# On the Edge Metric Dimension of Different Families of Möbius Networks

Authors: Bo Deng; Muhammad Faisal Nadeem; Muhammad Azeem
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

For an ordered subsetQe of vertices in a simple connected graph G, a vertex x∈V distinguishes two edges e1,e2∈E, if dx,e1≠dx,e2. A subset Qe having minimum vertices is called an edge metric generator for G, if every two distinct edges of G are distinguished by some vertex of Qe. The minimum cardinality of an edge metric generator for G is called the edge metric dimension, and it is denoted by dimeG. In this paper, we study the edge resolvability parameter for different families of Möbius ladder networks and we find the exact edge metric dimension of triangular, square, and hexagonal Möbius ladder networks.


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## Body

## 1. Introduction

A simple connected graphG=V,E with vertex set and edge set V and E, respectively. For two vertices a1,a2∈V, the distance da1,a2 between vertices a1 and a2 is the count of edges between them. A vertex v∈V is said to distinguish two vertices a1 and a2, if dv,a1≠dv,a2. A set Q⊂V is called a resolving set of G, if any pair of distinct vertices of G is distinguished by some element of Q. A resolving set of minimum cardinality is named as metric basis, and its cardinality is the metric dimension of G, denoted by dimG. A vertex v∈V and an edge e=a1a2∈E, and the distance between v and e is defined as de,v=minda1,v,da2,v. A vertex x∈V distinguishes two edges e1,e2∈E, if dx,e1≠dx,e2. A subset Qe having minimum vertices from a connected graph G is an edge resolving set for G, if every two distinct edges of G are distinguished by some vertex of Qe. The minimum cardinality of an edge resolving set for G is called the edge metric dimension and is denoted by dimeG.In 1975, the idea of metric dimension was delivered by Slater [1], he named the metric generators as locating sets which relates to the problem of uniquely recognizing the position of intruders in networks. On the same idea, in 1976 the concept of metric dimension of a graph was independently introduced by Harary and Melter in [2], and these time metric generators were named as resolving sets. One can think that instead of distinguishing two distinct vertices of graph according to chosen subset of vertices, two edges can be distinct with the same subset of vertices. For this concept, Kelenc et al. [3] introduced a new parameter named as the edge metric dimension. In this, they used graph metric to identify each pair of edges by the distance of graph to a chosen subset of vertices.As far as the idea of metric dimension is extensively studied and used in different fields of science as applications, Chartrand et al. in [4] relate the metric dimension of graph with the drug discovery and pharmacological activity, Khuller et al. in [5] try to put thinking that a robot can be shifted from Euclidean space to graph structure and left his thought as an application of metric dimension in robot navigation, and the metric dimension of Hamming graphs leads Chvátal to the analysis of mastermind games and opens the doors for researchers to view application of metric dimension in complex digital games [6], and in [7, 8], Erdös and Lindström assume that the metric dimension can be used in various coin-weighing problems. Resolving sets have served as inspiration for many theoretical studies of graphs.The edge metric dimension is the natural generalization of resolving set, and readers are directed towards the interesting literature containing the metric dimension of different classes of graphs for example, Ali et al. [9, 10] studied the metric dimension of Möbius networks and some other cycle-related graphs, Kuziak et al. [11] discussed the strong metric dimension of graphs, Liu et al. [12, 13] discussed the metric dimension of cocktail party, Toeplitz, and jellyfish graphs, and some cycle-related graphs were studied with the concept of metric dimension by Ahmad et al. [14]. Recently, the edge metric dimension becomes a very common topic in resolvability and a lot of families of graphs are studied. Koam and Ahmad [15] studied edge metric dimension of barycentric subdivision of Cayley graph. The convex polytope graph was discussed by Zhang and Gao in [16] and Ahsan et al. in [17]. Yang et al. discussed some chemical structures related to wheel graphs in [18]. Raza and Bataineh did comparative analysis between metric and edge metric dimension in [19]. Moreover, some interesting study of edge metric dimension can be found in [20, 21], where Okamoto et al. studied the local metric dimension and Yero briefly discussed the definition of metric and its related concepts. Mixed metric dimension is another type of dimension which satisfies the conditions of metric and edge metric dimensions simultaneously. Mixed metric dimension of different families of graphs is studied and gives their exact values, such as Raza et al. studied the mixed metric dimension different rotationally symmetric graphs and gave their exact values [22], Raza and Ji computed the mixed metric dimension of the generalize Petersen graph Pn,2 [23], and results on mixed metric dimension of some path-related graphs are discussed by Raza et al. in [24].In general, the edge metric dimension of a graph is NP-hard [3]. There is no general relation between metric and edge metric dimension of graphs but, in [3], Kelenc et al. inquired about the families of graphs which have dimG=dimeG, dimG<dimeG, and dimG>dimeG. In this paper, we find the edge metric dimension of different families of Möbius ladder networks and give a comparison between the metric and edge metric dimension of these families, in the response of this question.Definition 1.
The minimum number of edgesh between two vertices a1,a2 of a cycle (sub) graph is called as h-size gap between a1,a2.


