#### Abstract

A family of connected graphs is a family with constant metric dimension if dim() is finite and does not depend upon the choice of in . In this paper, we show that the graph and the graph obtained from the antiprism graph have constant metric dimension.

#### 1. Notation and Preliminary Results

For a connected graph , the distance between two vertices is the length of a shortest path between them. A vertex of a graph is said to resolve two vertices, and , of if . Let be an ordered set of vertices of and let be a vertex of . The representation of the vertex with respect to denoted by is the -tuple . If distinct vertices of have distinct representations with respect to , then is called a resolving set, for [1]. A resolving set of minimum cardinality is called a metric basis for , and the cardinality of this set is the metric dimension of , denoted by .

For a given ordered set of vertices of a graph , the th component of is if and only if . Thus, to show that is a resolving set it suffices to verify that for each pair of distinct vertices .

Caceres et al. [2] found the metric dimension of fan , and Javaid et al. [3] found the metric dimension of Jahangir graph .

In [1], Chartrand et al. proved that a graph has metric dimension if and only if it is a path; hence paths on vertices constitute a family of graphs with constant metric dimension. They also showed that cycles with vertices also constitute such a family of graphs as their metric dimension is . In [2], Caceres et al. proved that Prisms are the trivalent plane graphs obtained by the cartesian product of the path with a cycle . In [4], Javaid et al. proved that this family has constant metric dimension. In [4], Javaid et al. also proved that the antiprism graph , constitutes a family of regular graphs with constant metric dimension as , for every .

In this paper, we extend this study to antiprism-related graphs (see Figure 1). The graph is defined as follows: for each vertex , of the outer cycle of the antiprism graph, we introduce a new vertex , and join to and , , with taken modulo . Thus, . Here are the inner cycle vertices, are the outer cycle vertices, and are adjacent vertices to the outer cycle. We define the graph as follows: for each vertex , of the outer cycle of the antiprism graph, we introduce a new vertex and join to , . Thus, . Here, are the inner cycle vertices, are the outer cycle vertices, are the pendant vertices adjacent to the outer cycle vertices.

**(a)**

**(b)**

#### 2. Antiprism-Related Graphs with Constant Metric Dimension

In this section, we show that , have constant metric dimension.

Theorem 1. *Let be an integer, then .*

*Proof * *Case 1. *Let , , . Consider the set . We show that is a resolving set for . We find the representations of the vertices of with respect to :
We note that there are no two vertices having the same representation, which implies that .

To prove the theorem, it is sufficient to show that . Contrarily, assume that there exists a resolving set with . We have the following possibilities.(1)Both vertices of belong to . Without loss of generality, we suppose that one resolving vertex is and the other is . For , we have . For , , a contradiction.(2)Both vertices of belong to . Without loss of generality, we suppose that one resolving vertex is and the other is . For , we have . For , , a contradiction.(3)Both vertices of belong to . We suppose that one resolving vertex is and the other is . For , we have . For , , a contradiction.(4)One vertex belong to and the other belong to . Consider one resolving vertex is and the other is . For , we have . For , , a contradiction.(5)One vertex belong to and the other belong to . Consider one resolving vertex is and the other is . For , we have . For , a contradiction.(6)One vertex belong to and the other belong to . Consider one resolving vertex is and the other is . For this we have , a contradiction.

Hence, from above it follows that there is no resolving set with two vertices for . Therefore, , which implies that .*Case 2.* Let , . Consider the set . We show that is a resolving set for . We compute the representations of the vertices of with respect to :
We observe that there are no two vertices having the same representation, which implies that .

We now show that . Suppose contrarily with . Proceeding on the same way as in Case 1, it can be shown that no such can be a resolving set. Finally, from Cases 1 and 2, we get , which completes the proof.

Theorem 2. *Let be an integer, then .*

*Proof * *Case 1.* Let , , . Consider the set . We show that is a resolving set for . We compute the representations of the vertices of with respect to :

We observe that there are no two vertices having the same representations implying that .

To prove the theorem, it is sufficient to show that . Contrarily, assume that there exists a resolving set with . We have the following possibilities.(1)Both vertices of belong to . Without loss of generality, we suppose that one resolving vertex is and the other is . For , we have . For , a contradiction.(2)Both vertices of belong to . Without loss of generality, we suppose that one resolving vertex is and the other is . For , we have . For , a contradiction.(3)Both vertices of belong to . We suppose that one resolving vertex is and the other is . For , we have . For , a contradiction.(4)One vertex belong to and the other vertex belong to . Consider one resolving vertex is and the other is . For , we have . For , a contradiction.(5)One vertex belong to and the other vertex belong to . Consider one resolving vertex is and the other is . For , we have . For , a contradiction.(6)One vertex belong to and the other vertex belong to . Consider one resolving vertex is and the other is . For , we have . For , a contradiction.

Hence, from above it follows that there is no resolving set with two vertices for implying that .*Case 2.* For , , . Consider the set . We show that is a resolving set for . We compute the representations of the vertices of with respect to :

We observe that there are no two vertices having the same representation. Thus, .

We now show that . Assume that . Proceeding on the same lines as in Case 1, we can show that . Finally, from Cases 1 and 2, we get .

#### 3. Conclusion

In this paper, we have studied the metric dimension of some families of graphs which are the extension of the antiprism graph. We have seen that the metric dimension of these graphs is finite and does not depend on the order of the graph and only three appropriately chosen vertices suffice to resolve all the vertices of these graphs.

#### Acknowledgments

The authors are indebted to the anonymous referees for their many valuable comments and suggestions on the earlier version of this paper. This research was partially supported by FAST-NU, Peshawar, and by the Higher Education Commission of Pakistan.