# Computing Edge Version of Resolvability and Double Resolvability of a Graph

Authors: Muhammad Ahmad; Zohaib Zahid; Tabasam Rashid; Juan Luis Garcia Guirao
Journal: Journal of Chemistry
 (2022)

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## Abstract

The field of graph theory is extensively used to investigate structure models in biology, computer programming, chemistry, and combinatorial optimization. In order to work with the chemical structure, chemists require a mathematical form of the compound. The chemical structure can be depicted using nodes (which represent the atom) and links (which represent the many types of bonds). As a result, a graph theoretic explanation of this problem is to give representations for the nodes of a graph such that different nodes have unique representations. This graph theoretic study is referred to as the metric dimension. In this article, we have computed the edge version of the metric dimension and doubly resolving sets for the family of cycle with chordCnt for n≥6 and 2≤t≤⌊n/2⌋.


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## Body

## 1. Introduction

In research domains where networking is a basic and fundamental study block, graph theory is the most fundamental way to study and use these sciences. Consider the following scenario: (1) in network methods, database design, computer science, web document clustering, mobile phone networks, image processing, and resource management. (2) Only a few examples of biology include structures in cell biology, cell-sample sequencing, and bioinformatics. (3) In operations research, words such as minimal sum colouring, optimization, game theory, task, and time table scheduling are all employed. (4) Complex simulated structures of atoms, molecular bonds, chemoinformatics, and molecular descriptors are some of the study blocks in chemistry.Graph invariants are excellent tools for analysing the abstract structures of graphs in a variety of disciplines. Nowadays, there are several distinct types of metric generators in graphs, each of which is explored both theoretically and practically, depending on its popularity or uses. One of them is the classical metric dimension which is used to detect devices in computer network difficulties.For a connected graphG=VG,EG, the distance da,b between two vertices a and b is the minimum of the lengths of the a−b paths of G. A vertex e∈VG is said to resolve two vertices a and b if da,e≠db,e. Suppose an ordered set of vertices WG =  wμ|1≤μ≤k⊆VG and x be a vertex of G, then the representation rx|WG of x with respect to WG is the k-vector dx,wμμ=1k, also known as the distance vector. The set WG is called a resolving set for graph G, if any vertex of G has unique distance vectors with respect to WG. A metric basis for graph G is a resolving set with a minimal number of entries. The number of entries in the metric basis is called its metric dimension (MD) and it is usually denoted by dimG.Combinatorics gave rise to the concept of the MD, and for the first time, Slater [1] proposed the problem of investigating the MD. Later, for the connected graph G, Harary and Melter [2] suggested the same idea. The focus of the study was on a notion known as metric representation, which is a way of representing vertex locations in a graph. The resolving sets were initially created to find the precise position of an interrupter in a network, but later, Chartrand and Zhang [3] presented various applications in the fields of biology, chemistry, and robotics. The notion of trilateration may be applied to the MD of graphs from a two-dimensional real plane. For instance, three satellites in orbit use distances to calculate the precise location of an object on earth in the global positioning system (GPS). Several applications in chemical structure [4] and robot navigation [5] have been developed for the theoretical investigation of the MD. The MD of hamming graphs is linked to many coin weighing difficulties covered in [6, 7], as well as a detailed investigation of the game Mastermind given in [8]. This idea has been thoroughly investigated in a variety of domains, including combinatorial optimization [9], geographic routing protocols [10], and network discovery and verification [11].Determining the precise value of the MD of arbitrary graphs is a computationally difficult problem. As a result, certain