## 2. Edge Metric Dimension of Möbius Ladder Network

Möbius ladderMLψ is built by a grid of ψ×1, and this grid is twisted at 180∘; now, paste the extreme most left and right paths of vertices as seen in Figure 1. It contains ψ-horizontal cycles of order four. The metric dimension of MLψ is three [9], and in our first result, we prove that the edge metric dimension of MLψ is four.Figure 1
Möbius ladder networkMLΨ.Theorem 1.
LetMLψ be a Möbius ladder network with ψ≥3. Then,(1)dimeMLψ=4.Proof.
Consider the edge resolving setQe=ρ1,ρ2,ρψ+3/2,ρψ+1, to prove that the dimeMLψ=4, first of all, we show that dimeMLψ≤4, and for this claim, following are the distances of all edges with respect to edge resolving set:(2)dρωρψ+ω,ρ1=ω−1,ifω=1,2,…,ψ+22,ψ+32−ω+ψ−12,ifω=ψ+32,…,ψ,dρωρω+1,ρ1=ω−1,ifω=1,2,…,ψ+22;ψ+42−ω+ψ−12,ifω=ψ+42,…,ψ.dρψ+ωρψ+ω+1,ρ1=ω,ifω=1,2,…,ψ2;ψ−12−ω+ψ+22,ifω=ψ+22,…,ψ−1.dρ2ψρ1,ρ1=0,dρ2ψρ1,ρ2=2,dρ2ψρ1,ρψ+3/2=ψ2,dρωρψ+ω,ρ2=1,ifω=1;ω−2,ifω=2,3,…,ψ+52;ψ2−ω+ψ+72,ifω=ψ+72,…,ψ.dρωρω+1,ρ2=0,ifω=1;ω−2,ifω=2,3,…,ψ+42;ψ+62−ω+ψ−12,ifω=ψ+62,…,ψ.dρψ+ωρψ+ω+1,ρ2=1,ifω=1;ω−1,ifω=2,3,…,ψ+32;ψ+52−ω+ψ−12,ifω=ψ+52,…,ψ−1.dρωρψ+ω,ρ/ψ+3/2=ψ2,ifω=1;ψ−12−ω+2,ifω=2,3,…,ψ+32;ω−ψ+52+1,ifω=ψ+52,…,ψ.dρωρω+1,ρψ+3/2=ψ−12−ω+1,ifω=1,2,…,ψ+12;ω−ψ+32,ifω=ψ+32,…,ψ,dρψ+ωρψ+ω+1,ρψ+3/2=ψ+12−ω+1,ifω=1,2,…,ψ+12;ω−ψ+32+1,ifω=ψ+32,…,ψ−1,dρωρψ+ω,ρψ+1=ω−1,ifω=1,2,…,ψ+22;ψ+42−ω+ψ−12,ifω=ψ+42,…,ψ,dρωρω+1,ρψ+1=ω,ifω=1,2,…,ψ2;ψ−12−ω+ψ+22,ifω=ψ+22,…,ψ,dρψ+ωρψ+ω+1,ρψ+1=ω−1,ifω=1,2,…,ψ+22;ψ−12−ω+ψ+42,ifω=ψ+42,…,ψ−1,dρ2ψρ1,ρψ+1=1.
From the above given representation of all edges with respect to edge resolving setQe, no two edges have the same representation, and it proved that dimeMLψ≤4.
Now, on contrary,dimeMLψ=3 implies that the cardinality of edge metric generator Qe′ is three, following is some discussion for this contradiction:Case 1: if the first two vertices with zero-size gap and last two vertices with any arbitrary size of gap are selected in the edge metric generatorQe′=ρ1,ρ2,ρω with 3≤ω≤ψ, then it implies the same representations in the edges dρ1ρψ+1|Qe′=dρ1ρ2ψ|Qe′, and it is concluded that we cannot take these types of vertices in the edge metric generator with cardinality three.Case 2: if all three vertices with any arbitrary size of gap are selected andQe′=ρω,ρj,ρk with 1≤ω,j,k≤ψ, then it implies the same representations in the edges which are dρ1ρψ+1|Qe′=dρ1ρ2ψ|Qe′, and it is also concluded that we cannot take these types of vertices in the edge metric generator with cardinality three.Case 3: now, the vertices with indexψ+2≤ω,j,k≤2ψ with any of the gap-size are chosen and Qe′=ρω,ρj,ρk, then it implies the same representations in the edges which are dρ1ρ2|Qe′=dρ1ρψ+1|Qe′, and again, we conclude that we cannot take these types of vertices in the edge metric generator with cardinality three.Case 4: now, the verticesρω,ρj,ρk∈VMLψ are chosen with any of the size of gap, then there exist an edge from lower horizontal edges and another edge from upper horizontal edges having the same representations to each other with respect to the selected edge metric generator, i.e., dρpρp+1|Qe′=dρψ+pρψ+p+1|Qe′ where 1≤p≤ψ−1, and finally, we concluded that with any gap-size in the edge metric generator with cardinality three is not possible, which implies the contradiction that dimeMLψ≠3. Moreover, this proves the double inequality which is(3)dimeMLψ=4.


