# Characterization of (Molecular) Graphs with Fractional Metric Dimension as Unity

Authors: Muhammad Javaid; Muhammad Kamran Aslam; Abdulaziz Mohammed Alanazi; Meshari M. Aljohani
Journal: Journal of Chemistry
 (2021)

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## Abstract

Distance-based dimensions provide the foreground for the identification of chemical compounds that are chemically and structurally different but show similarity in different reactions. The reason behind this similarity is the occurrence of a setS of atoms and their same relative distances to some ordered set T of atoms in both compounds. In this article, the aforementioned problem is considered as a test case for characterising the (molecular) graphs bearing the fractional metric dimension (FMD) as 1. For the illustration of the theoretical development, it is shown that the FMD of path graph is unity. Moreover, we evaluated the extremal values of fractional metric dimension of a tetrahedral diamond lattice.


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## Body

## 1. Introduction

Day by day, the nexus of chemistry is progressing by the advancements in drug discovery, formation of chemical compounds, and development of testing kits for the diagnosis of different diseases and medical anomalies. Besides different concepts that arose as a result of the emergence of cheminformatics, distance-based dimensions also have their stake in this concern. Assume that, in a graphℂ, the shortest path between the 2 vertices s,t is given by ds,t. Let S=s1,s2,s3,…,sk⊆Vℂ and u∈Vℂ; then, the k-tuple metric form of S in terms of u is given by ru|S=du,s1,du,s2,du,s3,…,du,sk. The set S becomes a resolving set having k elements for a graph ℂ if each pairs of vertices in ℂ bears distinct k-tuple metric forms. The resolving set with minimum cardinality in ℂ forms its metric basis, and its cardinality represents its metric dimension.The terminology of resolving sets was introduced by Slatter [1, 2] by naming them as locating sets. Harary and Melter [3] personally discovered these terminologies and called them as the metric dimension of ℂ. Afterward, many researchers have studied different graph structures for the calculation of metric dimensions. The results for the metric dimensions of path, cycle, Peterson, and generalized Peterson graphs can be found in [4–6]. For various results on metric dimensions of graphs, we refer to [7–9] and [10]. Chartrand et al. [11] employed metric dimension to find the solution of an integer programming problem (IPP). Subsequently, Currie and Oellermann introduced the concept of fractional metric dimension (FMD) and obtained the solution of IPP with higher accuracy [12]. Arumugam and Mathew [13] after discovering the hidden properties of FMD formally defined it. Since then, many researchers have tried their luck in this area by attacking different graph structures. The results for the FMD of graph structures as obtained from Cartesian, hierarchial, corona, lexicographic, and comb product of connected graph structures can be seen in [14–16] and [17, 18]. Recently, Liu et al. [19] calculated the fractional metric dimension of the generalized Jahangir graph J5,k and Raza et al. calculated the FMD of a metal organic network [20, 21]. Alisyah et al. presented the concept of local fractional metric dimension (LFMD) and found the LFMD of the corona product of two connected networks [22]. Liu et al. calculated the LFMD of rotationally symmetric and planar networks [23]. Recently, Javaid et al. calculated the bounds for the LFMD of connected and cycle-related networks in [24, 25].Johnson [26, 27] employed the concept of metric dimension for creating proficiency of large datasets of chemical graph structures. The mathematical study of chemical structures concerns the development of mathematical classification of chemical compounds. The graph-theoretic version of chemical compounds naturally exists. Despite having different chemical and structural aspects, two chemical compounds show similar behaviour during the reactions. The reason behind this peculiarity is the existence of certain common substructures within these compounds. If in two compounds, the elements of the set S of atoms and the elements of the ordered set T are relatively equidistant, then we call these compounds to be similar or equivalent [28]. Finding a T with minimum cardinality such that the ordered lists associated with every two distinct vertices of S are distinct has applications to classification problems in chemistry, as described in [11].In this article, we are going to characterise the (molecular) graphs with FMDs as unity. As a test case, we have considered the allotropic form of carbon called by tetrahedral diamond developed by Ali et al. This article propels in the following manner: Section1 is for introduction, Section 2 is devoted for the applications of FMD in chemistry, Section 3 is for preliminaries, Section 4 concerns with the development of a tool for the characterization of graphs with FMD as 1, and Section 5 deals with the resolving neighbourhood sets of TDn. In Section 6, we have calculated the FMD of TDn. Section 7 gives the conclusion.


## 2. Applications in Chemistry

In a molecular graph, atoms are denoted by nodes and bond between them by edges. The fraternity of chemists and pharmacists is always in search of finding out chemical compounds in some collection bearing physiochemical properties in common at some particular places. This objective is achieved by the identification of the substructure having the smallest number of atoms. In graph theory, this problem is the same as finding the FMD of the graph under consideration. In this way, druggists and chemists will be able to capture the aforementioned features of these compounds and comprehend whether they are responsible for some pharmacological activity for a newly developed drug. For more on the applications like these, see [11].


## 3. Preliminaries

Forc∈Vℂ and a,b⊆Vℂ, a,b is said to be resolved by c if da,c≠db,c. The set formed by the pair of nodes comprising nodes such as c is called resolving neighbourhood. The resolving neighbourhood (RN) of a,b is mathematically given by Ra,b=c∈Vℂ|da,c≠db,c.Suppose a connected networkℂVℂ,Eℂ having order p. A function τ:Vℂ⟶0,1 is known as the resolving function (RF) of ℂ if τRa,b≥1∀a,b∈Vℂ, where τRx,y=∑z∈Rx,yτz. An RF η of ℂ is known as a minimal resolving function (MRF) if any function ϕ:Vℂ⟶0,1 such that ϕ≤η and ϕz≠ηz for at least one z∈Vℂ that is not an RF of ℂ. Then, the FMD of the network ℂ is given by dimfℂ=minη:ηis the MRF ofℂ, where η=∑z∈Vℂηz [13].


