Discrete Dynamics in Nature and Society

Volume 2018, Article ID 4531958, 10 pages

https://doi.org/10.1155/2018/4531958

## The Metric Dimension of Some Generalized Petersen Graphs

^{1}Institute of Computing Science and Technology, Guangzhou University, Guangzhou 510006, China^{2}Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran^{3}School of Information Science and Engineering, Chengdu University, Chengdu 610106, China^{4}School of Mathematics and Physics, Anhui Jianzhu University, Hefei 230601, China

Correspondence should be addressed to Jia-Biao Liu; moc.361@daoabaijuil

Received 26 March 2018; Accepted 9 July 2018; Published 1 August 2018

Academic Editor: Xiaohua Ding

Copyright © 2018 Zehui Shao et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

The distance between two distinct vertices and in a graph is the length of a shortest -path in . For an ordered subset of vertices and a vertex in , the code of with respect to is the ordered -tuple . The set is a resolving set for if every two vertices of have distinct codes. The* metric dimension* of is the minimum cardinality of a resolving set of . In this paper, we first extend the results of the metric dimension of and and study bounds on the metric dimension of the families of the generalized Petersen graphs and . The obtained results mean that these families of graphs have constant metric dimension.

#### 1. Introduction

Let be a connected graph with vertex set and edge set . The distance between two distinct vertices and in , denoted by , is the length of a shortest -path. For positive integer and a vertex , the -neighborhood of is the set . For an ordered subset of vertices and a vertex in , the code of with respect to is the ordered -tuple . The set is a resolving set [1] (or locating set [2]) for if every two vertices of have distinct codes. The* metric dimension* of , denoted by , is the minimum cardinality of a resolving set of . A resolving set containing a minimum number of vertices is called a basis for [3].

Graph theory is a powerful tool to model the real world applications such as physical-chemical property testing [4, 5]. Motivated by the problem of uniquely determining the location of an intruder in a network, the concept of metric dimension of a graph was introduced by Slater in [6], where the metric generators were called locating sets. The concept of metric dimension of a graph was also introduced by Harary and Melter in [1], where metric generators were called resolving sets. Applications of this invariant to the navigation of robots in networks are discussed in [7] and applications to chemistry in [8, 9]. This graph parameter was studied further in a number of other papers including, for instance, [10–19]. Several variations of metric generators including resolving dominating sets [20], independent resolving sets [21], local metric sets [22], strong resolving sets [23], mixed metric dimension [24], and -metric dimension [25] have since been introduced and studied.

We observe from definition that the property of a given set of vertices of a graph to be a resolving set of can be tested by investigating the vertices of because every vertex is the unique vertex of whose distance from is 0. If , we say that vertex distinguishes vertices and .

For natural numbers and , where , a generalized Petersen graph is a graph with vertex set , where and , and edge set , where , , and , where subscripts are taken modulo (see [2, 26]). We observe that, for each ,If and , then clearly and for each .

Javaid et al. [27] proved that for and posed the following problem.

*Problem 1. *Is the generalized Petersen graphs , for and , a family of graphs with constant metric dimension?

Some partial answers are given to aforementioned problem as follows.

Theorem 2 (see [28]). *For , when and if .*

Theorem 3 (see [29]). * *

In [30, 31], it was showed that

Theorem 4. *(i) if .**(ii) *

In [32], it was proved that

Theorem 5. *For , if , when for even , and otherwise.*

In this paper, we first extend the results of Theorems 3 and 5.

We make use of the following result in this paper.

Theorem 6 (see [7]). *If is a graph of order , diameter and metric dimension , then .*

#### 2. Main Results

Next result extends Theorem 3.

Lemma 7. *Let be a connected graph and let or for each . Then .*

*Proof. *Clearly, for any and for any we haveSuppose, to the contrary, that is a resolving set of . Since or , we deduce from (7) and the Pigeonhole principle that there exist two vertices such that , a contradiction.

Theorem 8. *For ,**(i) If or , then .**(ii) If , then .*

*Proof. *If , then let . The code of with respect to in is presented in Table 1 yielding .

Now, we show that . Suppose, to the contrary, there exists a resolving set of . First let . We may assume w.l.o.g. that . By (1), we have . For each , we have . By the Pigeonhole principle, we have for some and this leads to a contradiction. Now let . Assume without loss of generality that and for some . If , then , and if , then , a contradiction. Thus, and so .

If , then let . The code of with respect to in is presented in Table 2 showing that .

Next, we show that . Suppose, to the contrary, there exists a resolving set of . As above, we may assume that . We may assume w.l.o.g. that and for some . If , then , , and if , then we have , , a contradiction.

If , then let . The code of with respect to in is presented in Table 3 yielding .

Since , we deduce from Theorem 6 that . Thus, .

If , then let . The code of with respect to in is presented in Table 4 implying that .

Since , it follows from Theorem 6 that . Hence, .

If , then let . The code of with respect to in is presented in Table 5. This implies that .

Analogous to the proof of the case , we can obtain the desired lower bound with a more complicated analysis. Also it can be verified by computer search.

If , we can verify the results by computer. If and , we have and for any . Now by Lemma 7, we have . Now, the proof is complete.