relevant bounds for various graph classes have been discovered. For example, Shao et al. evaluated the MD bounds for generalised Petersen graphs [12]. Chartrand et al. [4] categorised all graphs with MD n−1, n−2, and 1. Ahmad et al. [13] determined the minimal resolving sets for chorded cycles and kayak paddle graphs. The MD for various classes of convex polytopes is calculated by Imran and Baig [14, 15]. The MD of wheel related graphs (Jahangir graphs, respectively) were determined by Buczkowski et al. [16] (Tomescu and Javaid, respectively [17]). More details are discussed in [18, 19].Finding the source of an epidemic in wide area networks can be a difficult task. For example, the only information utilized to identify a source of virus propagating throughout the network is the infection time of the subset of nodes termed as sensors. These sensors could be used to keep track of how long they have been infected. The only thing left is to figure out which sensors are needed to ensure the infection origin is precisely located. The solution to this difficulty is the property termed as double metric dimension [20, 21].It may be simple to determine the mysterious virus source if the complete virus spreading process can be observed. Unfortunately, due to the high cost of data collection, the entire process may be too expensive. A doubly resolving sensor set may accurately identify infection sources even if the initial time of virus spread is unknown.Caceres et al. [22] introduced the concept of doubly resolving sets (DRSs) in graphs during the investigation of the MD of the Cartesian product of graphs. For g1,g2,h1,h2∈VG, g1,g2 are said to doubly resolve h1,h2 if dh1,g1−dh1,g2≠dh2,g1−dh2,g2. A subset D is said to be a doubly resolving set (DRS) if some of its vertices doubly resolve every two vertices in G. In other words, for every pair of vertices g1,g2∈VG there exist rg1|D−rg2|D≠m.g for some m∈ℤ and g be a vector of length D with all entries equal to 1. A doubly basis of graph G is a DRS for G with a minimum count, and its count is known as the double metric dimension (or doubly resolving number) of G and is indicated by ψG. Every DRS is obviously a resolving set; this implies ψG≥dimG.Kratica et al. and Khuller et al. both demonstrated the NP-hardness ofψG and dimG in [5, 23], respectively. For the first time, Chen et al. [24] reported the explicit lower and upper bounds for the minimal DRS problem. Trees, cycles, and complete graphs were found to have a doubly resolving number by Caceres et al. [22]. Minimal DRSs for several families of Harary graphs have been established by Ahmad et al. [25]. MD and minimal DRSs were determined to be the same in various families of circulant graphs in [26]. Cangalovic et al. [27] and Kratica et al. [28] computed the doubly resolving number of prism graphs and hamming graphs, respectively. The doubly resolving number for several convex polytopes have been calculated in [29]. In [30], Pan et al. investigated the doubly resolving number for various types of convex polytopes. Liu et al. [31] studied the MD and minimal DRSs for the cocktail party graphs and jellyfish graphs. Further details are discussed in [32, 33].Caceres et al. [22] found an upper limit in terms of dimG and ψJ for the MD of the Cartesian product of graphs G and J. As a result, DRSs are essential when investigating Cartesian products of graphs. The idea of obtaining upper bounds of MD in the Cartesian product of graphs inspired us to work on DRSs of various graph families.In this work, we investigated a newly defined variant of metric generators called the edge version of metric and double metric generators in order to contribute to the knowledge on these distance-related parameters in graphs. In classical metric dimension, the distance between vertices was of more importance, and in the newly defined variants, the distance between the edges has been given more weightage. Also, the edge version of doubly resolving sets (EVDRSs) is used to find the upper bound of the edge version of metric dimension (EVMD) in the same manner as in the classical metric dimension case. However, both variants are useful in many fields, for instance, to find the location of an intruder in a network, problems of pattern recognition, image processing, as well as the source of epidemics in social networks and in contagious diseases, etc.


