# Metric Dimension of Some Generalized Families of Toeplitz Graphs

Authors: Muhammad Faisal Nadeem; Shaojian Qu; Ali Ahmad; Muhammad Azeem
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

The metric dimension of a graphG is the selection of the minimum possible number of vertices such that each vertex of the graph G is distinctively defined by its vector of distances to the set of selected vertices. It was proved that the problem of determining the metric dimension of a graph is NP-hard. In this paper, the metric dimension of Toeplitz graphs with two and three generators is discussed and the exact values are found. Also, two conjectures about the exact metric dimension of Toeplitz graphs are given.


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## Body

## 1. Introduction

Computation is the procedure of evaluating by mathematical or logical methods. The implementation of an algorithm is a well-recognized example of computation. A problem is considered naturally challenging if its solution takes substantial resources, no matter what the algorithm applied. In many daily life problems, we face difficult circumstances to handle and tackle these problems; most important of them is developing an algorithm that reduces the cost and, most importantly, the time to understand and solve the problem. The performance requirements of the solution of numerous problems, such as radar tracking, robot navigation, weather forecast, signal and image processing, numerical solutions of some differential equations, and several integral equations, are far from the capability of a single-core processing system. Parallel systems network breaks a single task into similar problems that are completed simultaneously, significantly reducing the application time. In a network, it is challenging, costly, and, more importantly, time taking to map all nodes and links between them in a large network; due to these issues nowadays, the most frequently used method is to attain a local view of the constructed networks from several positions and link them together to attain a good estimation for the network. The metric dimension is beneficial to overcome and resolve this type of problem [1, 2].Harary and Melter were first to study problem of finding the metric dimension of a graph [3]. Computational complexity and NP-hardness of the metric dimension problem were discussed in [4, 5]. Metric dimension has various practical applications in our daily life which motivate the researchers, and it has been widely studied. Metric dimension is extensively applied in the various fields such as robot navigation [6], pharmaceutical chemistry [7], weighing problems [8], computer networks [9], combinatorial optimization [10], image processing, facility location problems, and sonar and coastguard loran [11]; for further details, see [12, 13]. Due to its variety of applications, the concept of metric dimension is widely used to solve many difficult problems easily. Metric dimension of chemical structures is discussed in [14]; graphs of order n>11 and diameter 2 with partition dimension n−3 are evaluated in [15]; lattice structure of chemical nanotube is studied in [16]; sharp bounds for the metric dimension of chemical cellulose structure are found in [17]; metric dimension of Möbius structure is discussed in [18]; recently, the authors in [19] studied coronoid chemical network and proved that the structure has constant metric dimension; for the generalized version of this concept, we refer to [20–22]. For detail study on the metric dimension of some classes of graphs, see [23–26].The distance between two different verticesμ1,μ2∈VG of a graph G is the length of shortest path between them; it is denoted by dμ1,μ2, where VG is set of vertices of G. A subset ℵ of VG gratifying the property that for any two different vertices μ1,μ2∈VG, we have γ∈ℵ such that dμ1,γ≠dμ2,γ is said to be the resolving set for G and dimG denotes the least possible cardinality of ℵ which we called the metric dimension of G [27, 28]. The results given below in the form of lemma and theorems are useful in finding the metric basis and metric dimension for any graph. Note that the selection of suitable basis elements (vertices) also referred as landmarks [6] is the fundamental of finding the metric dimension.Lemma 1 (see [29]).
Letℵ be a resolving set for a connected graph G and μ1,μ2∈VG. If dμ1,γ=dμ2,γ for all vertices γ∈VG∖μ1,μ2, then μ1,μ2∩ℵ≠∅.Theorem 1 (see [7]).
LetG be a simple connected graph of order n≥2. Then,(a)
dimG=1 if and only if G=Pn(b)
dimG=n−1 if and only if G=Kn(c)
dimG=n−2 if and only if G=Kr,s, where r,s≥1 and n≥4Theorem 2 (see [30]).
LetG be a simple connected graph with metric dimension 2 and let μ,ν⊆VG be a metric basis in G, then the degree of both μ and ν is at most 3 and there exists a unique shortest path between μ and ν.


## 2. Toeplitz Matrices and Their Applications

Toeplitz matrices also referred to asT matrices are named after Otto Toeplitz. These matrices have the same fixed entries along their diagonals. The structure of T matrix is incredibly fascinating due to its many well-endowed theoretical properties; however, at the exact moment, it is also very vital for the extraordinary effect that it has in various applications. T matrices and their solution appear in several theoretical and applicative fields of sciences, such as in different mathematical models of all those types of problems in which a certain kind of shift invariance arises in the form of space or time. These types of shift invariances that appear in problems also reflect in the structure of the T matrix. The Toeplitz matrix structure may appear single entry-wise in a single-dimensional problem, block-wise in a two-dimensional problem, or maybe at many nested levels in a multidimensional problem. T matrices also play their role in linear prediction, east squares estimations, time series analysis, image restoration, error control coding, signal and image processing, spectrum analysis, queueing theory, and self-regression filter design. In the processing of digital signals, Toeplitz matrices are among the most generally applied special matrices. The shift of a matrix’s entries with a zero mean arbitrary stationary system one of the frequently utilized signals is a Toeplitz matrix due to its symmetric structure. A Toeplitz matrix may be finite or infinite depending on the characteristics of the problem that is modeled [31, 32].


### 2.1. Toeplitz Graphs

A Toeplitz graph denoted byTn is a graph obtained from the Toeplitz adjacency matrix. The Toeplitz graph TnJ1,J2,…,Js with vertex set μ1,μ2,μ3,…,μn the edge between two vertices μi and μj, 1≤i≤j≤n, exists if and only if j−i=Jς for some ς, 1≤ς≤s. The following theorem gives a lower bound on the number of components of a Toeplitz graph; we will consider only those cases when the Toeplitz graph has exactly one component.Theorem 3 (see [33]).
IfgcdJ1,J2,…,Js>1, then TnJ1,J2,…,Js is disconnected.


## 2.1. Toeplitz Graphs

A Toeplitz graph denoted byTn is a graph obtained from the Toeplitz adjacency matrix. The Toeplitz graph TnJ1,J2,…,Js with vertex set μ1,μ2,μ3,…,μn the edge between two vertices μi and μj, 1≤i≤j≤n, exists if and only if j−i=Jς for some ς, 1≤ς≤s. The following theorem gives a lower bound on the number of components of a Toeplitz graph; we will consider only those cases when the Toeplitz graph has exactly one component.Theorem 3 (see [33]).
IfgcdJ1,J2,…,Js>1, then TnJ1,J2,…,Js is disconnected.