### 3.1. Construction of Tetrahedral Diamond

The tetrahedral diamond graph is ann-dimensional lattice, comprising ni layers where 1≤i≤n. Figures 1 and 2 show TDn for 3≤n≤5.Figure 1
Tetrahedral diamond lattice with 3 (a) and 4 (b) layers.Figure 2
Tetrahedral diamond lattice with 5 layers.Eachni layer is having ni2 vertices, ni−2ni−1/2 hexagons, and three pendent edges. The vertices of each layer are denoted by vjni where 1≤j≤ni. The first layer is isomorphic to K1, and layer two is isomorphic to K1.3, whereas for 1≤i≤n, each ni−1 layer is the subgraph of the ni-th layer. Hence, the graph formed by each layer is denoted by Snini. Similarly, following are the subgraphs found to be in all the layers: Snjni,p, Pni−ni−1ni,p, Psni,p, and K1ni,p, where 1≤j,s≤ni−1 and p describes their position that can be top, top right, top left, bottom, bottom right, bottom left, middle, middle right, middle left, and bottom denoted by t,tr,tl,b,br,bl,m,mr, and ml, respectively. Figure 3 shows all the subsets of TDn.Figure 3
Subsets of the tetrahedral diamond lattice.It can be seen from the figure that, in each layer,v1ni is adjacent to v2ni+1, vni2−ni−12ni is adjacent to vni2−ni−12−1ni+1, and vni2ni is adjacent to vni2−1ni+1. Apart from them, every vertex with an odd label in the ni−1 layer is adjacent to the vertex with an even label in the ni layer and vice versa.


## 3.1. Construction of Tetrahedral Diamond

The tetrahedral diamond graph is ann-dimensional lattice, comprising ni layers where 1≤i≤n. Figures 1 and 2 show TDn for 3≤n≤5.Figure 1
Tetrahedral diamond lattice with 3 (a) and 4 (b) layers.Figure 2
Tetrahedral diamond lattice with 5 layers.Eachni layer is having ni2 vertices, ni−2ni−1/2 hexagons, and three pendent edges. The vertices of each layer are denoted by vjni where 1≤j≤ni. The first layer is isomorphic to K1, and layer two is isomorphic to K1.3, whereas for 1≤i≤n, each ni−1 layer is the subgraph of the ni-th layer. Hence, the graph formed by each layer is denoted by Snini. Similarly, following are the subgraphs found to be in all the layers: Snjni,p, Pni−ni−1ni,p, Psni,p, and K1ni,p, where 1≤j,s≤ni−1 and p describes their position that can be top, top right, top left, bottom, bottom right, bottom left, middle, middle right, middle left, and bottom denoted by t,tr,tl,b,br,bl,m,mr, and ml, respectively. Figure 3 shows all the subsets of TDn.Figure 3
Subsets of the tetrahedral diamond lattice.It can be seen from the figure that, in each layer,v1ni is adjacent to v2ni+1, vni2−ni−12ni is adjacent to vni2−ni−12−1ni+1, and vni2ni is adjacent to vni2−1ni+1. Apart from them, every vertex with an odd label in the ni−1 layer is adjacent to the vertex with an even label in the ni layer and vice versa.


## 4. Characterization of Graphs with FMD as Unity

In this section of the article, we are giving generic criteria for identifying graphs with FMD as 1. These criteria have been shaped up as a theorem given below.Theorem 1.
Letℂ be a connected graph and Ra,b be a resolving neighbourhood set of the pair of vertices a,b in ℂ. If ∩Ra,b≠Φ, then(1)dimfracℂ=1,where Vℂ≥3.Proof.
Assume thatR=Ra,b is an arbitrary resolving neighbourhood set for a,b⊂Vℂ and Y=∩R. Now, we define the function ψ:Vℂ⟶0,1 as c=∑x∈Yψx and 1−c=∑x∈R−Y∩Xψx, where c is a real number that approaches to 1 and X=Vℂ. For a,b∈Vℂ and c⟶1,(2)ψR=∑x∈Rψx=∑x∈Yψx+∑x∈R−Y∩Xψx,=c+w1−c≥1,where w=VℂVℂ−1/2. It implies that ψ is a resolving function. To check that ψ is a minimal resolving function, assume that there is another minimal resolving function τ such that τ≤ψ. By definition, τx<ψx for some x∈X. Now, for some resolving neighbourhood set R, we have(3)τR=∑x∈Yτx+∑x∈R−Y∩Xτx<∑x∈Yψx+∑x∈R−Y∩Xψx,=c+1−c=1.
Consequently,τR<1 which implies that τ is not a resolving function. Thus, ψ is a minimal resolving function. Let ψ¯ be another minimal resolving function of ℂ. Now, we have the following possibilities:(i)
ψ¯x<ψx∀x∈X(ii)
ψ¯x≥ψx∀x∈X(iii)
ψ¯x<ψx for some x∈XCase 1.
Ifψ¯x<ψx for all x∈X, then for each resolving neighbourhood set R, ψ¯R<1⇒ψ¯ is not a resolving function; therefore, this case does not hold.Case 2.
Ifψ¯x≥ψx for all x∈X, then we have the following subcases:Subcase A: for1≤r,s≤w and Rr−Y∩Rs−Y=Φ, we have(4)ψ¯=∑x∈Yψ¯x+∑x∈R−Y∩Xψ¯x>∑x∈Yψx+∑x∈R−Y∩Xψx,=c+w1−c=ψ.Asc⟶1, dimfracℂ=ψ=c+w1−c=1.Subcase B: for1≤r,s≤w and Rr−Y∩Rs−Y≠Φ, we have(5)ψ¯=∑x∈Yψ¯x+∑x∈R¯tψ¯x,whereR¯t=Rt∩X−∩j=1t−1Rt−Y. Then,(6)ψ¯=∑x∈Yψ¯x+∑x∈R¯tψ¯x≥∑x∈Yψx+∑x∈R¯tψx,=c+∑x∈R¯tψx.So,(7)dimfracℂ=c+∑x∈R¯tψx≤c+∑x∈R−Y∩Xψx,=c+w1−c,=1.Thus,dimfracℂ≤1. But, by definition, dimfracℂ≥1. Therefore,(8)dimfracℂ=1.Case 3.
Ifψ¯x<ψx for some x∈X, this case is a consequent of the abovementioned two cases (Case I and II); therefore, we have dimfracℂ=1.
Consequently, from Case 1–3, we arrive at the following conclusion:(9)dimfracℂ=1.
Using the result presented above, we are now going to prove the following fact:Proposition 1.
Suppose that, for anyn≥3, G≅Pn; then, dimfracG=1.Proof.
Case 1: forn=3: the resolving neighbourhood sets for the current case are R1=Ra1,a2=a1,a2,a3, R2=Ra2,a3=a1,a2,a3, and R3=Ra1,a3=a1,a3. It can be seen that ∩t=13Rt=a1,a3≠Φ. Therefore, from Theorem 1, we arrive at the conclusion thatdimfracP3=1.Case 2: forn≥4: the resolving neighbourhood sets of Pn are Rai,ai+p=VPn−a2i+p/2 and Rai,ai+s=Vℙn, where p,s≥1, 1≤i≤n, p≡0mod2, and s≡1mod2.
It can be seen that∩t=1n2R=a1,an≠Φ. Therefore, from Theorem 1, it implies that(10)dimfracPn=1.Remark 1.
The abovementioned proposition strengthens the result proved in [13].


