EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 4, 2023, 2368-2383 ISSN 1307-5543 – ejpam.com Published by New York Business Global Geodetic Roman Dominating Functions in a Graph Rona Jane G. Fortosa1,∗, Sergio R. Canoy, Jr.1,2 1 Department of Mathematics and Statistics, College of Science and Mathematics, 2 Center for Graph Theory, Algebra, and Analysis- PRISM, MSU-Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. Let G be a connected graph. A function f : V (G) → {0, 1, 2} is a geodetic Roman dominating function (or GRDF) if every vertex u for which f(u) = 0 is adjacent to at least one vertex v for which f(v) = 2 and V1 ∪ V2 is a geodetic set in G. The weight of a geodetic Roman dominating function f , denoted by ωgR G (f), is given by ωgR G (f) = ∑ v∈V (G) f(v). The minimum weight of a GRDF on G, denoted by γgR(G), is called the geodetic Roman domination number of G. In this paper, we give some properties of geodetic Roman domination and determine the geodetic Roman domination number of some graphs. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Geodetic set,geodetic dominating set, Roman dominating function, Roman domination number, geodetic Roman dominating function, geodetic Roman domination number 1. Introduction Roman domination was inspired by the strategies for defending the Roman Empire against invaders, as presented by Stewart [22] and ReVelle and Rosing [20]. Motivated by this strategy, Cockayne, Dreyer and Hedetniemi introduced the concept of Roman domination in 2004 [12]. Roman domination in a graph is a well studied concept under the topic of domination. As a protection strategy involving field armies, the Roman domination concept ensures that an unsecured location is made secured by sending an army to the location from an adjacent secured location subject to the constraint that one army must be left behind in the secured location. Other applications of the concept and some of its variations can be found in [1], [2], [3], [4], [5], [10], [12], [15], [16], [17], and [19]. Another variant of domination is the concept geodetic domination which was introduced by Buckley, Harary and Quintas [6]. Geodesics refers to the shortest paths between two vertices in a graph. The concept of geodesics is closely related to the ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i4.4962 Email addresses: ronajane.fortosa@g.msuiit.edu.ph (R.J. Fortosa), sergio.canoy@g.msuiit.edu.ph (S. Canoy) https://www.ejpam.com 2368 © 2023 EJPAM All rights reserved. R. Fortosa, S. Canoy Jr. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2368-2383 2369 notion of distance in a graph. As a matter of fact, the distance between two vertices is defined as the length of the shortest geodesic between them. In simple terms, the concept represents the minimum number of edges that must be traversed to travel from one vertex to another. Geodetic sets and geodetic domination have plenty of applications and researchers continue to investigate various concepts involving them. Some studies on geodetic sets and related concepts can be found in [7], [8], [11], [13], [14], [18], [21], [23] and [24]. In this study, we introduce the concept of geodetic Roman domination, a concept which combines the concepts of geodetic set and Roman domination. Geodetic Roman domination as a protection strategy (involving of field armies) guarantees, in addition to what the Roman domination requires, that every unsecured location lies along a shortest path between two secured locations. 2. Terminologies and Notations Let G be a connected graph. For vertices u and v in G, a u-v geodesic is any shortest path in G joining u and v. The length of a u-v geodesic is called the distance dG(u, v) between u and v. For every two vertices u and v of G, the symbol IG[u, v] is used to denote the set consisting of u and v and the vertices lying on any of the u-v geodesics. The set IG(u, v) is the set IG[u, v] \ {u, v}. The geodetic closure of a subset S of G is the set IG[S] = ∪u,v∈SIG[u, v]. Also, IG(S) = ∪u,v∈SIG(u, v). The open neighborhood of u ∈ V (G) is given by NG(u) = {v ∈ V (G) : vu ∈ E(G). The closed neighborhood of u is the set NG[u] = NG(u) ∪ {u}. If X ⊆ V (G), the open neighborhood of X is the set NG(X) = ∪u∈XNG(u). The closed neighborhood of X is the set NG[X] = NG(X)∪X. The