EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 1, Article Number 5602 ISSN 1307-5543 – ejpam.com Published by New York Business Global On Total Double Italian Domination in Graphs Sheryl Jane L. Sumbalan1,2,∗, Sheila M. Menchavez1,2, Ferdinand P. Jamil1,2 1 Department of Mathematics and Statistics, College of Science and Mathematics, Mindanao State University-Iligan Institute of Technology, 9200 Iligan City, Philippines 2 Center for Mathematical and Theoretical Physical Sciences (CMTPS), Premier Research Institute of Science and Mathematics (PRISM), Mindanao State University - Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. For a simple graph G = (V (G) , E (G)), a total double Italian dominating function is a function f : V (G) → {0, 1, 2, 3} with properties that every vertex v ∈ V (G) with f (v) ∈ {0, 1},∑ u∈N [v] f (u) ≥ 3 and every vertex v ∈ V (G) with f(v) ̸= 0 has a neighbor u with f(u) ̸= 0. The weight of a total double Italian dominating function is the sum ωG (f) = ∑ v∈V (G) f (v) ≥ 3 and the minimum weight of all the total double Italian dominating functions on a graph G is the total double Italian domination number, denoted by γtdI (G). In this paper we explore further the concept of total double Italian domination. We characterize graphs G with smaller values for γtdI(G). Also, we characterize the total double Italian dominating function on the join, corona, edge corona, and complementary prism of graphs. Exact values or bounds are also determined for their respective total double Italian domination number. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Total double Italian dominating function, Total double Italian domination number. 1. Introduction Since its introduction in 2004 by Cockayne et al. [12], Roman domination is one of the most well-studied concepts in graph theory. For a comprehensive understanding of its origins, historical development, and significance in the field along with recent advances, we refer to [1, 2, 10, 13, 16–19, 21, 22, 24, 25]. Building on the foundations of Roman domination, Chellali et al. [13] introduced a broader concept known as Italian domination (also referred as Roman-{2} domination). Meanwhile, Beeler et al. [10] extended the idea even further by developing the notion of double Roman domination, a stronger variant that inspires new research in the field. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i1.5602 Email addresses: sheryljane.sumbalan@g.msuiit.edu.ph (S.J. Sumbalan), sheila.menchavez@g.msu.iit.edu.ph (S. Menchavez), ferdinand.jamil@g.msuiit.edu.ph (F. Jamil) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 2 of 18 In 2020, Mojdeh et al. [19] introduced the concept of double Italian domination (or Roman {3}-domination) which is an optimization of the double Roman domination. In the same year, Shao et al. [26] initiated the study of total double Italian domination and have shown its relationship to other domination parameters. This present paper investigates further the total double Italian domination, particularly in graphs under the join, corona, edge corona and complementary prism of graphs. Throughout this paper, all graphs considered are undirected, finite and simple. See [3, 4, 9] for all the basic graph terminologies that are not defined but used in this paper. For a graph G = (V (G), E(G)), the open neighborhood of a vertex v ∈ V (G), denoted by NG(v), consists of all the vertices adjacent to v and its closed neighborhood, denoted by NG[v], is the open neighborhood of v together with vertex v. The degree of v, de- noted by degG(v), degG(v) = |NG(v)|. The minimum degree, δ(G) of G is the minimum degree among the vertices of G. The maximum degree of G, denoted by ∆(G), is the maximum degree among the vertices of G. For S ⊆ V (G), NG(S) = ∪v∈SNG(v) and NG[S] = S ∪NG(S). Let G and H be graphs with disjoint vertex sets. The join of graphs G and H is the graph G +H with vertex set V (G +H) = V (G) ∪ V (H) and edge set E(G +H) = E(G)∪E(H)∪{uv : u ∈ V (G)∧ v ∈ V (H)}. The corona of G and H, G ◦H, is the graph obtained by taking one copy of G and |V (G)| copies of H and then joining the ith vertex of G to every vertex of the ith copy of H. The edge corona, denoted by G ⋄H, of G and H is a graph obtained by taking one copy of G and |E(G)| copies of H and joining each of the end vertices u and v of each edge uv of G to every vertex of the copy Huv of H. The complementary prism, denoted GG, is formed from the disjoint union of G and its complement G by adding a perfect matching between corresponding vertices of G and G. The gluing of G and H along a common subgraph K is the graph G ⊔K H by combining G and H through K. Graphs C4 ⊔P3 C4 and C4 ⊔K2 K3 are given in Figure 1. We refer to [5] for a detailed information on the gluing of graphs. A set S ⊆ V (G) is a dominating set of G if NG[S] = V (G). The domination number of G, denoted by γ(G), is the smallest cardinality of a dominating set of G. A set S of vertices in a graph G is called a total dominating set if NG(S) = V (G). The total domination number γt(G) of G is the minimum cardinality of a total dominating set of vertices in