## 5. Resolving Neighbourhood Sets ofTDn

In this section, we present some the results regarding the resolving neighbourhood sets ofTDn. Lemma 1 deals with the resolving neighbourhoods of TDn having minimum cardinality followed by Lemma 2 and Lemma 3 that are concerned with resolving neighbourhood sets of maximum cardinalities.Lemma 1.
Suppose thatℂ≅TDn is an n-dimensional tetrahedral diamond lattice. Then, the minimum resolving neighbourhood sets are as follows:(a)
Forn≥4, n≡0mod2, α=n/2,β=α+1, γ=2α−1, and η=2β−1, Rt=Ra1α,a1β=Raγα,aηβ=Raα2α,aβ2β=nn2+3n+2/12 and ∪t=13Rt=nn2+3n+2/6.(b)
Forn≥3, n≡1mod2,α=n−1/2,β=n+1/2, γ=n+3/2, η=2α−1, λ=2β−1, and μ=2γ−1, Rt=Ra1α,a1β=Ra1β,a1γ=Raηα,aλβ=Raλβ,aμγ=Raα2α,aβ2β=Raβ2β,aγ2γ=n3+3n2+5n+3/12 and ∪t=16Rt=n3+6n2+11n−30/6.Proof.
(a)
The resolving neighbourhood sets ofRa1α,a1β,Raγα,aηβ, and Raα2α,aβ2β are Ra1α,a1β=Vℂ−∪j=βnSj2j−Sj−22j,t, Raγα,aηβ=Vℂ−∪j=βnSj2j−Sj−22j,bl, and Raγ2α,aη2β=Vℂ−∪j=βnSj2j−Sj−22j,br. We note that Rt=Ra1α,a1β=Raγα,aηβ=Raα2α,aα+12β=nn2+3n+2/12, ∪t=13Rt=∪j=1αSj2j∪∪j=α+1nSj−α2j,t∪Sj−α2j,br∪Sj−α2j,bl, and ∪t=13Rt=nn2+3n+2/6.(b)
The resolving neighbourhood sets ofRa1α,a1β,Ra1β,a1γ,Raηα,aλβ,Raλβ,aμγ,Raα2α,aβ2β,Raβ2β,aγ2γ are Ra1α,a1β=Vℂ−∪j=βnSj2j−Sj−22j,t, Raγα,aηβ=Vℂ−∪j=βnSj2j−Sj−22j,bl, and Raα2α,aβ2β=Vℂ−∪j=βnSj2j−Sj−22j,br and Ra1β,a1γ=Vℂ−∪j=γnSj2j−Sj−22j,t, Raλβ,aμγ=Vℂ−∪j=γnSj2j−Sj−22j,bl, and Raβ2β,aγ2γ=Vℂ−∪j=γnSj2j−Sj−22j,br .We note that Rt=Ra1α,a1β=Ra1β,a1γ=Raηα,aλβ=Raλβ,aμγ=Raα2α,aβ2β=Raβ2β,aγ2γ=n3+3n2+5n+3/12,∪t=16Rt=∪j=1γSj2j∪∪j=γ+1nSj−γ2j,t∪Sj−γ2j,br∪Sj−γ2j,bl, and ∪t=16Rt=n3+6n2+11n−30/6.Lemma 2.
Suppose thatℂ≅TDn is an n-dimensional tetrahedral diamond lattice with n≥3 and n≡1modn. Then,(a)
For1≤α≤n,β=α+1, γ=2α−1, and η=2β−1, Rt<Ra1α,a1β=Raγα,aηβ=Raα2α,aβ2β and Ra1α,a1β∩∪t=16Rt=Raγα,aηβ∩∪t=16Rt=Raα2α,aβ2β∩∪t=16Rt≥Rt(b)
For1≤α≤n, β=α+1 and 2≤γ≤2α−1, Rt<Raγα,aγβ and Raγα,aγβ∪t=16Rt≤Rt(c)
For2≤α≤n, β=2α−1,γ=2α+2,η=α2−3, and μ=α2, Rt<Raβα,aγα=Raηα,aμα and Raβα,aγα∪t=16Rt=Raii,ai+1i+1∪t=16Rt=Raηα,aμα∪t=16Rt≥Rt(d)
For anyaαα,aββ∈Eℂ, γ≥1, η≥2, γ≡1mod2, and η≡0mod2, Rt<Raαα,aββ=Raαα,aα+γα+η=Vℂ=n2n2+3n+1/6 and Raαα,aββ∩∪t=16Rt≥RtProof.
(a)
The resolving neighbourhood sets ofRa11,a12,a11,a32,a11,a42, a1α,a1β, and Raγα,aηβ are Ra1α,a1β=Vℂ−∪j=βnSj2j−Sj−12j,t, Raγα,aηβ=Vℂ−∪j=2βSj2j−Sj−12j,bl, and Raα2α,aβ2β=Vℂ−∪j=βnSj2j−Sj−12j,br, respectively.SinceRa1α,a1β=Raγα,aηβ=Raα2α,aβ2β=2n3+3n2+n+6αn+1α−n−6αβα−β−1/6, thus Rt=2n3+3n2+n+6αn+1α−n−6αβα−β−1/6. Therefore, Ra1α,a1β∩∪t=16Rt=Raγα,aηβ∩∪t=16Rt=Raα2α,aβ2β∩∪t=16Rt≥Rt.(b)
First of all, we introduce a notation for simplification.(11)μ=γ2,forn≡0modn,γ−12,forn≡1modn.The resolving neighbourhood forRaγα,aγβ is Raγα,aγβ=Vℂ−∪j=α−μ+1αSj2j∪∪j=βnSj2j,t∪∪j=μ+1μ+n−αSj2j,t. It can be seen that Raγα,aγβ=n−1−4μ+13+6μ+12−μ+2b3−3b2+b+2μ+n−α+13−μ+n−α+12−α/6>Rt. Therefore, Ra32,a42∪t=16Rt≥Rt. Also, by the symmetry of the network, Rt<Raγα,aγβ and Raγα,aγβ∪t=16Rt≥Rt.(c)
The resolving neighbourhood forRaβα,aγα and Raηα,aμα are Raβα,aγα=Vℂ−S11∪∪j=3αPjj,l∪∪j=α+1nPjj,l−Sj−α2j,l and Raηα,aμα=Vℂ−S11∪∪j=3αPjj,r∪∪j=α+1nPjj,r−Sj−α2j,r. It can be seen that Raβα,aγα=Raηα,aμα=2n+13−3n+12+2n+8−7α+62α−2n−α+1+2n−α+13−3n−α+12/6. Hence, Rt<2n+13−3n+12+2n+8−7α+62α−2n−α+1+2n−α+13−3n−α+12/6 and Raβα,aγα∪t=16Rt=Raηα,aμα∪t=16Rt≥Rt.(d)
The resolving neighbourhood setsRaαα,aββ and Raαα,aaα+γα+η are Raαα,aββ=Vℂ=Raαα,aaα+γα+η. Clearly, Raαα,aββ=Raαα,aaα+γα+η=nn+12n+1/6>Rt. Also, Raαα,aββ∪t=16Rt=Raαα,aaα+γα+η∪t=16Rt≥Rt.Lemma 3.
Suppose thatℂ≅TDn is an n-dimensional tetrahedral diamond lattice with n≥4 and n≡0modn. Then,(a)
For1≤α≤n,β=α+1, γ=2α−1, and η=2β−1, Rt<Ra1α,a1β=Raγα,aηβ=Raα2α,aβ2β and Ra1α,a1β∩∪t=13Rt=Raγα,aηβ∩∪t=13Rt=Raα2α,aβ2β∩∪t=13Rt≥Rt(b)
For1≤α≤n, β=α+1, and 2≤γ≤2α−1, Rt<Raγα,aγβ and Raγα,aγβ∪t=13Rt≤Rt(c)
For2≤α≤n, ,β=2α−1,γ=2α+2,η=α2−3, and μ=α2, Rt<Raβα,aγα=Raηα,aμα and Raβα,aγα∪t=13Rt=Raii,ai+1i+1∪t=13Rt=Raηα,aμα∪t=13Rt≥Rt(d)
For anyaαα,aββ∈Eℂ, γ≥1, η≥2, γ≡1mod2, and η≡0mod2, Rt<Raαα,aββ=Raαα,aα+γα+η=Vℂ=n2n2+3n+1/6 and Raαα,aββ∩∪t=13Rt≥RtProof.
The proof is the same as that of Lemma2.


