On minimal edge version of doubly resolving sets of a graph

In this paper, we introduce the edge version of doubly resolving set of a graph which is based on the edge distances of the graph. As a main result, we computed the minimum cardinality $\psi_E$ of edge version of doubly resolving sets of family of $n$-sunlet graph $S_n$ and prism graph $Y_n$.


Introduction and Preliminaries
Let us take a graph G = (V (G), E(G)), which is simple, connected and undirected, where its vertex set is V (G) and edge set is E(G).The cardinality of metric basis is represented by dim(G), and is called metric dimension of G.
In [19], Slater introduced the idea of resolving sets and also in [10], Harary and Melter introduced this concept individually.Different applications of this idea has been introduced in the fields like network discovery and verification [1], robot navigation [15] and chemistry.
The introduction of doubly resolving sets is given by Caceres et al. (see [3]) by presenting its connection with metric dimension of the cartesian product G G of the graph G.In case of some convex polytopes, hamming and prism graphs, the minimal doubly resolving sets has been obtained in [13], [14] and [4] respectively.
Clearly, if a and b doubly resolve and thus a or b resolve u 1 and v 1 , this shows that a doubly resolving set is also a resolving set, which implies dim(G) ≤ ψ(G) for all graphs G. Finding ψ(G) and dim(G) are NP-hard problems proved in [11,12].Since, the line graph L(G) of a graph G is defined as, the graph whose vertices are the edges of G, with two adjacent vertices if the corresponding edges have one vertex common in G.In mathematics, the metric properties of line graph have been studied to a great extent (see [2,5,6,17,18]) and in chemistry literature, its significant applications have been proved (see [7,8,9]).In [16], the edge version of metric dimension have been introduced, which is defined as: Definitions 1.1.
(1) The edge distance d E (f, g) between two edges f, g ∈ E(G) is the length of a shortest path between vertices f and g in the line graph L(G).
(2) If d E (e, f ) = d E (e, g), then the edge e ∈ E(G) is said to edge resolve two edges f and g of E(G).
(3) Suppose that The following theorems in [16] are important for us.
Theorem 1.2.Let S n be the family of n-sunlet graph then 3, if n is odd.
( Note that every edge version of doubly resolving set is an edge version of resolving set, which implies dim E (G) ≤ ψ E (G) for all graphs G.

2.
The edge version of doubly resolving sets for family of n-sunlet graph S n .
The family of n-sunlet graph S n is obtained by joining n pendant edges to a cycle graph C n (see Figure 1).
For our purpose, we label the inner edges of S n by {e i : ∀ 0 ≤ i ≤ n − 1} and the pendent edges by {f i : ∀ 0 ≤ i ≤ n − 1} as shown in Figure 1.
As motivated by the Theorem 1.2, we obtain Furthermore, we will show that ψ E (S n ) = 3 for n ≥ 4.
In order to calculate the edge distances for family of n-sunlet graphs S n , consider the line graph L(S n ) as shown in Figure 2.
Define S i (e 0 ) = {e ∈ E(S n ) : d(e 0 , e) = i}.For ψ E (S n ) with n ≥ 4, we can locate the sets S i (e 0 ) that are represented in the Table 1.It is clearly observed from Figure 2 that From the above mentioned sets S i (e 0 ), it is clear that they can be utilized to define the edge distances between two arbitrary edges of E(S n ) in the subsequent way.
The symmetry in Figure 2 shows that d E (e i , e j ) = d E (e 0 , e |j−i| ) for 0 ≤ |j − i| ≤ n − 1.
If n = 2k, where k ≥ 2, we have If n = 2k + 1 where k ≥ 2, we have As a result, if we know the edge distance d E (e 0 , e) for any e ∈ E(S n ), then one can recreate the edge distances between any two edges from E(S n ).
Proof.As we know that for n = 2k, ψ E (S n ) ≥ 2. So it is necessary to prove that each of the subset D E of edge set E(S n ) such that |D E | = 2 is not an edge version of doubly resolving set for S n .In Table 2, seven possible types of set D E are presented and for each of them the resultant non-edge doubly resolved pair of edges from edge set E(S n ) is found.
To verify, let us take an example, the edges e k , e k+1 are not edge doubly resolved by any two edges of the set {e 0 , e i ; k < i ≤ n − 1}.Obviously, for k < i ≤ n − 1, we have Now from Table 3, as e 0 ∈ D * E , so the first edge version of metric coordinate of the vector of e 0 ∈ S i (e 0 ) is equal to 0. For each i ∈ {1, 2, 3, . . ., k}, one can easily check that there are no two edges Also, for each i, j ∈ {1, 2, 3, . . ., k}, i = j, there are no two edges h 1 ∈ S i (e 0 ) and In this manner, the set D * E = {e 0 , e 1 , e k } is the minimal edge version of doubly resolving set for S n with n = 2k, k ≥ 2 and hence Lemma 2.2 holds.