G. A dominating set of G of cardinality γ(G) is referred to as γ-set of G. A total dominating set ofG of cardinality γt(G) is called a γt-set ofG. We refer to [6, 11, 14, 15, 20] for the fundamental concepts, some recent developments and applications of domination and total domination in graphs. For a positive integer k, a set D ⊆ V (G) is called a k-dominating set if each v ∈ V (G) \D is adjacent to at least k vertices in D. The k-domination number γk(G) is then defined to be the smallest cardinality of a k-dominating set of G. M. Chellali et al. in [8] presented an outstanding survey of results in k-domination in graphs. A subset S ⊆ V (G) is a vertex cover of G if for every edge uv ∈ E(G), u ∈ S or v ∈ S. S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 3 of 18 The smallest cardinality of a vertex cover is the vertex cover number of G, and is denoted by β(G). Excellent references for vertex cover include [7, 23]. A double Italian dominating function (or DIDF ) of G is a function f : V (G) → {0, 1, 2, 3} having the property that for every vertex v ∈ V (G), if f(v) ∈ {0, 1}, then ∑ u∈N [v] f(u) ≥ 3. The weight of a DIDF is the sum ωG(f) = ∑ v∈V (G) f(v), and the minimum weight of a DIDF f is the double Italian domination number, denoted by γdI(G). A DIDF function f : V (G) → {0, 1, 2, 3} is a total double Italian dominating function (or TDIDF ) of G if for each v ∈ V (G) with f(v) ̸= 0, there exists u ∈ V (G) such that f(u) ̸= 0 and uv ∈ E(G). The minimum weight of a TDIDF of G is the total double Italian domination number of G, and is denoted by γtdI(G). We write f ∈ TDIDF (G) to mean that f is TDIDF of G. Any TDIDF f of G with weight γtdI(G) is referred to as γtdI -function of G. For a function f : V (G) → {0, 1, 2, 3}, let (V0, V1, V2, V3) be the ordered partition induced by f , where Vi = {v ∈ V (G) : f(v) = i} for i ∈ {0, 1, 2, 3}. Then we can write f = (V0, V1, V2, V3). The weight of f is defined by ωG(f) = |V1|+ 2|V2|+ 3|V3|. More precisely, f = (V0, V1, V2, V3) ∈ TDIDF (G) if each of the following holds: (i) For each v ∈ V0, at least one of the following holds: (a) |V1 ∩NG(v)| ≥ 3; (b) |V1 ∩NG(v)| ≥ 1 and |V2 ∩NG(v)| ≥ 1; (c) |V2 ∩NG(v)| ≥ 2; (d) |V3 ∩NG(v)| ≥ 1. (ii) For each v ∈ V1, at least one of the following holds: (a) |V1 ∩NG(v)| ≥ 2; (b) |(V2 ∪ V3) ∩NG(v)| ≥ 1. (iii) For each v ∈ V1 ∪ V2 ∪ V3, |(V1 ∪ V2 ∪ V3) ∩NG(v)| ≥ 1. 2. Preliminary results Proposition 1. [26] If G is a connected graph of order n ≥ 2, then γtdI(G) ≥ 3 and γtdI(G) = 3 if and only if G has at least two vertices of degree ∆(G) = n− 1. Proposition 2. [26] If G has only one vertex of degree ∆(G) = n− 1, then γtdI(G) = 4. Theorem 1. [26] If G is a graph with δ(G) = δ ≥ 2, then γtdI(G) ≤ |V (G)|+ 2− δ, and this bound is sharp. Proposition 3. Let G be a nontrivial connected graph of order n ≥ 4. Then γtdI(G) = 4 if and only if one of the following holds: S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 4 of 18 (i) G has exactly one vertex of degree ∆(G) = n− 1; (ii) γ(G) ≥ 2 and G has a 3-dominating set D with |D| = 4 and δ(⟨D⟩) ≥ 2. Proof : Suppose that γtdI(G) = 4. If γ(G) = 1, then by Proposition 1 and Proposition 2, G contains exactly one vertex of degree n − 1, and (i) holds. Suppose that γ(G) ≥ 2. Let f = (V0, V1, V2, V3) be a γtdI -function of G. Since γ(G) ̸= 1, V2 = V3 = ∅ and |V1| = 4. Put D = V1. Then δ(⟨D⟩) ≥ 2. If V0 = ∅, then G = C4. If V0 ̸= ∅, then 3 ≤ |NG(v) ∩ V1| ≤ 4 for every v ∈ V0. In any case, D is a 3-dominating set of G. Thus, (ii) holds. Conversely, if (i) holds, then the desired result follows from Proposition 2. Suppose (ii) holds. By Proposition 1 and Proposition 2, γtdI(G) ≥ 4. On the other hand, since f = (V (G) \ D,D,∅,∅) is a total double Italian dominating function on G, γtdI(G) ≤ |D| = 4. Therefore, γtdI(G) = 4. ■ In Statement (ii) of Proposition 3, it is not necessary that γ3(G) = 4. To see this, note that if G = C4 +K3, then γtdI(G) = 4 while γ3(G) = 3. It can be verified that if G is connected of order n = 4, then γtdI(G) ̸= 5. Proposition 4. Let G be a nontrivial connected graph of order n ≥ 5. Then γtdI(G) = 5 if and only if one of the following holds: (i) G ∈ {C5,K2,3,K3 ⊔K2 C4,K2 + (K1 ∪K2)} (see Figure 1); (ii) n > 5, γ(G) ≥ 2, G does not contain a 3-dominating set D with |D| = 4 and δ(⟨D⟩) ≥ 2, and one of the following holds: (a) G contains a 3-dominating set D with |D| = 5 and δ(⟨D⟩) ≥ 2; (b) G contains a 2-dominating set D with |D| = 3 and ⟨D⟩ is connected. (c) G contains a 2-dominating set D with |D| = 4 such that there exists v ∈ D for which uv ∈ E(G) for all u ∈ V (G) \D with |NG(u)∩D| = 2. Moreover, ⟨D⟩ is connected and xv ∈ E(G) for every x ∈ D \ {v} with |NG(x) ∩D| = 1. Proof : Suppose that γtdI(G) = 5. By Proposition 1 and Proposition 3, γ(G) ≥ 2 and G does not contain a 3-dominating set D with |D| = 4 and δ(⟨D⟩) ≥ 2. If n = 5, then G ∈ {C5,K2,3,K3⊔K2C4,K2+(K1 ∪K2)}. Assume that n > 5. Let f = (V0, V1, V2, V3) be a γtdI function of G. By Proposition 1 and Proposition 3, V3 = ∅. Consider the following cases: Case 1: Suppose that |V1| = 5 and |V2| = 0. Since n > 5, V0 ̸= ∅. Then 3 ≤ |NG(u) ∩ V1| ≤ 5 for every u ∈ V0. Hence, D = V1 is a 3-dominating set on G. Since |NG(v) ∩ V1| ≥ 2 for every v ∈ V1, δ(⟨D⟩) ≥ 2 and (ii)(a) holds. Case 2: Suppose that |V1| = 1 and |V2| = 2. Since n > 5, V0 ̸= ∅. Thus D = V1 ∪ V2 is a 2-dominating set of G. Moreover, since f ∈ TIDF (G), ⟨D⟩ is connected. Therefore, (ii)(b) holds. S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 5 of 18 Case 3: Suppose that |V1| = 3 and |V2| = 1. Put V2 = {v}. Then for each u ∈ V0, either |N(u) ∩ V1| = 3 or 1 ≤ |N(u) ∩ V1| ≤ 2 and uv ∈ E(G). Hence, |N(u) ∩ (V1 ∪ V2)| ≥ 2. Thus, D = V1 ∪ V2 is a 2-dominating set of G. Observe that, if |N(u) ∩ (V1 ∪ V2)| = 2, then uv ∈ E(G). By the definition of f , ⟨V1 ∪ V2⟩ has no isolated vertex. If u′ ∈ V1 such that d⟨S⟩(u ′) = 1, then u′v ∈ E(G). Thus, (ii)(c) holds. Conversely, if G ∈ {C5,K2,3,K3 ⊔K2 C4,K2 + (K1 ∪K2)}, then γtdI(G) = 5. Now, suppose that n > 5, γ(G) ≥ 2, G does not contain a 3-dominating set D for which |D| = 4 and δ(⟨D⟩) ≥ 2. Then by Proposition 3, γtdI(G) ≥ 5. Assume that (ii)(a) holds. Let V0 = V (G) \ D, V1 = D, V2 = ∅, and V3 = ∅. Then f = (V0, V1, V2, V3) is a TDIDF on G with ωG(f) = 5. This means that γtdI(G) = 5. Assume (ii)(b) holds. Take x ∈ D. Let V0 = V (G) \D, V1 = {x}, V2 = D \ {x}, and V3 = ∅. Then f = (V0, V1, V2, V3) is a TDIDF on G with ωG(f) = 5. Hence, γtdI(G) = 5. Assume (ii)(c) holds. Let v ∈ D. Then f = (V0, V1, V2, V3) where V0 = V (G) \D, V1 = D \ {v}, V2 = {v}, and V3 = ∅ is a TDIDF on G with ωG(f) = 5. Thus, γtdI(G) = 5. ■ C4 ⊔P3 C4 = K3,2 K3 ⊔K2 C4 K2 + (K1 ∪K2) Figure 1: Examples of graphs G with γtdI(G) = 5 3. On the join of graphs In this section, we denote by f |G the restriction of the function f on the subgraph G of a graph H. The following proposition characterizes all TDIDF on the join of two nontrivial connected graphs. Proposition 5. Let G and H be nontrivial connected graphs. Then f = (V0, V1, V2, V3) is a TDIDF on (G+H) if and only if one of the following holds: (i) f |G ∈ TDIDF (G); (ii) f |H ∈ TDIDF (H); (iii) f |G /∈ TDIDF (G), f |H /∈ TDIDF (H), and each of the following holds: (a) For every v ∈ V0 ∪ V1, (1) ωH(f |H) ≥ 3− f |G(NG[v]), whenever v ∈ V (G) and f |G(NG[v]) < 3; (2) ωG(f |G) ≥ 3− f |H(NH [v]), whenever v ∈ V (H) and f |H(NH [v]) < 3. (b) For every v ∈ V1 ∪ V2 ∪ V3, S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 6 of 18 (1) (V1 ∪ V2 ∪ V3) ∩ V (H) ̸= ∅, whenever v ∈ V (G) and NG(v) ⊆ V0; (2) (V1 ∪ V2 ∪ V3) ∩ V (G) ̸= ∅, whenever v ∈ V (H) and NH(v) ⊆ V0. Proof : Let f = (V0, V1, V2, V3) be a function on V (G +H). Assume that (i) holds for f . Let v ∈ (V0 ∪ V1) ∩ V (G+H). If v ∈ V (G), then 3 ≤ f |G(NG[v]) = ∑ x∈NG[v] f(x) ≤ ∑ x∈NG+H [v] f(x) = f(NG+H [v]). Suppose that v ∈ V (H). Since f |G is a TDIDF on G, Proposition 1 implies that ωG(f |G) ≥ 3. Hence, f(NG+H [v]) = ∑ x∈NG+H [v] f(x) = ∑ x∈NG+H [v]\V (G) f(x) + ∑ x∈V (G) f(x) ≥ 3. Now, let v ∈ V1 ∪ V2 ∪ V3. If v ∈ V (G), then since f |G ∈ TDIDF (G), there exists u ∈ ((V1 ∪ V2 ∪ V3) ∩ V (G)) \ {v} such that vu ∈ E(G) ⊆ E(G +H). If v ∈ V (H), then ∅ ̸= (V1∪V2∪V3)∩V (G) ⊆ NG+H(v). Thus, f ∈ TDIDF (G+H). Similarly, if condition (ii) holds for f , then f ∈ TDIDF (G + H). Suppose (iii) holds. Let v ∈ V0 ∪ V1. If v ∈ V (G) such that f |G(NG[v]) < 3, then by (iii)(a), ωH(f |H) ≥ 3 − f |G(NG[v]). Since f(NG+H [v]) = f |G(NG[v]) + ωH(f |H), f(NG+H [v]) ≥ 3. Similarly, if v ∈ V (H) with f |H(NH [v]) < 3 then f(NG+H [v]) ≥ 3. Since v is arbitrary, f(NG+H [v]) ≥ 3 for each v ∈ V0 ∪ V1. Let u ∈ V1 ∪ V2 ∪ V3. If u ∈ V (G) with NG(u) ⊆ V0, then by (iii)(b), (V1 ∪ V2 ∪ V3) ∩ V (H) ̸= ∅. This means that NG+H(u) ∩ (V1 ∪ V2 ∪ V3) ̸= ∅. Similarly, NG+H(u) ∩ (V1 ∪ V2 ∪ V3) ̸= ∅ for each u ∈ V (H) with NH(u) ⊆ V0. Thus, ⟨V1 ∪ V2 ∪ V3⟩ has no isolated vertex. Therefore, f ∈ TDIDF (G+H). Conversely, suppose that f ∈ TDIDF (G+H). Suppose neither (i) nor (ii) holds for f , i.e., f |G /∈ TDIDF (G+H) and f |H /∈ TDIDF (G+H). Since f |G /∈ TDIDF (G+H), either there exists v ∈ [(V0 ∪ V1) ∩ V (G)] with f |G(NG[v]) < 3 or ⟨(V1 ∪ V2 ∪ V3) ∩ V (G)⟩ has an isolated vertex or both. Assume that there exists v ∈ [(V0 ∪ V1) ∩ V (G)] with f |G(NG[v]) < 3. Since f ∈ TDIDF (G+H), ωH(f |H) ≥ 3− f |G(NG[v]). Thus, (iii)(a(1)) holds. Similarly, (iii)(a(2)) follows. On the other hand, assume that ⟨(V1∪V2∪V3)∩V (G)⟩ has an isolated vertex. Let u ∈ [(V1 ∪ V2 ∪ V3) ∩ V (G)] such that NG(v) ⊆ V0. Since ⟨V1 ∪ V2 ∪ V3⟩ is isolated vertex-free, (V1 ∪ V2 ∪ V3) ∩ V (H) ̸= ∅. Thus, (iii)(b(1)) holds. Similarly, (iii)(b(2)) follows. ■ Corollary 1. Let G and H be nontrivial connected graphs. Then 3 ≤ γtdI(G+H) ≤ min{6, γtdI(G), γtdI(H)}. (1) Proof : The lower bound follows immediately from Propositon 1. To show the upperbound, take u ∈ V (G) and v ∈ V (H). Then f = (V (G) \ {u, v},∅,∅, {u, v}) is a TDIDF on G + H, with ωG+H(f) = 6. Thus, γtdI(G + H) ≤ 6. Let f = (V0, V1, V2, V3) be a γtdI - function of G. By Proposition 5, g = (V0 ∪ V (H), V1 ∪ V (H), V2 ∪ V (H), V3 ∪ V (H)) is a TDIDF on G with ωG+H(f) = ωG(f). This means that γtdI(G+H) ≤ γtdI(G). Similarly, γtdI(G+H) ≤ γtdI(H). ■ S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 7 of 18 Corollary 2. Let G and H be nontrivial connected graphs. If