Abstract
Let be a connected graph. A subset of vertices of is said to be a resolving set of , if for any two vertices and of there is at least a member of such that . The minimum number that any subset of vertices with is a resolving set for , is called the metric dimension threshold, and is denoted by . In this paper, the concept of metric dimension threshold is introduced according to its application in some real-word problems. Also, the metric dimension threshold of some families of graphs and a characterization of graphs of order for which the metric dimension threshold equals 2, , and are given. Moreover, some graphs with equal the metric dimension threshold and the standard metric dimension of graphs are presented.
1. Introduction
Throughout this paper, all graphs considered are assumed to be finite, simple, undirected, and connected. For a graph , the vertex set and the edge set of are denoted by and , respectively. We remind that all notations and terminologies are standard here and taken mainly from the standard books of graph theory. For instance as usual, we denote the path, the cycle, the star, and the complete graph on vertices by , , , and , respectively. The distance between two vertices and , denoted by , is the number of edges in a shortest path from and . Let be a graph with . The eccentricity is the distance between and a vertex farthest from in . The diameter of is the greatest eccentricity among the vertices of . For an ordered subset of vertices and a vertex in a connected graph , the metric -representation of is the vector . The set is a resolving set for if every two vertices of have distinct -representations. Particularly, a vertex resolves a pair of vertices if . The minimum cardinality of a resolving set of is called metric dimension of and denoted by . This concept was first introduced by Slater in 1975 [22]. To see more details about metric dimension of graphs, we recommend references [2, 5, 9, 26].
Also, several versions of the standard metric dimension have been introduced and studied. In the following, we will discuss some of these versions.
In 2004, Sebö and Tannier introduced and studied a stronger version of the standard metric dimension which is called the strong metric dimension [21]. In the standard metric dimension, a resolving set uniquely specifies the position of vertices; however, this set cannot distinguish distances in the graph. Sebö and Tannier’s goal was to introduce a set called a strong resolving set so that in addition to vertices, they also distinguish distances. For the sake of it, they said, one vertex strongly resolves vertices and , if there exist the smallest path between ( respectively ) and such that ( respectively ) belongs to the path. As well, in [11], Okamoto et al. introduced a special version of the standard metric dimension. In this way, instead of all vertices in the standard metric dimension, only adjacent vertices have a different representation from the metric generator. The cardinality of the smallest set which resolves every two adjacent vertices is called the local metric dimension. Researchers have had significant results on this concept. Among the interesting correlations obtained between the local metric dimension and the independent set, it is shown that is an upper bound for the local metric dimension of a graph with order . The classification of graphs that apply to the above boundary equation is still open. For more results, the reader can refer to papers [4, 17, 24].
In addition, other versions of the metric dimension had been introduced, including -metric dimension, adjacency dimension, resolving partitions, edge metric dimension, mixed metric dimension, and fractional metric dimension [1, 7, 8, 13–15].
Graph theory has many applications in various sciences, including chemistry, to study some related parameters: see, for example, [18, 19]. Also, there are many applications for this invariant including error correcting codes, robot navigation, privacy in social networks, chemistry, pattern recognition, coin weighing, image processing, and locating intruders in networks [3, 5, 6, 12, 16, 20, 22, 25]. In situations where the points in the (submarine, drone, etc.) robot’s movement space are fixed and without damage, a metric generator set is used to locate the robot (submarine, drone, etc.). Now, consider the situation in which the points in the robot movement space are vulnerable. In this case, some vertices of could damage and unusable. Therefore, there is a need to replace with another metric generator set such as . On the other hand, the problem of finding a metric generator set is -hard, and as a result, finding takes a long time. This reason inspired us to use the following concept in such situations.
For any connected graph , we define metric dimension threshold, denoted by , to be the minimum number such that any subset of vertices with is a resolving set for . In this paper, we determine the metric dimension threshold for some well-known family of graphs. We also characterize graphs of order for which is equal to 2, , or . Finally, we present some graphs with the property that .
2. Basic Results
In this section, we determine the metric dimension threshold of paths, cycles, and the Petersen graph. Also, we characterize all graphs with order such that or .
Clearly, for every connected graph with order , we have . It is especially suitable for the robot if the resolving set is of size . In fact, finding the graph with the property that is on attention.
Remarks 1. Let be a connected graph.(i)Letbe a positive integer. For every subsetwith, we have, for, and then, for everywith, we have, for.(ii)Assume that. Then, it follows from (i) that any subsetof vertices ofwithis resolving.
