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Symmetry 2018, 10(10), 439; https://doi.org/10.3390/sym10100439

Article
Secure Resolving Sets in a Graph
by 1,* and
1
Research Scholar, Register No. 10445, The M.D.T. Hindu College, Tirunelveli 627 010 Affiliated to Manonmaniam Sundaranar University, Abishekapatti, Tirunelveli 627 012, Tamilnadu, India
2
Principal, Department of Mathematics, The M.D.T. Hindu College, Tirunelveli 627 010, India
*
Author to whom correspondence should be addressed.
Received: 5 September 2018 / Accepted: 19 September 2018 / Published: 27 September 2018

## Abstract

:
Let G = (V, E) be a simple, finite, and connected graph. A subset S = {u1, u2, , uk} of V(G) is called a resolving set (locating set) if for any xV(G), the code of x with respect to S that is denoted by CS (x), which is defined as CS (x) = (d(u1, x), d(u2, x), .., d(uk, x)), is different for different x. The minimum cardinality of a resolving set is called the dimension of G and is denoted by dim(G). A security concept was introduced in domination. A subset D of V(G) is called a dominating set of G if for any v in V – D, there exists u in D such that u and v are adjacent. A dominating set D is secure if for any u in V – D, there exists v in D such that (D – {v}) ∪ {u} is a dominating set. A resolving set R is secure if for any sV – R, there exists rR such that (R – {r}) ∪ {s} is a resolving set. The secure resolving domination number is defined, and its value is found for several classes of graphs. The characterization of graphs with specific secure resolving domination number is also done.
Keywords:
resolving set; domination; secure resolving set and secure resolving domination

## 1. Introduction

Let G = (V, E) be a simple, finite, and connected graph. Let S = {u1, u2, , uk} on which the ordering (u1, u2, , uk) is imposed. For any wV (G), the ordered k-tuples r (w |S) = (d(u1, w), d(u2, w), …., d(uk, w)) is known as the metric description of w with respect to S. The set S is called a resolving set of G if r (u|S) = r (w |S) implies u = w for all u, wV(G). A resolving set of G of minimum cardinality is called a minimum resolving set or a basis, and the cardinality of a minimum resolving set is called the dimension of G, which is denoted by dim(G) [1].
The idea of locating sets in a connected graph is already available in the literature [2,3]. Slater initiated the concept of locating sets (resolving sets) and a reference set (metric dimension) nearly four decades ago. Later, Harary and Melter found the above-mentioned theory [4] independently. They adopted the term metric dimension for locating number. Several papers have been published on resolving sets, resolving dominating sets, independent resolving sets, etc.
Security is a concept that is associated with several types of sets in a graph. For example, a dominating set D of G is secure set if for any v ∈ V − D there exist u ∈ D such that (D − {u}) ∪ {v} is a dominating set [5,6]. Secure independent sets, secure equitable sets etc., have been defined and discussed. In this paper, secure resolving sets and secure resolving dominating sets are introduced and studied.
In this paper, G refers to a simple, finite, and connected graph. The abbreviations used in this paper are as follows:
• SR set: Secure resolving set
• SRD set: Secure resolving dominating set
The role of symmetry in the following study:
Regarding the symmetry role, a complete graph has vertex transitivity, which is a symmetry. $K m , n$ has vertices that have degree symmetry in the partite sets. One more important aspect of symmetry is present in the concept of the SR set, as well as in SRD sets. In SR sets, every vertex has the opportunity of being a member of a resolving set. Thus, a symmetry is achieved in the presence of vertices. A similar thing happens in domination. In practical application, in any Executive Council, equal opportunity is to be given to all of the members of the General Council for inclusion in the Executive Council. Thus, the spirit of symmetry is present in the form of equality. That is, there is a symmetry in the treatment of vertices.

## 2. Secure Resolving Dimension

Definition 1.
A subset T of G is a SR set of G if T is resolving and for any xV − T, there exists yT such that (T − {y}) ∪ {x} is a resolving set of G. The minimum cardinality of a SR set of G is known as the secure resolving dimension of G, and is marked by sdim(G).
Remark 1.
The existence of a SR set is guaranteed.For, in any graph, the vertex set V(G) is a secure set as well as a resolving set.
Remark 2.
dim(G) ≤ sdim (G).

