EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 5984 ISSN 1307-5543 – ejpam.com Published by New York Business Global Independent Double Roman Domination Stability in Graphs Seyed Mahmoud Sheikholeslami1,∗, Mina Esmaeili1, Jamil J. Hamja2,5, Cris L. Armada3,4, Imelda S. Aniversario5 1 Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, I.R. Iran 2 Department of Mathematics, College of Arts and Sciences, MSU - Tawi-Tawi College of Technology and Oceanography, 7500 Tawi-Tawi, Philippines 3 Vietnam National University Ho Chi Minh City, Linh Trung Ward, Thu Duc City, Ho Chi Minh City, Vietnam 4 Department of Applied Mathematics, Faculty of Applied Science, Ho Chi Minh City University of Technology (HCMUT), 268 Ly Thuong Kiet, District 10, Ward 14, Ho Chi Minh City, Vietnam 5 Department of Mathematics and Statistics, College of Science and Mathematics, MSU - Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. An independent double Roman dominating function (IDRD-function) on a graph G is a function f : V (G) → {0, 1, 2, 3} having the property that (i) if f(v) = 0, then the vertex v must have at least two neighbors assigned 2 under f or one neighbor w with f(w) = 3, and if f(v) = 1, then the vertex v must have at least one neighbor w with f(w) ≥ 2, and (ii) the subgraph induced by the vertices with positive weight under f is edgeless. The weight of an IDRD-function is the sum of its function values over all vertices, and the independent double Roman domination number (IDRD-number) idR(G) is the minimum weight of an IDRD-function on G. The idR-stability (i−dR- stability, i+dR-stability) of G, denoted by stidR(G) (st−idR(G), st+idR(G)), is defined as the minimum size of a set of vertices whose removal changes (decreases, increases) the independent double Roman domination number. In this paper, we first determine the exact values on the idR-stability of some special classes of graphs, and then present some bounds on stidR(G). In addition, for a tree T with maximum degree ∆, we show that stidR(T ) = 1 and st−idR(T ) ≤ ∆, and characterize the trees that achieve the upper bound. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Roman domination, double Roman domination, independent Ro- man domination, independent double Roman domination, independent double Roman domination stability ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5984 Email addresses: s.m.sheikholeslami@azaruniv.ac.ir (S. M. Sheikholeslami), minaesmaeeli1999@gmail.com (M. Esmaeili), jamilhamja@msutawi-tawi.edu.ph (J. J. Hamja), cris.armada@hcmut.edu.vn (C. L. Armada), imelda.aniversario@g.msuiit.edu.ph (I. S. Aniversario) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) S. M. Sheikholeslami et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5984 2 of 16 1. Introduction Roman domination, introduced by Cockayne et al. [1] in 2004 and inspired by earlier work of Stewart [2] and ReVelle et al. [3], has since seen various extensions. One notable variant is the [k]-Roman domination [4], which generalizes the original Roman domination number γR(G), recovered when k = 1 [5–7]. Beeler et al. [8] further refined this concept through double Roman domination, establishing bounds and complexity results, later ex- panded by Abdollahzadeh Ahangar et al. [9]. To incorporate independence, Maimani et al. [10] introduced independent double Roman domination, providing bounds and relation- ships to related parameters, and inspiring studies on I[k]RD for k = 1, 2 [11–14]. The concept of domination stability in graphs was introduced by Bauer et al. [15] in 1983. This notion has since been extended to various domination parameters. In 2016, Rad et al. [16] advanced this line of research by proving that the γ-stability problem is NP-hard, even when restricted to bipartite graphs. Continuing this direction, Zhuang et al. [17] determined the exact values of the γdR-stability number for several special classes of graphs and established bounds on stγdR(G). In particular, for a tree T with maximum de- gree ∆, they showed that st−γdR(T ) ≤ ∆, and characterized the trees that attain this bound. Motivated by these developments, this paper explores the independent double Roman domination stability of graphs. We determine exact values for special graph classes, es- tablish bounds on stidR(G), and characterize extremal cases. For trees, we show that stidR(T ) = 1 and st−idR(T ) ≤ ∆, fully characterizing trees attaining this bound. 