## 3. Results on the Metric Dimension of Toeplitz Graphs

Liu et al. asked open problems about the metric dimension for different classes of Toeplitz graphs [34]. In this paper, we gave positive answer to their open problems and found the exact values of the metric dimension of Tn2,β, Tn3,β, and Tn1,2,β, where β≥10.


### 3.1. Metric Dimension ofTn2,β

In this subsection, we will find the metric dimension of Toeplitz graphTn2,β, where 1 and β are its generators; here, β must be odd by Theorem 3. We show that the dimension is two if n≤2β and three otherwise.Theorem 4.
LetTn2,β be a Toeplitz graph, with n≥β+3 and β≥3odd. Then,(1)dimTn2,β=2ifn≤2β3ifn>2β.Proof.
We will split our proof in two cases.Case A.n≤2β.Letℵ=μ1,μ2⊂VTn2,β; now, we will show that the subset ℵ resolves all the vertices of Tn2,β. To prove this, we give representation of all the vertices of Tn2,β, as follows:(2)rμκ|ℵ=β+12κ2,1κβ+12,forκ=1,2κ2−νβ2+ν2,β+12−κ2+1+νβ+12+κβ+5,forκ≡1,3,…,β+2modββ−12−κ2+2+β+12u+u2,κ2−uβ2−1+u,forκ≡2,4,…,β+1modβ,whereν=κ−3/β and u=κ−4/β. Since all the representations are different for any two distinct vertices, then dimTn2,β≤2.On the other hand, we know thatTn2,β is not a path graph; hence, by [7], dimTn2,β=2.Case B.n>2β.In this case, we show that only three elements in the basis set are enough to resolve all the vertices of Toeplitz graphTn2,β. We claim that the subset ℵ=μ1,μ2,μ3 of VTn2,β resolves all the vertices of Tn2,β. To prove our claim, the representation of all the vertices of Tn2,β is given as follows:(3)rμκ|ℵ=β+12κ2,1κβ+12,β−12κ−12+1,forκ=1,2κ2−νβ2+ν2,β+12−κ2+1+νβ+12+κβ+5+y,κ2−νβ2+ν2−1,forκ≡1,3,…,β+2modβ2+β−12−κ2+β+12u+u2,κ2−uβ2−1+u,β+12−κ2+2+uβ+12+u2+z,forκ≡2,4,…,β+1modβ,whereν=κ−3/β, u=κ−4/β, y=cc−1κ=νβ+3>2β0otherwise, and z=cc−1κ=uβ+4>β0otherwise.
Again, there are no two vertices having the same representations; hence,(4)dimTn2,β≤3.
Now, for the converse, we have to prove thatdimTn2,β≥3. On contrary, suppose that dimTn2,β=2, and all possibilities for the resolving set ℵ=μi,μj1≤i,j≤n are discussed. In every case, there exist at least two vertices which has the same representation.(i)
ℵ=μn−β+1,μn−β+i. If 2≤ieven≤β−1, then rμ2|ℵ=μβ+4|ℵ=1,β+1/2, and if 3≤iodd≤β−1, then rμ1|ℵ=μβ+3|ℵ=β+1/2,β+1/2−κ2−β/2+2 where κ2=n−β+i.(ii)
ℵ=μn−β+2,μn−β+i. rμ2|ℵ=μβ+4|ℵ=μβ+3|ℵ=2,κ2−β/2 where 4≤ieven≤β−1 and κ2=n−β+i.(iii)
ℵ=μn−β+3,μn−β+i. rμ1|ℵ=μβ+3|ℵ=μβ+3|ℵ=β+1/2,β+1/2−κ2−β−4/2+2 where 5≤iodd≤β−1 and κ2=n−β+i.(iv)
ℵ=μ1,μi.(5)rμ2|ℵ=μβ+4|ℵ=β+12,β+12−i2+1,fori≡1mod3,β+12,i−22,fori≡0mod3,where3≤i≤β.(v)
ℵ=μ1,μ2.(6)rμβ+3|ℵ=μ2β+1|ℵ=2,β+12.(vi)
ℵ=μ2,μi.(7)rμ1|ℵ=μβ+3|ℵ=β+12,i2,fori≡1mod3,β+12,β+12−i2+1,fori≡0mod3,where3≤i≤β.(vii)
ℵ=μβ−1,μβ. rμ1|ℵ=μβ+3|ℵ=2,β/2.(viii)
ℵ=μβ−3,μn−1. rμ1|ℵ=μβ+3|ℵ=β+1/2,3.(ix)
ℵ=μ2,μ2β+1. rμ1|ℵ=μβ+3|ℵ=β+1/2,β+1/2.(x)
ℵ=μ3,μn−1. rμ2|ℵ=μβ+4|ℵ=β+1/2,3.(xi)
ℵ=μβ,μn. rμ1|ℵ=μβ+3|ℵ=2,β−1/2.(xii)
ℵ=μβ,μn−1. rμ1|ℵ=μ2β−1|ℵ=β−1/2,β+1/2.(xiii)
ℵ=μβ,μn−β. rμ1|ℵ=μβ+3|ℵ=β−1/2,1.(xiv)
ℵ=μβ,μn−β+1. rμ1|ℵ=μβ+3|ℵ=β−1/2,β+1/2.(xv)
ℵ=μβ,μn−β+2. rμ2|ℵ=μβ+4|ℵ=β−1/2,β+1/2.
Hence, in all possible cases, there is no subset of vertices with two elements which resolve the vertices ofTn2,t; thus,(8)dimTn2,t≥3.
From Inequalities 1 and 2, we have(9)dimTn2,t=3.


