Resolvability in Subdivision of Circulant Networks Cn1,k

Circulant networks form a very important and widely explored class of graphs due to their interesting and wide-range applications in networking, facility location problems, and their symmetric properties. A resolving set is a subset of vertices of a connected graph such that each vertex of the graph is determined uniquely by its distances to that set. A resolving set of the graph that has the minimum cardinality is called the basis of the graph, and the number of elements in the basis is called the metric dimension of the graph. In this paper, the metric dimension is computed for the graph Gn1,k constructed from the circulant graph Cn1,k by subdividing its edges. We have shown that, for k=2, Gn1,k has an unbounded metric dimension, and for k=3 and 4, Gn1,k has a bounded metric dimension.


Introduction
Resolvability of graphs becomes an important parameter in graph theory due to its wide applications in different branches of mathematics, such as facility location problems, chemistry, especially molecular chemistry [1], the method of positioning robot networks [2], the optimization problem in combinatorics [3], applications in pattern recognition and image processing [4], and the problems of sonar and Coast Guard LORAN [5].
e resolvability of graphs depends on the distances in graphs. e distance between two vertices in a connected graph is the smallest distance connecting those two vertices. e representation of a vertex u with respect to the set W is denoted by r(u, W) and is defined as a k-tupple (d(u,w 1 ), \dots, d(u,w n )), where w 1 , \dots, w n \in W. e set W is called the resolving set [1] or sometimes locating set [5] if each vertex of the graph has a unique representation with respect to W. A resolving set of the graph that has the minimum cardinality is called the basis of the graph, and the number of elements in the basis is called the metric dimension of the graph, generally denoted by β(G).
Motivated by the problem of uniquely determining the location of an intruder in a network, the concept of metric dimension was first introduced by Slater in [5,6] and studied independently by Harary and Melter in [7]. Applications of this invariant to the navigation of robots in networks are discussed in [2], and applications to chemistry are given in [1], while applications to the problem of pattern recognition and image processing, some of which involve the use of hierarchical data structures, are given in [4].
A family of connected graphs F is said to have a bounded metric dimension if the metric dimension of each graph in F is bounded above by a positive integer. Otherwise, F has an unbounded metric dimension.
It is important to note that to determine the graph has a bounded metric dimension is an NP-complete problem [19]. Some bounds for this parameter, in terms of the diameters of the graph, are given in [2], and it was shown in [1,2,4,20] that the metric dimension of the tree can be determined efficiently. However, it is highly unlikely to determine the dimension of the graph unless the graph belongs to such family for which the distance between vertices can be computed systematically.
Geometrically, by subdividing an edge, we mean to insert a new vertex in the edge such that the existing edge is divided into two edges. e subdivision of the graph G is a graph obtained after performing a sequence of edge subdivision. Subdivision of graphs is an important tool to determine whether the graph is planar or not. In [24], plane graphs are characterized using subdivision as follows: A necessary and sufficient condition of a graph to be planar is that each of its subdivision is planar A necessary and sufficient condition of a graph to be planar is if it does not contain a subdivision of K 5 or K 3,3 In this paper, we have investigated the resolvability of subdivision of circulant graph C n [1, k] for k ≥ 2. It is shown that, for k � 2, this class has an unbounded metric dimension, and for k � 3 and 4, it has a bounded metric dimension.