there exists a γtdI-function f = (V0, V1, V2, V3) of G+H such that either f |G is a TDIDF on G or f |H is a TDIDF on H then γtdI(G+H) = min{γtdI(G), γtdI(H)}. Proof : By Corollary 1, γtdI(G + H) ≤ min{γtdI(G), γtdI(H)}. Assume, WLOG, f |G = (V0 ∩ V (G), V1 ∩ V (G), V2 ∩ V (G), V3 ∩ V (G)) is a TDIDF on G. Then γtdI(G) ≤ ωG(f |G) ≤ ωG+H(f). This implies that γtdI(G +H) ≥ min{γtdI(G), γtdI(H)}. Therefore, γtdI(G+H) = min{γtdI(G), γtdI(H)}. ■ Proposition 6. Let G and H be nontrivial connected graphs of orders n and m, respec- tively. Then γtdI(G+H) = 3 if and only if one of the following holds: (i) γtdI(G) = 3; (ii) γtdI(H) = 3; (iii) G and H each contains at least one vertex of degree n− 1 and m− 1, respectively. Proof : Suppose that γtdI(G+H) = 3. By Proposition 1, G+H has at least two vertices of degree n + m − 1. Take u, v ∈ V (G + H) for which dG+H(u) = n + m − 1 and dG+H(v) = n + m − 1. If u, v ∈ V (G), then both u and v have degree n − 1. By Proposition 1, (i) holds. Similarly, if u, v ∈ V (H), then (ii) holds. If u ∈ V (G) and v ∈ V (H), then (iii) holds. Conversely, if (i) or (ii) holds, then Equation (1) in Corollary 1 yields γtdI(G+H) = 3. Suppose (iii) holds. Then G + H contains at least two vertices of maximum degree ∆(G+H) = (m+ n)− 1. By Proposition 1, γtdI(G+H) = 3. ■ Proposition 7. Let G and H be nontrivial connected graphs of orders n and m, respec- tively. Then γtdI(G+H) = 4 if and only if one of the following holds: (i) ∆(H) ≤ m− 2 and G has exactly one vertex of degree n− 1; (ii) ∆(G) ≤ n− 2 and H has exactly one vertex of degree m− 1; (iii) γ(G) = 2 and γ(H) = 2. (iv) γ(G) ≥ 2 and γ(H) ≥ 2 and one of the following holds: (a) G or H has a 3-dominating set D with |D| = 4 and δ(⟨D⟩) ≥ 2; (b) G or H has a 2-dominating set D such that |D| = 3 and ⟨D⟩ is connected; Proof : Assume that γtdI(G + H) = 4. If G + H has exactly one vertex v for which degG+H(v) = m + n − 1, then (i) or (ii) holds. Otherwise, by Proposition 3, γ(G) ≥ 2, γ(H) ≥ 2 and G + H has a 3-dominating set D with |D| = 4 and δ(⟨D⟩) ≥ 2. Let DG = D ∩ V (G) and DH = D ∩ V (H). Suppose first that |DG| = 2 = |DH |. For each x ∈ V (G) \ DG, since D is a 3-dominating set of G + H, there exists u ∈ DG for which ux ∈ E(G). Thus, DG is a dominating set of G. Since γ(G) ≥ 2, γ(G) = |DG| = 2 S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 8 of 18 Similarly, γ(H) = 2. Thus, (iii) holds. Next, if D ⊆ V (G) or D ⊆ V (H), then (iv)(a) holds. Finally, WLOG suppose that |DG| = 3 and |DH | = 1. Clearly, DG is a 2-dominating set of G. Since δ(⟨D⟩) ≥ 2, ⟨DG⟩ is connected and (iv)(b) holds. Conversely, note that each of the conditions implies that one of the conditions in Proposition 3 is satisfied for G+H. Thus, γtdI(G+H) = 4. ■ In view of Proposition 4, if G and H are nontrivial graphs of orders n and m, respec- tively with n+m = 5, then γtdI(G+H) = 5 if and only if G+H ∈ {K2,3,K2+(K1 ∪K2)}. Proposition 8. Let G and H be nontrivial graphs of orders n and m, respectively, such that m+ n > 5. Then γtdI(G+H) = 5 if and only if each of the following holds: (i) γ(G) ≥ 2 and γ(H) ≥ 2, but γ(G) and γ(H) cannot be both equal to 2; (ii) Neither G nor H contains a 2-dominating set D for which |D| = 3 and ⟨D⟩ is connected; (iii) Neither G nor H has a 3-dominating set D with |D| = 4 and δ(⟨D⟩) ≥ 2; (iv) One of the following holds: (a) min{γtdI(G), γtdI(H)} = 5; (b) G or H has a 2-dominating set D for which |D| = 4 and ⟨D⟩ is connected; (c) G or H has a dominating set D with |D| = 3; (d) γ(G) = 2 and γ(H) ≥ 3; (e) γ(H) = 2 and γ(G) ≥ 3. Proof : Suppose that γtdI(G + H) = 5. By Proposition 6, γ(G) ≥ 2 and γ(H) ≥ 2. Moreover, by Proposition 7, γ(G) and γ(H) cannot be both equal to 2; neither G nor H contains a 2-dominating set D for which |D| = 3 and ⟨D⟩ is connected; and neither G nor H has a 3-dominating set D with |D| = 4 and δ(⟨D⟩) ≥ 2. Now we claim that α = min{γtdI(G), γtdI(H)} ≥ 5. Suppose not. Then α = 4. WLOG, assume γtdI(G) = 4. By Proposition 3, G has a 3-dominating set D with |D| = 4 and δ(⟨D⟩) ≥ 2, a contradiction. This establishes our claim. If α = 5, then (iv)(a) holds. Suppose that α > 5. In view of Proposition 4, one of the following cases holds: Case 1: G + H contains a 3-dominating set D with |D| = 5 and δ(⟨D⟩) ≥ 2. Let DG = D ∩ V (G) and DH = D ∩ V (H). Since α > 5, 1 ≤ |DG| ≤ 4 and 1 ≤ |DH | ≤ 4. If either |DG| = 4 or |DH | = 4, then (iv)(b) holds. If either |DG| = 3 or |DH | = 3, then (iv)(c) holds. Case 2: G + H contains a 2-dominating set D with |D| = 3 and ⟨D⟩ is connected. If DG = D ∩ V (G) and DH = D ∩ V (H), then 1 ≤ |DG| ≤ 2 and 1 ≤ |DH | ≤ 2. If |DG| = 2 and |DH | = 1, then DG is a dominating set of G and γ(G) = 2. By (i), γ(H) ≥ 3, and (iv)(d) holds. Similarly, if |DG| = 1 and |DH | = 2, then (iv)(e) holds. S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 9 of 18 Case 3: G+H contains a 2-dominating set D with |D| = 4 such that there exists v ∈ D for which uv ∈ E(G+H) for all u ∈ V (G+H) \D with |NG+H(u) ∩D| = 2. Moreover, ⟨D⟩ is connected and xv ∈ E(G + H) for every x ∈ D \ {v} with |NG+H(x) ∩ D| = 1. If DG = D ∩ V (G) and DH = D ∩ V (H), then 1 ≤ |DG| ≤ 3 and 1 ≤ |DH | ≤ 