EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 6040 ISSN 1307-5543 – ejpam.com Published by New York Business Global Geodetically Undominated Vertices in a Graph Sergio R. Canoy, Jr.1,2, Jesica M. Anoche1,2,∗ 1 Department of Mathematics and Statistics, College of Science and Mathematics, MSU-Iligan Institute of Technology, 9200 Iligan City, Philippines 2 Center of Mathematical and Theoretical Physical Sciences-PRISM, MSU-Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. Let G = (V (G), E(G)) be a simple undirected graph. If γg(G) is the geodetic dom- ination number of G and S ⊆ V (G) such that |S| < γg(G), then definitely, there is at least one vertex of G that is not geodetically dominated by S, that is, not dominated by any vertex in S or not in any geodesic of any two vertices in S. If k is a positive integer with k ≤ γg(G) − 1 and S ⊆ V (G) with |S| = γg(G)− k, then the number ζgk(S) given by ζgk(S) = |V (G) \Ng G[S]|, where Ng G[S] = NG[S]∩ IG[S], is called the k-geodetic domination defect of S in G. The k-geodetic dom- ination defect of G is denoted and given by ζgk(G) = min{ζgk(S) : S ⊆ V (G) and |S| = γg(G)− k}. In this paper, we study this newly defined parameter for some known classes of graphs. Moreover, we determine some sharp bounds of the parameter. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Geodetic set, domination, geodetic domination, k-geodetic domination defect 1. Introduction Domination is a fundamental concept in graph theory with numerous applications across various fields, including network design, resource allocation, and social network analysis (see [1] and [2]). The domination number γ(G) of a graph G refers to the smallest number of vertices required to dominate all the vertices of G. In other words, it is the minimum cardinality of a set S of vertices such that every vertex in the graph is either in the set S or is adjacent to at least one vertex in S. If a set S of vertices has cardinality strictly less than γ(G), then there will exist vertices in the graph that are not dominated by any vertex in the set S. Recently, Das et al. [3] introduced and studied the notion of k- domination defect of a graph, where k is a positive integer strictly less than the domination ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.6040 Email addresses: sergio.canoy@g.msuiit.edu.ph (S. Canoy, Jr.) jesica.anoche@g.msuiit.edu.ph (J. Anoche) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) S. Canoy, Jr., J. Anoche / Eur. J. Pure Appl. Math, 18 (2) (2025), 6040 2 of 16 number of the graph. In their study, the authors established various bounds on the k- domination defect of a graph, based on its maximum degree, domination number, and other parameters. Subsequent studies on this topic (see [4–6]) focused on characterizing k-domination defect sets and determining the k-domination defect in the join, corona, edge corona, and composition of two graphs. Recently, a new variant of the domination defect was introduced and explored in [7]. Some variations of the standard domination utilize the concept of geodetic set (see [8–12]). The associated parameter, called geodetic number of a graph, was introduced by Harary et al. [13]. Geodetic number and geodetic domination were considered in [14], [15], [16], [17], [18], and [19]. Note that if G is a graph and S is a set of vertices of G with cardinality strictly less than the geodetic domination number γg(G) of G, then there is at least one vertex outside S that is not geodetically dominated, that is, has no neighbor in S or is not in any shortest path joining any two vertices in S. In this paper, we introduce the notion k-geodetic domination defect and study it for some classes of graphs. For the motivation of the study, consider a prison facility with numerous prisoners who need to undergo routine assessments such as behavior evaluations or security checks. The warden needs to ensure that every prisoner is evaluated by a jail guard. This situation can be modeled by constructing a graph, where each vertex represents a prisoner or a jail guard, and an edge between a prisoner and a jail guard indicates that the jail guard is responsible for evaluating the prisoner. Moreover, to ensure visibility and security, it is required that every prisoner must be on a shortest path connecting two jail guards. However, due to a lack of personnel and financial support, the required minimum number of jail guards to do the task may not always be met. Furthermore, it may happen that during the assessment, a designated jail guard may be absent and, subsequently, unable to perform his or her task. As a result, some prisoners may not be evaluated in the manner expected. Determining the number of unevaluated prisoners when a designated team of jail guards does not meet the required minimum number of members could assist the management to act accordingly. This situation led us to introduce the concept of geodetic domination defect in a graph. 