## 3. Edge Metric Dimension of Hexagonal Möbius Ladder Network

Hexagonal Möbius ladderHMLψ is built in [25], it can be constructed by dividing each horizontal edge of a square grid by inserting a new vertex, it becomes a grid of ψ×1 with each cycle having order six, and now, twist this grid at 180∘ and paste the extreme most left and right paths of vertices as shown in Figure 2. This graph contains ψ-horizontal cycles of order six. The metric dimension of the hexagonal Möbius ladder network is three [25]. In this section, we proved that edge metric dimension is also three for the hexagonal Möbius ladder network.Figure 2
Hexagonal Möbius ladder networkHMLψ.Theorem 2.
LetHMLψ be a hexagonal Möbius ladder network with ψ≥2. Then,(4)dimeHMLψ=3.Proof.
Consider the edge resolving setQe=ρ1,ρ2,ρ2ψ when ψ=2,3, Qe=ρ1,ρψ+1,ρ2ψ when ψ≥4even, and Qe=ρ1,ρψ,ρ2ψ when ψ≥5odd. To show that the dimeHMLψ=3, we will use the method of double inequality. For dimeHMLψ≤3, following are the distances of all edges with respect to edge resolving set.
Distances of all vertices with respect toρ2 are as follows:(5)dρωρψ+ω,ρ2=1,ifω=1,andψ=2;ω−2,ifω=3,andψ=2;1,ifω=1,andψ=3;ω−2,ifω=3,5,andψ=3,dρωρω+1,ρ2=0,ifω=1;ω−2,ifω=2,3,…,ψ+2;2ψ+2−ω,ifω=ψ+3,ψ+4,…,2ψ,dρ2ψ+ωρ2ψ+ω+1,ρ2=2,ifω=1,2;ω−1,ifω=3,4,…,ψ+1;2ψ+2−ω,ifω=ψ+2,ψ+3,…,2ψ−1.dρ4ψρ1,ρ2=2,dρ4ψρ1,ρ1=0,dρωρψ+ω,ρ2ψ=1,ifω=1,3andψ=2;1,ifω=1,5andψ=3;3,ifω=ψ=3.dρωρψ+ω,ρ1=ω−1,ifω=1,3,…,ψ+1,andψeven;2ψ+1−ω,ifω=ψ+3,ψ+5