2. Terminology and Notation All graphs considered in this article are finite, undirected, and simple. Let G = (V,E) be a graph of order |V (G)| = n. For any vertex v ∈ V (G), the open neighborhood of v is the set N(v) = {u ∈ V | uv ∈ E(G)} and the closed neighborhood of v is the set N [v] = N(v) ∪ {v}. We denote the degree of a vertex v in a graph G by degG(v), or simply by deg(v) if the graph G is clear from the context. Let δ(G) and ∆(G) denote the minimum and maximum degrees, respectively, of vertices in G. We call a vertex of degree one a leaf, and its (unique) neighbor a support vertex. A support vertex is said to be strong if it has at least two leaf neighbors, otherwise, it is called weak. A complete graph on n vertices is denoted by Kn, while a complete bipartite graph with partite sets of size p and q is denoted by Kp,q. We write Pn for the path of order n, Cn for the cycle of length n, and Kn for the graph with n vertices and no edges. The distance dG(u, v) between two vertices u and v in a connected graph G is the length of a shortest u–v path in G, while the diameter, diam(G), is the maximum distance among all pairs of vertices in G. A tree is an acyclic connected graph. A star is the graph K1,m, where m ≥ 1; the vertex of degree m is called the center of the star. A double star Sr,t is formed from two disjoint stars K1,r and K1,t by adding an edge joining their center vertices. A S. M. Sheikholeslami et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5984 3 of 16 rooted tree T distinguishes one vertex r, called the root. For each vertex v ̸= r in T , the parent of v is the neighbor of v on the unique r–v path, while a child of v is any other neighbor of v. A descendant of v is a vertex u ̸= v such that the unique r–u path contains v. Thus, every child of v is a descendant of v. Let D(v) denote the set of descendants of v, and let D[v] = D(v) ∪ {v}. The depth of v, denoted depth(v), is the largest distance from v to a vertex in D(v). The maximal subtree at u is the subtree of T induced by D[u], and is denoted by Tu. Let k be a positive integer, and let f : V (G) → {0, 1, 2, . . . , k + 1} be a function that assigns labels from the set {0, 1, . . . , k+1} to the vertices of a graph G. The active neighbor- hood AN(v) of a vertex v ∈ V (G) with respect to f is the set of all vertices w ∈ N(v) such that f(w) ≥ 1. Let AN [v] = {v}∪AN(v). A [k]-Roman dominating function, abbreviated as [k]RDF, is a function f : V (G) → {0, 1, . . . , k+ 1} satisfying the condition that for any vertex v ∈ V (G) with f(v) < k, it holds that ∑ u∈N [v] f(u) ≥ |AN(v)| + k. The weight of a [k]RDF is defined as ω(f) = ∑ v∈V (G) f(v), and the [k]-Roman domination number γ[kR](G) of G is the minimum weight of a [k]RDF on G. A function f achieving this min- imum is called a γ[kR](G)-function. For a [k]RDF f on G, let V f i = {v ∈ V (G) | f(v) = i} for all i ∈ {0, 1, . . . , k + 1}. Consequently, any [k]RDF f can be represented by the tuple (V f 0 , V f 1 , . . . , V f k+1), where the superscript f may be omitted from V f i when no confusion arises. A double Roman dominating function (DRDF) on a graph G = (V,E) is a function f : V → {0, 1, 2, 3} having the property that if f(v) = 0, then the vertex v must have at least two neighbors assigned 2 under f or one neighbor u with f(u) = 3, and if f(v) = 1, then the vertex v must have at least one neighbor u with f(u) ≥ 2. The weight of a DRDF is the sum of its function values over all vertices, and the double Roman dom- ination number γdR(G) is the minimum weight of a DRDF on G. The double Roman domination stability, or just γdR-stability, of a graph G is the minimum size of a set of vertices whose removal changes the double Roman domination number. We denote the γdR-stability of G by stγdR(G). The decreasing γdR-stability of G, denoted by st−γdR(G), is defined as the minimum size of a set of vertices whose removal decreases the double Roman domination number. For the null graph N0, which is the unique graph having no vertices and hence has order zero, we let st−γdR(N0) = 0. With this consideration, the decreasing γdR-stability of a non-null graph is always defined. For example, st−γdR(K1) = 1. The increasing γdR-stability of G, denoted by st+γdR(G), is defined as the minimum size of a set of vertices whose removal increases the double Roman domination number, if such a set exists. Clearly, stγdR(G) = min { st−γdR(G), st+γdR(G) } . An independent [k]-Roman dominating function, abbreviated as I[k]RDF, is a [k]- Roman dominating function f such that the subgraph induced by the vertices with positive weight under f is edgeless. The minimum weight of an I[k]RDF on a graph G is called the independent [k]-Roman domination number, denoted by i[k]R(G), which we refer to as the S. M. Sheikholeslami et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5984 4 of 16 I[k]RD-number. By definition, we have γ[k]R(G) ≤ i[k]R(G). (1) An independent double Roman dominating function (IDRD-function) on a graph G is a function f : V (G) → {0, 1, 2, 3} having the properties that (i) if f(v) = 0, then the vertex v must have at least two neighbors assigned 2 under f , or one neighbor w with f(w) = 3, and if f(v) = 1, then the vertex v must have at least one neighbor w with f(w) ≥ 2; and (ii) the subgraph induced by the vertices with positive weight under f is edgeless. The weight of an IDRD-function is the sum of its function values over all vertices, and the independent double Roman domination stability, or simply the idR-stability, of a graph G is the minimum size of a set of vertices whose removal changes the independent double Roman domination number. We denote the idR-stability of G by stidR(G). The i−dR- stability of G, denoted by st−idR(G), is defined as the minimum size of a set of vertices whose removal decreases the independent double Roman domination number, and the i+dR-stability of G, denoted by st+idR(G), is defined as the minimum size of a set of vertices whose removal increases the independent double Roman domination number, if such a set exists. If there is no set of vertices in G whose removal increases the independent double Roman domination number, then we set st+idR(G) = ∞. Clearly, stidR(G) = min{st−idR(G), st+idR(G)}. 