Metric and Edge Metric Dimension of Zigzag Edge Coronoid Fused with Starphene

Sunny Kumar Sharma1,a, Vijay Kumar Bhat1,∗, Hassan Raza2,b, and Karnika Sharma1,c

1School of Mathematics, Shri Mata Vaishno Devi University, Katra-182320, J & K, India.2Business School, University of Shanghai for Science and Technology, Shanghai 200093, China.asunnysrrm94@gmail.com, ∗vijaykumarbhat2000@yahoo.com, bhassan raza783@yahoo.com,c19dmt001@smvdu.ac.in

Abstract Let Γ = (V,E) be a simple connected graph. d(α, ε) = min{d(α,w), d(α, d} computesthe distance between a vertex α ∈ V (Γ) and an edge ε = wd ∈ E(Γ). A single vertex α is said torecognize (resolve) two different edges ε1 and ε2 from E(Γ) if d(α, ε2) 6= d(α, ε1}. A subset of distinctordered vertices UE ⊆ V (Γ) is said to be an edge metric generator for Γ if every pair of distinctedges from Γ are recognized by some element of UE . An edge metric generator with a minimumnumber of elements in it, is called an edge metric basis for Γ. Then, the cardinality of this edgemetric basis of Γ, is called the edge metric dimension of Γ, denoted by edim(Γ). The concept ofstudying chemical structures using graph theory terminologies is both appealing and practical. Itenables chemical researchers to more precisely and easily examine various chemical topologies andnetworks. In this article, we investigate a fascinating cluster of organic chemistry as a result of thismotivation. We consider a zigzag edge coronoid fused with starphene and find its minimum vertexand edge metric generators.

MSC(2020): 05C12, 05C90.

Keywords: Resolving set, starphene, hollow coronoid structure, metric dimension, independentset

1 Introduction

The theory of chemical graphs is the part of graph theory that deals with chemistry and mathe-matics. The study of various structures related to chemicals from the perspective of graphs is thesubject of chemical graph theory. Chemical structures that are complicated and large in size aredifficult to examine in their natural state. Then chemical graph theory is employed to make thesecomplex chemical structures understandable. The molecular graph is a graph of a chemical structurein which the atoms are the vertices and the edges reflect the bonds between the atoms.

The physical attributes of a chemical structure are studied using a unique mathematical represen-tation in which every atom (vertex) has its own identification or position within the given chemicalstructure. A few atoms (vertices) are chosen for this unique identification of the whole vertex set sothat the set of atoms has a unique location to the selected vertices. This idea is known as metricbasis in graph theory [26] and a resolving set (metric generator) in applied graph theory [10]. If anyelement of a metric generator fails (crashes), the entire system can be shut down, to address suchproblems the concept of fault tolerance in metric generators was introduced by Hernando et al. [12].

Next, one can think that, rather than obtaining a unique atomic position, bonds could be utilizedto shape the given structure, to address this Kelenc et al. [15] proposed and initiated the study of anew variant of metric dimension in non-trivial connected graphs that focuses on uniquely identifyingthe graph's edges, called the edge metric dimension (EMD). Similar to the concept of fault tolerancein resolving sets, the idea of fault tolerance in edge resolving sets has also been introduced by Liuet al. [20].

The researchers are motivated by the fact that the metric dimension has a variety of practicalapplications in everyday life and so it has been extensively investigated. Metric dimension is utilizedin a wide range of fields of sciences, including robot navigation [16], geographical routing protocols[19], connected joints in network and chemistry [5], telecommunications [3], combinatorial optimiza-tion [21], network discovery and verification [3] etc. NP-hardness and computational complexity forthe resolvability parameters are addressed in [11, 18].

An organic compound with the chemical formula C6H6 is known as benzene. Many commercial,research and industrial operations use it as a solvent. Benzene is a key component of gasoline and

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can be found in crude oil. Dyes, detergents, resins, plastics, rubber lubricants, medicines, pesticides,and synthetic fibers are all made from it. When benzene rings are linked together, they form largerpolycyclic aromatic compounds known as polyacenes.

The word coronoid was coined by Brunvoll et al. [4], due to its possible relationship with ben-zenoid. A coronoid is a benzenoid that has a hole in the middle. Coronoid is a polyhex system thathas its origin in organic chemistry. The zigzag-edge coronoids, denoted by HCa,b,c, as shown in Fig.1(i), can be considered as a structure obtained by fusing six linear polyacenes segments into a closedloop. This structure is also known as a hollow coronoid [17]. Next, starphenes, denoted by SPa,b,c,are the two-dimensional polyaromatic hydrocarbons with three polyacene arms joined by a singlebenzene ring, as shown in Fig. 1(ii). They can be utilized as logic gates in single-molecule elec-tronics. Furthermore, as a type of 2D polyaromatic hydrocarbon, starphenes could be a promisingmaterial for organic electronics, such as organic light-emitting diodes (OLEDs) or organic field-effecttransistors [1]. A composite benzenoid obtained by fusing a zigzag-edge coronoid HCa,b,c with astarphene SPa,b,c is depicted in Fig. 2. We denote this system by FCSa,b,c.

The metric dimension was investigated for numerous chemical structures because of several ap-plication of this parameter in chemical sciences. [14] discuss the vertex resolvability of V C5C7 andH-Napthalenic nanotubes. [23] determines the minimum resolving sets for silicates star networks,[22] set upper bounds for the minimum resolving sets of cellulose network, and [13] discuss themetric dimension of 2D lattice of Boron nanotubes (Alpha). Similarly, other variants of metricdimension, such as EMD, fault-tolerant metric dimension (FTMD), fault-tolerant edge metric di-mension (FTEMD), etc have been studied for different graph families and chemical structures.

Azeem and Nadeem [2], studied metric dimension, EMD, FTMD, and FTEMD for polycyclicaromatic hydrocarbons. Sharma and Bhat [24, 25], studied metric dimension and EMD for someconvex polytope graphs. Koam et al. [17] studied the metric dimension and FTMD of hollow coro-noid structures. The metric dimension and these recently introduced concepts have been studied bymany authors for different graph families. For instance, path graphs, cycle graphs, prism graphs,wheel-related graphs, tadpole graphs, cycle with chord graphs, kayak paddle graphs, etc. But stillthere are several chemical graphs for which the metric dimension and the EMD has not been foundyet. Such as the graph FCSa,b,c. Thus, this paper aims to compute the metric dimension and theEMD of FCSa,b,c.

The present paper is organized as follows. In Sect. 2 theory and concepts related to metricdimension, EMD, and independence in their respective metric generators have been discussed. InSect. 3 we study the metric dimension and independence in the vertex metric generator of FCSa,b,c.Sect. 4 gives the edge metric dimension of FCSa,b,c. Finally, the conclusion and future work of thispaper is presented in section 5.

2 Preliminaries

In this section, we discuss some basic concepts, definitions, and existing results related to the metricdimension, edge metric dimension, and independent (vertex and edge) metric generators of graphs.

Suppose Γ = (V,E) is a non-trivial, connected, simple, and finite graph with the edge set E(Γ)and the vertex set V (Γ). We write E instead of E(Γ) and V instead of V (Γ) throughout themanuscript when there is no scope for ambiguity. The topological distance (geodesic) between twovertices a and w in Γ, denoted by d(a,w), is the length of a shortest a−w path between the verticesa and w in Γ.

Degree of a vertex: The number of edges that are incident to a vertex of a graph H is known asits degree (or valency) and is denoted by dα. The minimum degree and the maximum degree of Γare denoted by δ(Γ) and ∆(Γ), respectively.

Independent set: [24] An independent set is a set of vertices in Γ, in which no two verticesare adjacent.

Metric Dimension: [26] If for any three vertices α, β, γ ∈ V (Γ), we have d(α, β) 6= dG(α, γ),

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then the vertex α is said to recognize (resolve or distinguish) the pair of vertices β, γ (β 6= γ) inV (Γ). If this condition of resolvability is fulfilled by some vertices comprising a subset U ⊆ V (Γ) i.e.,every pair of different vertices in the given undirected graph Γ is resolved by at least one elementof U , then U is said to be a metric generator (resolving set) for Γ. The metric dimension of thegiven graph Γ is the minimum cardinality of a metric generator U , and is usually denoted by dim(Γ).The metric generator U with minimum cardinality is the metric basis for Γ. For an ordered subsetof vertices U = {a1, a2, a3, ..., ak}, the k-code (representation or coordinate) of vertex j in V (Γ) is;

γ(j|Rm) = (d(a1, j), d(a2, j), d(a3, j), ..., d(ak, j))

Then we say that, the set U is a metric generator for Γ, if γ(a|Rm) 6= γ(w|Rm), for any pair ofvertices a,w ∈ V (Γ) with a 6= w.

