Secure Resolving Sets in a Graph

: Let G = (V, E) be a simple, ﬁnite, and connected graph. A subset S = { u 1 , u 2 , ... , 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(u 1 , x), d(u 2 , 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.


Introduction
Let G = (V, E) be a simple, finite, and connected graph. Let S = {u 1 , u 2 , . . . , u k } on which the ordering (u 1 , u 2 , . . . , u k ) is imposed. For any w ∈ V (G), the ordered k-tuples r (w |S) = (d(u 1 , w), d(u 2 , 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 Theorem 1. sdim(G) = 2, G is a tree if and only if G = P n (n ≥ 3).
Conversely, suppose n ≥ 4. Suppose there are two pendant vertices v 1 , v 2 adjacent with w of G. Take a vertex t, which is adjacent to w (t = v 1 , v 2 ). {v 1 , w} is not a resolving set, since t and v 2 will not have distinct codes with respect to {v 1 , w}. Assume that {v 1 , v 2 } is a resolving set of G. Then, it is not secure, since {v 1 , w} and {v 2 , w} are not resolving sets. Suppose {v 1 , t} is resolving. Then, it is not secure (since {v 1 , w}, {w, t} are not resolving sets of G). Let {a, b} be a resolving set of G. a, b / ∈ {v 1 , v 2 , w}. Then, {a, b} is not secure, since neither {w, b} nor {w, a} is a resolving set of P n (since d(v 1 , a) = d(w, a) + 1 = d(v 2 , 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 = P 3 . Let M = {m 1 , m 2 }. Then, M is a resolving set of C n . It can be verified that {m 1 , m i } is a resolving set where 3 ≤ i ≤ n.

Case (ii): n = 2k.
Then, m 1 and m k are diametrically opposite vertices. Let M = {m 1 , m 2 }. Clearly, {m 1 , m 2 } is a resolving set of C n . It can be substantiated that {m 1 , m i } is resolving when 3 ≤ i ≤ n, i = k. Also, {m k , m 2 } is a resolving set of C n . Therefore, sdim (G) = 2 = dim(G).

Secure Resolving Domination Number
Definition 2. Let U be a subset of G. U = {u 1 , u 2 , . . . ., 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.

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.

Proposition 1.
Let γ s be the minimum cardinality of a secure dominating set of G. Then, max {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. 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, . 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)  ( 1 ) 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 Since S is resolving, S must contain two of the pendant vertices. If S contains u 2 , then u 6 and the remaining pendant vertices are not resolved. If S contains u 6 , then u 2 and the remaining pendant vertices are not resolved. If S contains u 4 , then u 5 and u 3 are not resolved. Therefore, either S contains u 5 and two of the pendant vertices or u 3 and two of the pendant vertices. If S contains u 5 and two of the pendant vertices, then the remaining pendant vertex is not resolved. Therefore, the resolving dominating set contains u 1 . Therefore, the possibilities of the resolving dominating sets are {u 1 , u 5 , u 7 , u 8 }, {u 1 , u 5 , u 8 , u 9 }, {u 1 , u 5 , u 7 , u 9 }, {u 1 , u 3 , u 7 , u 8 }, {u 1 , u 3 , u 8 , u 9 }, and {u 1 , u 3 , u 7 , u 9 }. None of these is secure, since u 2 and u 6 cannot be replaced in all of these sets. Therefore, 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).
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.
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, 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: 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 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 by removing a vertex of X, since such a resulting set is not a dominating set. Therefore, G = P 4 . If both pendants r and t are removed, then the resulting set is K 3 , for which If r, s, and t are present and r is adjacent with s, then the graph is: 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 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, 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 Here, s cannot enter X, since resolution fails. If r and t are adjacent, then the graph H 3 is as follows: 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.
In H 3 , s cannot enter X, since domination fails. If r, s, and t are mutually adjacent, then the graph H 4 is as follows: 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.
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 = P 4 . (ii) r and t are available, s is not present and r and t are adjacent. We get G = C 4 . (iii) r and s are alone present and r and s adjacent. We get C 4 with a diagonal. (iv) r and s are alone and present, and they are not adjacent. We get K 3 with a pendant vertex. Thus, in this case, G = P 3 , P 4 , C 4 , C 4 with a diagonal and K 3 with a pendant vertex.
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: 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: For security in H 5 , r 1 cannot enter {p, q} by removing p or q, since domination fails. Therefore, only one of r and r 1 can be present. Similarly, one of s and s 1 can be present. Therefore, the graphs are 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: Symmetry 2018, 10, x FOR PEER REVIEW 8 of 11 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.
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.
In this case, w1 cannot enter X by replacing p, q, since domination fails.
In graph H 9 , w cannot enter X = {p, q}. In H 6 , if w enters X by removing p or q, then the resulting set is not resolving, although it is dominating. In graph H 7 , if w enters X, then for domination, q should be replaced by w. However, the resulting set is not resolving. Same is the graph H 8 . In graphs H 11 , H 12 , and H 13 , w cannot enter X, since resolution fails. In graph H 14 , w cannot enter X, since domination fails. 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.
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.
In this case, w1 cannot enter X by replacing p, q, since domination fails.

Subcase (ii): s2 is exist.
Only one of r, r 2 , and r 3 can be present, since {p, q} is an SRD set. Similarly, only one of s and s 2 can be present. If any number of edges among the vertices r 3 , r, x 2 , s, and s 2 are inserted, then w 1 cannot enter X by replacing p or q, since domination fails.

Subcase (i): r 3 is present.
In this case, w 1 cannot enter X by replacing p, q, since domination fails.

Subcase (ii): s 2 is exist.
In this case, w 2 cannot enter X by replacing p, q. (since domination fails).

Subcase (iii):
One of r, r 2 , and s is present.
Then, the graphs are as follows: 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 H 15 , either w 1 or w 2 cannot enter X by replacing p, q, since resolution fails. In H 16 , w 1 cannot enter X, since domination fails.
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 H 17 , w 1 cannot enter X, since resolution fails. In H 18 , w 1 cannot enter X, since domination fails.

Subcase (iv):
Only one of r, r 2 is present, and none of s, s 2 is present.
Then, the graphs are as follows: Symmetry 2018, 10, x FOR PEER REVIEW 9 of 11 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.
Proof. We follow the proof given in proposition 3.1 [6]. Construct a graph G from the path P 3l − 1 : v 1 , v 2 , . . . , 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 P 2 : 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: Symmetry 2018, 10, x FOR PEER REVIEW 9 of 11 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 y2 is present. Then, G = P4. □ . . , 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 3l−2 3 = l vertices to dominate V-{v 3l − 1 }. As a result, γ r (G) ≥ l + m. However, K = {v 2 , v 5 , . . . , 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. 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.
Clearly, S is a resolving set, since d (u, v) = 1, d(u, w) ≥ 2. Therefore, S 1 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, S 2 = (S ∪ {w}) − {x} is a dominating set of G. d(u, v) ≥ 2 and d(x, w) = 1. Therefore, S 2 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: Suppose that (S − N(u)) = . 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 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.

Discussion and Conclusions
A study of SR sets and SRD sets is initiated in this paper. Further work may be done on (i) where v is adjacent with some or all of x 1 , x 2 , . . . , 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 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 H 1 or of the form H 2 , where u and v are made adjacent in H 1 . However, H 1 and H 2 have an (n − 2) secure dominating set of G, which is a contradiction. Suppose that (S − N(u)) = . 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 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.

Discussion and Conclusions
A study of SR sets and SRD sets is initiated in this paper. Further work may be done on (i)