,…,2ψ−1,andψeven;ω−1,ifω=1,3,…,ψ,andψodd;2ψ+1−ω,ifω=ψ+2,ψ+4,…,2ψ−1,andψodd,dρωρω+1,ρ1=ω−1,ifω=1,2,…,ψ+1;2ψ+1−ω,ifω=ψ+2,ψ+3,…,2ψ.dρ2ψ+ωρ2ψ+ω+1,ρ1=ω,ifω=1,2,…,ψ;2ψ−ω,ifω=ψ+1,…,2ψ−1.dρωρψ+ω,ρψ=ψ−ω,ω=1,3,…,2ψ−1,dρωρω+1,ρψ=ψ−ω−1,ifω=1,2,…,ψ−1;ω−ψ,ifω=ψ,ψ+1,…,2ψ.dρ2ψ+ωρ2ψ+ω+1,ρψ=ψ−ω,ifω=1,2,…,ψ−1;ω−ψ+1,ifω=ψ,ψ+1,…,2ψ−1,dρ4ψρ1,ρψ=dρ4ψρ1,ρψ+1=ψ,dρ4ψρ1,ρ2ψ=2,dρωρψ+ω,ρψ+1=ψ−ω+1,ifω=1,3,…,ψ+1;ω−ψ−1,ifω=ψ+3,ψ+5,…,2ψ−1,dρωρω+1,ρψ+1=ψ−ω,ifω=1,2,…,ψ;ω−ψ−1,ifω=ψ+1,…,2ψ.dρ2ψ+ωρ2ψ+ω+1,ρψ+1=ψ−ω+1,ifω=1,2,…,ψ;ω−ψ,ifω=ψ+1,ψ+2,…,2ψ−1,dρωρψ+ω,ρ2ψ=1,ifω=1;ω+1,ifω=3,5,…,ψ−1,andψ≥4even;2ψ−ω,ifω=ψ+1,ψ+3,…,2ψ−1,andψ≥4even;ω+1,ifω=3,5,…,ψ−2,andψ≥5odd;2ψ−ω,ifω=ψ,ψ+2,…,2ψ−1,andψ≥5odd,dρωρω+1,ρ2ψ=ω+1,ifω=1,2,…,ψ−1;2ψ−1−ω,ifω=ψ,ψ+1,…,2ψ−1;0,ifω=2ψ,dρ2ψ+ωρ2ψ+ω+1,ρ2ψ=ω+1,ifω=1,2,…,ψ−1;2ψ−ω,ifω=ψ,ψ+1,…,2ψ−2;2,ifω=2ψ−1.
It is clear to see that the representations of all edges with respect to edge resolving setQe are distinct, and it is proved that dimeHMLψ≤3.
Now, fordimeHMLψ≥3, the contradiction method implies that dimeHMLψ=2.
If two vertices are chosen with any arbitrary size of gap andQe′=ρω,ρj with 1≤ω,j≤4ψ, then it implies the same representations in the edges which are either dρpρψ+p|Qe′=dρ2ψ+qρ2ψ+q+1|Qe′ where 1≤p,q≤2ψ−1 and p=odd or dρpρp+1|Qe′=dρ2ψ+pρ2ψ+p+1|Qe′ where 1≤p≤2ψ−1, and it is also concluded that we cannot take vertices in edge metric generator with cardinality two. So, it is not possible that dimeHMLψ=2, which implies that dimeHMLψ=3.