## 6. Fractional Metric Dimension ofTDn

In this section, the FMD ofTDn is calculated and the criterion of their evaluation is devised by the following result.Theorem 2.
Ifℂ≅TDn is an n-dimensional tetrahedral diamond lattice with n≥3 and n≡1mod2, then(12)1<dimfracℂ≤2n3+6n2+11n−30n3+3n2+5n+3.Proof.
Case 1: whenn=3.The resolving neighbourhood sets are as shown in Tables1–4.The resolving neighbourhood sets that are equal due to symmetry are given by the following:Now, foruv∈Eℂ and 47≤L≤62, we have(13)Ruv=Vℂ.In the same manner, the pairwise resolving neighbourhood sets that equalsVℂ are as follows:As we can see, Table3 shows the resolving neighbourhood sets of TD3 having the maximum cardinality of 13 and ∪L=6384R´L=VTD3. Table 4, on the other hand, shows the resolving neighbourhood sets with minimum cardinality of 6 and ∪t=16Rt=VTD3. Suppose that, for 63≤L≤84, R´L=γ, for 1≤t≤6, Rt=λ, η=∪L=6384R´L=VTD3=13, and δ=∪t=16Rt=VTD3=13. Now, we define a mapping κ:VTD3⟶0,1 such that κa=1/13 for all a∈∪L=6384R´L; assigning the value of 1/13 to all the elements of ∪L=6384R´L and summing up all the labels, we get κ=∑a∈∪L=6384R´Lκa=13/13=1; thus, dimfracTD3=κ=1, as all the R´L for 63≤L≤84 are all having VTD3 in common. Similarly, we define another mapping τ:VTD3⟶0,1 such that τ=1/λ=1/6 for all a∈∪t=16Rt, and giving labels to the elements of ∪t=16Rt and later on summing them up, we get τ=∑a∈∪t=16Rtτa=13/6=2.33; hence, dimfracℂ<2.33 as all the Rt for 1≤t≤6 are pairwise overlapping. Therefore, we arrive at the following conclusion:(14)1<dimfracTD3≤2.33.Case 2: whenn≥5.The required minimum resolving neighbourhood sets areRa1α,a1β,Ra1β,a1γ,Raηα,aλβ,Raλβ,aμγ, Raα2α,aβ2β, and Raβ2β,aγ2γ, where α=n−1/2, β=n+1/2, γ=n+3/2, η=2α−1, λ=2β−1, and μ=2γ−1. Lemma 1 clarifies that Rt=Ra1α,a1β=Ra1β,a1γ=Raηα,aλβ=Raλβ,aμγ=Raα2α,aβ2β=Raβ2β,aγ2γ=n3+3n2+5n+3/12≤R´a,b for all a,b∈Vℂ and ∪t=16Rt=∪j=1γSj2j∪∪j=γ+1nSj−γ2j,t∪Sj−γ2j,br∪Sj−γ2j,bl. Also, the resolving neighbourhood sets with maximum cardinality of Vℂ, as clarified by Lemma 2, are R´1=Raαα,aaα+γα+η and R´1=Raαα,aββ and R´1=Raαα,aaα+γα+η, respectively. Moreover, ∪L=12R´L=Vℂ. Let Raαα,aaα+γα+η=Raαα,aββ=γ, Rt=λ, ∪L=12R´L=η, and ∪t=16Rt=δ. To find the minimum value for the dimfracℂ, we define a mapping κ:Vℂ⟶0,1 such that(15)κa=1κ,fora∈∪L=12R´L,0,fora∈Vℂ−∪L=12R´L,where γ=η=n2n2+3n+1/6. Assigning the labels to the elements of ∪∪L=12R´L and summing them up, we get κ=∑a∈∪L=12R´L1/γ=1.
Similarly, for the maximum value ofdimfracℂ, we define another mapping τ:Vℂ⟶0,1 such that(16)τa=1λ,fora∈∪t=16Rt,0,fora∈Vℂ−∪t=16Rt.
It can be seen thatτ is a resolving function for ℂ with n≥3 because τRu,v≥1∀u,v∈Vℂ. On the contrary, assume that there is another resolving function ρ such that ρu≤τu, for at least one u∈Vℂρu≠τu. As a consequence, ρRu,v<1, where Ru,v is a resolving neighbourhood of ℂ with minimum cardinality λ. It shows that ρ is not a resolving function which is a contradiction. Therefore, τ is a minimal resolving function that attains minimum τ for ℂ. Since all the Rt have nonempty intersection, there is another minimal resolving function of τ¯ of ℂ such that τ¯≤τ. Hence, assigning 1/λ to the vertices of ℂ in ∪t=13Rt and calculating the summation of all the weights, we get(17)dimfracℂ=∑t=1δ1λ≤2n3+6n2+11n−30n3+3n2+5n+3.
In the end, we arrive at the following finding:(18)1<dimfracℂ≤2n3+6n2+11n−30n3+3n2+5n+3.Table 1