## 2. Preliminaries

The line graphLG of an undirected graph G is the graph whose vertices are in one-one correspondence with the edges of G, and two vertices of the line graph are adjacent if and only if the corresponding edges in graph G are adjacent. The edge distances can be calculated using the same way as the vertex distances. The EVMD is defined as follows [34].Definition 1.
(1)
The edge distancedEGg,h between two edges g and h of EG is the minimum of the lengths of the g−h paths in LG.(2)
An edgef∈EG is said to resolve two edges g and h of EG if dEGf,g≠dEGf,h.(3)
Suppose an ordered set of edgesNEG=gi|1≤i≤k⊆EG and f be an edge of G, then the edge version of representation rEGf|NEG of f with respect to NEG is the k-vector dEGf,gii=1k, also known as the edge distance vector.(4)
The setNEG is called an edge version of resolving set for graph G, if any edge of G has unique edge distance vector with respect to NEG.(5)
An edge version of the metric basis of graphG is an edge version of the resolving set having a minimal number of edges. The dimEG represents the cardinal number of the edge version of the metric basis and is known as the edge version of the metric dimension of G.The EVDRSs presented in [35], is defined as follows.Definition 2.
(1)
IfdEGf1,f−dEGf1,g≠dEGf2,f−dEGf2,g, the edges f and g of the graph G with size EG≥2 are supposed to edge doubly resolve edges f1 and f2 of the graph G.(2)
An ordered subsetDEG⊆EG is called an edge version of doubly resolving set, if some of its edges g≠h∈DEG edge doubly resolve every two edges in EG. In other words, for every pair of distinct edges g,h∈EG, all coordinates of the vector rEGg|DEG−rEGh|DEG cannot be same.(3)
A doubly basis in the line graphLG is an edge version of doubly resolving set for G with minimum count, and its count is termed as the edge version of double metric dimension of G and indicated by ψEG.It is important to note that the MD and DRSs of the line graph of a graph (particularly edges uniquely identifying edges) explored here are not the same as the parameter edge MD described in [36], which is vertices uniquely identifying edges.Any EVDRS is also an edge version of resolving set, implying thatdimEG≤ψEG for all graphs G. The MD of line graphs has been extensively examined in mathematics [37, 38], and their importance in chemistry has been demonstrated [39–41]. Chartrand et al. studied the EVMD of path graphs in [4]. The EVMD was investigated in detail by Caceres et al. [22]. Eroh et al. used G as bouquet and wheel graphs to determine dimEG in [42]. For the prism and n-sunlet graphs, Ruby et al. [34] established the EVMD. Lv et al. determined the EVMD for the family of circulant graphs in [43]. The dimEG and ψEG of prism, n-sunlet, and kayak paddle graphs have been established by Ahmad et al. (further details are discussed in [35, 44]). Liu et al. explored the dimEG and ψEG of the necklace graph in [45]. Layer-sun graphs and their line graphs have also been explored in [46], for the minimal DRS problem. Further details are discussed in [47, 48].The following is the structure of the rest of the article.We discussed the graph in Section3 and computed the dimEG for the chorded cycles Cnt. Section 4 calculates the ψEG for the chorded cycles Cnt. We conclude this paper at Section 5 by expressing an opinion.


