EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 1, 2023, 44-61 ISSN 1307-5543 – ejpam.com Published by New York Business Global On The Global Distance Roman Domination of Some Graphs Giovannie Entero1,∗, Stephanie Espinola2 1 Department of Mathematics and Statistics, College of Arts and Sciences, University of Southeastern Philippines, Bo. Obrero, Davao City, Davao del Sur, Philippines 2 Department of Mathematics and Statistics, Faculty, College of Arts and Sciences, University of Southeastern Philippines, Bo. Obrero, Davao City, Davao del Sur, Philippines Abstract. Let k ∈ Z+. A k − distance Roman dominating function (kDRDF ) on G = (V,E) is a function f : V → {0, 1, 2} such that for every vertex v with f(v) = 0, there is a vertex u with f(u) = 2 with d(u, v) ≤ k. The function f is a global k − distance Roman dominating function (GkDRDF ) on G if and only if f is a k − distance Roman dominating function (kDRDF ) on G and on its complement G. The weight of the global k − distance Roman dominating function (GkDRDF ) f is the value w(f) = ∑ x∈V f(x). The minimum weight of the global k − distance Roman dominating function (GkDRDF ) on the graph G is called the global k − distance Roman domination number of G and is denoted as γk gR(G). A γk gR(G) − function is the global k − distance Roman dominating function on G with weight γk gR(G). Note that, the global 1 − distance Roman domination number γ1 gR(G) is the usual global Roman domination number γgR(G), that is, γ1 gR(G) = γgR(G). The authors initiated this study. In this paper, the authors obtained and established the following results: preliminary results on global distance Roman domination; the global distance Roman domination onKn,Kn, Pn, and Cn; and, some bounds and characterizations of global distance Roman domination over any graphs. 2020 Mathematics Subject Classifications: 05A18, 05C07, 05C12, 05C22, 05C38, 05C40, 05C62, 05C69, 05C70, 05C76, 05C78, 97K30 Key Words and Phrases: Classical Domination, Roman Domination, Distance Domination, Global Domination, Distance Roman Domination, Global Roman Domination 1. Introduction Mathematics plays a vital role in various fields. One of the important areas in math- ematics is graph theory which is mainly used in structural models. Graph theory is an interesting branch of mathematics when it comes to research. In mathematics and com- puter science, graph theory is the study of graphs which are mathematical structures used ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i1.4478 Email addresses: giovannieentero.ge@gmail.com (G. Entero), saomega@usep.edu.ph (S. Espinola) https://www.ejpam.com 44 © 2023 EJPAM All rights reserved. G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 45 to model pairwise relations between objects. There are several areas in graph theory in which extensive research activities grow fast−one of which is Domination. Domination is a classical and an interesting topic in the theory of graphs as well as one of the most active areas of research in this discipline. The increasing interest in this area is partly explained by the diversity of its applications to both theoretical and real-world problems. Domination comes with one of its most famous variants called Roman Domination−the defense strategy (a.k.a. protection strategy) employed by Emperor Constantine the Great to defend the Roman Empire when it was under a certain attack. It was traced back that, in the 4th century A. D., when the Roman Empire was under attack during the period of Emperor Constantine the Great, he had the requirement that any army or a legion could be sent from its home to defend a neighboring location only if there was a second army which would stay and protect the home [12]. Thus, there are two types of armies − traveling and stationary. The first type of legion was particularly skilled agile combatants who could be promptly deployed to an adjacent province for defending against any potential attack. The latter would behave as a local force permanently located in the given province. In addition, no legion could ever depart a province in order to defend another one if such action leaves the base province unprotected [7]. Translating this strategy into the language of graph theory, each vertex with no army must have a neighboring vertex with a traveling army. Stationary armies then dominate their own vertices and a vertex with two armies is dominated by its stationary army and its open neighborhood is dominated by the traveling army [12]. Cockayne et al. [4] introduced a variant of domination called Roman domination suggested by the recent article in Scientific American by Ian Stewart, entitled “Defend the Roman Empire” as mentioned in the previous paragraph. According