# On the Upper Bounds of Fractional Metric Dimension of Symmetric Networks

Authors: Muhammad Javaid; Muhammad Kamran Aslam; Jia-Bao Liu
Journal: Journal of Mathematics
 (2021)

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## Abstract

Distance-based numeric parameters play a pivotal role in studying the structural aspects of networks which include connectivity, accessibility, centrality, clustering modularity, complexity, vulnerability, and robustness. Several tools like these also help to resolve the issues faced by the different branches of computer science and chemistry, namely, navigation, image processing, biometry, drug discovery, and similarities in chemical compounds. For this purpose, in this article, we are considering a family of networks that exhibits rotationally symmetric behaviour known as circular ladders consisting of triangular, quadrangular, and pentagonal faced ladders. We evaluate their upper bounds of fractional metric dimensions of the aforementioned networks.


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## Body

## 1. Introduction

A networkℵ comprises of two collections of distinct objects, which are set containing nodes/vertices Vℵ and set of objects that forms connection among those nodes/vertices Eℵ, where Eℵ⊆Vℵ×Vℵ. The cardinality of Vℵ constitutes the order of ℵ denoted by v, and size of ℵ is the cardinality of Eℵ denoted by e. For a,b∈Vℵ, da,b represents the distance of shortest path between a and b. In order to have an in depth study of graph theory’s strata, please refer to [1–3].Day by day technological advancements are restructuring our lives by replacing manpower with robots, devices, and machinery. In the same instance, the necessity of restructuring is also making it inevitable for us to utilize these machineries which are keeping in view cost effectiveness and serving the large amount of ques in production line, health services, and public dealing areas with minimum number of them. Here, distance-based parameters come to fulfill the aforementioned objectives. ConsiderM=m1,m2,…,mk⊆Vℵ, then M becomes an ordered set of vertices whenever some ordering is being imposed on them. The k-tuple metric form of M with respect to a is represented by ra|M=da,m1,da,m2,da,m3,…,da,mk, where da,mt for 1≤t≤k shows the distance of a from mt. M changes into resolving set if for any pair of distinct vertices a,b∈Vℵ−M such that ra|M≠rb|ℵ. The resolving set having minimum number of vertices in ℵ presents the metric dimension (MD) of ℵ denoted by λℵ [4, 5].For any connected network, Slater [6, 7] laid the foundation of resolving set by calling it as locating set. After studying these terminologies at their own, Harary and Melter named it as the MD of a network. Since then, a number of researchers considered different families of networks for the computation of MD. The results regarding the families of networks with constant and bounded MD can be found in [8]. Chartrand et al. proved the results regarding the metric dimensions of path and cycle in [4]. Similarly, ℙn,2,Am, and ℂm2 as networks bearing constant, MD has been shown in [8]. Moreover, in [9], the MD of the generalized Petersen network is proved to be bounded. Also the MD of the Jahangir network and infinite regular networks has been studied in [10, 11], respectively.Chartrand et al. [4] employed MD to acquire the solution of integer programming problem (IPP). Later on, Currie and Oellermann [12] established the concept of fractional metric dimension (FMD) and acquired the higher accuracy solution. Fehr et al. [13] utilized FMD to obtain the optimal solution of a certain linear programming relaxation problem. Arumugam and Mathew [14] uncovered some peculiar features of FMD. Afterwards, a bunch of results appeared related to the FMD of several networks that are the resultants of graph products, namely, Cartesian, hierarchial, corona, and lexicographic comb products (see [14–18]). Furthermore, the FMD of the generalized Jahangir network and permutation network can be found in [19, 20], respectively. In a recent time, Raza et al. [21] evaluated the FMD of metal organic networks.Aisyah et al. defined the notion of local fractional metric dimension (LFMD) and calculated for corona product of two networks [22]. Similarly, Liu et al. [23] calculated the LFMD of rotationally symmetric and planar networks. More recently, Javaid et al. evaluated the sharp extremal values of LFMD of connected networks [24]. Results regarding the LFMD of cycle related networks can be found in [25]. In this article, we have considered a family of network bearing rotational symmetry known by circular ladders consisting of different triangular, quadrangular, and pentagonal faces. This article propels in the following manners: Section 1 is introduction, Section 2 covers the role of MD in Robotics and Chemistry, and Section 3 is of preliminaries. In Section 4, we will discuss the idea behind the creation of the circular ladders under consideration. Section 5, contains the main results, and Section 6 closes this article with some conclusions and future directions.


## 2. Applications

### 2.1. Robotics

The ever increasing demand of networking enlarged the study of distance-based dimensions. All such tools are used to allocate an appropriate region to an interpolar for its effective usage [6, 26]. The placement of robots at different sites such as restaurants, production units, and public health facilities come under study by using one such tool in [27]. The effective rectification of example and picture handling and handling of informational structures using the aforementioned parameters has been given in [28]. The parameter like these has not only made it possible to serve restaurants, factories, and public service areas with a minimum number of robots but also truncated the delay in their serving. To find more applications on this topic, we refer to [29, 30].


### 2.2. Chemistry

A chemical compound in graph theoretic form is regarded as a molecular graph having nodes as atoms and links between them as bonds between them [4]. In this picturesque form of a compound along with distance-based parameters give chemists an opportunity to remove discrepancies in a chemical structure and to find the sites showing similar properties in them. All such strategies can be found in [5, 6, 28, 31].


## 2.1. Robotics

The ever increasing demand of networking enlarged the study of distance-based dimensions. All such tools are used to allocate an appropriate region to an interpolar for its effective usage [6, 26]. The placement of robots at different sites such as restaurants, production units, and public health facilities come under study by using one such tool in [27]. The effective rectification of example and picture handling and handling of informational structures using the aforementioned parameters has been given in [28]. The parameter like these has not only made it possible to serve restaurants, factories, and public service areas with a minimum number of robots but also truncated the delay in their serving. To find more applications on this topic, we refer to [29, 30].


## 2.2. Chemistry

A chemical compound in graph theoretic form is regarded as a molecular graph having nodes as atoms and links between them as bonds between them [4]. In this picturesque form of a compound along with distance-based parameters give chemists an opportunity to remove discrepancies in a chemical structure and to find the sites showing similar properties in them. All such strategies can be found in [5, 6, 28, 31].


## 3. Preliminaries

A vertexc∈ℵ is said to resolve a pair a,b in ℵ if da,c≠db,c. For a,b∈Vℵ, then the resolving neighbourhood (RN) of the pair a,b is given by Ra,b=c∈Vℵ|da,c≠db,c.Assume a connected networkℵVℵ,Eℵ bearing v as its order. A function κ:Vℵ⟶0,1 is called resolving function (RF) of ℵ if each RN set Ra,b, κRa,b=∑c∈Ra,bκc≥1. An RF of ℵ is known as minimal RF if any function ϕ:Vℵ⟶0,1 such that ϕ≤κ and ϕc≠κc for at least one c∈Vℵ, which is not an RF of ℵ. The FMD of the network ℵ is given by dimfℵ=minκ:κis minimal RF ofℵ, where κ=∑c∈Vℵκc [14].


## 4. Construction of Networks

Here, we give the construction of networks covered in this article. A cycleℂ3 is a network bearing the regularity of 2, 3 vertices, and 2 faces known as triangle. On the other hand, ℂ4 is a network having the regularity of 2 with 4 vertices and 2 faces. A 3-faced quadrangle is obtained after joining any pair of nonadjacent vertices of it. Figure 1(a) shows the planar quadrangles having 2 and 3 faces. A network ℂ5 bearing the regularity of 2, 5 nodes, and 2 faces is known as pentagon. The 3-faced pentagons are formed by joining any two nonadjacent nodes of 2-faced pentagon, and 4-faced pentagons are formed by forming links between any two pairs of nonadjacent nodes of 3-faced pentagon. Figure 1(b) illustrates some of the possible 2-faced, 3-faced, and 4-faced planar pentagons.Figure 1
Quadrangles (a) and pentagons (b).
(a)(b)


### 4.1. Triangular Circular Ladder

The triangular circular ladder (TCL)ℵ≅Tm is created by joining each edge a11a21,a21a31,…,am−11am1,am1a11 of a cycle ℂm with the corresponding nodes a12,a22,…,am2, where m≡0mod2. Thus, each step of TCL consists of 2-faced triangle and its vertex and edge sets are as follows: Vℵ=ar1,ar2|1≤r≤m, Eℵ=ar1ar+11|1≤r≤m∧am+11=a11∪ar1ar2,ar1ar−12|1≤r≤m∧ao2=am2 with v=2m and e=3m respectively. Figure 2 shows TCL.Figure 2
TCLTm.