3. Preliminary Results In this section we will investigate simple results. Remark 1. Let G be a nontrivial connected graph with γdR(G) = idR(G). Then st−γdR(G) ≤ st−idR(G). Moreover, If st+γdR(G) < ∞, then st+idR(G) ≤ st+γdR(G). Maimani et al. [10] observed that for any graph G and any idR(G)-function f = (V0, V1, V2, V3) we have V1 = ∅. Proposition 1. Let G be a graph and v be a vertex of G. If G′ is obtained from G by adding a star K1,t with t ≥ 2 and joining v to a leaf of K1,t, then idR(G ′) = idR(G) + 3. Proof. Let V (K1,t) = {u, u1, u2, . . . , ut}, where u is the center of the star and u1, . . . , ut are its leaves. Suppose v ∈ V (G) is connected to the leaf u1 in G′, i.e., vu1 ∈ E(G′). First, we show that idR(G′) ≤ idR(G)+3. Let f be an IDRD-function on G of minimum weight, i.e., an idR(G)-function. Define an extension f ′ on G′ by: f ′(x) =  f(x), if x ∈ V (G), 3, if x = u, 0, if x ∈ {u1, u2, . . . , ut}. Clearly, f ′ is a valid IDRD-function on G′, and its weight is idR(G) + 3. Thus, idR(G′) ≤ idR(G) + 3. S. M. Sheikholeslami et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5984 5 of 16 Now we show that idR(G′) ≥ idR(G)+3. Let f = (V0,∅, V2, V3) be an idR(G ′)-function. If f(u1) = 0, then to double Roman dominate the vertices u and u2, . . . ut we must have f(u) + f(u2) + · · · + f(ut) = 3. On the other hand, the function f restricted to G is an IDRD-function of G implying that idR(G ′) ≥ idR(G) + 3. Assume that f(u1) ≥ 2. Then we must have f(u) = 0 and f(ui) = 2 for 2 ≤ i ≤ t. If f(u1) = 2, then to double Roman dominate of v, it must have a neighbor w with f(w) ≥ 2 and the function g defined on G by g(w) = min{3, f(w) + 1} and g(x) = f(x) for other vertices, is an IDRD-function of G of weight at most idR(G′)−3 leading to idR(G ′) ≥ idR(G)+3. Assume that f(u1) = 3. If v has a neighbor assigned at least two under f , then as before we get idR(G′) ≥ idR(G)+3. Hence let f(N [v] − {u1}) = 0. Then the function g defined on G by g(v) = 2 and g(x) = f(x) for other vertices, is an IDRD-function of G of weight at most idR(G ′) − 3, leading to idR(G ′) ≥ idR(G)+ 3. Thus, in all cases, we have that idR(G′) ≥ idR(G)+ 3, and since the reverse inequality was already established, we have idR(G ′) = idR(G) + 3. 4. Exact values and bounds In this section, we obtain the independent double Roman domination stability for some classes of graphs and present various bounds for this parameters. The proof of the next Propositions can be found in [10]. Proposition 2. For n ≥ 1, idR(Pn) = γdR(Pn) = { n if n ≡ 0 (mod 3) n+ 1 if n ≡ 1, 2 (mod 3). Proposition 3. For n ≥ 3, idR(Cn) = γdR(Cn) = { n if n ≡ 0, 2, 3, 4 (mod 6) n+ 1 if n ≡ 1, 5 (mod 6). Also Zhuang [17] determined the double Roman domination stability of paths and cycles as follows. Proposition 4. For n ≥ 2, st−γdR(Pn) = { 1 if n ≡ 1, 2 (mod 3) 2 if n ≡ 0 (mod 3). Proposition 5. For n ≥ 2, st+γdR(Pn) = { ∞ if n ≡ 1, 2 (mod 3) 1 if n ≡ 0 (mod 3). Proposition 6. For n ≥ 3, st−γdR(Cn) = { 1 if n ≡ 1, 4, 5 (mod 6) 2 otherwise. Proposition 7. For n ≥ 3, st+γdR(Cn) = ∞. We first determine the idR-stability for paths. Proposition 8. For n ≥ 2, st−idR(Pn) = { 1 if n ≡ 1, 2 (mod 3) 2 if n ≡ 0 (mod 3). S. M. Sheikholeslami et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5984 6 of 16 Proof. The result is trivial for n ≤ 3, so we assume that n ≥ 4. Let P = [v1, v2, . . . , vn] be a path on n vertices. If n ≡ 1, 2 (mod 3), then using Proposition 2 we have idR(Pn − vn) = idR(Pn−1) ≤ n < idR(Pn). Hence, st−idR(Pn) = 1. Now assume that n ≡ 0 (mod 3). First we show that st−idR(Pn) ≥ 2. By Proposition 4, we have st−γdR(Pn) = 2. It follows from Remark 1 that st−idR(Pn) ≥ 2. On the other hand, by Proposition 2 we have idR(Pn− {vn, vn−1}) = idR(Pn−2) = n− 1 < idR(Pn) that yields st−idR(Pn) ≤ 2. Thus st−idR(Pn) = 2 and the proof is complete. Proposition 9. For n ≥ 2, st+idR(Pn) = { ∞ if n ≡ 1, 2 (mod 3) 1 if n ≡ 0 (mod 3). Proof. The result is trivial for n = 2, so we consider the case for n ≥ 3. Let P = [v1, v2, . . . , vn] be a path on n vertices. If n ≡ 0 (mod 3), then by Proposition 2, we have idR(Pn − v2) = idR(P1) + idR(Pn−2) = 2 + (n− 2 + 1) = n+ 1 > idR(Pn), and so st+idR(Pn) = 1. Assume that n ≡ r (mod 3) where r ∈ {1, 2}. By contradiction, we may assume that there exists a n′ ≡ r (mod 3) such that st+idR(Pn′) is an integer m. Let S be a set of vertices such that idR(Pn′) < idR(Pn′ − S). Let Pn1 , Pn2 , . . . , Pnk be the components of Pn′ − S. Then by Proposition 2, we