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.
The following theorem extends the result of Theorem 5.
Theorem 9. Let be the graph with ; then if , then .
Proof. If , let . The code of with respect to in is presented in Table 6 showing that .
If , let . The code of with respect to in is presented in Table 7 showing that .
Note that the diameter of is 4; by Theorem 6, we have .
If , let . The code of with respect to in is presented in Table 8 showing that .
Note that the diameter of is 4; by Theorem 6, we have .
If , let . The code of with respect to in is presented in Table 9 showing that .
Note that the diameter of is 5; by Theorem 6, we have .
If , let . The code of with respect to in is presented in Table 10 showing that .
Note that the diameter of is 6; by Theorem 6, we have .
Theorem 10. Let be the graph with ; then
Proof.
Case 1 ( or (mod 2)). If , let . The code of with respect to in is presented in Table 11 showing that .
If , let . The code of with respect to in is presented in Table 12 showing that .
If , let . Then the codes of the outer vertices are and the codes of the inner vertices are , where It can be verified that there are no two vertices on the outer cycle with the same codes, and there are no two vertices in the inner cycle and outer cycle with the same codes. Moreover, no two vertices in the inner cycle have same codes. Hence, is a resolving set of for even . This means that for even .
Case 2 ( (mod 2) and ). Let . Then the codes of the outer vertices are , whereThen the codes of the inner vertices are , whereIt can be verified that there are no two vertices on the outer cycle with the same codes, and there are no two vertices in the inner cycle and outer cycle with the same codes. Moreover, no two vertices in the inner cycle have same codes. Hence, is a resolving set of for odd . This means that for odd .
Theorem 11. Let be the graph with ; then
Proof.
Case 1 ( (mod 2)). If , let . The code of with respect to in is presented in Table 13 showing that .
If , let . Then the codes of the outer vertices are and the codes of the inner vertices are , where It can be verified that there are no two vertices on the outer cycle with the same codes, and there are no two vertices in the inner cycle and outer cycle with the same codes. Moreover, no two vertices in the inner cycle have same codes. Hence, is a resolving set of for even . This means that for even .
Case 2 ( (mod 2) and ). If , let . The code of with respect to in is presented in Table 14 showing that .
Note that the diameter of is 5; by Theorem 6, we have .
If , we can confirm that by an exhaustive search. Let . The code of with respect to in is presented in Table 15 showing that .
If , let . Then the codes of the outer vertices are and the codes of the inner vertices are , where It can be verified that there are no two vertices on the outer cycle with the same codes, and there are no two vertices in the inner cycle and outer cycle with the same codes. Moreover, no two vertices in the inner cycle have same codes. Hence, is a resolving set of for odd . This means that for odd .
Now, we will show that . Note that for any , say , we have (see Figure 1) and for any , say , we have (see Figure 2). By Lemma 7, we have .
Data Availability
The resolving sets of some generalized Petersen graphs can also be found in https://www.researchgate.net/publication/324182258_Resolving_sets_of_some_generalized_Petersen_graphs_providing_the_corresponding_upper_bounds_for_metric_dimension?.
Disclosure
The authors confirm that the paper has been read and approved by all named authors and that there are no other persons who satisfied the criteria for authorship but are not listed. The authors further confirm that the order of authors listed in the paper has been approved by all of them.
Conflicts of Interest
The authors wish to confirm that there are no known conflicts of interest associated with this paper and there has been no significant financial support for this work that could have influenced its outcome.
Acknowledgments
This work was supported by the National Key Research and Development Program under Grants 2017YFB0802300, the Key Project of the Sichuan Provincial Department of Education under Grants 17ZA0079 and 18ZA0118, the Soft Scientific Research Foundation of Sichuan Provincial Science and Technology Department under Grant 2018ZR0265, and Applied Basic Research (Key Project) of Sichuan Province under Grant 2017JY0095.