Metric Dimension of Subdivision of Circulant
Graph C n [1, k] for k ≥ 2 e circulant graphs are an important class of graphs that can be used in local area networks.
A circulant graph on n vertices and m parameters a 1 , . . . , a m , where each parameter a i is at most half of n, is e graph G n [a 1 , . . . , a k ] is a graph obtained from C n [a 1 , a 2 , ..., a m ] by subdividing all the edges of C n [a 1 , a 2 , ..., a m ] except the edges between vertices v i and v i+1 .
In this paper, the resolvability of G n [1, k] is investigated. Let u i be the added vertex in each of the edge v i v i+k . us, the graph G n [1, k] has 2n vertices and 3n edges. Let x i and x j be two vertices of G n [1, k]; then, the gap between vertices x i and In the following theorem, it is shown that the metric dimension of the graph G n [1,2] is unbounded.
Proof. Let W � x i : 1 ≤ i ≤ q be a minimum resolving set of G n [1,2]. We have two cases: either then v i+2 also belongs to W because, otherwise, v i+2 and u i+1 will have the same representation.
Claim 2. If x i � u i for some i, then u i+3 must belong to W because, otherwise, v i+3 and u i+2 will have the same representation.
Both these case imply that the two consecutive vertices in W can have at most distance 3. us, the gap between two vertices of W is at most 3. Since vertices presented on the outer cycle are n, therefore, q ≥ n/3 ⌊ ⌋. Hence, To prove the upper bound, consider the set [1,2]. e construction of W shows that every vertex in W determines a gap of size 3. For be the set of vertices determined by the two consecutive vertices u i and u i+3 . It is enough to show that every vertex in S is uniquely determined by some vertices in W.
e vertices v i and v i+2 are the only vertices in G n [1,2] that are at distance 1 from , and u i−2 are the only vertices in G n [1,2] that are at distance 2 from u i . e vertices v i−1 and u i−2 also have the same distance 5 from u i+3 , but they can be resolved by the vertex u 4 n/3 ⌊ ⌋−1 . e vertices v i+1 and u i+2 also have the same distance 3 from u i+3 , but they can be resolved by the vertex u i+6 . e vertex v i+3 is the unique vertex in G n [1,2] is shows that every vertex in the set S is uniquely determined by some vertices in the set W. us, W becomes a resolving set, and From equations (2) and (3), we have β(G n [1,2]) � n/3 ⌊ ⌋. In the next results, it is shown that the graph G n [1, k] has constant metric dimension for 3 ≤ k ≤ 4.
Proof. Let W � u 1 , u 3 , u 5 be the set of vertices in G n [1,3]. It is enough to show that every vertex of the graph G n [1,3]  Discrete Dynamics in Nature and Society For the remaining vertices, we have is shows that every vertex of the graph G n [1,3] is determined uniquely by some of the vertices in W. Hence, W becomes a resolving set, and Now, to compute the lower bound, suppose, on contrary, that W is a minimum resolving set of G n [1,3] and if j ≡ 0, 1(mod 3), then

If W Contains Both
Vertices from u i . One can suppose without losing any generality that W � u 1 , u j : 2 ≤ j ≤ n . However, in this case, if j ≡ 0(mod 3), then Discrete Dynamics in Nature and Society and if j ≡ 1, 2(mod 3), 2.3. If One Each from v i and u j Belongs to W. One can suppose without losing any generality that W � v i , u j : However, in this case, if j ≡ 0(mod 3), then and if j ≡ 1, 2(mod 3), us, there is no resolving set of G n [1,3] having two vertices. is implies that From (7) and (14), we get 4 be the set of vertices in G n [1,4]. It is enough to show that every vertex of the graph G n [1,4] is uniquely determined by some vertices in W. For this, the representations of each vertex are calculated as follows. e representations of outer vertices v i : 1 ≤ i ≤ n of G n [1,3] are calculated as follows: If i � (n + 1)/2 ⌊ ⌋ + 1, , for n � 6k + 1 and k ≥ 3, , for n � 6k + 3 and k ≥ 2, , for n � 6k + 4 and k ≥ 2.
is shows that every vertex of the graph G n [1,4] is uniquely determined by some of the vertices in W. Hence, W become a resolving set, and Let n � 6k + l and l ∈ 0, 1, 3, 4 { }. We show that there is no resolving set of G n [1,4] with three elements. Suppose, on contrary, that W is a minimum resolving set of G n [1,4] of cardinality 3. We have the following possibilities to choose the vertices of W. □
2.6. If Two Vertices from u i and One Vertex from v j Belong to W. One can suppose without losing any generality that W � v 1 , u i , u j , where 1 ≤ i ≤ 3k and i ≤ j ≤ 6k.