have γdR(Pn′) = idR(Pn′) < idR(Pn′ − S) = k∑ i=1 idR(Pni) = k∑ i=1 γdR(Pni) = γdR(Pn′ − S), a contradiction with Proposition 5. Thus st+idR(Pn) = ∞ when n ≡ r (mod 3) where r ∈ {1, 2}. The following result is an immediate consequence of Propositions 8 and 9. Corollary 1. For n ≥ 2, stidR(Pn) = 1. Next we determine the independent double Roman domination stability of cycles. Theorem 1. For n ≥ 3, st−idR(Cn) = { 1 if n ≡ 1, 4, 5 (mod 6) 2 otherwise. Proof. Since idR(Cn) = γdR(Cn) for n ≥ 3 and idR(Pn) = γdR(Pn) for n ≥ 1, we deduce from Proposition 6 that st−idR(Cn) = 1 when n ≡ 1, 4, 5 (mod 6). Assume that n ≡ r (mod 6) where r ∈ {0, 2, 3}. By Proposition 3, we have idR(Cn) = n. It follows from Proposition 6 and Remark 1 that st−idR(Cn) ≥ 2. On the other hand, we note that Cn − {v1, v5} is a disjoint union of two paths P3 and Pn−5 and Proposition 2 leads to idR(Cn − {v1, v5}) = 3 + idR(Pn−5) ≤ 3 + (n− 5) + 1 < n = idR(Cn). Thus st−idR(Cn) = 2 and the proof is complete. Theorem 2. For n ≥ 3, st+idR(Cn) = ∞. S. M. Sheikholeslami et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5984 7 of 16 Proof. By contradiction, we may assume that there exists an integer n′ ≥ 3 such that st+idR(Cn′) is an integer m. Let S be a set of vertices such that idR(Cn′) < idR(Cn′ −S). If Pn1 , Pn2 , . . . , Pnk are the components of Cn′ − S, then by Proposition 3 we have γdR(Cn′) = idR(Cn′) < idR(Cn′ − S) = k∑ i=1 idR(Pni) = k∑ i=1 γdR(Pni) = γdR(Cn′ − S), a contradiction with Proposition 7. Thus st+idR(Cn) = ∞. Corollary 2. For n ≥ 3, stidR(Cn) = st−idR(Cn). One can observe that for n ≥ 2, idR(Kn) = idR(K1,n−1) = 3 and for 1 ≤ r ≤ t, idR(Sr,t) = { 5 if r = 1 3 + 2r if r ≥ 2. From the above Observations, we can easily obtain the following conclusions. Corollary 3. For n ≥ 2, stidR(Kn) = st−idR(Kn) = n− 1 and st+idR(Kn) = ∞. Corollary 4. For n ≥ 3, stidR(K1,n−1) = st+idR(K1,n−1) = 1 and st−idR(K1,n−1) = n− 1. Corollary 5. For 1 ≤ r ≤ t with t ≥ 2, stidR(Sr,t) = st−idR(Sr,t) = 1 and st+idR(Sr,t) = 1 when r < t and st+idR(Sr,t) = 2 when r = t. In what follows we give some bounds for the IDRD-stability of a graph. Since for any graph G of order at least 2, idR(G) ≥ 3 with equality if and only if ∆(G) = n − 1, the proof of this is trivial. Remark 2. If G is a graph of order n ≥ 2, then st−idR(G) ≤ n− 1 with equality if and only if ∆(G) = n− 1. Proposition 10. If G ̸= Kn is a connected graph of order n ≥ 3 with idR(G) ≥ 4, then st−idR(G) ≤ n− ω(G)− 1 where ω(G) is the clique number of G. Proof. Let S be a maximum clique in G. Since G is connected and G ̸= Kn, there exists a vertex x ∈ V (G) \ S such that x is adjacent to a vertex in S, say y. Now the function f defined on G[S∪{x}] by f(x) = 3 and f(z) = 0 for the remaining vertices, is an IDRD-function of G[S ∪ {x}]. This implies that st−idR(G) ≤ n− (|S|+ 1) = n− ω(G)− 1, as desired. Proposition 11. Let G be a graph of order n ≥ 2. Then stidR(G) ≤ δ(G) + 1. In particular, this bound is sharp for graphs with isolated vertices. S. M. Sheikholeslami et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5984 8 of 16 Proof. Let x be a vertex of G with minimum degree δ(G). Assume that G′ = G−N(x) and G′′ = G − N [x] and f is an idR(G)-function. If deg(x) = 0, then f(x) = 2 and we have idR(G ′′) = idR(G) − 2. Thus, stidR(G) ≤ δ(G) + 1. So we assume that deg(x) ≥ 1. If idR(G ′) ̸= idR(G), then stidR(G) ≤ δ(G) < δ(G) + 1. Let idR(G ′) = idR(G) and g is an idR(G ′)-function. Since x is an isolated vertex in G′, g(x) = 2, and we have idR(G ′) = idR(G ′′) + 2, that is, idR(G′′) = idR(G)− 2, and so stidR(G) ≤ δ(G) + 1. Proposition 12. Let G be a connected graph of order n with idR(G) ≥ 4, then stidR(G) ≤ n−∆(G)− 1. Proof. If ∆(G) = 2, then G is the path Pn or a cycle Cn. By Corollary 1 or Corollary 2, we are done. Let ∆(G) ≥ 3 and x be a vertex with maximum degree ∆(G). Since idR(G) ≥ 4, we have X = V (G)−N [x] ̸= ∅. Clearly, the function g defined on G−X by g(x) = 3 and g(y) = 0 for the remaining vertices, is an IDRD-function of G −X, and so stidR(G) ≤ n−∆(G)− 1. Combining Propositions 11 and 12, the following result follows. Corollary 6. Let G be a graph with idR(G) ≥ 4, then stidR(G) ≤ min{δ(G) + 1, n−∆(G)− 1}. 5. Graphs G with large idR-stability In this section we characterize graphs G with stidR(G) ∈ {n− 1, n− 2, n− 3, n− 4}. Proposition 13. Let G be a connected graph of order n ≥ 2. Then stidR(G) = n − 1 if and only if G = Kn. Proof. If G = Kn, then clearly stidR(G) = n− 1. Now, we prove the necessity. Let G be a connected graph of order n ≥ 2 with stidR(G) = n − 1. By Proposition 11, we have n− 1 = stidR(G) ≤ δ(G)+1, that is δ(G) ≥ n− 2. If δ(G) = n− 1, then G is the complete graph Kn, as desired. Assume that δ(G) = n − 2. If idR(G) ≥ 4, then by Proposition 12, we obtain ∆(G) ≤ n − 1 − stidR(G) = 0 contradicting the connectivity of G. Thus, idR(G) = 3. Since δ(G) = n − 2, G has two non-adjacent vertices u and v and it follows from idR(G[{u, v}]) = 4 that n − 1 = stidR(G) ≤ n − 2, a contradiction. This completes the proof. Proposition 14. Let G ̸= Kn be a