≤ 2γ∗t1,2(G). Theorem 3.8. (lexicographic product of graphs) Let G and H be connected graphs and f = (V0, V1, V2) a function on V (G[H]). Let A, B and C be subsets of V (G) and let Ax, Bx and Cx be subsets of V (H) such that V0 = ∪x∈A ({x} ×Ax), V1 = ∪x∈B ({x} ×Bx) and V2 = ∪x∈C ({x} × Cx). Then f ∈ hID(G[H]) if and only if each of the following holds: (i) B ∪ C is a hop dominating set of G; (ii) For each x ∈ A for which C ∩NG(x, 2) = ∅, one of the following holds: (a) |B ∩NG(x, 2)| ≥ 2; (b) B ∩NG(x, 2) = {w} such that |Bw| ≥ 2; (c) x ∈ B ∪ C, B ∩NG(x, 2) = {w} with |Bw| = 1 and Bx ∪ Cx is a PND-set of H; (d) x ∈ B ∪ C, B ∩NG(x, 2) = ∅ and the restriction f |⟨{x}×V (H)⟩ of f on ⟨{x} × V (H)⟩ is a PNDI-function of ⟨{x} × V (H)⟩. Proof : Assume that f ∈ hID(G[H]). Then V1 ∪ V2 is a hop dominating set of G[H]. This implies that B ∪ C is a hop dominating set of G, and (i) holds. Next, to prove (ii), let x ∈ A for which C ∩NG(x, 2) = ∅. We consider the following cases: Case 1: Suppose that B ∩NG(x, 2) = ∅. Since B ∪C hop-dominates A, x ∈ B ∪C. Put Tx = ⟨{x}×V (H)⟩. Let y ∈ Ax. Then |V2∩NG[H]((x, y), 2)| ≥ 1 or |V1∩NG[H]((x, y), 2)| ≥ 2. If (u, v) ∈ V2 ∩ NG[H]((x, y), 2), then x = u so that (u, v) ∈ V x 2 = V2 ∩ V (Tx) and (x, y)(u, v) /∈ E(Tx). On the other hand, if (u, v), (w, z) ∈ V1 ∩ NG[H]((x, y), 2), then u = w = x so that (u, v), (w, z) ∈ V x 1 = V1 ∩ V (Tx) and (x, y)(u, v), (x, y)(w, z) /∈ E(Tx). Since y is arbitrary, f |Tx = (V x 0 , V x 1 , V x 2 ) is a PNDI -function of Tx, where V x 0 = V0∩V (Tx). This proves (ii)(d). Case 2: Suppose that B ∩ NG(x, 2) ̸= ∅. If |B ∩ NG(x, 2)| ≥ 2, then (ii)(a) is done. Note that such holds particularly when x /∈ B ∪ C, y ∈ Ax and we have (u, v), (w, z) ∈ V1 ∩NG[H]((x, y), 2) with u ̸= w. Assume that |B ∩ NG(x, 2)| = 1, say B ∩ NG(x, 2) = {w}. If |Bw| ≥ 2, then (ii)(b) holds. Note that this readily follows if x /∈ B ∪ C. Now suppose that |Bw| = 1. Since C ∩ NG(x, 2) = ∅, necessarily x ∈ B ∪ C. We claim that Bx ∪ Cx is a PND-set of H. Let y ∈ V (H) \ (Bx ∪ Cx) = Ax \ (Bx ∪ Cx). Then (x, y) ∈ V0 \ (V1 ∪ V2). There exists (a, b) ∈ V2 ∩NG[H]((x, y), 2) or there exist distinct (a, b), (s, t) ∈ V1 ∩NG[H]((x, y), 2). The former implies that b ∈ Cx and by /∈ E(H). Since |Bw| = 1, the latter implies that a ∈ Bx and by /∈ E(H) or s ∈ Bx and ty /∈ E(H). Accordingly, Bx ∪Cx is a PND-set of H. This proves (ii)(d). S.R. Jr. Canoy, F.P. Jamil and S.M. Menchavez / Eur. J. Pure Appl. Math, 16 (4) (2023), 2431-2449 2446 Conversely, suppose that conditions (i) and (ii) all hold for f . Let (x, y) ∈ V0. Then x ∈ A. If u ∈ C∩NG(x, 2), then for any v ∈ Cu, (u, v) ∈ V2∩NG[H]((x, y), 2). Now assume that C ∩NG(x, 2) = ∅. It is straightforward to show that if (ii)(a) or (ii)(b) holds for x, then |V1 ∩NG[H]((x, y), 2)| ≥ 2. Suppose that (ii)(c) holds for x. Let B ∩NG(x, 2) = {w} and let z ∈ Bw. If t ∈ Cx for which ty /∈ E(H), then (x, t) ∈ V2 ∩NG[H]((x, y), 2). On the other hand, if t ∈ Bx for which ty /∈ E(H), then (x, t) and (w, z) are distinct vertices in V1∩NG[H]((x, y), 2). Finally, suppose that (ii)(d) holds for x. Put Tx = ⟨{x}×V (H)⟩ and define V x i = Vi ∩ V (Tx) for i = 0, 1, 2. Then f |Tx = (V x 0 , V x 1 , V x 2 ). Since (x, y) ∈ V x 0 and f |Tx is a PNDI -function of Tx, there exists (x, v) ∈ V x 2 with (x, y)(x, v) /∈ E(Tx) or there exist distinct (x, v), (x, z) ∈ V x 1 such that (x, y)(x, v), (x, y)(x, z) /∈ E(Tx). The former implies that (x, v) ∈ V2 ∩ NG[H]((x, y), 2), while the latter implies that (x, v), (x, z) ∈ V1 ∩NG[H]((x, y), 2). Therefore, f ∈ hID(G[H]). ■ Corollary 3.9. Let G and H be nontrivial connected graphs where H is noncomplete. Then γhI(G[H]) ≤ min{2|S ∩NG(S, 2)|+ pndI(H)|S \NG(S, 2)| : S ∈ HD(G)}, and this bound is sharp. Proof : Put αH(G) = min{2|S ∩ NG(S, 2)| + pndI(H)|S \ NG(S, 2)| : S ∈ HD(G)}. Let S ⊆ V (G) be a hop dominating set of G. For each x ∈ S \NG(S, 2), let fx = (V x 0 , V x 1 , V x 2 ) be a pndI -function of ⟨{x} × V (H)⟩. By Lemma 2.8, since ⟨{x} × V (H)⟩ is noncomplete, we assume that V x 2 ̸= ∅ for each x ∈ S \NG(S, 2). Pick y ∈ V (H). Define the following sets: • V2 = [ ∪x∈S∩NG(S,2){(x, y)} ] ∪ [ ∪x∈S\NG(S,2)V x 2 ] ; • V1 = ∪x∈S\NG(S,2)V x 1 ; and • V0 = V (G[H]) \ (V1 ∪ V2). Let f = (V0, V1, V2). As in Theorem 3.8, write V0 = ∪x∈A ({x} ×Ax), V1 = ∪x∈B ({x} ×Bx) and V2 = ∪x∈C ({x} × Cx). Since V x 2 ̸= ∅ for each x ∈ S \ NG(S, 2), C = S and B = S \ NG(S, 2). Thus B ∪ C is a hop dominating set of G. Let x ∈ A with C ∩ NG(x, 2) = ∅. Since C is a hop dominating set of G, x ∈ C \ NG(C, 2) = B. Note that if B ∩ NG(x, 2) ̸= ∅ and u ∈ B ∩ NG(x, 2), then u ∈ C ∩ NG(x, 2), a contra- diction. Thus, B ∩ NG(x, 2) = ∅. Since f |⟨{x}×V (H)⟩ = fx for each x ∈ C \ NG(C, 2), f ∈ hID(G[H]) by Theorem 3.8. Therefore, γhI(G[H]) ≤ 2|V2|+ |V1| = 2|S ∩NG(S, 2)|+ ∑ u∈S\NG(S,2) [2|V u 2 |+ |V u 1 |] = 2|S ∩NG(S, 2)|+ pndI(H)|S \NG(S, 2)|. Since S is arbitrary, γhI(G[H]) ≤ αH(G). To show the sharpness of the upperbound, consider the graph G in Figure 6. Verify REFERENCES 2447 • • • • • • • • • • • • • • • ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... ........................................................................................................................... .................................... ........................................................................................................................... ................................................................................................................................ .................................... .................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ...... .................................... ........................................................................................................................... .................................... ........................................................................................................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... ........... ...... .................................... .................................... ................................................................................................................................ .................................... ........................................................................................................................... .................................... x y z G Figure 6: Graph G showing sharpness of the bound in Corollary 3.9 that for n ≥ 3, γhI(G[Pn]) = 7. Note that pndI(Pn) = 3 while the set S = {x, y, z} is a hop dominating set of G with |S ∩ NG(S, 2)| = 2 and |S \ NG(S, 2)| = 1. In this case, αH(G) = 2(2) + pndI(Pn)(1) = 7. ■ Strict inequality in Corollary 3.9 can also be attained. Note that for n ≥ 3, γhI(C5[Pn]) = 5 while αPn(C5) = 6. The same example also shows that αH(G) need not be determined by a γh-set S of G. If C5 = [x1, x2, x3, x4, x5, x1], then S = {x1, x2, x4} is a hop dominating set but not a γh-set of C5. However, αPn(C5) = 2|S∩NC5(S, 2)|+pndI(Pn)|S\NC5(S, 2)| = 6. 4. Conclusion It turned out that the hop Italian domination is directly related to both the hop Roman domination and the 2-hop domination. More precisely, γhI(G) ≤ min{γhR(G), γ2h(G)} for all graphs G. More interestingly, it is shown that, in fact, the difference γhR(G)− γhI(G) can be made arbitrary large, and that any pair of positive integers a and b with 4 ≤ a ≤ b are realizable as the hop Italian domination number and the 2-hop domination number, respectively, of some connected graph. Finally, for graphs under the complementary prism, join, corona and lexicographic product of graphs, the hop Italian domination number is expressible in terms of the hop Italian domination numbers or of the pndI numbers of its factors. Acknowledgements This project is fully supported by the MSU-Iligan Institute of Technology through the Office of the Vice Chancellor for Research and Extension (OVCRE). References [1] R.A. Beeler, T.W. Haynes and S.T. Hedetniemi, Double Roman domination. Discrete Applied Mathematics, 211:23-29, 2016. [2] C. Berge, Theorie des graphes et ses applications, Dunod, Paris, 1958. 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