Metric Dimension Threshold of Graphs

LetG be a connected graph. A subset S of vertices ofG is said to be a resolving set ofG, if for any two vertices u and v ofG there is at least amemberw of S such that d(u, w)≠ d(v, w).eminimumnumber t that any subset S of verticesGwith |S| t is a resolving set forG, is called the metric dimension threshold, and is denoted by dimth(G). 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 G of order n for which the metric dimension threshold equals 2, n − 2, and n − 1 are given. Moreover, some graphs with equal the metric dimension threshold and the standard metric dimension of graphs are presented.


Introduction
roughout this paper, all graphs considered are assumed to be nite, simple, undirected, and connected. For a graph G, the vertex set and the edge set of G are denoted by V(G) and E(G), 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 n vertices by P n , C n , S n , and K n , respectively. e distance between two vertices u and v, denoted by d(x, y), is the number of edges in a shortest path from u and v. Let G be a graph with v ∈ V(G). e eccentricity e(v) is the distance between v and a vertex farthest from v in G. e diameter diam(G) of G is the greatest eccentricity among the vertices of G. For an ordered subset S v 1 , v 2 , . . . , v k of vertices and a vertex v in a connected graph G, the metric S-representation of v is the vector r(v|S) (d(v, v 1 ), . . . , d(v, v k )). e set S is a resolving set for G if every two vertices of G have distinct S-representations. Particularly, a vertex w ∈ V(G) resolves a pair of vertices v, u ∈ V(G) if d(v, w) ≠ d(u, w). e minimum cardinality of a resolving set of G is called metric dimension of G and denoted by dim(G). is concept was rst 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 speci es 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 a strongly resolves vertices b and c, if there exist the smallest path between b( respectively c) and a such that c( respectively b) 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 di erent representation from the metric generator. e cardinality of the smallest set which resolves every two adjacent vertices is called the local metric dimension. Researchers have had signi cant results on this concept. Among the interesting correlations obtained between the local metric dimension and the independent set, it is shown that n − β(G) is an upper bound for the local metric dimension of a graph G with order n.
e 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].
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 S 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 S could damage and unusable. erefore, there is a need to replace S with another metric generator set such as S′. On the other hand, the problem of finding a metric generator set is NP-hard, and as a result, finding S ′ takes a long time. is reason inspired us to use the following concept in such situations.
For any connected graph G, we define metric dimension threshold, denoted by dim th (G), to be the minimum number t such that any subset S of vertices G with |S| � t is a resolving set for G. In this paper, we determine the metric dimension threshold for some well-known family of graphs. We also characterize graphs G of order n for which dim th (G) is equal to 2, n − 2, or n − 1. Finally, we present some graphs G with the property that dim(G) � dim th (G).

Basic Results
In this section, we determine the metric dimension threshold of paths, cycles, and the Petersen graph. Also, we characterize all graphs G with order n such that dim th (G) � n − 1 or n − 2.
Clearly, for every connected graph G with order n, we have dim(G) ≤ dim th (G) ≤ n − 1. It is especially suitable for the robot if the resolving set S is of size dim(G). In fact, finding the graph G with the property that dim(G) � dim th (G) is on attention.  For every subsetS⊆V(G)with|S| � r, we haver(v|S) ≠ r(u|S), forv, u ∈ V(G), and then, for everyS⊆V(G)with|S| ≥ r, we haver(v|S) ≠ r(u|S), forv, u ∈ V(G).
For a vertex a of G, we use N(a) to denote the set of its neighbours. Also, for a set S of vertices of graph G, we set N(S) � ∪ a∈S N(a).
Proof. At first, we claim that dim th (P n ) ≠ 1. To see this, we consider the set S: � N(a), where a is a pendant vertex in P n . It is not hard to see that S is not a resolving set of P n .
Let S: � a, b { } be an arbitrary subset of vertices of P n with two elements. Since the set included a pendant vertex of P n is a resolving set, we may assume that a and b are not pendant vertices. Consider arbitrary vertices v and u of P n , and assume that d(v, a) � d(u, a) � k. Let P and P ′ denote the paths between a and the pendant vertices P n . Clearly, both v and u cannot belong to P or P ′ . Without loss of generality, we may assume that v ∈ P, and u ∈ P ′ . Suppose that b ∈ P. If b belongs to the path between v and a, via erefore, r(v|S) ≠ r(u|S), and the proof is complete.
Proof. Assume first that m � 2n. We claim that dim th (C m ) ≠ 2. To see this, suppose that S: � a, b { }, for some vertices a and b of G with d(a, b) � n. en, N(a) � v, u { }, and so r(v|S) � r(u|S).
In the following, we show that dim th (C m ) � 3. To achieve this, let S � a, b, c { } be an arbitrary subset of the vertices of C m . Also, assume that v and u are arbitrary vertices of C m such that d(v, a) � d(u, a) and a) for some vertices u and v of C m . By using a method similar to that we used in to above paragraph, one can easily check that