## 4. Edge Metric Dimension of Triangular Ladder Network

The ladder network can be built by the cross product of two path graphsLψ=Pψ×P2. Triangular ladder TLψ is built by inserting new edges of Lψ vertices ρωρω+1 where ω is even indices. This graph contains 2ψ−2-cycles of order three shown in Figure 3. Following is the edge metric generator of this network.Figure 3
Triangular ladder networkTLψ.Theorem 3.
LetTLψ be a triangular ladder network with ψ≥3. Then,(6)dimeTLψ=4.Proof.
LetQe=ρ1,ρ2,ρ2ψ−2,ρ2ψ−1 be the edge resolving set. To show that the dimeTLψ=4, we will prove first for dimeTLψ≤4, following are the representations of all edges with respect to edge resolving set:(7)dρωρω+1,ρ1=ω−12,ifω=1,3,…,2ψ−3;ω+22,ifω=2,4,…,2ψ−4,dρωρω+2,ρ1=ω−12,ifω=1,3,…,2ψ−3;ω+22,ifω=2,4,…,2ψ−4,dρωρω+1,ρ2=ω−12,ifω=1,3,…,2ψ−1;ω−22,ifω=2,4,…,2ψ−2,dρωρω+2,ρ2=1,ifω=1;ω−12,ifω=3,5,…,2ψ−3;ω−22,ifω=2,4,…,2ψ−2,dρωρω+1,ρ2ψ−2=ψ−2−ω−12,ifω=1,3,…,2ψ−1;2ψ−ω−22,ifω=2,4,…,2ψ−2;dρωρω+2,ρ2ψ−2=2ψ−ω−32,ifω=1,3,…,2ψ−5;1,ifω=2ψ−3;0,ifω=2;ω−42,ifω=4,6,…,2ψ−2,dρωρω+1,ρ2ψ−1=2ψ−ω−12ifω=1,3,…,2ψ−1;2ψ−ω−22ifω=2,4,…,2ψ−2;dρωρω+2,ρ2ψ−1=2ψ−ω−12,ifω=1,3,…,2ψ−3;2ψ−ω2,ifω=2,4,…,2ψ−2.
Above given representations of all edges with respect to edge resolving setQe are unique, and it is proved that dimeTLψ≤4. Now, for dimeTLψ≥4, choose on contrary which implies that dimeTLψ=3. Analogously, the cardinality of edge metric generator Qe′ is three, following is some discussion for this contradiction:Case 1: if all vertices with any arbitrary size of gap are selected andQe′=ρ1,ρω,ρj with 3≤ω,jodd≤2ψ−1, then it implies the same representations in the edges which are dρ2ρ3|Qe′=dρ3ρ4|Qe′, and it is concluded that we cannot take vertices like this in the edge metric generator with cardinality three.Case 2: if all vertices with any arbitrary size of gap are selected andQe′=ρω,ρj,ρk with 1≤ω,j,kodd≤2ψ−1, then it implies the same representations in the edges which are either dρ1ρ3|Qe′=dρ2ρ3|Qe′ or dρpρp+1|Qe′=dρqρq+2|Qe′ where 2≤p,qeven≤2ψ, and it is also concluded that we cannot take vertices in the edge metric generator with cardinality three.Case 3: if all vertices with any arbitrary size of gap are selected andQe′=ρω,ρj,ρk with 2≤ω,j,keven≤2ψ, then it implies the same representations in the edges which are dρ1ρ2|Qe′=dρ2ρ3|Qe′, and it is also concluded that we cannot take vertices in the edge metric generator with cardinality three.Case 4: now, the verticesρω,ρj,ρk∈VTLψ without choosing the size of gap, then there exist an edge from joining edges and another one edge from upper horizontal edges having the same representation to each other with respect to the decided edge metric generator, i.e., dρpρp+1|Qe′=dρqρq+2|Qe′ where 2≤p,qeven≤2ψ, and finally, we concluded that we cannot take any type of vertices with any gap-size in the edge metric generator with cardinality three, which implies the contradiction that dimeTLψ≠3, Moreover, this proves the double inequality which is(8)dimeTLψ=4.