Independent metric generator (IMG): [25] A set of distinct ordered vertices U in Γ is saidto be an IMG for Γ if U is both independent as well as a metric generator.

Edge Metric Dimension: [15] The topological distance between a vertex a and an edge ε = bw isgiven as d(a, ε) = min{d(a,w), d(a, b)}. The vertex α is said to recognize (resolve or distinguish) thepair of edges ε1, ε2 with ε1 6= ε2) in E(Γ). If this condition of edge resolvability is fulfilled by somevertices comprising a subset UE ⊆ V (Γ) i.e., every pair of different edges in the given undirectedgraph Γ is resolved by at least one element of UE , then UE is said to be an edge metric generator(EMG) for Γ. The edge metric dimension of the graph Γ is the minimum cardinality of an ERS UE ,and is usually denoted by edim(Γ). The edge metric generator (EMG) UE with minimum cardinalityis the edge metric basis (EMB) for Γ. For an ordered subset of vertices UE = {b1, b2, b3, ..., bk}, thek-edge code (coordinate) of an edge ε in E(Γ) is;

γE(ε|UE) = (d(b1, ε), d(b2, ε), d(b3, ε), ..., d(bk, ε))

Then we say that, the set UE is an EMG for Γ, if γE(ε1|UE) 6= γ(ε2|UE), for any pair of edgesε1, ε2 ∈ V (Γ) with ε1 6= ε2.

Independent edge metric generator (IEMG): [25] A set of distinct vertices U iE (ordered)in Γ is said to be an IEMG for Γ if U iE is both independent as well as a edge metric generator.

Figure 1: HCa,b,c and SPa,b,c

Pn, Cn, and Kn denotes respectively the path graph, cycle graph and the complete graph on nvertices. Then the following results are helpful in obtaining the metric and the edge metric dimensionof a graph.

Proposition 1. [15] For n ≥ 3, we have dim(Pn) = edim(Pn) = 1, dim(Cn) = edim(Cn) = 2, anddim(Kn) = edim(Kn) = n− 1.

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3 Metric Dimension of FCSa,b,c

In this section, we obtain the metric dimension and IVMG for FCSa,b,c.

The fused hollow coronoid with starphene structure FCSa,b,c comprises of six sides in which threesides (a, b, c) are symmetric to other three sides (a, b, c) as shown in Fig. 2. This means thatFCSa,b,c has three linear polyacenes segments consist of a, b, and c number of benzene rings. Itconsists of 3a + 3b + 3c − 11 number of faces having six sides, two faces having 4a + 2b + 2c − 18sides, a face having 4b + 4c − 18 sides, and a face having 4a + 4b + 4c − 6 sides. FCSa,b,c has6(a + b + c − 6) number of vertices of degree two and 6(a + b + c − 3) number of vertices of de-gree three. From this, we find that δ(FCSa,b,c) = 2 and ∆(FCSa,b,c) = 3. The vertex set andthe edge set of FCSa,b,c, are denoted by V (FCSa,b,c) and E(FCSa,b,c) respectively. Moreover, thecardinality of edges and vertices in FCSa,b,c is given by |E(FCSa,b,c)| = 3(5a + 5b + 5c − 21) and|V (FCSa,b,c)| = 6(2a+ 2b+ 2c− 9), respectively. The edge and vertex set of FCSa,b,c are describeas follows: V (FCSa,b,c) = {p1,d, p2,d|1 ≤ d ≤ 2a − 1} ∪ {q1,d, q2,d|1 ≤ d ≤ 2c − 1} ∪ {r1,d, r2,d|1 ≤d ≤ 2b − 1} ∪ {s1,d, s2,d|1 ≤ d ≤ 2a − 3} ∪ {u1,d, u2,d|1 ≤ d ≤ 2b − 3} ∪ {t1,d, t2,d|1 ≤ d ≤2c− 3} ∪ {p3,d, s3,d|1 ≤ d ≤ 2a− 5} ∪ {q3,d, t3,d|1 ≤ d ≤ 2c− 5} ∪ {r3,d, u3,d|1 ≤ d ≤ 2b− 5}

and

E(FCSa,b,c) = {p1,dp1,d+1, p2,dp2,d+1|1 ≤ d ≤ 2a − 2} ∪ {q1,dq1,d+1, q2,dq2,d+1|1 ≤ d ≤ 2c − 2} ∪{r1,dr1,d+1, r2,dr2,d+1|1 ≤ d ≤ 2b−2}∪{s1,ds1,d+1, s2,ds2,d+1|1 ≤ d ≤ 2a−4}∪{u1,du1,d+1, t2,dt2,d+1|1 ≤d ≤ 2b − 4} ∪ {t1,dt1,d+1, u2,du2,d+1|1 ≤ d ≤ 2c − 4} ∪ {p1,2ds1,2d−1, p2,2ds2,2d−1|1 ≤ d ≤ a − 1} ∪{q1,2dt1,2d−1, q2,2du2,2d−1|1 ≤ d ≤ c−1}∪{r1,2du1,2d−1, r2,2dt2,2d−1|1 ≤ d ≤ b−1}∪{p3,dp3,d+1, s3,ds3,d+1

|1 ≤ d ≤ 2a− 6} ∪ {q3,dq3,d+1, t3,dt3,d+1|1 ≤ d ≤ 2c− 6} ∪ {r3,dr3,d+1, u3,du3,d+1|1 ≤ d ≤ 2b− 6} ∪{p3,2d−1s3,2d−1|1 ≤ d ≤ a − 2} ∪ {q3,2d−1t3,2d−1|1 ≤ d ≤ c − 2} ∪ {r3,2d−1u3,2d−1|1 ≤ d ≤ b − 2} ∪{p3,1r3,1, q3,1u3,1, s3,1t3,1} ∪ {p3,2a−5t2,j−3, s3,2a−5u2,2, q3,2c−5u1,2c−4, t3,2c−5s2,2a−4, r3,2b−5s1,2a−4,u3,2b−5t1,2} ∪ {p1,1q2,1, s1,1u2,1, p1,iq1,1, s1,i−2t1,1, q1,jr1,1, t1,j−2u1,1, r1,kp2,i, u1,k−2s2,l, p2,1r2,k,s2,1t2,k−2, r2,1q2,j , t2,1u2,j−2}.

We name the vertices on the cycle p1,1, ..., p1,i, q1,1, ..., q1,jr1,1, ..., r1,k, p2,1, ..., p2,i, q2,1, ..., q2,jr2,1, ...,r2,k as the outer pqr-cycle vertices, the vertices on the cycle s1,1, ..., s1,i−2, r3,k−4, ..., r3,1, p3,1, ...,p3,i−4, t2,j−2, ..., t2,1 as the vertices of first interior cycle, the vertices on the cycle t1,1, ..., t1,j−2, u1,1,..., u1,k−2, q3,j−4, ..., q3,1, u3,1, ..., u3,k−4 as the vertices of second interior cycle, and the vertices onthe cycle u2,1, ..., u2,k−2, s2,1, ..., s2,i−2, t3,j−4, ..., t3,1, s3,1, ..., s3,i−4 as the vertices of third interiorcycle in FCSa,b,c. In vertices, p1,i, p2,i, q1,j , q2,j , r1,k, and r2,k, the indices i = 2a − 1, j = 2c − 1and k = 2b− 1. In the next result, we determine the metric dimension of FCSa,b,c.

Theorem 1. For positive integers a, b, c ≥ 4, we have dim(FCSa,b,c) = 3.

Proof. In order to show that dim(FCSa,b,c) ≤ 3, we construct a metric generator for FCSa,b,c. LetU = {p1,1, r1,1, r2,k} be a set of distinct vertices from FCSa,b,c. We claim that U is a vertex metricgenerator for FCSa,b,c. Now, to obtain dim(FCSa,b,c) ≤ 3, we can give metric coordinate to everyvertex of FCSa,b,c with respect to the set U . For the vertices {υ = p1,d|1 ≤ d ≤ 2a− 1}, the set ofvertex metric coordinates is as follow:P1 = {γ(υ|U) = (d−1, 2a+2c−d−1, 2b+2c−2)|d = 1}∪{γ(υ|U) = (d−1, 2a+2c−d−1, 2b+2c+d− 5)|2 ≤ d ≤ 2a− 3} ∪ {γ(υ|U) = (d− 1, 2a+ 2c− d− 1, 2a+ 2b+ 2c− 9)|d = 2a− 2} ∪ {γ(υ|U) =(d− 1, 2a+ 2c− d− 1, 2a+ 2b+ 2c− 8)|d = 2a− 1}.

