Barycentric Subdivision of Cayley Graphs With Constant Edge Metric Dimension

A motion of a robot in space is represented by a graph. A robot change its position from point to point and its position can be determined itself by distinct labelled landmarks points. The problem is to determine the minimum number of landmarks to find the unique position of the robot, this phenomena is known as metric dimension. Motivated by this a new modification was introduced by Kelenc. In this paper, we computed the edge metric dimension of barycentric subdivision of Cayley graphs <inline-formula> <tex-math notation="LaTeX">$Cay(\mathbb {Z}_{\alpha }\oplus \mathbb {Z}_{\beta })$ </tex-math></inline-formula>, for every <inline-formula> <tex-math notation="LaTeX">$\alpha \ge 6, \beta \ge 2$ </tex-math></inline-formula> and an observation is made that it has constant edge metric dimension and only three carefully chosen vertices can appropriately suffice to resolve all the edges of barycentric subdivision of Cayley graphs <inline-formula> <tex-math notation="LaTeX">$Cay(\mathbb {Z}_{\alpha }\oplus \mathbb {Z}_{\beta })$ </tex-math></inline-formula>.


I. INTRODUCTION
Slater [1] and Harary el at [2] independently introduced the concept of metric dimension. Metric dimension has contributed in different real world applications like navigation of robots [3], wireless communications and sensor networks [4], pattern recognition and image processing [5], and above of all it has most applications in chemistry that are discussed in [6]- [8]. A lot of interesting work has been done on metric dimension for several types of graphs that can be seen in [9]- [19].
It is denoted r(e|R). The minimum cardinality of R is called the edge metric dimension and is denoted by edim(G) [20].
The associate editor coordinating the review of this manuscript and approving it for publication was Yilun Shang .
It is observed that in some graphs metric generator and edge metric generator are same. That is why it is misunderstood that any edge metric generator R is same as standard metric generator. But the matter of fact is that there are a few families of graphs in which such coincidence happen. Kelenc et al. [20] explained a detailed comparison between the edge metric generator and standard metric generator. In this paper, they also showed the edim(G) is a NP-hard problem and determined that the edim(G) of grid graph is 2. Kelenc et al. [20] proved that the edim(W 1,n ) of wheel graph W 1,n is n for n = 3, 4 and n − 1 for n ≥ 5. They also determined the edge metric dimension of path, cycle, complete graph, complete bipartite, Fan graphs, cartesian product of cycles and bounds for some families of graphs.
An operation to split an edge into two edges by inserting a new vertex into the interior of an edge is called subdivision an edge. If this operation is applied on a sequence on edges of graph then it is called subdividing a graph G and resulting graph is known as subdivision of the graph G. The subdivision is used to convert a pseudograph graph into a simple graph. If subdivision operation is performed on all edge of the graph G, then this subdivision is called the barycentric subdivision of G. Gross and Yellen [22] proved the results that the barycentric subdivision of any graph is a simple and bipartite graph. In this paper, we discussed the edge metric dimension of barycentric subdivision of Cayley graphs Cay(Z α ⊕ Z β ).

