Abstract
For a connected graph , let be a vertex and be an edge. The distance between the vertex and the edge is given by . A vertex distinguishes two edges if A well-known graph invariant related to resolvability of graph edges, namely, the edge resolving set, is studied for a family of -regular graphs. A set of vertices in a connected graph is an edge metric generator for if every two edges of are distinguished by some vertex of . The smallest cardinality of an edge metric generator for is called the edge metric dimension and is denoted by . As a main result, we investigate the minimum number of vertices which works as the edge metric generator of double generalized Petersen graphs, . We have proved that when and when .
1. Introduction
The concept of the resolving set or the locating set was introduced by Slater [1] and then by Harary and Melter [2] separately. A resolving set of a given graph is the subset of the vertex set of the given graph so that the vertices of the graph can be located at a unique location by the distances of them from the vertices of . The cardinality of a such a smallest set for is called the metric dimension of , denoted by [1, 2]. Resolvability in graphs has diverse applications related to the navigation of robots in networks [3–5], pattern identification, and image processing. It has also many applications in pharmaceutical chemistry and drugs [6–8]. In [9], some interesting connections of metric generators of graphs with the mastermind game and coin weighing problem have been presented. This concept was used by Slater [1] to determine the location of an intruder in a network in a unique way. However, if there is an intruder that cannot be accessed by its nodes or vertices rather by using the links between them called edges, then, in this case, the location of an intruder in a network need not to be identified. Thus, in this situation, there is need to define extra condition to overcome the problem.
For a simple, undirected, and connected graph , the vertex set and edge set of is denoted by and , respectively. The vertices and the edges of can be resolved or determined with the help of distance parameter. The distance between a vertex and an edge in a graph is given by . Any two edges and are said to be resolved (identified) by a vertex of if and only if implies . A set of vertices is an edge resolving set for if and only if every two distinct edges of are resolved by some vertex in . The edge metric dimension of is the cardinality of such a smallest set , denoted by [10].
2. Literature Review
Motivated by this, Kelenc et al. [10] recently defined the concept of edge resolvability in graphs and studied its different properties. In their work, they extended the study of the resolving set to the edge resolving set. There are several graphs for which the metric dimension is equal to the edge metric dimension, the metric dimension is less than the edge metric dimension, and the metric dimension is greater than the edge metric dimension [10]. Moreover, exact values for the edge metric dimension of many classes of graphs were computed, while the upper and lower bounds of many other graphs have been given. The results of the mixed metric dimensions and the edge metric dimensions for some families of graphs are given in [11, 12]. Peterin and Yero [13] computed exact formulae for different products of graphs. Zubrilina [14] categorized some properties of edge metric dimension of graphs. Moreover, we refer the reader to the work [15–19] where some related results on this topic can be found.
3. Double Generalized Petersen Graph
The double generalized Petersen graph is a cubic graph, for and , with vertex set and edge set . The exterior cycle of the contains vertices in which every vertex is adjacent to the vertex to make spokes of type . Similarly, the inner cycle of the contains the vertices , in which every vertex is adjacent to the vertex to make spokes of type [20]. Throughout the study, the indices will be taken under modulo . For and , the double generalized Petersen graph is shown in Figure 1.
4. Edge Metric Dimension of
In this section, we investigate the edge metric dimension of by proving its upper and lower bounds.
4.1. Upper Bound
For a set , the code of any edge is the -vector:
Equivalently, the set is an edge resolving set for if and only if each edge in has the unique code with respect to . That is, is an edge metric for if and only if two edges and imply . Next, first four lemmas provide the upper bound for the edge metric dimension of .
Lemma 1. For and , we have
Proof. Let . Define and . Codes of all the edges of with respect to are given in Tables 1–10.
From Tables 1–10, it can be seen that each edge of has the unique code w. r. t. . So, is an edge resolving set for and .
Lemma 2. For and , we have
Proof. Let . Define and . Codes of all the edges of with respect to are given in Tables 11–20.
From Tables 11–20, it can be observed that each edge of has unique codes w. r. to . So, is an edge resolving set for and .
Lemma 3. For and , we have
Proof. Let . Define and . Codes of all the edges in with respect to are given in Tables 21–29.
From Tables 21–29, we observe that each edge of has unique code w. r. t. . Hence, is an edge resolving set for and .
Lemma 4. For and , we have
Proof. Let . Define and . Codes of outermost edges, inner edges, and spokes of with respect to are given in Tables 30–39.
From Tables 30–39, we observe that each edge of has unique code w. r. t. . Thus, is an edge resolving set for and
4.2. Lower Bound
For all -regular graphs, the following lower bound for the edge metric dimension of any connected graph was explored in [21].
Lemma 5 (see [21]). If is a connected -regular graph, then .
As we know that are 3-regular graphs and , so the next result holds consequently.
Lemma 6. For any double generalized Petersen graph , we have
The next result provides the lower bound for the edge metric dimension of whenever
Lemma 7. For and , we have
Proof. We now prove that the cardinality of any minimum edge resolving set is . On the contrary, suppose that the cardinality of is , by Lemma 5. To prove that the existence of such edge resolving set is not possible, we have the following claims. Define , and let the vertex set of be , where , , , and .