to the mentioned authors, the Roman dominating function (RDF ) on the graph G = (V,E) is the function f : V → {0, 1, 2} satisfying the condition that every vertex v for which f(v) = 0 is adjacent to at least one vertex u for which f(u) = 2. Roman domination also comes with various varieties, one of its variants that caught the authors’ attention is Distance Roman Domination initiated by Aram et al. [2]. Aram et al. [2] defined that the k − distance Roman dominating function (kDRDF) on the graph G = (V,E) is the function f : V → {0, 1, 2} satisfying the conditions that for every vertex v for which f(v) = 0, there is a vertex u for which f(u) = 2 and d(u, v) ≤ k, where d(u, v) is the distance from u to v. Additionally, the weight of kDRDF f is the value w(f) = ∑ u∈V f(u) and the minimum weight of the kDRDF on G will be the k − distance Roman domination number and is denoted by γkR(G), where k ∈ Z+. All graphs considered in this paper are all finite, simple, and undirected. Let G = (V,E) be a finite, simple, and undirected graph. The graphG has vertex set V = V (G) and edge set E = E(G). Further, let the order of the graph G be p, that is, |V | = |V (G)| = p and the size be q, that is, |E| = |E(G)| = q. The authors defined the Global Distance Roman Domination on graphs as follows: The function f is a global k − distance Roman dominating function (GkDRDF ) on G if and only if f is a k − distance Roman dominating function (kDRDF ) on G and on its complement G. The weight of the global k − distance Roman dominating function G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 46 (GkDRDF) f is the value w(f) = ∑ x∈V f(x). The minimum weight of the global k − distance Roman dominating function (GkDRDF) on the graph G is called the global k − distance Roman domination number of G and is denoted by γkgR(G). A γkgR(G)−function is the GkDRDF on G with weight γkgR(G). The GkDRDF f : V → {0, 1, 2} can be represented by the ordered partition (V f 0 , V f 1 , V f 2 ) of V induced by f , where f is the given function and V f i = {v ∈ V |f(v) = i and i = 0, 1, 2}. Observe that, there is a one-to-one correspondence between the function f : V → {0, 1, 2} and the ordered partition (V f 0 , V f 1 , V f 2 ) of V induced by f . Hence, we may write f = (V f 0 , V f 1 , V f 2 ). In this representation, its weight can be computed as w(f) = |V f 1 |+ 2|V f 2 |. Note that, the global 1 − distance Roman domination number γ1gR(G) is the usual global Roman domination number γgR(G), that is, γ1gR(G) = γgR(G). It is worth noting that, since we are dealing with simple graphs, the distance of each vertex, say u ∈ V , to itself is zero, that is, d(u, u) = 0 while the distance of two different vertices say u, v ∈ V , coming from different components of graph G is assigned to be ∞, that is, d(u, v) = ∞, where u and v belong to different components of G. 2. Terminologies and Notations To better understand the scope of this study, we will be needing the following defini- tions and some related literature. The distance between vertices u and v in graph G, denoted by d(u, v), is the length of the shortest path from vertex u to vertex v in graph G. The eccentricity of vertex u on graph G is the maximum distance from vertex u to any other vertex, say vertex v, in graph G and is denoted by ecc(u) = max{d(u, v) : v ∈ V (G)}. The radius of graph G is the minimum eccentricity taken over all vertices of graph G and is denoted as rad(G) = min{ecc(u) : u ∈ V (G)} and the diameter of graph G is the maximum eccentricity taken over all vertices of graph G and is denoted as diam(G) = max{ecc(u) : u ∈ V (G)}. [8] The degree of a vertex v of the graph G is the number of edges incident with v in G and is denoted by deg(v). The maximum degree of the graph G, denoted by ∆(G), is the maximum degree for every vertex in G, that is, ∆(G) = max{deg(v) : v ∈ V (G)}. The minimum degree of the graph G, denoted by δ(G), is the minimum degree for every vertex in G, that is, δ(G) = min{deg(v) : v ∈ V (G)}. [3] The neighbourhood (or open neighbourhood) of a vertex v, denoted by N(v), is the set of vertices adjacent to v, that is, N(v) = {x ∈ V (G) : vx ∈ E(G)}. The closed neighbourhood of a vertex v, denoted by N [v], is simply the set {v} ∪N(v). Given a set S of vertices, we define the neighbourhood of S, denoted by N(S), to be the union of the neighbourhoods of the vertices in S. Similarly, the closed neighbourhood of S, denoted by N [S], is defined to be S ∪N(S). [11] Let k ∈ Z+. The k − degree of a vertex v in graph G, denoted as degk,G(v), is defined to be degk,G(v) = |{u ∈ V (G)|u ̸= v and d(u, v) ≤ k}|. Let k ∈ Z+. The maximum k − degree of the graph G, denoted by ∆k(G), is the maximum k − degree taken over all vertices