3. If |DG| = 3 and |DH | = 1, then whether v ∈ V (G) or v ∈ V (H), DG is a dominating set of G. Similarly, if |DH | = 3 and |DG| = 1, then DH is a dominating set of H. This implies (ii)(c). Suppose that |DG| = 2 = |DH |. Then either γ(G) = 2 and γ(H) ≥ 3 or γ(H) = 2 and γ(G) ≥ 3. Conversely, assume that statements (i)-(iii) hold. Suppose that (iv)(a) holds. WLOG, assume γtdI(G) = 5. In view of Proposition 5, γtdI(G+H) ≤ γtdI(G) = 5. Since statements (i)-(iii) hold, Proposition 6 and Proposition 7 imply that γtdI(G+H) ≥ 5. Suppose that (iv)(b) holds, say G has a 2-dominating set D for which |D| = 4 and ⟨D⟩ is connected. Pick v ∈ V (H). Then D ∪ {v} is a 3-dominating set of G+H with δ(⟨D ∪ {v}⟩) ≥ 2. By Proposition 4, γtdI(G +H) = 5. Suppose that (iv)(c) holds, say G has a dominating set D with |D| = 3. Choose v ∈ V (H). Then D ∪ {v} satisfies Proposition 4(ii)(c). Thus, γtdI(G + H) = 5. Suppose that (iv)(d) holds. Then D ∪ {v} is a 2-dominating set of cardinality 3 and ⟨D ∪ {v}⟩ is connected. Thus, γtdI(G + H) = 5. Similarly, if (iv)(e) holds, then γtdI(G+H) = 5. ■ In view of the above results, in particular, if γ(G) ≥ 5 and γ(H) ≥ 5, then γtdI(G + H) = 6. 4. On the corona of graphs Let G and H be connected graphs. We adapt the notation Hv used in [16] to denote the copy of H whose vertices is joined to v ∈ V (G). Proposition 9. Let G be a nontrivial connected graph and H be any graph without isolated vertices, and let f = (V0, V1, V2, V3) be a function on V (G◦H). Then f ∈ TDIDF (G◦H) if and only if each of the following holds: (i) For each v ∈ V0 ∩ V (G), f |Hv ∈ TDIDF (Hv) ; (ii) For each v ∈ V1 ∩ V (G), f(NHv [u]) ≥ 2 for all u ∈ (V0 ∪ V1) ∩ V (Hv); (iii) For each v ∈ V2 ∩ V (G), f(NHv(u)) ≥ 1 for all u ∈ V0 ∩ V (Hv); (iv) For each v ∈ V3 ∩ V (G) for which NG(v) ⊆ V0, f(V (Hv)) ≥ 1. Proof : Note first that Hv admits a TDIDF for every v ∈ V (G). Assume f ∈ TDIDF (G◦ H). For each v ∈ V (G), NG◦H [u] = {v} ∪NHv [u] (2) so that f(NG◦H [u]) = f(v) + f(NHv [u]) (3) S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 10 of 18 for all u ∈ V (Hv). It follows from Equation (2) and Equation (3) that if v ∈ V0, then f |Hv(NHv [u]) = f(NG◦H [u]) ≥ 3 for all u ∈ (V0 ∪ V1) ∩ V (Hv) and NG◦H(u) ∩ [V1 ∪ V2 ∪ V3] = NHv(u) ∩ [(V1 ∪ V2 ∪ V3) ∩ V (Hv)] for all u ∈ (V1 ∪ V2 ∪ V3) ∩ V (Hv). Thus, f |Hv ∈ TDIDF (Hv) and (i) holds. Similarly, (ii) and (iii) follow immediately from Equation (3). Statement (iv) follows from the fact that ⟨V1∪V2∪V3⟩ has no isolated vertex. Conversely, suppose conditions (i) - (iv) hold for f . First, let u ∈ V0, and let v ∈ V (G) for which u ∈ V (Hv + v). Suppose that v ∈ V0. If u ̸= v, then by (i), f(NG◦H [u]) = f |Hv(NHv [u]) ≥ 3. Suppose that u = v. Statement (i) implies that f(NG◦H [u]) ≥ f |Hv(V (Hv)) ≥ 3. Suppose that v ∈ V (G) \ V0. Then u ∈ V0 ∩ V (Hv) and v ∈ NG◦H [u]. If v ∈ V3, then f(NG◦H [u]) ≥ f(v) = 3. If v ∈ V1 ∪ V2, then by (ii) and (iii), f(NG◦H [u]) = f(v) + f(NHv [u]) ≥ 3. Similar arguments show that if u ∈ V1, then f(NG◦H [u]) ≥ 3. Now, let v ∈ V1 ∪ V2 ∪ V3. Consider the following cases: Case 1: Suppose v ∈ V (G). If V0∩V (Hv) = ∅, then for each u ∈ V (Hv), u ∈ V1∪V2∪V3 and uv ∈ E(G ◦H). Suppose that V0 ∩ V (Hv) ̸= ∅, say w ∈ V0 ∩ V (Hv). If v ∈ V1 ∪ V2, then by Conditions (ii) and (iii), there exists u ∈ [(V1 ∪ V2 ∪ V3) ∩ V (Hv)] for which uw ∈ E(Hv). Incidentally, uv ∈ E(G ◦ H). Suppose that v ∈ V3. Then either there exists u ∈ [(V1 ∪V2 ∪V3)∩NG(v)] or, by Condition (iv), there exists u ∈ V (Hv) for which f(u) ≥ 1. In this case, uv ∈ E(G ◦H). Case 2: Suppose v ∈ V (Hx) for some x ∈ V (G). If x ∈ V1 ∪V2 ∪V3, then we are done. If x ∈ V0, then since f |Hx ∈ TDIDF (Hx) (Condition (i)), there exists u ∈ [(V1 ∪ V2 ∪ V3) ∩ V (Hv)] for which uv ∈ E(Hx), which means uv ∈ E(G ◦H). ■ Corollary 3. Let G be a nontrivial connected graph of order n, and let H be any graph. Then γtdI(G ◦H) = 3n. Proof : Since f = (∪x∈V (G)V (Hx),∅,∅, V (G)) ∈ TDIDF (G ◦H), γtdI(G◦H) ≤ 3n. To get the other inequality, first, suppose that H has an isolated vertex, say x. For each v ∈ V (G), let xv denote the isolated vertex of Hv being identified with x. Let f = (V0, V1, V2, V3) be a γtdI -function of G ◦H. Since xv is an endvertex in Hv + v, f(V (Hv + v)) ≥ f(v) + f(xv) ≥ 3 for all v ∈ V (G). This yields γtdI(G ◦H) ≥ ∑ v∈V (G) f(V (Hv + v) ≥ 3n. Next, suppose that H has no isolated vertices, and let f = (V0, V1, V2, V3) be a γtdI - function of G ◦H. Let v ∈ V (G). Clearly, if v ∈ V3, then f(V (Hv + v)) ≥ 3. If v ∈ V0, then f |Hv ∈ TDIDF (Hv) by Proposition 9 (i). Thus, f(V (Hv + v)) = f |Hv(V (Hv)) ≥ 3. Suppose that v ∈ V1 ∪ V2. If V0 ∩ V (Hv) = ∅, then since |V (Hv)| ≥ 2, f(V (Hv)) ≥ 2 so S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 11 of 18 that f(V (Hv + v)) = f(v) + f(V (Hv)) ≥ 3. On the other hand, if V0 ∩ V (Hv) ̸= ∅, say u ∈ V0 ∩ V (Hv), then Proposition 9(ii) and Proposition 9(iii) yield f(V (Hv + v)) ≥ f(v) + f(NHv [u]) ≥ 3. Therefore, γtdI(G ◦H) = ωG◦H(f) = ∑ v∈V (G) f(V (Hv + v)) ≥ 3n. ■ 5. On the edge corona of graphs We adapt the following notations from [21]. Given graphs G and H, we write Huv to denote the copy of H that is being joined