For the vertices {υ = q1,d|1 ≤ d ≤ 2c− 1}, the set of vertex metric coordinates is as follow:Q1 = {γ(υ|U) = (2a + d − 2, 2c − d, 2a + 2b + 2c − 7)|d = 1} ∪ {γ(υ|U) = (2a + d − 2, 2c − d, 2a +2b+ 2c− 8)|d = 2}∪ {γ(υ|U) = (2a+ d− 2, 2c− d, 2a+ 2b+ 2c− d− 4)|3 ≤ d ≤ 2c− 2}∪ {γ(υ|U) =(2a+ d− 2, 2c− d, 2a+ 2b− 1)|d = 2c− 1}.

For the vertices {υ = r1,d|1 ≤ d ≤ 2b− 1}, the set of vertex metric coordinates is as follow:R1 = {γ(υ|U) = (2a+ 2c− 2, d− 1, 2a+ 2b− d− 1)|d = 1} ∪ {γ(υ|U) = (2a+ 2c+ d− 5, d− 1, 2a+2b−d−1)|2 ≤ d ≤ 2c−3}∪{γ(υ|U) = (2a+2b+2c−9, d−1, 2a+2b−d−1)|d = 2b−2}∪{γ(υ|U) =(2a+ 2b+ 2c− 8, d− 1, 2a+ 2b− d− 1)|d = 2b− 1}.

For the vertices {υ = p2,d|1 ≤ d ≤ 2a− 1}, the set of vertex metric coordinates is as follow:P2 = {γ(υ|U) = (d, 2a + 2b − d − 2, 2b + 2c − 1)|d = 1} ∪ {γ(υ|U) = (d, 2a + 2b − d − 2, 2b + 2c +

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d − 4)|2 ≤ d ≤ 2a − 3} ∪ {γ(υ|U) = (d, 2a + 2b − d − 2, 2a + 2b + 2c − 8)|d = 2a − 2} ∪ {γ(υ|U) =(d, 2a+ 2b− d− 2, 2a+ 2b+ 2c− 7)|d = 2a− 1}.

For the vertices {υ = q2,d|1 ≤ d ≤ 2c− 1}, the set of vertex metric coordinates is as follow:Q2 = {γ(υ|U) = (d, 2a + 2c − 1, 2b + 2c − d − 2)|d = 1} ∪ {γ(υ|U) = (d, 2a + 2c + d − 4, 2b + 2c −d − 2)|2 ≤ d ≤ 2c − 3} ∪ {γ(υ|U) = (d, 2a + 2b + 2c − 8, 2b + 2c − d − 2)|d = 2c − 2} ∪ {γ(υ|U) =(d, 2a+ 2b+ 2c− 7, 2b+ 2c− d− 2)|d = 2c− 1}.

For the vertices {υ = r2,d|1 ≤ d ≤ 2b− 1}, the set of vertex metric coordinates is as follow:R2 = {γ(υ|U) = (2c + d − 1, 2a + 2b + 2c − 8, 2b − d − 1)|d = 1} ∪ {γ(υ|U) = (2c + d − 1, 2a +2b + 2c − 9, 2b − d − 1)|d = 2} ∪ {γ(υ|U) = (2c + d − 1, 2a + 2b + 2c − d − 5, 2b − d − 1)|3 ≤ d ≤2b− 2} ∪ {γ(υ|U) = (2c+ d− 1, 2a+ 2b− 2, 2b− d− 1)|d = 2b− 1}.

For the vertices {υ = s1,d|1 ≤ d ≤ 2a− 3}, the set of vertex metric coordinates is as follow:S1 = {γ(υ|U) = (d + 1, 2a + 2c − d − 3, 2b + 2c + d − 5)|1 ≤ d ≤ 2a − 5} ∪ {γ(υ|U) = (d + 1, 2a +2c−d−3, 2a+2b+2c−11)|d = 2a−4}∪{γ(υ|U) = (d+1, 2a+2c−d−3, 2a+2b+2c−10)|d = 2a−3}.

For the vertices {υ = t1,d|1 ≤ d ≤ 2c− 3}, the set of vertex metric coordinates is as follow:T1 = {γ(υ|U) = (2a + d − 2, 2c − d, 2a + 2b + 2c − 9)|d = 1} ∪ {γ(υ|U) = (2a + d − 2, 2c − d, 2a +2b+ 2c− 10)|d = 2} ∪ {γ(υ|U) = (2a+ d− 2, 2c− d, 2a+ 2b+ 2c− d− 6)|3 ≤ d ≤ 2c− 3}.

For the vertices {υ = u1,d|1 ≤ d ≤ 2b− 3}, the set of vertex metric coordinates is as follow:U1 = {γ(υ|U) = (2a + 2c + d − 5, d + 1, 2a + 2b − d − 3)|1 ≤ d ≤ 2b − 5} ∪ {γ(υ|U) = (2a + 2b +2c−11, d+1, 2a+2b−d−3)|d = 2b−4}∪{γ(υ|U) = (2a+2b+2c−10, d+1, 2a+2b−d−3)|d = 2b−3}.

For the vertices {υ = s2,d|1 ≤ d ≤ 2a− 3}, the set of vertex metric coordinates is as follow:S2 = {γ(υ|U) = (2b + 2c + d − 4, 2a + 2b − d − 4, d + 2)|1 ≤ d ≤ 2a − 5} ∪ {γ(υ|U) = (2a + 2b +2c−10, 2a+2b−d−4, d+2)|d = 2a−4}∪{γ(υ|U) = (2a+2b+2c−9, 2a+2b−d−4, d+2)|d = 2a−3}.

For the vertices {υ = t2,d|1 ≤ d ≤ 2c− 3}, the set of vertex metric coordinates is as follow:T2 = {γ(υ|U) = (d + 2, 2a + 2c + d − 4, 2b + 2c − d − 4)|1 ≤ d ≤ 2c − 5} ∪ {γ(υ|U) = (d + 2, 2a +2b+2c−10, 2b+2c−d−4)|d = 2c−4}∪{γ(υ|U) = (d+2, 2a+2b+2c−9, 2b+2c−d−4)|d = 2c−3}.

For the vertices {υ = u2,d|1 ≤ d ≤ 2b− 3}, the set of vertex metric coordinates is as follow:U2 = {γ(υ|U) = (2c+ d− 1, 2a+ 2b+ 2c− 10, 2b− d− 1)|d = 1} ∪ {γ(υ|U) = (2c+ d− 1, 2a+ 2b+2c− 11, 2b− d− 1)|d = 2} ∪ {γ(υ|U) = (2c+ d− 1, 2a+ 2b+ 2c− d− 7, 2b− d− 1)|3 ≤ d ≤ 2b− 3}.

For the vertices {υ = p3,d|1 ≤ d ≤ 2a− 5}, the set of vertex metric coordinates is as follow:P3 = {γ(υ|U) = (2a+ 2c− d− 6, 2a+ 2c+ d− 6, 2a+ 2b− d− 4)|1 ≤ d ≤ 2a− 5}.

For the vertices {υ = q3,d|1 ≤ d ≤ 2c− 5}, the set of vertex metric coordinates is as follow:Q3 = {γ(υ|U) = (2a+ 2b+ d− 7, 2b+ 2c− d− 7, 2a+ 2c− d− 5)|1 ≤ d ≤ 2c− 5}.

For the vertices {υ = r3,d|1 ≤ d ≤ 2b− 5}, the set of vertex metric coordinates is as follow:R3 = {γ(υ|U) = (2a+ 2b− d− 7, 2b+ 2c− d− 5, 2a+ 2b+ d− 7)|1 ≤ d ≤ 2b− 5}.

For the vertices {υ = s3,d|1 ≤ d ≤ 2a− 5}, the set of vertex metric coordinates is as follow:S3 = {γ(υ|U) = (2a+ 2c− d− 5, 2a+ 2c+ d− 7, 2a+ 2b− d− 5)|1 ≤ d ≤ 2a− 5}.

For the vertices {υ = t3,d|1 ≤ d ≤ 2c− 5}, the set of vertex metric coordinates is as follow:T3 = {γ(υ|U) = (2a+ 2c+ d− 6, 2b+ 2c− d− 6, 2a+ 2c− d− 6)|1 ≤ d ≤ 2c− 5}.

For the vertices {υ = u3,d|1 ≤ d ≤ 2b− 5}, the set of vertex metric coordinates is as follow:U3 = {γ(υ|U) = (2a+ 2b− d− 6, 2b+ 2c− d− 6, 2a+ 2b+ d− 6)|1 ≤ d ≤ 2b− 5}.