connected graph of order n ≥ 3. Then stidR(G) = n−2 if and only if G = Kn − e. Proof. If G = Kn − e, then clearly stidR(G) = n − 2. Now, we prove the necessity. Let G ̸= Kn be a connected graph of order n ≥ 3 with stidR(G) = n − 2. If idR(G) ≥ 4, then by Proposition 12, we have ∆(G) ≤ 1 which contradicts the connectivity of G. So idR(G) = 3. If G has two pair of non-adjacent vertices u, v and x, y, then idR(G[x, y, u, v]) ≥ 5 if |{x, y, u, v}| = 3 and idR(G[x, y, u, v]) ≥ 4 if |{x, y, u, v}| = 4. This leads to the contradiction n− 2 = stidR(G) ≤ n− 3. Therefore, by Proposition 13, we are done. S. M. Sheikholeslami et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5984 9 of 16 Proposition 15. Let G be a connected graph of order n ≥ 4 and G ̸∈ {Kn,Kn− e}. Then stidR(G) = n−3 if and only if G ∈ {P4, C4,K1,3,K1,3+ e, 3K1∨Kn−3,Kn−3∨ (K2∪K1)}. Proof. The sufficiency is straightforward to check. To prove the necessity, let G be a connected graph of order n ≥ 4 such that G ̸∈ {Kn,Kn − e} and stidR(G) = n − 3. Obviously, ∆(G) ≥ 2. If stidR(G) = 1, then n = 4. It is easy to verify that G ∈ {P4, C4,K1,3,K1,3 + e} as desired. Hence, we assume that stidR(G) ≥ 2. If idR(G) ≥ 4, then Proposition 12 and our earlier assumption leads to ∆(G) = 2. Combining this with the condition that G is connected, we have that G is a path or a cycle. Since stidR(Pn) = 1, it follows from stidR(G) ≥ 2 that G is a cycle. Combining Corollary 2 and the condition stidR(G) = n − 3, we obtain that n = 5 which is a contradiction. Hence, we assume that idR(G) = 3. It follows from this and the fact stidR(G) = n − 3 that G has exactly n − 3 universal vertices. Let x, y, z be the vertices of G that are not universal. It follows that G[{x, y, z}] = 3K1 or G[{x, y, z}] = K2 ∪ K1. Thus, G = 3K1 ∨ Kn−3 or G = Kn−3 ∨ (K2 ∪K1). This completes the proof. Proposition 16. Let G be a connected graph of order n ≥ 6. Then stidR(G) = n−4 if and only if G ∈ {P6, C6,Kn−4∨H} where H ∈ {P4, C4, 2K2, 4K1,K2∪2K1, P3∪K1,K3∪K1}. Proof. The sufficiency is straightforward to check. To prove the necessity, let G be a connected graph of order n ≥ 6 with stidR(G) = n−4. Clearly ∆(G) ≥ 2 and stidR(G) ≥ 2. First let idR(G) ≥ 4. It follows from Proposition 11 that n− 4 = stidR(G) ≤ δ + 1 and so δ ≥ n− 5. Combining this with Proposition 12, we obtain n− 5 ≤ δ ≤ ∆ ≤ 3. (2) If ∆(G) = 2, then n ∈ {6, 7} and G is a path or a cycle of order n. It follows from Corollaries 1 and 2 that G ∈ {P6, C6}. Henceforth we assume that ∆(G) = 3. By (2) we obtain n ∈ {6, 7, 8}. Let v ∈ V (G) be a vertex with maximum degree 3 with N(v) = {v1, v2, v3} and let f = (V0,∅, V2, V3) be an idR(G)-function. Since G is a connected graph, we assume, without loss of generality that u ∈ N(v1)−N [v]. If idR(G) ≥ 6, then the function g defined on G[N [v] ∪ {u}] with f(v) = 3, f(u) = 2, f(v1) = f(v2) = f(v3) = 0 is an IDRDF of weight less that ω(f) and so n − 4 = stidR(G) ≤ n − ∆(G) − 2 which leads to the contradiction ∆(G) ≤ 2. Thus, we have idR(G) ∈ {4, 5}. Then either |V2| = 2 or |V2| = |V3| = 1. Since each vertex in V0 must be adjacent to a vertex with wight 3 or two vertices with weight 2, we have certainly ∆(G) ≥ 4 which is a contradiction. Assume now that idR(G) = 3. It follows from this and the fact stidR(G) = n−4 that G has exactly n−4 universal vertices. Let x, y, z, w be the vertices of G that are not universal vertex, that is ∆(G[{x, y, z, w}]) ≤ 2. There are seven graphs of order 4 with maximum degree at most two, that is G[{x, y, z, w}] ∈ {P4, C4, 2K2, 4K1,K2∪2K1, P3∪K1,K3∪K1}. Thus G = Kn−4 ∨H where H ∈ {P4, C4, 2K2, 4K1,K2 ∪ 2K1, P3 ∪K1,K3 ∪K1} and the proof is complete. At the end of this section, we present a Nordhaus-Gaddum type inequality for the sum of the independent double Roman domination stability of a graph G and its complement G. S. M. Sheikholeslami et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5984 10 of 16 Theorem 3. Let G be a graph of order n ≥ 2. Then stidR(G) + stidR(G) ≤ n. Proof. Since n ≥ 2, we have min{idR(G), idR(G)} ≥ 3. If idR(G) = 3 (the case idR(G) = 3 is similar), then G has a universal vertex, and so G has an isolated vertex. Using Remark 2 and noting that stidR(G) = 1, we obtain stidR(G) + stidR(G) ≤ n. Now suppose that min{idR(G), idR(G)} ≥ 4. Since ∆(G) + ∆(G) ≥ n − 1, we may assume, without loss of generality that ∆(G) ≥ (n − 1)/2. Applying Propositions 11 and 12, we obtain stidR(G) + stidR(G) ≤ (n−∆(G)− 1) + (δ(G) + 1) ≤ (n−∆(G)− 1) + (n−∆(G)) = 2n− 2∆− 1 ≤ n, as desired. 