of graph G, that is, ∆k(G) = max{degk,G(v) : v ∈ V (G)}. The minimum G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 47 k − degree of the graph G, denoted by δk(G), is the minimum k − degree taken over all vertices of graph G, that is, δk(G) = min{degk,G(v) : v ∈ V (G)}. [2] The k − neighbourhood (or open k − neighbourhood) of a vertex v in graph G, denoted by Nk,G(v), is the set of vertices (different from v) adjacent to v in G within distance k, that is, Nk,G(v) = {u ∈ V (G) : u ̸= v and d(u, v) ≤ k}, where k ∈ Z+. The closed k − neighbourhood of a vertex v, denoted by Nk,G[v], is simply the set {v} ∪ Nk,G(v). Given a set S of vertices, we define the k − neighbourhood of S, denoted by Nk,G(S), to be the union of the k − neighbourhoods of the vertices in S within distance k with respect to graph G, where k ∈ Z+. Similarly, the closed k − neighbourhood of S, denoted by Nk,G[S], is defined to be S ∪Nk,G(S). [2] Let v ∈ S ⊆ V . Then, u is called a private neighbour of v with respect to S, denoted by u is an S−pn of v, if u ∈ N(v)−N(S−{v}). An S−pn of v is external if it is in V −S. The pn(v, S) = N(v)−N(S−{v}) of all S− pn’s of v is called the private neighbourhood set of v with respect to S. Equivalently, pn(v, S) = {u ∈ V |N(u) ∩ S = {v}}. [4] The complement G of a graph G is the graph with vertex set V (G) such that two vertices are adjacent in G if and only if these vertices are not adjacent in G. This means that both G and its complement G have the same vertices, but G has precisely the edges that G lacks. [8] The join of two graphs G and H, denoted by G + H, is the graph with vertex set V (G + H) = V (G) ∪ V (H) and the edge set E(G + H) = E(G) ∪ E(H) ∪ {uv : u ∈ V (G) and v ∈ V (H)}. [8] Let G and H be graphs of orders p1 and p2, respectively. Then the graph obtained by taking one copy of G of order p1 and p1 copies of H and then connecting the ith vertex of G to every vertex of the ith copy of H (ith means first, second, third and so on) is called corona, denoted by G ◦H. The order and the size of the corona G ◦H are p1 + p1p2 and q1+p1q2+p1p2, respectively, where q1 and q2 are the sizes of graphs G and H, respectively. [5] A cartesian product, denoted by G × H, of two graphs G and H, is the graph with vertex set V (G × H) = V (G) × V (H) and edge set E(G × H) satisfying the following conditions: (u1, v1)(u2, v2) ∈ E(G×H) if and only if either u1 = u2 and v1v2 ∈ E(H) or v1 = v2 and u1u2 ∈ E(G). [3] The lexicographic product of graphs G and H is the graph G[H] with a vertex set V (G[H]) = {(u, v) : u ∈ V (G) and v ∈ V (H)} and with edge set E(G[H]) = {(u, v)(w, x) : uw ∈ E(G) or u = w and vx ∈ E(H)}. [6] 3. Basic Concepts Definition 3.1. [5] A set of vertices S ⊆ V (G) is a dominating set for graph G = (V (G), E(G)) if every vertex not in S is adjacent to at least one vertex in S. The dom- ination number of graph G is the cardinality of any minimum (smallest) dominating set in G and is denoted by γ(G). Definition 3.2. [14] Let k ∈ Z+. A set D ⊆ V (G) is a distance k − dominating set of G if each x ∈ V (G)\D is within distance k from some vertex of D. The minimum G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 48 cardinality taken over all distance k − dominating sets of graph G is called the distance k − domination number of G and is denoted by γk(G). Definition 3.3. [4] A Roman dominating function (RDF ) on a graph G = (V,E) is a function f : V → {0, 1, 2} satisfying the condition that every vertex v for which f(v) = 0 is adjacent to at least one vertex u for which f(u) = 2. The weight of the Roman dominating function (RDF ) f is the value w(f) = ∑ x∈V f(x). The minimum weight of the Roman dominating function of the graph G is called the Roman domination number of G and is denoted as γR(G). Definition 3.4. [9] A dominating set D of G = (V,E) is a global dominating set if D is also a dominating set of the complement G of G. The minimum cardinality taken over all global dominating sets of G is called the global domination number of G and is denoted by γg(G). Definition 3.5. [1] A k − distance Roman dominating function (kDRDF ) on G = (V,E) is a function f : V → {0, 1, 2} such that for every vertex v with f(v) = 0, there is a vertex u with f(u) = 2 with distance of at most k from each other. The weight of the distance Roman dominating function f is the value w(f) = ∑ x∈V f(x). The minimum weight of a k − distance Roman dominating function on the graph G is called the k − distance Roman domination number of G and is denoted as γkR(G). Definition 3.6. [12] A global Roman dominating function (GRDF ) on the graph G = (V,E) is a function f : V → {0, 1, 2} such that f is an RDF for both G and its complement G. The weight of the global Roman dominating