degree of a vertex v in G is given by degG(v) = |NG(u)|. A vertex of a connected graph G is an extreme or simplicial vertex if its open neighborhood induces a complete subgraph of G. The set of extreme vertices of G is denoted by Ext(G). A set S ⊆ V (G) is said to be a dominating set of a graph G if for every vertex v ∈ V (G) \ S there exists an element of w ∈ S such that vw ∈ E(G), i.e., N [S] = V (G). The smallest cardinality of a dominating set in G is called the domination number of G and is denoted by γ(G). Any dominating set in G with cardinality γ(G) is called a γ-set in G. A set S of vertices in a graph G is a geodetic set if IG[S] = V (G). The minimum cardinality of a geodetic set in G, denoted by g(G), is the geodetic number of G. A set S ⊆ V (G) is called a geodetic dominating set if S is both a geodetic and a dominating set. The minimum cardinality of a geodetic dominating set in G, denoted by γg(G), is the geodetic domination number of G. Any geodetic dominating set in G with cardinality γg(G) is called a γg-set in G. A set S ⊆ V (G) of a graph G is 2-path closure absorbing if for each x ∈ V (G) \ S there exist u, v ∈ S such that dG(u, v) = 2 and x ∈ IG(u, v). The minimum cardinality of a 2-path closure absorbing set in G is denoted by ρ2(G). Any 2-path closure absorbing set in G with cardinality ρ2(G) is called a ρ2-set. A function f : V (G) → {0, 1, 2} is a Roman dominating function (or just RDF) if R. Fortosa, S. Canoy Jr. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2368-2383 2370 every vertex u for which f(u) = 0 is adjacent to at least one vertex v for which f(v) = 2. The weight of an RDF f is given by ωG(f) = ∑ v∈V (G) f(v). The Roman domination number of a graph G, denoted by γR(G), is the minimum weight of an RDF on G. Any RDF f on G with ωG(f) = γR(G) is called a γR-function. If f is an RDF on G and Vi = {v ∈ V (G) : f(v) = i} for i ∈ {0, 1, 2}, then we denote f by f = (V0, V1, V2). In this case, ωG(f) = |V1|+ 2|V2|. A Roman dominating function f = (V0, V1, V2) on G is a geodetic Roman dominating function (or GRDF) if V1 ∪ V2 is a geodetic set in G. The weight of a geodetic Roman dominating function f = (V0, V1, V2) on G is given by ωgR G (f) = |V1|+2|V2|. The minimum weight of a GRDF on G, denoted by γgR(G), is called the geodetic Roman domination number of G. Any GRDF f on G with ωgR G (f) = γgR(G) is called a γgR-function. The join of two graphs G and H, denoted by G + H, is the graph with V (G+H) = V (G)∪V (H) and E(G+H) = E(G)∪E(H)∪{uv : u ∈ V (G) and v ∈ V (G)}, where “∪” refers to a disjoint union of sets. 3. Known Results We state some results that will be needed in this study. Remark 1. [9] Every geodetic set in a graph contains the extreme vertices. Remark 2. [11] Let n be a positive integer. Then (i) γg(Cn) = ⌈n3 ⌉ (ii) γg(Pn) = ⌈n+2 3 ⌉. Theorem 1. [13] Let G be a connected graph of order n ≥ 2. Then the following hold: (i) γg(G) = 2 if and only if there exists a geodetic set S = {u, v} of G such that d(u, v) ≤ 3. (ii) γg(G) = n if and only if G is the complete graph on n vertices. (iii) γg(G) = n−1 if and only if there exists a vertex v in G such that V (G)\{v} ⊆ NG(v) and G \ v is the union of at least two complete graphs. Remark 3. [24] Every 2-path closure absorbing set in a connected graph G is a dominating set in G. 4. Results Proposition 1. Let G be a graph of order n and let f = (V0, V1, V2) be a γgR-function. Then each of the following statements holds: (i) V1 ∪ V2 contains all the extreme vertices of G. R. Fortosa, S. Canoy Jr. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2368-2383 2371 (ii) |V0| = 0 if and only if |V2| = 0. (iii) If |V0| = 0, then γgR(G) = n. (iv) If |V1| = 0, then V2 is γg-set of G and γgR(G) = 2γg(G). Proof. (i) By Remark 1, V1 ∪ V2 contains all the extreme vertices of G. (ii) Suppose |V0| = 0. Suppose further that |V2| ≠ 0. Define g = (∅, V (G),∅). Then ωgR G (g) = n = |V1| + |V2| < |V1| + 2|V2| = γgR(G), a contradiction. Thus, |V2| = 0. The converse is clear. (iii) Suppose |V0| = 0. Then |V2| = 0 by (ii). Hence, γgR(G) = |V1| = n. (iv) Suppose |V1| = 0. Then V2 is a geodetic dominating set of G. Suppose V2 is not a γg-set. Let D be a γg-set of G. Then |D| < |V2|. Define h = (V (G)\D,∅, D). Then h is a GRDF on G. Hence, ωgR G (h) = 2|D| < 2|V2| = γgR(G), a