# Secure Resolving Sets in a Graph

^{1}

^{2}

^{*}

## Abstract

**:**

_{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.

## 1. Introduction

_{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].

- SR set: Secure resolving set
- SRD set: Secure resolving dominating set

## 2. Secure Resolving Dimension

**Definition**

**1.**

**Remark**

**1.**

**Remark**

**2.**

## 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}(a_{1}, a_{2}, …, a_{m})) = $\{\begin{array}{l}\text{}dim({K}_{m}({a}_{1},{a}_{2},\dots ,{a}_{m}))+1\\ if\text{}{a}_{i}\ge 2\text{}for\text{}atleast\text{}one\text{}i\\ \text{}dim\text{}({K}_{m}({a}_{1},{a}_{2},\dots ,{a}_{m}))\\ if\text{}{a}_{i}=1\text{}for\text{}all\text{}i.\end{array}$where (K_{m}(a_{1}, a_{2}, …, 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}) = {x_{1}, x_{2}, …, x_{m}}.

**Illustration**

**1.**

_{5.}

_{1}, u

_{3}}. Then, H is resolving, and for any u ∈ V − H, there exists v ∈ H such that (H − {v}) ∪ {u}) is a resolving set of C

_{5}. It can be easily seen that sdim(G) = 2.

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

**Observation**

**1.**

_{n}. However, sdim(P

_{n}) = 2, which is a contradiction. Therefore, sdim(G) ≥ 2.

**Observation**

**2.**

_{1}or P

_{2}.

**Theorem**

**1.**

_{n}(n ≥ 3).

**Proof.**

_{n}(n ≥ 3), then sdim(G) = 2.

_{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}. □

**Theorem**

**2.**

_{n}) = dim(C

_{n}) = 2.

**Proof.**

_{n}) = {m

_{1}, m

_{2}, …, m

_{n}}.

**Case**

**(i):**

_{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):**

_{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).

**Remark**

**3.**

**Proof.**

_{1}, u

_{2}, …., u

_{k}} be a basis of G. Let W = {u

_{1}, u

_{2}, …, 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.**

_{n}or K

_{1, n −}

_{1}.

**Proof.**

_{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}+ (K

_{1}∪ 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).

_{a}+ (K

_{1}∪ K

_{b}), (a, b ≥ 1). When a = 1 and b = 1, G = P

_{3}and sdim(P

_{3}) = 2 and dim(P

_{3}) = 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.**

_{2}. Then, sdim(T) ≤ sdim(TK

_{2}) ≤ sdim(T) + 1.

**Proof.**

_{2}, T

_{1}, and T

_{2}be the transcripts of T in G. Let X be a basis of T, and let X

_{1}= {x

_{1}, x

_{2}, …, x

_{k}} and X

_{2}= {y

_{1}, y

_{2}, …, y

_{k}} be the basis of T

_{1}and T

_{2}respectively, corresponding to X. Let S = X

_{1}∪ {y

_{1}}. Then, S is a SR set of G. Therefore, sdim(G) ≤ sdim(T) + 1. Let V

_{1}, V

_{2}be the vertex sets of T

_{1}and T

_{2}respectively

_{.}Then, V(G) = V

_{1}∪ V

_{2}. Let X be a secure basis of G. Let X

_{1}= X ∩ V

_{1}, X

_{2}= X ∩ V

_{2}. Let S

_{1}⊆ V (T

_{1}) be the union of X

_{1}and the set X’

_{2}consisting of those vertices of V

_{1}corresponding to X

_{2}. Then, S

_{1}is a SR set of T

_{1}. Therefore, sdim(T) = sdim(T

_{1}) ≤ | S

_{1}| = |X

_{1}∪ X’

_{2}| ≤ |X

_{1}| + | X’

_{2}| = |X| = sdim(G). Hence, the theorem. □

**Corollary**

**1.**

**Proof.**

_{1, 2}

^{n}

^{+ 1}. sdim(T) = 2

^{n}

^{+ 1}. Let G = T K

_{2}. 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.**

_{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.**

## 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. - ${\gamma}_{sr}({P}_{n})=\{\begin{array}{ll}2\text{}& if\text{}n=3,4\\ \lceil \frac{n}{3}\rceil +1& if\text{}n\ge 5.\end{array}$
- ${\gamma}_{sr}({C}_{\mathrm{n}})=\{\begin{array}{ll}2\text{}& if\text{}n=3,4\\ \lceil \frac{n}{3}\rceil +1& if\text{}n\ge 5.\end{array}$
- ${\gamma}_{sr}({K}_{{a}_{1},{a}_{2},\dots ,{a}_{m}})=({a}_{1}+{a}_{2}+\dots +{a}_{m})-m.$
- $\begin{array}{l}\\ {\gamma}_{sr}({K}_{m}({a}_{1},{a}_{2},\dots ,{a}_{m}))=\{\begin{array}{ll}m+{a}_{k+1}+\dots +{a}_{m}-k\text{\hspace{0.17em}}\hfill & \hfill \\ if\text{\hspace{0.17em}}{a}_{1}=\dots ={a}_{k}=1\text{\hspace{0.17em}}{a}_{i}\ge 2,k+1\le i\le m\hfill & \hfill \\ m\text{}\mathrm{if}\text{}ai=1\text{}for\text{}all\text{}i\text{\hspace{0.17em}}\hfill & \hfill \end{array}\end{array}$

_{m}(a

_{1}, a

_{2}, …, 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}) = {x

_{1}, x

_{2}, …, x

_{m}}.