Now, from these sets of vertex metric codes for the graph FCSa,b,c, we find that |P1| = |P2| = 2a−1,|Q1| = |Q2| = 2c − 1, |R1| = |R2| = 2b − 1, |S1| = |S2| = 2a − 3, |T1| = |T2| = 2c − 3,|U1| = |U2| = 2b − 3, |P3| = |S3| = 2a − 5, |Q3| = |T3| = 2c − 5, and |R3| = |U3| = 2b − 5. We seethat the sum of all of these cardinalities is equal to |V (FCSa,b,c)| and which is 6(2a+ 2b+ 2c− 9).Moreover, all of these sets are pairwise disjoint, which implies that dim(FCSa,b,c) ≤ 3. To complete

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the proof, we have to show that dim(FCSa,b,c) ≥ 3. To show this, we have to prove that there existsno vertex metric generator U for FCSa,b,c such that |U | ≤ 2. Since, the graph FCSa,b,c is not apath graph, so the possibility of a singleton vertex metric generator for FCSa,b,c is ruled out [5].Next, suppose on the contrary that there exists an edge resolving set U with |U | = 2. Therefore,we have the following ten cases to be discussed (for the contradictions, the naturals a, b, and c are≥ 5):

Figure 2: FCSa,b,c

Case(I) When U = {a, b}, where a and b are the vertices from the outer pqr-cycle of FCSa,b,c.

• Suppose U = {p1,1, p1,d}, p1,d (2 ≤ d ≤ i). Then γ(q1,2|U) = γ(t1,2|U), for 2 ≤ d ≤ i − 1;γ(u1,2|U) = γ(r1,2|U) when d = i, a contradiction.

• Suppose U = {p1,1, q1,d}, q1,d (1 ≤ d ≤ j). Then γ(r1,2|U) = γ(u1,2|U), for d = 1; γ(p1,i|U) =γ(s1,i−2|U) when 2 ≤ d ≤ j, a contradiction.

• Suppose U = {p1,1, r1,d}, r1,d (1 ≤ d ≤ k). Then γ(p1,i|U) = γ(s1,i−2|U) for d = 1; γ(p1,3|U) =γ(s1,1|U), for 2 ≤ d ≤ k, a contradiction.

• Suppose U = {p1,1, p2,d}, p2,d (1 ≤ d ≤ i). Then γ(s1,1|U) = γ(q2,2|U) for 1 ≤ d ≤ i − 3;γ(p1,3|U) = γ(s1,1|U), for i− 2 ≤ d ≤ i, a contradiction.

• Suppose U = {p1,1, r2,d}, r2,d (1 ≤ d ≤ k). Then γ(s2,2|U) = γ(p2,2|U) for d = 1; γ(q2,j |U) =γ(t2,j−2|U) for 2 ≤ d ≤ k, a contradiction.

• Suppose U = {p1,1, q2,d}, q2,d (1 ≤ d ≤ j). Then γ(s2,2|U) = γ(p2,2|U) for 1 ≤ d ≤ j, acontradiction.

Case(II) When U = {a, b}, where a and b are the vertices from the first interior cycle of FCSa,b,c.

• Suppose U = {s1,1, s1,d}, s1,d (2 ≤ d ≤ i − 2). Then γ(q2,2|U) = γ(t2,2|U), for 2 ≤ d ≤ i − 1,a contradiction.

• Suppose U = {s1,1, r3,d}, r3,d (1 ≤ d ≤ k − 5). Then γ(p1,2|U) = γ(t2,1|U), for 1 ≤ d ≤ k − 5,a contradiction.

• Suppose U = {s1,1, t2,d}, t2,d (1 ≤ d ≤ j − 2). Then γ(p1,2|U) = γ(s1,2|U), for 1 ≤ d ≤ j − 2,a contradiction.

• Suppose U = {s1,1, p3,d}, p3,d (1 ≤ d ≤ i− 5). Then γ(s2,2|U) = γ(p2,2|U), for 1 ≤ d ≤ 2a− 3,a contradiction.

Case(III) When U = {a, b}, where a and b are the vertices from the second interior cycle ofFCSa,b,c.

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• Suppose U = {u1,1, u1,d}, u1,d (1 ≤ d ≤ k − 2). Then γ(p2,i−1|U) = γ(s2,i−3|U), for 1 ≤ d ≤k − 2, a contradiction.

• Suppose U = {u1,1, t1,d}, t1,d (1 ≤ d ≤ j − 2). Then γ(r1,1|U) = γ(r1,3|U), for 1 ≤ d ≤ j − 2,a contradiction.

• Suppose U = {u1,1, u3,d}, u3,d (1 ≤ d ≤ k−5). Then γ(p1,i|U) = γ(p1,i−2|U), for 1 ≤ d ≤ k−5,a contradiction.

• Suppose U = {u1,1, q3,d}, q3,d (1 ≤ d ≤ j−5). Then γ(p1,i|U) = γ(p1,i−2|U), for 1 ≤ d ≤ j−5,a contradiction.

Case(IV) When U = {a, b}, where a and b are the vertices from the third interior cycle ofFCSa,b,c.

• Suppose U = {s2,1, s2,d}, s2,d (2 ≤ d ≤ i−2). Then γ(r2,k|U) = γ(r2,k−2|U), for 2 ≤ d ≤ i−2,a contradiction.

• Suppose U = {s2,1, u2,d}, u2,d (1 ≤ d ≤ k− 2). Then γ(p2,1|U) = γ(p2,3|U), for 1 ≤ d ≤ k− 2,a contradiction.

• Suppose U = {s2,1, s3,d}, s3,d (1 ≤ d ≤ i−5). Then γ(q2,j |U) = γ(q2,j−2|U), for 1 ≤ d ≤ i−5,a contradiction.

• Suppose U = {s2,1, t3,d}, t3,d (1 ≤ d ≤ j−5). Then γ(q2,j |U) = γ(q2,j−2|U), for 1 ≤ d ≤ j−5,a contradiction.

Case(V) When U = {a, b}, where a is in outer pqr-cycle and b is in the first interior cycle ofFCSa,b,c.

• Suppose U = {p1,1, s1,d}, s1,d (1 ≤ d ≤ i− 2). Then γ(r2,2|U) = γ(u2,2|U), for 1 ≤ d ≤ i− 2,a contradiction.

• Suppose U = {p1,1, t2,d}, t2,d (1 ≤ d ≤ j − 2). Then γ(r2,2|U) = γ(u2,2|U), for 1 ≤ d ≤ j − 2,a contradiction.

• Suppose U = {p1,1, r3,d}, r3,d (1 ≤ d ≤ k − 5). Then γ(q1,4|U) = γ(t1,4|U), for 1 ≤ d ≤ k − 5,a contradiction.

• Suppose U = {p1,1, p3,d}, p3,d (1 ≤ d ≤ i− 5). Then γ(q1,4|U) = γ(t1,4|U), for 1 ≤ d ≤ i− 5,a contradiction.

Case(VI) When U = {a, b}, where a is in outer pqr-cycle and b is in the second interior cycleof FCSa,b,c.

• Suppose U = {p1,1, t1,d}, t1,d (1 ≤ d ≤ j − 2). Then γ(u1,2|U) = γ(r1,2|U), for 1 ≤ d ≤ j − 2,a contradiction.

• Suppose U = {p1,1, u1,d}, u1,d (1 ≤ d ≤ k−2). Then γ(t1,1|U) = γ(u3,k−5|U), for 1 ≤ d ≤ k−2,a contradiction.

• Suppose U = {p1,1, u3,d}, u3,d (1 ≤ d ≤ k − 5). Then γ(q1,4|U) = γ(t1,4|U), for 1 ≤ d ≤ k − 5,a contradiction.

• Suppose U = {p1,1, q3,d}, q3,d (1 ≤ d ≤ j − 5). Then γ(u3,2|U) = γ(r3,1|U), for 1 ≤ d ≤ j − 5,a contradiction.

Case(VII) When U = {a, b}, where a is in outer pqr-cycle and b is in the third interior cycle ofFCSa,b,c.

• Suppose U = {p1,1, u2,d}, u2,d (1 ≤ d ≤ k− 2). Then γ(s2,2|U) = γ(p2,2|U), for 1 ≤ d ≤ k− 2,a contradiction.

• Suppose U = {p1,1, s3,d}, s3,d (1 ≤ d ≤ i− 5). Then γ(u3,4|U) = γ(r2,4|U), for 1 ≤ d ≤ i− 5,a contradiction.