### 4.2. Quadrangular Circular Ladders

Form≥3, the quadrangular circular ladder (QCL) ℚm1 (prism) is a cubic network which is the resultant of the Cartesian product P2×ℂm, with v=2m and e=3m. Its sets of vertices and edges are, respectively, given by the following: Vℚm1=ar1,ar2|1≤r≤m, Eℚm1=ar1ar+11|1≤r≤m∧am+11=a11∪ar1ar2,ar2ar+12|1≤r≤m∧am+12=a12.The two possible QCLsℵ≅ℚm2 and ℵ≅ℚm3 having each step of 3-faced pentagons are formed by forming edges either ar1ar−12 or ar−11ar2 in quadrangular circular ladder ℚm1. The sets of vertices and edges of these networks are, respectively, given as follows: Vℚm2=ar1,ar2|1≤r≤m and Eℚm2=ar1ar+11,ar1ar2,ar2ar+12|1≤r≤m∧am+11=a11,am+12=a12∪ar1ar−12|1≤r≤m∧a02=am2 and Vℚm3=ar1,ar2|1≤r≤m and Eℚm3=ar1ar+11,ar1ar2,ar2ar+12|1≤r≤m∧am+11=a11,am+12=a12∪ar1ar+12|1≤r≤m∧am+12=a12 with v=2m and e=4m.Figure3 illustrates ℚm1, ℚm2, and ℚm3. It can be observed from Figure 3 that ℚ3≅ℚm2.Figure 3
QCLs: (a)ℚm1, (b) ℚm2, and (c) ℚm3.
(a)(b)(c)


### 4.3. Pentagonal Circular Ladders

The each step of a pentagonal circular ladder (PCL)ℙm1 consists of 2-faced pentagon, and it is created by inserting a new vertex ar3 between the vertices ar2 and ar+12 of ℚm1, where Vℙm1 and Eℙm1 are given by Vℙm1=ar1,ar2,ar3|1≤r≤m and Eℙm1=ar1ar+11|1≤r≤m∧am+11=a11∪ar1ar2,ar2ar3,ar2ar−13|1≤r≤m∧a03=am3 with v=3m and e=4m, respectively.Figure4 illustrates ℙm1.Figure 4
ℙm1.The three possible PCLsℵ≅ℙm2,ℙm3,ℙm4, each step of which is having 3-faced pentagons are formed by creating edges either ar2ar+12,ar1ar+12, or ar1ar−12 (shown by magenta colour in Figure 5) in pentagonal circular ladder ℙm1. Figure 5 illustrates ℙm2,ℙm3, and ℙm4.Figure 5
PCLs: (a)ℙm2, (b) ℙm3, and (c) ℙm4.
(a)(b)(c)As seen from Figure5, ℙm4≡ℙm3. The sets Vℙm2 and Eℙm2 are given as follows: Vℙm2=ar1,ar2,ar3|1≤r≤m and Eℙm2=ar1ar+11,ar2ar+12|1≤r≤m∧am+11=a11,am+12=a12∪ar1ar2,ar2ar3,ar2ar−13|1≤r≤m∧a03=am3, Vℙm3=ar1,ar2,ar3|1≤r≤m and Eℙm3=ar1ar+11,ar1ar−12|1≤r≤m∧am+11=a11,a02=am2∪ar1ar2,ar2ar3,ar2ar−13|1≤r≤m∧a03=am3, Vℙm4=ar1,ar2,ar3|1≤r≤m and Eℙm4=ar1ar+11,ar−11ar2|1≤r≤m∧am+11=a11,a01=am1∪ar1ar2,ar2ar3,ar2ar−13|1≤r≤m∧a03=am3 with v=3m and e=5m.The three possible PCLsℵ≅ℙm5,ℙm6 and ℙm7 whose each step is having 4-faced pentagons are created by forming edge ar1ar−12 (shown by cyan colour in Figure 6) in ℙm2, edge ar1ar−13 in ℙm2, edges ar−11ar2 in ℙm2, or edges ar1ar3 and ar1zt−1 in the plane network ℙm1. Figure 6(a) illustrates ℙm5, 6(b) illustrates ℙm6, and 6(c) illustrates ℙm7, respectively.Figure 6
PCLs: (a)ℙm5, (b) ℙm6, and (c) ℙm7.
(a)(b)(c)It can be seen from Figure6 that ℙm6≅ℙm5.The sets of vertices and edges of these networks areVℙm5=ar1,ar2,ar3|1≤r≤mEℙm5=ar1ar+11,ar2ar+12,ar1ar−12|1≤r≤m∧am+11=a11,am+12=a12,a02=am2∪ar1ar2,ar2ar3,ar+12ar3|1≤r≤m, Vℙm6=ar1,ar2,ar3|1≤r≤m, Eℙm6=ar1ar+11,ar2ar+12,ar1ar−12|1≤r≤m∪ar1ar2,ar2ar3,ar2ar−13|1≤r≤m∧a03=am3, Vℙm7=ar1,ar2,ar3|1≤r≤m, Eℙm7=ar1ar+11,ar1ar3,ar1ar−13|1≤r≤m∧a03=am3∪ar1ar2,ar2ar3,ar2ar−13|1≤r≤m∧a03=am3 with v=3m and e=6m, respectively.


## 4.1. Triangular Circular Ladder

The triangular circular ladder (TCL)ℵ≅Tm is created by joining each edge a11a21,a21a31,…,am−11am1,am1a11 of a cycle ℂm with the corresponding nodes a12,a22,…,am2, where m≡0mod2. Thus, each step of TCL consists of 2-faced triangle and its vertex and edge sets are as follows: Vℵ=ar1,ar2|1≤r≤m, Eℵ=ar1ar+11|1≤r≤m∧am+11=a11∪ar1ar2,ar1ar−12|1≤r≤m∧ao2=am2 with v=2m and e=3m respectively. Figure 2 shows TCL.Figure 2
TCLTm.


## 4.2. Quadrangular Circular Ladders

Form≥3, the quadrangular circular ladder (QCL) ℚm1 (prism) is a cubic network which is the resultant of the Cartesian product P2×ℂm, with v=2m and e=3m. Its sets of vertices and edges are, respectively, given by the following: Vℚm1=ar1,ar2|1≤r≤m, Eℚm1=ar1ar+11|1≤r≤m∧am+11=a11∪ar1ar2,ar2ar+12|1≤r≤m∧am+12=a12.The two possible QCLsℵ≅ℚm2 and ℵ≅ℚm3 having each step of 3-faced pentagons are formed by forming edges either ar1ar−12 or ar−11ar2 in quadrangular circular ladder ℚm1. The sets of vertices and edges of these networks are, respectively, given as follows: Vℚm2=ar1,ar2|1≤r≤m and Eℚm2=ar1ar+11,ar1ar2,ar2ar+12|1≤r≤m∧am+11=a11,am+12=a12∪ar1ar−12|1≤r≤m∧a02=am2 and Vℚm3=ar1,ar2|1≤r≤m and Eℚm3=ar1ar+11,ar1ar2,ar2ar+12|1≤r≤m∧am+11=a11,am+12=a12∪ar1ar+12|1≤r≤m∧am+12=a12 with v=2m and e=4m.Figure3 illustrates ℚm1, ℚm2, and ℚm3. It can be observed from Figure 3 that ℚ3≅ℚm2.Figure 3
QCLs: (a)ℚm1, (b) ℚm2, and (c) ℚm3.
(a)(b)(c)


## 4.3. Pentagonal Circular Ladders

The each step of a pentagonal circular ladder (PCL)ℙm1 consists of 2-faced pentagon, and it is created by inserting a new vertex ar3 between the vertices ar2 and ar+12 of ℚm1, where Vℙm1 and Eℙm1 are given by Vℙm1=ar1,ar2,ar3|1≤r≤m and Eℙm1=ar1ar+11|1≤r≤m∧am+11=a11∪ar1ar2,ar2ar3,ar2ar−13|1≤r≤m∧a03=am3 with v=3m and e=4m, respectively.Figure4 illustrates ℙm1.Figure 4
ℙm1.The three possible PCLsℵ≅ℙm2,ℙm3,ℙm4, each step of which is having 3-faced pentagons are formed by creating edges either ar2ar+12,ar1ar+12, or ar1ar−12 (shown by magenta colour in Figure 5) in pentagonal circular ladder ℙm1. Figure 5 illustrates ℙm2,ℙm3, and ℙm4.Figure 5
PCLs: (a)ℙm2, (b) ℙm3, and (c) ℙm4.
(a)(b)(c)As seen from Figure5, ℙm4≡ℙm3. The sets Vℙm2 and Eℙm2 are given as follows: Vℙm2=ar1,ar2,ar3|1≤r≤m and Eℙm2=ar1ar+11,ar2ar+12|1≤r≤m∧am+11=a11,am+12=a12∪ar1ar2,ar2ar3,ar2ar−13|1≤r≤m∧a03=am3, Vℙm3=ar1,ar2,ar3|1≤r≤m and Eℙm3=ar1ar+11,ar1ar−12|1≤r≤m∧am+11=a11,a02=am2∪ar1ar2,ar2ar3,ar2ar−13|1≤r≤m∧a03=am3, Vℙm4=ar1,ar2,ar3|1≤r≤m and Eℙm4=ar1ar+11,ar−11ar2|1≤r≤m∧am+11=a11,a01=am1∪ar1ar2,ar2ar3,ar2ar−13|1≤r≤m∧a03=am3 with v=3m and e=5m.The three possible PCLsℵ≅ℙm5,ℙm6 and ℙm7 whose each step is having 4-faced pentagons are created by forming edge ar1ar−12 (shown by cyan colour in Figure 6) in ℙm2, edge ar1ar−13 in ℙm2, edges ar−11ar2 in ℙm2, or edges ar1ar3 and ar1zt−1 in the plane network ℙm1. Figure 6(a) illustrates ℙm5, 6(b) illustrates ℙm6, and 6(c) illustrates ℙm7, respectively.Figure 6
PCLs: (a)ℙm5, (b) ℙm6, and (c) ℙm7.
(a)(b)(c)It can be seen from Figure6 that ℙm6≅ℙm5.The sets of vertices and edges of these networks areVℙm5=ar1,ar2,ar3|1≤r≤mEℙm5=ar1ar+11,ar2ar+12,ar1ar−12|1≤r≤m∧am+11=a11,am+12=a12,a02=am2∪ar1ar2,ar2ar3,ar+12ar3|1≤r≤m, Vℙm6=ar1,ar2,ar3|1≤r≤m, Eℙm6=ar1ar+11,ar2ar+12,ar1ar−12|1≤r≤m∪ar1ar2,ar2ar3,ar2ar−13|1≤r≤m∧a03=am3, Vℙm7=ar1,ar2,ar3|1≤r≤m, Eℙm7=ar1ar+11,ar1ar3,ar1ar−13|1≤r≤m∧a03=am3∪ar1ar2,ar2ar3,ar2ar−13|1≤r≤m∧a03=am3 with v=3m and e=6m, respectively.