□
Recall that, the Kneser graph KG(r, s) is the graph whose vertices are in a 1-1 correspondence with the r-sized subsets of 1, 2, 3, . . . , s { }, and there is an edge between two vertices if their corresponding r-subsets are disjoint [10]. e Petersen graph KG (2,5) has shown in Figure 1. Proof. Consider the subset S: , v 5 is a subset of P, which is not a resolving set for P and seeks a contradiction. Since diam(P) � 2, two vertices u and v can be selected in the following cases: Since P is a 3-arc transitive graph [10], without loss of generality, we may assume that v � (1) { }. Now, any other vertex that we consider as v 5 resolves vertices v and u, which is a contradiction. Case 2. d(v, u) � 1. In this situation, by using a method similar to that we used in Case 1, one can obtain a contradiction.
Proof. (⇒) Assume that dim th (G) � n − 1. en, there exist two vertices v and u of G such that V(G) − v, u { } is not a resolving set. Hence, the distance of v and u from vertices of V(G) − v, u { } is the same. We claim that there exists a vertex w with u ≠ w in V(G) such that w is adjacent to v. To achieve this aim, suppose that, for any vertex w with u ≠ w in V(G), we have w≁v and seek a contradiction. Since G is a connected graph, we have v ∼ u.
us, for any vertex erefore, dim th (G) � n − 1, and the proof is complete. □ e following corollaries are immediate from eorem 4.

Corollary 2.
ere is an infinite number of graph G with dim th (G) � n − 1 .

Corollary 3.
Let G be a complete multipartite graph with order n . en, dim th (G) � n − 1.

Corollary 4. Let G be a graph with order
e sets V 0 , V 1 , V 2 , . . . , V k are called distance partite sets.

Theorem 5.
Let G be a graph with order n ≥ 4 . en, dim th (G) � n − 2if and only ifGsatisfies in the following conditions: Proof. If dim th (G) � n − 2, then there exist vertices v, u, and which is a contradiction. Similarly, it is impossible to have v, u { }⊆N(w). Hence, v ∈ N(w) and u ∉ N(w).

Journal of Mathematics
For (a2), it is enough to show that N(v) ∩ N(u) ≠ ∅. Assume that N(v) ∩ N(u) � ∅, and we seek a contradiction. Since G is connected, we may have a vertex { }, then there exists a vertex y ∈ A such that y ∈ N(u). us, y ∉ N(v). Hence, y resolves vertices v and u, which is a contradiction.
For (a3), suppose that w is not a pendant vertex. Hence, w has some neighbours in A, and so it is easy to show that w is not a member of N(v) ∩ N(u). However, if w ∈ N(V 1 )∖N(V j ) or w ∈ N(V 2 )∖N(V j ), for 3 ≤ j ≤ e(u), one can see that the vertices in A keep their distance from v and u. Suppose, contrary to our claim, that w ∈ N(V i ), for an integer i with 3 ≤ i ≤ e(u). en, there exists vertex z ∈ V i such that w is adjacent to z. Now, if i � 3, then d(z, u) � 3. Since d(z, v) � 2, we have d(y, u) ≠ d(y, v), which is impossible. If i ≥ 5, then the smallest path between u and z contains v. us, z resolves v and u, a contradiction. If i � 4, then we have two cases. First, suppose that v is adjacent to u. en, d(z, u) � 3, and so d(z, u) ≠ d(z, v). Finally, if v is not adjacent to u, one can check that d(z, u) � 4. erefore, in both cases, A is a resolving set, whenever w ∈ N(V i ), for 3 ≤ i ≤ e(u), which is a contradiction.
For the converse, assume that G satisfies the conditions (a) and (b). In view of eorem 4, the statement (b) implies where v, u, and w are the vertices described in (a). One can easily check that r(v|A) � r(u|A). is implies that dim th (G) > n − 3. Hence, dim th (G) � n − 2, and the proof is complete. □ Now, we consider the family of graphs G of order n with the property that dim th (G) � n − 2. Clearly, the smallest one is P 4 . For a positive integer k, let G k be the graph with V(G k ) � a 1 , a 2 , . . . , a k ∪ v, u, w { } and E(G k ) � a 1 v, a 2 v, . . . , a k v} ∪ a 1 u, a 2 u, . . . , a k u ∪ vw { }. By matching G k to G in the previous theorem, one can check that dim th (G k ) � k + 1. us, we have the following corollary.

