Abstract

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 constructed from the circulant graph by subdividing its edges. We have shown that, for , has an unbounded metric dimension, and for and , has a bounded metric dimension.

1. 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].

The resolvability of graphs depends on the distances in graphs. The distance between two vertices in a connected graph is the smallest distance connecting those two vertices. The representation of a vertex with respect to the set is denoted by and is defined as a -tupple (d(,1), \dots, d(,)), where , \dots, \in . The set is called the resolving set [1] or sometimes locating set [5] if each vertex of the graph has a unique representation with respect to . 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 .

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 is said to have a bounded metric dimension if the metric dimension of each graph in is bounded above by a positive integer. Otherwise, has an unbounded metric dimension.

If every graph has a constant metric dimension, then is said to have constant metric dimension. A connected graph has if and only if is the path [1]; cycles have metric dimension for every . Also, generalized Peterson graphs , antiprism , and circulant graphs are families of graphs with constant metric dimension. The families of graphs having constant metric dimension are studied in [8â€“23].

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. The subdivision of the graph 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:â€‰A necessary and sufficient condition of a graph to be planar is that each of its subdivision is planarâ€‰â€‰A necessary and sufficient condition of a graph to be planar is if it does not contain a subdivision of or

In this paper, we have investigated the resolvability of subdivision of circulant graph for . It is shown that, for , this class has an unbounded metric dimension, and for and , it has a bounded metric dimension.

2. Metric Dimension of Subdivision of Circulant Graph for

The circulant graphs are an important class of graphs that can be used in local area networks.

A circulant graph on vertices and parameters , where each parameter is at most half of , is denoted by . If are vertices of , then there is an edge between two vertices and if and only if is one of . The parameters of are called generators of .

The graph is a graph obtained from by subdividing all the edges of except the edges between vertices and .

In this paper, the resolvability of is investigated. Let be the added vertex in each of the edge . Thus, the graph has vertices and edges. Let and be two vertices of ; then, the gap between vertices and is defined to be , where .

In the following theorem, it is shown that the metric dimension of the graph is unbounded.

Theorem 1. For ,

Proof. Let be a minimum resolving set of . We have two cases: either or , for some â–¡.

Claim 1. If for some , then also belongs to because, otherwise, and will have the same representation.

Claim 2. If for some , then must belong to because, otherwise, and will have the same representation.
Both these case imply that the two consecutive vertices in can have at most distance . Thus, the gap between two vertices of is at most . Since vertices presented on the outer cycle are , therefore, . Hence,To prove the upper bound, consider the set of vertices of . The construction of shows that every vertex in determines a gap of size .
For , let be the set of vertices determined by the two consecutive vertices and . It is enough to show that every vertex in is uniquely determined by some vertices in .â€‰The vertices and are the only vertices in that are at distance from , but and .â€‰â€‰The vertices , , , , and are the only vertices in that are at distance from . The vertices and also have the same distance from , but they can be resolved by the vertex . The vertices and also have the same distance from , but they can be resolved by the vertex . The vertex is the unique vertex in such that and .â€‰â€‰The vertex is the unique vertex in such that and .This shows that every vertex in the set is uniquely determined by some vertices in the set . Thus, becomes a resolving set, andFrom equations (2) and (3), we have .
In the next results, it is shown that the graph has constant metric dimension for .

Theorem 2. For ,

Proof. Let be the set of vertices in . It is enough to show that every vertex of the graph is determined uniquely by some of the vertices in . For this, the representations of each vertex are calculated as follows.
The vertices have representations , , , , , , , respectively.
The representations of the remaining vertices of are calculated as follows:The vertices of have representations , , , , , , respectively.
For the remaining vertices, we haveThis shows that every vertex of the graph is determined uniquely by some of the vertices in . Hence, becomes a resolving set, andNow, to compute the lower bound, suppose, on contrary, that is a minimum resolving set of of cardinality . We have the following possibilities to choose the vertices of .â–¡

2.1. If Contains Both Vertices from

One can suppose without losing any generality that . However, in this case, if , thenand if , then

2.2. If Contains Both Vertices from

One can suppose without losing any generality that . However, in this case, if , thenand if ,

2.3. If One Each from and Belongs to

One can suppose without losing any generality that . However, in this case, if , thenand if ,

Thus, there is no resolving set of having two vertices. This implies that

From (7) and (14), we get

Theorem 3. For , if , then

Proof. Let be the set of vertices in . It is enough to show that every vertex of the graph is uniquely determined by some vertices in . For this, the representations of each vertex are calculated as follows.
The representations of outer vertices of are calculated as follows:If ,If ,If ,The vertices have representations , , , , , , respectively.
The representations of the remaining vertices of are calculated as follows:If ,If ,If ,This shows that every vertex of the graph is uniquely determined by some of the vertices in . Hence, become a resolving set, andLet and . We show that there is no resolving set of with three elements. Suppose, on contrary, that is a minimum resolving set of of cardinality . We have the following possibilities to choose the vertices of .â–¡

2.4. If Only

One can suppose without losing any generality that , where and . In this case, the vertices of that have the same representation for all choices of the resolving set are given as follows:

2.5. If Two Vertices from and One Vertex from Belong to

One can suppose without losing any generality that , where and .

In this case, the vertices of that have the same representation for all choices of the resolving set are given as follows:

2.6. If Two Vertices from and One Vertex from Belong to

One can suppose without losing any generality that , where and .

In this case, the vertices of that have the same representation for all choices of the resolving set are given as follows:

2.7. If Contains All Three Vertices from

One can suppose without losing any generality that is a resolving set, where and .

In this case, the vertices of that have the same representation for all choices of the resolving set are given as follows:

Thus, there is no resolving set of having three elements. Hence,

From equations (25) and (30), we get

Theorem 4. For , if , then

Proof. Let be the set of vertices in . It is enough to show that every vertex of the graph is uniquely determined by some vertices in . For this, the vertices have representations , , , , , , , , , , , respectively.
Let . Then, the representations of the vertices of with respect to and the vertex are calculated as follows:The representations of the vertices of are calculated as follows: , , , , , , , , , .
The remaining vertices of have the following representation with respect to , and the vertex is calculated as follows:One can easily verify that each vertex of has unique representation with respect to . Hence, is a resolving set, andNow, to prove the lower bound, it is sufficient to show that there is no resolving set of with three elements. Suppose, on contrary, that is a minimum resolving set of of cardinality . DefineIt is easy to see that these are disjoint subsets of the vertex set of . We make the following claims.

Claim 3. Let be an arbitrary vertex of the resolving set that does not belong to . Then,