contradiction. Thus V2 is a γg-set in G and γgR(G) = 2|V2| = 2γg(G). Proposition 2. For any graph G, 1 ≤ γg(G) ≤ γgR(G) ≤ min{n, 2γg(G)}. Proof. Let f = (V0, V1, V2) be a γgR-function. Then V1 ∪ V2 is a geodetic dominating set of G. Hence, 1 ≤ γg(G) ≤ |V1| + |V2| ≤ |V1| + 2|V2| = γgR(G). Now, let V ′ 0 = V ′ 2 = ∅ and V ′ 1 = V (G). Then g = (V ′ 0 , V ′ 1 , V ′ 2) is a GRDF on G and γgR(G) ≤ |V ′ 1 | = n. Finally, let S be a γg-set of G. Define h = (V ′′ 0 , V ′′ 1 , V ′′ 2 ) by setting V ′′ 2 = S, V ′′ 0 = V (G) \ S, and V ′′ 1 = ∅. Then h is a GRDF on G and γgR(G) ≤ ωgR G (G) = 2|V ′′ 2 | = 2γg(G). Therefore, 1 ≤ γg(G) ≤ γgR(G) ≤ min{n, 2γg(G)}. Theorem 2. Let G be any graph of order n. Then each of the following statements holds. (i) γgR(G) = 1 if and only if G = K1. (ii) γgR(G) = 2 if and only if G = K2 or G = K2. (iii) γgR(G) = 3 if and only if G ∈ {K3,K3,K1 ∪K2} or G = K2+H for some graph H of order n− 2. Proof. Let f = (V0, V1, V2) be a γgR-function on G. (i) Assume that γgR(G) = 1. Then |V1| = 1 and |V0| = 0. Hence G = K1. The converse is clear. (ii) Suppose γgR(G) = 2. Then ωgR G (f) = |V1| + 2|V2| = 2. Suppose |V2| = 1. Then |V1| = 0 and ∅ ̸= V0 = V (G) \ V2 ⊆ NG(V2). This implies that V2 is not a geodetic set in G, a contradiction. Hence, |V2| = 0 and |V1| = n. It follows that G = K2 or G = K2. R. Fortosa, S. Canoy Jr. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2368-2383 2372 Conversely, if G = K2 or G = K2, γgR(G)=2. (iii) Suppose γgR(G) = 3. Then |V1|+ 2|V2| = 3. Hence |V2| ≤ 1. Consider the following cases: Case 1: |V2| = 0 Then |V0| = 0 and |V1| = n = 3. This implies that G ∈ {P3,K3,K3,K1 ∪K2}. Case 2: |V2| = 1 Then |V1| = 1. Let V1 = {x} and V2 = {y}. Then V0 = V (G) \ {x, y} ⊆ NG(y). Since V1 ∪ V2 is a geodetic set, xy /∈ E(G). Let w ∈ NG(x). Then w ∈ V0. This implies that dG(x, y) = 2. Since V0 ⊆ IG(x, y), V0 = NG(x) ∩ NG(y). Let H = ⟨V0⟩. Then G = ⟨{x, y}⟩+H (isomorphic to K2 +H). For the converse, suppose that G ∈ {K3,K3,K1 ∪ K2}. Then γgR(G) = 3. Next, suppose that G = K2 + H for some graph H. Let V (K2) = {p, q} and let V0 = V (H), V1 = {p}, and V2 = {q}. Then g = (V0, V1, V2) is a GRDF on G. It follows that γgR(G) ≤ ωgR G (g) = 3. By (i) and (ii), γgR(G) = 3. Lemma 1. Let G be a graph of order n. Then γg(G) = n if and only if every component of G is complete. Proof. Suppose γg(G) = n. If G is connected, then G is complete by Theorem 1(ii). Suppose G is disconnected with components G1, G2, . . . , Gk. Suppose G has a component Gj that is not complete. Then γg(Gj) < |V (Gj)| by Theorem 1(ii). Hence, γg(G) = ∑k i=1 γg(Gi) < n, a contradiction. Thus, every component of G is complete. The converse is clear. Theorem 3. Let G be a graph of order n. Then γg(G) = γgR(G) if and only if every component of G is complete. Proof. Suppose γg(G) = γgR(G). Let f = (V0, V1, V2) be a γgR-function on G. Then γg(G) ≤ |V1| + |V2| ≤ |V1| + 2|V2| = γgR(G). Since γg(G) = γgR(G), it follows that γg(G) = |V1| + |V2| = |V1| + 2|V2|. Consequently, |V2| = 0, |V0| = 0, and |V1| = |V (G)|. Thus, γg(G) = n. By Lemma 1, every component of G is complete. For the converse, suppose that every component of G is complete. Then γg(G) = n by Lemma 1. Thus, γgR(G) = n by Proposition 2. Proposition 3. Let n be a positive integer. Then (i) γgR(Cn) =  3, if n = 3 2n 3 , if n ≡ 0(mod 3) 2n+1 3 , if n ≡ 1(mod 3) 2n+2 3 , if n ≡ 2(mod 3) R. Fortosa, S. Canoy Jr. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2368-2383 2373 (ii) γgR(Pn) =  1, if n = 1 2n+3 3 , if n ≡ 0(mod 3) 2n+4 3 , if n ≡ 1(mod 3) 2n+2 3 , if n ≡ 2(mod 3) Proof. (i) Clearly, γgR(C3) = 3. Let n ≥ 4 and let Cn = [v1, v2, . . . , vn, v1]. Consider the following cases: Case 1: n ≡ 0(mod 3) Let n = 3r for some positive integer r. Let V1 = ∅, V2 = {v1, v4, v7, . . . , v3r−2}, and V0 = V (Cn) \ V2. Then f = (V0, V1, V2) is a GRDF on Cn. Hence, γgR(Cn) ≤ ωCn(f) = 2|V2| = 2n 3 . Let g = (V ′ 0 , V ′ 1 , V ′ 2) be a γgR-function on Cn. Since γgR(G) ≤ 2n 3 , it follows that V ′ 2 ̸= ∅. Suppose |V ′ 1 | = k. Then k + 2|V ′ 2 | ≤ 2n 3 . This implies that |V ′ 2 | ≤ r − k 2 and |V ′ 0 | = n − (|V ′ 1 | + |V ′ 2 |) ≥ 2r − k 2 . Suppose that k ≥ 1. Then |V ′ 2 | ≤ r − k 2 implies that |V ′ 0 | ≤ 2|V ′ 2 | ≤ 2r − k. This contradicts the fact that |V ′ 0 | ≥ 2r − k 2 > 2r − k. Therefore, k = 0. By