Corollary 5.
ere are an infinite number of graphs G with dim th (G) � n − 2 .
For vertices v and u of a graph G, we shall use F(v, u) to denote the set of those vertices w ∈ V(G) with the property that d(w, u) � d(w, v). Now, we define F(G) to be the greatest |F(v, u)| for u, v ∈ V(G). Hence, One can easily check that 1 ≤ F(G) ≤ n − 2. In the next theorem, we present a lower bound for dim th (G) in terms of F(G). □ It is obvious that the above upper bound is sharp. For instance, if P n is a path of length n, then F(G) + 1 � dim th (G) � 2. Moreover, all graphs G with the property that dim th (G) � n − 1 satisfy in the equality F(G) � n − 2. In fact, eorem 4 states that F(G) � n − 2 if and only if dim th (G) � n − 1. So, in view of the previous theorem, we can express the following proposition.

Graphs with dim(G)dim th (G)
In Corollary 4, we show that the only graph G of order n with dim(G) � dim th (G) � n − 1 is the complete graph K n . In this section, we want to investigate graphs G of order n with the property that dim(G) � dim th (G) � k, for k ∈ 2, n − 2 { }. Moreover, we characterize all graphs G with dim th (G) � 2.
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 n and metric dimension n − 2. In the following, we will use the results in [6,23] to obtain a characterization of graphs G of order n with dim(G) � dim th (G) ∈ 2, n − 2 { }.

Theorem 8.
If G is a graph with order n ≥ 4 and dim(G) � n − 2 , then dim th (G) � n − 1 .
en, G is a multipartite graph and dim th (G) � n − 1. Now, assume that G � K s + (K 1 ∪ K t )(s, t ≥ 1). If s � t � 1, then n < 4, which is a contradiction. Suppose that s ≥ 2 or t ≥ 2. Without loss of generality, we may assume that s ≥ 2. Hence, for two arbitrary vertices a, b ∈ V(K s ), we have ere is no graph G of order n with property that dim(G) � dim th (G) � n − 2 .  (1 ≤ i, j ≤ n) , and k and ℓ with 1 ≤ k ≤ e(v i ) and 1 ≤ ℓ ≤ e(v j ) , we have |V ik ∩ V jℓ | ≤ 1 if and only if G ∈ P n , C 2n−1 , for a n ≥ 2 .
Proof. Only one implication requires proof. Suppose that G is a graph of order n ≥ 4. We claim that deg(v i ) ≤ 2, for every i with 1 ≤ i ≤ n. Suppose, contrary to our claim, that there 3 . en, N(v i )⊆V i1 . We need to consider the following four cases for G[N(v i )].
In this situation, similar to that used in Case 3, one can obtain a contradiction.

So, for any
is implies that G is isomorphic to P n or C n , for some positive integer n.
In the following, we show that if G � C n , then n is an odd integer. On the contrary, suppose that G � C 2k , for an in- erefore, |V 11 ∩ V (k+1)(k−1) | > 1, which is a contradiction, and the proof is complete. □ Theorem 10. Let G be a graph. en, dim th (G) � 2 if and only if G is isomorphic to one of the graphs P n or C 2n−1 , for a positive integer n ≥ 2 .
Proof. Let V(G) � v 1 , v 2 , . . . , v m . First, we suppose that G is isomorphic to one of the graphs P n or C 2n−1 , for a positive integer n ≥ 2. By Lemma 1, each v i and v j (1 ≤ i, j ≤ m), k and ℓ with 1 ≤ k ≤ e(v i ) and 1 ≤ ℓ ≤ e(v j ), implies that |V ik ∩ V jℓ | ≤ 1. By the way of contradiction, assume that there exist v i and v j (1 ≤ i ≤ m) and k, ℓ which 1 ≤ k ≤ e(v i ) and 1 ≤ ℓ ≤ e(v j ) such that |V ik ∩ V jℓ | > 1. Hence, there exist distinct vertices u 1 and u 2 in V ik ∩ V jℓ . Since u 1 , u 2 ∈ V ik and u 1 , u 2 ∈ V jℓ , we have d(u 1 , v i ) � d(u 2 , v i ) � k and d(u 1 , v j ) � d(u 2 , v j ) � ℓ. is means that set v i , v j can not to be a resolving set of G, which is a contradiction. e converse implication follows almost immediately from eorems 1 and 2.  We end the paper with the following problem.
Open Problem. Let k be a positive integer 3 ≤ k ≤ n − 3. Characterize the class of graphs G with the property that dim(G) � dim th (G) � k.

Data Availability
No data were used to support this study.

Conflicts of Interest
e authors declare that they have no conflicts of interest.