2. Terminology and Notation For any two vertices u and v in an undirected connected graph G, the distance dG(u, v) is the length of a shortest path joining u and v. Any u-v path of length dG(u, v) is called a u-v geodesic. The diameter of G, denoted by diam(G), is the maximum distance between any two vertices in G. The distance between two subsets A and B of V (G) is given by dG(A,B) = min{dG(a, b) : a ∈ A and b ∈ B}. The open neighborhood of a vertex u is the set NG(u) consisting of all vertices v which are adjacent to u. The closed neighborhood of u is NG[u] = NG(u) ∪ {u}. For any A ⊆ V (G), NG(A) = ⋃ v∈A NG(v) is called the open neighborhood of A and NG[A] = NG(A) ∪ A is called the closed neighborhood of A. A vertex v of G is isolated if |NG(v)| = 0. The set containing all the isolated vertices of G is denoted by I(G). If C ⊆ V (G), then the induced subgraph ⟨C⟩ is the graph with vertex-set S. Canoy, Jr., J. Anoche / Eur. J. Pure Appl. Math, 18 (2) (2025), 6040 3 of 16 C and uv ∈ E(⟨C⟩) whenever u, v ∈ C and uv ∈ E(G). A set S ⊆ V (G) is a dominating set of G if NG[S] = V (G). The smallest cardinality of a dominating set of G, denoted by γ(G), is called the domination number of G. A dominating set S of G with |S| = γ(G), is called a γ-set of G. For every two vertices u and v in G, the symbol IG [u, v], is the set interval containing u, v and all vertices lying in some u-v geodesic. The geodetic closure of a set S ⊆ V (G), denoted by IG[S], is the union of the intervals IG[u, v], where u, v ∈ S. The set S is a geodetic set in G if IG[S] = V (G). The smallest cardinality among all geodetic sets in G, denoted by g(G), is called the geodetic number of G. A geodetic set of cardinality g(G) is called a g-set of G. A set S ⊆ V (G) is a geodetic dominating set in G if it is both a dominating and a geodetic set. The geodetic domination number γg(G) of G is the minimum cardinality among all geodetic dominating sets in G. Any geodetic dominating set of G with cardinality γg(G) is called a γg-set. Let G be a non-trivial graph of order n and let 1 ≤ k < γg(G). Let S ⊆ V (G) with cardinality |S| = γg(G) − k and let Ng G[S] = NG[S] ∩ IG[S], the set of geodetically dominated set of vertices of G. The set V (G) \Ng G[S] is called the k-geodetic domination defect set of S and the k-geodetic domination defect of S in G is ζgk(S) = |V (G)\Ng G[S]| = n−|Ng G[S]|. The minimum cardinality of a k-geodetic domination defect set in G, denoted by ζgk(G), is called the k-geodetic domination defect of G, i.e., ζgk(G) = min{ζgk(S) : S ⊆ V (G) with |S| = γg(G)− k}. A set S ⊆ V (G) of cardinality γg(G) − k for which |V (G) \ Ng G[S]| = ζgk(G) is called a ζgk -set of G. Thus, 〈 Ng G[S] 〉 , the subgraph induced by Ng G[S], is a subgraph of G with n− ζgk(G) vertices and geodetic domination number γg − k. Consider the graph G in Figure 1. Then R = {a, b, x, y} is a γg-set of G, i.e., γg(G) = 4. If k = 1, then D1 = {a, b, x} is a ζg1 -set of G. Since Ng G[D1] = {a, b, c, d, u, v, x}, it follows that ζg1 (G) = ζg1 (D1) = |V (G)| − |Ng G[D1]| = 8− 7 = 1. The set {x, y, c} is not a ζg1 -set of G because Ng G[{x, y, c}] = {x, y, c, d, u, v}, i.e., ζg1 ({x, y, c}) = 8 − 6 = 2. If k = 2, then D2 = {a, x} is a ζg2 -set of G and Ng G[D2] = {a, x, c, d, u, v}. Hence, ζg2 (G) = 8 − 6 = 2. It is easy to verify that the sets {a, y} and {c, v} are not ζg2 -sets of G. Finally, if k = 3, then any 1-element subset D3 of V (G) is a ζg3 -set of G. Since Ng G[D3] = D3, it follows that ζ g 3 (G) = |V (G)| − |Ng G[D3]| = 8− 1 = 7. ................................................................................................................ ................................................................................................................ ................................................................................................................ .................................... ................................................................................................................ ................................................................................................................ ................................................................................................................ .