## 3. Secure Resolving Dimension for Some Well-Known Graphs

• sdim (Kn) = n − 1 = dim (Kn)
• sdim (K1, n) = n > dim (K1, n)
• sdim (Km,n) = m + n − 2 = dim(Km,n) (m, n ≥ 2)
• sdim (Pn) = 2 > dim (Pn) = 1 (n ≥ 3)
• sdim(Cn) = 2 = dim(Cn)
• sdim (Km (a1, a2, …, am)) =
where (Km(a1, a2, …, am)) is the multi-star graph formed by joining ai 1 (1 ≤ i ≤ m) pendant vertices to each vertex xi of a complete graph Km with V(Km) = {x1, x2, …, xm}.
Illustration 1.
Consider C5.
let H = {u1, u3}. Then, H is resolving, and for any u ∈ V − H, there exists v ∈ H such that (H − {v}) ∪ {u}) is a resolving set of C5. It can be easily seen that sdim(G) = 2.

## 4. Secure Resolving Dimension for Special Classes of Graphs

Observation 1.
Let order of G ≥ 3. Suppose sdim (G) = 1. Then, dim (G) = (since sdim(G) ≥ dim(G)). Therefore, G = Pn. However, sdim(Pn) = 2, which is a contradiction. Therefore, sdim(G) ≥ 2.
Observation 2.
sdim(G) = 1 if and only if G = P1or P2.
Theorem 1.
sdim(G) = 2, G is a tree if and only if G = Pn (n ≥ 3).
Proof.
If G = Pn (n 3), then sdim(G) = 2.
Conversely, suppose n 4. Suppose there are two pendant vertices v1, v2 adjacent with w of G. Take a vertex t, which is adjacent to w (t ≠ v1, v2). {v1, w} is not a resolving set, since t and v2 will not have distinct codes with respect to {v1, w}. Assume that {v1, v2} is a resolving set of G. Then, it is not secure, since {v1, w} and {v2, w} are not resolving sets. Suppose {v1, t} is resolving. Then, it is not secure (since {v1, w}, {w, t} are not resolving sets of G). Let {a, b} be a resolving set of G. a, b ∉ {v1, v2, w}. Then, {a, b} is not secure, since neither {w, b} nor {w, a} is a resolving set of Pn (since d(v1, a) = d(w, a) + 1 = d(v2, a)). Therefore, no vertex of G supports two or more pendant vertices. Suppose that w is a vertex of G that supports one pendant vertex and there exists at least two neighbors of w having degrees greater than or equal to two. Then, we will not get any resolving set with cardinality two containing w. Therefore, any vertex of G with a pendant neighbor has at most one neighbor of degree greater than or equal to two. Therefore, G is a path. Suppose that n = 3. Since G is acyclic and connected, G = P3. □
Theorem 2.
sdim(Cn) = dim(Cn) = 2.
Proof.
Let V(Cn) = {m1, m2, …, mn}.
Case (i):
n = 2k + 1.
Let M = {m1, m2}. Then, M is a resolving set of Cn. It can be verified that {m1, mi} is a resolving set where 3 ≤ i ≤ n.
Case (ii):
n= 2k.