For a vertex of , we use to denote the set of its neighbours. Also, for a set of vertices of graph , we set .
Theorem 1. For any , we have .
Proof. At first, we claim that . To see this, we consider the set , where is a pendant vertex in . It is not hard to see that is not a resolving set of .
Let be an arbitrary subset of vertices of with two elements. Since the set included a pendant vertex of is a resolving set, we may assume that and are not pendant vertices. Consider arbitrary vertices and of , and assume that . Let and denote the paths between and the pendant vertices . Clearly, both and cannot belong to or . Without loss of generality, we may assume that , and . Suppose that . If belongs to the path between and , via , we have . Otherwise, in the case that , we have . This means that , for any position of . Therefore, , and the proof is complete.
Corollary 1. Let be a graph. Then, if and only if .
Theorem 2. For each integer , we have
Proof. Assume first that . We claim that . To see this, suppose that , for some vertices and of with . Then, , and so .
In the following, we show that . To achieve this, let be an arbitrary subset of the vertices of . Also, assume that and are arbitrary vertices of such that and . Thus, . Suppose the path between and includes (respectively, ) are denoted by (respectively, ). We have and . Hence, or . In any cases, it is easy to check that . Therefore, .
Now, let , and be an arbitrary subset of vertices of . Suppose that for some vertices and of . By using a method similar to that we used in to above paragraph, one can easily check that . Hence, . Since , we have , which completes the proof.
Recall that, the Kneser graph is the graph whose vertices are in a 1–1 correspondence with the -sized subsets of , and there is an edge between two vertices if their corresponding -subsets are disjoint [10]. The Petersen graph has shown in Figure 1.
Theorem 3. If is the Petersen graph, then .
Proof. Consider the subset of vertices of . Then, , where and . This implies that .
Suppose that is a subset of , which is not a resolving set for and seeks a contradiction. Since , two vertices and can be selected in the following cases:(i)Case 1. . Since is a 3-arc transitive graph [10], without loss of generality, we may assume that and . Suppose that is an arbitrary vertex of . If , then it is easy to see that . Otherwise, we have . Now, when , we have or . Also whenever , we have . The only common neighbour of and is . Assume that , and we are looking for elements for which there are other candidate members for . Members of must be selected with the following conditions:(i)Thus, or , for . The vertices obtained in this way are , , and . Now, any other vertex that we consider as resolves vertices and , which is a contradiction. Case 2. . In this situation, by using a method similar to that we used in Case 1, one can obtain a contradiction.
Theorem 4. Let be a connected graph of order . Then, if and only if there exist vertices and of such that .
Proof. Assume that . Then, there exist two vertices and of such that is not a resolving set. Hence, the distance of and from vertices of is the same. We claim that there exists a vertex with in such that is adjacent to . To achieve this aim, suppose that, for any vertex with in , we have and seek a contradiction. Since is a connected graph, we have . Thus, for any vertex , we have which is a contradiction. This means that . Since the distance of any vertex of from is 1, we have .
Assume that there exist vertices and of such that . Clearly, cannot be a resolving set of . Thus, . Therefore, , and the proof is complete.
The following corollaries are immediate from Theorem 4.
Corollary 2. There is an infinite number of graph with .
Corollary 3. Let be a complete multipartite graph with order . Then, .
Corollary 4. Let be a graph with order . Then, if and only if.
The partition of is called a distance partition with reference to the vertex if and contain those vertices which are at distance from for . The sets are called distance partite sets.
Theorem 5. Let be a graph with order . Then, if and only ifsatisfies in the following conditions:(a)(a1)There exist vertices, andofsuch that(a2)(a3)The vertexis a pendant vertex orforand, wheres are the distance partite sets ofreference to(b)For any verticesand,.
Proof. If , then there exist vertices , and of with property that is not a resolving set. Thus, there exist two vertices of with the same -representation. Assume that . Since the set must resolves vertices and , we have . Suppose that . Then, . For every , since , we have . This implies that which is a contradiction. Similarly, it is impossible to have . Hence, and .
For (a2), it is enough to show that . Assume that , and we seek a contradiction. Since is connected, we may have a vertex such that . If , then, for any member of , as , we have , a contradiction. Otherwise, if , then there exists a vertex such that . Thus, . Hence, resolves vertices and , which is a contradiction.