### 3.2. Metric Dimension ofTn3,β

In this subsection, we will find the metric dimension of Toeplitz graphTn3,β. We are interested only in connected Toeplitz graphs; therefore, we will consider only two cases, i.e., when β=3t+1 and β=3t+2. For β=3t, the Toeplitz graph Tn3,β has three components. The appropriate selection of landmarks is crucial in determining the metric dimension of Tn3,β.Theorem 5.
LetTn3,β be a Toeplitz graph, where n≥β+5 and β≥8. Then,(10)dimTn3,β=2,ifn=4+β,5+β3,ifn>β+5.,Proof.
We split our proof in two parts. In first part, we split representation of verticesr1 and r2 and in second part r1, r2, and r3. In both parts, the equations of r1μκ|ℵ, r2μκ|ℵ, and r3μκ|ℵ depend on the choice of κ; in some choices, we have the same case and different cases for some other choices of κ. First, we will discuss same cases for the choice of κ to show the representations of vertices and then different cases.
In the first part of the proof, only two vertices appropriately selected are enough to resolve all the vertices inTn3,β, where n=4+β,5+β. Let ℵ=μ1,μ2 be the resolving set. Two different representations of vertices r1 and r2 areCaseA1. β≡2mod3.Here,j1=2β−3i1/4, i1=β−5−β/3, ζ=κ/j1+1, and α1=κ−j1+β−1/β.(1)
Ifκ≡1,4,7,…,β−1modβ, then r1μκ|ℵ=κ/3−β/3κ/β+κ/β+1/3 and r2μκ|ℵ=β+4/3−κ/3+κ/ββ+4/3−κ/β+1/3.(2)
Ifκ≡β+2,β+5,…,j1modβ, then r1μκ|ℵ=κ/3−β/3α1+α1+1/3 and r2μκ|ℵ=κ/β+2+κ/3−β+1/3−ζi1+ζ+1/3iii(3)
Ifκ=2,5,…,β−3i1+3/4, then r1μκ|ℵ=β+4/3+κ/3 and r2μκ|ℵ=κ/3.(4)
Ifκ≡0modβ, where κ,β≥11, then r1μκ|ℵ=β+7/3+κ/β−1 and r2μκ|ℵ=κ−2/3−β−2/3κ−1/β+κ−1/β+2/3.(5)
Ifκ=β−3 where β≥14, then r1μκ|ℵ=κ+13/3 and r2μκ|ℵ=κ−2/3.For remaining values ofκ, the following three different cases will be used to show the representations of vertices.(6)
Ifκ≡3,6,…,β−2modβ, then r1μκ|ℵ=β+1/3−κ/3+1−β+4/3κ/β+κ/β+2/3.(7)
Ifκ=3,6,…,β−2, then r2μκ|ℵ=β+1/3+κ/3−1.(8)
Ifκ≡3,6,…modβ and β+3≤κ≤2β−2, then r2μκ|ℵ=κ/3−3β−6/3+β+1/3−ηβ−2/3+η/3 where η=κ−β/β.CaseA2. β≡1mod3.Here,j2=2β−3i2/4−1, i2=β−7−β−2/3, and α2=κ−j2+β−1/β.(1)
Ifκ≡2,5,…,β−2modβ, then r1μκ|ℵ=β+2/3−κ/3+β+2/3κ/β+κ/β+2/3 and r2μκ|ℵ=κ/3−κ/ββ−4/3−κ/β+2/3.(2)
Ifκ=β−4, β≥14, then r1μκ|ℵ=β+11κ+4/3β and r2μκ|ℵ=κ/β−4+2.For remaining values ofκ, the following different cases arise.(3)
Ifκ≡0,1,4,7,…,β−3modβ, then r1μκ|ℵ=κ/3−ββ−2/3−β+2/3.(4)
Ifκ≡β+3,β+6,…,j2modβ, then r1μκ|ℵ=κ/3−α2β−2/3−α2+2/3.(5)
Ifκ=3,6,…,β−3i2+3/4−1, then r1μκ|ℵ=β+5/3+κ/3−1.(6)
Ifκ≡β−1modβ and β≥13, then r1μκ|ℵ=β+8/3+κ/β−1−1vii′.(7)
Ifκ=β−4, β≥14, then r1μκ|ℵ=β+11κ+4/3β.(8)
Ifκ=1,4,7,…,β−3, then r2μκ|ℵ=β+2/3+κ/3.(9)
Ifκ≡1,4,…modβ and β+1≤κ≤2β−3, then r2μκ|ℵ=β−1/3+κ−β−1/3−κ−β/3β−4/3−η/3.(10)
Ifκ≡t,β+3,…,j2modβ, then r2μκ|ℵ=β+8/3−κ−β/3+κ−1/j2β+2/3+κ−1/j2/3.(11)
Ifκ=3,6,…,β−3, then r2μκ|ℵ=β+2/3−κ/3+1. x′′ If κ≡β−1modββ≥11, then r2μκ|ℵ=κ/β−1+1xi″κ=β−4, β≥14, and then r2μκ|ℵ=κ/β−4+2.
In both cases, all the representations of the vertices are different; hence,dimTn3,β≤2. Converse is trivial as the graph is not a path graph; hence, dimTn3,β=2.
For second part of the proof, letℵ=μ1,μ2,μ3 be subset of VTn3,β. We will show that ℵ resolves all the vertices of Tn3,β. The representation of all vertices μκ with respect to resolving set ℵ can be written in the following form: rμκ|ℵ=r1μκ|ℵ,r2μκ|ℵ,r3μκ|ℵ.CaseB1. β≡2mod3.i′If κ≡1,4,7,…,β−1modβ, then r1μκ|ℵ=κ/3−β/3κ/β+κ/β+1/3.ii′ If κ≡β+2,β+5,…,j1modβ, then r1μκ|ℵ=κ/3−β/3α1+α1+1/3. iii′If κ=2,5,…,β−3i1+3/4, then r1μκ|ℵ=β+4/3+κ/3. iv′ If κ≡3,6,…,β−2modβ, then r1μκ|ℵ=β+1/3−κ/3+1−β+4/3κ/β+κ/β+2/3. v′ If κ≡0modβ; κandβ≥11, then r1μκ|ℵ=β+7/3+κ/β−1. vi′ If κ=β−3 and β≥14, then r1μκ|ℵ=κ+13/3.i′ If κ≡1,4,7,…,β−1modβ, then r2μκ|ℵ=β+4/3−κ/3+κ/ββ+4/3−κ/β+1/3. ii′ If κ≡β+2,β+5,…,j1modβ, then r2μκ|ℵ=κ/β+2+κ/3−β+1/3−ζi1+ζ+1/3. iii′ If κ=2,5,…,β−3i1+3/4, then r2μκ|ℵ=κ/3. iv′ If κ=3,6,…,β−2, then r2μκ|ℵ=β+1/3+κ/3−1. v′ If κ≡3,6,…modβ, β+3≤κ≤2β−2, then r2μκ|ℵ=κ/3−3β−6/3+β+1/3−ηβ−2/3+η/3. vi′ If κ≡0modβ, κ,β≥11, then r2μκ|ℵ=κ−2/3−β−2/3β+β+2/3. vii″ If κ=β−3, β≥14, then r2μκ|ℵ=κ−2/3.i′ If κ=1,4,7,…,β−1, r3μκ|ℵ=β+1/3+κ/3. ii′ If κ≡β+2,β+5,…,j1modβ, then r3μκ|ℵ=β+7/3−κ−β−2/3+α1β+4/3−α1+2/3. iii′ If κ=2,5,…,β−3i1+3/4, then r3μκ|ℵ=β+4/3−κ/3. iv′ If κ≡3,6,…,β−2modβ, then r3μκ|ℵ=κ/3−1−ξβ−2/3+ξ+2/3 where ξ=κ−1/β−2. v′ If κ≡1,4,…modβ, β+1≤κ≤2β−1, then r3μκ|ℵ=κ/3−1−ηβ−2/3+η+2/3. vi′ If κ≡0modβκandβ≥11, then r3μκ|ℵ=κ+β/β. vii″ If κ=β−3, β≥14, then r3μκ|ℵ=κ/β−3+2CaseB2. β≡1mod3,β≥10i′ If κ≡0,1,4,7,…,β−3modβ, then r1μκ|ℵ=κ/3−ββ−2/3−β+2/3. ii′ If κ≡β+3,β+6,…,j2modβ, then r1μκ|ℵ=κ/3−α2β−2/3−α2+2/3. iii′ If κ≡2,5,…,β−2modβ, then r1μκ|ℵ=β+2/3−κ/3+β+2/3κ/β+κ/β+2/3. iv′ If κ=3,6,…,β−3i1+3/4−1, r1μκ|ℵ=β+5/3+κ/3−1. v′ If κ≡β−1modββ≥13, r1μκ|ℵ=β+8/3+κ/β−1−1. vi′ If κ=β−4, β≥14, then r1μκ|ℵ=β+11κ+4/3βi″ If κ=1,4,7,…,β−3, then r2μκ|ℵ=β+2/3+κ/3. ii″ If κ≡1,4,…modβ with β+1≤κ≤2β−3, then r2μκ|ℵ=β−1/3+κ−β−1/3−κ−β/3β−4/3−η/3. iii″ If κ≡t,β+3,…,j2modβ, then r2μκ|ℵ=β+8/3−κ−β/3+κ−1/j2β+2/3+κ−1/j2/3. iv″ If κ≡2,5,…,β−2modβ, then r2μκ|ℵ=κ/3−κ/ββ−4/3−κ/β+2/3. v″ If κ=3,6,…,β−3, then r2μκ|ℵ=β+2/3−κ/3+1. vi″ If κ≡β−1modββ≥11, then r2μκ|ℵ=κ/β−1+1. vii″ If κ=β−4, β≥14, then r2μκ|ℵ=κ/β−4+2.i‴ If κ≡0,1,4,7,…,β−3modβ, then r3μκ|ℵ=β+5/3−κ/3+ββ+2/3+β+1/3. i‴ If κ≡β+3,β+6,…,j2modβ, then r3μκ|ℵ=κ−β/3−α2β−4/3+α2/3. i‴ If κ=2,5,…,β−2, then r3μκ|ℵ=β+2/3+κ/3. i‴ If κ=3,6,…,β−3i1+3/4−1, then r3μκ|ℵ=κ/3−1. i‴ If κ≡β−1modβ, β≥13, then r3μκ|ℵ=κ−3/3−ββ−4/3+β/3. i‴ If κ=β−4, β≥14, then r3μκ|ℵ=κ−3/3. i‴ If κ≡2,5,…modβ with β+2≤κ≤2β−2. rμκ|ℵ=β−1/3+κ−β−2/κ−β−2−ηβ−2/3−η/3.
As, all the representations of different vertices are distinct. This shows thatℵ=μ1,μ2,μ3 resolves the vertices of Tn3,β; hence,(11)dimTn3,β≤3.
For converse, we will show thatdimTn3,β=2 is not possible and discuss all possible cases.(i)
Ifℵ=μ1,μ2, then rμ3|ℵ=μβ+6|ℵ=β+5/3,β+2/3.(ii)
Ifℵ=μn−β+i,μn−β+j, where 1≤i,j≤β, i≠j, and β≡2mod3, β≥8rμ1|ℵ=μβ+4|ℵ=(12)β+κ1+23,κ23−1,Ifi≡0mod3β+κ1+23,β−23−κ23+2,Ifi≡1mod3β+κ1+23,β−23−κ23+2,Ifi≡2mod3,whereκ1=n−β+i and κ2=n−β+j.In following each case, we have two subcases to discuss.(iii)
Ifℵ=μ1,μi, where 3≤i≤β, then we have the following two subcases:Forβ≡2mod3 and β≥8, we have(13)rμ2|ℵ=μβ+5|ℵ=β+43,β+43+i3−1,Ifi≡0mo3β+43,β+13−i3+1,Ifi≡1mod3β+43,i3,Ifi≡2mod3.And, forβ≡1mod3 and β≥10, we get(14)rμ2|ℵ=μβ+5|ℵ=β+43,β+43−i3+1,Ifi≡0mod3β+43,β+53+i3−1,Ifi≡1mod3β+43,i3,Ifi≡2mod3(iv)
Ifℵ=μj,μi,3≤i≤β,2≤j≤β again, we have following two subcases:Forβ≡2mod3 and β≥8, we have(15)rμ1|ℵ=μβ+4|ℵ=β+43,β+13−i3+1,Ifi≡0mod3β+43,β+43+i3−1,Ifi≡1mod3β+43,β+73+i3−1,Ifi≡2mod3And, forβ≡1mod3, where β≥10, we have(16)rμ2|ℵ=μβ+5|ℵ=β+43,β+53+i3−1,Ifi≡0mod3β+43,i3,Ifi≡1mod3β+43,β+53+i3−1,Ifi≡2mod3(v)
Ifℵ=μ1,μn−β+i, where 1≤i≤β, then for β≡2mod3 and β≥8, (17)rμ2|ℵ=μβ+5|ℵ=β+43,β−43−κ2−93,Ifi≡0mod3β+43,β+83+κ2−73,Ifi≡1mod3β+43,β+73+κ23,Ifi≡2mod3And, forβ≡1mod3, where β≥10,(18)rμ2|ℵ=μβ+5|ℵ=β+23,2−κ2−63,i≡0mod3β+23,β+83+κ2−73,i≡1mod3β+23,κ23−1,i≡2mod3whereκ2=n−β+i.(vi)
Finally, ifℵ=μj,μn−β+i, where 1≤i≤β,2≤j≤β, then we have following two subcases:Forβ≡2mod3, where β≥8rμ1|ℵ=μβ+4|ℵ=(19)β−κ1+63,β−κ2+46,Ifi≡0mod3β−κ1+63,κ23,Ifi≡1mod3β−κ1+63,β+73−κ2−83−1,Ifi≡2mod3And, forβ≡1mod3, where β≥10(20)rμ2|ℵ=μβ+5|ℵ=β+κ13,β+83−κ2−93,Ifi≡0mod3β+κ13,κ23,Ifi≡1mod3β+κ13,β−κ2+43,Ifi≡2mod3whereκ1=j and κ2=n−β+i
Hence, from above representations, it follows that there is no resolving set with two vertices. So,(21)dimTn3,β=3.