## 5. Main Results

### 5.1. Triangular Circular Ladder

Lemma 1.
Letℵ≅Tm be TCL with m≥6 and m≡0mod2. For 1≤r,l≤m, p≥1, s≥2p≡1mod2 and s≡0mod2, we have(a)
Rl=Ral1,al2=Rr=Rar1,ar−12=m+1 and ∪l=1mRl∪∪r=1mRr=2m, where a02=am2.(b)
Rr<Ru2m−1 for 1≤u≤8, where R¯1=Rar1,ar+p1,R¯2=Rar1,ar+s1,R¯3=Rar2,ar+s2,R¯4=Rar2,ar+p2,R¯5=Rar1,ar+p2,R¯6=Rar1,ar+s2,R¯7=Rar1,ar−p2,R¯8=Rar1,ar−s2 and R¯u∩∪r=1mRr≥Rr with am+11=a11.Proof.
(a)
The RNs ofar1,ar2 and ar1,ar−12 are Rl=Ral1,al2=VTm−ah1|h≡l+1,l+2,…,l+m/2modm∪ah2|h≡l+1,l+2,…,l+m/2−1modm and Rr=Rar1,ar−12=VTm−ah1|h≡r−1,r−2,…,r−m/2+1modm∪ah2|h≡r−2,r−3,…,r−m/2modm. We note that ∪l=1mRl=VTm, ∪r=1mRr=VTm and ∪l=1mRr∪∪r=1mRr=2m.(b)
The RNs ofR¯u for 1≤u≤8 are R¯1=Rar1,ar+p1=VTm−ah2|h≡r+p−1/2,r+p+m−1/2modm=R¯4=Rar2,ar+p2, R¯2=Rar1,ar+s1=VTm−ah2|h≡r+s/2,r+m+s/2modm=R¯3=Rar2,ar+s2, R¯5=Rar1,ar+p2=VTm−ah1|h≡r+p+1/2modm∪ah2|h≡r+m+p−1/2modm, R¯6=Rar1,ar+s2=VTm−ah1|h≡r+m+s/2modm∪ah2|h≡r+s/2modm, R¯7=Rar1,ar−p2=VTm−ah1|h≡r−p+1/2modm∪ah2|h≡r+m−p−1/2modm and R¯8=Rar1,ar−s2=VTm−ah1|h≡r−s/2modm∪ah2|h≡r+m+s−2/2modm, respectively.
Clearly,R¯u=2m−1. Since Rr=m+1<R¯u, then R¯u∩∪r=1mRr=2m−1≥Rr.Theorem 1.
Ifℵ≅Tm with m≥6 and m≡0mod2. Then, dimfTm≤2m/m+1.Proof.
Case 1. Whenm=6.The RNs are given as follows.We have seen from the above that Table1 shows the RNs with maximum cardinality that is 10 and Table 2 shows the RNs with minimum cardinality that is 7, which is less than the cardinalities of all other RNs Rr¯ for 1≤t≤6 that is 10. Moreover, ∪r=112Rr=VT6 which implies ∪r=112Rr=12 and Ru¯∩∪r=112Rr≥Rr, where 1≤u≤24.Now, we define a functionκ:VT6⟶0,1 such that κar1=κar2=1/7. As we can see, Rr for 1≤r≤12 of T6 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of T6 such that κ¯≤κ. As a result, dimfT6≤∑r=1121/7=12/7.Case 2.m≥8.According to Lemma1 the RNs with minimum cardinality are Rar1,ar2 and Rar1,ar−12 having the cardinality of m+1 and R¯u∩∪r=1mRr≥Rr, where 1≤u≤8.Letη=∪r=1mRr=2m and λ=Rr=m+1. Now, we define a function κ:VTm⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈VTm−∪r=1mRr. We can see that κ is a resolving function for Tm because κRa,b≥1∀a,b⊂Vℵ. Suppose that there is another resolving function ρ, such that ρx≤κx, for at least one x∈VTmρx≠κx. As a result, ρRa,b<1, where Ra,b is an RN of Tm having the minimum cardinality of λ. This shows that ρ is not a resolving function which is contradiction. Hence, κ is a minimal resolving function that achieves minimum κ for Tm. As a matter of fact Rr for 1≤r≤m of Tm are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of Tm such that κ¯≤κ. As a result, we arrive at the following: dimfTm≤∑r=1η1/λ=2m/m+1.