function f is the value w(f) = ∑ x∈V f(x). The minimum weight of the global Roman dominating function on the graph G is called the global Roman domination number of G and is denoted as γgR(G). Definition 3.7. Let k ∈ Z+. A k − distance Roman dominating function (kDRDF ) on G = (V,E) is a function f : V → {0, 1, 2} such that for every vertex v with f(v) = 0, there is a vertex u with f(u) = 2 such that d(u, v) ≤ k. The function f is a global k − distance Roman dominating function (GkDRDF ) on G if and only if f is a k − distance Roman dominating function (kDRDF ) on G and on its complement G. Definition 3.8. Let k ∈ Z+. The weight of the global k − distance Roman dominating function (GkDRDF ) f is the value w(f) = ∑ x∈V f(x). The minimum weight of the global k − distance Roman dominating function (GkDRDF ) on the graph G is called the global k − distance Roman domination number of G and is denoted as γkgR(G). A γkgR(G) − function is a GkDRDF with weight γkgR(G). Remark 3.9. [12] For any n − vertex graph G, 2 ≤ γgR(G) ≤ n. Theorem 3.10. [10] Let G be a simple graph of order n. Then γ(G) = n if and only if G ≡ Kn. Proposition 3.11. [15] For any graph G of order n, γ(G) = γk(G) if and only if every vertex in G has degree 0. (Such a graph is denoted as G = Kn, the complement of the complete graph of order n) G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 49 Remark 3.12. [2] Let k ≥ 1 be an integer. For n − vertex graphs, always γkR(G) ≤ n, with equality when G ∼= Kn. Proposition 3.13. [13] (i) For a graph G with p vertices, γg(G) = p if and only if G = Kp or Kp. (ii) γg(Km,n) = 2 for all m,n ≥ 1. (iii) γg(C4) = 2, γg(C5) = 3 and γg(Cn) = {n 3 }, for n ≥ 6. (vi) γg(Pn) = 2 for n = 2, 3 and γg(Pn) = {n 3 } for n ≥ 4. Proposition 3.14. [12] Let G be any graph. Then γg(G) = γgR(G) if and only if G = Kn. 4. Results and Discussions All throughout this paper, we will be using either of the notations fk or f to denote a function. However, for general cases where the distance k ∈ Z+ is explicit, we will be using the notation f to denote a function. Nevertheless, for the cases where the distance k ∈ Z+ must be specified, we will be using the notations fk to refer to a function with respect to such distances. Additionally, for each k ∈ Z+, we may have at least one fk (resp., f) on just a single graph, that is, for a particular distance, we can define several global distance Roman dominating functions over a given graph. 4.1. Preliminary Results on Global Distance Roman Domination Remark 4.1. Let k ∈ Z+. Given the global k − distance Roman dominating function (GkDRDF) fk : V → {0, 1, 2} on graph G = (V,E), for all k ∈ Z+, we have the following facts: (i) function fk can be represented by the ordered partition (V fk 0 , V fk 1 , V fk 2 ) of V induced by fk, where V fk i = {v ∈ V |fk(v) = i and i = 0, 1, 2}; (ii) from (i), there is a one-to-one correspondence between fk : V → {0, 1, 2} and the ordered partition (V fk 0 , V fk 1 , V fk 2 ) of V induced by fk; and, (iii) from (ii), fk can be written as fk = (V fk 0 , V fk 1 , V fk 2 ). Proof. Let k ∈ Z+. Suppose we have the global k − distance Roman dominating function (GkDRDF) fk : V → {0, 1, 2} on graph G = (V,E). • There is nothing to prove in part (i). • For part (ii), we note that fk can be expressed as follows fk = {(uj , f(uj)) : uj ∈ V and f(uj) ∈ {0, 1, 2}}. G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 50 Now, let us partition (ordered partition) the function fk in terms of images of each uj ∈ V treating fk as the set of ordered pairs. So, we can have the cells f0 k , f 1 k , and f2 k in which fk = f0 k ∪ f1 k ∪ f2 k , where f0 k = {(uj , 0) : uj ∈ V }, f1 k = {(uj , 1) : uj ∈ V }, and f2 k = {(uj , 2) : uj ∈ V }. and f i k ∩ f l k = ∅ for all i ̸= l and i, l ∈ {0, 1, 2} which means that f0 k ∩ f1 k ∩ f2 k = ∅. Now, for the ordered partition (V fk 0 , V fk 1 , V fk 2 ) of V induced by fk, where V fk i = {v ∈ V |fk(v) = i and i = 0, 1, 2}, we have V fk 0 = {uj ∈ V : fk(uj) = 0}, V fk 1 = {uj ∈ V : fk(uj) = 1}, and V fk 2 = {uj ∈ V : fk(uj) = 2}. which also means that V fk i ∩ V fk l = ∅ for all i ̸= l and i, l ∈ {0, 1, 2} and V fk 0 ∩ V fk 1 ∩ V fk 2 = ∅. Hence, by matching class f0 k to class V fk 0 , class f1 k to class V fk 1 , and class f2 k to class V fk 2 , we can see that there is a one-to-one correspondence between fk : V → {0, 1, 2} and the ordered partition (V fk 0 , V fk 1 , V fk 2 ) of V induced by fk. This concludes the proof of part (ii). • For part (iii), the proof is straightforward. Therefore, we can say now that Remark 4.1 is true and valid. Remark 4.2. For any graph G = (V,E) of order n, there exists a function f : V → {0, 1, 2} satisfying the conditions that for every vertex v for which f(v) = 