Resolving neighbourhood sets ofTD3.
Resolving neighbourhood setsElementsR1=Ra11,a21Vℂ−a22,a32,a42∪a43,a53,a63,a83,a93R2=Ra11,a23Vℂ−a12,a22,a42∪a13,a23,a73,a83,a93R3=Ra11,a42Vℂ−a12,a22,a32∪a13,a23,a33,a43,a53R4=Ra21,a31Vℂ−a23,a33,a43,a53,a63,a73,a83,a93R5=Ra32,a53Vℂ−a13,a23,a33,a43,a63,a73,a83,a93R6=Ra42,a93Vℂ−a13,a23,a33,a43,a53,a63,a73,a83R1´=Ra12,a32Vℂ−a11∪a22,a42∪a33,a83,a93R2´=Ra12,a42Vℂ−a11∪a22,a32∪a43,a53,a73R3´=Ra12,a33Vℂ−a32∪a13,a23,a73,a83,a93R4´=Ra12,a53Vℂ−a32∪a33,a83,a93R5´=Ra12,a63Vℂ−a32,a42∪a33,a73R6´=Ra12,a73Vℂ−a42∪a13,a23,a33,a43,a53R7´=Ra12,a93Vℂ−a42∪a43,a53,a73R8´=Ra22,a23Vℂ−a12∪a13,a23,a33,a73R9´=Ra22,a43Vℂ−a32−a13,a23,a73,a83,a93R10´=Ra22,a83Vℂ−a42∪a13,a23,a33,a43,a53R11´=Ra32,a13Vℂ−a12∪a33,a83,a93R12´=Ra32,a33Vℂ−a12∪a43,a53,a63,a83,a93R13´=Ra32,a53Vℂ−a13,a23,a33,a43,a63,a73,a83,a93R14´=Ra32,a63Vℂ−a42∪a13,a23,a33,a43,a53R15´=Ra32,a73Vℂ−a42∪a33,a63R16´=Ra32,a93Vℂ−a42∪a43,a53,a73R17´=Ra42,a13Vℂ−a12∪a43,a53,a73R18´=Ra42,a33Vℂ−a12,a32∪a63,a73,a73R19´=Ra42,a53Vℂ−a32∪a13,a23,a63R20´=Ra42,a63Vℂ−∪a13,a23,a83,a93R21´=Ra42,a73Vℂ−a12∪a43,a53,a63,a83,a93R22´=Ra11,a63Vℂ−a32,a42∪a13,a23R23´=Ra11,a73Vℂ−a12,a42∪a43,a53R24´=Ra32,a42Vℂ−a11∪a12,a22∪a13,a23,a63R25´=Ra33,a93Vℂ−a11∪a22∪a63,a73R26´=Ra43,a73Vℂ−a11∪a22∪a33,a63R27´=Ra53,a63Vℂ−a11∪a12,a22,a32∪a13,a23,a33,a43R28´=Ra63,a93Vℂ−a11∪a12,a22,a42∪a13,a23,a73,a83R29´=Ra12,a32Vℂ−a11∪a22,a42∪a33,a83,a93R30´=Ra12,a42Vℂ−a11∪a22,a32∪a43,a53,a73R31´=Ra13,a33Vℂ−a11∪a12,a22,a42∪a23,a73,a83R32´=Ra13,a63Vℂ−a11∪a22∪a33,a73R43´=Ra13,a73Vℂ−a11∪a12,a22,a32∪a23,a33,a43,a53R33´=Ra23,a43Vℂ−a11∪a22,a42∪a33,a83,a93R34´=Ra23,a83Vℂ−a11∪a22,a32∪a43,a53,a73R35´=Ra53,a73Vℂ−a11∪a22∪a33,a63Table 2
Resolving neighbourhood sets ofTD3 equal due to symmetry.
Resolving neighbourhood setsElementsR36´=Ra11,a13Ra22,a23R38´=Ra11,a53Ra22,a43R40´=Ra33,a73Ra32,a42R42´=Ra63,a73Ra12,a32R44´=Ra33,a63Ra12,a42R46´=Ra13,a93Ra12,a42R37´=Ra11,a33Ra22,a43R39´=Ra11,a93Ra22,a83,R41´=Ra43,a83Ra32,a42R43´=Ra73,a93Ra12,a32R45´=Ra13,a53Ra12,a32Table 3
Resolving neighbourhood sets ofTD3 equal to VTD3.
Resolving neighbourhood setsElementsR63´=Ra11,a23VℂR65´=Ra11,a83VℂR67´=Ra12,a83VℂR69´=Ra22,a33VℂR71´=Ra22,a63VℂR73´=Ra22,a93VℂR75´=Ra32,a43VℂR77´=Ra42,a23VℂR79´=Ra42,a83VℂR81´=Ra13,a83VℂR83´=Ra23,a63VℂR64´=Ra11,a43VℂR66´=Ra12,a43VℂR68´=Ra22,a13VℂR70´=Ra22,a53VℂR72´=Ra22,a73VℂR74´=Ra32,a23VℂR76´=Ra32,a83VℂR78´=Ra42,a43VℂR80´=Ra13,a43VℂR82´=Ra23,a53VℂR84´=Ra23,a93VℂTable 4
Resolving neighbourhood sets ofTD3 with minimum cardinality.
Resolving neighbourhood setsElementsR1=Ra11,a21Vℂ−a22,a32,a42∪a43,a53,a63,a83,a93R2=Ra11,a23Vℂ−a12,a22,a42∪a13,a23,a73,a83,a93R3=Ra11,a42Vℂ−a12,a22,a32∪a13,a23,a33,a43,a53R4=Ra21,a31Vℂ−a23,a33,a43,a53,a63,a73,a83,a93R5=Ra32,a53Vℂ−a13,a23,a33,a43,a63,a73,a83,a93R6=Ra42,a93Vℂ−a13,a23,a33,a43,a53,a63,a73,a83Theorem 3.
Ifℂ≅TDn is an n-dimensional tetrahedral diamond lattice with n≥4 and n≡0mod2, then(19)1<dimfracℂ≤2.Proof.
Case 1: whenn=4.The resolving neighbourhood sets are as shown in Tables4–6.The resolving neighbourhood sets that are equal due to symmetry are given by the following:For109≤L≤149 and uv∈Vℂ, we have(20)RL=Ruv=Vℂ.Similarly, the pairwise resolving neighbourhood sets that equalsVℂ are as follows:As we can see, Table7 shows the resolving neighbourhood sets of ℂ having the maximum cardinality of 30 and ∪L=150240R´L=Vℂ. Table 8, on the other hand, shows the resolving neighbourhood sets with a minimum cardinality