### 3.3. Metric Dimension ofTn1,2,β

In this section, we study the metric dimension ofTn1,2,β, where 1, 2, and β are its generators and β≥10 is even.Theorem 6.
LetTn1,2,β be a Toeplitz graph, with n≥β+2 and β≥10. Then, dimTn1,2,β=3.Proof.
Letℵ=μ2,μ6,μa, where a=2β/2+3/2 be a resolving set. We will show that ℵ resolves all the vertices of Tn1,2,β. The representation of all vertices with respect to resolving set ℵ can be written in the following form: rμκ|ℵ=r1μκ|ℵ,r2μκ|ℵ,r3μκ|ℵ, where r1,r2, and r3 are representations of vertex μκ with vertices μ2,μ6, and μa, respectively.(i)
Ifκ≡3,4,…,β−3modβ, then r1μκ|ℵ=κ−1/2−β−2/2κ/2.(ii)
Ifκ≡0,1,2,β−2,β−1modβ, then r1μκ|ℵ=3−κ−β+2/2+β+2/2κ/β+3+1/κ.i′ If κ≡0,1modβ, then(22)r2μκ|ℵ=3κ−β−22κβ+1+κ−β/22,Ifβ=10,123κ+5κ−1β−4κβ+1,Ifβ≥14ii′ If κ≡2,3,…,6modβ, then r2μκ|ℵ=2−κ−2/2+β+2/2κ/β.iii′ If κ≡7,8,…,β−1modβ, then r2μκ|ℵ=κ−5/2−β−2/2κ/β.i′′ If κ=1,2, then r3μκ|ℵ=β/4, and if κ=3,4, then r3μκ|ℵ=β+2/4−κ+3.ii′′ If κ≡0,…,4,2a,2a+1,…,β−1modβ and κ≥2ar3μκ|ℵ=κ−2a/2−β−2/2κ−5/β+z. Where z is 0 if κ even and 1 otherwise.iii′′ If κ≡5,…,β−3modβ, then r3μκ|ℵ=β−6/2−κ−5/2+β+2/2κ/β−z. Here, again z is 0 if κ is odd and 1 if κ is even.
As, all the representations of different vertices are distinct. So,dimTn1,2,β≤3.
For converse, we have only one possible subsetℵ=μ1,μn according to Theorem 2.
There are four possible cases for the choice ofn as follows:(1)
Ifn≡2,4,6,…,β−4modβ, then rμ2|ℵ=μβ+1|ℵ=1,n−β/2−β+2/2n−β/β.(2)
Ifn≡3,5,7,…,β−3modβ, then rμ3|ℵ=μβ+1|ℵ=1,n−β/2−β+2/2n−8/β.(3)
Ifn≡β−1modβ, then rμ2|ℵ=μ3|ℵ=,1,3+n/β.(4)
Ifn≡β−2modβ, then rμ8|ℵ=μ9|ℵ=4,n−10/β+n/β.(5)
Ifn≡0modβ, then rμ2|ℵ=μ3|ℵ=1,n/β.(6)
Ifn≡1modβ, then rμ6|ℵ=μ7|ℵ=3,4−3β/2+n−β+1n−2β/β.
It shows that there does not exist a resolving set of two vertices inTn1,2,β. Hence,(23)dimTn1,2,β=3.


## 3.1. Metric Dimension ofTn2,β

In this subsection, we will find the metric dimension of Toeplitz graphTn2,β, where 1 and β are its generators; here, β must be odd by Theorem 3. We show that the dimension is two if n≤2β and three otherwise.Theorem 4.
LetTn2,β be a Toeplitz graph, with n≥β+3 and β≥3odd. Then,(1)dimTn2,β=2ifn≤2β3ifn>2β.Proof.
We will split our proof in two cases.Case A.n≤2β.Letℵ=μ1,μ2⊂VTn2,β; now, we will show that the subset ℵ resolves all the vertices of Tn2,β. To prove this, we give representation of all the vertices of Tn2,β, as follows:(2)rμκ|ℵ=β+12κ2,1κβ+12,forκ=1,2κ2−νβ2+ν2,β+12−κ2+1+νβ+12+κβ+5,forκ≡1,3,…,β+2modββ−12−κ2+2+β+12u+u2,κ2−uβ2−1+u,forκ≡2,4,…,β+1modβ,whereν=κ−3/β and u=κ−4/β. Since all the representations are different for any two distinct vertices, then dimTn2,β≤2.On the other hand, we know thatTn2,β is not a path graph; hence, by [7], dimTn2,β=2.Case B.n>2β.In this case, we show that only three elements in the basis set are enough to resolve all the vertices of Toeplitz graphTn2,β. We claim that the subset ℵ=μ1,μ2,μ3 of VTn2,β resolves all the vertices of Tn2,β. To prove our claim, the representation of all the vertices of Tn2,β is given as follows:(3)rμκ|ℵ=β+12κ2,1κβ+12,β−12κ−12+1,forκ=1,2κ2−νβ2+ν2,β+12−κ2+1+νβ+12+κβ+5+y,κ2−νβ2+ν2−1,forκ≡1,3,…,β+2modβ2+β−12−κ2+β+12u+u2,κ2−uβ2−1+u,β+12−κ2+2+uβ+12+u2+z,forκ≡2,4,…,β+1modβ,whereν=κ−3/β, u=κ−4/β, y=cc−1κ=νβ+3>2β0otherwise, and z=cc−1κ=uβ+4>β0otherwise.
Again, there are no two vertices having the same representations; hence,(4)dimTn2,β≤3.
Now, for the converse, we have to prove thatdimTn2,β≥3. On contrary, suppose that dimTn2,β=2, and all possibilities for the resolving set ℵ=μi,μj1≤i,j≤n are discussed. In every case, there exist at least two vertices which has the same representation.(i)
ℵ=μn−β+1,μn−β+i. If 2≤ieven≤β−1, then rμ2|ℵ=μβ+4|ℵ=1,β+1/2, and if 3≤iodd≤β−1, then rμ1|ℵ=μβ+3|ℵ=β+1/2,β+1/2−κ2−β/2+2 where κ2=n−β+i.(ii)
ℵ=μn−β+2,μn−β+i. rμ2|ℵ=μβ+4|ℵ=μβ+3|ℵ=2,κ2−β/2 where 4≤ieven≤β−1 and κ2=n−β+i.(iii)
ℵ=μn−β+3,μn−β+i. rμ1|ℵ=μβ+3|ℵ=μβ+3|ℵ=β+1/2,β+1/2−κ2−β−4/2+2 where 5≤iodd≤β−1 and κ2=n−β+i.(iv)
ℵ=μ1,μi.(5)rμ2|ℵ=μβ+4|ℵ=β+12,β+12−i2+1,fori≡1mod3,β+12,i−22,fori≡0mod3,where3≤i≤β.(v)
ℵ=μ1,μ2.(6)rμβ+3|ℵ=μ2β+1|ℵ=2,β+12.(vi)
ℵ=μ2,μi.(7)rμ1|ℵ=μβ+3|ℵ=β+12,i2,fori≡1mod3,β+12,β+12−i2+1,fori≡0mod3,where3≤i≤β.(vii)
ℵ=μβ−1,μβ. rμ1|ℵ=μβ+3|ℵ=2,β/2.(viii)
ℵ=μβ−3,μn−1. rμ1|ℵ=μβ+3|ℵ=β+1/2,3.(ix)
ℵ=μ2,μ2β+1. rμ1|ℵ=μβ+3|ℵ=β+1/2,β+1/2.(x)
ℵ=μ3,μn−1. rμ2|ℵ=μβ+4|ℵ=β+1/2,3.(xi)
ℵ=μβ,μn. rμ1|ℵ=μβ+3|ℵ=2,β−1/2.(xii)
ℵ=μβ,μn−1. rμ1|ℵ=μ2β−1|ℵ=β−1/2,β+1/2.(xiii)
ℵ=μβ,μn−β. rμ1|ℵ=μβ+3|ℵ=β−1/2,1.(xiv)
ℵ=μβ,μn−β+1. rμ1|ℵ=μβ+3|ℵ=β−1/2,β+1/2.(xv)
ℵ=μβ,μn−β+2. rμ2|ℵ=μβ+4|ℵ=β−1/2,β+1/2.
Hence, in all possible cases, there is no subset of vertices with two elements which resolve the vertices ofTn2,t; thus,(8)dimTn2,t≥3.
From Inequalities 1 and 2, we have(9)dimTn2,t=3.