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• Suppose U = {p1,1, s2,d}, s2,d (1 ≤ d ≤ i− 2). Then γ(s1,1|U) = γ(q2,2|U), for 1 ≤ d ≤ i− 5;γ(s3,2|U) = γ(p3,1|U), for i− 4 ≤ d ≤ i− 2, a contradiction.

• Suppose U = {p1,1, t3,d}, t3,d (1 ≤ d ≤ j − 5). Then γ(s3,2|U) = γ(p3,1|U), for 1 ≤ d ≤ j − 5,a contradiction.

Case(VIII) When U = {a, b}, where a is in first interior cycle and b is in the second interiorcycle of FCSa,b,c.

• Suppose U = {s1,1, t1,d}, t1,d (1 ≤ d ≤ j − 2). Then γ(t2,2|U) = γ(q2,2|U), for 1 ≤ d ≤ j − 2,a contradiction.

• Suppose U = {s1,1, u3,d}, u3,d (1 ≤ d ≤ k − 5). Then γ(p1,2|U) = γ(t2,1|U), for 1 ≤ d ≤ k − 5,a contradiction.

• Suppose U = {s1,1, u1,d}, u1,d (1 ≤ d ≤ k− 2). Then γ(s3,2|U) = γ(p3,1|U), for 1 ≤ d ≤ k− 2,a contradiction.

• Suppose U = {s1,1, q3,d}, q3,d (1 ≤ d ≤ j − 5). Then γ(s3,2|U) = γ(p3,1|U), for 1 ≤ d ≤ j − 5,a contradiction.

Case(IX) When U = {a, b}, where a is in first interior cycle and b is in the third interior cycleof FCSa,b,c.

• Suppose U = {s1,1, u2,d}, u2,d (1 ≤ d ≤ k− 2). Then γ(p1,1|U) = γ(p1,3|U), for 1 ≤ d ≤ k− 2,a contradiction.

• Suppose U = {s1,1, s2,d}, s2,d (1 ≤ d ≤ i− 2). Then γ(u3,2|U) = γ(r3,1|U), for 1 ≤ d ≤ i− 2,a contradiction.

• Suppose U = {s1,1, t3,d}, t3,d (1 ≤ d ≤ j − 5). Then γ(u3,2|U) = γ(r3,1|U), for 1 ≤ d ≤ j − 5,a contradiction.

• Suppose U = {s1,1, s3,d}, s3,d (1 ≤ d ≤ i− 5). Then γ(r2,4|U) = γ(u2,4|U), for 1 ≤ d ≤ i− 5,a contradiction.

Case(X) When U = {a, b}, where a is in second interior cycle and b is in the third interior cycleof FCSa,b,c.

• Suppose U = {u1,1, u2,d}, u2,d (1 ≤ d ≤ k− 2). Then γ(r3,3|U) = γ(u3,2|U), for 1 ≤ d ≤ k− 2,a contradiction.

• Suppose U = {u1,1, s2,d}, s2,d (1 ≤ d ≤ i − 2). Then γ(q1,j−1|U) = γ(t1,j−3|U), for 1 ≤ d ≤i− 2, a contradiction.

• Suppose U = {u1,1, t3,d}, t3,d (1 ≤ d ≤ j − 5). Then γ(s3,2|U) = γ(p3,1|U), for 1 ≤ d ≤ j − 5,a contradiction.

• Suppose U = {u1,1, s3,d}, s3,d (1 ≤ d ≤ i− 5). Then γ(t3,2|U) = γ(q3,1|U), for 1 ≤ d ≤ i− 5,a contradiction.

As a result, we infer that for FCSa,b,c, there is no vertex metric generator U such that |U | = 2.Therefore, we must have |U | ≥ 3 i.e., dim(FCSa,b,c) ≥ 3. Hence, dim(FCSa,b,c) = 3, whichconcludes the theorem.

In terms of minimum an IVMG, we have the following result

Theorem 2. For a, b, c ≥ 4, the graph FCSa,b,c has an IVMG with cardinality three.

Proof. To show that, for zigzag edge coronoid fused with starphene FCSa,b,c, there exists an IVMGU i with |U i| = 3, we follow the same technique as used in Theorem 1.

Suppose U i = {p1,1, r1,1, r2,k} ⊂ V (FCSa,b,c). Now, by using the definition of an independentset and following the same pattern as used in Theorem 1, it is simple to show that the set of verticesU i = {p1,1, r1,1, r2,k} forms an IVMG for FCSa,b,c with |U i| = 3, which concludes the theorem.

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4 Edge Metric Dimension of FCSa,b,c

In this section, we obtain the metric dimension and IVMG for FCSa,b,c.

Theorem 1. For positive integers a, b, c ≥ 4, we have edim(FCSa,b,c) = 3.

Proof. In order to show that edim(FCSa,b,c) ≤ 3, we construct an edge metric generator forFCSa,b,c. Let UE = {p1,1, r1,1, r2,k} be a set of distinct vertices from FCSa,b,c. We claim thatUE is an edge metric generator for FCSa,b,c. Now, to obtain edim(FCSa,b,c) ≤ 3, we can give edgemetric coordinate to every edge of FCSa,b,c with respect to UE . For the edges {η = p1,dp1,d+1|1 ≤d ≤ 2a− 2}, the set of edge metric coordinates is as follow:P1 = {γE(η|UE) = (d − 1, 2a + 2c − d − 2, 2b + 2 − 3)|d = 1} ∪ {γE(η|UE) = (d − 1, 2a + 2c − d −2, 2b+2c+d−5)|2 ≤ d ≤ 2a−4}∪{γE(η|UE) = (d−1, 2a+2c−d−2, 2a+4b−8)|2a−3 ≤ d ≤ 2a−2}.

For the edges {η = q1,dq1,d+1|1 ≤ d ≤ 2c− 2}, the set of edge metric coordinates is as follow:Q1 = {γE(η|UE) = (2a+d−2, 2c−d−1, 2a+2b+2c−8)|1 ≤ d ≤ 2}∪{γE(η|UE) = (2a+d−2, 2c−d−1, 2a+2b+2c−d−5)|3 ≤ d ≤ 2c−3}∪{γE(η|UE) = (2a+d−2, 2c−d−1, 2a+2b−2)|d = 2c−2}.

For the edges {η = r1,dr1,d+1|1 ≤ d ≤ 2b− 2}, the set of edge metric coordinates is as follow:R1 = {γE(η|UE) = (2a+ 2c− 3, d− 1, 2a+ 2b− d− 1)|d = 1} ∪ {γE(η|UE) = (2a+ 2c+ d− 5, d−1, 2a+2b−d−1)|2 ≤ d ≤ 2b−4}∪{γE(η|UE) = (2a+2b+2c−9, d−1, 2a+2b−d−1)|2b−3 ≤ d ≤ 2b−2}.

For the edges {η = p2,dp2,d+1|1 ≤ d ≤ 2a− 2}, the set of edge metric coordinates is as follow:P2 = {γE(η|UE) = (2b+ 2c− 2, 2a+ 2b− d− 1, d)|d = 1} ∪ {γE(η|UE) = (2b+ 2c+ d− 4, 2a+ 2b−d− 1, d)|2 ≤ d ≤ 2a− 4} ∪ {γE(η|UE) = (2a+ 2b+ 2c− 8, 2a+ 2b− d− 1, d)|2a− 3 ≤ d ≤ 2a− 2}.

For the edges {η = q2,dq2,d+1|1 ≤ d ≤ 2c− 2}, the set of edge metric coordinates is as follow:Q2 = {γE(η|UE) = (d, 2a + 2c − 2, 2b + 2c − d − 3)|d = 1} ∪ {γE(η|UE) = (d, 2a + 2c + d − 4, 2b +2c−d−3))|2 ≤ d ≤ 2c−4}∪{γE(η|UE) = (d, 2a+ 2b+ 2c−8, 2b+ 2c−d−3))|2c−3 ≤ d ≤ 2c−2}.

For the edges {η = r2,dr2,d+1|1 ≤ d ≤ 2b− 2}, the set of edge metric coordinates is as follow:R2 = {γE(η|UE) = (2c + d − 1, 2a + 2b + 2c − 9, 2b − d − 2)|1 ≤ d ≤ 2} ∪ {γE(η|UE) = (2c + d −1, 2a+4b−d−6, 2b−d−2)|3 ≤ d ≤ 2b−3}∪{γE(η|UE) = (2c+d−1, 2a+2b−3, 2b−d−2)|d = 2b−2}.

For the edges {η = s1,ds1,d+1|1 ≤ d ≤ 2a− 4}, the set of edge metric coordinates is as follow:S1 = {γE(η|UE) = (d + 1, 2a + 2c − d − 4, 2b + 2c + d − 5)|1 ≤ d ≤ 2a − 6} ∪ {γE(η|UE) =(d+ 1, 2a+ 2c− d− 4, 2a+ 4b− 11)|2a− 5 ≤ d ≤ 2a− 4}.