## 3. The Edge Version of Metric Dimension for the Family of Cycle with ChordCnt

A graph constructed from a cycleCn by combining two vertices whose distance in the cycle is t, where 2≤t≤n−2, generates the family of cycle with chord Cnt (Figure 1).Figure 1
The graph ofC93.We label the chord ofCnt by e0 and the outer edges of Cnt by \{fı: ∀0≤1≤n−1\}, as demonstrated in Figure 1. The graph Cnt is also called the graph Cnn−t. Thus, it is sufficient to consider t≤⌊n/2⌋ for given value of n. The following result was recently computed for the MD of a cycle with chord Cnt.Theorem 1 (see [13]).
The metric dimension of the graphCntis 2, forn≥4and2≤t≤⌊n/2⌋.The following theorem computes the constantdimECnt:Theorem 2.
The edge version of the metric dimension of the graphCntis 2, forn≥4and2≤t≤⌊n/2⌋.Proof.
In order to calculate the EVMD, we prove the following four cases.Case 1.
Letn be odd and t be even.
ConsiderNE=f0,ft+1⊂ECnt, and for each edge of Cnt, the edge version of representation with respect to NE is given as follows.
Here,rEe0|NE=1,1 is the edge version of the representation of e0 with respect to NE, and the edge version of the representations of the edges fl are as follows:(1)rEfl|N=0,2,ifl=0;l,l+1,if1≤l≤t2;t−l+2,t−l+1,ift+22≤l≤t;2,0,ifl=t+1;l−t+1,l−t−1,ift+2≤l≤n+t−12;l−t−1,l−t−1,ifl=n+t+12;n−l,n−l+2,ifn+t+32≤l≤n−1.Case 2.
Letn and t be even.
ConsiderNE=f0,ft+1⊂ECnt, and for each edge of Cnt, the edge version of representation with respect to NE is given as follows.
Here,rEe0|NE=1,1 is the edge version of the representation of e0 with respect to NE, and the edge version of the representations of the edges fl are as follows:(2)rEfl|NE=0,2,ifl=0;l,l+1,if1≤l≤t2;t−l+2,t−l+1,ift+22≤l≤t;2,0,ifl=t+1;n+t−22−l,n+t−62−l,ifn+t−82≤l≤n+t−22;n+2t+22−l,l−t−1,ifn+t2≤l≤n+t+22;n+2t+22−l,n+2t+62−l,ifn+t+42≤l≤n−1.Case 3.
Letn and t be odd.
ConsiderNE=f0,ft+1⊂ECnt, and for each edge of Cnt, the edge version of representation with respect to NE is given as follows.
Here,rEe0|NE=1,1 is the edge version of the representation of e0 with respect to NE, and the edge version of the representations of the edges fl are as follows:(3)rEfl|NE=0,2,ifl=0;l,l+1,if1≤l≤t−12;l,l,ifl=t+12;t−l+2,t−l+1,ift+32≤l≤t;2,0,ifl=t+1;l−t+1,l−t−1,ift+2≤l≤n+t−22;n−l,l−t−1,ifn+t2≤l≤n+t+22;n−l,n−l+2,ifn+t+42≤l≤n−1.Case 4.
Letn be even and t be odd.
ConsiderNE=f0,ft+1⊂ECnt, and for each edge of Cnt, the edge version of representation with respect to NE is shown as follows.
Here,rEe0|NE=1,1 is the edge version of the representation of e0 with respect to NE, and the edge version of the representations of the edges fl are as follows:(4)rEfl|NE=0,2,ifl=0;l,l+1,if1≤l≤t−12;l,l,ifl=t+12;t−l+2,t−l+1,ift+32≤l≤t;2,0,ifl=t+1;l−t+1,l−t−1,ift+2≤l≤n+t−12;l−t−1,l−t−1,ifl=n+t+12;n−l,n−l+2,ifn+t+32≤l≤n−1.
We can see that the edge version of all the edges in the graphCnt is unique, implying that dimECnt≤2. But the abovementioned graph Cnt is not a path graph implies that dimECnt≤2. As a result, we can conclude that dimECnt is 2 in all four cases.