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ... .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ... .................................... ................................................................................................................................................... .................................... .................................... ............ ........... ........... ........... ........... ........... ........... ........... ........... ........... .................................... .................................... a b x y c u d v Figure 1: Graph G with γg(G) = 4, ζg1 (G) = 1, ζg2 (G) = 2, and ζg3 (G) = 7 S. Canoy, Jr., J. Anoche / Eur. J. Pure Appl. Math, 18 (2) (2025), 6040 4 of 16 3. Results Theorem 1 ([20]). Let n be positive integer. Then each of the following holds. (i) For a complete graph Kn, γg(Kn) = n. (ii) For a star graph K1,n−1, γg(K1,n−1) = n− 1. (iii) For a complete bipartite graph Km,n with m,n ≥ 2, γg(Km,n) = min{m,n, 4}. (iv) For a wheel graph Wn, γg(Wn) = ⌈n−1 2 ⌉, n ≥ 5. (v) For a cycle Cn on n vertices, we have γg(Cn) = ⌈n3 ⌉, n ≥ 6. (vi) For a path Pn on n vertices, γg(Pn) = ⌈n+2 3 ⌉. (vii) For the Petersen graph P , γg(P ) = 4. Remark 1. Let G1, G2, · · · , Gr be the components of a graph G. Then each of the following holds: (i) γg(G) = ∑r j=1 γg(Gj). (ii) If Aj ⊆ V (Gj) for each j ∈ [r] = {1, 2, · · · , r} and A = ∪r j=1Aj, then Ng G[A] = ∪r j=1N g G[Aj ] (a disjoint union). Theorem 2. Let G1, G2, · · · , Gr be the components of graph G and let ζg1 (Gi) be the 1-geodetic domination defect of Gi for each i ∈ [r] = {1, 2, · · · , r}. Then ζg1 (G) = min{ζg1 (Gi) : i ∈ [r]}. Proof. Let γg(Gi) and γg(G) be the geodetic domination numbers of Gi and G, re- spectively. By Remark 1(i), γg(G) = ∑r j=1 γg(Gi). For each i ∈ [r], let Di be a ζg1 -set of Gi. Then |Di| = γg(Gi) − 1 and ζg1 (Gi) = |V (Gi) − Ng G[Di]|. Let j ∈ [r] be such that ζg1 (Gj) = min{ζg1 (Gi) : i ∈ [r]}. Let Si be a γg-set in Gi for each i ∈ [r] and let S = (∪i∈[r]\{j}Si) ∪Dj . Then |S| = ∑ i∈[r]\{j} |Si|+ |Dj | = γg(G)− 1 and, by Remark 1(ii), |Ng G[S]| = |Ng Gj [Dj ]|+ ∑ i∈[r]\{j} |Ng Gi [Si]| = |V (Gj)| − ζg1 (Gj) + ∑ i∈[r]\{j} |V (Gi)| = r∑ i=1 |V (Gi)| − ζg1 (Gj). S. Canoy, Jr., J. Anoche / Eur. J. Pure Appl. Math, 18 (2) (2025), 6040 5 of 16 Thus, in G, ζg1 (S) = |V (G)| − |Ng G[S]| = ζg1 (Gj). We claim that ζg1 (S) is the minimum among all subsets of V (G) with cardinality γg(G) − 1. To this end, suppose there exists Q ⊆ V (G) such that |Q| = γg(G) − 1 and ζg1 (Q) < ζg1 (S). Let Q = Q1 ∪ Q2 ∪ · · · ∪ Qr where Qi ⊆ V (Gi) for each i ∈ [r]. Since |Q| = γg(G)−1, at least one Qt is not a geodetic dominating set of Gt by Remark 1(i). Thus, |Qt| = γg(Gt) − 1 and ζg1 (Qt) ≥ ζg1 (Gt) ≥ ζg1 (Gj). Hence, ζg1 (Q) = |V (G)| − |Ng G[Q]| = r∑ i=1 (|V (Gi)| − |Ng Gi [Qi]|) ≥ |V (Gt)| − |Ng Gt [Qt]| = ζg1 (Qt) ≥ ζg1 (Gj) = ζg1 (S), contrary to the assumption that ζg1 (Q) < ζg1 (S). Therefore, ζ g 1 (G) = ζg1 (S) = ζg1 (Gj). Theorem 3. Let G be a graph with I(G) ̸= ∅ and suppose |I(G)| = r. Then ζgj (G) = j for every j ∈ [r] = {1, 2, · · · , r} and ζgk(G) = r+ζgk−r(G ′) for every k ∈ {r+1, · · · , γg(G)−1}, where G′ = ⟨V (G) \ I(G)⟩. Proof. Let I(G) = {v1, v2, · · · , vr} and let S be a γg-set in G. Then I(G) ⊆ S. Let j ∈ [r]. Then D = S \ {v1, v2, · · · , vj} is a ζgj -set in G and |Ng G[D]| = |Ng G[S]| − |Ng G[{v1, v2, · · · , vj}]| = |V (G)| − j. Hence, ζgj (G) = |V (G)| − (|V (G)| − j) = j. Next, let k ∈ {r+1, · · · , γg(G)−1}. Then S0 = S\I(G) is γg-set inG′ = ⟨V (G) \ I(G)⟩. Hence, γg(G ′) = γg(G)−r. Since k ≤ γg(G)−1, k−r ≤ γg(G)−(r+1) < γg(G)−r. Let S′ be a ζgk−r-set in G′. Then |S′| = (γg(G)−r)−(k−r) = γg(G)−k and ζgk−r(G ′) = |V (G′)|− |Ng G′ [S′]| = (|V (G)| − r) − |Ng G[S ′]|. This implies that |V (G)| − |Ng G[S ′]| = r + ζgk−r(G ′). Therefore, since S′ is also a ζgk -set in G, ζgk(G) = r + ζgk−r(G ′). Theorem 4. If G is a graph of order n ≥ 2 and k ≤ γg(G)− 1, then 1 ≤ ζgk(G) ≤ n− 1. Proof. Let S be a ζgk -set in G. Then |S| = γg(G) − k. Hence, V (G) \Ng G[S] ̸= ∅. It follows that ζgk(G) = |V (G)| − |Ng G[S]| ≥ 1. Also, since |Ng G[S]| ≥ 1, ζgk(G) = |V (G)| − |Ng G[S]| ≤ n− 1. This proves the assertion. Theorem 5. Let G be a non-trivial graph