of 10 and ∪t=13Rt=Vℂ−a23,a33,a43,a63,a73,a83∪a44,a104,a114,a124. Suppose that, for 150≤L≤240, R´L=γ, for 1≤t≤3, Rt=λ, η=∪L=6384R´L=VTD3=30, and δ=∪t=16Rt=VTD3=20. Now, we define a mapping κ:VTD3⟶0,1 such that κa=1/30 for all a∈∪L=6384R´L, and assigning the value of 1/30 to all the elements of ∪L=6384R´L and summing up all the labels, we get κ=∑a∈∪L=6384R´Lκa=30/30=1; thus, dimfracℂ=κ=1, as all the R´L for 150≤L≤240 are all having VTD3 in common. Similarly, we define another mapping τ:VTD3⟶0,1 such that τ=1/λ=1/10 for all a∈∪t=13Rt, and giving labels to the elements of ∪t=13Rt and latter on summing them up, we get τ=∑a∈∪t=13Rtτa=20/10≤2; hence, dimfracℂ<2 as all the Rt for 1≤t≤6 are pairwise overlapping. Therefore, we arrive at the following conclusion:(21)1<dimfracTD3≤2.Case 2: whenn≥6.The required minimum resolving neighbourhood sets areRa1α,a1β,Raγα,aηβ, and Raα2α,aβ2β, where α=n/2, β+1, γ=2α−1, and η=2β−1. As it is evident from Lemma 1, Rt=Ra1α,a1β=Raγα,aηβ=Raα2α,aβ2β=nn2+3n+2/12≤‖acuteRa,b| for all a,b∈Vℂ and ∪t=13Rt=∪j=1αSj2j∪∪j=α+1nSj−α2j,t∪Sj−α2j,br∪Sj−α2j,bl. Moreover, Ru,v∩∪t=13Rt≥Rt for all u,v∈Vℂ. Also, by Lemma 3 (d), the resolving neighbourhood sets with maximum cardinality of Vℂ=n2n2+3n+1/6 are R´1=Raαα,aββ and R´2=Raαα,aα+γα+η with ∪L=12R´L=Vℂ. Let Rt=λ, R´L=γ, ∪L=12R´L=η, and ∪t=13Rt=δ; to find the minimum value of dimfracℂ, we define a mapping κ:Vℂ⟶0,1 such that(22)ka=1k,fora∈∪L=13Rt,0,fora∈Vℂ−∪t=13RL.where γ=η=n2n2+3n+1/6. Assigning the labels to the elements of ∪∪L=12R´L and summing them up, we get κ=∑a∈∪L=12R´L1/γ=1.
Similarly, for the maximum value ofdimfracℂ, we define another mapping τ:Vℂ⟶0,1 such that(23)τa=1λ,fora∈∪t=13Rt,0,fora∈Vℂ−∪t=13Rt.
It can be seen thatτ is a resolving function for ℂ with n≥3 because τRu,v≥1∀u,v∈Vℕ. On the contrary, we assume that there is another resolving function ρ such that ρu≤τu, for at least one u∈Vℂρu≠τu. As a consequence, ρRu,v<1, where Ru,v is a resolving neighbourhood of ℂ with minimum cardinality κ. It shows that ρ is not a resolving function which is a contradiction. Therefore, τ is a minimal resolving function that attains minimum τ for ℂ. Since 1≤t≤3Rt has a nonempty intersection, thus there exists another minimal resolving function τ¯ of ℂ such that τ¯≤τ. Thus, assigning 1/λ to the vertices of ℂ in ∪t=13Rt and calculating the summation of all the weights, we get dimfracℂ=∑t=1δ1/λ≤12nn2+3n+2/6nn2+3n+2=2. Therefore, we arrive at the following result:(24)1<dimfracℂ≤2.Table 5
Resolving neighbourhood sets ofTD4 that are not equal to each other.
Resolving neighbourhood setsElementsR1=Ra21,a31Vℂ−a23,a33,a43,a53,a63,a73,a83,a93∪a44,a54,a64,a74,a84,a104,a114,a124,a134,a144,a154,a164R2=Ra32,a53Vℂ−a13,a23,a33,a43,a63,a73,a83,a93∪a14,a24,a34,a44,a94,a104,a114,a124,a134,a144,a154,a164R3=Ra42,a93Vℂ−a13,a23,a33,a43,a53,a63,a73,a83∪a14,a24,a34,a44,a54,a64,a74,a84,a94,a104,a114,a124,a134R1´=Ra11,a21Vℂ−a22,a32,a42∪a43,a53,a63,a83,a93∪a64,a74,a84,a124,a134,a154,a164R2´=Ra11,a23Vℂ−a12,a22,a42∪a13,a23,a73,a83,a93∪a41,a42,a49,a410,a414,a415,a416R3´=Ra11,a42Vℂ−a12,a22,a32∪a13,a23,a33,a43,a53∪a14,a24,a34,a44,a54,a64,a74R4´=Ra12,a32Vℂ−a11∪a22,a42∪a33,a83,a93∪a34,a44,a54,a114,a154,a164R5´=Ra12,a42Vℂ−a11∪a22,a32∪a43,a53,a73∪a64,a74,a94,a104,a114,a144R6´=Ra12,a33Vℂ−a32∪a13,a23,a73,a83,a93∪a14,a24,a94,a104,a144,a154,a164R12´=Ra12,a53Vℂ−a32∪a33,a83,a93∪a34,a44,a54,a114,a154,a164R13´=Ra12,a63Vℂ−a32,