Secure Resolving Sets in a Graph

Abstract

Let G = (V, E) be a simple, finite, and connected graph. A subset S = {u₁, u₂, _…, u_k} of V(G) is called a resolving set (locating set) if for any x ∈ V(G), the code of x with respect to S that is denoted by C_S (x), which is defined as C_S (x) = (d(u₁, x), d(u₂, x), .., d(u_k, 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 s ∈ V – R, there exists r ∈ R 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.

1. Introduction

Let G = (V, E) be a simple, finite, and connected graph. Let S = {u₁, u₂, _…, u_k} on which the ordering (u₁, u₂, _…, u_k) is imposed. For any w ∈ V (G), the ordered k-tuples r (w |S) = (d(u₁, w), d(u₂, w), …., d(u_k, 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, w ∈ V(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 x ∈ V − T, there exists y ∈ T 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 (K_n) = n − 1 = dim (K_n)
• sdim (K_{1, n}) = n > dim (K_{1, n})
• sdim (K_m,n) = m + n − 2 = dim(K_m,n) (m, n ≥ 2)
• sdim (P_n) = 2 > dim (P_n) = 1 (n ≥ 3)
• sdim(C_n) = 2 = dim(C_n)
• sdim (K_m (a₁, a₂, …, a_m)) = $\{\begin{array}{l}dim(K_m(a_1,a_2,\dots ,a_m))+1\\ if{a}_i\ge 2foratleastonei\\ dim(K_m(a_1,a_2,\dots ,a_m))\\ if{a}_i=1foralli.\end{array}$
  where (K_m(a₁, a₂, …, a_m)) is the multi-star graph formed by joining a_i ≥ 1 (1 ≤ i ≤ m) pendant vertices to each vertex x_i of a complete graph K_m with V(K_m) = {x₁, x₂, …, x_m}.

Illustration 1.

Consider C_5.

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 = P_n. However, sdim(P_n) = 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 = P_n (n ≥ 3).

Proof. 

If G = P_n (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 P_n (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(C_n) = dim(C_n) = 2.

Proof. 

Let V(C_n) = {m₁, m₂, …, m_n}.

Case (i):

n = 2k + 1.

Let M = {m₁, m₂}. Then, M is a resolving set of C_n. It can be verified that {m₁, m_i} is a resolving set where 3 ≤ i ≤ n.

Case (ii):

n= 2k.

Then, m₁ and m_k are diametrically opposite vertices. Let M = {m₁, m₂}. Clearly, {m₁, m₂} is a resolving set of C_n. It can be substantiated that {m₁, m_i} is resolving when 3 ≤ i ≤ n, i ≠ k. Also, {m_k, m₂} is a resolving set of C_n. 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, …., u_k} be a basis of G. Let W = {u₁, u₂, …, u_k, 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 = K_n or K_{1, n −1}.

Proof. 

Let G = K_n or K_{1,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 = K_n. Suppose that dim(G) is n − 2. Then, G = K_{a, b} (a, b≥ 1), $\mathrm{Ka}+\overline{\mathrm{Kb}}$ (b ≥ 2, a ≥ 1), K_a + (K₁ ∪ K_b) (b, a ≥ 1). Suppose that G = K_a,b (a, b ≥ 2), sdim(G) = dim(G) = a + b − 2 [1]. Suppose that G = $\mathrm{Ka}+\overline{\mathrm{Kb}}$ (b ≥ 2, a ≥ 1), then dim(G) = a + b − 2. If a = 1, then $\mathrm{Ka}+\overline{\mathrm{Kb}}$ = K_1,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 = K_a + (K₁∪ K_b), (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 = K_{1, 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, …, x_k} and X₂ = {y₁, y₂, …, y_k} 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 S₁ ⊆ V (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, $\frac{sdim(G)}{sdim(T)}<ϵ$ where T is a connected induced subgraph of G.

Proof. 

Let T = K_{1, 2}^{n + 1}. sdim(T) = 2^{n + 1}. Let G = T K₂. Then, we get a graph G containing T as an induced subgraph [1]. Further, sdim(G) ≤ 2n. Therefore, $\frac{sdim(G)}{sdim(T)}\le \frac{2n}{2^{n+1}}\to 0$ as n → 0. Hence, the corollary. □

5. Secure Resolving Domination Number

Definition 2.

Let U be a subset of G. U = {u₁, u₂, …., u_k} 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 γ_sr(G).

Remark 4.

V is a SRD set of G.

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

• γ_sr(K_n) = n − 1, n ≥ 2.
• γ_sr (K_{1, n – 1}) = n − 1, n ≥ 2.
• ${\gamma }_{sr}(P_n)=\{\begin{array}{ll}2& ifn=3,4\\ \lceil \frac{n}{3}\rceil +1& ifn\ge 5.\end{array}$
• ${\gamma }_{sr}(C_{\mathrm{n}})=\{\begin{array}{ll}2& ifn=3,4\\ \lceil \frac{n}{3}\rceil +1& ifn\ge 5.\end{array}$
• ${\gamma }_{sr}(K_{a_1,a_2,\dots ,a_m})=(a_1+a_2+\dots +a_m)-m.$
• $\begin{array}{l}\\ {\gamma }_{sr}(K_m(a_1,a_2,\dots ,a_m))=\{\begin{array}{ll}m+a_{k+1}+\dots +a_m-k\hfill & \hfill \\ if{a}_1=\dots =a_k=1a_i\ge 2,k+1\le i\le m\hfill & \hfill \\ m\mathrm{if}ai=1foralli\hfill & \hfill \end{array}\end{array}$

where (K_m(a₁, a₂, …, a_m)) is the multi-star graph formed by joining a_i ≥ 1 (1 ≤ i ≤ m) pendant vertices to each vertex x_i of a complete graph K_m with V(K_m) = {x₁, x₂, …, x_m}.