with the end vertices of the edge uv ∈ E(G) in the edge corona G ⋄ H. Moreover, we denote by Huv + uv the subgraph of G ⋄ H corresponding to the join Huv + ⟨u, v⟩, u, v ∈ V (G). For f = (V0, V1, V2, V3) on V (G ⋄H), we write for each i, j ∈ {0, 1, 2, 3}, Eij = {uv ∈ E(G) : either u ∈ Vi and v ∈ Vj or u ∈ Vj and v ∈ Vi}. Proposition 10. Let G be nontrivial connected graph and H be any graph without isolated vertices. Let f = (V0, V1, V2, V3) be a function on V (G). Then f ∈ TDIDF (G ⋄H) if and only if each of the following holds: (i) For each uv ∈ E00, f |Huv ∈ TDIDF (Huv); (ii) For each uv ∈ E01, f(NHuv [w]) ≥ 2 for all w ∈ (V0 ∪ V1) ∩ V (Huv); (iii) For each uv ∈ E11 ∪ E02, f(NHuv [w]) ≥ 1 for all w ∈ V0 ∩ V (Huv); (iv) For each uv ∈ E03 with v ∈ V3 and NG(v) ⊆ V0, we have V (Huv) \ V0 ̸= ∅. Proof : Since H has no isolated vertices, Huv admits a TDIDF for each uv ∈ E(G). If f ∈ TDIDF (G ⋄ H), then properties (i)-(iii) follow immediately from the fact that for each uv ∈ E(G), f(NG⋄H [w]) = f(u) + f(v) + f(NHuv [w]) for all w ∈ V (Huv). While property (iv) is clear from the definition of f . Conversely, suppose that conditions (i)-(iv) hold for f . Let w ∈ V0 and uv ∈ E(G) for which w ∈ V (Huv + uv). Clearly, if uv ∈ [E03 ∪ E12 ∪ E13 ∪ E22 ∪ E23 ∪ E33], then f(NG⋄H [w]) ≥ 3. We proceed with the following cases: Case 1: Suppose that uv ∈ E00. Then f |Huv ∈ DIDF (Huv) by (i). If w = u or w = v, then f(NG⋄H [w]) ≥ f |Huv(V (Huv)) ≥ 3. If u ̸= w ̸= v, then f(NG⋄H [w]) = f |Huv(NHuv [w]) ≥ 3. S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 12 of 18 Case 2: Suppose that uv ∈ E01 ∪ E02, and assume u ∈ V0. If V0 ∩ V (Huv) = ∅, then w = u and since Huv has no isolated vertices, f(V (Huv) ≥ 2. Thus, f(NG⋄H [w]) ≥ f(v) + f(V (Huv)) ≥ 3. Suppose that V0 ∩ V (Huv) ̸= ∅, say z ∈ V0 ∩ V (Huv). If w = u, then by (ii) and (iii), f(NG⋄H [w]) ≥ f(v) + f(NHuv [z]) ≥ 3. If w ∈ V (Huv), then f(NG⋄H [w]) ≥ f(v) + f(NHuv [w]) ≥ 3. Case 3: Suppose that uv ∈ E11. Then w ∈ V (Huv) and by (iii), f(NG⋄H [w]) ≥ f(u) + f(v) + f(NHuv)[w]) ≥ 3. Following similar arguments, f(NG⋄H [w]) ≥ 3 for all w ∈ V1. Finally, let v ∈ V1 ∪ V2 ∪ V3. First, suppose that v ∈ V (G). If NG(v) ⊈ V0, then there exists x ∈ V1 ∪ V2 ∪ V3 such that xv ∈ E(G ⋄ H). Suppose that NG(v) ⊆ V0, and let u ∈ NG(v). Then uv ∈ E01 ∪ E02 ∪ E03. In view of conditions (ii)-(iv), f(V (Huv) ≥ 1. Thus, there exists u ∈ [(V1 ∪ V2 ∪ V3) ∩ V (Huv)] for which uv ∈ E(G ⋄H). Next, suppose that v ∈ V (Hxy) for some xy ∈ E(G). If x /∈ V0, then x ∈ V1 ∪ V2 ∪ V3 with xv ∈ E(G ⋄ H). Similarly, if y /∈ V0, then y ∈ V1 ∪ V2 ∪ V3 with yv ∈ E(G ⋄ H). Suppose that xy ∈ E00. By (i), f |Hxy ∈ TDIDF (Huv) so that there exists u ∈ V (Hxy) for which f(u) = f |Hxy(u) > 0 and uv ∈ E(G ⋄H). The argument above implies that f ∈ TDIDF (G ⋄H). ■ For the purpose of the next result, we define for any D ⊆ V (G), E(G,D) = {uv ∈ E(G) : both u, v /∈ D}. Let G = P5 = [v1, v2, v3, v4, v5]. If S1 = {v1, v4} and S2 = {v1, v2}, then E(G,S1) = ∅ and E(G,S2) = {v3v4, v4v5}. Corollary 4. Let G be a nontrivial connected graph of order n. Then for any graph H, 3 ≤ γtdI(G ⋄H) ≤ n+ β(G). Moreover, if H has no isolated vertices, then 3 ≤ γtdI(G ⋄H) ≤ min{n+ β(G), θ(G ⋄H)} where θ(G ⋄H) = 3γt(G) + γtdI(H)min{|E(G,S)| : S is a γt-set of G}, and these bounds are sharp. S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 13 of 18 Proof : The lower bound follows immediately from Proposition 1. Let S ⊆ V (G) be a β-set of G. Then f = (V0, V1, V2, V3) ∈ TDIDF (G ⋄ H), where V0 = ∪uv∈E(G)V (Huv), V1 = V (G) \ S, V2 = S and V3 = ∅. Thus, γtdI(G ⋄H) ≤ 2|S|+ n− |S| = n+ β(G). Assume thatH has no isolated vertices. ThenHuv admits a TDIDF . Let S be a γt-set of G. Suppose fuv = (V uv 0 , V uv 1 , V uv 2 , V uv 3 ) is a γtdI -function of Huv for each uv ∈ E(G,S). Define f = (V0, V1, V2, V3) on G ⋄H, where V0 = (V (G) \ S) ∪ ( ∪uv∈E(G,S)V uv 0 ) ∪ ( ∪uv∈E(G)\E(G,S)V (Huv) ) , V1 = ∪uv∈E(G,S)V uv 1 , V2 = ∪uv∈E(G,S)V uv 2 , and V3 = S ∪ ( ∪uv∈E(G,S)V uv 3 ) . Note that E01 = E02 = E11 = ∅ and (V1 ∪ V2) ∩ V (G) = ∅. Moreover, V3 ∩ V (G) = S is a γt-set of G so that NG(x) ⊈ V0 for each x ∈ V3 ∩ V (G). Hence, we only need to satisfy condition (i) in Proposition 10. Let uv ∈ E00. Then f |Huv = fuv ∈ TDIDF (Huv). THus, f ∈ TDIDF (G ⋄ H) with ωG⋄H(g) = 3γ(G) + γtdI(H)|E(G,S)|. It follows that, γtdI(G ⋄H) ≤ θ(G ⋄H). For the sharpness, note first that for the left-hand side, γtdI(P2 ⋄ H) = 3 for any H. For the right-hand side, consider the following graphs. If G = C4 of order n = 4, then for any H, γtdI(G ⋄ H) = 6 = n + β(G). On the other hand, if G is the graph in Figure 2, then for any H with γtdI(H) = 3, γtdI(G ⋄H) = 15 = θ(G ⋄H). ■ Figure 2: Example of a graph G for which γtdI(G ⋄H) = θ(G ⋄H) Strict inequality in Corollary 4 may also be attained. Consider, for example, the star graph G = K1,7. For any graph H without isolated vertices, γtdI(G ⋄H) = 4 < 6 = min{n+ β(G), θ(G ⋄H)}. 