a42∪a33,a73∪a34,a44,a54,a94,a104,a144R14´=Ra12,a73Vℂ−a42∪a13,a23,a33,a43,a53∪a14,a24,a34,a44,a54,a64,a74R15´=Ra12,a93Vℂ−a42∪a43,a53,a73∪a64,a74,a94,a104,a114,a144R16´=Ra22,a23Vℂ−a12∪a43,a53,a63,a83,a93∪a64,a74,a84,a124,a134,a154,a164R17´=Ra13,a14Vℂ−a24,a34,…,a164R18´=Ra13,a34Vℂ−a33,a43,a53,a63∪a14,a24,…,a74R19´=Ra13,a54Vℂ−a33,a43,a53,a63∪a104,a144,a154,a164R20´=Ra13,a74Vℂ−a43,a63∪a44,a114,a154,a164R21´=Ra13,a84Vℂ−a43,a83∪a44,a104,a144R22´=Ra13,a94Vℂ−a63,a73,a83,a93∪a14,a24,…,a74R23´=Ra13,a114Vℂ−a33,a43,a53,a73,a83,a93∪a34,a94R24´=Ra13,a144Vℂ−a63,a73,a83,a93∪a44,a54,a64,a74R25´=Ra13,a164Vℂ−a63,a83∪a64,a74,a104,a114R26´=Ra23,a24Vℂ−a13∪a44,a54,…,a84,a104…,a104R27´=Ra23,a44Vℂ−a33,a43,a53,a63,∪a14,a24,a94,a104,a144,a154,a164R28´=Ra23,a64Vℂ−a43,a63∪a44,a114,a154,a164R29´=Ra23,a104Vℂ−a63,a73,a83,a93,∪a14,a24,a34,a44,a54,a64,a74R30´=Ra23,a124Vℂ−a43,a53,a83,a93∪a44,a104,a144R31´=Ra23,a154Vℂ−a63,a83∪a54,a64,a104,a114R32´=Ra33,a14Vℂ−a13∪a34,a154,a164R33´=Ra33,a54Vℂ−a53∪a14,a24,a34,a44,…,a94,…,a164R34´=Ra33,a74Vℂ−a53∪a54,a154,a164R35´=Ra33,a84Vℂ−a43,a53,a63,a83,a93∪a44,a54,a104,a114,a143R36´=Ra33,a94Vℂ−a13,a73,a83,a93∪a84,a114,a124,a134R37´=Ra33,a114Vℂ−a63,a73,a83,a93∪a14,a24,…,a74R38´=Ra33,a134Vℂ−a63,a83∪a64,a74,a104,a114R39´=Ra33,a144Vℂ−a83∪a84,a114,a124,a134R40´=Ra33,a164Vℂ−a83∪a84,a124R41´=Ra43,a24Vℂ−a23,a73∪a44,a54,a154,a164R42´=Ra43,a44Vℂ−a13,a23,a33,a73∪a64,a74,a84,a124,a134,a154,a164R43´=Ra43,a64Vℂ−a53∪a14,a24,a34,a44,…,a94,…,a164R44´=Ra43,a104Vℂ−a83,a93∪a44,a54R45´=Ra43,a154Vℂ−a73,a83∪a114,a124R46´=Ra53,a14Vℂ−a23,a73∪a44,a114,a154,a164R47´=Ra53,a34Vℂ−a23,a73∪a54,a124,a134R48´=Ra53,a54Vℂ−a22∪a13,a23,a33,a73∪a14,a24,…,a54,a94,a104,a114,a144R49´=Ra53,a74Vℂ−a14,a24,…,a64,a84…,a164R50´=Ra53,a84Vℂ−a63,a73,a83,a93∪a14,a24,a34,a44,a54,a64,a74R51´=Ra53,a94Vℂ−a83,a93∪a44,a124,a134R52´=Ra53,a114Vℂ−a83,a93∪a44,a124,a134R53´=Ra53,a134Vℂ−a73,a83∪a14,a24,a34,a44,a84R54´=Ra53,a144Vℂ−a13,a23,a83∪a34,a44,a84R55´=Ra53,a164Vℂ−a73,a83∪a14,a24,a114,a124R56´=Ra63,a14Vℂ−a12∪a23∪a44,a54,a104,a154R57´=Ra93,a164Vℂ−a14,a24…,a154R58´=Ra32,a42Vℂ−a11∪a12,a22,a32∪a13,a23,a33,a43∪a14,a24,a34,a44,a84R59´=Ra53,a63Vℂ−a11∪a12,a22,a32∪a13,a23,a33,a43∪a14,a24,a34,a44,a84R60´=Ra24,a124Vℂ−a11∪a12∪a33,a73∪a44,a104R61´=Ra23,a43Vℂ−a11∪a22,a42∪a33,a83,a3,a93∪a34,a44,a54,a114,a154,a164R62´=Ra23,a83Vℂ−a11∪a22,a32∪a43,a53,a3,a73∪a64,a74,a94,a104,a114,a144R63´=Ra53,a73Vℂ−a11∪a22∪a33,a63∪a34,a44,a154,a164R64´=Ra33,a93Vℂ−a11∪a22∪a63,a73∪a64,a74,a94,a104R65´=Ra13,a33Vℂ−a11∪a12,a22,a42∪a23,a73,a83∪a34,a104,a144,a154,a164R66´=Ra13,a73Vℂ−a11∪a12,a22,a32∪a23,a33,a43,a53∪a44,a54,a64,a74,a94R67´=Ra14,a34Vℂ−a11∪a12,a22,a42∪a13,a23,a73,a83,a93∪a24,a94,a104,a144,a154,a164R68´=Ra14,a74Vℂ−a11∪a12,a22,a32∪a13,a23,a33,a43,a53∪a24,a34,a44,a54,a64,a74R69´=Ra63,a93Vℂ−a11∪a12,a22,a42∪a13,a23,a73,a83∪a14,a24,a94,a104,a134R70´=Ra53,a93Vℂ−a11∪a12,a22∪a13,a23,a63∪a14,a24,a114,a124R71´=Ra54,a144Vℂ−a11∪a12,a22∪a13,a23,a63∪a14,a24,a114,a124R72´=Ra74,a164Vℂ−a11∪a12,a22∪a13,a23,a63∪a14,a24,a114,a124R73´=Ra11,a14Vℂ−a23,a33,a73∪a64,a74,a84,a124,a134,a154,a164R74´=Ra11,a74Vℂ−a33,a43,a63∪a14,a24,a94,a104,a144,a154,a164R75´=Ra11,a94Vℂ−a63,a73,a83∪a14,a24,a34,a44,a54,a64,a74R76´=Ra23,a94Vℂ−a63,a73,a83∪a14,a24,a34,a44,a54,a64,a74R77´=Ra93,a144Vℂ−a12∪a13,a23,a33,a73∪a14,a24,…,a54,a94,a104,a114,a144R78´=Ra23,a24Vℂ−a13∪a44,a54,…,a84,a104,…,a164Table 6
Resolving neighbourhood sets ofTD4 that are equal due to symmetry.