### 5.2. Quadrangular Circular Ladder

Lemma 2.
Letℵ≅ℚm1 be the 2-faced QCL with m≥6and m≡0mod2. For 1≤l,r≤m, p≥3, s≥2p≡1mod2 and s≡0mod2. Then,(a)
Rl=Ral1,al+12=Rr=Rar1,ar−12=m and ∪l=1mRl∪∪r=1mRr=2m.(b)
Forp≥3, s≥2, Rr<R¯1=Rar1,ar+s1=R¯2=Rar2,ar+s2=R¯3=Rar1,ar+p2=R¯4=Rar1,ar−p2=2m−2 and R¯u∩∪r=1mRr=2m, where 1≤u≤4.(c)
Forp≥3, s≥0, Rr<R¯5=Rar1,ar+p1=R¯6=Rar2,ar+p2=R¯7=Rar1,ar+s2=R¯8=Rar1,ar−s2=2m and R¯u∩∪r=1mRr=2m, where 5≤u≤8.Proof.
(a)
The RNs ofar1,ar+12 and ar1,ar−12 are given by Rl=Ral1,al+12=Vℚm1−ah1|h≡l+1,l+2,…,l+m/2modm∪ah2|h≡l,l−1,…,l−m/2+1modm and Rr=Rar1,ar−12=Vℚm1−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2|h≡r,r−1,…,r+m/2−1modm, respectively. We note that ∪l=1mRl=Vℚm1, ∪r=1mRr=Vℚm1 and ∪l=1mRl∪∪r=1mRr=2m.(b)
The RNs ofR¯u for 1≤u≤4 are as follows: R¯1=Rar1,ar+s1=Vℚm1−ah1|h≡r−s/2,r+s/2modm∪ah2|h≡r−s/2,r+s/2modm=R¯2=Rar2,ar+s2, R¯3=Rar1,ar+p2=Vℚm1−ah1|h≡r+p+1/2,r+p+m−1/2modm∪ah2|h≡r+p−1/2,r+m−p+1/2modm and R¯4=Rar1,ar−p2=Vℚm1−ah1|h≡r+p,r−p+1/2modm∪ah2|h≡r+p+1/2,r−p−1/2modm respectively. Clearly, R¯u=2m−2. Since Rr=2m−2<R¯u, then R¯u∩∪r=1mRr=2m−2≥Rr, it can be easily seen that ∪r=1mRr=Vℚm1 and ∪r=1mRr=2m.(c)
The RNs ofar1,ar+p1, ar2,ar+p2, ar1,ar+s2 and ar1,ar−p1 are R¯5=Rar1,ar+p1=R¯6=Rar2,ar+p2=R¯7=Rar1,ar+s2=R¯8=Rar1,ar−p1=VG. We can see that R¯u=2m and R¯u∩∪r=1mRr=2n≥Rr.Theorem 2.
Ifℵ≅ℚm1 with m≥6 and m≡0mod2. Then, dimfℚm1≤2.Proof.
Case 1. Whenm=6.The RNs are given as follows.We can see that Tables3 and 4 show the RNs with maximum cardinalities, that is, 12 and 8, respectively, whereas, Table 5 shows the RNs with minimum cardinality, that is, 6, and ∪r=112Rr=Vℚ61 which implies ∪r=112Rr=12 and Ru¯∩∪r=112Rr≥Rr, where 1≤u≤36.Here, we define a functionϕ:Vℚ61⟶0,1 such that κar1=κar2=1/6. Also Rr for 1≤r≤12 of ℚ61 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℚ61 such that κ¯≤κ. As a result, dimfℚ61≤∑r=1121/6=2.Case 2. Whenm≥6.We have seen from Lemma2 that the RNs with minimum cardinality of m are Rar1,ar2 and Rar1,ar−12, respectively.Letη=∪r=1mRr=2m and λ=Rr=2m. Then, we define a function κ:Vℚm1⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈Vℚm1−∪r=1mRr. We see that κ is a resolving function for ℚm1 because κRa,b≥1∀a,b⊂Vℵ. Now, suppose on contrary that there is another resolving function ρ, such that ρu≤ϕu, for at least one u∈Vℚm1ρu≠κu. As a result, ρRa,b<1, where Ra,b is an RN of ℚm1 having minimum cardinality λ. This shows that ρ is not a resolving function which is a contradiction to our supposition. Therefore, κ is a minimal resolving function that achieves minimum κ for ℚm1. Since all the RNs of ℚm1 are equal, hence ∃ another minimal resolving function κ¯ of ℚm1 such that κ¯≤κ. Assigning 1/λ to all the vertices of ℚm1 and calculating their sum, we get: dimfℚm1≤∑r=1η1/m=2.Lemma 3.
Letℵ≅ℚm2 be the 3-faced QCL with m≥6 and m≡0mod2. For 1≤l,r≤m, p≥3, s≥2p≡1mod2 and s≡0mod2. Then(a)
Rl=Ral1,al−12=Rr=Rar1,ar2=m with a02=am2, ∪l=1mRl=2m, ∪r=1mRr=2m and ∪l=1mRl∪∪r=1mRr=2m.(b)
Rr<R¯1=Rar1,ar+p1=R¯2=Rar1,ar+s1=R¯3=Rar2,ar+s2=R¯4=Rar1,ar+p2=R¯5=Rar1,ar+s2=R¯6=Rar1,ar−p2=R¯7=Rar1,ar−s2 and R¯u∩∪r=1mRr≥Rr with am+11=a11 and 1≤u≤8.Proof.
(a)
The RNs ofar1,ar−12 and ar1,ar2 are Rar1,ar−12=Vℚm2−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2|h≡r,r+1,…,r+m/2−1modm and Rar1,ar2=Vℚm2−ah1|h≡r+1,r+2,…,r+m/2modm∪ah2|h≡r−1,r−2,…,r−m/2modm respectively.We note that∪r=1mRr=Vℚm2 and ∪r=1mRr=3m.(b)
The RNs ofar1,ar+p1,ar1,ar+s1,ar2,ar+s2,ar1,ar+p2,ar1,ar+s2,ar1,ar−p1 and ar1,ar−s2 are given by: R¯1=Rar1,ar+p1=Vℚm2−ah2|h≡r+p−1/2,r+m+p−1/2modm, R¯2=Rar1,ar+s1=Vℚm2−ah1|h≡r+s/2,r+m+s/2modm, R¯3=Rar2,ar+s2=Vℚm2−ah2|h≡r+s/2,r+m+s/2modm, R¯4=Rar1,ar+p2=Vℚm2−ah1|h≡r+p+1/2modm∪ah2|h≡r−p−1/2modm, R¯5=Rar1,ar+s2=Vℚm2−ah1|h≡r+m+s/2modm∪ah2|h≡r+m+s/2modm, R¯6=Rar1,ar−p2=Vℚm2−ah1|h≡r−p+1/2modm∪ah2|h≡r+m+s/2modm, R¯7=Rar1,ar−s2=Vℚm2−ah1|h≡r−s/2modm∪ah2|h≡r−s/2modm, R¯8=Rar2,ar+p2=Vℚm2−ah1|h≡r+p+1/2,r+m+p+1/2modm.Clearly,R¯u=2m−1. Since Rar1ar−12=Rar1ar2=n<Rar1ar+11 then Rar1ar+11∩∪r=1mRr≥Rr.Theorem 3.
Letℵ≅ℚm2 with m≥6 and m≡0mod2. Then, dimfℚm2≤2.Proof.
Case 1. Whenm=6.The RNs are given as follows.Tables6 and 7 given above shows the RNs with cardinalities of 10 and 6 respectively. Clearly, the table shows the RNs bearing the minimum cardinality of 6 and ∪r=112Rr=Vℚ62 which implies ∪r=112Rr=12 and Ru¯∩∪r=112Rr≥Rr, where 1≤u≤52.Now, we define a functionκ:Vℚ62⟶0,1 such that κar1=κar2=1/6. Moreover, Rr for 1≤r≤12 of ℚ62 are pairwise overlapping; thus, ∃ is another minimal resolving function κ¯ of ℚ62 such that κ¯≤κ. As a result, dimfℚ62≤∑r=1121/6=2.Case 2.m≥8It is clear from Lemma3 that the RNs bearing the minimum cardinality of m in ℚm2 are Rl=Ral1,al2 and Rr=Rar1,ar−12 respectively, with R¯u∩∪r=1mRr=2n≥Rr.Letη=∪r=1mRr=2m and λ=Rr=m. Then we define a mapping κ:Vℚm2⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈Vℚm2−∪r=1mRr. We see that κ is a resolving function for ℚm1 because κRa,b≥1∀a,b⊂Vℵ. Now suppose on contrary that there is another resolving function ρ, such that ρu≤ϕu, for at least one u∈Vℚm1ρu≠κu. As a result, ρRa,b<1, where Ra,b is a RN of ℚm1 having minimum cardinality λ. This shows that ρ is not a resolving function which is a contradiction to our supposition. Therefore, κ is a minimal resolving function that achieves minimum κ for ℚm1. Since all the RNs of ℚm2 are equal, hence ∃ another minimal resolving function κ¯ of ℚm2 such that κ¯≤κ\enleadertwodots Assigning 1/λ to all the vertices of ℚm2 and calculating their sum, we get dimfℚm2≤∑r=1η1/m=2.Corollary 1.
Letℵ≅ℚm3 with m≥6 and m≡0mod2. Then, dimfℚm3≤2.Proof.
The result is evident asℚm3≅ℚm2.