For the edges {η = t1,dt1,d+1|1 ≤ d ≤ 2c− 4}, the set of edge metric coordinates is as follow:T1 = {γE(η|UE) = (2a+ d, 2c− d− 1, 2a+ 2b+ 2c− 10)|1 ≤ d ≤ 2}{γE(η|UE) = (2a+ d, 2c− d−1, 2a+ 2b+ 2c− d− 7)|3 ≤ d ≤ 2c− 4}.

For the edges {η = u1,du1,d+1|1 ≤ d ≤ 2b− 4}, the set of edge metric coordinates is as follow:U1 = {γE(η|UE) = (2a + 2c + d − 5, d + 1, 2a + 2b − d − 4)|1 ≤ d ≤ 2b − 6} ∪ {γE(η|UE) =(2a+ 2b+ 2c− 11, d+ 1, 2a+ 2b− d− 4)|2b− 3 ≤ d ≤ 2b− 4}.

For the edges {η = s2,ds2,d+1|1 ≤ d ≤ 2a− 4}, the set of edge metric coordinates is as follow:S2 = {γE(η|UE) = (2b + 2c + d − 4, 2a + 2b − d − 5, d + 2)|1 ≤ d ≤ 2a − 6} ∪ {γE(η|UE) =(2a+ 2b+ 2c− 10, 2a+ 2b− d− 5, d+ 2)|2a− 5 ≤ d ≤ 2a− 4}.

For the edges {η = t2,dt2,d+1|1 ≤ d ≤ 2c− 4}, the set of edge metric coordinates is as follow:T2 = {γE(η|UE) = (d + 2, 2a + 2c + d − 4, 2b + 2c − d − 7)|1 ≤ d ≤ 2b − 6} ∪ {γE(η|UE) =(d+ 2, 2a+ 2b+ 2c− 10, 2b+ 2c− d− 7)|2b− 5 ≤ d ≤ 2c− 4}.

For the edges {η = u2,du2,d+1|1 ≤ d ≤ 2b− 4}, the set of edge metric coordinates is as follow:U2 = {γE(η|UE) = (2c + d − 1, 2a + 2b + 2c − 11, 2b − d − 2)|1 ≤ d ≤ 2} ∪ {γE(η|UE) =(2c+ d− 1, 2a+ 4b− d− 8, 2b− d− 2)|3 ≤ d ≤ 2b− 4}.

For the edges {η = p3,dp3,d+1|1 ≤ d ≤ 2a− 6}, the set of edge metric coordinates is as follow:

9

P3 = {γE(η|UE) = (2a+ 2c− d− 7, 2b+ 2c+ d− 6, 2a+ 2b− d− 7)|1 ≤ d ≤ 2a− 6}.

For the edges {η = q3,dq3,d+1|1 ≤ d ≤ 2c− 6}, the set of edge metric coordinates is as follow:Q3 = {γE(η|UE) = (2a+ 2b+ d− 7, 2b+ 2c− d− 8, 2a+ 2c− d− 6)|1 ≤ d ≤ 2c− 6}.

For the edges {η = r3,dr3,d+1|1 ≤ d ≤ 2b− 6}, the set of edge metric coordinates is as follow:R3 = {γE(η|UE) = (2a+ 2b− d− 8, 2b+ 2c− d− 6, 2a+ 2b+ d− 7)|1 ≤ d ≤ 2b− 6}.

For the edges {η = s3,ds3,d+1|1 ≤ d ≤ 2a− 6}, the set of edge metric coordinates is as follow:S3 = {γE(η|UE) = (2a+ 2c− d− 6, 2b+ 2c+ d− 7, 2a+ 2b− d− 8)|1 ≤ d ≤ 2a− 6}.

For the edges {η = t3,dt3,d+1|1 ≤ d ≤ 2c− 6}, the set of edge metric coordinates is as follow:T3 = {γE(η|UE) = (2a+ 2b+ d− 6, 2b+ 2c− d− 7, 2a+ 2c− d− 7)|1 ≤ d ≤ 2c− 6}.

For the edges {η = u3,du3,d+1|1 ≤ d ≤ 2b− 6}, the set of edge metric coordinates is as follow:U3 = {γE(η|UE) = (2a+ 2b− d− 7, 2b+ 2c− d− 7, 2a+ 2b+ d− 6)|1 ≤ d ≤ 2b− 6}.

For the edges {η1 = p1,iq1,1, η2 = s1,i−2t1,1, η3 = r1,1q1,j , η4 = t1,ju1,1, η5 = r1,kp2,i, η6 = u1,k−2s2,i−2, η7 =p2,1r2,k, η8 = s2,1u2,k−2, η9 = r2,1q2,j , η10 = u2,1t2,j−2, η11 = p1,1q2,1, η12 = u1,1t2,1}, the set of edgemetric coordinates is as follow:V1 = {γE(η1|UE) = (2a−2, 2c−1, 2a+4b−8), γE(η2|UE) = (2a−2, 2c−1, 2a+4b−10), γE(η3|UE) =(2a+ 2c− 3, 0, 2a+ 2b− 2), γE(η4|UE) = (2a+ 2c− 5, 2, 2a+ 2b− 4), γE(η5|UE) = (2a+ 2b+ 2c−10, 2b − 2, 2a − 1), γE(η6|UE) = (2a + 2b + 2c − 8, 2b − 2, 2a − 1), γE(η7|UE) = (2b + 2c − 2, 2a +2b − 3, 0), γE(η8|UE) = (2b + 2c − 4, 2a + 2b − 5, 2), γE(η9|UE) = (2c − 1, 2a + 2b + 2c − 10, 2b −2), γE(η10|UE) = (2c−1, 2a+2b+2c−8, 2b−4), γE(η11|UE) = (0, 2a+2c−2, 2a+2c−3), γE(η12|UE) =(2, 2a+ 2c− 4, 2b+ 2c− 5)}.

For the edges {η = p1,2ds1,2d−1|1 ≤ d ≤ a− 1}, the set of edge metric coordinates is as follow:PS1 = {γE(η|UE) = (2d − 1, 4a + 2c − 2d − 12, 4a + 2c + 2d − 18)|1 ≤ d ≤ a − 2} ∪ {γE(η|UE) =(2d− 1, 4a+ 2c− 2d− 12, 2a+ 4b− 10)|d = a− 1}.

For the edges {η = q1,2dt1,2d−1|1 ≤ d ≤ c− 1}, the set of edge metric coordinates is as follow:QT1 = {γE(η|UE) = (4a+ 2d− 13, 4c− 2d− 8, 2a+ 2b+ 2c− 9)|d = 1} ∪ {γE(η|UE) = (4a+ 2d−13, 4c− 2d− 8, 4a+ 2b+ 2c− 2d− 15)|1 ≤ d ≤ c− 1}.

For the edges {η = r1,2du1,2d−1|1 ≤ d ≤ b− 1}, the set of edge metric coordinates is as follow:RU1 = {γE(η|UE) = (4a + 2c + 2d − 16, 2d − 1, 4a + 2b − 2d − 12)|1 ≤ d ≤ b − 2} ∪ {γE(η|UE) =(4a+ 2c+ 2d− 16, 2d− 1, 2a+ 2b+ 2c− 10)|d = b− 1}.

For the edges {η = p2,2ds2,2d−1|1 ≤ d ≤ a− 1}, the set of edge metric coordinates is as follow:PS2 = {γE(η|UE) = (4b + 2c + 2d − 13, 4a + 2b − 2d − 13, 2d − 1)|1 ≤ d ≤ a − 2} ∪ {γE(η|UE) =(2a+ 2b+ 2c− 9, 4a+ 2b− 2d− 13, 2d− 1)|d = a− 1}.

For the edges {η = r2,2du2,2d−1|1 ≤ d ≤ b− 1}, the set of edge metric coordinates is as follow:RU2 = {γE(η|UE) = (4c+ 2d− 10, 2a+ 2b+ 2c− 10, 4b− 2d− 8)|d = 1} ∪ {γE(η|UE) = (4c+ 2d−10, 4a+ 2b− 2d− 16, 4b− 2d− 8)|2 ≤ d ≤ b− 1}.

For the edges {η = q2,2dt2,2d−1|1 ≤ d ≤ c− 1}, the set of edge metric coordinates is as follows:QT2 = {γE(η|UE) = (2d, 4a + 2c + 2d − 15, 4c + 2b − 2d − 11)|1 ≤ d ≤ c − 2} ∪ {γE(η|UE) =(2d, 2a+ 2b+ 2c− 8, 4c+ 2b− 2d− 11)|d = c− 1}.