0 there exists at least one vertex u for which f(u) = 2 with d(u, v) ≤ k such that f is a k − distance Roman dominating function (kDRDF ) on G and on its complement G, where k ∈ Z+. Then such a function f on G with minimum weight also exists. We call the function f : V → {0, 1, 2} as the global k − distance Roman dominating function (GkDRDF ) on G. Proof. Suppose that G is a graph of order n. Assume that the vertex set of graph G is V (G) = {u1, u2, u3, . . . , un−1, un}. Let f : V (G) → {0, 1, 2} be a function on G and let k ∈ Z+. For all k ∈ Z+, since f = (V f 0 (G), V f 1 (G), V f 2 (G)), we let V f 0 (G) = V f 2 (G) = ∅ and V f 1 (G) = {u1, u2, u3, . . . , un−1, un} = V (G), which means that, f(ui) = 1, where ui ∈ V (G) and i = 1, 2, 3, . . . , n−1, n, for all k ∈ Z+. Hence, by Definition 3.7, the function f is a global k − distance Roman dominating function (GkDRDF ) on G. Furthermore, since the existence of the GkDRDF on any graph is now guaranteed and since the order of graph G is finite, it follows that the existence of the GkDRDF with minimum weight on any graph is also guaranteed. This proves Remark 4.2. Remark 4.3. For all k ∈ Z+, if fk can be written as fk = (V fk 0 , V fk 1 , V fk 2 ), then its weight w(fk) can be computed as w(fk) = |V fk 1 |+ 2|V fk 2 |. G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 51 Proof. Let k ∈ Z+. We suppose that fk can be written as fk = (V fk 0 , V fk 1 , V fk 2 ). Now, for all k ∈ Z+, since V = V fk 0 ∪V fk 1 ∪V fk 2 with |V | = |V fk 0 |+ |V fk 1 |+ |V fk 2 | and since, from Definition 3.8, w(fk) = ∑ uj∈V fk(uj), we have w(fk) = ∑ uj∈V fk(uj) = ∑ uj∈V fk 0 fk(uj) + ∑ uj∈V fk 1 fk(uj) + ∑ uj∈V fk 2 fk(uj) = 0 + (1)(|V fk 1 |) + (2)(|V fk 2 |) = |V fk 1 |+ 2|V fk 2 |. that is, w(fk) = |V fk 1 |+2|V fk 2 |, for all k ∈ Z+. Therefore, for all k ∈ Z+, if fk can be written as fk = (V fk 0 , V fk 1 , V fk 2 ), then its weight w(fk) can be computed as w(fk) = |V fk 1 |+2|V fk 2 |. Remark 4.4. Let k ∈ Z+. For any graph G, γkR(G) ≤ γkgR(G). Proof. Let f be a γkgR(G) − function of G and let k ∈ Z+. Then f is a k − distance Roman dominating function of G. Thus, γkR(G) ≤ γkgR(G) for all k ∈ Z+. Remark 4.5. For any graph G, γkgR(G) = γkgR(G), for all k ∈ Z+. Proof. By saying global, it constitutes the given graph, say graph G, together with its complement G. Thus, for all k ∈ Z+, γkgR(G) accounts γkR(G) and γkR(G) simultaneously and also, since G = G, γkgR(G) accounts γkR(G) and γkR(G) = γkR(G) simultaneously. Therefore, for any graph G, γkgR(G) = γkgR(G), for all k ∈ Z+. 5. The Global Distance Roman Domination on Special Graphs 5.1. The Global Distance Roman Domination on Empty Graph Kn and on Complete Graph Kn Theorem 5.1. Let G ∼= Kn, where Kn is the null graph (empty graph) of order n. Then, for all k ∈ Z+, γkgR(G) = n. Proof. Assume that G ∼= Kn is of order n. Suppose that we have a function fk mapping the vertex set V (G) of graph G to the set {0, 1, 2}, that is, fk : V (G) −→ {0, 1, 2}, for all k ∈ Z+. Let fk be defined by fk(ui) = 1, for all ui ∈ V (G). This mapping will give a weight of w(fk) = n, for all k ∈ Z+ and this is minimum. This can be easily verified. Corollary 5.2. Let G ∼= Kn, where Kn is the complete graph of order n. Then, for any k ∈ Z+, γkgR(G) = n. Proof. The proof follows from Theorem 5.1. G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 52 5.2. The Global Distance Roman Domination on Path Graph Pn Proposition 5.3. Let G ∼= Pn, where Pn is the path graph of order n. For all k ∈ Z+, γkgR(G) =  for n ≤ 4 :{ n, if n = 1, 2, 3 and ∀ k ∈ Z+ ⌊ 4 2k⌋+ 2, if n = 4 and ∀ k ∈ Z+ for n > 4 : 2⌊ n ∆k+1⌋, if n ≡ 0 (mod (∆k + 1)) and 1 ≤ k < rad(G) 2⌊ n ∆k+1⌋+ 1, if n ≡ 1 (mod (∆k + 1)) and 1 ≤ k < rad(G) 2⌊ n ∆k+1⌋+ 2, if n ̸≡ 0, 1 (mod (∆k + 1)) and 1 ≤ k < rad(G) 2, if k ≥ rad(G). 5.3. The Global Distance Roman Domination on Cycle Graph Cn Proposition 5.4. Let G ∼= Cn, where Cn is the cycle graph of order n ≥ 3. For all k ∈ Z+, γkgR(G) =  for n ≤ 5 : n, if n = 3, 4 and ∀ k ∈ Z+ n, if n = 5 and k = 1 2, otherwise for n > 5 : 2n 2k+1 , if n ≡ 0 (mod (2k + 1)) and 1 ≤ k < diam(G) 2⌊ n 2k+1⌋+ 1, if n ≡ 1 (mod (2k + 1)) and 1 ≤ k < diam(G) 2⌊ n 2k+1⌋+ 2, if n ̸≡ 0, 1 (mod (2k + 1)) and 1 ≤ k < diam(G) 2, if k ≥ diam(G). 