of order n. Then ζg1 (G) = 1 if and only if there exists v ∈ V (G) such that γg(G− v) = γg(G)− 1. Proof. Suppose ζg1 (G) = 1 and let S be a ζg1 -set in G. Then |S| = γg(G) − 1 and ζgk(G) = |V (G)\Ng G[S]| = 1. Let v ∈ V (G)\Ng G[S]. Then Ng G[S] = V (G)\{v}. Therefore, γg(G− v) = γg( 〈 Ng G[S] 〉 ) = γg(G)− 1. Conversely, let v ∈ V (G) such that γg(G−v) = γg(G)−1. Then there exists D ⊆ V (G) with |D| = γg(G)−1 andNg G[D] = V (G)\{v}. This implies that ζg1 (G) = |V (G)\Ng G[D]| = |{v}| = 1. Therefore, ζg1 (G) = 1. S. Canoy, Jr., J. Anoche / Eur. J. Pure Appl. Math, 18 (2) (2025), 6040 6 of 16 Lemma 1. Let G be a non-trivial graph of order n. If k = γg(G)−1, then ζgk(G) = n−1. Proof. Let k = γg(G)− 1 and let S be a ζgk -set of G. Then |S| = 1, say, S = {x} and ζgk(G) = ζgk(S) = n− |Ng G[S]|. Since x ∈ NG[S] and IG[S] = S, it follows that Ng G[S] = S. It follows that ζgk(G) = n− |Ng G[S]| = n− 1. Lemma 2. Let G be a graph and S ⊆ V (G). If every component of ⟨S⟩ is complete, then Ng G[S] = S. Proof. Let H1, H2, · · · , Ht be the components of ⟨S⟩. Then S = ∪t i=1V (Hi). By Remark 1(ii), Ng G[S] = ∪t i=1N g Gi [V (Hi)] = ∪t i=1V (Hi) = S. This proves the assertion. Theorem 6. If Kn is a complete graph on n vertices, where n ≥ 2, and 1 ≤ k ≤ n − 1, then ζgk(Kn) = k. Proof. Let k be a positive integer with k ≤ γg(Kn)− 1. Since γg(Kn) = n, k ≤ n− 1. Let S be a ζgk -set of Kn. Then S is a clique, |S| = n − k and ζgk(G) = n − |Ng G[S]|. Therefore, by Lemma 2, ζgk(G) = n− (n− k) = k. Let G be a graph and let S be a subset of V (G). Then the set I2G(S) is given by I2G(S) = {x ∈ V (G) \ S : x ∈ IG(y, z) for some y, z ∈ S with dG(y, z) = 2}. If G is non-trivial and k is a positive integer with k ≤ γg(G)− 1, then the number λk 2(G) is given by λk 2(G) = max{|I2G(S)| : S ⊆ V (G) and |S| = γg(G)− k}. Remark 2. Let G be a graph and let k be a postive integer with k ≤ γg(G)− 1. If S is a (γg(G)− k)-element subset of V (G) and λk 2(G) = |I2G(S)|, then S need not be a ζgk -set in G. To see this, consider graph G in Figure 2. The set D = {a, b, f} is a γg-set in G. Hence, γg(G) = 3. Let k = 1. Consider S = {b, d} and S′ = {b, f}. Then Ng G[S] = {b, c, d, e} and Ng G[S ′] = {b, c, d, e, f}. Thus, ζg1 (S) = 6 − 4 = 2 > 1 = ζg1 (S ′). This implies that S′ is a ζ1-set in G. Since dG(b, f) = 3, it follows that |I2G(S′)| = 0. It is easy to see that λk 2(G) = |I2G(S)| = 2 where S is not a ζ1-set in G. S. Canoy, Jr., J. Anoche / Eur. J. Pure Appl. Math, 18 (2) (2025), 6040 7 of 16 ................................................................................................................ ......... ........ ........ ........ ........ ........ ........ ........ ........ ... .................................... ................................................................................................................ ......... ........ ........ ........ ........ ........ ........ ........ ........ ... .................................... ............................................................................ ........................................................................ ................................................................................................................ .................................... a b c d e f Figure 2: Graph G with γg(G) = 3, ζg1 (G) = 1, and ζg2 (G) = 5 Theorem 7. Let G be a non-trivial graph and let S ⊆ V (G). Then the following state- ments hold: (i) S ∪ I2G(S) ⊆ Ng G[S]. (ii) If diam(G) = 2, then Ng G[S] = S ∪ I2G(S). (iii) If diam(G) = 2, k is a positive integer with k ≤ γg(G)− 1, and S is a ζgk -set in G, then λk 2(G) = |I2G(S)|. Proof. (i) Note that if x ∈ I2G(S), then x ∈ Ng G[S]. Hence, I2G(S) ⊆ Ng G[S]. Since S ⊆ Ng G[S], it follows that S ∪ I2G(S) ⊆ Ng G[S]. (ii) Suppose diam(G) = 2. Let x ∈ Ng G[S]. Then x ∈ NG[S] ∩ IG[S]. If x ∈ S, then x ∈ S ∪ I2G(S). If x /∈ S, then there exist y, z ∈ S such that x ∈ IG(y, z). Since diam(G) = 2, it follows that dG(y, z) = 2. Thus, x ∈ I2G(S). Therefore, N g G[S] ⊆ S∪I2G(S). With (i), we get the desired equality Ng G[S] = S ∪ I2G(S). (iii) Let S′ be a (γg(G)−k)-element subset of V (G). By part (ii), Ng G[S ′] = S∪I2G(S). It follows that ζgk(S ′) = n − γg(G) + k − |I2G(S′)|. Since S is a ζgk -set in G, ζgk(S) = n − γg(G) + k − |I2G(S)| ≤ ζgk(S ′). This implies that |I2G(S′)| ≤ |I2G(S)|. Since S′ was arbitrarily chosen, we have λk 2(G) = |I2G(S)|. Theorem 8. Let G be a non-trivial graph of order n and let k be a positive integer with k ≤ γg(G)− 