**Illustration**

**2.**

_{3}(1, 1, 1). let N = {u

_{1}, u

_{2}, u

_{3}}. Then N is a secure, dominating, and resolving set of K

_{3}(1, 1, 1). It can be easily seen that. γ

_{sr}(K

_{3}(1, 1, 1)) = 3.

**Proposition**

**1.**

_{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.**

_{sr}(G) ≤ γ

_{s}(G) + dim(G). The first inequality is obvious. □

**Remark**

**5.**

**Illustration**

**3.**

_{1}, u

_{4}} is a minimum dominating set, {u

_{5}, u

_{7}, u

_{8}} is a minimum resolving set of G). γ

_{r}(G) = 4 (since {u

_{1}, u

_{5}, u

_{7}, u

_{8}} is a minimum resolving dominating set of G. {u

_{1}, u

_{4}, u

_{5}, u

_{7}, u

_{8}} 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 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, 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 {u

_{1}, u

_{4}, u

_{3}} 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.**

_{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.**

_{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.**

**Proof.**

_{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.**

_{sr}(G) = 2. Then, the number of connected graphs with γ

_{sr}(G) = 2 has an order of at most 11.

**Proof.**

**Observation**

**5.**

_{sr}(G) = 2, the order of G is not more than 4.

**Proof.**

_{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):**

_{4}. If both pendants r and t are removed, then the resulting set is K

_{3}, for which γ

_{sr}(G) = 2. That is, G = K

_{3}.

_{3}is as follows:

_{3}, s cannot enter X, since domination fails. If r, s, and t are mutually adjacent, then the graph H

_{4}is as follows:

_{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.

**Case**

**(ii):**

_{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:

_{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.

**Case**

**(iii):**

_{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):**

_{3}is present.

_{1}cannot enter X by replacing p, q, since domination fails.

**Subcase**

**(ii):**

_{2}is exist.

_{2}cannot enter X by replacing p, q. (since domination fails).

**Subcase**

**(iii):**

_{2}, and s is present.

_{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.

_{17}, w

_{1}cannot enter X, since resolution fails. In H

_{18}, w

_{1}cannot enter X, since domination fails.

**Subcase**

**(iv):**

_{2}is present, and none of s, s

_{2}is present.

_{19}, w

_{1}cannot enter X, since resolution fails. In H

_{20}, w

_{2}cannot enter X, since domination fails. In H

_{21}, w

_{1}cannot enter X, since domination fails. Similarly, if only one of s and s

_{2}is present, and none of r, r

_{1}, and r

_{2}is present, then w

_{2}cannot enter X. Therefore, G = P

_{4}.

**Subcase**

**(v):**

_{2}, x

_{3}, y, and y

_{2}is present. Then, G = P

_{4}. □

**Corollary**

**2.**

_{sr}(G) = 2 if and only if G = P

_{4}, P

_{3}, C

_{3}, C

_{4}, and K

_{3}with a pendant vertex and K

_{4}− {e}.

**Proposition**

**2.**

_{sr}(G) = n.

**Proof.**

_{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:

_{1}, v

_{2}, …, v

_{3l-1}}, T = {x

_{1}, x

_{2}, …, x

_{m}}, and W = {y

_{1}, y

_{2}, …, y

_{m}}. Then, γ(G) = l and dim(G) = m (since {v

_{2}, v

_{5}, …, 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 = {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. □

**Theorem**

**4.**

_{sr}(G) = n − 1 if and only if G = K

_{n}or K

_{1, n − 1.}

**Proof.**

_{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. □

_{1}= (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, 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.

_{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.

_{i,}z) ≠ 1. Therefore, (S − {x

_{i}}) ∪ {u} is resolving. Therefore, there exists an (n − 2) SRD set of G, which is a contradiction.

_{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.

_{n}.

## 7. Discussion and Conclusions

## Author Contributions

## Funding

## Conflicts of Interest

## References

- Chartrand, G.; Eroh, L.; Johnson, M.A.; Oellermann, O.R. Resolvability in graphs and the metric dimension of a graph. Discret. Appl. Math.
**2000**, 105, 99–113. [Google Scholar] [CrossRef] - Slater, P.J. Leaves of trees. In Proceedings of the 6th Southeast Conference on Combinatorics, Graph Theory and Computing, Boca Raton, FL, USA, 17–20 February 1975; pp. 549–559. [Google Scholar]
- Slater, P.J. Dominating and reference sets in graphs. J. Math. Phys.
**1988**, 22, 445–455. [Google Scholar] - Brigham, R.C.; Chartrand, G.; Dutton, R.D.; Zhang, P. Resolving Domination in Graphs. Math. Bohem.
**2003**, 128, 25–36. [Google Scholar] - Harary, F.; Melter, R.A. On the metric dimension of graph. Ars Comb.
**1976**, 24, 191–195. [Google Scholar] - Cockayne, E.J.; Favaron, O.; Mynhardt, C.M. Secure domination, weak roman domination and forbidden subgraph. Bull. Inst. Comb. Appl.
**2003**, 39, 87–100. [Google Scholar]

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Subramanian, H.; Arasappan, S.
Secure Resolving Sets in a Graph. *Symmetry* **2018**, *10*, 439.
https://doi.org/10.3390/sym10100439

**AMA Style**

Subramanian H, Arasappan S.
Secure Resolving Sets in a Graph. *Symmetry*. 2018; 10(10):439.
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**Chicago/Turabian Style**

Subramanian, Hemalathaa, and Subramanian Arasappan.
2018. "Secure Resolving Sets in a Graph" *Symmetry* 10, no. 10: 439.
https://doi.org/10.3390/sym10100439