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. γ_sr (K₃(1, 1, 1)) = 3.

Proposition 1.

Let γ_s be the minimum cardinality of a secure dominating set of G. Then, max {γ_s(G), dim(G), γ_r(G)} ≤ γ_sr(G) ≤ γ_s(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, γ_sr(G) ≤ γ_s(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). γ_r(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, γ_sr(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, γ_sr(G) ≥ 5. Hence, γ_sr(G) = 5. Thus, γ(G) < dim(G) < γ_r(G) < γ_sr(G). Also, γ_sr(G) = γ(G) + dim(G). γ_s(G) = 3, since G has no secure dominating set with two vertices, and {u₁, u₄, u₃} is a secure dominating set. Therefore, γ_sr(G) = 5 < γ_s(G) + dim(G) = 6 and max {γ_s(G), dim(G), γ_r(G)} = 4 < γ_sr(G) = 5.

Remark 6.

When G = K_n, γ(G) = 1, γ_s(G) = 1, dim(G) = n − 1, γ_r(G) = n − 1, γ_sr(G) = n − 1. Therefore, max {γ_s(G), dim(G), γ_r(G)} = γ_sr(G).

Observation 3.

γ_sr(G) ≥ g(m, d), where g(m, d) = min$\left\{t:t+{\displaystyle {\sum }_{i=1}^t\left(\begin{array}{c}t\\ i\end{array}\right)}{(d-1)}^{t-i}\ge m\right\}$, 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 γ_sr(G) ≥ γ_r(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 γ_sr(G) = k. From corollary 2.2 [6] $m\le k+{\displaystyle {\sum }_{i=1}^k\left(\begin{array}{l}k\\ i\end{array}\right)}{(d-1)}^{l-i}$. γ(G) ≤ γ_sr(G ) = k. Therefore, the diameter of G is not more than 3k − 1. Therefore, $m\le k+{\displaystyle {\sum }_{i=1}^k\left(\begin{array}{l}k\\ i\end{array}\right)}{(3k-2)}^{k-i}$. Therefore, there are only finitely many connected graphs with γ_sr(G) = k. □

Remark 7.

Suppose that γ_sr(G) = 2. Then, the number of connected graphs with γ_sr(G) = 2 has an order of at most 11.

Proof. 

By the above observation, $n\le 2+{\displaystyle {\sum }_{i=1}^2\left(\begin{array}{c}2\\ i\end{array}\right)}{(6-2)}^{2-i}=2+\left(\begin{array}{c}2\\ 1\end{array}\right)4+\left(\begin{array}{c}2\\ 2\end{array}\right)4=2+8+1=11.$

In fact, the above bound for n can be improved. □

Observation 5.

For any G with γ_sr(G) = 2, the order of G is not more than 4.

Proof. 

Let γ_sr(G) = 2. Let X = {p, q} be a γ_sr—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 γ_sr(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.

γ_sr(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 γ_sr(G) = n.

Proof. 

We follow the proof given in proposition 3.1 [6]. Construct a graph G from the path P_{3l − 1}: v₁, v₂, …, v_{3l − 1} of order 3l − 1. Join m—pairs of vertices x_j, y_j, 1 ≤ j ≤ m and join x_j and y_j for each j. Consider, F_j—a copy of the path P₂: x_j y_j. Join the vertex of F_j, 1 ≤ j ≤ m to the vertex v_{3t − 1}. For l = m = 2, the graph is as follows:

Let V = {v₁, v₂, …, v_3l-1}, T = {x₁, x₂, …, x_m}, and W = {y₁, y₂, …, y_m}. Then, γ(G) = l and dim(G) = m (since {v₂, v₅, …, v_{3l − 1}} is dominating, and T is a basis of G. Each resolving set of G has at least one vertex from each set, {x_j, y_j}, 1 ≤ j ≤ m. All of the vertices x_j, y_j, and v_{3l − 1} are dominated by them. We need at least $\frac{3l-2}{3}$= l vertices to dominate V-{v_{3l − 1}}. As a result, γ_r(G) ≥ l + m. However, K = {v₂, v₅, …, v_{3l − 1}} ∪ X is a resolving dominating set for G. Hence, γ_r(G) ≤|K| = l + m. Therefore, γ_r(G) = l + m. Clearly, K is a SRD set of G. Therefore, γ_sr(G) ≤ l + m. However, γ_sr(G) ≥ γ_r(G) = l + m. Therefore, γ_sr(G) = l + m = n. □

Theorem 4.

Let G be a graph of order n ≥ 2. γ_sr(G) = n − 1 if and only if G = K_n or K_{1, n − 1.}

Proof. 

γ_sr(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, γ_sr(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, γ_sr(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, …, x_k. If G is a star, then γ_sr(G) = n − 1. If G is not a star, then the above graph has γ_sr(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(x_i, z) ≠ 1. Therefore, (S − {x_i}) ∪ {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 K_n.

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).