II. THE EDGE METRIC DIMENSION OF BARYCENTRIC SUBDIVISION OF CAYLEY GRAPHS
Let G be a semigroup, and K = ∅ ⊆ G. The Cayley graph = Cay(G, K ) is defined as: such that kx = y for some k ∈ K . The Cayley graph of a group G is symmetric or undirected if and only if K = K −1 .
The Caylay graphs = Cay(Z α ⊕ Z β ), α ≥ 3, β ≥ 2, is a graph which can be obtained as the cartesian product of a path with β vertices with a cycle on α vertices. The vertex set and edge set of defined as: denote the number of vertices, edges of the Cayley graphs , respectively.
The barycentric subdivision graph BS( ) can be obtained by adding a new vertex ( In the next theorem, we proved that the metric dimension of the barycentric subdivision BS( ) of is constant and only three vertices appropriately chosen suffice to resolve all the vertices of the BS( ). Selection of appropriate basis vertices (also refereed as landmarks in [3]) is core of the problem. For our purpose, we call the sets of points as Theorem 1: Let BS( ) be the barycentric subdivision of Cayley graphs ; then edim(BS( )) = 3 for every α ≥ 6, β ≥ 2.
Proof 1: The above equality can be proved using double inequalities as following: Case 1. When α is even.
, we show that R is a resolving set for BS( ) in this case. For representations of any edge of E(BS( )) with respect to R.
Representations for the edges of BS( ) are It can be observed that there are no two edges have same representations implying that edim(BS( )) ≤ 3.
On the other side edim(BS( )) ≥ 3. Suppose on contrary that edim(BS( )) = 2, then there are the following possibilities to be discussed.
(1) Both vertices belong to the set H 1 or H 2 or H 3 . Here are the following subcases.
• When both vertices belong to the set H 1 = {(a i , b j ) : 1 ≤ i ≤ α, 1 ≤ j ≤ β} and on the same level. Without loss of generality, we can suppose that one resolving vertex is (a 1 , b j ). Suppose that the second resolving vertex is (a k , b j ), , a contradiction. When both vertices belong to the set H 1 = {(a i , b j ) : 1 ≤ i ≤ α, 1 ≤ j ≤ β} and on the different level. Without loss of generality, we can suppose that one resolving vertex is (a 1 , b j ). Suppose that the second resolving vertex is (a k , b s ), (1 ≤ k ≤ α 2 + 1, 1 ≤ j < s ≤ β). Then for k = 1, 1 ≤ j < s ≤ β, we have r((a 1 , b j )(c 1 , d j )|{(a 1 , b j ), (a 1 , b s )}) = r((a 1 , b j )(c α , d j )|{(a 1 , b j ), (a 1 , b s )}) = (0, 2(s − j)), and for 2 ≤ k ≤ α 2 + 1, 1 ≤ j < s ≤ β, we have r ((a 1 , b j • When both vertices belong to the set H 2 = {(c i , d j ) : 1 ≤ i ≤ α, 1 ≤ j ≤ β} and on the same level. Without loss of generality, we can suppose that one resolving vertex is (c 1 , d j ). Suppose that the second resolving vertex is (c k , d j ), , a contradiction. When both vertices belong to the set H 2 = {(c i , d j ) : 1 ≤ i ≤ α, 1 ≤ j ≤ β} and on the different level. Without loss of generality, we can suppose that one resolving vertex is (c 1 , d j ). Suppose that the second resolving vertex is (c k , d s ), • When both vertices belong to the set and on the same level. Without loss of generality, we can suppose that one resolving vertex is (u 1 , v j ). Suppose that the second resolving vertex is (u k , v j ), When both vertices belong to the set H 3 = {(u i , v j ) : 1 ≤ i ≤ α, 1 ≤ j ≤ β − 1} and on the different level. Without loss of generality, we can suppose that one resolving vertex is (u 1 , v j ). Suppose that the second resolving vertex is (u k , v s ), − 1)), a contradiction.
(2) When both vertices do not belong to the same set. The subcases are as follows: • When one vertex belong to the set H 1 = {(a i , b j ) : 1 ≤ i ≤ α, 1 ≤ j ≤ β} and the second vertex belong to the set There are two possibilities.
• When one vertex belong to the set H 2 = {(c i , d j ) : 1 ≤ i ≤ α, 1 ≤ j ≤ β} and the second vertex belong to the set H 3 = {(u i , v j ) : 1 ≤ i ≤ α, 1 ≤ j ≤ β − 1}. Without loss of generality, we can suppose that one resolving vertex is (c 1 , d j ) ⊆ H 2 . Suppose that the second resolving vertex is (u k , v s ) ⊆ H 3 , (1 ≤ k ≤ α 2 + 1, 1 ≤ j ≤ β, 1 ≤ s ≤ β − 1). Due to the symmetry of the graph the behavior of the inner most cycle and the outer most cycle is same. Therefore, it is sufficient to discuss the representation of the vertices for