1. Then ζgk(G) ≤ n− γg(G) + k − λk 2(G) and equality holds if k = γg(G)− 1 or 1 ≤ diam(G) ≤ 2. Proof. Let k be a positive integer with k ≤ γg(G)−1 and let S be a (γg(G)−k)-element subset of V (G) such that λk 2(G) = |I2G(S)|. By Theorem 7(i), S ∪ I2G(S) ⊆ Ng G[S]. Hence, ζgk(G) ≤ ζgk(S) = n− |Ng G[S]| ≤ n− (|S|+ |I2G(S)|) = n− γg(G) + k − λk 2(G). Suppose k = γg(G)−1. Then ζgk(G) = n−1 by Lemma 1. In this case, λk 2 = 0. Hence, ζgk(G) = n− 1 = n− γg(G) + k − λk 2(G). Suppose 1 ≤ diam(G) ≤ 2 and let S be a ζgk -set of G. If diam(G) = 1, then every component of G is complete and γg(G) = n. Hence, if S′ is any (γg(G)− k)-element set in S. Canoy, Jr., J. Anoche / Eur. J. Pure Appl. Math, 18 (2) (2025), 6040 8 of 16 G, then |I2G(S′)| = 0. It follows that λk 2(G) = 0. Since Ng G[S] = S by Lemma 2, we have ζgk(G) = ζgk(S) = n− (γg(G)− k) = n− γg(G) + k − λk 2(G) = k. Next, suppose that diam(G) = 2. Then Ng G[S] = S ∪ I2G(S) by Theorem 7(ii) and λk 2(G) = |I2G(S)| by Theorem 7(iii) . Thus, ζgk(G) = ζgk(S) = n− γg(G) + k − |I2G(S)| = n− γg(G) + k − λk 2(G). This proves the assertion. Remark 3. Strict inequality in Theorem 8 is possible. To see this, consider graph H = P8 in Figure 3. By Theorem 1(vi), γg(H) = 4. Let P8 = [v1, v2, · · · , v8] and let k = 1. Let S1 = {v1, v4, v7}. Then |S| = 3 and Ng H [S1] = {v1, v2, · · · , v7}. Hence, ζg1 (S1) = 1. It follows that S1 is a ζg1 -set of H. It can be verified that λ1 2(G) = 2. Hence, ζg1 (H) = ζg1 (S1) = 1 < 3 = 8− γg(H) + 1− λ1 2(G). If k = 2, then S2 = {v1, v4} is a ζg2 -set of H and Ng H [S2] = {v1, v2, v3, v4}. Also, λ2 2(H) = 1. Thus, ζg2 (H) = 4 < 5 = 8− γg(H) + 2− λ2 2(G). ................................................................................................................ ................................................................................................................ ................................................................................................................ ................................................................................................................ ................................................................................................................ ................................................................................................................ ................................................................................................................ .................................... v1 v2 v3 v4 v5 v6 v7 v8 Figure 3: H = P8 and γg(H) = 4 Corollary 1. For a star graph K1,n, ζgk(K1,n) = { n if k = n− 1 k if k < n− 1. Proof. Let V (K1,n) = {v0, w1, w2, · · · , wn} where v0 has degree n. By Theorem 1(ii), γg(K1,n) = n. Thus, k ≤ n− 1. If k = n− 1, then by Lemma 1, ζgk(K1,n) = n+1− 1 = n. Suppose k < n− 1. Choose an (n− k)-element set S = {w1, w2, · · · , wn−k}. Then S is a ζgk -set of K1,n and, Theorem 7(iii), λk 2(K1,n) = |I2K1,n (S)| = 1. Hence, ζgk(K1,n) = ζgk(S) = n+ 1− n+ k − 1 = k by Theorem 8. Corollary 2. For a wheel graph Wn of order n, where n ≥ 5, ζgk(Wn) =  n− 1 if k = γg(Wn)− 1 2k + 1 if n is odd 2k if n is even . S. Canoy, Jr., J. Anoche / Eur. J. Pure Appl. Math, 18 (2) (2025), 6040 9 of 16 Proof. Let V (Wn) = {w1, w2, · · ·wn} where w1 has degree n − 1. By Theorem 1(iv), γg(Wn) = ⌈n−1 2 ⌉. If k = ⌈n−1 2 ⌉ − 1, then ζgk(Wn) = n − 1 by Lemma 1. Let 1 ≤ k < ⌈n−1 2 ⌉ − 1. If n is odd, then γg(Wn) = n−1 2 . Choose the (n−1 2 − k)-element set S = {w2, w4, · · · , wn−2k−1}. Then S is a ζgk -set of Wn and λk 2(Wn) = n−2k−1 2 . By Theorem 8, ζgk(Wn) = ζgk(S) = n − n−1 2 + k − n−2k−1 2 = 2k + 1. Hence, ζgk(Wn) = 2k + 1. If n is even, then choose the (n2 − k)-element set S = {w2, w4, · · · , wn−2k}. Then S is a ζgk -set of Wn and λk 2(Wn) = n−2k 2 . By Theorem 8, ζgk(Wn) = ζgk(S) = n− n 2 + k − n−2k 2 = 2k. Corollary 3. For the Petersen graph P , ζgk(P ) =  4 if k = 1 7 if k = 2 9 if k = 3. Proof. Let V (P ) = {w1, w2, · · · , w10}. By Theorem 1(vii), γg(P ) = 4. If k = 3, then ζg3 (P ) = 9 by Lemma 1. Let 1 ≤ k ≤ 2. Choose the (4−k)-element set Sk = {w1, w2, w4−k} as shown in Figure 4. Then Sk is a ζgk -set of P . Thus, λ1 2(P ) = |I2P (S1)| = 3 and λ2 2(P ) = |I2P (S2)| = 1 by Theorem 7(iii). Therefore, by Theorem 8, ζg1 (P ) = 10− 4 + k − 3 = 4 ............. ............ ............ ............ ............ ............ ............ ............ ............ ............ ............ ............ ............ ............ ............ .. .................................... ....................................................................................................................................................................................... .................................... .......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ..... .................................... ..................................................................................................................................... .................................... ...................................................................................................................................................... .................................... .................................... ......................................................................................................................................................................... .............. ............. ............. ............. ............. ............. ............. ............. ............. ............. .. .... ................................ .......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ... .................................... ........................................................................................................................................... .................................... ..................................................................................................................................... .................................... ...................................................................................................................................... .................................... ......... ........ ........ ........ ........ ........ ........ ..... .................................... .................................... .............................................................. ........................................................................ ...................................................... .................................... .................................... .......... ......... ......... ......... ......... ........ .................................... .................................... w8 w6 w7 w3w2 w1 w5 w4 w10w9 P : • • • ............. ............ ............ ............ ............ ............ ............ ............ ............ ............ ............ ............ ............ ............ ............ .. .................................... ....................................................................................................................................................................................... .................................... .......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ..... .................................... ..................................................................................................................................... .................................... ...................................................................................................................................................... .................................... .................................... ......................................................................................................................................................................... .............. ............. ............. ............. ............. ............. ............. ............. ............. ............. .. .... ................................ .......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ... .................................... ........................................................................................................................................... .................................... ..................................................................................................................................... .................................... ...................................................................................................................................... .................................... ......... ........ ........ ........ ........ ........ ........ ..... .................................... .................................... .............................................................. ........................................................................ ...................................................... .................................... .................................... .......... ......... ......... ......... ......... ........ .................................... .................................... w8 w6 w7 w3w2 w1 w5 w4 w10w9 P : • • Figure 4: S1 = {w1, w2, w3} and S2 = {w1, w2} and ζg2 (P ) = 10− 4 + 2− 1 = 7. This proves the assertion. Theorem 9 ([9]). If G = Km1,m2,··· ,mr is a complete multipartite graph with 2 ≤ m1 ≤ m2 ≤ · · · ≤ mr, where r ≥ 2, then γg(G) =  2 if m1 = 2 3 if m1 = 3 4 if m1 ≥ 4. Theorem 10. For a complete multipartite graph G = Km1,m2,··· ,mr where r ≥ 2 and 2 ≤ m1 ≤ m2 ≤ · · · ≤ mr, we have S. Canoy, Jr., J. Anoche / Eur. J. Pure Appl. Math, 18 (2) (2025), 6040 10 of 16 ζgk(G) =  n− 1 if k = γg(G)− 1 1 if m1 = 3 and k = 1 m1 + k − 4 if m1 ≥ 4 and 1 ≤ k ≤ 2. where n = ∑r j=1mj. Proof. Let Q1, Q2, · · · , Qr be the partite sets of G with |Qj | = mj for each j ∈ [r]. Consider the following cases: Case 1: k = γg(G)− 1. By Lemma 1, ζgk(G) = n− 1. Case 2: m1 = 3 and k = 1. By Theorem 9, γg(G) = 3. Then any 2-element subset S of V (Q1) is a ζgk -set of G and λk 2(G) = ∑r j=2mj . By Theorem 8, ζgk(G) = ζgk(S) = ∑r j=1mj − 3 + 1− ∑r j=2mj = 1. Case 3: m1 ≥ 4 and 1 ≤ k ≤ 2. By Theorem 9, γg(G) = 4. Choose an (4 − k)-element set S′ = {{q11, q21, q 4−k 1 } where qt1 ∈ Q1 and t ∈ {1, 2, 4−k}. Then S′ is a ζgk -set of G and λk 2(G) = ∑r j=2mj . By Theorem 8, ζgk(G) = ζgk(S) = r∑ j=1 mj − 4 + k − r∑ j=2 mj = m1 + k − 4. This proves the assertion. The next result is a consequence of Theorem 10. Corollary 4. For a complete bipartite graph Km,n where 2 ≤ m ≤ n, ζgk(Km,n) =  m+ n− 1 if k = γg(G)− 1 1 if m = 3 and k = 1 m+ k − 4 if m ≥ 4 and 1 ≤ k ≤ 2. S. Canoy, Jr., J. Anoche / Eur. J. Pure Appl. Math, 18 (2) (2025), 6040 11 of 16 Theorem 11. For a path Pn with n ≥ 2 vertices, ζgk(Pn) =  3k − 2 if n = 2 and k = 2 or n = 3r + 2 and k ≤ r + 1 3k − 1 if n = 3r and k ≤ r 3k if n = 3r + 1 and k ≤ r. Proof. Let Pn = [v1, v2, · · · , vn] and let r ≥ 1. By Theorem 1(vi), γg(Pn) = ⌈n+2 3 ⌉. If n = 2, then γg(P2) = 2. Thus, ζg1 (P2) = 1. Let n ≥ 3. Consider the following cases: Case 1: n = 3r. By Theorem 1(vi), γg(Pn) = ⌈n+2 3 ⌉ = r + 1. Here, k ≤ r and γg(Pn) − k = r − k + 1. Consider an (r − k + 1)-element set S1 = {v1, v4, · · · , v3r−3k+1}. The set S1 is a ζgk -set of Pn and Ng G[S1] = {v1, v2, v3, v4, · · · , v3r−3k+1}. Therefore, ζgk(Pn) = ζgk(S1) = n− |Ng G[S1]| = 3r − (3r − 3k + 1) = 3k − 1. Case 2: n = 3r + 1. By Theorem 1(vi), γg(Pn) = ⌈n+2 3 ⌉ = r + 1. Then k ≤ r. Choose an (r − k + 1)-element set S2 = {v1, v4, · · · , v3r−3k+1}. Then S2 is a ζgk -set of Pn and Ng G[S2] = {v1, v2, v3, v4, · · · , v3r−3k+1}. Thus, ζgk(Pn) = ζgk(S2) = n− |Ng G[S2]| = (3r + 1)− (3r − 3k + 1) = 3k. Case 3: n = 3r + 2. By Theorem 1(vi), γg(Pn) = ⌈n+2 3 ⌉ = r + 2. Then k ≤ r + 1 and γg(Pn)− k = r − k + 2. Consider an (r − k + 2)-element set S3 = {v1, v4, · · · , v3r−3k+4}. The set S3 is a ζgk -set of Pn and Ng G[S3] = {v1, v2, v3, v4, · · · , v3r−3k+4}. Therefore, ζgk(Pn) = ζgk(S3) = n− |Ng G[S3]| = 3r + 2− (3r − 3k + 4) = 3k − 2. S. Canoy, Jr., J. Anoche / Eur. J. Pure Appl. Math, 18 (2) (2025), 6040 12 of 16 Theorem 12. If Cn is the cycle with n vertices and k ≤ γg(Cn)− 1, then ζgk(Cn) =  3k + 2 if n = 3r, r ≥ 2 and k = γg(Cn)− 1 or r is odd, r ≥ 3, and k = γg(Cn)− 2 3k + 1 if n = 3r + 2, r ≥ 2 and k = γg(Cn)− 1 or r is odd, r ≥ 3, and k = γg(Cn)− 2 3k if n = 3r, r ≥ 4 and, 1 ≤ k ≤ γg(Cn)− 3 or n = 3r, r is even, r ≥ 4, and k = γg(Cn)− 2 or n = 3r + 1 and k = γg(Cn)− 1 or n = 3r + 1, r is even, and k = γg(Cn)− 2 3k − 2 if n = 3r + 1, r ≥ 3, and 1 ≤ k ≤ γg(Cn)− 3 or n = 3r + 1, r is odd, r ≥ 3, and k = γg(Cn)− 2 or n = 5 and k = 2 3k − 1 if n = 3r + 2, r ≥ 3 and k = 1 or n = 3r + 2, r is even and k = γg(Cn)− 2 or n = 3r + 2, r ≥ 4 and 2 ≤ k ≤ γg(Cn)− 3 or n = 5 and k = 1 k if n = 3. Proof. Let Cn = [v1, v2, · · · , vn, v1] and let k ≤ γg(Cn)− 1. If n = 3, then ζgk(C3) = k by Theorem 6. If n = 5, then ζg1 (C5) = 2 = 3k− 1 and ζg2 (C5) = 4 = 3k− 2. Let n = 4 or n ≥ 6. Consider the following cases: Case 1: n = 3r where r ≥ 2. By Theorem 1(v), γg(Cn) = ⌈n3 ⌉ = r. Thus, k ≤ r−1. Consider the following subcases: Subcase 1: k = γg(Cn)− 1 = r − 1. By Lemma 1, ζgk(Cn) = n− 1 = 3r − 1 = 3k + 2. Subcase 2: r is odd and k = γg(Cn)− 2 = r − 2. Let D2 = {v1, v4}. Then D2 is a ζgk -set of Cn and Ng G[D2] = {v1, v2, v3, v4}. Thus, ζgk(Cn) = ζgk(D2) = 3r − |Ng G[D2]| = 3r − 4 = 3k + 