For the edges {η = p3,2d−1s3,2d−1|1 ≤ d ≤ a − 2}, the set of edge metric coordinates is as fol-low:PS3 = {γE(η|UE) = (4a+ 2c− 2d− 15, 4b+ 2c+ 2d− 16, 4a+ 2b− 2d− 16)|1 ≤ d ≤ a− 2}.

For the edges {η = q3,2d−1t3,2d−1|1 ≤ d ≤ c − 2}, the set of edge metric coordinates is as fol-lows:QT3 = {γE(η|UE) = (4a+ 2b+ 2d− 18, 4b+ 2c− 2d− 14, 4a+ 2c− 2d− 15)|1 ≤ d ≤ c− 2}.

For the edges {η = r3,2d−1u3,2d−1|1 ≤ d ≤ b − 2}, the set of edge metric coordinates is as fol-

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low:RU3 = {γE(η|UE) = (4a+ 2b− 2d− 16, 2b+ 4c− 2d− 13, 4a+ 2b+ 2d− 18)|1 ≤ d ≤ b− 2}.

For the edges {η1 = t2,j−3p3,i−5, η2 = u2,2s3,i−5, η3 = s2,i−3t3,j−5, η4 = u1,k−3q3,j−5, η5 = t1,2u3,k−5, η6 =s1,i−3r3,k−5, η7 = p3,1r3,1, η8 = u3,1q3,1, η9 = s3,1t3,1}, the set of edge metric coordinates is as follow:V2 = {γE(η1|UE) = (2c − 2, 2a + 2b + 2c − 11, 2b − 1), γE(η2|UE) = (2c, 2a + 2b + 2c − 12, 2b −3), γE(η3|UE) = (2a+2b+2c−11, 2b−1, 2a−2), γE(η4|UE) = (2a+2b+2c−12, 2b−3, 2a), γE(η5|UE) =(2a − 1, 2c − 2, 2a + 2b + 2c − 11), γE(η6|UE) = (2a − 3, 2c, 2a + 2b + 2c − 12), γE(η7|UE) =(2b+ 2b− 8, 2b+ 2c− 8, 2a+ 2b− 7), γE(η8|UE) = (2a+ 2b− 7, 2b+ 2c− 8, 2a+ 2b− 6), γE(η9|UE) =(2a+ 2c− 6, 2b+ 2c− 7, 2a+ 2b− 8)}.

Now, from these sets of edge metric codes for the graph FCSa,b,c, we find that |P1| = |P2| = 2a−2,|Q1| = |Q2| = 2c − 2, |R1| = |R2| = 2b − 2, |S1| = |S2| = 2a − 4, |T1| = |T2| = 2c − 4,|U1| = |U2| = 2b − 4, |P3| = |S3| = 2a − 6, |Q3| = |T3| = 2c − 6, |R3| = |U3| = 2b − 6,|PS1| = |PS2| = a− 1, |RU1| = |RU2| = b− 1, |QT1| = |QT2| = c− 1, |PS3| = a− 2, |QT3| = c− 2,|RU3| = b− 2, |V1| = 12, and |V2| = 9. We see that the sum of all of these cardinalities is equal to|E(FCSa,b,c)| and which is 3(5a+5b+5c−21). Moreover, all of these sets are pairwise disjoint, whichimplies that edim(FCSa,b,c) ≤ 3. To complete the proof, we have to show that edim(FCSa,b,c) ≥ 3.To show this, we have to prove that there exists no edge metric generator UE for FCSa,b,c suchthat |UE | ≤ 2. Since, the graph FCSa,b,c is not a path graph, so the possibility of a singleton edgemetric generator for FCSa,b,c is ruled out [15]. Next, suppose on the contrary that there exists anedge metric generator UE with |UE | = 2. Therefore, we have the following cases to be discussed (forthe contradictions, the naturals a, b, and c are ≥ 5):

Case(I) When UE = {a, b}, where a and b are the vertices from the outer pqr-cycle of FCSa,b,c.

• Suppose UE = {p1,1, p1,d}, p1,d (2 ≤ d ≤ i). Then γ(u3,k−5r3,k−5|UE) = γ(r3,k−5r3,k−6|UE),for 2 ≤ d ≤ i, a contradiction.

• Suppose UE = {p1,1, q1,d}, q1,d (1 ≤ d ≤ j). Then γ(r1,2u1,1|UE) = γ(u1,1u1,2|UE), for1 ≤ d ≤ j − 1; γ(p1,is1,i−2|UE) = γ(s1,i−2s1,i−3|UE) when d = j, a contradiction.

• Suppose UE = {p1,1, r1,d}, r1,d (1 ≤ d ≤ k). Then γ(p3,1s3,1|UE) = γ(p3,1r3,1|UE), for1 ≤ d ≤ k, a contradiction.

• Suppose UE = {p1,1, p2,d}, p2,d (1 ≤ d ≤ i). Then γ(t1,1q3,1|UE) = γ(q3,1q3,2|UE), for 1 ≤ d ≤i, a contradiction.

• Suppose UE = {p1,1, r2,d}, r2,d (1 ≤ d ≤ k). Then γ(s2,1s2,2|UE) = γ(s2,1p2,2|UE), for1 ≤ d ≤ k − 1; γ(s1,1t2,1|UE) = γ(q2,2t2,1|UE), for d = k, a contradiction.

• Suppose UE = {p1,1, q2,d}, q2,d (1 ≤ d ≤ j). Then γ(s2,1s2,2|UE) = γ(s2,1p2,2|UE), for1 ≤ d ≤ j, a contradiction.

Case(II) When UE = {a, b}, where a and b are the vertices from the first interior cycle ofFCSa,b,c.

• Suppose UE = {s1,1, s1,d}, s1,d (2 ≤ d ≤ i − 2). Then γ(t2,1t2,2|UE) = γ(t2,1q2,2|UE), for2 ≤ d ≤ i− 1, a contradiction.

• Suppose UE = {s1,1, r3,d}, r3,d (1 ≤ d ≤ k − 5). Then γ(q1,4t1,3|UE) = γ(t1,3t1,4|UE), for1 ≤ d ≤ k − 5, a contradiction.

• Suppose UE = {s1,1, t2,d}, t2,d (1 ≤ d ≤ j − 2). Then γ(p1,2s1,1|UE) = γ(s1,1s1,2|UE), for1 ≤ d ≤ j − 2, a contradiction.

• Suppose UE = {s1,1, p3,d}, p3,d (1 ≤ d ≤ i − 5). Then γ(q3,1t3,1|UE) = γ(t3,1t3,2|UE), for1 ≤ d ≤ 2a− 3, a contradiction.

Case(III) When UE = {a, b}, where a and b are the vertices from the second interior cycle ofFCSa,b,c.

• Suppose UE = {u1,1, u1,d}, u1,d (1 ≤ d ≤ k − 2). Then γ(u1,1r1,2|UE) = γ(u1,1t1,j−2|UE), for1 ≤ d ≤ k − 2, a contradiction.

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• Suppose UE = {u1,1, t1,d}, t1,d (1 ≤ d ≤ j − 2). Then γ(u1,1u1,2|UE) = γ(u1,1r1,2|UE), for1 ≤ d ≤ j − 2, a contradiction.

• Suppose UE = {u1,1, u3,d}, u3,d (1 ≤ d ≤ k−5). Then γ(s1,i−4p1,i−3|UE) = γ(s1,i−4s1,i−5|UE),for 1 ≤ d ≤ k − 5, a contradiction.

• Suppose UE = {u1,1, q3,d}, q3,d (1 ≤ d ≤ j − 5). Then γ(p3,1s3,1|UE) = γ(s3,1s3,2|UE), for1 ≤ d ≤ j − 5, a contradiction.

Case(IV) When UE = {a, b}, where a and b are the vertices from the third interior cycle ofFCSa,b,c.

• Suppose UE = {s2,1, s2,d}, s2,d (2 ≤ d ≤ i−2). Then γ(u2,k−2u2,k−3|UE) = γ(u2,k−2r2,k−1|UE),for 2 ≤ d ≤ i− 2, a contradiction.

• Suppose UE = {s2,1, u2,d}, u2,d (1 ≤ d ≤ k − 2). Then γ(s2,1s2,2|UE) = γ(s2,1p2,2|UE), for1 ≤ d ≤ k − 2, a contradiction.

• Suppose UE = {s2,1, s3,d}, s3,d (1 ≤ d ≤ i−5). Then γ(r1,k−3u1,k−4|UE) = γ(u1,k−4u1,k−5|UE),for 1 ≤ d ≤ i− 5, a contradiction.