Resolving neighbourhood setsEqualityResolving neighbourhood setsEqualityR79´=Ra13,a134Ra13,a84R80´=Ra33,a63Ra12,a42R81´=Ra54,a84Ra12,a42R82´=Ra75,a105Ra12,a42R83´=Ra63,a73Ra12,a32R84´=Ra134,a144Ra12,a32R85´=Ra22,a43Ra12,a33R86´=Ra22,a83Ra12,a73R87´=Ra32,a33Ra22,a23R88´=Ra42,a73Ra22,a23R89´=Ra32,a63Ra12,a73R90´=Ra42,a63Ra12,a33R91´=Ra54,a114Ra53,a63R92´=Ra54,a114Ra53,a63R93´=Ra74,a134Ra53,a63R94´=Ra114,a144Ra63,a93R95´=Ra124,a154Ra63,a93R96´=Ra84,a164Ra63,a93R97´=Ra54,a144Ra53,a93R98´=Ra54,a144Ra53,a93R99´=Ra74,a134Ra53,a63R100´=Ra84,a164Ra63,a93R101´=Ra24,a104Ra13,a73R102´=Ra24,a44Ra13,a33R103´=Ra33,a34Ra23,a24R104´=Ra34,a114Ra13,a73R105´=Ra94,a114Ra13,a33R106´=Ra63,a114Ra93,a144R107´=Ra53,a54Ra93,a144R108´=Ra33,a34Ra23,a24Table 7
Resolving neighbourhood sets ofTD4 that are equal to VTD4.
Resolving neighbourhood setsElementsResolving neighbourhood setsElementsR150´=Ra11,a23VℂR151´=Ra11,a43VℂR152´=Ra11,a83VℂR153´=Ra11,a24VℂR154´=Ra11,a44VℂR155´=Ra11,a64VℂR156´=Ra11,a104VℂR157´=Ra11,a124VℂR158´=Ra11,a154VℂR159´=Ra12,a23VℂR160´=Ra12,a43VℂR161´=Ra12,a83VℂR162´=Ra22,a13VℂR163´=Ra22,a33VℂR164´=Ra22,a53VℂR165´=Ra22,a63VℂR166´=Ra22,a73VℂR167´=Ra22,a93VℂR168´=Ra32,a23VℂR169´=Ra32,a43VℂR170´=Ra32,a83VℂR171´=Ra42,a23VℂR172´=Ra42,a43VℂR173´=Ra42,a83VℂR174´=Ra13,a43VℂR175´=Ra13,a83VℂR176´=Ra23,a53VℂR177´=Ra23,a63VℂR178´=Ra23,a93VℂR179´=Ra14,a44VℂR180´=Ra14,a104VℂR181´=Ra14,a64VℂR182´=Ra14,a124VℂR183´=Ra14,a154VℂR184´=Ra24,a54VℂR185´=Ra24,a114VℂR186´=Ra24,a144VℂR187´=Ra24,a74VℂR188´=Ra24,a84VℂR189´=Ra24,a134VℂR190´=Ra24,a164VℂR191´=Ra34,a64VℂR192´=Ra34,a124VℂR193´=Ra34,a154VℂR194´=Ra44,a74VℂR195´=Ra44,a84VℂR196´=Ra44,a134VℂR197´=Ra44,a164VℂR198´=Ra11,a24VℂR199´=Ra11,a44VℂR200´=Ra11,a64VℂR201´=Ra11,a124VℂR202´=Ra11,a154VℂR203´=Ra12,a24VℂR204´=Ra12,a44VℂR205´=Ra12,a64VℂR206´=Ra12,a124VℂR207´=Ra12,a44VℂR208´=Ra12,a64VℂR209´=Ra12,a124VℂR210´=Ra22,a14VℂR211´=Ra22,a34VℂR212´=Ra22,a54VℂR213´=Ra22,a74VℂR214´=Ra22,a84VℂR215´=Ra22,a114VℂR216´=Ra22,a134VℂR217´=Ra22,a164VℂR218´=Ra32,a24VℂR219´=Ra32,a44VℂR220´=Ra32,a64VℂR221´=Ra32,a104VℂR222´=Ra32,a124VℂR223´=Ra32,a154VℂR224´=Ra42,a24VℂR225´=Ra42,a44VℂR226´=Ra42,a64VℂR227´=Ra42,a104VℂR228´=Ra42,a124VℂR229´=Ra42,a154VℂR230´=Ra13,a24VℂR231´=Ra13,a44VℂR232´=Ra13,a64VℂR233´=Ra13,a124VℂR234´=Ra13,a154VℂR235´=Ra23,a14VℂR236´=Ra23,a34VℂR237´=Ra23,a54VℂR238´=Ra23,a74VℂR239´=Ra23,a84VℂR240´=Ra23,a114VℂTable 8
Resolving neighbourhood sets ofTD4 that are not equal to each other.
Resolving neighbourhood setsElementsR1=Ra21,a31Vℂ−a23,a33,a43,a53,a63,a73,a83,a93∪a44,a54,a64,a74,a84,a104,a114,a124,a134,a144,a154,a164R2=Ra32,a53Vℂ−a13,a23,a33,a43,a63,a73,a83,a93∪a14,a24,a34,a44,a94,a104,a114,a124,a134,a144,a154,a164R3=Ra42,a93Vℂ−a13,a23,a33,a43,a53,a63,a73,a83∪a14,a24,a34,a44,a54,a64,a74,a84,a94,a104,a114,a124,a134


## 7. Conclusions

We conclude our discussion by the following remarks:(i)
In this article, we have made a characterization of graphs having the FMD as unity(ii)
It is computed that the FMD of the path is 1 that strengthens the result proved in [13](iii)
We have calculated the extremal values of FMD ofTDn as (i) for n≡0mod2, 1<dimfracℂ≤2 and (ii) for n≡1mod2, 1<dimfracℂ≤2n3+6n2+11n−30/n3+3n2+5n+3(iv)
Now, we close our discussion with the open problem that investigates the families of graphs other thanPn having FMD as unity



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*Source: 9910572-2021-06-28.xml*