6. Trees In this section, we determine the idR(T )-stability, the idR+(T )-stability and the idR−(T )- stability for trees. From Proposition 9, we know that st+idR(T ) cannot be bounded. Theorem 4. For every tree T of order n ≥ 2, stidR(T ) = 1. Proof. If diam(T ) ≤ 2, then T is a star K1,n−1, and we have stidR(T ) = 1. If diam(T ) = 3, then T is a double star Sr,t for some 1 ≤ r ≤ t and one can easily deduce from idR(Sr,t) = 3 + 2r that stidR(Sr,t) = 1. If ∆ = 2, then T = Pn and we are done by Corollary 1. Hence we may assume that diam(T ) ≥ 4 and ∆ ≥ 3. By contradiction, we assume that there exists a tree T such that stidR(T ) ≥ 2. We choose such a tree with smallest order. First, we claim that T has no strong support vertex. Let T has a strong support vertex y with leaf neighbors y1, y2, . . . , yk. Then the vertices y1, y2, . . . , yk are isolated vertices in T ′ = T −y and any idR(T ′)-function certainly assigns 2 to each yi. Now reassigning y1, y2, . . . , yk the value 0 and y the value 3 provides an IDRD-function of T of weight less than idR(T ′) and this leads to a contradiction. Henceforth, we may assume that T has no strong support vertex. Let P = x1x2 . . . xt be a longest path in T and root the tree T at the vertex xt. Let f = (V0,∅, V2, V3) be an idR(T )-function such that f(x3) is maximized. Since T has no strong support vertex, deg(x2) = 2 and each child of x3 with depth 1 has degree 2. We consider the cases: Case 1. x3 has a child z with depth 0, that is z is a leaf neighbor of x3. It is easy to verify that f(x1)+f(x2)+f(x3)+f(z) ≥ 5. We have two situations. First note that if f(x4) = 0, then we may assume that f(x3) = 3, f(x1) = 2 and f(z) = f(x2) = 0. This implies that idR(T−x1) < idR(T ), which leads to a contradiction. Now, let f(x4) ≥ 2, then we may assume that f(x2) = 3, f(z) = 2 and f(x3) = f(x1) = 0. This implies that idR(T − z) < idR(T ), which leads to a contradiction. Case 2. x3 has a child u2 ̸= x2 with depth 1. Let u1 be the leaf neighbor of u2. It is easy to verify that f(x1) + f(x2) + f(x3) + S. M. Sheikholeslami et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5984 11 of 16 f(u2) + f(u1) ≥ 6. We consider two cases. If f(x4) = 0, then we may assume that f(x3) ≥ 2, f(x1) = f(u1) = 2 and f(u2) = f(x2) = 0. Then the function g defined on T − x1 by g(x3) = 3, g(v) = f(v) for each vertex v ∈ V (T − x1) − {x3}, is an IDRD- function of T − x1 of weight at most ω(f) − 1 and so idR(T − x1) < idR(T ) leading to a contradiction. Now, if f(x4) ≥ 2, then we may assume that f(x1) = f(x3) = f(u1) = 0 and f(u2) = f(x2) = 3. Then the function g defined on T − x1 by g(x2) = 2, g(v) = f(v) for each vertex v ∈ V (T − x1) − {x2}, is an IDRD-function of T − x1 of weight at most ω(f)− 1 and so idR(T − x1) < idR(T ) leading to a contradiction. Case 3. deg(x3) = 2. Now, we take any vertex w ∈ V (T )−{x1, x2, x3}, and let T1, T2, . . . , Tk be the components of T − w. Without loss of generality, assume that the pendent star x1x2x3 is contained in T1 (Note that if w = x4, then T1 is a path P3). Denote T ′ = T − {x1, x2, x3} and T ′ 1 = T1 − {x1, x2, x3}. By Proposition 1, we have idR(T ) = idR(T ′) + 3 and idR(T1) = idR(T ′ 1) + 3. It conclude that idR(T ′ −w) = idR(T ′ 1) + idR(T2) + · · ·+ idR(Tk) = idR(T1) + idR(T2)+ · · ·+ idR(Tk)−3 = idR(T −w)−3 = idR(T )−3 = idR(T ′), that is, stidR(T ′) ≥ 2, contradicting the choice of T . Therefore, stidR(T ) = 1 for any tree of order n ≥ 2. This completes the proof. Finally, we will establish an upper bound on the st−idR(T ) of a tree. Moreover, we characterize the trees that achieve the upper bound. For this purpose, we define two families of trees as follows. For integers k ≥ 1 and ∆ ≥ 2, let Tk,∆ be a tree obtained from the k copies of the star K1,∆, say S1, S2, . . . , Sk, by adding k − 1 edges between the leaves of these stars, so that the resulting graph is a connected graph with maximum degree ∆. Let Tk,∆ be the family of all such trees Tk,∆, and T∆ = ⋃ k≥1 Tk,∆. Moreover, let Lk,∆ be a graph obtained from a tree Tk,∆ and a path P2 by joining a vertex of the P2 to a leaf of some star Si (i ∈ {1, 2, . . . , k}) of Tk,∆, so that the resulting graph is a tree with maximum degree ∆. Let Lk,∆ be the family of all such trees Lk,∆, and L∆ = ⋃ k≥1 Lk,∆. It is observed in [17] that if T ∈ Tk,∆, then γdR(T ) = 3k and T has a unique γdR(T )- function f that assigns 3 to the central vertex of each Si, and 0 to the leaves of each Si for i = 1, 2, . . . , k. Also, it is shown in [17] that if T ∈ Lk,∆, then γdR(T ) = 3k + 3. Lemma 1. If T ∈ Tk,∆, then idR(T ) = 3k. Furthermore, T has a unique idR(T )-function f that assigns 3 to the central vertex of each Si, and 0 to the leaves of each Si for i = 1, 2, . . . , k. Proof. First note that idR(T ) ≥ γdR(T ) = 3k. On the other hand, the unique γdR(T )-function f is an independent double Roman dominating function of T and therefore idR(T ) ≤ γdR(T ) = 3k. Thus, idR(T ) = 3k. It follows from idR(T ) = 3k that any idR(T )- function is a γdR-function of T and since T has a unique γdR(T )-function, we deduce that f is the unique idR(T )-function. Lemma 2. If T ∈ Lk,∆, then idR(T ) = 3k + 3. S. M. Sheikholeslami et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5984 12 of 16 Proof. First note that idR(T ) ≥ γdR(T ) = 3k + 3. On the other hand, assigning 3 to the center of stars S1, . . . , Sk and a vertex of P2 and 0 to other vertices, provides an IDRD- function of weight 3k+3 leading to idR(T ) ≤ γdR(T ) = 3k+3. Thus, idR(T ) = 3k+3. Lemma 3. If T ∈ T∆, then st−idR(T ) = ∆. Proof. For k ≥ 1 and ∆ ≥ 2, let T ∈ Tk,∆. We proceed by induction on