Proposition 1(iv) and Remark 2(i), γgR(Cn) = ωgR Cn (g) = 2n 3 . Case 2: n ≡ 1(mod 3) Let n = 3s+1 for some positive integer s. Let V1 = {v3s}, V2 = {v1, v4, v7, . . . , v3s−2}, and V0 = V (G) \ (V1 ∪ V2). Thus f = (V0, V1, V2) is a GRDF in Cn. Hence, γgR(Cn) ≤ ωgR Cn (f) = 2n+1 3 . Suppose g = (V ′ 0 , V ′ 1 , V ′ 2) is a γgR-function on Cn. Since γgR(G) ≤ 2n+1 3 , it follows that V ′ 2 ̸= ∅. Suppose |V ′ 1 | = k. Then k + 2|V ′ 2 | ≤ 2n+1 3 . Thus |V ′ 2 | ≤ s − 1 2(k − 1) and |V ′ 0 | = n − (|V ′ 1 | + |V ′ 2 |) ≥ 2s − 1 2(k − 1). If k = 0, then |V ′ 2 | ≤ s + 1 2 and |V ′ 0 | ≥ 2s + 1 2 . Hence, |V ′ 2 | ≤ s and |V ′ 0 | ≥ 2s + 1. This is not possible. Hence k ≥ 1. Suppose k ≥ 2. Then |V ′ 2 | ≤ s − 1 2(k − 1) implies that |V ′ 0 | ≤ 2s − (k − 1). However, |V ′ 0 | ≥ 2s − 1 2(k − 1) > 2s − (k − 1), a contradiction. Therefore, k = 1 and γgR(Cn) = ωgR Cn (g) = 2n+1 3 . Case 3: n ≡ 2(mod 3) Let n = 3t+2 for some positive integer t. Let V1 = {v3t, v3t+1}, V2 = {v1, v4, v7, . . . , v3t−2}, and V0 = V (G) \ (V1 ∪ V2). Thus f = (V0, V1, V2) is a GRDF in Cn. Hence, γgR(Cn) ≤ ωgR Cn = 2n+2 3 . Let g = (V ′ 0 , V ′ 1 , V ′ 2) be a γgR-function on Cn. Since γgR(G) ≤ 2n+2 3 , it follows that |V ′ 2 | ̸= ∅. Suppose |V ′ 1 | = k. Then k + 2|V ′ 2 | ≤ 2n+2 3 . Thus |V ′ 2 | ≤ t + 2−k 2 and |V ′ 0 | = n − (|V ′ 1 | + |V ′ 2 |) ≥ 2t + 2−k 2 . If k = 0, then |V ′ 2 | ≤ t + 1 and |V ′ 0 | ≥ 2t + 1. Hence, |V ′ 2 | ≤ t and |V ′ 0 | ≥ 2t + 1. If k = 1, then |V ′ 2 | ≤ t + 1 2 and |V ′ 0 | ≥ 2t + 1 2 . Hence, |V ′ 2 | ≤ t and |V ′ 0 | ≥ 2t + 1, which is not possible. Thus, k ̸= 1. Suppose k ≥ 3. Then |V ′ 2 | ≤ t+ 2−k 2 implies that |V ′ 0 | ≤ 2|V ′ 2 | ≤ 2t+2−k. However, |V ′ 0 | ≥ 2t+ 2−k 2 > 2t+2−k, a contradiction. Therefore, k = 0 or k = 2. If k = 0, by Proposition 1(iv) and Remark R. Fortosa, S. Canoy Jr. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2368-2383 2374 2(i), γgR(Cn) = ωgR Cn (g) = 2n+2 3 . If k = 2, then g is of the same type as the function f defined earlier. Hence, γgR(Cn) = ωgR Cn (g) = 2n+2 3 . (ii) Let Pn = [v1, v2, . . . , vn]. Clearly, γgR(P1) = 1. Suppose n ≥ 2. Consider the following cases: Case 1: n ≡ 0(mod 3) Let n = 3r for some positve integer r. Let V2 = {v1, v3, . . . , v3r−2}, V1 = {v3r} and V0 = V (Pn) \ (V1 ∪ V2). Thus f = (V0, V1, V2) is a GRDF on Pn. Hence, γgR(Pn) ≤ ωgR Pn (f) = |V1|+ 2|V2| = 1 + 2(n3 ) = 2n+3 3 . Let g = (V ′ 0 , V ′ 1 , V ′ 2) be a γgR-function. Since γgR(G) ≤ 2n+3 3 , it follows that V ′ 2 ̸= ∅. Suppose |V ′ 1 | = k. Then k + 2|V ′ 2 | ≤ 2n+3 3 . Thus |V ′ 2 | ≤ r − 1 2(k − 1) and |V ′ 0 | = n − (|V ′ 1 | + |V ′ 2 |) ≥ 2r − 1 2(k + 1). Suppose k = 0. Then |V ′ 2 | ≤ r + 1 2 and |V ′ 0 | ≥ 2r − 1 2 . This implies that |V ′ 2 | ≤ r and |V ′ 0 | ≥ 2r. Since |V ′ 1 | = 0, |V ′ 0 | < 2|V ′ 2 | (as v1 ∈ V ′ 2 or vn ∈ V ′ 2 ; hence, at least one of them has only one neighbor in V ′ 0). Thus, |V ′ 2 | ≤ r implies that |V ′ 0 | < 2r. This contradicts the fact that |V ′ 0 | ≥ 2r. Suppose k = 2. Then |V ′ 2 | ≤ r− 1 2 and |V ′ 0 | ≥ 2r− 3 2 . This implies that |V ′ 2 | ≤ r−1 and |V ′ 0 | ≥ 2r−1. This is not possible. Suppose k ≥ 4. Then |V ′ 2 | ≤ r− 1 2(k−1) implies that |V ′ 0 | ≤ 2r−(k−1). However, |V ′ 0 | ≥ 2r − 1 2(k + 1) > 2r − (k − 1), a contradiction. Thus, k = 1 or k = 3. If k = 1, then g is of the same type as the function f defined earlier. Hence γgR(Pn) = 2n+3 3 . If k = 3, then we may consider h = (V ′′ 0 , V ′′ 1 , V ′′ 2 ) where V ′′ 1 = {v1, v2, v3r}, V ′′ 2 = {v4, v7, . . . , v3r−2} and V ′′ 0 = V (Pn) \ (V ′′ 1 ∪ V ′′ 2 ). Hence, h is a GRDF on Pn and ωgR Pn (h) = 2n+3 3 . Case 2: n ≡ 1(mod 3) Let n = 3s + 1 for some positive integer s. Let V1 = ∅ V2 = {v1, v4, v7, . . . , v3s+1}, and V0 = V (Pn) \ (V1 ∪ V2). Thus f = (V0, V1, V2) is a GRDF in Pn. Hence, γgR(Pn) ≤ ωgR Pn = 2(n+2 3 ) = 2n+4 3 . Let g = (V ′ 0 , V ′ 2 , V ′ 2) be a γgR-function on Pn. Since γgR(Pn) ≤ 2n+4 3 , it follows that V ′ 2 ̸= ∅. Suppose that |V ′ 1 | = k. Then k + 2|V ′ 2 | ≤ 2n+4 3 . Thus |V ′ 2 | ≤ s − 1 2(k − 2) and |V ′ 0 | = n− (|V ′ 1 |+ |V ′ 2 |) ≥ 2s− k 2 . Suppose that k = 1. Then |V ′ 2 | ≤ s+ 1 2 and |V ′ 0 | ≥ 2s− 1 2 . This implies that |V ′ 2 | ≤ s and |V ′ 0 | ≥ 2s. Since |V ′ 1 | = 1, |V ′ 0 | < 2|V ′ 2 | (as v1 ∈ V ′ 2 or vn ∈ V ′ 2 ; hence, at least one of them has only one neighbor in V ′ 