• Suppose UE = {s2,1, t3,d}, t3,d (1 ≤ d ≤ j−5). Then γ(r1,k−3u1,k−4|UE) = γ(u1,k−4u1,k−5|UE),for 1 ≤ d ≤ j − 5, a contradiction.

Case(V) When UE = {a, b}, where a is in outer pqr-cycle and b is in the first interior cycle ofFCSa,b,c.

• Suppose UE = {p1,1, s1,d}, s1,d (1 ≤ d ≤ i − 2). Then γ(t1,1q1,2|UE) = γ(t1,1t1,2|UE), for1 ≤ d ≤ i− 2, a contradiction.

• Suppose UE = {p1,1, t2,d}, t2,d (1 ≤ d ≤ j − 2). Then γ(u2,1u2,2|UE) = γ(u2,1r2,2|UE), for1 ≤ d ≤ j − 2, a contradiction.

• Suppose UE = {p1,1, r3,d}, r3,d (1 ≤ d ≤ k − 5). Then γ(q1,4t1,3|UE) = γ(t1,3t1,4|UE), for1 ≤ d ≤ k − 5, a contradiction.

• Suppose UE = {p1,1, p3,d}, p3,d (1 ≤ d ≤ i − 5). Then γ(q3,1t3,1|UE) = γ(t3,1t3,2|UE), for1 ≤ d ≤ i− 5, a contradiction.

Case(VI) When UE = {a, b}, where a is in outer pqr-cycle and b is in the second interior cycleof FCSa,b,c.

• Suppose UE = {p1,1, t1,d}, t1,d (1 ≤ d ≤ j − 2). Then γ(u1,1u1,2|UE) = γ(u1,1r1,2|UE), for1 ≤ d ≤ j − 2, a contradiction.

• Suppose UE = {p1,1, u1,d}, u1,d (1 ≤ d ≤ k−2). Then γ(u1,k−2s2,i−2|UE) = γ(r1,k−2u1,k−2|UE),for 1 ≤ d ≤ k − 2, a contradiction.

• Suppose UE = {p1,1, u3,d}, u3,d (1 ≤ d ≤ k − 5). Then γ(q1,4t1,3|UE) = γ(t1,3t1,4|UE), for1 ≤ d ≤ k − 5, a contradiction.

• Suppose UE = {p1,1, q3,d}, q3,d (1 ≤ d ≤ j−5). Then γ(u1,k−2u1,k−3|UE) = γ(u1,k−3u1,k−4|UE),for 1 ≤ d ≤ j − 5, a contradiction.

Case(VII) When UE = {a, b}, where a is in outer pqr-cycle and b is in the third interior cycleof FCSa,b,c.

• Suppose UE = {p1,1, u2,d}, u2,d (1 ≤ d ≤ k − 2). Then γ(s2,1s2,2|UE) = γ(s2,1p2,2|UE), for1 ≤ d ≤ k − 2, a contradiction.

• Suppose UE = {p1,1, s3,d}, s3,d (1 ≤ d ≤ i − 5). Then γ(q3,1t3,1|UE) = γ(t3,1t3,2|UE), for1 ≤ d ≤ i− 5, a contradiction.

• Suppose UE = {p1,1, s2,d}, s2,d (1 ≤ d ≤ i− 2). Then γ(r1,kr1,k−1|UE) = γ(r1,k−1r1,k−2|UE),for 1 ≤ d ≤ i− 2, a contradiction.

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• Suppose UE = {p1,1, t3,d}, t3,d (1 ≤ d ≤ j − 5). Then γ(r1,kr1,k−1|UE) = γ(r1,k−1r1,k−2|UE),for 1 ≤ d ≤ j − 5, a contradiction.

Case(VIII) When UE = {a, b}, where a is in first interior cycle and b is in the second interiorcycle of FCSa,b,c.

• Suppose UE = {s1,1, t1,d}, t1,d (1 ≤ d ≤ j − 2). Then γ(q2,2t2,1|UE) = γ(t2,1t2,2|UE), for1 ≤ d ≤ j − 2, a contradiction.

• Suppose UE = {s1,1, u3,d}, u3,d (1 ≤ d ≤ k − 5). Then γ(q3,1q3,2|UE) = γ(q3,1t3,1|UE), for1 ≤ d ≤ k − 5, a contradiction.

• Suppose UE = {s1,1, u1,d}, u1,d (1 ≤ d ≤ k−2). Then γ(r1,k−1u1,k−2|UE) = γ(u1,k−2s2,i−2|UE),for 1 ≤ d ≤ k − 2, a contradiction.

• Suppose UE = {s1,1, q3,d}, q3,d (1 ≤ d ≤ j−5). Then γ(u1,k−2u1,k−3|UE) = γ(u1,k−3u1,k−4|UE),for 1 ≤ d ≤ j − 5, a contradiction.

Case(IX) When UE = {a, b}, where a is in first interior cycle and b is in the third interior cycleof FCSa,b,c.

• Suppose UE = {s1,1, u2,d}, u2,d (1 ≤ d ≤ k − 2). Then γ(p1,2s1,1|UE) = γ(s1,1s1,3|UE), for1 ≤ d ≤ k − 2, a contradiction.

• Suppose UE = {s1,1, s2,d}, s2,d (1 ≤ d ≤ i− 2). Then γ(r1,kr1,k−1|UE) = γ(r1,k−1r1,k−2|UE),for 1 ≤ d ≤ i− 2, a contradiction.

• Suppose UE = {s1,1, t3,d}, t3,d (1 ≤ d ≤ j − 5). Then γ(r1,kr1,k−1|UE) = γ(r1,k−1r1,k−2|UE),for 1 ≤ d ≤ j − 5, a contradiction.

• Suppose UE = {s1,1, s3,d}, s3,d (1 ≤ d ≤ i − 5). Then γ(q3,1t3,1|UE) = γ(t3,1t3,2|UE), for1 ≤ d ≤ i− 5, a contradiction.

Case(X) When UE = {a, b}, where a is in second interior cycle and b is in the third interiorcycle of FCSa,b,c.

• Suppose UE = {u1,1, u2,d}, u2,d (1 ≤ d ≤ k − 2). Then γ(q2,jq2,j−1|UE) = γ(q2,j−1q2,j−2|UE),for 1 ≤ d ≤ k − 2, a contradiction.

• Suppose UE = {u1,1, s2,d}, s2,d (1 ≤ d ≤ i − 2). Then γ(r1,2u1,1|UE) = γ(u1,1t1,j−2|UE), for1 ≤ d ≤ i− 2, a contradiction.

• Suppose UE = {u1,1, t3,d}, t3,d (1 ≤ d ≤ j − 5). Then γ(s3,1s3,2|UE) = γ(p3,1s3,1|UE), for1 ≤ d ≤ j − 5, a contradiction.

• Suppose UE = {u1,1, s3,d}, s3,d (1 ≤ d ≤ i − 5). Then γ(u2,1u2,d|UE) = γ(u2,2u2,3|UE), for1 ≤ d ≤ i− 5, a contradiction.

As a result, we infer that for FCSa,b,c, there is no edge resolving set UE such that |UE | = 2.Therefore, we must have |UE | ≥ 3 i.e., edim(FCSa,b,c) ≥ 3. Hence, edim(FCSa,b,c) = 3, whichconcludes the theorem.

In terms of minimum IEMG, we have the following result

Theorem 2. For a, b, c ≥ 4, the graph FCSa,b,c has an IEMG with cardinality three.

Proof. To show that, for zigzag edge coronoid fused with starphene FCSa,b,c, there exists an IEMGU iE with |U iE | = 3, we follow the same technique as used in Theorem 3.

Suppose U iE = {p1,1, r1,1, r1,k} ⊂ V (FCSa,b,c). Now, by using the definition of an independentset and following the same pattern as used in Theorem 1, it is simple to show that the set of verticesU iE = {p1,1, r1,1, r2,k} forms an IEMG for FCSa,b,c with |U iE | = 3, which concludes the theorem.

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5 Conclusions

In this paper, we have studied the minimum vertex and edge metric generators for the zigzag edgecoronoid fused with starphene FCSa,b,c structure. For positive integers a, b, c ≥ 4, we have provedthat dim(FCSa,b,c) = edim(FCSa,b,c) = 3 (a partial response to the question raised recently in[15]). We also observed that the vertex and edge metric generators for FCSa,b,c are independent.In future, we will try to obtain the other variants of metric dimension (for instance, fault-tolerantmetric dimension (vertex and edge), mixed metric dimension, etc) for the graph FCSa,b,c.

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