### 5.3. Pentagonal Circular Ladders

Lemma 4.
Letℵ≅ℙm1 be the 2-faced QCL with m≥10 and m≡0mod2. For 1≤l,r≤m, p≥3, s≥2p≡1mod2 and s≡0mod2. Then,(a)
Rl=Ral1,al−23=Rr=Rar1,ar+13=5m/2−2 with am+13=a13, ∪l=1mRl=∪r=1mRr=3m and ∪l=1mRl∪∪r=1mRr=3m.(b)
Rr<R¯1=Rar2,ar3=R¯2=Rar2,ar−13=5m/2 with a03=am3 and Ru∩∪r=1mRr≥Rr.(c)
Rr<R¯3=Rar1,ar+11=R¯4=Rar1,ar+p1=R¯5=Rar2,ar+12=R¯6=Rar2,ar+p2=R¯7=Rar3,ar+s3=R¯8=Rar1,ar+s2=3m−2 and R¯u∩∪r=1mRr≥Rr with am+11=a11.(d)
Rr<R¯9=Rar1,ar2=R¯10=Rar1,ar+s1=R¯11=Rar3,ar+13=R¯12=Rar3,ar+p3=R¯13=Rar2,ar+s2=R¯14=Rar1,ar3 and R¯u∩∪r=1mRr≥Rr.(e)
Rr<R¯15=Rar1,ar+12=R¯16=Rar1,ar−12 and R¯u∩∪r=1mRr≥Rr.Proof.
(a)
The RNs ofar1,ar+13 and al1,al−22 are Rl=Rar1,ar+13=Vℙm1−ah1|h≡r+2,r+3,…,r+m/2modm∪ar+42∪ah3|h≡r,r+4modm and R2=Ral1,al−23=Vℙm1−Vℙm1−ah1|h≡r+2,r+3,…,r+m/2modm∪ar−42∪ah3|h≡r,r−5modm, respectively. We can see that Rr=5m/2−2 and ∪r=1mRr=∪l=1mRl=Vℙm1 and ∪l=1mRl∪∪r=1mRr=3m.(b)
The RNs ofar2,ar−13 and ar2,ar3 are R¯1=Rar2,ar−13=Vℙm1−ah1|h≡r−1,r−2,…,r−m/2modm and R¯1=Rar2,ar3=Vℵ−ah1|h≡r+1,r+2,…,r+m/2modm mod m}, respectively.Since for1≤u≤2R¯u>Rr=5m/2−1, thus R¯u∩∪r=1mRr=3m.(c)
The RNs ofar1,ar+11,ar1,ar+p1,ar2,ar+12,ar2,ar+p2,ar3,ar+s3 and ar1,ar+s2 are R¯3=Rar1,ar+11=Vℙm1−ah3|h≡r,r+m/2modm=R¯5=Rar2,ar+12, R¯4=Rar1,ar+p1=Vℙm1−ah3|h≡r+p−1/2,r+p+m−1/2modm=R¯6=Rar2,ar+p2, R¯7=Rar3,ar+s3=Vℙm1−ah3|h≡r+s/2,r+m+s/2modm and R¯8=Rar1,ar+s2=Vℙm1−ah1|h≡r+s/2modm∪ah3|h≡r+m+s−2/2modm, respectively.Clearly, for3≤u≤8,R¯u=3m−2. Since Rar2,ar3|=5m/2<Rar1ar+11, then R¯u∩∪r=1mRr=3m−2≥Rr.(d)
The RN ofar1,ar2 is R¯9=Rar1,ar2=Vℙm1−ah2|h≡r−1,r+1modm∪ah3|h≡r−2,r+1modm. R¯10=Rar1,ar+s1=Vℙm1−ah1||h≡r+s/2r+m+s/2modm∪ah2||h≡r+s/2,r+m+s/2modm,R¯11=Rar3,ar+13=ar1,ar+21, R¯12=Rar3,ar+p3=R¯13=Rar2,ar+s2=Rar1,ar+s1, R¯14=Rar1,ar3=Vℙm1−ah2|h≡r,r+2modm∪ah3|h≡r−1,r+2modm.Clearly, for9≤u≤14,Rr=5m/2<R¯u and R¯u∩∪r=1mRr≥Rr.(e)
The RNs ofar1,ar+12 and ar1,ar−12 are R¯15=Rar1,ar+12=Vℙm1−ah1|h≡r+1,r+2,…,r+m/2mod5∪ah2|h≡r+m/2modm and R¯16=Rar1,ar−12=Vℙm1−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2h≡r−m/2modm, respectively. We can see that Rr=5m/2 and R¯u∪r=1mRr≥Rr.Theorem 4.
Ifℵ≅ℙm1 with m≥8 and m≡0mod2. Then, dimfℙm1≤6m/5m−2for6≤m≤8,6m/5m−4form≥10.Proof.
Case 1. (i) Whenm=6.The RNs are given as follows.We can see that Tables8 and 9 show the RNs with maximum cardinality of 16 and 15, respectively, whereas both Tables 10 and 11 show the RNs with minimum cardinality of 14. Moreover, ∪r=145Rr=Vℙ61 which implies ∪r=145Rr=18 and Rr¯∩∪r=124Rr≥Rr.Now, we define a mappingκ:Vℙ61⟶0,1 such that κar1=κar2=κar3=1/14. It is observed that Rr for 1≤r≤45 of ℙ61 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙ61 such that κ¯≤κ. As a result, dimfℙ61≤∑r=1161/14=18/14=9/7.(ii)
Whenm=8.The RNs are given as follows.We can see that Tables12–14 shows the RNs with maximum cardinalities that is 22 and 20, respectively, whereas Table 15 shows the RNs with minimum cardinality of 19. Moreover, ∪r=116Rr=Vℙ81 which implies ∪r=116Rr=19 and Rr¯∩∪r=124Rr≥Rr.Now, we define a mappingκ:Vℙ81⟶0,1 such that κar1=κar2=κar3=1/19. It is observed that Rr for 1≤r≤16 of ℙ81 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙ81 such that κ¯≤κ. As a result, dimfℙ81≤∑r=1161/19=24/19.(iii)
For6≤m≤8.We have seen thatdimfℙ61≤9/7 whereas dimfℙ81≤24/19. In generic form, we can write it as dimf≤6m/5m−2.Case 2. Whenm=10.The RNs are given as follows.We can see that Tables16–18 show the RNs with maximum cardinalities, that is, 28 and 26 and 25, respectively, whereas Table 19 shows the RNs with minimum cardinality of 23. Moreover, ∪r=116Rr=Vℙ101 which implies ∪r=116Rr=23 and Rr¯∩∪r=124Rr≥Rr.Now, we define a mappingκ:Vℙ101⟶0,1 such that κar1=κar2=κar3=1/23. It is observed that Rr for 1≤r≤16 of ℙ101 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙ101 such that κ¯≤κ. As a result, dimfℙ101≤∑r=1231/19≤30/23.Case 3.m≥12.From Lemma4, we can see that the RNs with minimum cardinality of 5m/2−2 are Rar1,ar+13 and Rar1,ar−23, respectively. Moreover, for 1≤u≤15,Ru∩∪r=1mRr≥Rr.Letη=∪r=1mRr=3m and λ=Rr=5m/2−2. Then, we define a mapping κ:Vℙm1⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈Vℙm1−∪r=1mRr. We note that κ is a resolving function for ℙm1 because κRa,b≥1∀a,b∈Vℙm1. Now, assume that there is another resolving function ρ, such that ρu≤κu, for at least one u∈Vℙm1ρu≠κu. As a result, ρRa,b<1, where Ra,b is a RN of ℙm1 bearing minimum cardinality λ. This shows that ρ is not a resolving function a contradiction. Therefore, κ is a minimal resolving function for ℙm1 that achieves minimum κ. As a matter of fact Rr for 1≤t≤2m of ℙm1 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙm1 such that κ¯≤κ. Therefore, assigning 1/λ to all the vertices of ℙm1 and calculating their sum, we get: dimfℙm1≤∑r=1η1/λ=3m/5m/2−2=6m/5m−4.Lemma 5.
Letℵ≅ℙm2 be the 3-faced QCL with m≥6 and m≡0mod2. For 1≤l,r≤m, p≥3, s≥2p≡1mod2 and s≡0mod2. Then,(a)
Rl=Ral2,al3=Rr=Rar2,ar−13=3m/2+1, ∪l=1mRl, ∪r=1mRr=3m and ∪l=1mRl∪∪r=1mRr=3m.(b)
Rr<R¯1=Rar1,ar+13=R¯2=Rar1,ar−13 and R¯u∩∪r=1mRr≥Rr.(c)
Rr<R¯3=Rar1,ar+11=R¯4=Rar1,ar+p1=R¯5=Rar2,ar+12=R¯6=Rar2,ar+p2=R¯7=Rar3,ar+s3=R¯8=Rar1,ar+s2=R¯9=Rar1,ar+s3=R¯10=Rar1,ar+p3 and R¯u∩∪r=1mRr≥Rr with am+11=a11 and am+12=a12.(d)
Rr<R¯11=Rar2,ar+p3=R¯12=Rar2,ar+s3 and R¯u∩∪r=1mRr≥Rr.(e)
Rr<R¯13=Rar1,ar2 and R¯u∩∪r=1mRr≥Rr.Proof.
(a)
The RNs ofar2ar3 and ar2ar−13 are Rar2ar3=Vℙm2−ah1|h≡r+1,r+2,…,r+m/2modm∪ah2|h≡r+1,r+2,…,r+m/2modm∪ah3|h≡r+1,r+2,…,r+m/2−1modm and Rar2ar−13=Vℙm2−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2|h≡r−1,r−2,…,r−m/2modm∪ah3|h≡r−2,r−3,…,r−m/2+1modm. We note that ∪r=1mRr=Vℙm2 and ∪r=1mRr=3m.(b)
The RNs ofar1,ar−13 and ar1,ar−13 are R¯1=Rar1,ar+13=Vℙm2−ah1|h≡r+2,r+3,…,r+m/2modm∪ah3|h≡r,r+m/2modm and R¯2=Rar1,ar−13=Vℙm2−ah1|h≡r−2,r−3,…,r−m/2modm∪ah3|h≡rmodm respectively. It is clear from the above that Rr<R¯u and R¯u∩∪r=1mRr≥Rr.(c)
The RNs ofar1,ar+11,ar1,ar+p1,ar2,ar+12,ar2,ar+p2,ar3,ar+s3,ar1,ar+s2,ar1,ar+s3 and ar1,ar+p3 are R¯3=Rar1,ar+11=Vℙm2−ah3|h≡r,r+m/2modm=R¯5=Rar2,ar+12, R¯4=Rar1,ar+p1=Vℙm2−ah3|h≡r+p−1/2,r+p+m−1/2modm=R¯6=Rar2,ar+p2, R¯7=Rar3,ar+s3=Vℙm2−ah3|h≡r+s/2,r+m+s/2modm, R¯8=Rar1,ar+s2=Vℙm2−ah3|h≡r+s/2,r+s+m/2modm, R¯9=Rar1,ar+s3=Vℙm2−ah1|h≡r+s+2/2modm∪ah3|h≡r+s+m/2modm, R¯10=Rar1,ar+p3=Vℙm2−ah1|h≡r+p+m+1/2modm∪ah3|h≡r+p−1/2modm, Clearly, R¯u=3m−2. Since Rr=3m/2+1<R¯u, then R¯u∩∪r=1mRr=3m−2≥Rr.(d)
The RNs ofar2,ar+p3 and ar2,ar+s3 are R¯11=Rar2,ar+p3=Vℙm2−ah1|h≡r+p+1/2modm∪ah2|h≡r+p+1/2modm∪ah3|h≡r+p+m+1/2modm and R¯12=Rar2,ar+s3=Vℙm2−ah1|h≡r+m+s/2modm∪ah2|h≡r+m+s/2modm∪ah3|h≡r+s/2modmClearly,R¯u=3m−1. Since Rr=3m/2+1<R¯u, then R¯u∩∪r=1mRr≥Rr.(e)
The RN ofar1,ar2 is R¯13=Rar1,ar2=Vℙm2.Clearly,Rar2ar3=3m/2+1<Rar1ar2 and Rar1ar2∩∪r=1mRr=3n≥Rr.Theorem 5.
Ifℵ≅ℙm2 with m≥6 and m≡0mod2. Then, dimfℙm2≤6m/3m+2.Proof.