6. Some Bounds of the Global Distance Roman Domination Lemma 6.1. Let k ∈ Z+. Given any graph G, as k −→ ∞, γkgR(G) is decreasing, that is, γ1gR(G) ≥ γ2gR(G) ≥ γ3gR(G) ≥ · · · ≥ γk−1 gR (G) ≥ γkgR(G) ≥ γk+1 gR (G) ≥ . . . . Proof. Let k ∈ Z+. Note that, if k = 1, then it is clear that γ1gR(G) = γgR(G). Now, given k > k−1, we have ∆k(G) ≥ ∆k−1(G) and it follows that, γkgR(G) ≤ γk−1 gR (G). Hence, in general, given · · · > k + 1 > k > k − 1 > · · · > 3 > 2 > 1, we have · · · > ∆k+1(G) ≥ ∆k(G) ≥ ∆k−1(G) > · · · > ∆3(G) ≥ ∆2(G) ≥ ∆1(G) ≥ and so, we can say that, γ1gR(G) ≥ γ2gR(G) ≥ γ3gR(G) ≥ · · · ≥ γk−1 gR (G) ≥ γkgR(G) ≥ γk+1 gR (G) ≥ . . . . This completes the proof. G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 53 Theorem 6.2. Let k ∈ Z+. For any graph G of order n ≥ 1, γkgR(G) ≤ γgR(G). Proof. Suppose that k ∈ Z+ and let G be any graph of order n ≥ 1. Since γ1gR(G) = γgR(G) and by Lemma 6.1, we have · · · ≤ γkgR(G) ≤ γk−1 gR (G) ≤ · · · ≤ γ3gR(G) ≤ γ2gR(G) ≤ γ1gR(G) = γgR(G), that is, · · · ≤ γkgR(G) ≤ γk−1 gR (G) ≤ · · · ≤ γ3gR(G) ≤ γ2gR(G) ≤ γgR(G), and hence, by transitivity, we have γkgR(G) ≤ γgR(G). So, for all k ∈ Z+, γkgR(G) ≤ γgR(G). This completes the proof. Remark 6.3. The upper bound in Theorem 6.2 is sharp and the strict inequality is attained. Proof. The proof is clear. Theorem 6.4. Let k ∈ Z+. For any graph G = (V (G), E(G)) of order |V (G)| ≥ 2, 2 ≤ γkgR(G) ≤ 2|V (G)|. Proof. Let G = (V (G), E(G)) be any graph of order |V (G)| ≥ 2 and let k ∈ Z+. We let fk to be an arbitrary global k − distance Roman dominating function (GkDRDF ) on graph G, where k ∈ Z+. For all k ∈ Z+, by Remark 4.3, w(fk) = |V fk 1 (G)| + 2|V fk 2 (G)|. Since we are talking about number of elements, |V (G)|, |V fk 0 (G)|, |V fk 1 (G)|, |V fk 2 (G)| ≥ 0 and so as the weight w(fk) of fk. Now, since |V (G)| ≥ 2, it follows that, γkgR(G) ≥ 2 (1) and since γkgR(G) is the minimum weight taken over all GkDRDF on graph G, for all fk on G and for all k ∈ Z+, we have γkgR(G) ≤ w(fk) ≤ 2|V (G)| and thus, by transitivity, we have γkgR(G) ≤ 2|V (G)| (2) Hence, from (1) and (2), for all k ∈ Z+, we have 2 ≤ γkgR(G) ≤ 2|V (G)|. This completes the proof. Theorem 6.5. Let k ∈ Z+. For any graph G of order n, γkgR(G) ≤ n, with the equality when G ∼= Kn,Kn. G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 54 Proof. Let k ∈ Z+ and let G be any graph of order n. For all k ∈ Z+, by Theorem 6.2, we have γkgR(G) ≤ γgR(G) (1) and by Remark 3.9, we have 2 ≤ γgR(G) ≤ n (2) Thus, for all k ∈ Z+, from (1) and (2), we obtain γkgR(G) ≤ γgR(G) ≤ n, and by transitivity, we get γkgR(G) ≤ n. Moreover, the equality γkgR(G) = n, when G ∼= Kn,Kn, is an immediate consequence of Theorem 5.1 and Corollary 5.2, and hence, the upper bound n is sharp. This completes the proof. 7. Some Characterizations of the Global Distance Roman Domination Theorem 7.1. Let G be the class of connected graphs whose complements are also connected. For any graph G ∈ G of order n ≥ 4, γkgR(G) = 2 if and only if k ≥ max{diam(G), diam(G)}, where k ∈ Z+. Proof. Suppose that G denote the class of connected graphs whose complements are also connected. Let G ∈ G be any graph of order n ≥ 4 and let its vertex set be V (G) = {u1, u2, u3, . . . , un−1, un}. Since G ∈ G, with n = |V (G)| ≥ 4, the distances and eccentricities in graph G are finite. Thus, diam(G), diam(G) < ∞. Now, let us consider the following: (=⇒) Assume that γkgR(G) = 2. We let fk : V (G) −→ {0, 1, 2} be a γkgR(G)− function, where k ∈ Z+. This implies that, w(fk) = ∑ ui∈V (G) fk(ui) = γkgR(G) = 2, where i = 1, 2, 3, . . . , n − 1, n. This implies further that, for k ∈ Z+, we may define the γkgR(G)− function fk as fk = (V fk 0 (G), V fk 1 (G), V fk 2 (G)), where there exists a vertex uj ∈ V (G) = V (G) such that V fk 0 (G) = {ui ̸=j ∈ V (G) : i ∈ {1, 2, 3, . . . , n}} = V (G)\{uj}, V fk 1 (G) = ∅, and V fk 2 (G) = {uj}, for all k ∈ Z+. Thus, for all ui ̸=j ∈ V (G) = V (G), where i ∈ {1, 2, 3, . . . , n}, and for all k ∈ Z+, we have fk(ui ̸=j) = 0 and fk(uj) = 2 and this implies further that, for a distance k ∈ Z+, we have d(ui ̸=j , uj) ≤ k (1) which means that the distances of vertices ui ̸=j from vertex uj is at most k. Rewriting (1), we get d(uj , ui ̸=j) ≤ k (2) G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 55 Since we are talking about distances here, we have d(uj , ui ̸=j), k > 0, and so, from (2), with respect to G and G, we have max{d(uj , ui ̸=j) : ui ∈ V (G) where i ∈ {1, 2, 3, . . . , n}} ≤ max{k} which implies that max{d(uj , ui ̸=j) : ui ∈ V (G) where i ∈ {1, 2, 3, . . . , n}} ≤ k (3) From (3), ecc(uj) ≤ k. (4) Since uj is the only vertex in the set partition V fk 2 (G) (note that, V fk 2 (G)=V fk 2 (G)) at distance k ∈ Z+, for all i ̸= j, ecc(ui ̸=j) ≤ ecc(uj) (5) Hence, from (4) and (5), we have ecc(ui ̸=j) ≤ k (6) Thus, for all ui ∈ V (G) = V (G), including vertex uj , from (4) and (6), (ui) ≤ k, that is, with respect to graph G and G, ecc(u1), ecc(u2), ecc(u3), . . . , ecc(un) ≤ k (7) Taking the maximum of both sides of (7), we get max{ecc(u1), ecc(u2), ecc(u3), . . . , ecc(un)} ≤ max{k} which implies that max{ecc(u1), ecc(u2), ecc(u3), . . . , ecc(un)} ≤ k (8) Now, from (8), we know that max{ecc(u1), ecc(u2), ecc(u3), . . . , ecc(un)} = max{diam(G), diam(G)} (9) and