6. On the complementary prism of graphs Proposition 11. Let G be any graph. Then S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 14 of 18 (i) γtdI(GG) = 3 if and only if G = K1; (ii) γtdI(GG) ≥ 6 whenever G is nontrivial, and this lower bound is sharp. Proof : Statement (i) follows immediately from Proposition 1. Assume G is nontrivial. Suppose that γtdI(GG) = 4. Then γ(GG) ̸= 1. By Proposition 3, γ(GG) ≥ 2 and GG has a 3-dominating set D with |D| = 4 and δ(⟨D⟩) ≥ 2. If |D ∩ V (G)| ≥ 3, then there exists u ∈ D ∩ V (G) for which u /∈ D. For this u, |D ∩NGG(u)| ≤ 2, a contradiction. A similar contradiction is attained if |D ∩ V (G)| ≥ 3. Suppose that |D ∩ V (G)| = 2 = |D ∩ V (G)|. Since δ(⟨D⟩) ≥ 2, ⟨D⟩ = C4, which is impossible. Thus, γtdI(GG) ̸= 4. Suppose that γtdI(GG) = 5. In view of Proposition 4, it suffices to consider the following cases: Case 1: GG has a 3-dominating set D with |D| = 5 and δ(⟨D⟩) ≥ 2. If D ⊆ V (G) and u ∈ D, then |D ∩ NGG(u)| ≤ 1, a contradiction. Similarly, a contradiction is attained if D ⊆ V (G). Since δ(⟨D⟩) ≥ 2, |D ∩V (G)| ≠ 4 and |D ∩V (G)| ≠ 4. Assume, WLOG, that |D ∩ V (G)| = 3 and |D ∩ V (G)| = 2. Since δ(⟨D⟩) ≥ 2, there exist u, v ∈ D ∩ V (G) such that D ∩ V (G) = {u, v} and u v ∈ E(G). If w ∈ (D ∩ V (G)) \ {u, v}, then w /∈ D and |D ∩NGG(w)| = 1, a contradiction. Case 2: GG has a 2-dominating set D with |D| = 3 and ⟨D⟩ is connected. Following a similar argument, D ∩ V (G) ̸= ∅ and D ∩ V (G) ̸= ∅. Assume |D ∩ V (G)| = 2 and |D ∩ V (G)| = 1, say D ∩ V (G) = {u, v} and D ∩ V (G) = {w}. Since ⟨D⟩ is connected, either u = w and D∩NGG(v) = {v} or v = w and D∩NGG(u) = {u}, which is impossible. Case 3: GG contains a 2-dominating set D with |D| = 4 such that there exists v ∈ D for which uv ∈ E(GG) for all u ∈ V (GG) \ D with |NGG(u) ∩ D| = 2. Moreover, ⟨D⟩ is connected and xv ∈ E(GG) for every x ∈ D \ {v} with |NGG(x) ∩ D| = 1. If |D ∩ V (G)| ≥ 3, then since ⟨D⟩ is connected, there exists u ∈ D ∩ V (G) such that u /∈ D and |NGG(u) ∩ D| = 1, a contradiction. Thus, |D ∩ V (G)| = 2 and |D ∩ V (G)| = 2. Assume, v ∈ D ∩ V (G). Let D ∩ V (G) = {x, v}. Since ⟨D⟩ is connected, D = {x, v, x, v}, xv /∈ E(GG), and NGG(x) ∩D = {x}. This is a contradiction. The above contradictions imply that γtdI(GG) ̸= 5. Finally, observe that if G = P2, then γtdI(GG) = γtdI(P4) = 6, showing that the bound provided in (ii) is sharp. ■ The following proposition is clear. Proposition 12. Let G be a nontrivial connected graph. Then f = (V0, V1, V2, V3) is a TDIDF on GG if and only if each of the following holds: (i) For each v ∈ (V0 ∪ V1) ∩ V (G), either f |G(NG[v]) ≥ 3 or f |G(NG[v]) < 3 and 3− f |G(NG[v]) ≤ f(v); (ii) For each v ∈ (V0 ∪ V1) ∩ V (G), either f |G(NG[v]) ≥ 3 or f |G(NG[v]) < 3 and 3− f |G(NG[v]) ≤ f(v); (iii) For each v ∈ V (G) ∩ (V1 ∪ V2 ∪ V3), v /∈ V0 whenever NG(v) ⊆ V0; S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 15 of 18 (iv) For each v ∈ V (G) ∩ (V1 ∪ V2 ∪ V3), v /∈ V0 whenever NG(v) ⊆ V0. 2 v4 1 v1 1 v2 1 v3 1 v1 1 v3 1 u1 1 u2 1 u3 1 u4 1 u2 1 u1 1 u4 1 u3 0v2 0 v4 Figure 3: The complementary prisms P4P 4 and C4C4 Proposition 13. Let G be a nontrivial connected graph of order n. Then 6 ≤ γtdI(GG) ≤ 3n. (4) Moreover, if γ(G) ̸= 1 and γ(G) ̸= 1, then 1 + max{γtdI(G), γtdI(G)} ≤ γtdI(GG) ≤ 2n. (5) These bounds are sharp. Proof : The left-hand inequality in Inequality 4 is a reiteration of Proposition 11. By Propo- sition 12, the function f = (V (G),∅,∅, V (G)) ∈ TDIDF (GG). Thus, γtdI(GG) ≤ 3|V3| = 3n and the Inequality 4 holds. Now, suppose that γ(G) ̸= 1 and γ(G) ̸= 1. Let V0 = ∅ = V2 = V3 and V1 = V (G) ∪ V (G), and let f = (V0, V1, V2, V3). For each v ∈ V (G), since v is not an isolated vertex, there exists u ∈ V (G) for which uv ∈ E(G). Thus, {v, u} ⊆ NG[v] so that f(NG[v]) ≥ 2. If f(NG[v]) = 2, then 3 − f(NG[v]) = 1 ≤ f(v). Thus, Condi- tion (i) in Proposition 12 holds. Similarly, since G has no isolated vertex, Proposition 12(ii) holds. Since V0 = ∅, Conditions (iii) and (iv) of Proposition 12 also hold. Thus, f ∈ TDIDF (GG). Therefore, γtdI(GG) ≤ 2n. WLOG assume that γtdI(G) ≥ γtdI(G). Let f be a γtdI -function of GG. If V (G) ⊆ V0, then V (G) ⊆ V3 and ωGG(f) = 3|V (G)|. However, by Theorem 1, γtdI(GG) ≤ |V (GG)|+ 2− δ(GG). Since G and G have no isolated vertices, the least possible value of δ(GG) is 2. Hence, γtdI(GG) ≤ 2|V (G)| < 3|V (G)|, a contradiction. Suppose f |G is a TDIDF onG. Since V (G) ⊈ V0, (V1 ∪ V2 ∪ V3) ∩ V (G) ̸= ∅ and ω(f |G) ≥ 1. Thus, ω(f |G) + 1 ≤ ω(f |G) + ω(f |G) = ωGG(f) S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 16 of 18 Suppose f |G is not a TDIDF on G. Let A = {v ∈ V0 ∩ V (G) : 0 ≤ f(NG(v)) ≤ 1}, B = {v ∈ V (G) ∩ V0 : f(NG(v)) = 2}, C = {v ∈ V (G) ∩ V1 : 0 ≤ f(NG(v)) ≤ 1}, D = {v ∈ V (G) ∩ (V2 ∪ V3) : NG(v) ⊆ V0 \B}. Now, let X ⊆ NG(D) ∩ V0 be the smallest set that dominates D. Then |X| ≤ |D|. Define a function g on V (G) as follows: g(x) =  0, if x ∈ (V (G) ∩ V0) \ (A ∪B ∪X); 1, if x ∈ [(V (G) ∩ V1) \ C] ∪B ∪X; 2, if x ∈ (V (G) ∩ V2) ∪A ∪ C; 3, if x ∈ (V (G) ∩ V3), Then g is a TDIDF on G with ωG(g) = |V (G) ∩ V1| − |C|+ |B ∪X|+ 2|V (G) ∩ V2|+ 2|A|+ 2|C| + 3|V (G) ∩ V3| = |V (G) ∩ V1|+ 2|V (G) ∩ V2|+ 3|V (G) ∩ V3|+ 2|A|+ |C|+ |B ∪X| = ωG(f |G) + 2|A|+ |C|+ |B ∪X|. We claim that ωG(f |G) − 1 ≤ ωGG(f). Put A1 = {v ∈ V0 ∩ V (G) : f(NG(v)) = 0}, A2 = {v ∈ V0 ∩ V (G) : f(NG(v)) = 1}, C1 = {v ∈ V (G) ∩ V1 : f(NG(v)) = 0}, and C2 = {v ∈ V (G) ∩ V1 : f(NG(v)) = 1}. Then A1 ∪ A2 = A and C1 ∪ C2 = C. Now, we denote by S = {v ∈ V (G) : v ∈ S} for each S ⊆ V (G). Note that, A1 ⊆ V (G) ∩ V3, A2 ⊆ V (G)∩(V2∪V3), B ⊆ V (G)∩(V1∪V2∪V3), C1 ⊆ V (G)∩(V2∪V3), C2 ⊆ V (G)∩(V1∪V2∪V3), and D ⊆ V (G) ∩ (V1 ∪ V2 ∪ V3). WLOG, assume that A2 ⊆ V (G) ∩ V2, B ⊆ V (G) ∩ V1, C1 ⊆ V (G) ∩ V2, C2 ⊆ V (G) ∩ V1, and D ⊆ V (G) ∩ V1. Then C2 ∩D = ∅, C2 ∩ B = ∅ and A2 ∩ C1 = ∅. Thus, ωG(f |G) = |V (G) ∩ V1|+ 2|V (G) ∩ V2|+ 3|V (G) ∩ V3| ≥ 3|A1|+ 2|A2|++2|C1|+ |C2|+ |B ∪D| = 2|A|+ |C|+ |A1|+ |C1|+ |B ∪D|. Since |X| ≤ |D|, ωG(f |G) ≥ 2|A|+ |C|+ |B ∪X|+ 1. Hence, γtdI(GG) = ωGG(f) = ωG(f |G) + ωG(f |G) ≥ ωG(f |G) + 2|A|+ |C|+ |B ∪X|+ 1 = ωG(g) + 1. Therefore, γtdI(GG) ≥ γtdI(G) + 1. If G = Kn, then γtdI(GG) = 3n. If G = P4, then G = P4 and γtdI(GG) = 1 + max{γtdI(G), γtdI(G)}. And if G = Cn on n = 4 vertices, then γtdI(GG) = 2n. Therefore, the inequalities in Inequality 4 and Inequality 5 are sharp. ■ S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 17 of 18 Acknowledgements This research is fully funded by the Department of Science and Technology (DOST) under the Accelerated Science and Technology Human Resource Development Program (ASTHRDP) and the Office of the Vice Chancellor for Research and Enterprise (OVCRE), MSU-Iigan Institute of Technology, Philippines, through the Premier Research Institute of Science and Mathematics (PRISM). References [1] R. J. Fortosa, S. R. Canoy, and F. P. Jamil. Convex Roman dominating functions on graphs under some binary operations. European Journal of Pure and Applied Mathematics, 17(2):1335–1351, 2024. [2] J. B. Cariaga and F. P. Jamil. On double Roman dominating functions in graphs. European Journal of Pure and Applied Mathematics, 16(2):847–863, 2023. [3] G. Chartrand and L. Lesniak. Graphs and Digraphs: Third Edition. Chapman and Hall, London, 1996. [4] G. Chartrand and O. R. Oellermann. Applied and Algorithmic Graph Theory. McGraw-Hill, London, 1996. [5] A. Chilelli and J. Jun. Gluing of graphs and their Jacobians. Involve, 16(3):389–407, 2023. [6] E. Cockayne and S. Hedetniemi. Towards a theory of domination in graphs. Networks, 7(3):247–261, 1977. [7] B. Escoffier, L. Gourves, and J. Monnot. Complexity and approximation results for the connected vertex cover problem in graphs and hypergraphs. Journal of Discrete Algorithms, 8(1):36–49, 2010. [8] M. Chellali, O. Favaron, A. Hansberg, and L. Volkmann. k-Domination and k- independence in graphs: A Survey. Graphs and Combinatorics, 28:1–55, 2012. [9] F. Harary. Graph Theory. Massachusetts: Addison-Wesley Publication Company, USA, 1969. [10] R. A. Beeler, T. W. Haynes, and S. T. Hedetniemi. Double roman domination. Discrete Applied Mathematics, 211:23–29, 2016. [11] W. J. Desormeaux, T. W. Haynes, and M. A. Henning. An extremal problem for total domination stable graphs upon edge removal. Discrete Applied Mathematics, 159:1048–1052, 2011. [12] E. J. Cockayne, P. A. Dreyer, S. M. Hedetniemi, and S. T. Hedetniemi. Roman domination in graphs. Discrete Mathematics, 278:11–22, 2004. [13] M. Chellali, T. W. Haynes, S. T. Hedetniemi, and A. A. McRae. Roman {2}- domination. Discrete Applied Mathematics, 211:22–28, 2016. [14] T. W. Haynes, S. T. Hedetniemi, and P. J. Slater. Fundamentals of Domination in Graphs. Marcel Dekker, Inc., New York, 1998. [15] M. Henning and A. Yeo. Total Domination in Graphs. Springer, 2013. S.J.L. Sumbalan, S.M. Menchavez, F.P. Jamil / Eur. J. Pure Appl. Math, 18 (1) (2025), 5602 18 of 18 [16] S. R. Canoy, F. P. Jamil, and S. M. Menchavez. Hop Italian domination in graphs. European Journal of Pure and Applied Mathematics, 16(4):2431–2449, 2023. [17] A. Mohannad and D. A. Mojdeh. On the total restrained double Italian domination. Journal of Algebra and Related Topics, 12(1):105–126, 2024. [18] A. Mohannad and D. A. Mojdeh. On the (total) restrained double Italian dom- ination of central of graphs. Discrete Mathematics, Algorithms and Applications, 16(8):2350102, 2024. [19] D. A. Mojdeh and L. Volkmann. Roman {3}-domination (double Italian domination). Discrete Applied Mathematics, 283:555–564, 2020. [20] P. Dankelmann, D. Day, D. Erwin, S. Mukwembi, and H. Swart. Domination with exponential decay. Discrete Mathematics, 309:5877 – 5883, 2009. [21] L. Paleta and F. Jamil. More on perfect Roman domination in graphs. European Journal of Pure and Applied Mathematics, 13(3):529–548, 2020. [22] L. Paleta and F. Jamil. On perfect Italian domination in graphs. Discrete Mathe- matics, Algorithms and Applications, 16, 2023. [23] P. Roushini Leely Pushpam and C. Suseendran. Secure vertex cover of a graph. Discrete Mathematics, Algorithms and Applications, 9(2):1750026, 2017. [24] J. M. Rivera and F. Jamil. Total Roman domination in the join, corona and com- plementary prism of graphs. Asia-Pacific Journal of Science, Mathematics and En- gineering, 7(2):37–48, 2021. [25] A. Khodkar, D. A. Mojdeh, B. Samadi, and I. G. Yero. Covering Italian domination in graphs. Discrete Applied Mathematics, 304:324–331, 2021. [26] D. A. Mojdeh Z. Shao. Total Roman 3-domination in graphs. Symmetry, 12(2):268, 2020.