## 4. The Edge Version of Doubly Resolving Sets for the Family of Cycle with ChordCnt

This section will focus to calculate the EVDRSs for the family of cycle with chordCnt.Theorem1 implies that ψECnt≥2, for n≥4. We will also demonstrate that ψECnt=3 for n≥4.To calculate the edge distances for the family of cycles with chordCnt, the following steps are carried out.DefineSle0=e∈ECnt:dEe0,e=l be the edge set in ECnt at distance l from e0. Table 1 can be easily formulated for Sle0, and it will be used to calculate the distances between any two edges in ECnt.Table 1
Sle0 for Cnt.
ntlSle0EvenEven1{f0, f1, ft, ft+1}2≤l≤t/2{fl, ft−l+1, ft+l, fn−l+1}t+2/2≤l≤n−t/2{ft+l, fn−l+1}OddEven1{f0, f1, ft, ft+1}2≤l≤t/2{fl, ft−l+1, ft+l, fn−l+1}t+2/2≤l≤n−t−1/2{ft+l, fn−l+1}n−t+1/2fn+t+1/2EvenOdd1{f0, f1, f3, f4}632{f2, f5}>6≥31{f0, f1, ft, ft+1}2≤l≤t−1/2{fl, ft−l+1, ft+l, fn−l+1}t+1/2{fl, ft+l, fn−l+1}t+3/2≤l≤n−t−1/2{ft+l, fn−l+1}n−t+1/2fn+t+1/2OddOdd1{f0, f1, ft, ft+1}2≤l≤t−1/2{fl, ft−l+1, ft+l, fn−l+1}t+1/2{fl, ft+l, fn−l+1}t+3/2≤l≤n−t/2{ft+l, fn−l+1}The symmetry ofCnt, where n≥4, shows that if n and t are even, we have the following:(5)dEfl,fm=dEe0,fm−l−1,ifm−l=0;dEe0,fm−l,if1≤m−l≤t2;dEe0,fm−l+1,ift+22≤m−l≤n+t−22;dEe0,fm−l,ifm−l=n+t2;dEe0,fm−l−1,ifn+t+22≤m−l≤n−1.Ifn is odd and t is even, we have the following:(6)dEfl,fm=dEe0,fm−l−1,ifm−l=0;dEe0,fm−l,if1≤m−l≤t2;dEe0,fm−l+1,ift+22≤m−l≤n+t−12;dEe0,fm−l−1,ifn+t+12≤m−l≤n−1.Ifn is even and t is odd, we have the following:(7)dEfl,fm=dEe0,fm−l−1,ifm−l=0,dEe0,fm−l,if1≤m−l≤t+12,dEe0,fm−l+1,ift+32≤m−l≤n+t−12,dEe0,fm−l−1,ifn+t+12≤m−l≤n−1.Ifn and t are odd, we have the following:(8)dEfl,fm=dEe0,fm−l−1,ifm−l=0,dEe0,fm−l,if1≤m−l≤t+12,dEe0,fm−l+1,ift+32≤m−l≤n+t−22,dEe0,fm−l,ifm−l=n+t2,dEe0,fm−l−1,ifn+t+22≤m−l≤n−1.As an outcome, if we know the distancedEe0,e for each e∈ECnt, we can rebuild the distances between any two edges in ECnt.Lemma 1.
ψECnt>2, for allnandt, wheren≥6and2≤t≤⌊n/2⌋.Proof.
We know that for alln and t, ψECnt>2. Therefore, we need to explain that every subset DE⊆ECnt of cardinal number 2 is not an EVDRS for Cnt. Every set DE, as well as the nonedge doubly resolved pairs of edges from ECnt, are listed in Table 2. Let us explain that the edges e0,f0 are not edge doubly resolved by any two edges of the set e0,fn+2/2;l=n+2/2. It is obvious that for l=n+2/2, we have
dEe0,e0=0, dEe0,f0=1, dEe0,fn+2/2=n−4/2 and dEf0,fn+2/2=dEe0,fn+2/2+1=n−2/2. So, dEe0,e0−dEe0,f0=dEe0,fn+2/2−dEf0,fn+2/2=−1, that is, the set e0,fn+2/2;l=n+2/2 is not an EVDRS of Cnt. All other forms of the set DE in Table 2 can be examined and confirmed to have nonedge doubly resolved pairs of edges as well.Table 2
Nonedge doubly resolved pairs ofCnt for n≥6 and 2≤t≤n/2.