0). Thus, |V ′ 2 | ≤ s implies that |V ′ 0 | < 2s. This contradicts the fact that |V ′ 0 | ≥ 2s. Suppose that k = 3. Then |V ′ 2 | ≤ s− 1 2 and |V ′ 0 | ≥ 2s− 3 2 . This implies that |V ′ 2 | ≤ s− 1 and |V ′ 0 | ≥ 2s− 1. This is not possible. Suppose k ≥ 5. Then |V ′ 2 | ≤ s − 1 2(k − 2) implies that |V ′ 0 | ≤ 2s − (k − 2). However, |V ′ 0 | ≥ 2s − k 2 > 2s − (k − 2), a contradiction. Thus, k = 0 or k = 2 or k = 4. If k = 0, by Proposition 1(iv) and Remark 2(ii), γgR(Pn) = 2n+4 3 . If k = 2, then we may consider j = (V ′′ 0 , V ′′ 1 , V ′′ 2 ) where V ′′ 1 = {v1, v3s+1}, V ′′ 2 = {v3, v6, v9, . . . , v3s} and V ′′ 0 = V (Pn) \ (V ′′ 1 ∪ V ′′ 2 ). Hence, j is a GRDF on Pn and ωgR Pn (j) = 2n+4 3 . If k = 4, then we may consider l = (V ∗ 0 , V ∗ 1 , V ∗ 2 ) where V ∗ 1 = {v1, v2, v3s, v3s+1}, V ∗ 2 = {v4, v7, . . . , v3s−1} and V ∗ 0 = V (Pn) \ (V ∗ 1 ∪ V ∗ 2 ). Hence, l is a GRDF on Pn and ωgR Pn (l) = 2n+4 3 . Therefore, γgR(Pn) = 2n+4 3 . R. Fortosa, S. Canoy Jr. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2368-2383 2375 Case 3: n ≡ 2(mod 3) Let n = 3t+2 for some positive integer t. Define V1 = {v1, v3t+2}, V2 = {v3, v6, v9, . . . , v3t} and V0 = V (G) \ (V1 ∪ V2). Thus, f = (V0, V1, V2) is a GRDF in Pn. Hence, γgR(Pn) ≤ ωgR G (f) = |V1|+ 2|V2| = 3 + 2(n−2 3 ) = 2n+2 3 . Let g = (V0, V1, V2) be a γgR-function on Pn. Since γgR(G) ≤ 2n+2 3 , it follows that V ′ 2 ̸= ∅. Suppose that |V ′ 1 | = k. Then k + 2|V ′ 2 | ≤ 2n+2 3 . Thus, |V ′ 2 | ≤ t − 1 2(k − 2) and |V ′ 0 | = n− (|V ′ 1 |+ |V ′ 2 |) ≥ 2t− 1 2(k− 2). If k = 0, then |V ′ 2 | ≤ t+1 and |V ′ 0 | ≥ 2t+1. This is not possible. Suppose that k = 1. Then |V ′ 2 | ≤ t + 1 2 and |V ′ 0 | ≥ 2t + 1 2 . This implies that |V ′ 2 | ≤ t and |V ′ 0 | ≥ 2t + 1. This is also not possible. Suppose that k ≥ 3. Then |V ′ 2 | ≤ t− 1 2(k−2) implies that |V ′ 0 | ≤ 2t−(k−2). However, |V ′ 0 | ≥ 2t− 1 2(k−2) > 2t−(k−2), a contradiction. Thus, k = 2 and γgR(Pn) = 2n+2 3 . Theorem 4. Let G1, . . . , Gk (k ≥ 2) be the components of G. Then γgR(G) = k∑ j=1 γgR(Gj). Proof. Let G1, . . . , Gk be the components of G. For each j ∈ {1, 2, . . . , k}, let gj = (V j 0 , V j 1 , V j 2 ) be a γgR-functions of Gj . Let V0 = ∪k j=1V j 0 , V1 = ∪k j=1V j 1 , and V2 = ∪k j=1V j 2 . Then g = (V0, V1, V2) is a GRDF on G. Hence, γgR(G) ≤ ωgR G (g) = |V1|+ 2|V2| = k∑ j=1 γgR(Gj). Next, suppose that f = (V0, V1, V2) is a γgR-function on G. Then fj = (V j 0 , V j 1 , V j 2 ), where V j 0 = V0 ∩ V (Gj), V j 1 = V1 ∩ V (Gj), and V j 2 = V2 ∩ V (Gj), is a GRDF on Gj for each j ∈ {1, 2, . . . k}. Thus, γgR(Gj) ≤ ωgR Gj (fj) for all j ∈ {1, 2, . . . , k}. Hence,∑k j=1 γgR(Gj) ≤ γgR(G). This establishes the desired equality. Proposition 4. Let G be a graph of order n. If γg(G) = n− 1, then γgR(G) = n. Proof. Suppose γg(G) = n − 1. Let f = (V0, V1, V2) be a γgR-function on G. Since V1 ∪ V2 is a geodetic set, V1 ∪ V2 = V (G) or V1 ∪ V2 = V (G) \ {x} for some x ∈ V (G). If V1 ∪ V2 = V (G), then |V0| = 0. Thus, |V2| = 0 and |V1| = n. Hence, γgR(G) = n. Suppose V1 ∪ V2 = V (G) \ {x} for some x ∈ V (G). Then V0 = {x}. Since f is a γgR-function, |V2| = 1. Therefore, γgR(G) = |V1|+ 2|V2| = (n− 2) + 2 = n. Corollary 1. For any positive integer n, γgR(K1,n) = n. The next result follows from Theorem 3, Corollary 1, and Theorem 4. Corollary 2. Let G be a graph of order n. If every component of G is either complete or a star, then γgR(G) = n. R. Fortosa, S. Canoy Jr. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2368-2383 2376 Proposition 5. If G is a graph of order n ≥ 5 and γgR(G) = n, then G has no induced subgraph P5. Proof. Suppose G has an induced subgraph P5 = [v1, v2, v3, v4, v5]. Define V0 = {v2, v4}, V2 = {v3} and V1 = V (G) \ {v2, v3, v4}. Then V1 ∪ V2 is a geodetic dominating set in G and V0 ⊆ NG(v3). This implies that f = (V0, V1, V2) is a GRDF on G. Thus, ωgR G (f) = |V1| + 2|V2| = (n − 3) + 2(1) = n − 1, a contradiction. Therefore, G is P5-free. The converse of Proposition 5 is not true. The cycle C5 has no induced subgraph P5 but γgR(C5) = 4 ̸= 5 by Proposition 3. Proposition 6. Let G be a connected graph such that γg(G) ̸= γgR(G). Then γgR(G) = γg(G) + 1 if and only if one of the following holds: (i) There exists a vertex v in G such that V (G) \ {v} ⊆ NG(v) and G \ v is the union of at least two complete graphs. (ii) There