Then, m1 and mk are diametrically opposite vertices. Let M = {m1, m2}. Clearly, {m1, m2} is a resolving set of Cn. It can be substantiated that {m1, mi} is resolving when 3 ≤ i ≤ n, i ≠ k. Also, {mk, m2} is a resolving set of Cn. Therefore, sdim (G) = 2 = dim(G).
Remark 3.
sdim(G) ≤ 1 + dim(G).
Proof.
sdim(G)≥ dim(G). Suppose that dim(G) < sdim(G). Let T= {u1, u2, …., uk} be a basis of G. Let W = {u1, u2, …, uk, v}. Then, W is a SR set of G. Therefore, sdim(G) ≤ k + 1. However, sdim(G) > dim(G) = k. Therefore, sdim(G) ≤ 1 + dim(G). Hence the remark. □
Theorem 3.
sdim(G) = n − 1 if and only if G = Kn or K1, n −1.
Proof.
Let G = Kn or K1,n − 1. Then, sdim(G) = n − 1. Suppose that sdim(G) = n − 1. Then, dim(G) is n − 1 or n − 2. If dim(G) = n − 1, then G = Kn. Suppose that dim(G) is n − 2. Then, G = Ka, b (a, b≥ 1), $Ka + Kb ¯$ (b 2, a ≥ 1), Ka + (K1 ∪ Kb) (b, a 1). Suppose that G = Ka,b (a, b 2), sdim(G) = dim(G) = a + b − 2 [1]. Suppose that G = $Ka + Kb ¯$ (b 2, a 1), then dim(G) = a + b − 2. If a = 1, then $Ka + Kb ¯$ = K1,b. In this case, sdim(G) = b and dim(G) = b − 1. If a ≥ 2, then sdim(G) = a + b − 2 = dim(G).
Let G = Ka + (K1∪ Kb), (a, b ≥ 1). When a = 1 and b = 1, G = P3 and sdim(P3) = 2 and dim(P3) = 1. Clearly, G is a star. When a = 1 and b ≥ 2, sdim(G) = a + b − 1 = dim(G). Suppose that a > 1 and b = 1. Then, sdim(G) = a = dim(G). Suppose that a, b > 1. Then, sdim(G) = a + b − 1= dim(G). Except when G is a star, sdim(G) = dim(G) = n − 2. Therefore, G = K1, n − 1. □
Theorem 3.
Let T be a connected graph. Let G = TK2. Then, sdim(T) ≤ sdim(TK2) ≤ sdim(T) + 1.
Proof.
Refer to Theorem 7 [1]. Let G = T K2, T1, and T2 be the transcripts of T in G. Let X be a basis of T, and let X1 = {x1, x2, …, xk} and X2 = {y1, y2, …, yk} be the basis of T1 and T2 respectively, corresponding to X. Let S = X1 {y1}. Then, S is a SR set of G. Therefore, sdim(G) ≤ sdim(T) + 1. Let V1, V2 be the vertex sets of T1 and T2 respectively. Then, V(G) = V1 ∪ V2. Let X be a secure basis of G. Let X1 = X ∩ V1, X2 = X ∩ V2. Let S1V (T1) be the union of X1 and the set X’2 consisting of those vertices of V1 corresponding to X2. Then, S1 is a SR set of T1. Therefore, sdim(T) = sdim(T1) ≤ | S1| = |X1 X’2| ≤ |X1| + | X’2| = |X| = sdim(G). Hence, the theorem. □
Corollary 1.
Letϵ> 0. Then, $s d i m ( G ) s d i m ( T ) < ϵ$ where T is a connected induced subgraph of G.
Proof.
Let T = K1, 2n + 1. sdim(T) = 2n + 1. Let G = T K2. Then, we get a graph G containing T as an induced subgraph [1]. Further, sdim(G) ≤ 2n. Therefore, $s d i m ( G ) s d i m ( T ) ≤ 2 n 2 n + 1 → 0$ as n → 0. Hence, the corollary. □