2. Subcase 3: r ≥ 4 and 1 ≤ k ≤ γg(Cn)− 3. Let k ≤ ⌊γg(Cn)−2 2 ⌋ and let D3 = {v1, v4, · · · , v3r−3k−2}. Then D3 is a ζgk -set of S. Canoy, Jr., J. Anoche / Eur. J. Pure Appl. Math, 18 (2) (2025), 6040 13 of 16 Cn and Ng G[D3] = {v1, v2, v3, v4, · · · , v3r−3k−2, v3r−3k−1, v3r}. Thus, ζgk(Cn) = ζgk(D3) = n − |Ng G[D3]| = 3r − [(3r − 3k − 2) + 2] = 3k. Next, let ⌊γg(Cn)−2 2 ⌋ < k ≤ γg(Cn) − 3. Choose an (r− k)-element set D4 = {v⌈ 3r+2 2 ⌉, v1, v4, · · · , v3r−3k−5}. Then D4 is a ζgk -set of Cn and Ng G[D4] = {v1, v2, v3, v4, · · · , v3r−3k−5, v3r−3k−4, v⌈ 3r+2 2 ⌉−1, v⌈ 3r+2 2 ⌉, v⌈ 3r+2 2 ⌉+1, v3r}. This implies that ζgk(Cn) = ζgk(D4) = n− |Ng G[D4]| = 3r − [(3r − 3k − 4) + 4] = 3k. Subcase 4: r ≥ 4 is even and k = γg(Cn)− 2 = r − 2. Let D5 = {v1, v 3r+2 2 }. Then D5 is a ζgk -set of Cn and Ng G[D5] = {v1, v2, v 3r 2 , v 3r+2 2 , v 3r+4 2 , v3r}. Hence, ζgk(Cn) = ζgk(D5) = 3r − |Ng G[D5]| = 3r − 6 = 3k. Case 2: n = 3r + 1. By Theorem 1(v), γg(Cn) = ⌈n3 ⌉ = r+1. Thus, k ≤ r. Consider the following subcases: Subcase 1: k = γg(Cn)− 1 = r. By Lemma 1, ζgk(Cn) = n− 1 = 3r + 1− 1 = 3r = 3k. Subcase 2: r is even and k = γg(Cn)− 2 = r − 1. Let D6 = {v1, v4}. Then D6 is a ζgk -set of Cn and Ng G[D6] = {v1, v2, v3, v4}. Hence, ζgk(Cn) = ζgk(D6) = 3r + 1− |Ng G[D6]| = 3r + 1− 4 = 3k. Subcase 3: r ≥ 3 and 1 ≤ k ≤ γg(Cn)− 3 . Let k ≤ ⌊γg(Cn)−2 2 ⌋ and D8 = {v1, v4, · · · , v3r−3k+1}. Then D8 is a ζgk -set of Cn and Ng G[D8] = {v1, v2, v3, v4, · · · , v3r−3k+1, v3r−3k+2, v3r+1}. This implies that ζgk(Cn) = ζgk(D8) = n− |Ng G[D8]| = 3r+1− [(3r− 3k+2)+1] = 3k− 2. Hence, ζgk(Cn) = 3k − 2. Next, let ⌊γg(Cn)−2 2 ⌋ < k ≤ γg(Cn) − 3. Choose an (r − k + 1)- element set D9 = {v⌈ 3r+3 2 ⌉, v1, v4, · · · , v3r−3k−2}. Then D9 is a ζgk -set of Cn and Ng G[D9] = {v1, v2, v3, v4, · · · , v3r−3k−2, v3r−3k−1, v⌈ 3r+3 2 ⌉−1, v⌈ 3r+3 2 ⌉, v⌈ 3r+3 2 ⌉+1, v3r+1}. This implies that ζgk(Cn) = ζgk(D9) = n− |Ng G[D9]| = 3r+1− [(3r− 3k− 1)+ 4] = 3k− 2. Hence, ζgk(Cn) = 3k − 2. S. Canoy, Jr., J. Anoche / Eur. J. Pure Appl. Math, 18 (2) (2025), 6040 14 of 16 Subcase 4: r ≥ 3 is odd and k = γg(Cn)− 2 = r − 1. Let D10 = {v1, v 3r+3 2 }. Then D10 is a ζgk -set of Cn and Ng G[D10] = {v1, v2, v 3r+1 2 , v 3r+3 2 , v 3r+5 2 , v3r+1}. Hence, ζgk(Cn) = ζgk(D10) = 3r + 1− |Ng G[D10]| = 3r + 1− 6 = 3k − 2. Case 3: n = 3r + 2 where r ≥ 2. By Theorem 1(v), γg(Cn) = ⌈n3 ⌉ = r+1. Thus, k ≤ r. Consider the following subcases: Subcase 1: r ≥ 3 and k = 1. LetD11 = {v1, v4, · · · , v3r−2} be an (r)-element set. ThenNg G[D11] = {v1, v2, · · · , v3r−2, v3r−1, v3r+2}. Thus, D11 is a ζg1 -set of Cn. Hence, ζg1 (Cn) = ζg1 (D11) = n− |Ng G[D11]| = 3r + 2− [(3r − 1) + 1] = 2 = 3k − 1. Subcase 2: k = γg(Cn)− 1 = r. By Lemma 1, ζgk(Cn) = n− 1 = 3r + 2− 1 = 3r + 1 = 3k + 1. Subcase 3: r is odd and k = γg(Cn)− 2 = r − 1. Let D12 = {v1, v4}. Then D12 is a ζgk -set of Cn and Ng G[D12] = {v1, v2, v3, v4}. Hence, ζgk(Cn) = ζgk(D12) = 3r + 2− |Ng G[D12]| = 3r + 2− 4 = 3k + 1. Subcase 4: r is even and k = γg(Cn)− 2 = r − 1. LetD14 = {v1, v 3r+4 2 } is a ζgk -set of Cn andNg G[D14] = {v1, v2, v 3r+2 2 , v 3r+4 2 , v 3r+6 2 , v3r+2}. Hence, ζgk(Cn) = ζgk(D14) = 3r + 2− |Ng G[D14]| = 3r + 2− 6 = 3k − 1. Subcase 5: r ≥ 4 and 2 ≤ k ≤ γg(Cn)− 3 . Let k ≤ ⌊γg(Cn)−2 2 ⌋ and let D15 = {v1, v4, · · · , v3r−3k+1}. Then D15 is a ζgk -set of Cn and Ng G[D15] = {v1, v2, v3, v4, · · · , v3r−3k+1, v3r−3k+2, v3r+2}. Thus, ζgk(Cn) = ζgk(D15) = n−|Ng G[D15]| = 3r+2− [(3r−3k+2)+1] = 3k−1. Next, let ⌊γg(Cn)−2 2 ⌋ < k ≤ γg(Cn)−3. Choose an (r − k + 1)-element set D16 = {v⌈ 3r+5 2 ⌉, v1, v4, · · · , v3r−3k−2}. Then D16 is a ζgk -set of Cn and Ng G[D16] = {v1, v2, v3, v4, · · · , v3r−3k−2, v3r−3k−1, v⌈ 3r+5 2 ⌉−1, v⌈ 3r+5 2 ⌉, v⌈ 3r+5 2 ⌉+1, v3r+2}. S. Canoy, Jr., J. Anoche / Eur. J. Pure Appl. Math, 18 (2) (2025), 6040 15 of 16 This implies that ζgk(Cn) = ζgk(D16) = n− |Ng G[D16]| = 3r + 2− [(3r − 3k − 1) + 4] = 3k − 1. This shows that the assertion holds. 4. Conclusion In this paper, we introduced the concept of k-geodetic domination defect of a graph and computed its values for several well-known graphs. Additionally, we established some sharp bounds of this parameter. It is recommended that further investigation of this newly defined parameter be done especially on graphs resulting from some graph operations. Acknowledgements The authors would like to thank the referees for the comments and suggestions they offered us which led to the improvement of the paper. 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