exists a vertex v in G and S ⊆ V (G) such that S ⊆ NG(v) and V (G) \ S is a γg-set in G. Proof. Suppose γg(G) + 1 = γgR(G). Let f = (V0, V1, V2) be a γgR-function. Consider the following cases: Case 1: γg(G) < |V1|+ |V2| Then γg(G) + 1 ≤ |V1| + |V2| ≤ |V1| + 2|V2| ≤ γgR(G). The assumption that γgR(G) = γg(G) + 1 implies that |V2| = 0. By Proposition 1(ii), |V0| = 0 and γgR(G) = n. It follows that γg(G) = n− 1. By Theorem 1(iii), (i) follows. Case 2: γg(G) = |V1|+ |V2| Then γg(G) + 1 = |V1| + |V2| + 1 = |V1| + 2|V2| = γgR(G). Hence, |V2| = 1 and |V1| = γg(G) − 1. This implies that |V0| = |V (G) \ (V1 ∪ V2)| = n − γg(G). Let V2 = {v} and S = V0. Then S ⊆ NG(v). Moreover, V (G) \ S = V1 ∪ V2 is a γg-set because it is a geodetic set and |V1 ∪ V2| = γg(G). Therefore (ii) holds. For the converse, suppose first that (i) holds. Let S = V (G) \ {v}. Let w ∈ S. Since G\v = ⟨S⟩ is the union of at least two complete graphs, the component C ofG\v containing w as a vertex is complete. This implies that S = Ext(G). Now, let C1 and C2 be distinct components of G \ v and let x ∈ V (C1) and y ∈ V (C2). Then v ∈ IG(x, y). Hence, S = Ext(G) is the unique γg-set of G and γg(G) = n − 1. By Proposition 4, we have γgR(G) = n = γg(G) + 1. Next, suppose that (ii) holds. Let V0 = S, V2 = {v} and V1 = V (G) \ (S ∪ {v}). Then V1 ∪ V2 = V (G) \ S is a γg-set of G and V0 ⊆ NG(v). It follows that g = (V0, V1, V2) is a GRDF on G and γgR(G) ≤ ωgR G (g) = |V1|+ 2|V2| = γg(G)− 1 + 2 = γg(G) + 1. Since γg(G) < γgR(G), γg(G) + 1 ≤ γgR(G). Thus, γgR(G) = γg(G) + 1. R. Fortosa, S. Canoy Jr. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2368-2383 2377 Theorem 5. Let G = Kn1,...,nk be the complete k-partite graph with 1 ≤ n1 ≤ n2 . . . ≤ nk and |{nj : nj ̸= 1}| ≥ 2. Then γgR(G) = min{n(G) + 1, 6}, where n(G) = min{nj : nj ≥ 2}. Proof. Let Sn1 , Sn2 , . . . , Snk be the partite sets in G and let n(G) = min{nj : nj ≥ 2}. Suppose n(G) = 2. Then γgR(G) = 3 = n(G) + 1, by Theorem 2(iii). Next, suppose that n(G) ≥ 3. Pick u ∈ Sn. Let V2 = {u}, V0 = V (G) \ Sn(G), and V1 = Sn(G) \ {u}. Then f = (V0, V1, V2) is a GRDF on G. This implies that γgR(G) ≤ ωgR G (f) = (n(G)− 1) + 2 = n(G) + 1. Next, let V ∗ 2 = {x, y}, V ∗ 1 = {w, z}, and V ∗ 0 = V (G) \ (V ∗ 1 ∪ V ∗ 2 ) where x,w ∈ Snr and y, z ∈ Snt where nr ̸= 1 and nt ̸= 1. Then f ′ = (V ∗ 0 , V ∗ 1 , V ∗ 2 ) is a GRDF on G and γgR(G) ≤ ωgR G (f ′) = |V ∗ 1 |+ 2|V ∗ 2 | = 2 + 2(2) = 6. Therefore, γgR(G) ≤ min{n(G) + 1, 6}. Now, let g = (V ′′ 0 , V ′′ 1 , V ′′ 2 ) be a γgR-function on G. Suppose that γgR(G) < n(G) + 1 ≤ 6. Then γgR(G) = ωgR G (g) = |V ′′ 1 | + 2|V ′′ 2 | < n(G) + 1. This implies that |V ′′ 2 | ≤ 2. If |V ′′ 2 | = 0, then |V ′′ 0 | = 0 and |V ′′ 1 | = ∑k i=1 ni ≥ 6, a contradiction. Suppose that |V ′′ 2 | = 1, say V ′′ 2 = {v′′}. We may assume that v′′ ∈ Sn(G). Then Sn(G) \ {v′′} ⊆ V ′′ 1 . This implies that n(G) + 1 = |Sn(G) \ {v′′}|+ 2|V ′′ 2 | ≤ |V ′′ 1 |+ 2|V ′′ 2 | < n(G) + 1, a contradiction. Suppose now that |V ′′ 2 | = 2. Suppose |V ′′ 1 | = 1. Then n(G) = 5. Let V ′′ 2 = {p, q} and V ′′ 1 = {s}. Since V ′′ 1 ∪ V ′′ 2 is a geodetic set, at least two of the vertices p, q, and s belong to the same partite set, say Sni where i ∈ {1, 2, . . . , k}. Choose any z ∈ Sni \ {p, q, s}(such z exists because nj ≥ n(G) = 5). Then z /∈ IG({p, q, s}), a contradiction. Suppose |V ′′ 1 | = 0. Then V ′′ 2 ⊆ Snj for some j ∈ {1, 2, . . . , k}. Let w ∈ Snj \ V ′′ 2 . Then w ∈ V ′′ 0 \ NG(V ′′ 2 ), a contradiction. Hence, γgR(G) ≥ n(G) + 1. The same argument can be used to show that γgR(G) ≥ 6 if 6 ≤ n(G) + 1. Accordingly, γgR(G) = min{n(G) + 1, 6}. Example 1. For any two integers m,n ≥ 2, γgR(Km,n) = min{m+ 1, n+ 1, 6}. The next result shows that every pair of positive integers (both at least 4) are realizable as the geodetic domination number and geodetic Roman domination number of a connected graph. Theorem 6. Let a and b be positive integers such that 4 ≤ a ≤ b ≤ 2a. Then there exists a connected graph G such that γg(G) = a and γgR(G) = b. Proof. Consider the following cases: Case 1. a = b. Let G = Ka. Then γg(G) = γgR(G) = a. R. Fortosa, S. Canoy Jr. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2368-2383 2378 Case 2. a < b. Subcase 1. b = a+ 1. Let G = K1,a. Then γg(G) = a and γgR(G) = a+ 1 = b. Subcase 2. b = 2a− 1. Let m = b− a = a− 1 and let G = P3m. Then γg(P3m+1) = m+ 1 = a by Remark 2(ii), and by Proposition 3(ii), γgR(P3m) = 2m+ 1 = 2a− 1 = b. Subcase 3. b = 2a. Let G = C3a. Then γg(G) = γg(C3a) = ⌈3a3 ⌉ = a by Remark 2(i) and by Proposition 3(i), γgR(G) = 2a. Subcase 4. a+ 2 ≤ b < 2a− 1 Then 2a − b − 1 ≥ 1, i.e., 2a − b ≥ 2. Let m = b − a. Consider the graph G in Figure 1 obtained from P3m−2 = [v1, v2, v3, v4, . . . , v3(m+1)−2] by adding the edges v3(m+1)−2wj for each j ∈ {1, 2, . . . , 2a − b − 1}. Let S1 = {v1, v4, . . . , v3(m+1)−2}. Then S1 is a γg-set in P3(m+1)−2. Hence, S = {v1, v4, . . . , v3(m+1)−2, w1, w2, . . . , w2a−b−1} is a γg-set in G and γg(G) = |S| = (m+1)+2a−b−1 = (b−a+1)+2a−b−1 = a. Suppose 2a−b−1 = 1. Then b = 2a−2, m = a−2. Then G = P3(m+1)−1. By Remark 2(ii), γg(P3(m+1)−1) = m+2 = a and by Proposition 3(ii), γgR(G) = γgR(P3(m+1)−1) = 2(m + 1) = b. Next, suppose that 2a− b− 1 ≥ 2. If S1 ∩ V2 = ∅, then γgR(G) = γgR(P3m+1) + 2a− b− 1 = 2(m+ 1) + a− (m+ 1) = m+ a = b. Suppose S1 ∩ V2 ̸= ∅. Since f is a γgR-function on G, |S1 ∩ V2| = 1 and v3m+1 ∈ V0. Let v3m+2 ∈ S1∩V2. It is routine to show that g = (V ′ 0 , V ′ 1 , V ′ 2) where V ′ 1 = V1 \ (S1 \{v3m+2}), V ′ 2 = V2 and V ′ 0 = V0 is a γgR-function on P3m+2. Then γgR(G) = γgR(P3m+2) + 2a− b− 2 = 2(m+ 1) + a− (m+ 2) = b. G : v1 v2 v3 v4 · · · v3(m+1)−2 w2a−b−1 ... w1 w2 w3 w4 w5 Figure 1: A graph G with γg(G) = a and γgR(G) = b This proves the assertion. Corollary 3. Let n be a positive integer with n ≥ 2. Then there exists a connected graph G such that γgR(G) − γg(G) = n. In other words, the difference γgR(G) − γg(G) can be made arbitrarily large. Proposition 7. Let G and H be non-complete graphs. Then 3 ≤ γgR(G+H) ≤ 6. R. Fortosa, S. Canoy Jr. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2368-2383 2379 Proof. Since G+H /∈ {K1,K2}, γgR(G+H) ≥ 3, by (i) and (ii) of Theorem 2. Pick u, v ∈ V (G) and x, y ∈ V (H) such that uv /∈ E(G) and xy /∈ E(H). Let V1 = {v, y}, V2 = {u, x} and V0 = V (G+H) \ (V1 ∪ V2). Then f = (V0, V1, V2) is a GRDF on G+H. Hence, γgR(G+H) ≤ ωgR G+H(f) = 6. Lemma 2. Let G and H be non-complete graphs and let S = SG∪SH , where SG ⊆ V (G) and SH ⊆ V (H), be a geodetic set in G+H. Then each of the following statements holds. (i) If |SG| ≥ 2 and |SH | ≤ 1, then SG is a 2-path closure absorbing set in G. (ii) If |SH | ≥ 2 and |SG| ≤ 1, then SH is a 2-path closure absorbing set in H. Proof. Suppose |SG| ≥ 2 and |SH | ≤ 1. If SG = V (G), then we are done. Suppose that SG ̸= V (G) and let v ∈ V (G) \SG. Since S is a geodetic set in G+H and |SH | ≤ 1, there exist p, q ∈ SG such that v ∈ IG+H(p, q). This implies that dG(p, q) = 2 and v ∈ IG(p, q). Hence, SG is a 2-path closure absorbing set in G, showing that (i) holds. Similarly, (ii) holds. Theorem 7. Let G and H be non-complete graphs. Then γgR(G+H) = 3 if and only if G ∈ {K2,K2 +G1} or H ∈ {K2,K2 +H1} for some graphs G1 and H1. Proof. Suppose γgR(G+H) = 3. Since G and H are non-complete graphs and G+H is a connected graph, G+H = K2+F for some non-complete graph F by Theorem 2(iii). Let K2 = {a, b}. Then a, b ∈ V (G) or a, b ∈ V (H). We may assume that a, b ∈ V (G). Then G = K2 or G = K2 +G1 where G1 = ⟨V (G) \ {a, b}⟩. Conversely, if G = K2, then γgR(G + H) = 3. If G = K2 + G1 for some graph G1, then G+H = K2+(G1+H). By Theorem 2(iii) , γgR(G+H) = 3. The same conclusion holds when H ∈ {K2,K2 +H1} for some graph H1. Theorem 8. Let G and H be non-complete graphs. Then γgR(G+H) = 3 if and only if ρ2(G) = 2 or ρ2(H) = 2. Proof. Suppose γgR(G+H) = 3. By Theorem 7, ρ2(G) = 2 or ρ2(G) = 2. Conversely, suppose that ρ2(G) = 2 say S = {x, y} is a 2-path closure absorbing set in G. If |V (G)| = 2, then G = K2. Suppose G ̸= K2. Then for all u ∈ V (G) \ {x, y}, dG(x, y) = 2 and u ∈ IG(x, y). This implies that G = {x, y}+G1 for some graph G1. By Theorem 2(iii), γgR(G+H) = 3. Similarly, if ρ2(H) = 2, then γgR(G+H) = 3. Theorem 9. Let G and H be non-complete graphs. Then γgR(G+H) = 4 if and only if one of the following conditions holds: (i) ρ2(H) ̸= 2 and there exists a ρ2-set {x, y, z} in G such that V (G)\{x, y, z} ⊆ NG(x). (ii) ρ2(G) ̸= 2 and there exists a ρ2-set {x, y, z} in H such that V (H)\{x, y, z} ⊆ NH(x). R. Fortosa, S. Canoy Jr. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2368-2383 2380 Proof. Suppose γgR(G + H) = 4. Let f = (V0, V1, V2) be a γgR-function on G + H. Then |V1| + 2|V2| = 4. Suppose |V2| = 0. Then |V1| = |V (G + H)| = 4. Since G and H are non-complete graphs and γgR(G +H) ̸= 3, this is not possible. Suppose |V2| = 2, say V2 = {v, w}. Then V2 ⊆ V (G) or V2 ⊆ V (H), since V2 is a geodetic set of G + H. Suppose that V2 ⊆ V (G). By Lemma 2, V2 is a 2-path closure absorbing set. Hence, ρ2(G) = 2. By Theorem 8, this implies that γgR(G + H) = 3, a contradiction. Hence, |V2| = 1 and |V1| = 2. Assume first that V2 = {x} ⊆ V (G). Let V1 = {y, z}. Since γgR(G +H) ̸= 3, ρ2(G) ̸= 2 and ρ2(H) ̸= 2. Hence, V1 ⊆ V (G). Since f is a GRDF on G+H, V (G) \ {x, y, z} ⊆ NG(x) and {x, y, z} is a ρ2-set in G. This shows that (i) holds. Similarly, (ii) holds if V2 = {x} ⊆ V (H). Conversely, suppose (i) holds. By the preceding result, γgR(G + H) ̸= 3. Thus, γgR(G + H) ≥ 4. Let V2 = {x}, V1 = {y, z} and V0 = V (G + H) \ (V1 ∪ V2). Then f = (V0, V1, V2) is a GRDF on G + H. Hence, γgR(G + H) ≤ ωgR G+H(f) = 4. Therefore, γgR(G+H) = 4. The same conclusion holds if (ii) holds. Theorem 10. Let G and H be non-complete graphs such that γgR(G+H) /∈ {3, 4}. Then γgR(G+H) = 5 if and only if one of the following holds: (i) γ(G) = 1 and ρ2(H) = 3 (ii) γ(H) = 1 and ρ2(G) = 3 (iii) There exists nonadjacent vertices v, w ∈ V (G) and x, y ∈ V (H) such that V (G) \ {v, w} ⊆ NG(v). (iv) There exists nonadjacent vertices v, w ∈ V (G) and x, y ∈ V (H) such that V (H) \ {x, y} ⊆ NH(x). (v) There exist v, w, x, y ∈ V (G) such that V (G) \ {v, w, x, y} ⊆ NG(v). (vi) There exist v, w, x, y ∈ V (H) such that V (H) \ {v, w, x, y} ⊆ NH(v). (vii) There exist v, w, x ∈ V (G) such that V (G) \ {v, w, x} ⊆ NG({v, w}). (viii) There exist v, w, x ∈ V (H) such that V (H) \ {v, w, x} ⊆ NH({v, w}). Proof. Let G and H be non-complete graphs such that γgR(G+H) ̸= {3, 4}. Suppose that γgR(G+H) = 5. Let f = (V0, V1, V2) be a γgR-function onG+H. Then |V1|+2|V2| = 5. Suppose that |V2| = 0. Then |V1| = |V (G + H)| = 5. Since G and H are non-complete graphs and γgR(G+H) ̸= {3, 4}, this is not possible. Suppose |V2| = 1, say V2 = {v}. Then |V1| = 3. Assume that V2 = {v} ⊆ V (G). If V1 ⊆ V (H). Then V (G) \ {v} ⊆ NG(v). This implies that γ(G) = 1. Since V1 is a 2-path closure absoring set in H and γgR(G+H) ̸= 3, V1 is a ρ2-set in H. Hence, ρ2(H) = 3 and (i) holds. Suppose |V1 ∩ V (G)| = 1, say w ∈ v ∩ V (G). Then V (G) \ {v, w} ⊆ NG(v). Since H is non-complete and ρ2(H) ̸= 2, vw /∈ E(G). Since ρ2(G) ̸= 2, xy /∈ E(H). This shows that (iii) holds. Next, suppose that REFERENCES 2381 |V1 ∩ V (G)| ≥ 2. Since γgR(G+H) ̸= 4, it follows that V1 ⊆ V (G), i.e. |V1 ∩ V (G)| = 3. Clearly, V (G) \ {v, w, x, y} ⊆ NG(v), showing that (v) holds. Suppose now that |V2| = 2, say V2 = {v, w}. Then |V1| = 1. Let V1 = {x}. Assume that V2∩V (G) ̸= ∅, say v ∈ V (G). Since ρ2(G) ̸= 2 and ρ2(H) ̸= 2, {v, w, x} ⊆ V (G). Hence {v, w, x} is a ρ2-set in G and V (G) \ {v, w, x} ⊆ NG({v, w}). This shows that (vii) holds. Similarly, (ii) or (iv) or (vi) or (viii) holds. The converse is clear. Conclusion This study introduced the notion of geodetic Roman domination. Some properties of geodetic Roman dominating functions were explored and the geodetic Roman domination numbers of certain graphs were determined. It was also shown that any pair of positive integers (subject to a constraint) are realizable as the geodetic domination number and geodetic Roman domination number of a connected graph. This newly defined variant of Roman domination may be investigated further for other graphs including those ones resulting from some binary operations of graphs. One may also try exploring the relationship between this variant and the other variations of Roman domination. Acknowledgements The authors would like to thank the referees for the invaluable assistance they gave us through their comments and suggestions which led to the improvement of the paper. The authors are also grateful to the Department of Science and Technology - Accelerated Science and Technology Human Resource Development Program (DOST-ASTHRDP)- Philippines and MSU-Iligan Institute of Technology, Philippines for funding this research. References [1] M. Adabi, E. Ebrahimi Targhi, N. Jafari Rad, and M. Saied Moradi. 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