Case 1. Whenm=6.The RNs are given as follows.Tables20–22 show the RNs with maximum cardinality of 18, 16, and 14, respectively. On the other hand, Table 23 bears RNs with the minimum cardinality of 10. Also, it is observed that ∪r=112Rr=Vℙ62 which implies ∪r=112Rr=18 and Rr¯∩∪r=16Rr≥Rr.Now, we define a functionκ:Vℙ62⟶0,1 such that κar1=κar2=κar3=1/10. As Rr for 1≤t≤12 of ℙ62 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙ62 such that κ¯≤κ. As a result, dimfℙ62≤∑r=1181/10=9/5.Case 2.m≥8.As Lemma5 clears the fact that among the RNs of ℙm2, those that have the minimum cardinality of 3m/2+1 are Rar2,ar3 and Rar2,ar−13, respectively. Also, R¯u∩∪r=1mRr≤Rr, where 1≤u≤17.Letη=∪r=1mRr=3m and λ=Rr=3m/2+1. Then, we define a function κ:Vℙm2⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈Vℙm2−∪r=1mRr. It is seen that κ is a resolving function for ℙm2 because κRa,b≥1∀a,b⊂Vℙm2. Now, suppose on contrary that there is another resolving function ρ, such that ρu≤κu, for at least one u∈Vℙm2ρu≠κu. As a result, ρRa,b<1, where Ra,b is an RN of ℙm2 having minimum cardinality λ. This shows that ρ is not a resolving function, a contradiction. Thus, κ is a minimal resolving function for ℙm2 which attains minimum κ. Also, Rr for 1≤t≤2m of ℙm2 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙm2 such that κ¯≤κ. Therefore, assigning 1/λ to all the vertices of ℙm2 and calculating their sum, we get dimfℙm2≤∑r=1η1/λ=3m/3m+2/2=6m/3m+2.Lemma 6.
Letℵ≅ℙm3 be the 3-faced QCL with m≥6 and m≡0mod2. For 1≤r≤m, p≥3, s≥2p≡1mod2 and s≡0mod2. Then,(a)
Rl=Rar1,ar2=Rr=2m−1, ∪r=1mRl=3m and ∪r=1mRr=3m.(b)
Rr<R¯1=Rar1,ar+11=R¯2=Rar1,ar+p1=R¯3=Rar2,ar+12=R¯4=Rar2,ar+p2=R¯5=Rar3,ar+13=R¯6=Rar3,ar+p3=R¯7=Rar1,ar+s1=R¯8=Rar1,ar+13=R¯9=Rar1,ar−13=R¯10=Rar1,ar3 and R¯u∩∪r=1mRr≥Rr with am+11=a11 and am+12=a12.(c)
Rr<R¯11=Rar2,ar+s2=R¯12=Rar3,ar+s3=R¯13=Rar1,ar+p3=R¯14=Rar1,ar+p3R¯u∩∪r=1mRr≥Rr.(d)
Rr<R¯15=Rar2,ar+s2=R¯16=Rar3,ar+s3=R¯17=Rar1,ar+p3=R¯18=Rar1,ar+p3R¯u∩∪r=1mRr≥Rr.(e)
Rr<R¯19=Rar1,ar−12 and Rar1,ar−12∩∪r=1mRr≥Rr.(f)
Rr<R¯20=Rar2,ar−13 and Rar2,ar−13∩∪r=1mRr≥Rr.(g)
Rr<R¯21=Rar2,ar3 and Rar2,ar3∩∪r=1mRr≥Rr.Proof.
(a)
The RN ofar1,ar2 is Rar1ar2=Vℙm3−ah1|h≡r+1,r+2,…,r+m/2modm∪ah2|h≡r+1,r+2,…,r+m/2−1modm∪ah3|h≡r+1,r+2modm.We note that∪r=1mRr=Vℙm3 and ∪r=1mRr=3m.(b)
The RNs ofar1,ar+11,ar1,ar+p1,ar2,ar+12,ar2,ar+p2,ar3,ar+13,ar3,ar+p3,ar1,ar+s1,ar1,ar+13, ar1,ar−13, and ar1,ar3 are R¯1=Rar1,ar+11=Vℙm3−ah2|h≡r,r+m/2modm∪ah3|h≡r,r+m/2modm, R¯2=Rar1,ar+p1=Vℙm3−ah2|h≡r+p−1/2,r+p+m−1/2modm∪ah3|h≡r+p−1/2,r+p+m−1/2modm, R¯3=Rar2,ar+12=Vℙm3−ah1|h≡r+1,r+m+2/2modm∪ah3|h≡r,r+m/2modm, R¯4=Rar2,ar+p2=Vℙm3−ah1|h≡r+p+1/2,r+p+m+1/2modm∪ah3|h≡r+p+1/2,r+p+m+1/2modm, R¯5=Rar3,ar+13=Vℙm3−ah1|h≡r+1,r+m+2/2modm∪ah2|h≡r+1,r+m+2/2modm, R¯6=Rar3,ar+p3=Vℙm3−ah1|h≡r+p+1/2,r+p+m+1/2modm∪ah2|h≡r+p+1/2,r+p+m+1/2modm, R¯7=Rar1,ar+s1=Vℙm3−ah1|h≡r+s/2,r+s+m/2modm∪ah3|h≡r+s−2/2,r+s+m−2/2modm=R¯8=Rar1,ar+13, R¯9=Rar1,ar−13=Vℙm3−ah1|h≡r−s/2,r−s+m/2modm∪ah3|h≡r−s−2/2,r−s+m−2/2modm, R¯10=Rar1,ar3=Vℙm3−ah2|h≡r,r+2modm∪ah3|h≡r−1,r+2modm, Clearly, R¯u=3m−4. Since Rr=3m/2+1<R¯u, then R¯u∩∪r=1mRr=3m−2≥Rr.(c)
The RN ofar1,ar+p2, ar1,ar+s2, ar1,ar−p2, and ar1,ar−s2 are R¯11=Rar1,ar+p2=VG−ah1|h≡r+p+1/2modm∪ah2|h≡r+p+m−1/2modm∪ah3|h≡r+p−1/2modm, R¯12=Rar1,ar+s2=VG−ah1|h≡r+m+s/2modm∪ah2|h≡r+s/2modm∪ah3|h≡r+m+s−2/2modm, R¯13=Rar1,ar−p2=VG−ah1|h≡r−p+1/2modm∪ah2|h≡r−p+m−1/2modm∪ah3|h≡r−p−1/2modm, R¯14=Rar1,ar+s2=VG−ah1|h≡r−m+s/2modm∪ah2|h≡r−s/2modm∪ah3|h≡r−m+s−2/2modm, Clearly, R¯u=3m−1. Since Rr=3m/2+1<R¯u, then R¯u∩∪r=1mRr=3m−2≥Rr.(d)
The RNs ofar2,ar+s2, ar3,ar+s3, ar1,ar+p3, and ar1,ar+p3 are R¯15=ar2,ar+s2=VG−ah2|r≡r+s/2,r+s+m−2/2modm, R¯16=ar3,ar+s3=VG−ah3|r≡r+s/2,r+s+m−2/2modm, R¯17=ar1,ar+s3=VG−ah1|r≡r+s+2/2,r+s+m+2/2modm, R¯18=ar1,ar+p3=VG−ah2|r≡r+p+1/2modm∪ah1|r≡r+p+m+1/2,modm, Clearly, R¯u=3m−2. Since Rr=3m/2+1<R¯u, then R¯u∩∪r=1mRr=3m−2≥Rr.(e)
The RNs ofar1,ar+12 and ar1,ar−12 are R¯19=Rar1ar−12=Vℙm3−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2|h≡r−2modm∪ah3|h≡r−3modm.Clearly,Rar1ar2=2m−1<Rar1ar−12 and Rar1ar−12∩∪r=1mRr=5m−4/2≥Rr.(f)
The RN ofar2ar−13 is given by R¯20=Rar2ar−13=Vℙm3−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2|h≡r−1modm.Clearly,Rar1ar2=2m−1<Rar2ar−13 and Rar2ar−13∩∪r=1mRr=5m−2/2≥Rr.(g)
The RN ofar2ar3 is given by R¯21=Rar2,ar3=Vℙm3−ah2|h≡r−1,r−2,…,r−m/2+1modm.Clearly,Rar1ar2=2m−1<Rar2,ar3 and Rar2ar−13∩∪r=1mRr=5m+2/2≥Rr.Theorem 6.
Ifℵ≅ℙm3 with m≥6 and m≡0mod2. Then, dimfℙm3≤3m/2m−1.Proof.
Case 1. Whenm=6The RNs are given as follows.We note that the cardinalities of RNs as shown in Tables24–26 are 16, 14, and 13, respectively. Similarly, the cardinality of RNs given in Table 27 is 11 which is less than the least among all the RNs. Moreover, ∪r=16Rr=Vℙ63 which implies ∪r=16Rr=18 and Rr¯∩∪r=16Rr≥Rr.Now, we define a functionκ:Vℙ63⟶0,1 such that κar1=κar2=κar3=1/11. Furthermore, Rr for 1≤t≤6 of ℙ63 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙ63 such that κ¯≤κ. As a result, dimfℙ63≤∑r=1181/11=18/11.Case 2.m≥8We have seen from Lemma6Rar1,ar2=2m−1≤Ra,b for all a,b⊂Vℙm3 andRa,b∩∪r=1mRr≥Rr.Letη=∪r=1mRr=3m and λ=Rr=2m−1. Then, we define a function κ:Vℙm3⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈Vℙm3−∪r=1mRr. We see that κ is a resolving function for ℙm3 because κRa,b≥1∀uv∈Eℙm3. Now, suppose on contrary that there is another resolving function ρ, such that ρu≤κu, for at least one u∈Vℙm3ρu≠κu. As a result, ρRa,b<1, where Ra,b is an RN of ℙm3 having the minimum cardinality of λ . This shows that ρ is not a resolving function, a contradiction. Thus, κ is a minimal resolving function for ℙm3 that achieves minimum κ. Moreover, Rr for 1≤r≤m of ℙm3 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙm3 such that κ¯≤κ. Therefore, assigning 1/λ to all the vertices of ℙm3 and calculating their sum, we get dimfℙm3≤∑r=1η1/λ=3m/2m−1=3m/2m−1.Corollary 2.
Ifℵ≅ℙm4 with m≥6 and m≡0mod2. Then, dimfℙm4≤3m/2m−1.Proof.
The above assertion is valid asℙm4≅ℙm3.Lemma 7.
Letℵ≅ℙm5 be the 4-faced QCL with m≥6 and m≡0mod2. For 1≤l,r≤m, p≥3, s≥2p≡1mod2 and s≡0mod2. Then,(a)
Rl=Ral1,al2=Rr=Rar2,ar−13=3m/2+1 with a03=am3,a02=am2, ∪l=1mRl=3m, ∪r=1mRr=3m and ∪l=1mRl∪∪r=1mRr=3m,(b)
Rr<R¯1=Rar1,ar3=R¯2=Rar2,ar+13=R¯3=Rar2,ar−13 and R¯u∩∪r=1mRr≥Rr.(c)
Rr<R¯4=Rar1,ar+s1=R¯5=Rar2,ar+12=R¯6=Rar2,ar+p2=R¯7=Rar1,ar+s3 with am+11=a11 and R¯u∩∪r=1mRr≥Rr.(d)
Rr<R¯8=Rar1,ar+11=R¯9=Rar1,ar+p1=R¯10=Rar2,ar+p3=R¯11=Rar2,ar−p3 and R¯u∩∪r=1mRr≥Rr.(e)
Rr<R¯12=Rar2,ar+s2=R¯13=Rar1,ar+p2=R¯14=Rar1,ar+s2=R¯15=Rar1,ar+p3 and R¯u∩∪r=1mRr≥Rr.(f)
Rr<R¯16=Rar2,ar3 and Rar2,ar3∩∪r=1mRr≥Rr.Proof.
(a)
The RNs ofal1,al2 and ar2ar−13 are Ral1,al2=Vℙm5−ah1|h≡l+1,l+2,…,l+m/2modm∪ah2|h≡l−1,l−2,…,l−m/2modm∪ah3|h≡l−1,l−2,…,l−m/2+1modm and Rar2ar−13=Vℙm5−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2|h≡r−1,r−2,…,r−m/2modm∪ah3|h≡r−1,r−2,…,r−m/2+1modm, respectively. We note that ∪r=1mRr=Vℙm5 and ∪r=1mRr=3m.(b)
The RNs ofar1,ar3,ar2,ar+13 and ar2,ar−13 are R¯1=Rar1,ar3=Vℙm5−ah1|h≡r+2,r+3,…,r+m/2modm∪ah2|h≡rmodm∪ah3|h≡r−1modm, R¯2=Rar2,ar+13=Vℙm5−ah1|h≡r+2,r+3,…,r+m+2/2modm∪ah2|h≡rmodm∪ah3|h≡r+m/2modm and R¯3=Rar2,ar−13=Vℙm5−ah1|h≡r−2,r−3,…,r−m+2/2modm∪ah2|h≡rmodm∪ah3|h≡r−m/2modm, respectively. We can see that R¯u=5m/2+2. Since Rar1ar2=Rar2ar−13=3m/2+1<R¯u, then R¯u∩∪r=1mRr=3m−4≥Rr, where 1≤u≤3.(c)