For (a3), suppose that is not a pendant vertex. Hence, has some neighbours in , and so it is easy to show that is not a member of . However, if or , for , one can see that the vertices in keep their distance from and . Suppose, contrary to our claim, that , for an integer with . Then, there exists vertex such that is adjacent to . Now, if , then . Since , we have , which is impossible. If , then the smallest path between and contains . Thus, resolves and , a contradiction. If , then we have two cases. First, suppose that is adjacent to . Then, , and so . Finally, if is not adjacent to , one can check that . Therefore, in both cases, is a resolving set, whenever , for , which is a contradiction.
For the converse, assume that satisfies the conditions (a) and (b). In view of Theorem 4, the statement (b) implies that . Let be the subset of vertices , where , and are the vertices described in (a). One can easily check that . This implies that . Hence, , and the proof is complete.
Now, we consider the family of graphs of order with the property that . Clearly, the smallest one is . For a positive integer , let be the graph with and . By matching to in the previous theorem, one can check that . Thus, we have the following corollary.
Corollary 5. There are an infinite number of graphs with .
For vertices and of a graph , we shall use to denote the set of those vertices with the property that . Now, we define to be the greatest for . Hence,
One can easily check that . In the next theorem, we present a lower bound for in terms of .
Theorem 6. Let be a graph. Then, .
Proof. Suppose that , for vertices and of . Since the set does not resolve vertices and , the result follows.
It is obvious that the above upper bound is sharp. For instance, if is a path of length , then . Moreover, all graphs with the property that satisfy in the equality . In fact, Theorem 4 states that if and only if . So, in view of the previous theorem, we can express the following proposition.
Proposition 1. Let be a graph with . Then, .
3. Graphs with
In Corollary 4, we show that the only graph of order with is the complete graph . In this section, we want to investigate graphs of order with the property that , for . Moreover, we characterize all graphs with .
Graphs with metric dimension two were studied by Sudhakara and Hemanth Kumar in [23]. Also, Chartrand et al. [6] provided a characterization of graphs of order and metric dimension . In the following, we will use the results in [6, 23] to obtain a characterization of graphs of order with .
Theorem 7 [6, Theorem 4]. Let be a connected graph of order . Then, if and only if , , or .
Theorem 8. If is a graph with order and , then .
Proof. Let , or . Then, is a multipartite graph and .
Now, assume that . If , then , which is a contradiction. Suppose that or . Without loss of generality, we may assume that . Hence, for two arbitrary vertices , we have . Thus, by Theorem 4, .
Corollary 6. There is no graph of order with property that .
Theorem 9 [23, Theorem 6.1]. Let be a graph which is not a path with and be the distance partition of with reference to the vertex where is the eccentricity of . The metric dimension of is 2 if and only if there exist verticesandsuch thatfor everyandwithand.
Lemma 1. Let be a graph with and be the distance partition of with reference to the vertex . For every and , and and with and , we have if and only if , for a .
Proof. Only one implication requires proof. Suppose that is a graph of order . We claim that , for every with . Suppose, contrary to our claim, that there exists with such that . Let . Then, . We need to consider the following four cases for . Case 1. . Thus, . This implies that , which is a contradiction. Case 2. . Suppose that is adjacent to . Hence, . So we have , which is a contradiction. Case 3. . Without loss of generality, we may assume that . This means that , again, which is a contradiction. Case 4. . In this situation, similar to that used in Case 3, one can obtain a contradiction.So, for any with , we have . This implies that is isomorphic to or , for some positive integer .
In the following, we show that if , then is an odd integer. On the contrary, suppose that , for an integer with . Assume that and that . Thus, and . Therefore, , which is a contradiction, and the proof is complete.
Theorem 10. Let be a graph. Then, if and only if is isomorphic to one of the graphs or , for a positive integer .
Proof. Let . First, we suppose that is isomorphic to one of the graphs or , for a positive integer . By Lemma 1, each and , and with and , implies that . By the way of contradiction, assume that there exist and and which and such that . Hence, there exist distinct vertices and in . Since and , we have and . This means that set can not to be a resolving set of , which is a contradiction.
The converse implication follows almost immediately from Theorems 1 and 2.
The next corollary is obtained from Theorem 10, [6, Theorem 2], and Theorem 2.
Corollary 7. Let be a graph. Then, if and only if , for a positive integer .
We end the paper with the following problem.
Open Problem. Let be a positive integer . Characterize the class of graphs with the property that
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.