hence, from (8) and (9), we have max{diam(G), diam(G)} ≤ k which can be rewritten as k ≥ max{diam(G), diam(G)}. Hence, if γkgR(G) = 2, then k ≥ max{diam(G), diam(G)}, where k ∈ Z+. This proves the forward part of the theorem. (⇐=) Assume that k ≥ max{diam(G), diam(G)}, where k ∈ Z+. Hence, if k = max{diam(G), diam(G)} ∈ Z+, then there exists at least one vertex, say uj ∈ V (G) = V (G), such that Nk,G(uj) = ⋃ ui∈V (G) i∈{1,2,...,n} Nk,G(ui ̸=j) = V (G)\{uj} G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 56 and Nk,G(uj) = ⋃ ui∈V (G) i∈{1,2,...,n} Nk,G(ui ̸=j) = V (G)\{uj}. Thus, for this case, we will define a function fk over G and over G, respectively, as fk = (V fk 0 (G), V fk 1 (G), V fk 2 (G)) and fk = (V fk 0 (G), V fk 1 (G), V fk 2 (G)) where V fk 0 (G) = Nk,G(uj)\{uj} = V (G)\{uj}, V fk 1 (G) = ∅, and V fk 2 (G) = {uj}, and V fk 0 (G) = Nk,G(uj)\{uj} = V (G)\{uj}, V fk 1 (G) = ∅, and V fk 2 (G) = {uj}, which means that, with respect to G and G, for i ̸= j, fk(ui ̸=j) = 0 and fk(uj) = 2. Since all vertices ui ̸=j ∈ V (G) = V (G), for i ∈ {1, 2, . . . , n}, is open neighbours of uj ∈ V (G) = V (G), and, with respect to G and G, since fk(uj) = 2, fk(ui ̸=j) = 0 is permissible for fk to be called as k − distance Roman dominating function for G and for G. So, by Definition 3.7, the function fk is a global k − distance Roman dominating function on graph G with k = max{diam(G), diam(G)} ∈ Z+. Now, using Definition 3.8 to compute the weight of fk, we have w(fk) = ∑ ui∈V (G) i∈{1,2,...,n} fk(ui) = fk(u1) + fk(u2) + · · ·+ fk(uj−1) + fk(uj) + fk(uj+1) + · · ·+ fk(un) = 0 + 0 + · · ·+ 0 + 2 + 0 + · · ·+ 0 = 2. Hence, if k = max{diam(G), diam(G)}, then γkgR(G) = 2 and this follows from Theorem 6.4. Moreover, by just following the same arguments, we may generalize this fact as k ≥ max{diam(G), diam(G)}, where k ∈ Z+, and still get γkgR(G) = 2, for all k ≥ max{diam(G), diam(G)}. Thus, if k ≥ max{diam(G), diam(G)}, then γkgR(G) = 2. This proves the backward part of the theorem. Therefore, given the class G of connected graphs whose complement are also connected, for any graph G ∈ G of order n ≥ 4, γkgR(G) = 2 if and only if k ≥ max{diam(G), diam(G)}, where k ∈ Z+. Theorem 7.2. Let k ∈ Z+. For any graph G = (V (G), E(G)) of order n, γkgR(G) = γ(G) if and only if G ∼= Kn. G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 57 Proof. Let k ∈ Z+ and let G = (V (G), E(G)) be any graph of order n. Suppose that γkgR(G) = γ(G) and let f = (V f 0 (G), V f 1 (G), V f 2 (G)) be a γkgR(G) − function of G, where V f i (G) = {v ∈ V (G) : f(v) = i for i = 0, 1, 2} is the partition of the vertex set V (G) of G induced by the function f and V (G) = V f 0 (G) ∪ V f 1 (G) ∪ V f 2 (G). Note that, |V f 0 (G)|, |V f 1 (G)|, |V f 2 (G)| ≥ 0. Clearly, V f 1 (G) ∪ V f 2 (G) is a dominating set of G. Thus, since γ(G) is the minimum cardinality taken over all dominating sets of G, γ(G) ≤ |V f 1 (G) ∪ V f 2 (G)| = |V f 1 (G)|+ |V f 2 (G)|, and so, we have γ(G) ≤ |V f 1 (G)|+ |V f 2 (G)|. (1) Since f = (V f 0 (G), V f 1 (G), V f 2 (G)) is a γkgR(G) − function of G, γkgR(G) = |V f 1 (G)|+ 2|V f 2 (G)|. (2) Thus, since we assumed that γkgR(G) = γ(G), from (1) and (2), we have |V f 1 (G)| + 2|V f 2 (G)| = γkgR(G) = γ(G) ≤ |V f 1 (G)|+ |V f 2 (G)|, and hence, we have |V f 1 (G)|+ 2|V f 2 (G)| ≤ |V f 1 (G)|+ |V f 2 (G)|. (3) Since |V f 0 (G)|, |V f 1 (G)|, |V f 2 (G)| ≥ 0, by addition and subtraction properties of equal- ity, we have |V f 2 (G)| ≤ 0, that is, |V f 2 (G)| ≤ 0 and since |V f 2 (G)| ≥ 0, we may conclude that |V f 2 (G)| = 0. Thus, V f 0 (G) = ∅, and so, we are forced to have V f 1 (G) = V (G) which implies that |V f 1 (G)| = |V (G)| = n. Since f is a γkgR(G) − function of G, γkgR(G) = |V f 1 (G)| + 2|V f 2 (G)| = n + 0 = n which implies that γ(G) = n for we as- sumed that γkgR(G) = γ(G). Thus, in reference to Theorem 3.10, this shows that G ∼= Kn. Conversely, assume that G ∼= Kn. By Theorem 5.1, γkgR(G) = γkgR(Kn) = n and since G ∼= Kn, it follows that, γ(G) = γ(Kn) = n. Thus, since γkgR(G) = n and γ(G) = n, γkgR(G) = γ(G). This completes the proof. Theorem 7.3. Let k ∈ Z+. For any graph G = (V (G), E(G)) of order n, γkgR(G) = γk(G) if and only if G ∼= Kn. Proof. Let k ∈ Z+ and let G = (V (G), E(G)) be any graph of order n. Assume that γkgR(G) = γk(G) and let f = (V f 0 (G), V f 1 (G), V f 2 (G)) be a γkgR(G) − function of G, where V f i (G) = {v ∈ V (G) : f(v) = i for i = 0, 1, 2} is the partition of the vertex set V (G) of G induced by the function f and V (G) = V f 0 (G) ∪ V f 1 (G) ∪ V f 2 (G). Note that, |V f 0 (G)|, |V f 1 (G)|, |V f 2 (G)| ≥ 0. Clearly, V f 1 (G) ∪ V f 2 (G) is a k − distance dominating set of G, where k ∈ Z+. Hence, since γk(G) is