Table 4. Vectors of edge metric coordinates for S
Proof.The Table 4 demonstrate that the edge version of representations of S n in relation to the set D * E = {e 0 , e 1 , e k+1 } in a different way.
Now from Table 4, as e 0 ∈ D * E , so the first edge version of metric coordinate of the vector of e 0 ∈ S i (e 0 ) is equal to 0. Similarly for each i ∈ {1, 2, 3, . . ., k + 1}, one can easily find that there are no two edges h 1 , h 2 ∈ S i (e 0 ) such that r E (h 1 , D * E ) − r E (h 2 , D * E ) = 0. Likewise, for every i, j ∈ {1, 2, 3, . . ., k + 1}, i = j, there are no two edges h 1 ∈ S i (e 0 ) and Like so, the set D * E = {e 0 , e 1 , e k+1 } is the minimal edge version of doubly resolving set for S n with n = 2k + 1, k ≥ 2 and consequently Lemma 2.3 holds.
It is displayed from the whole technique that ψ E (S n ) = 3, for n ≥ 4. We state the resulting main theorem by using Lemma 2.2 and Lemma 2.3 as mentioned below; Theorem 2.4.Let S n be the n-sunlet graph for n ≥ 4. Then ψ E (S n ) = 3.

The edge version of doubly resolving sets for family of prism graph
A family of prism graph Y n is cartesian product graph C n × P 2 , where C n is cycle graph of order n and P 2 is a path of order 2 (see Figure 3).In order to calculate the edge distances for family of prism graphs Y n , consider the line graph L(Y n ) as shown in Figure 4.
For ψ E (Y n ) with n ≥ 6, we can locate the sets S i (f 0 ) that are represented in the Table 5.It is clearly observed from Figure 4 that S i (f 0 ) = ∅ for i ≥ k + 2. From the above mentioned sets S i (f 0 ), it is clear that they can be utilized to define the edge distance between two arbitrary edges of E(Y n ) in the subsequent way.
The symmetry in Figure 4 shows that If n = 2k, where k ≥ 3, we have If n = 2k + 1 where k ≥ 3, we have As a result, if we know the edge distance d E (f 0 , f ) for any f ∈ E(Y n ) then one can recreate the edge distances between any two edges from E(Y n ).6, as e 0 ∈ D * E , so the first edge version of metric coordinate of the vector of f 0 ∈ S i (f 0 ) is equal to 1.For each i ∈ {1, 2, 3, . . ., k + 1}, one can easily check that there are no two edges Also, for each i, j ∈ {1, 2, 3, . . ., k + 1}, i = j, there are no two edges h 1 ∈ S i (f 0 ) and Proof.The Table 7 demonstrate that the edge version of representations of Y n in relation to the set D * E = {e 0 , e k , g k+2 } in a different way.
Now from Table 7, as e 0 ∈ D * E , so the first edge version of metric coordinate of the vector of f 0 ∈ S i (f 0 ) is equal to 1. Similarly for each i ∈ {1, 2, 3, . . ., k + 1}, one can easily find that here are no two edges h 1 , h 2 ∈ S i (f 0 ) such that r E (h 1 , D * E ) − r E (h 2 , D * E ) = 0. Likewise, for every i, j ∈ {1, 2, 3, . . ., k + 1}, i = j, there are no two edges h 1 ∈ S i (f 0 ) and It is displayed from the whole technique that ψ E (Y n ) = 3, for n ≥ 6.We state the resulting main theorem by using Lemma 3.1 and Lemma 3.2 as mentioned below; Theorem 3.3.Let Y n be the prism graph for n ≥ 6.Then ψ E (Y n ) = 3.