Claim 1. The set can contain at most one vertex from either or . On contrary, let , then with out loss of generality, the third vertex of is either from or or or . Thus, we have following possibilities:(i)If , when , then when . For , we have the following edges with same codes: (ii)If , when , then when . For , the edges with same code: . Also when , we have .(iii)If , when and , then when . For , . When , . for . For , we have . For and , we have . When , . For and , we have . When , the edges with same codes: . Also when .(iv)If , when and , then when . For , . When , . If , then . For , the edges with same codes: . For and , the edges having same codes: and respectively.(v)If , when and , then for . When , . If , then . For , the following edges have same codes: . If , then . When , then .(vi)If , when and , then when and . If , then . For , we have . If , then . When , the edges have same codes are: . If , then . When and , then . For , . If , then . For and , the edges with same codes: and , respectively.
Claim 2. The set can contain at most one vertex either from or . Due to symmetry, it is enough to show that contains at most one vertex from . On contrary, let .(i)If , when , then for . If , then .(ii)If and , when and , then when . If , then . If , then . When , then . For and .If , then .(iii)If and ,when and , then when and . If , then . When , then . If , then edges having same codes: . For and , then edges with same codes: and , respectively.(iv)If , when , then for . If , then . For , we have .(v)If and when with , then when . If , then . For , we have . If , then edges with same codes: . For and and , then we have and .(vi)If , when and , then for and . If , then . If , then the following edges have same codes: . For , then . If , then . When , then .
Claim 3. If , when and , then for . If and , then . If and , then edges with same codes are: . When and , then . For , we have . If and , then . Moreover, for and , , for , and for , .(i)If , when and , then for and . If , the . For , we have following edges having same codes: . If , then . If and , then . When and , then . For , . If and , then . For and , and for , .(ii)If , when and , then for and . If and , then . For and , the edges with same codes are: . For and , . When and , then . For and , we have . If , then . For , we have the following edges with same codes: . If , then . For , and for , .(iii)If , when and , then If and and , then . For and , we have . When , then . If , then . The edges have same codes when . For and , we have . If , then . For , we have . When , then . For and and , and for and , .(iv)If , when and , then when . If , then . For , we have . The following edges and have same codes for , and when , respectively.(v)If , when and , then for . If and , then . When , then . For and and for , these are the edges and .(vi)If , when and , then when , and . If and , then . When , then . For , and for and and , .(vii)If , when and , then for and . If , then . When and and , then we have . For and , and when , then .Now, when , the remaining cases are as follows:
Claim 4. The set can contain at most one vertex from either or . On contrary, let , then with out loss of generality, the third vertex of is either from or or or . Thus, we have following possibilities: for . If , then . If when , then . For , then edges with same codes are: . When , then we have . If when , then and if when , then . For , edges having same codes are: .(i)If , when and , then If , then edges with same codes are: . For , . The edges and have same codes when . For when , and for , . Moreover, if , the edges having same codes are and .(ii)If , when and , then for . If , then . When , then edges having same codes are and . If , then we have . For when , then we have following edges with same codes and for , then and are the edges having same codes. If , then . For and for , it can be seen that, and , respectively.(iii)If , when and , then for . If and , then we have . When , then the edges have same codes with respect to . If , . When and , we have . When and , then . If , then . If and , then and .
Claim 5. The set contains at most one vertex either from or . Due to symmetry, it is enough to show that contains at most one vertex from . On the contrary, let .(i)If and , when and , then for . If and when , then . For and when , then we have . If , then . When , then edges having same codes are: . For , we have . If when and if when , then we have and , respectively.(ii)If and, when and , then For , we have . If , then . When , then . If , . For and when , then we have . If and when , then . When when , then . For when , then . Moreover, if , then .(iii)If and , when and , then for and , the edges having same codes are: . If , then . If , then . When and , the edges having same codes are: and , respectively.(iv)If , when and , then when . If , then . For and , then . If and and , then we have and .
Claim 6. If , when and , then for . When when , then . If when , then . For , we have . If when , then and when when , then . When and , the edges are . When and , the edges with same codes are: . If and , then . When and , then . For and , and .(i)If , when and , then if , then . When , then . For , we have . If when , then we have . When and , then . If , then . For and , . When , then . For when and when , and .(ii)If , when and , then If , then . If , then . For , we have . When and , then and when , and , then . If and , then the edges with same codes are: and if and , then . For and and when and , the edges having same codes are: and .(iii)If , when and , then for , . If , then . When , then . For and , the edges with same codes are: . If , then .(iv)If , when and , then if . For , . For , .(v)If , when and , then If , then we have . For , .(vi)If , when and , then for . if and , then . When and , then . When and , then . If and , then .(vii)If and with : for and , we have . Whenever , and . Whenever . For and , we have , . If , then .For and and for and , .
From all these claims, it yields that there is no such edge resolving set for . Therefore
From Lemmas 1–6, we have the following main result:
Theorem 1. For , we have
5. Conclusion
In this study, a family of double generalized Petersen graph has been considered in the context of edge resolvability. It has been investigated that minimum 4 vertices perform the edge resolvability in when (mod 4) and minimum 3 vertices perform the edge resolvability in when (mod 4).
Data Availability
All the required data are available within the article.
Conflicts of Interest
The authors declare that there are no conflicts of interests regarding the publication of this paper.