the minimum cardinality taken over all k − distance dominating sets of G, for each k ∈ Z+, γk(G) ≤ |V f 1 (G) ∪ V f 2 (G)| = |V f 1 (G)|+ |V f 2 (G)|, G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 58 that is, we have γk(G) ≤ |V f 1 (G)|+ |V f 2 (G)|. (1) Since f = (V f 0 (G), V f 1 (G), V f 2 (G)) is a γkgR(G) − function of G, γkgR(G) = |V f 1 (G)|+ 2|V f 2 (G)|. (2) Thus, since we assumed that γkgR(G) = γk(G), from (1) and (2), we have |V f 1 (G)| + 2|V f 2 (G)| = γkgR(G) = γk(G) ≤ |V f 1 (G)|+ |V f 2 (G)|, and hence, we have |V f 1 (G)|+ 2|V f 2 (G)| ≤ |V f 1 (G)|+ |V f 2 (G)|. (3) Since |V f 0 (G)|, |V f 1 (G)|, |V f 2 (G)| ≥ 0, by addition and subtraction properties of equality, we have |V f 2 (G)| ≤ 0, that is, |V f 2 (G)| ≤ 0 and since |V f 2 (G)| ≥ 0, we may conclude that |V f 2 (G)| = 0. Hence, V f 0 (G) = ∅, and so, we are forced to have V f 1 (G) = V (G) which implies that |V f 1 (G)| = |V (G)| = n. Since f is a γkgR(G) − function of G, γkgR(G) = |V f 1 (G)| + 2|V f 2 (G)| = n + 0 = n which implies that γk(G) = n for we assumed that γkgR(G) = γk(G). Hence, in reference to Proposition 3.11, this shows that G ∼= Kn. Conversely, suppose that G ∼= Kn. By Theorem 5.1, γkgR(G) = γkgR(Kn) = n and since G ∼= Kn, it follows that, γk(G) = γk(Kn) = n. Thus, since γkgR(G) = n and γk(G) = n, γkgR(G) = γk(G). This completes the proof. Theorem 7.4. Let k ∈ Z+. For any graph G = (V (G), E(G)) of order n, γkgR(G) = γR(G) if and only if G ∼= Kn. Proof. Let k ∈ Z+ and let G = (V (G), E(G)) be any graph of order n. Suppose that γkgR(G) = γR(G). By Remark 3.12, we have γkR(G) ≤ n and γkR(G) = n if and only if G ∼= Kn and this is true for all k ∈ Z+. So, γ1R(G) = γR(G) ≤ n and γ1R(G) = γR(G) = n if and only if G ∼= Kn. Thus, since γkgR(G) = γR(G) and γR(G) = n if and only if G ∼= Kn, it follows that γkgR(G) = n if and only if G ∼= Kn. Hence, by Remark 3.12, γkgR(G) = γR(G) = n if and only if G ∼= Kn and thus, the converse will then follows. This completes the proof. Theorem 7.5. Let k ∈ Z+. For any graph G = (V (G), E(G)) of order n, γkgR(G) = γg(G) if and only if G ∼= Kn or G ∼= Kn. Proof. Let k ∈ Z+ and let G = (V (G), E(G)) be any graph of order n. Suppose that γkgR(G) = γg(G) and let f = (V f 0 (G), V f 1 (G), V f 2 (G)) be a γkgR(G) − function of G, where V f i (G) = {v ∈ V (G) : f(v) = i for i = 0, 1, 2} is the partition of the vertex set V (G) of G induced by the function f and V (G) = V f 0 (G) ∪ V f 1 (G) ∪ V f 2 (G). Note that, |V f 0 (G)|, |V f 1 (G)|, |V f 2 (G)| ≥ 0. Clearly, V f 1 (G) ∪ V f 2 (G) is a global dominating set of G. Thus, since γg(G) is the minimum cardinality taken over all global dominating sets of G, γg(G) ≤ |V f 1 (G) ∪ V f 2 (G)| = |V f 1 (G)|+ |V f 2 (G)|, G. Entero, S. Espinola / Eur. J. Pure Appl. Math, 16 (1) (2023), 44-61 59 and so, we have γg(G) ≤ |V f 1 (G)|+ |V f 2 (G)|. (1) Since f = (V f 0 (G), V f 1 (G), V f 2 (G)) is a γkgR(G) − function of G, γkgR(G) = |V f 1 (G)|+ 2|V f 2 (G)|. (2) Thus, since we assumed that γkgR(G) = γg(G), from (1) and (2), we have |V f 1 (G)| + 2|V f 2 (G)| = γkgR(G) = γg(G) ≤ |V f 1 (G)|+ |V f 2 (G)|, and hence, we have |V f 1 (G)|+ 2|V f 2 (G)| ≤ |V f 1 (G)|+ |V f 2 (G)|. (3) Since |V f 0 (G)|, |V f 1 (G)|, |V f 2 (G)| ≥ 0, by addition and subtraction properties of equality, we have |V f 2 (G)| ≤ 0, that is, |V f 2 (G)| ≤ 0 and since |V f 2 (G)| ≥ 0, we may conclude that |V f 2 (G)| = 0. Hence, V f 0 (G) = ∅, and so, we are forced to have V f 1 (G) = V (G) which implies that |V f 1 (G)| = |V (G)| = n. Since f is a γkgR(G) − function of G, γkgR(G) = |V f 1 (G)| + 2|V f 2 (G)| = n + 0 = n which implies that γg(G) = n for we assumed that γkgR(G) = γg(G). Hence, in reference to Proposition 3.13, G ∼= Kn. Conversely, assume that G ∼= Kn or G ∼= Kn. By Theorem 5.1 and Corollary 5.2, γkgR(G) = γkgR(Kn) = n and γkgR(G) = γkgR(Kn) = n, respectively. Also, by Proposition 3.13, γg(G) = n if and only if G ∼= Kn or G ∼= Kn. Thus, the desired result follows. Theorem 7.6. Let k ∈ Z+. For any graph G = (V (G), E(G)) of order n, γkgR(G) = γkR(G) if and only if G ∼= Kn. Proof. Let k ∈ Z+ and let G = (V (G), E(G)) be any graph of order n. Suppose that γkgR(G) = γkR(G). By Remark 3.12, we have γkR(G) ≤ n and γkR(G) = n if and only if G ∼= Kn. Thus, since γkgR(G) = γkR(G) and γkR(G) = n if and only if G ∼= Kn, it follows that γkgR(G) = n if and only if G ∼= Kn. Hence, γkgR(G) = γkR(G) = n if and only if G ∼= Kn and thus, the converse will then follows. This completes the proof. Theorem 7.7. Let k ∈ Z+. For any graph G = (V (G), E(G)) of order n, γkgR(G) = γgR(G) if and only if G ∼= Kn or G ∼= Kn. Proof. Let k ∈ Z+ and let G = (V (G), E(G)) be any graph of order n. 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