ntDENonedge doubly resolved pairsEvenEvene0,fl, 0≤l≤n−2/2e0,fn/2−le0,fl, n/2≤l≤n−1e0,fl−n/2OddEvene0,fl, 0≤l≤n−2/2f0,fn−1e0,fl, n+1/2≤l≤n−1fl−n+1/2,fl−n−1/2EvenOdde0,fl, 0≤l≤n2f0,fn−1e0,fn+2/2, l=n+2/2e0,f0e0,fl, n+4/2≤l≤n−1f1,f2OddOdde0,fl, 0≤l≤n−1/2f0,fn−1e0,fn+1/2, l=n+1/2e0,f0e0,fl, n+3/2≤l≤n−1f2,fl−n−3/2Lemma 2.
ψECnt=3, for both n and t even, where n≥6 and 2≤t≤⌊n/2⌋.Proof.
To proveψECnt=3, for both t and n even, it is sufficient to find an EVDRS having cardinal number 3. Now, from Table 1 using the sets Sle0, Table 3 illustrates the edge distance vectors for each vertex of Cnt in relation to the set DE=e0,ft−2/2,fn+t+2/2.
From Table3, we can verify that if two edges h1,h2∈Sle0 for some l∈1,2,…,n−t/2, then rEh1,DE−rEh2,DE≠0. Thus, if there are two edges h1∈Sle0 and h2∈Sme0 for any l,m∈1,2,…,n−t/2, and l≠m, then rEh1,DE−rEh2,DE≠l−m. Therefore, the set DE=e0,ft−2/2,fn+t+2/2 is the minimal EVDRS. Thus, Lemma 2 holds.Table 3
Edge distance vectors forCnt, where n and t are even.
tlSle0DE=e0,ft−2/2,fn+t+2/2Even0e0=0,1,n−2/2ift=20,t−2/2,n−t/2ift≥41f01,t−2/2,n−t−2/2f1=1,1,n−4/2ift=21,t−4/2,n−t/2ift≥4ft=1,2,n/2ift=21,t/2,n−t+2/2ift≥4ft+1=1,2,n−2/2ift=21,t/2,n−t/2ift≥42≤l≤t/2fl=t/2,1,n−2/2ifl=t/2l,t−2/2−l,n−t−2/2+lifl≠t/2ft−l+1l,t+4/2−l,n−t/2+lft+ll,t−2/2+l,n−t+2/2−lfn−l+1l,t−4/2+l,n−t/2−lt+2/2≤l≤n−t/2ft+l=n−2/2,n−2/2,1ifl=n−2/2,t=2l,l+1,n/2−lifl≠n−2/2,t=2l,t−2/2+l,n−t+2/2−l,ift≥4fn−l+1l,t−4/2+l,n−t/2−lLemma 3.
ψECnt=3, for oddnand event, wheren≥6and2≤t≤⌊n/2⌋.Proof.
To proveψECnt=3, for odd n and even t, it is sufficient to find an EVDRS having cardinal number 3. Now, from Table 1 using the sets Sle0, Table 4 illustrates the edge distance vectors for each vertex of Cnt in relation to the set DE=e0,ft−2/2,fn+t+2/2.
From Table4, we can verify that if two edges h1,h2∈Sle0 for some l∈1,2,…,n−t+1/2, then rEh1,DE−rEh2,DE≠0. Thus, if there are two edges h1∈Sle0 and h2∈Sme0 for any l,m∈1,2,…,n−t+1/2, and l≠m, then rEh1,DE−rEh2,DE≠l−m. Therefore, the set DE=e0,ft−2/2,fn+t+1/2 is the minimal EVDRS. Thus, Lemma 3 holds.Table 4
Edge distance vectors forCnt, where n is odd and t is even.
tlSle0DE=e0,ft−2/2,fn+t+2/2Even0e0=0,1,n−1/2ift=20,t−2/2,n−t+1/2ift≥41f01,t−2/2,n−t−1/2f1=1,1,n−1/2ift=21,t−4/2,n−t+1/2ift≥4ft=1,2,n−1/2ift=21,t/2,n−t+1/2ift≥4ft+1=1,2,n−3/2ift=21,t/2,n−t−1/2ift≥42≤l≤t/2fl=t/2,1,n−1/2ifl=t/2l,t−2/2−l,n−t−1/2+lifl≠t/2ft−l+1l,t+4/2−l,n−t−1/2+lft+ll,t−2/2+l,n−t+1/2−lfn−l+1l,t−4/2+l,n−t+1/2−lt+2/2≤l≤n−t−1/2ft+l=l,l+1,n−1/2−lift=2l,t−2/2+l,n−t+1/2−lift≥4fn−l+1l,t−4/2+l,n−t+1/2−ln−t+1/2fn−t+1/2l,t−4/2+l,0Lemma 4.
ψECnt=3, for evennand oddt, wheren≥6and2≤t≤⌊n/2⌋.Proof.
ForC63, the minimal EVDRS is DE=e0,f1,f4. To prove ψECnt=3, for even n and odd t, it is sufficient to find an EVDRS having cardinal number 3. Now, from Table 1 using the sets Sle0, Table 5 illustrates the edge distance vectors for each vertex of Cnt (where n>6) in terms of the set DE=e0,ft−1/2,fn+t−1/2.