## 3.2. Metric Dimension ofTn3,β

In this subsection, we will find the metric dimension of Toeplitz graphTn3,β. We are interested only in connected Toeplitz graphs; therefore, we will consider only two cases, i.e., when β=3t+1 and β=3t+2. For β=3t, the Toeplitz graph Tn3,β has three components. The appropriate selection of landmarks is crucial in determining the metric dimension of Tn3,β.Theorem 5.
LetTn3,β be a Toeplitz graph, where n≥β+5 and β≥8. Then,(10)dimTn3,β=2,ifn=4+β,5+β3,ifn>β+5.,Proof.
We split our proof in two parts. In first part, we split representation of verticesr1 and r2 and in second part r1, r2, and r3. In both parts, the equations of r1μκ|ℵ, r2μκ|ℵ, and r3μκ|ℵ depend on the choice of κ; in some choices, we have the same case and different cases for some other choices of κ. First, we will discuss same cases for the choice of κ to show the representations of vertices and then different cases.
In the first part of the proof, only two vertices appropriately selected are enough to resolve all the vertices inTn3,β, where n=4+β,5+β. Let ℵ=μ1,μ2 be the resolving set. Two different representations of vertices r1 and r2 areCaseA1. β≡2mod3.Here,j1=2β−3i1/4, i1=β−5−β/3, ζ=κ/j1+1, and α1=κ−j1+β−1/β.(1)
Ifκ≡1,4,7,…,β−1modβ, then r1μκ|ℵ=κ/3−β/3κ/β+κ/β+1/3 and r2μκ|ℵ=β+4/3−κ/3+κ/ββ+4/3−κ/β+1/3.(2)
Ifκ≡β+2,β+5,…,j1modβ, then r1μκ|ℵ=κ/3−β/3α1+α1+1/3 and r2μκ|ℵ=κ/β+2+κ/3−β+1/3−ζi1+ζ+1/3iii(3)
Ifκ=2,5,…,β−3i1+3/4, then r1μκ|ℵ=β+4/3+κ/3 and r2μκ|ℵ=κ/3.(4)
Ifκ≡0modβ, where κ,β≥11, then r1μκ|ℵ=β+7/3+κ/β−1 and r2μκ|ℵ=κ−2/3−β−2/3κ−1/β+κ−1/β+2/3.(5)
Ifκ=β−3 where β≥14, then r1μκ|ℵ=κ+13/3 and r2μκ|ℵ=κ−2/3.For remaining values ofκ, the following three different cases will be used to show the representations of vertices.(6)
Ifκ≡3,6,…,β−2modβ, then r1μκ|ℵ=β+1/3−κ/3+1−β+4/3κ/β+κ/β+2/3.(7)
Ifκ=3,6,…,β−2, then r2μκ|ℵ=β+1/3+κ/3−1.(8)
Ifκ≡3,6,…modβ and β+3≤κ≤2β−2, then r2μκ|ℵ=κ/3−3β−6/3+β+1/3−ηβ−2/3+η/3 where η=κ−β/β.CaseA2. β≡1mod3.Here,j2=2β−3i2/4−1, i2=β−7−β−2/3, and α2=κ−j2+β−1/β.(1)
Ifκ≡2,5,…,β−2modβ, then r1μκ|ℵ=β+2/3−κ/3+β+2/3κ/β+κ/β+2/3 and r2μκ|ℵ=κ/3−κ/ββ−4/3−κ/β+2/3.(2)
Ifκ=β−4, β≥14, then r1μκ|ℵ=β+11κ+4/3β and r2μκ|ℵ=κ/β−4+2.For remaining values ofκ, the following different cases arise.(3)
Ifκ≡0,1,4,7,…,β−3modβ, then r1μκ|ℵ=κ/3−ββ−2/3−β+2/3.(4)
Ifκ≡β+3,β+6,…,j2modβ, then r1μκ|ℵ=κ/3−α2β−2/3−α2+2/3.(5)
Ifκ=3,6,…,β−3i2+3/4−1, then r1μκ|ℵ=β+5/3+κ/3−1.(6)
Ifκ≡β−1modβ and β≥13, then r1μκ|ℵ=β+8/3+κ/β−1−1vii′.(7)
Ifκ=β−4, β≥14, then r1μκ|ℵ=β+11κ+4/3β.(8)
Ifκ=1,4,7,…,β−3, then r2μκ|ℵ=β+2/3+κ/3.(9)
Ifκ≡1,4,…modβ and β+1≤κ≤2β−3, then r2μκ|ℵ=β−1/3+κ−β−1/3−κ−β/3β−4/3−η/3.(10)
Ifκ≡t,β+3,…,j2modβ, then r2μκ|ℵ=β+8/3−κ−β/3+κ−1/j2β+2/3+κ−1/j2/3.(11)
Ifκ=3,6,…,β−3, then r2μκ|ℵ=β+2/3−κ/3+1. x′′ If κ≡β−1modββ≥11, then r2μκ|ℵ=κ/β−1+1xi″κ=β−4, β≥14, and then r2μκ|ℵ=κ/β−4+2.
In both cases, all the representations of the vertices are different; hence,dimTn3,β≤2. Converse is trivial as the graph is not a path graph; hence, dimTn3,β=2.
For second part of the proof, letℵ=μ1,μ2,μ3 be subset of VTn3,β. We will show that ℵ resolves all the vertices of Tn3,β. The representation of all vertices μκ with respect to resolving set ℵ can be written in the following form: rμκ|ℵ=r1μκ|ℵ,r2μκ|ℵ,r3μκ|ℵ.CaseB1. β≡2mod3.i′If κ≡1,4,7,…,β−1modβ, then r1μκ|ℵ=κ/3−β/3κ/β+κ/β+1/3.ii′ If κ≡β+2,β+5,…,j1modβ, then r1μκ|ℵ=κ/3−β/3α1+α1+1/3. iii′If κ=2,5,…,β−3i1+3/4, then r1μκ|ℵ=β+4/3+κ/3. iv′ If κ≡3,6,…,β−2modβ, then r1μκ|ℵ=β+1/3−κ/3+1−β+4/3κ/β+κ/β+2/3. v′ If κ≡0modβ; κandβ≥11, then r1μκ|ℵ=β+7/3+κ/β−1. vi′ If κ=β−3 and β≥14, then r1μκ|ℵ=κ+13/3.i′ If κ≡1,4,7,…,β−1modβ, then r2μκ|ℵ=β+4/3−κ/3+κ/ββ+4/3−κ/β+1/3. ii′ If κ≡β+2,β+5,…,j1modβ, then r2μκ|ℵ=κ/β+2+κ/3−β+1/3−ζi1+ζ+1/3. iii′ If κ=2,5,…,β−3i1+3/4, then r2μκ|ℵ=κ/3. iv′ If κ=3,6,…,β−2, then r2μκ|ℵ=β+1/3+κ/3−1. v′ If κ≡3,6,…modβ, β+3≤κ≤2β−2, then r2μκ|ℵ=κ/3−3β−6/3+β+1/3−ηβ−2/3+η/3. vi′ If κ≡0modβ, κ,β≥11, then r2μκ|ℵ=κ−2/3−β−2/3β+β+2/3. vii″ If κ=β−3, β≥14, then r2μκ|ℵ=κ−2/3.i′ If κ=1,4,7,…,β−1, r3μκ|ℵ=β+1/3+κ/3. ii′ If κ≡β+2,β+5,…,j1modβ, then r3μκ|ℵ=β+7/3−κ−β−2/3+α1β+4/3−α1+2/3. iii′ If κ=2,5,…,β−3i1+3/4, then r3μκ|ℵ=β+4/3−κ/3. iv′ If κ≡3,6,…,β−2modβ, then r3μκ|ℵ=κ/3−1−ξβ−2/3+ξ+2/3 where ξ=κ−1/β−2. v′ If κ≡1,4,…modβ, β+1≤κ≤2β−1, then r3μκ|ℵ=κ/3−1−ηβ−2/3+η+2/3. vi′ If κ≡0modβκandβ≥11, then r3μκ|ℵ=κ+β/β. vii″ If κ=β−3, β≥14, then r3μκ|ℵ=κ/β−3+2CaseB2. β≡1mod3,β≥10i′ If κ≡0,1,4,7,…,β−3modβ, then r1μκ|ℵ=κ/3−ββ−2/3−β+2/3. ii′ If κ≡β+3,β+6,…,j2modβ, then r1μκ|ℵ=κ/3−α2β−2/3−α2+2/3. iii′ If κ≡2,5,…,β−2modβ, then r1μκ|ℵ=β+2/3−κ/3+β+2/3κ/β+κ/β+2/3. iv′ If κ=3,6,…,β−3i1+3/4−1, r1μκ|ℵ=β+5/3+κ/3−1. v′ If κ≡β−1modββ≥13, r1μκ|ℵ=β+8/3+κ/β−1−1. vi′ If κ=β−4, β≥14, then r1μκ|ℵ=β+11κ+4/3βi″ If κ=1,4,7,…,β−3, then r2μκ|ℵ=β+2/3+κ/3. ii″ If κ≡1,4,…modβ with β+1≤κ≤2β−3, then r2μκ|ℵ=β−1/3+κ−β−1/3−κ−β/3β−4/3−η/3. iii″ If κ≡t,β+3,…,j2modβ, then r2μκ|ℵ=β+8/3−κ−β/3+κ−1/j2β+2/3+κ−1/j2/3. iv″ If κ≡2,5,…,β−2modβ, then r2μκ|ℵ=κ/3−κ/ββ−4/3−κ/β+2/3. v″ If κ=3,6,…,β−3, then r2μκ|ℵ=β+2/3−κ/3+1. vi″ If κ≡β−1modββ≥11, then r2μκ|ℵ=κ/β−1+1. vii″ If κ=β−4, β≥14, then r2μκ|ℵ=κ/β−4+2.i‴ If κ≡0,1,4,7,…,β−3modβ, then r3μκ|ℵ=β+5/3−κ/3+ββ+2/3+β+1/3. i‴ If κ≡β+3,β+6,…,j2modβ, then r3μκ|ℵ=κ−β/3−α2β−4/3+α2/3. i‴ If κ=2,5,…,β−2, then r3μκ|ℵ=β+2/3+κ/3. i‴ If κ=3,6,…,β−3i1+3/4−1, then r3μκ|ℵ=κ/3−1. i‴ If κ≡β−1modβ, β≥13, then r3μκ|ℵ=κ−3/3−ββ−4/3+β/3. i‴ If κ=β−4, β≥14, then r3μκ|ℵ=κ−3/3. i‴ If κ≡2,5,…modβ with β+2≤κ≤2β−2. rμκ|ℵ=β−1/3+κ−β−2/κ−β−2−ηβ−2/3−η/3.
As, all the representations of different vertices are distinct. This shows thatℵ=μ1,μ2,μ3 resolves the vertices of Tn3,β; hence,(11)dimTn3,β≤3.
For converse, we will show thatdimTn3,β=2 is not possible and discuss all possible cases.(i)
Ifℵ=μ1,μ2, then rμ3|ℵ=μβ+6|ℵ=β+5/3,β+2/3.(ii)
Ifℵ=μn−β+i,μn−β+j, where 1≤i,j≤β, i≠j, and β≡2mod3, β≥8rμ1|ℵ=μβ+4|ℵ=(12)β+κ1+23,κ23−1,Ifi≡0mod3β+κ1+23,β−23−κ23+2,Ifi≡1mod3β+κ1+23,β−23−κ23+2,Ifi≡2mod3,whereκ1=n−β+i and κ2=n−β+j.In following each case, we have two subcases to discuss.(iii)
Ifℵ=μ1,μi, where 3≤i≤β, then we have the following two subcases:Forβ≡2mod3 and β≥8, we have(13)rμ2|ℵ=μβ+5|ℵ=β+43,β+43+i3−1,Ifi≡0mo3β+43,β+13−i3+1,Ifi≡1mod3β+43,i3,Ifi≡2mod3.And, forβ≡1mod3 and β≥10, we get(14)rμ2|ℵ=μβ+5|ℵ=β+43,β+43−i3+1,Ifi≡0mod3β+43,β+53+i3−1,Ifi≡1mod3β+43,i3,Ifi≡2mod3(iv)
Ifℵ=μj,μi,3≤i≤β,2≤j≤β again, we have following two subcases:Forβ≡2mod3 and β≥8, we have(15)rμ1|ℵ=μβ+4|ℵ=β+43,β+13−i3+1,Ifi≡0mod3β+43,β+43+i3−1,Ifi≡1mod3β+43,β+73+i3−1,Ifi≡2mod3And, forβ≡1mod3, where β≥10, we have(16)rμ2|ℵ=μβ+5|ℵ=β+43,β+53+i3−1,Ifi≡0mod3β+43,i3,Ifi≡1mod3β+43,β+53+i3−1,Ifi≡2mod3(v)
Ifℵ=μ1,μn−β+i, where 1≤i≤β, then for β≡2mod3 and β≥8, (17)rμ2|ℵ=μβ+5|ℵ=β+43,β−43−κ2−93,Ifi≡0mod3β+43,β+83+κ2−73,Ifi≡1mod3β+43,β+73+κ23,Ifi≡2mod3And, forβ≡1mod3, where β≥10,(18)rμ2|ℵ=μβ+5|ℵ=β+23,2−κ2−63,i≡0mod3β+23,β+83+κ2−73,i≡1mod3β+23,κ23−1,i≡2mod3whereκ2=n−β+i.(vi)
Finally, ifℵ=μj,μn−β+i, where 1≤i≤β,2≤j≤β, then we have following two subcases:Forβ≡2mod3, where β≥8rμ1|ℵ=μβ+4|ℵ=(19)β−κ1+63,β−κ2+46,Ifi≡0mod3β−κ1+63,κ23,Ifi≡1mod3β−κ1+63,β+73−κ2−83−1,Ifi≡2mod3And, forβ≡1mod3, where β≥10(20)rμ2|ℵ=μβ+5|ℵ=β+κ13,β+83−κ2−93,Ifi≡0mod3β+κ13,κ23,Ifi≡1mod3β+κ13,β−κ2+43,Ifi≡2mod3whereκ1=j and κ2=n−β+i
Hence, from above representations, it follows that there is no resolving set with two vertices. So,(21)dimTn3,β=3.