## 5. Secure Resolving Domination Number

Definition 2.
Let U be a subset of G. U = {u1, u2, …., uk} of V (G) is said to be a SRD set of G if U is a dominating set of G, U is resolving, and U is secure. The minimum cardinality of a SRD set of G is known as a secure resolving domination number of G, and is represented by γsr(G).
Remark 4.
V is a SRD set of G.

## 6. Secure Resolving Domination Number for Some Well-known Graphs

• γsr(Kn) = n − 1, n ≥ 2.
• γsr (K1, n – 1) = n − 1, n ≥ 2.
• $γ s r ( K a 1 , a 2 , … , a m ) = ( a 1 + a 2 + … + a m ) − m .$
where (Km(a1, a2, …, am)) is the multi-star graph formed by joining ai 1 (1 ≤ i ≤ m) pendant vertices to each vertex xi of a complete graph Km with V(Km) = {x1, x2, …, xm}.
Illustration 2.
Consider the following graph K3(1, 1, 1). let N = {u1, u2, u3}. Then N is a secure, dominating, and resolving set of K3(1, 1, 1). It can be easily seen that. γsr (K3(1, 1, 1)) = 3.
Proposition 1.
Let γs be the minimum cardinality of a secure dominating set of G. Then, max {γs(G), dim(G), γr(G)} ≤ γsr(G) ≤ γs(G) + dim(G).
Proof.
Let L be a minimum secure dominating set of G and W be a basis of G. Then, L ∪ W is a SRD set of G. Hence, γsr(G) ≤ γs(G) + dim(G). The first inequality is obvious. □
Remark 5.
P ∪ {u} is a SRD set of G, P is a minimum resolving dominating set of G.
Illustration 3.
Consider the given graph G.
Here, γ(G) = 2 and dim(G) = 3 (since {u1, u4} is a minimum dominating set, {u5, u7, u8} is a minimum resolving set of G). γr(G) = 4 (since {u1, u5, u7, u8} is a minimum resolving dominating set of G. {u1, u4, u5, u7, u8} is a SRD set of G. Let S be a minimum SRD set of G. Consequently, γsr(G) ≤ 5. Since S is resolving, S must contain two of the pendant vertices. If S contains u2, then u6 and the remaining pendant vertices are not resolved. If S contains u6, then u2 and the remaining pendant vertices are not resolved. If S contains u4, then u5 and u3 are not resolved. Therefore, either S contains u5 and two of the pendant vertices or u3 and two of the pendant vertices. If S contains u5 and two of the pendant vertices, then the remaining pendant vertex is not resolved. Therefore, the resolving dominating set contains u1. Therefore, the possibilities of the resolving dominating sets are {u1, u5, u7, u8}, {u1, u5, u8, u9}, {u1, u5, u7, u9}, {u1, u3, u7, u8}, {u1, u3, u8, u9}, and {u1, u3, u7, u9}. None of these is secure, since u2 and u6 cannot be replaced in all of these sets. Therefore, γsr(G) 5. Hence, γsr(G) = 5. Thus, γ(G) < dim(G) < γr(G) < γsr(G). Also, γsr(G) = γ(G) + dim(G). γs(G) = 3, since G has no secure dominating set with two vertices, and {u1, u4, u3} is a secure dominating set. Therefore, γsr(G) = 5 < γs(G) + dim(G) = 6 and max {γs(G), dim(G), γr(G)} = 4 < γsr(G) = 5.
Remark 6.
When G = Kn, γ(G) = 1, γs(G) = 1, dim(G) = n − 1, γr(G) = n − 1, γsr(G) = n − 1. Therefore, max {γs(G), dim(G), γr(G)} = γsr(G).
Observation 3.
γsr(G) ≥ g(m, d), where g(m, d) = min${ t : t + ∑ i = 1 t ( t i ) ( d − 1 ) t − i ≥ m }$, d is a diameter of G, the order of G is m ≥ 2, and d and m are positive integers with d < m. This follows from proposition 2.1 [6] and that γsr(G) ≥ γr(G).
Observation 4.
For every positive integer k, there are only finitely many connected graphs with secure resolving domination number k.
Proof.
Consider a graph G with order m ≥ 2 and γsr(G) = k. From corollary 2.2 [6] $m ≤ k + ∑ i = 1 k ( k i ) ( d − 1 ) l − i$. γ(G) ≤ γsr(G ) = k. Therefore, the diameter of G is not more than 3k − 1. Therefore, $m ≤ k + ∑ i = 1 k ( k i ) ( 3 k − 2 ) k − i$. Therefore, there are only finitely many connected graphs with γsr(G) = k.
Remark 7.
Suppose that γsr(G) = 2. Then, the number of connected graphs with γsr(G) = 2 has an order of at most 11.
Proof.
By the above observation, $n ≤ 2 + ∑ i = 1 2 ( 2 i ) ( 6 − 2 ) 2 − i = 2 + ( 2 1 ) 4 + ( 2 2 ) 4 = 2 + 8 + 1 = 11 .$
In fact, the above bound for n can be improved. □
Observation 5.
For any G with γsr(G) = 2, the order of G is not more than 4.
Proof.
Let γsr(G) = 2. Let X = {p, q} be a γsr—set of G. If d(p, q) ≥ 4, then p and q cannot dominate the point at a distance 2 from p in the shortest path joining p and q. Therefore, d(p, q) ≤ 3.
Case (i):
Distance between p and q is 1.