the number k. If k = 1, then T ∼= K1,∆ and the result follows from Corollary 4. This establishes the base case. Now, let k ≥ 2. If ∆ = 2, then T ∼= P3k and we are done by Proposition 8. Assume that ∆ ≥ 3 and that for any tree T ′ ∈ Tk′,∆ with 1 ≤ k′ < k we have st−idR(T ′) = ∆. Since T ∈ Tk,∆, T is the graph obtained from the k copies of the star K1,∆, say S1, S2, . . . , Sk, by adding k − 1 edges between the leaves of these stars so that the resulting graph is a connected graph with maximum degree ∆. Let xi be the central vertex of the star Si for each i ∈ {1, 2, . . . , k}. By Lemma 1, idR(T ) = 3k. By the construction of T , there is one of Si, say Sk, such that ∆ − 1 of its leaves are the leaves of T , and the remaining leaf, say u, has degree two in T . Clearly, u is adjacent to the central vertex xk of Sk, and a leaf, say v, of another star Si, say Sk−1. To show that st−idR(T ) ≤ ∆, let T1 = T − (V (Sk) − {u}), and define a function f1 : V (T1) → {0, 1, 2, 3} by f1(u) = 2, f1(xj) = 3 for j ∈ {1, 2, 3, . . . , k − 1}, and f1(x) = 0 for x ∈ V (T1) − {u, x1, x2, . . . , xk−1}. Clearly, f1 is an IDRD-function of T1 with weight 3k − 1, so we have st−idR(T ) ≤ ∆. Now, we show that st−idR(T ) ≥ ∆. Let S be a vertex set of T such that idR(T − S) ≤ 3k − 1. Among all such sets, we choose S to have minimum cardinality. We show that |S| ≥ ∆. Suppose, to the contrary, that |S| ≤ ∆−1. Let f = (V0,∅, V2, V3) be a idR(T−S)- function, and so ω(f) ≤ 3k − 1. Choose f such that f(xk) is as large as possible when xk /∈ S. Let T ′ = T − V (Sk), S′ = S ∩ V (T ′) and Wi = S ∩ V (Si) for i ∈ {1, 2, . . . , k}. Since T ′ ∈ Tk−1,∆, by Lemma 1, idR(T ′) = 3k − 3 and from the induction hypothesis we have st−idR(T ′) = ∆. Since |S| ≤ ∆ − 1, |V (Sk) − Wk| ≥ 2. To independent double Roman dominate the vertices in V (Sk)−Wk, we have ∑ x∈V (Sk)−Wk f(x) ≥ 2. First let ∑ x∈V (Sk)−Wk f(x) = 2. Then it is easy to verify that |V (Sk)−Wk| = 2, that is, S = Wk. Moreover, V (Sk)−Wk consists of u and xk, or u and a leaf-neighbour of xk other than u. If V (Sk)−Wk consists of u and xk, then T −S ∈ Lk−1,∆ and by Lemma 2, we have idR(T −S) = 3(k− 1)+3 which is a contradiction. Hence, we assume that V (Sk)−Wk consists of u and a leaf-neighbour y of xk. Then we must have f(y) = 2, f(u) = 0 and f(v) = 3. Then the restriction of f to V (T ′), say f ′, is an IDRD-function of T ′ which is not an idR(T ′)-function by Lemma 1. It follows that idR(T − S) = ω(f) = ω(f ′) + 2 ≥ (3(k − 1) + 1) + 2 = 3k leading to a contradiction. Assume that ∑ x∈V (Sk)−Wk f(x) ≥ 3. If |S ∩ {u, v}| ≥ 1 or f(v) ≥ 2 or f(u) = 0, then the restriction of f on V (T ′ − S′) is an IDRD-function of T ′ − S′ with weight at most 3k − 4 and so idR(T ′ − S′) ≤ 3k − 4. On the other hand, since |S′| ≤ |S| ≤ ∆ − 1 and st−idR(T ′) = ∆, we have idR(T ′ − S′) ≤ 3k − 4, a contradiction. So we may assume that |S ∩ {u, v}| = 0, f(v) = 0 and f(u) ≥ 2. By definition, we must have f(v) = 0. We distinguish two cases. S. M. Sheikholeslami et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5984 13 of 16 Case 1. f(u) = 2. It follows from ∑ x∈V (Sk)−Wk f(x) ≥ 3 and the fact V1 = ∅ that ∑ x∈V (Sk)−Wk f(x) ≥ 4. Since f(v) = 0, v has a neighbor w with f(w) ≥ 2. Define the function g on T ′ − S′ by g(w) = 3 and g(x) = f(x) for otherwise, is an IDRD-function T ′ − S′ of weight at most 3k − 4 which leads contradiction. Case 2. f(u) = 3. If ∑ x∈V (Sk)−Wk f(x) = 3, then we must have S = Wk and V (Sk) − S consists of u and xk. But then T ′ − S′ = T − S ∈ Lk−1,∆ and f is an IDRD-function of T ′ − S′ of weight 3k − 1, a contradiction with Lemma 2. Assume that ∑ x∈V (Sk)−Wk f(x) ≥ 4. It follows from V1 = ∅ that ∑ x∈V (Sk)−Wk f(x) ≥ 5. If v has a neighbor w with f(w) ≥ 2, then define the function g on T ′ − S′ by g(w) = 3 and g(x) = f(x) otherwise. Clearly, g is an IDRD-function T ′ − S′ of weight 3k − 5, a contradiction. Assume that all neighbors of v assigned 0 under f . Then the function g on T ′−S′ by f(v) = 2 and g(x) = f(x) otherwise, is an IDRD-function T ′ − S′ of weight 3k − 4, a contradiction again. Theorem 5. For every tree T of order n ≥ 3 with maximum degree ∆, st−idR(T ) ≤ ∆ with equality if and only if T ∈ T∆. Proof. If diam(T ) = 2, then T is the star K1,∆ and by Corollary 4, we have st−idR(T ) = ∆. If ∆ = 2, then T is the path Pn and the result is true by Proposition 8. Assume that diam(T ) ≥ 3 and ∆ ≥ 3. Let x1x2 . . . xd be a diametral path in G and root T at xd. We have d(x2) ≤ ∆. Let f = (V0,∅, V2, V3) be an idR(T )-function. Clearly, f(N [x2]) ≥ 3. If f(x3) ∈ V2 ∪ V3, then f(x2) = 0 and f(x1) = 2. The function g defined on T − x1 by g(x3) = max{3, f(x3)} and g(x) = f(x) otherwise, is an IDRD-function of the tree T − x1 of weight at most ω(f)− 1. So, st−idR(T ) = 1. Now, assume that f(x3) = 0. Then we have f(N [x2] − {x3}) = 3. If x3 has a neighbor u ̸= x2 with f(u) ≥ 2, then the function g on T −Tx2 by g(u) = min{3, f(u)+1} and g(x) = f(x) otherwise, is an IDRD-function of the tree T − Tx2 of weight at most ω(f)− 1 and so st−idR(T ) ≤ d(x2) ≤ ∆. Now, let f(x) = 0 