From Table5, we can verify that if two edges h1,h2∈Sle0 for some l∈1,2,…,n−t+1/2, then rEh1,DE−rEh2,DE≠0. Thus, if there are two edges h1∈Sle0 and h2∈Sme0 for any l,m∈1,2,…,n−t+1/2, and l≠m, then rEh1,DE−rEh2,DE≠l−m. Therefore, the set DE=e0,ft−1/2,fn+t−1/2 is the minimal EVDRS. Thus, Lemma 4 holds.Table 5
Edge distance vectors forCnt, where n is even and t is odd.
tlSle0DE=e0,ft−1/2,fn+t−1/2Odd0e00,t−1/2,n+t−1/21f01,t−1/2,n−t+1/2f11,t−3/2,n−t+1/2ft1,t+1/2,n−t−1/2ft+11,t+1/2,n−t−3/22≤l≤t−1/2fll,t−1/2−l,n−t−1/2+lft−l+1l,t+3/2−l,n−t−3/2+lft+ll,t−1/2+l,n−t−1/2−lfn−l+1l,t−3/2+l,n−t+3/2−lt+1/2fll,1,n−2/2ft+ll,t−2/2+l,n−t−1/2−lfn−l+1l,t−3/2+l,n−t+3/2−lt+3/2≤l≤n−t−1/2ft+ll,t−1/2+l,n−t−1/2−lfn−l+1l,t−3/2+l,n−t+3/2−ln−t+1/2fn+t+1/2l,t−3/2+l,1Lemma 5.
ψECnt=3, when bothnandtodd, wheren≥6and2≤t≤⌊n/2⌋.Proof.
ForC73, the minimal EVDRS is DE=e0,f1,f4. In order to prove ψECnt=3, for both n and t odd, it is sufficient to find an EVDRS of cardinality 3. Now, from Table 1 using the sets Sle0, Table 6 illustrates the edge distance vectors for each vertex of Cnt (where n>7) in terms of the set DE=e0,ft−1/2,fn+t−2/2.
From Table6, we can verify that if two edges h1,h2∈Sle0 for some l∈1,2,…,n−t/2, then rEh1,DE−rEh2,DE≠0. Thus, if there are two edges h1∈Sle0 and h2∈Sme0 for any l,m∈1,2,…,n−t/2, and l≠m, then rEh1,DE−rEh2,DE≠l−m. Therefore, the set DE=e0,ft−1/2,fn+t−2/2 is the minimal EVDRS. Thus, Lemma 5 holds.Table 6
Edge distance vectors forCnt, where n and t are odd.
tlSle0DE=e0,ft−1/2,fn+t−2/2Odd0e00,t−1/2,n−t−2/21f01,t−1/2,n−t/2f11,t−3/2,n−t/2ft1,t+1/2,n−t−2/2ft+11,t+1/2,n−t−4/22≤l≤t−1/2fll,t−1/2−l,n−t−2/2+lft−l+1l,t+3/2−l,n−t−4/2+lft+ll,t−1/2+l,n−t−2/2−lfn−l+1l,t−3/2+l,n−t+4/2−lt+1/2fll,1,n−3/2ft+ll,t−1/2+l,n−t−2/2−lfn−l+1l,t−3/2+l,n−t+4/2−lt+3/2≤l≤n−t/2ft+l=n−t/2,n−1/2,1ifl=n−t/2l,t−1/2+l,n−t−2/2−lifl≠n−t/2fn−l+1l,t−3/2+l,n−t+4/2−lIt is displayed from the whole technique thatψECnt=3 for n=4,5 and t=2. The set DE=e0,f0,f1 is the minimal EVDRS for C42 and C52. Using Lemma 2–Lemma 5, the main theorem is stated as follows:Theorem 3.
LetCntbe a family of cycle with chord. Then, ψECnt=3, forn≥4.


## 5. Conclusion

We computed the EVMDdimEG in this article by considering G as a family of cycles with the chord Cnt. The edge version of the double metric dimension ψECnt was also computed. The graph is interesting to study in this situation because the cardinality of minimal EVDRSs ψECnt and its EVMD dimECnt are not dependent on the number of vertices of these graphs.Here we end with the following open problem:Open problem 5.1. Find the classes of graphsG such that dimG=dimEG=ψEG



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