As every single vertex in V (G) − X is adjacent with either both p and q or one of them, the distances of the vertices in V (G) − X from p and q are (1, 2), (2, 1), and (1, 1). Then, G is as follows:
Here, s cannot enter X by removing a vertex of X, since such a resulting set is not a dominating set. Therefore, G = P4. If both pendants r and t are removed, then the resulting set is K3, for which γsr(G) = 2. That is, G = K3.
If r, s, and t are present and r is adjacent with s, then the graph is:
Here, r cannot enter X, since resolving fails.
If r, s, and t are existing, r and t are adjacent with s, then the graph is:
Here, s cannot enter X, since resolution fails.
If r and t are adjacent, then the graph H3 is as follows:
In H3, s cannot enter X, since domination fails. If r, s, and t are mutually adjacent, then the graph H4 is as follows:
In the above graph, s cannot enter X, since resolution fails. The remaining cases are: (i) s is not present, and r and t are non-adjacent. In this case, G = P4. (ii) r and t are available, s is not present and r and t are adjacent. We get G = C4. (iii) r and s are alone present and r and s adjacent. We get C4 with a diagonal. (iv) r and s are alone and present, and they are not adjacent. We get K3 with a pendant vertex. Thus, in this case, G = P3, P4, C4, C4 with a diagonal and K3 with a pendant vertex.
Case (ii):
d(p,q) = 2.
Since every vertex in V(G) − X is adjacent with at least one of p and q, the distances of the vertices 3 in V(G) − X from p and q are (1, 3), (3, 1), (1, 2), (2, 1), and (1, 1). Therefore, the graph is as follows:
For security in H5, r1 cannot enter {p, q} by removing p or q, since domination fails. Therefore, only one of r and r1 can be present. Similarly, one of s and s1 can be present. Therefore, the graphs are as follows:
In graph H9, w cannot enter X = {p, q}. In H6, if w enters X by removing p or q, then the resulting set is not resolving, although it is dominating. In graph H7, if w enters X, then for domination, q should be replaced by w. However, the resulting set is not resolving. Same is the graph H8. In graphs H11, H12, and H13, w cannot enter X, since resolution fails. In graph H14, w cannot enter X, since domination fails.
Case (iii):
d(p,q) = 3.
Since vertices in V(G) − X are adjacent with one of p and q, the distances of the vertices in V (G) − X from p and q are (1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (1, 4), and (4, 1).
Only one of r, r2, and r3 can be present, since {p, q} is an SRD set. Similarly, only one of s and s2 can be present. If any number of edges among the vertices r3, r, x2, s, and s2 are inserted, then w1 cannot enter X by replacing p or q, since domination fails.
Subcase (i):
r3 is present.
In this case, w1 cannot enter X by replacing p, q, since domination fails.
Subcase (ii):
s2 is exist.
In this case, w2 cannot enter X by replacing p, q. (since domination fails).
Subcase (iii):
One of r, r2, and s is present.
Then, the graphs are as follows:
In H15, either w1 or w2 cannot enter X by replacing p, q, since resolution fails. In H16, w1 cannot enter X, since domination fails.
In H17, w1 cannot enter X, since resolution fails. In H18, w1 cannot enter X, since domination fails.
Subcase (iv):
Only one of r, r2 is present, and none of s, s2 is present.
Then, the graphs are as follows:
In H19, w1 cannot enter X, since resolution fails. In H20, w2 cannot enter X, since domination fails. In H21, w1 cannot enter X, since domination fails. Similarly, if only one of s and s2 is present, and none of r, r1, and r2 is present, then w2 cannot enter X. Therefore, G = P4.
Subcase (v):
None of x, x2, x3, y, and y2is present. Then, G = P4. □
Corollary 2.
γsr(G) = 2 if and only if G = P4, P3, C3, C4, and K3with a pendant vertex and K4− {e}.
Proposition 2.
Let l ≥ 1, m ≥ 2, and n = l + m be three integers. Then, there exists G with γ(G) = l, dim(G) = q and γsr(G) = n.
Proof.
We follow the proof given in proposition 3.1 [6]. Construct a graph G from the path P3l 1: v1, v2, …, v3l − 1 of order 3l − 1. Join m—pairs of vertices xj, yj, 1 ≤ j ≤ m and join xj and yj for each j. Consider, Fj—a copy of the path P2: xj yj. Join the vertex of Fj, 1 ≤ j ≤ m to the vertex v3t 1. For l = m = 2, the graph is as follows:
Let V = {v1, v2, …, v3l-1}, T = {x1, x2, …, xm}, and W = {y1, y2, …, ym}. Then, γ(G) = l and dim(G) = m (since {v2, v5, …, v3l 1} is dominating, and T is a basis of G. Each resolving set of G has at least one vertex from each set, {xj, yj}, 1 ≤ j ≤ m. All of the vertices xj, yj, and v3l − 1 are dominated by them. We need at least $3 l − 2 3$= l vertices to dominate V-{v3l 1}. As a result, γr(G) ≥ l + m. However, K = {v2, v5, …, v3l 1} ∪ X is a resolving dominating set for G. Hence, γr(G) ≤|K| = l + m. Therefore, γr(G) = l + m. Clearly, K is a SRD set of G. Therefore, γsr(G) ≤ l + m. However, γsr(G) ≥ γr(G) = l + m. Therefore, γsr(G) = l + m = n.
Theorem 4.
Let G be a graph of order n ≥ 2. γsr(G) = n − 1 if and only if G = Kn or K1, n − 1.
Proof.
γsr(G) = n − 1. Consequently, no (n − 2) subset of V(G) is a SRD set of G. Suppose that there exists an (n − 2) resolving subset S of V (G) that is not a secure dominating set of G. Let V (G) − S = {u, v}. Suppose that S is not a dominating set of G. Since G is connected, exactly one of u and v is not dominated by S, say u. Clearly, u is a pendant of v. □
Claim: v is adjacent with every vertex of S.
Suppose that v is not adjacent with a vertex w of S. Let T’ = (S − {w})∪{v} = V (G) − {u, w}. Since G is connected and w is not adjacent with u and v, w is adjacent with some vertex of S. T’ is a dominating set of G. Therefore, there exists an (n − 2) subset that is a resolving and dominating set of G.
S1= (T’∪ {u}) − {v} is a dominating set of G. Clearly, S is a resolving set, since d (u, v) = 1, d(u, w) 2. Therefore, S1 is a secure resolving domination set of G. Therefore, γsr(G) ≤ n − 2, which is a contradiction. w is adjacent with some vertex x in S. Therefore, S2 = (S ∪ {w}) − {x} is a dominating set of G. d(u, v) 2 and d(x, w) = 1. Therefore, S2 is a secure resolving domination set of G. Therefore, γsr(G) ≤ n − 2, which is a contradiction. Suppose that S is a dominating set of G, but not a secure dominating set of G. Suppose that u cannot enter S by replacing a vertex of S. Then, any neighbor of u is either an isolate of S or has private neighbor v. Suppose that every neighbor x of u is an isolate of S. In this case, if u is not adjacent with v, then G is disconnected, which is a contradiction. If u is adjacent with v, then (S − N[u]) ∪ {v} is connected. Then, (S − {x}) ∪ {u} is a dominating set of G.
Suppose that (S − N(u)) = φ. Then, either G is a star or G is of the form: where v is adjacent with some or all of x1, x2, …, xk. If G is a star, then γsr(G) = n − 1. If G is not a star, then the above graph has γsr(G) ≤ n − 2, which is a contradiction.
Suppose that (S − N(u)) φ. Then, v is adjacent with at least one vertex, say z of (S − N(u)). d(v,z) = 1, d(xi, z) 1. Therefore, (S − {xi}) ∪ {u} is resolving. Therefore, there exists an (n − 2) SRD set of G, which is a contradiction.
Suppose that there exists a neighbor x of u which has private neighbor v. Let x be an isolate of S. Then, G is of the form H1 or of the form H2, where u and v are made adjacent in H1. However, H1 and H2 have an (n − 2) secure dominating set of G, which is a contradiction.
If x is not an isolate of S, then either G is complete, or G has (n − 2) SRD set of G, which is a contradiction. Similarly, v can enter S by replacing a vertex of S. Therefore, any (n − 2) resolving subset of V (G) is a secure dominating set of G, provided that G is not a star or G is not Kn.
Therefore, the theorem follows.

## 7. Discussion and Conclusions

A study of SR sets and SRD sets is initiated in this paper. Further work may be done on (i) conditions for the minimality of SR sets (SRD sets), (ii) uniform SR set (SRD set) (that is to find the least positive integer t such that every subset of V(G) of cardinality t is an SR set (SRD set)), (iii) a study of secure metric resolving sets (metric resolving dominating sets) in a graph, and (iv) secure independent resolving sets (secure independent resolving dominating sets).

## Author Contributions

H.S. first author is responsible for the preparation of an article, review, editing and visualization under the guidance of S.A.

## Funding

This research is carried out without any external financial support.

## Conflicts of Interest

The authors confirm no conflict of interest.

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