for each x ∈ N [x3]−{x2}. Then the function g defined on T − (N [x2]−{x3}) by g(x3) = 2 and g(x) = f(x) otherwise, is an IDRD-function of the tree T − (N [x2]− {x3}) of weight ω(f)− 1 and so st−idR(T ) ≤ d(x2) ≤ ∆. This proves the bound. Now we show that st−idR(T ) = ∆ if and only if T ∈ T∆. The sufficiency follows from Lemma 3. To prove the necessity, assume that st−idR(T ) = ∆. We proceed by induction on n. If diam(T ) = 2, then T is the star K1,∆ and clearly T ∈ T∆. If ∆ = 2, then T is the path P3k and the result is true by Proposition 8. This proves the base case. Suppose that for any tree T ′ of order 3 ≤ n′ < n with st−idR(T ) = ∆, we have T ′ ∈ T∆. Let T be a tree of order n with st−idR(T ) = ∆. As before, we can assume that diam(T ) ≥ 3 and ∆ ≥ 3. Corollary 5 implies that diam(T ) ≥ 4. Let f be an idR-function of T such that there is no vertex assigned 1 under f . Let [x1, x2, . . . , xd] be a diametral path in G and root T at xd. Using the above argument and the fact st−idR(T ) = ∆, we must have d(x2) = ∆. If f(x2) = 0, then clearly st−idR(T ) = 1 which is contradiction. Since f is an IDRD-function, it follows that f(x2) = 3, and thus f(x3) = 0. We claim that d(x3) = 2. By contradiction, assume that d(x3) ≥ 3. Let w be a neighbor of x3 different from x2 S. M. Sheikholeslami et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5984 14 of 16 and x4. Clearly, w is either a leaf or a support vertex. In the former case, it follows from f(x3) = 0 that f(w) = 2. In the latter case, by a similar argument as in above, we have that w has degree ∆ and f(w) = 3. In either case, remove all leaf-neighbor of x2 and denote the resulting tree by T ′. Then reassigning 2 to x2 provides an IDRD-function of T ′ leading to st−idR(T ) ≤ ∆− 1 which is contradiction. Thus, d(x3) = 2. By symmetry, we have d(xd−1) = ∆ and d(xd−2) = 2. If x3 = xd−2, then it is easy to see that idR(T ) = 6, and obviously idR(T − L(x2)) = 5 and so st−idR(T ) ≤ ∆ − 1, a contradiction. Hence x3 ̸= xd−2. Let T ′ = T − Tx3 . By Proposition 2, we have idR(T ) = idR(T ′)+3. We claim that st−idR(T ′) = ∆. By contradiction, assume that st−idR(T ′) ≤ ∆−1 and let S′ be a st−idR(T ′)-set. Clearly, any idR(T ′ − S′)-function can be extended to an IDRD-function of T − S′ by assigning a 3 to x2 and 0 to the neighbors of x2 and so idR(T − S′) ≤ idR(T ′ − S) + 3 < idR(T ) which contradicts the assumption st−idR(T ) = ∆. Hence, st−idR(T ′) = ∆ and by the induction hypothesis we have T ′ ∈ T∆. Thus, T ′ is a tree obtained from the k′ copies of the star K1,∆, say S1, S2, . . . , Sk′ , by adding k′ − 1 edges between the leaves of these stars so that the resulting graph is a connected graph with maximum degree ∆. It follows from ∆ ≥ degT (x4) = degT ′(x4) + 1 that x4 is a leaf of some star Si and so T ∈ T∆ and the proof is completed. S. M. Sheikholeslami et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5984 15 of 16 7. Open questions and problems We conclude this paper by mentioning some questions and problems suggested by this research. Problem 1. Characterize the connected graphs G of order n with idR(G) ≥ 4 and stidR(G) = n−∆(G)− 1. Problem 2. Is there a connected graph G of order n ≥ 2 such that stidR(G) = δ(G) + 1. Problem 3. Determine the independent double Roman domination stability of generalized Petersen graphs and Sierpinski graphs. 8. Conclusion In this paper, we have studied the independent double Roman domination stability. We determined exact values of the independent double Roman domination stability for special classes of graphs. Additionally, we established bounds on the idR-stability for gen- eral graphs. For trees, we proved that the idR-stability is always equal to 1, while the i−dR-stability is bounded above by the maximum degree ∆ of the tree. We also provided a complete characterization of the trees that attain this upper bound. These results con- tribute to the growing body of knowledge on Roman domination parameters and open avenues for further research on stability measures in more complex graph structures. Acknowledgements We sincerely thank the reviewers for their valuable comments and suggestions, which have significantly improved the quality of this paper. We acknowledge Ho Chi Minh City University of Technology (HCMUT), VNU-HCM for supporting this study. In addition, we extend our heartfelt appreciation to Mindanao State University – Tawi-Tawi College of Technology and Oceanography (MSU-TCTO) and Mindanao State University-Iligan Institute of Technology (MSU - IIT) for generously funding this work. References [1] E. J. Cockayne, P. A. Dreyer Jr., S. M. Hedetniemi, and S. T. Hedetniemi. Roman domination in graphs. Discrete Mathematics, 278(1-3):11–22, 2004. [2] I. Stewart. Defend the Roman Empire. Scientific American, 281(6):136–139, 1999. [3] C. S. ReVelle and K. E. Rosing. Defendens Imperium Romanum: a classical problem in military strategy. The American Mathematical Monthly, 107(7):585–594, 2000. S. M. Sheikholeslami et al. / Eur. J. Pure Appl. 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