## 5. Edge Metric Dimension of Triangular Möbius Ladder Network

Figure3 shows a triangular ladder network; now, twist this network at 180∘ and paste the extreme most left and right paths of vertices, and it will come up to a new type of graph named as triangular Möbius ladder graph TMLψ. This graph contains 2ψ−2-horizontal cycles of order three, which can be seen in Figure 4. Following results are the edge metric dimension of the triangular Möbius ladder network.Figure 4
Triangular Möbius ladder networkTMLψ.Theorem 4.
LetTMLψ be a triangular Möbius ladder network with ψ≥5. Then,(9)dimeTMLψ=4.Proof.
Consider the edge resolving setQe=ρ4,ρ5,ρ6,ρ7, when ψ=5, Qe=ρ1,ρ4,ρψ+2,ρψ+3, when ψ≥6even, and Qe=ρ3,ρψ−1,ρψ+2,ρ2ψ−2, and when ψ≥7odd. To show that the dimeTMLψ=4, we will use the method of double inequality, for dimeTMLψ≤4, following are the representations of all edges with respect to the edge resolving set.
Whenψ=5,(10)dρωρω+1,ρ4=1,ifω=1,2,ω−32,ifω=3,5,7,ω−42,ifω=4,6,8,dρ2ψ−2ρ2,ρ4=dρ2ψ−3ρ2,ρ4=dρ2ψ−1ρ1,ρ4=2,dρωρω+2,ρ4=2−ω,ifω=1,2,ω−12,ifω=3,5,ω−42,ifω=4,6,8,de1,ρ5=5−ω2ifω=1,3,5,7,4−ω2ifω=2,4,6,dρ2ψ−2ρ2,ρ5=dρ2ψ−1ρ1,ρ5=2,dρ2ψ−3ρ2,ρ5=1,dρωρω+2,ρ5=ω;ifω=1,2,0;ifω=3,5,1;ifω=4,6,dρωρω+1,ρ6=5−ω2ifω=1,3,5,7,6−ω2,ifω=2,4,6,dρ2ψ−2ρ2,ρ6=dρ2ψ−1ρ1,ρ6=1,dρ2ψ−3ρ2,ρ6=2,dρωρω+2,ρ6=2ω;ω=1,2,1;ω=3,5,0;ifω=4,6,dρωρω+1,ρ7=1;ifω=1,7−ω2;ifω=3,5,7,6−ω2;ifω=2,4,6,dρ2ψ−2ρ2,ρ7=dρ2ψ−1ρ1,ρ7=1,dρ2ψ−3ρ2,ρ6=0,dρωρω+2,ρ6=5−ω2;ifω=1,3,5,2;ifω=2,41;ifω=6.
Whenψ≥6even,(11)dρωρω+1,ρ1=ω−22;ifω=1,3,…,ψ−1,2ψ−ω−32;ifω=ψ+1,ψ+3,…,2ψ−3,ω2;ifω=2,4,…,ψ,2ψ−ω2;ifω=ψ+2,ψ+4,…,2ψ−4,dρ2ψ−2ρ2,ρ1=dρ2ψ−3ρ2,ρ1=1,dρ2ψ−1ρ1,ρ1=0,dρωρω+2,ρ1=ω−22;ifω=1,3,…,ψ−1,2ψ−ω−12;ifω=ψ+1,ψ+3,…,2ψ−5,ω2;ifω=2,4,…,ψ−2,2ψ−ω−22;ifω=ψ,ψ+2,…,2ψ−4,dρωρω+1,ρ4=1;ifω=1,2,ω−32;ifω=3,5,…,ψ+3,2ψ−ω+32;ifω=ψ+5,ψ+7,…,2ψ−3ω−42;ifω=4,6,…,ψ+2,2ψ−ω+22;ifω=ψ+4,ψ+6,…,2ψ−4.dρ2ψ−2ρ2,ρ4=dρ2ψ−3ρ2,ρ4=dρ2ψ−1ρ1,ρ4=2.
Whenψ≥7odd,
(12)dρωρω+2,ρ4=2−ω;ifω=1,2,ω−12;ifω=3,5,…,ψ+1,2ψ−ω+12;ifω=ψ+3,ψ+5,…,2ψ−5,ω−42;ifω=4,6,…,ψ+2,2ψ−ω+22;ifω=ψ+4,ψ+6,…,2ψ−4,dρωρω+1,ρψ+2=ψ−22;ifω=1,ψ−ω+12;ifω=3,5,…,2ψ−3,ψ−ω−22;ifω=2,4,…,2ψ−4,dρ2ψ−2ρ2,ρψ+2=dρ2ψ−3ρ2,ρψ+2=dρ2ψ−1ρ1,ρψ+2=ψ−42,dρωρω+2,ρψ+2=ψ−ω+12;ifω=1,3,…,ψ−1,ψ−ω+32;ifω=ψ+1,ψ+3,…,2ψ−5,ψ−ω2;ifω=2,4,…,ψ,ω−ψ−22;ifω=ψ+2,ψ+4,…,2ψ−4,dρωρω+1,ρψ+3=ψ−42;ifω=1,ψ−ω+32;ifω=3,5,…,2ψ−3,ψ−22;ifω=2,ψ−ω+42;ifω=4,6,…,2ψ−4,dρ2ψ−2ρ2,ρψ+3=dρ2ψ−1ρ1,ρψ+3=ψ−42,dρ2ψ−3ρ2,ρψ+3=ψ−62,dρωρω+2,ρψ+3=ψ−22;ifω=1,2,ψ−ω+12;ifω=3,5,…,2ψ−5,ψ−ω+22;ifω=4,6,…,ψ,1;ifω=ψ+2,ω−ψ−22;ifω=ψ+4,ψ+6,…,2ψ−4.