Conclusion
In this article, we computed the minimal edge version of doubly resolving sets and its cardinality ψ E (G) by considering G as a family of n-sunlet graph S n and prism graph Y n .In case of n-sunlet graphs, the graph is interesting to consider in the sense that its edge version of metric dimension dim E (S n ) is dependent on the parity of n for both even and odd cases.The cardinality ψ E (S n ) of minimal edge version of doubly resolving set of n-sunlet graph S n is independent from the parity of n.In the case of prism graph Y n , the edge version of metric dimension dim E (Y n ) and the cardinality ψ E (Y n ) of its minimal edge version of doubly resolving set are same for every n ≥ 6.
Open Problem 4.1.Compute edge version of doubly resolving sets for some generalized petersen graphs.
The order of a graph G is |V (G)| and the size of a graph G is |E(G)|.The distance d(a, b) between the vertices a, b ∈ V (G) is the length of a shortest path between them.If d(c, a) = d(c, b), then the vertex c ∈ V (G) is said to resolve two vertices a and b of V (G).Suppose that N = {n 1 , n 2 , . . ., n k } ⊆ V (G) is an ordered set and m is a vertex of V (G), then the representation r(m, N) of m with respect to N is the k-tuple d(m, n 1 ), d(m, n 2 ), . . ., d(m, n k ) .If different vertices of G have different representations with respect to N, then the set N is said to be a resolving set of G.The metric basis of G is basically a resolving set having minimum cardinality.
The doubly resolving sets create a valuable means for finding upper bounds on the metric dimension of graphs.The vertices a and b of the graph G with order |V (G)| ≥ 2 are supposed to doubly resolve vertices u 1 and v 1 of the graphG if d(u 1 , a) − d(u 1 , b) = d(v 1 , a) − d(v 1 , b).A subset D ofvertices doubly resolves G if every two vertices in G are doubly resolved by some two vertices of D. Precisely, in G there do not exist any two different vertices having the same difference between their corresponding metric coordinates with respect to D. A doubly resolving set with minimum cardinality is called the minimal doubly resolving set.The minimum cardinality of a doubly resolving set for G is represented by ψ(G).

Theorem 1 . 3 .Definitions 1 . 4 . ( 1 )
Let Y n be the family of prism graph then dim E (Y n ) = 3 for n ≥ 3.In this article, we proposed minimal edge version of doubly resolving sets of a graph G, based on edge distances of graph G as follows: The edges f and g of the graph G with size |E(G)| ≥ 2 are supposed to edge doubly resolve edges f 1 and f 2 of the graph ) Let D E = {f 1 , f 2 , . . ., f k } be an ordered set of the edges of G then if any two edges e = f ∈ E(G) are edge doubly resolved by some two edges of set D E then the set D E ⊆ E(G) is said to be an edge version of doubly resolving set of G.The minimum cardinality of an edge version of doubly resolving set of G is represented by ψ E (G).

Figure 3 .
Figure 3. Prism graph Y n The family of prism graph Y n consists of 4-sided faces and n-sided faces.For our purpose, we label the inner cycle edges of Y n by {e i : 0 ≤ i ≤ n − 1}, middle edges by {f i : 0 ≤ i ≤ n − 1} and the outer cycle edges by {g i : 1 ≤ i ≤ n − 1} as shown in Figure 3.

Figure 4 .
Figure 4. L(Y n ) of prism graph Y n As motivated by the Theorem 1.3, we obtain ψ E (Y n ) ≥ 3. Furthermore, we will show that ψ E (Y n ) = 3 for n ≥ 6.

Lemma 3 . 1 .
ψ E (Y n ) = 3, for n = 2k, k ≥ 3.Proof.The Table 6 demonstrate that edge version of representations ofY n in relation to the set D * E = {e 0 , e k−1 , f k+1 } in a different manner.Now from Table In this manner, the set D * E = {e 0 , e k−1 , f k+1 } is the minimal edge version of doubly resolving set for Y n with n = 2k, k ≥ 3 and hence Lemma 3.1 holds.Lemma 3.2.ψ E (Y n ) = 3, for n = 2k + 1, k ≥ 3.

h 2 ∈
S j (f 0 ) such that r E (h 1 , D * E )−r E (h 2 , D * E ) = i−j.Like so, the set D * E = {e 0 , e k , g k+2 } is the minimal edge version of doubly resolving set for Y n with n = 2k + 1, k ≥ 3 and consequently Lemma 3.2 holds.
is an ordered set and e is an edge of E(G), then the edge version of representation r E (e, N E ) of e with respect to N E is the k-tuple d E (e, f 1 ), d E (e, f 2 ), . . ., d E (e, f k ) .

Table 2 .
not an edge version of doubly resolving set of S n .Using this procedure we can verify all other non-edge doubly resolved pairs of edges for all other possible types of D E from Table2.Non-edge doubly resolved pairs of S n for n = 2k, k ≥ 2 Lemma 2.2.ψ E (S n ) = 3, for n = 2k, k ≥ 2.

Table 3 .
Table 3 demonstrate that edge version of representations of S n in relation to the set D * E = {e 0 , e 1 , e k } in a different manner.Vectors of edge metric coordinates for S n

Table 6 .
Vectors of edge metric coordinates for Y n

Table 7 .
Vectors of edge metric coordinates for Y n