The RNs ofar1,ar+s1,ar2,ar+p2 and ar1,ar+s3 are R¯4=Rar1,ar+s1=Vℙm5−ah2|h≡r+s/2,r+m+s/2modm∪ah3|h≡r+s−2/2,r+s+m−2/2modm,R¯5=Rar2,ar+12=Rar1,ar+21, for s≥4, R¯6=Rar2,ar+p2=Rar1,ar+s1, R¯7=Rar1,ar+s3=Vℙm5−ah1|h≡r+s+2/2,r+m+s/2modm∪ah2|h≡r+s/2,r+m+s/2modm, Clearly, R¯u=3m−4. Since Rar1,ar2=Rar2,ar−13=3m/2+1<R¯u, then R¯u∩∪r=1mRr=3m−4≥Rr.(d)
The RNs ofar1,ar+p1,ar2,ar+p3 and ar2,ar−p3 are R¯8=Rar1,ar+11=Vℙm5−ah2|h≡r,r+m/2modm∪ah3|h≡rmodm, R¯9=Rar1,ar+p1=Vℙm5−ah2|h≡r+p−1/2,r+p+m−1/2modm∪ah3|h≡r+p−1/2modm, R¯10=Rar2,ar+p3=Vℙm5−ah1|h≡r−1modm∪ah2|h≡r+p+1/2modm∪ah3|h≡r+p+m−1/2modm and R¯11=Rar2,ar−p3=Vℙm5−ah1|h≡r−1modm∪ah2|h≡r−p+1/2modm∪ah3|h≡r−p+m+1/2modm, respectively. We can see from these sets that R¯u=3m−1>Rr and R¯u∩∪r=1mRr=3m−4≥Rr.(e)
The RNs ofar2,ar+s2,ar1,ar+p2,ar1,ar+s2,ar1,ar+p3 and Rar1,ar+s3| are R¯12=Rar2,ar+s2=Vℙm5−ah2|h≡r+s/2,r+s+m/2modm, R¯13=Rar1,ar+p2=Vℙm5−ah2|h≡r+p−1/2modm∪ah3|h≡r+p+m−1/2modm, R¯14=Rar1,ar+s2=Vℙm5−ah1|h≡r+m+s/2modm∪ah3|h≡r+s−2/2modm, R¯15=Rar1,ar+p3=Vℙm5−ah1|h≡r+p+1/2modm∪ah3|h≡r+p−1/2modm, respectively. We note that R¯u=3m−2>Rr and R¯u∩∪r=1mRr=3m−2≥Rr.(f)
The RN ofar2,ar3 is R¯16=Rar2,ar3=Vℙm5−ah2|h≡r+1,r+2,…,r+m/2modm∪ah3|h≡r+1,r+2,…,r+m/2−1modm.Clearly,Rar1ar2=Rar2ar−13=3m/2+1<Rar2ar3 and Rar2ar3∩∪r=1mRr=2m+1≥Rr.Theorem 7.
Ifℵ≅ℙm5 with m≥6 and m≡0mod2. Then, dimfℙm5≤6m/3m+2.Proof.
Case 1. Whenm=6.The RNs are given as follows.In Tables28–31, the cardinalities of RNs are 16, 15, 14, and 13, respectively. In the same manner, Table 32 is having RNs withholding the cardinality of 10. Moreover, ∪r=112Rr=Vℙ65 which implies ∪r=112Rr=18 and Rr¯∩∪r=112Rr≥Rr.Now, we define a functionκ:Vℙ65⟶0,1 such that κar1=κar2=κar3=1/10. We have seen that Rr for 1≤t≤12 of ℙ65 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙ65 such that κ¯≤κ. As a result, dimfℙ65≤∑r=1181/10=9/5.Case 2.m≥8.It can be seen from Lemma7 that among all the RNs of ℙm5, Rar1ar2=Rar2ar−13=3m/2+1≤Ra,b for all a,b⊂Vℙm5 andRa,b∩∪r=1mRr=3m/2+1≤Rr.Letη=∪r=1mRr=3m and λ=Rr=3m/2+1. Then, we define a function κ:Vℙm5⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈Vℙm5−∪r=1mRr. It is found that κ is a resolving function for ℙm5 because κRa,b≥1∀a,b⊂Vℙm5. On contrary, assume that there is another resolving function ρ, such that ρu≤κu, for at least one u∈Vℙm5ρu≠κu. As a result, ρRa,b<1, where Ra,b is an RN of ℙm5 having the minimum cardinality of λ . This shows that ρ is not a resolving function, a contradiction. Thus, κ is a minimal resolving function for ℙm5 that achieves minimum κ. Moreover, Rr for 1≤t≤2m of ℙm5 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙm5 such that κ¯≤κ. Therefore, assigning 1/λ to all the vertices of ℙm5 and calculating their sum, we get dimfℙm5≤∑r=1η1/λ=3m/3m+2/2=6m/3m+2.Corollary 3.
Ifℵ≅ℙm6 with m≥6 and m≡0mod2. Then, dimfℙm6≤6m/3m+2.Proof.
This can be derived easily asℙm6≅ℙm5.Lemma 8.
Letℵ≅ℙm7 be the 4-faced QCL with m≥6 and m≡0mod2. For 1≤l,r≤m, p≥3, s≥2p≡1mod2 and s≡0mod2. Then,(a)
Rl=Ral1,al3=Rr=Rar1,ar−13, ∪l=1mRl=3m, ∪r=1mRr=3m and ∪l=1mRl∪∪r=1mRr=3m.(b)
Rr<R¯1=Rar1,ar+11=R¯2=Rar1,ar+p1=R¯3=Rar2,ar+12=R¯4=Rar2,ar+p2=R¯5=Rar1,ar+s2=R¯6=Rar1,ar+p2=R¯7=Rar2,ar+s3 with am+11=a11 and R¯u∩∪r=1mRr≥Rr.(c)
Rr<R¯8=Rar1,ar2=R¯9=Rar1,ar+s1=R¯10=Rar2,ar+s2=R¯11=Rar3,ar+13R¯12=Rar3,ar+p3 and R¯u∩∪r=1mRr≥Rr.(d)
Rr<R¯13=Rar1,ar+p3=R¯14=Rar1,ar+s3 and R¯u∩∪r=1mRr≥Rr.(e)
Rr<R¯15=Rar2,ar3 and Rar2ar3∩∪r=1mRr≥Rr.(f)
Rr<R¯16=Rar2,ar−13 and Rar2,ar−13∩∪r=1mRr≥Rr.(g)
Rr<R¯17=Rar2,ar+p3 and R¯17∩∪r=1mRr≥Rr.(h)
Rr<R¯18=Rar1,ar+12 and R¯18∩∪r=1mRr≥Rr.Proof.
(a)
The RNs ofal1al3 and ar1ar−13 are Rl=Ral1,al3=Vℙm7−ah1|h≡l+1,l+2,…,l+m/2modm∪ah2|h≡l,l+2,…,r+m/2modm∪ah3|t≡l+1,l+2,…,l+m/2modm and Rr=Rar1ar−13=Vℙm7−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2|h≡r,r−2,…,r−m/2modm∪ah3|t≡r−1,r−2,…,r−m/2modm, respectively. Clearly, Rr=3/2m+1 for 1≤r≤m.We note that∪r=1mRr=Vℙm7 and ∪r=1mRr=3m.(b)
The RNs ofar1,ar+11,ar1,ar+p1,ar2,ar+p2,ar2,ar+p2,ar1,ar+s2,ar1,ar+p2, and ar2,ar+s3| are R¯1=Rar1,ar+11=VG−ah3|h≡r,r+m/2modm=R¯3=Rar2,ar+12, R¯2=Rar1,ar+p1=VG−ah3|h≡r+p−1/2,r+p+m−1/2modm=R¯4=Rar2,ar+p2=R¯5=Rar1,ar+s2=VG−ah2|h≡r+s/2modm∪ah3|h≡r+m+s−2/2modm, R¯6=Rar1,ar+p2=VG−ah1|h≡r+p+1/2,r+p+m+1/2modm and R¯7=Rar2,ar+s3=VG−ah1|h≡r+s/2modm∪ah3|h≡r+m+s/2modm, respectively. We can see that Rr<R¯u=3m−2 and R¯u∩∪r=1mRr≥Rr.(c)
The RNs ofar1,ar2,ar1,ar+s1,ar2,ar+s2,ar3,ar+13and ar3,ar+p3 are R¯8=Rar1,ar2=Vℙm7−ah2|h≡r−1,r+1modm∪ah3|h≡r,r−1modm, R¯9=Rar1,ar+s1=Vℙm7−ah1|h≡r+s/2,r+s+m/2modm∪ah2|h≡r+s/2,r+s+m/2modm=R¯10=ar2,ar+s2, R¯11=Rar3,ar+13=Rar2,ar+22 and R¯12=Rar3,ar+p3=Rar2,ar+s2 where s≥4, respectively.Clearly,Rr=3m−4<Rar1,ar2 and Rar1,ar2∩∪r=1mRr=3m−4≥Rr.(d)
The RNs ofar1,ar+p3 and ar1,ar+s3 are R¯13=Rar1,ar+p3=VG−ah1|h≡r+p+1/2modm∪ah2|h≡r+p+1/2modm∪ah3|h≡r+p+m−1/2modm and R¯14=Rar1,ar+s3=VG−ah1|h≡r+m+s/2modm∪ah2|h≡r+m+s/2modm∪ah3|h≡r+s/2modm, respectively. Clearly, Rr=3m−1<Rar1ar2 and Rar1ar2∩∪r=1mRr=3m−4≥Rr.(e)
The RN ofar2,ar3 is R¯15=Rar2,ar3=Vℙm7−ah1|h≡r,r−1,…,r−m/2+1modm.Clearly,Rr=5m/2<Rar2ar3 and Rar2ar3∩∪r=1mRr=5m+2/2≥Rr.(f)
The RN ofar2,ar−13 is R¯16=Rar2,ar3=Vℙm7−ah1|h≡r,r+1,…,r+m/2−1modm.Clearly,Rr=5m/2<Rar2,ar3 and Rar2,ar3∩∪r=1mRr=5m+2/2≥Rr.(g)
The RN ofar2,ar+p3 is R¯17=Rar2,ar+p3=VG−ah1|h≡r+p+m−1/2modm. Clearly, Rr<R¯15=3m−1 and Rar2,ar+p3∩∪r=1mRr=5m+2/2≥Rr.(h)
The RN ofar1,ar+12 is R¯18=Rar1,ar+12=VG−ah1|h≡r+1,r+2,…,r+m/2modm∪ah2|h≡r+3modm∪ah3|h≡r,r+2modm. Clearly, Rr<R¯18=3m−1 and Rar2,ar+p3∩∪r=1mRr=5m+2/2≥Rr.Theorem 8.
Ifℵ≅ℙm7 with m≥6 and m≡0mod2. Then, dimfℙm7≤6m/3m+2.Proof.
Case 1. Whenm=6.The RNs are given as follows.Tables33–37 show the RNs of ℙ67 with cardinalities 16, 15, 14, 13, and 10, respectively. Moreover, in Table 37, ∪r=112Rr=Vℙ67 which implies ∪r=112Rr=18 and Ru¯∩∪r=112Rr≥Rr = , where 1≤u≤81.Now, we define a functionκ:Vℙ67⟶0,1 such that κar1=κar2=κar3=1/10. Furthermore, Rr for 1≤r≤12 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙ67 such that κ¯≤κ. As a result, dimfℙ67≤∑r=1181/10=9/5.Case 2.m≥8.As Lemma8 clears the fact that Rar2,ar3=Rar2,ar−13=3/2m+1≤Ra,bs for all a,b⊂Vℙm7 andRa,b∩∪r=1mRr≤Rr.Letη=∪r=1mRr=3m and λ=Rr=3/2m+1. Then, we define a function κ:Vℙm7⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈Vℙm7−∪r=1mRr. We find that κ is a resolving function for ℙm7 because κRa,b≥1∀a,b⊂Vℙm7. Assume on contrary that there is another resolving function ρ, such that ρu≤κu, for at least one u∈Vℙm7. As a result, ρRa,b<1, where Ra,b is an RN of ℙm7 having the minimum cardinality of λ. This shows that ρ is not a resolving function, a contradiction. Thus, κ is a minimal resolving function for ℙm7 that achieves minimum κ. Also Rr for 1≤t≤2m of ℙm7 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙm7 such that κ¯≤κ. Therefore, assigning 1/λ to all the vertices of ℙm7 and calculating their sum, we get dimfℙm7≤∑r=1η1/λ=3m/3/2m+1=6m/3m+2.