It is clear to see the distinct representations of all edges with respect to edge resolving setQe, and it is proved that dimeTMLψ≤4.
Now, fordimeTMLψ≥4, choose the contradiction method, and it implies that dimeTMLψ=3. Now, the cardinality of edge metric generator Qe′ is three, following is some discussion for this contradiction:Case 1: if three vertices with any arbitrary size of gap are selected andQe′=ρω,ρj,ρk with 1≤ω,j,kodd≤2ψ−3, then it implies the same distances in the edges which are either dρ1ρ2|Qe′=dρ1ρ2ψ−1|Qe′, and it is also concluded that we cannot take vertices in the edge metric generator with cardinality three.Case 2: if all vertices with any arbitrary size of gap can choose such asQe′=ρω,ρj,ρk with 2≤ω,j,keven≤2ψ−2, then it implies the same distances in the edges which are dρ1ρ2|Qe′=dρ2ρ3|Qe′, and it is also concluded that we cannot take vertices in the edge metric generator with cardinality three.Case 3: now, the verticesρω,ρj,ρk∈VTMLψ without choosing the size of gap, then there exist an edge from joining edges and another one edge from upper horizontal edges having the same distances to each other with respect to the decided edge metric generator, i.e., dρpρp+1|Qe′=dρpρp+2|Qe′ where 2≤peven≤2ψ−4 or dρpρp+2|Qe′=dρqρq+1|Qe′ where 1≤podd≤2ψ−5, and finally, we concluded that we cannot take any type of vertices with any gap-size in the edge metric generator with cardinality three, which implies the contradiction that dimeTMLψ≠3, Moreover, this proves the double inequality which is(13)dimeTMLψ=4.


## 6. Conclusion

In the response of the question proposed by Kelenc et al. [3], in this article, we discussed some networks which have dimG=dimeG. Hexagonal Möbius ladder has three metric and edge metric dimensions. Triangular ladder, triangular Möbius ladder, and Möbius ladder fall in the category which have dimG<dimeG. Moreover, there is no change in the metric dimension and edge metric dimension of triangular ladder graph after making it triangular Möbius ladder and have dimeG=dimG+1=4.



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