## 3.3. Metric Dimension ofTn1,2,β

In this section, we study the metric dimension ofTn1,2,β, where 1, 2, and β are its generators and β≥10 is even.Theorem 6.
LetTn1,2,β be a Toeplitz graph, with n≥β+2 and β≥10. Then, dimTn1,2,β=3.Proof.
Letℵ=μ2,μ6,μa, where a=2β/2+3/2 be a resolving set. We will show that ℵ resolves all the vertices of Tn1,2,β. The representation of all vertices with respect to resolving set ℵ can be written in the following form: rμκ|ℵ=r1μκ|ℵ,r2μκ|ℵ,r3μκ|ℵ, where r1,r2, and r3 are representations of vertex μκ with vertices μ2,μ6, and μa, respectively.(i)
Ifκ≡3,4,…,β−3modβ, then r1μκ|ℵ=κ−1/2−β−2/2κ/2.(ii)
Ifκ≡0,1,2,β−2,β−1modβ, then r1μκ|ℵ=3−κ−β+2/2+β+2/2κ/β+3+1/κ.i′ If κ≡0,1modβ, then(22)r2μκ|ℵ=3κ−β−22κβ+1+κ−β/22,Ifβ=10,123κ+5κ−1β−4κβ+1,Ifβ≥14ii′ If κ≡2,3,…,6modβ, then r2μκ|ℵ=2−κ−2/2+β+2/2κ/β.iii′ If κ≡7,8,…,β−1modβ, then r2μκ|ℵ=κ−5/2−β−2/2κ/β.i′′ If κ=1,2, then r3μκ|ℵ=β/4, and if κ=3,4, then r3μκ|ℵ=β+2/4−κ+3.ii′′ If κ≡0,…,4,2a,2a+1,…,β−1modβ and κ≥2ar3μκ|ℵ=κ−2a/2−β−2/2κ−5/β+z. Where z is 0 if κ even and 1 otherwise.iii′′ If κ≡5,…,β−3modβ, then r3μκ|ℵ=β−6/2−κ−5/2+β+2/2κ/β−z. Here, again z is 0 if κ is odd and 1 if κ is even.
As, all the representations of different vertices are distinct. So,dimTn1,2,β≤3.
For converse, we have only one possible subsetℵ=μ1,μn according to Theorem 2.
There are four possible cases for the choice ofn as follows:(1)
Ifn≡2,4,6,…,β−4modβ, then rμ2|ℵ=μβ+1|ℵ=1,n−β/2−β+2/2n−β/β.(2)
Ifn≡3,5,7,…,β−3modβ, then rμ3|ℵ=μβ+1|ℵ=1,n−β/2−β+2/2n−8/β.(3)
Ifn≡β−1modβ, then rμ2|ℵ=μ3|ℵ=,1,3+n/β.(4)
Ifn≡β−2modβ, then rμ8|ℵ=μ9|ℵ=4,n−10/β+n/β.(5)
Ifn≡0modβ, then rμ2|ℵ=μ3|ℵ=1,n/β.(6)
Ifn≡1modβ, then rμ6|ℵ=μ7|ℵ=3,4−3β/2+n−β+1n−2β/β.
It shows that there does not exist a resolving set of two vertices inTn1,2,β. Hence,(23)dimTn1,2,β=3.


## 4. Conjecture

In this section, we propose two conjectures regarding the metric dimension of the Toeplitz graph with two and three generators, andn is sufficiently large enough so the Toeplitz is connected and not a path graph.Conjecture 1.
LetTnα,β be a Toeplitz graph, with gcdα,β=1 and α≥3. Then,(24)dimTnα,β=2,3,4,…α−1,Ifn<2α+βα,Ifn≥2α+β.Conjecture 2.
IfTn1,2,β is a Toeplitz graph, with β≥10 odd, then dimTn1,2,β=3.


## 5. Conclusion

In this paper, metric dimension (location number) of Toeplitz graphs,Tn2,β, Tn3,β, and Tn1,2,β with β≥10 even, is studied by giving solution to the few open problems asked in [34]. We see that the metric dimension of studied Toeplitz graphs is constant and does not depend on the generators; we gave the exact values of the metric dimension in these cases. In the end, we give two conjectures on the metric dimension of Toeplitz graphs with two and three generators and believe that conjectures are correct.



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