## 5.1. Triangular Circular Ladder

Lemma 1.
Letℵ≅Tm be TCL with m≥6 and m≡0mod2. For 1≤r,l≤m, p≥1, s≥2p≡1mod2 and s≡0mod2, we have(a)
Rl=Ral1,al2=Rr=Rar1,ar−12=m+1 and ∪l=1mRl∪∪r=1mRr=2m, where a02=am2.(b)
Rr<Ru2m−1 for 1≤u≤8, where R¯1=Rar1,ar+p1,R¯2=Rar1,ar+s1,R¯3=Rar2,ar+s2,R¯4=Rar2,ar+p2,R¯5=Rar1,ar+p2,R¯6=Rar1,ar+s2,R¯7=Rar1,ar−p2,R¯8=Rar1,ar−s2 and R¯u∩∪r=1mRr≥Rr with am+11=a11.Proof.
(a)
The RNs ofar1,ar2 and ar1,ar−12 are Rl=Ral1,al2=VTm−ah1|h≡l+1,l+2,…,l+m/2modm∪ah2|h≡l+1,l+2,…,l+m/2−1modm and Rr=Rar1,ar−12=VTm−ah1|h≡r−1,r−2,…,r−m/2+1modm∪ah2|h≡r−2,r−3,…,r−m/2modm. We note that ∪l=1mRl=VTm, ∪r=1mRr=VTm and ∪l=1mRr∪∪r=1mRr=2m.(b)
The RNs ofR¯u for 1≤u≤8 are R¯1=Rar1,ar+p1=VTm−ah2|h≡r+p−1/2,r+p+m−1/2modm=R¯4=Rar2,ar+p2, R¯2=Rar1,ar+s1=VTm−ah2|h≡r+s/2,r+m+s/2modm=R¯3=Rar2,ar+s2, R¯5=Rar1,ar+p2=VTm−ah1|h≡r+p+1/2modm∪ah2|h≡r+m+p−1/2modm, R¯6=Rar1,ar+s2=VTm−ah1|h≡r+m+s/2modm∪ah2|h≡r+s/2modm, R¯7=Rar1,ar−p2=VTm−ah1|h≡r−p+1/2modm∪ah2|h≡r+m−p−1/2modm and R¯8=Rar1,ar−s2=VTm−ah1|h≡r−s/2modm∪ah2|h≡r+m+s−2/2modm, respectively.
Clearly,R¯u=2m−1. Since Rr=m+1<R¯u, then R¯u∩∪r=1mRr=2m−1≥Rr.Theorem 1.
Ifℵ≅Tm with m≥6 and m≡0mod2. Then, dimfTm≤2m/m+1.Proof.
Case 1. Whenm=6.The RNs are given as follows.We have seen from the above that Table1 shows the RNs with maximum cardinality that is 10 and Table 2 shows the RNs with minimum cardinality that is 7, which is less than the cardinalities of all other RNs Rr¯ for 1≤t≤6 that is 10. Moreover, ∪r=112Rr=VT6 which implies ∪r=112Rr=12 and Ru¯∩∪r=112Rr≥Rr, where 1≤u≤24.Now, we define a functionκ:VT6⟶0,1 such that κar1=κar2=1/7. As we can see, Rr for 1≤r≤12 of T6 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of T6 such that κ¯≤κ. As a result, dimfT6≤∑r=1121/7=12/7.Case 2.m≥8.According to Lemma1 the RNs with minimum cardinality are Rar1,ar2 and Rar1,ar−12 having the cardinality of m+1 and R¯u∩∪r=1mRr≥Rr, where 1≤u≤8.Letη=∪r=1mRr=2m and λ=Rr=m+1. Now, we define a function κ:VTm⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈VTm−∪r=1mRr. We can see that κ is a resolving function for Tm because κRa,b≥1∀a,b⊂Vℵ. Suppose that there is another resolving function ρ, such that ρx≤κx, for at least one x∈VTmρx≠κx. As a result, ρRa,b<1, where Ra,b is an RN of Tm having the minimum cardinality of λ. This shows that ρ is not a resolving function which is contradiction. Hence, κ is a minimal resolving function that achieves minimum κ for Tm. As a matter of fact Rr for 1≤r≤m of Tm are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of Tm such that κ¯≤κ. As a result, we arrive at the following: dimfTm≤∑r=1η1/λ=2m/m+1.


## 5.2. Quadrangular Circular Ladder

Lemma 2.
Letℵ≅ℚm1 be the 2-faced QCL with m≥6and m≡0mod2. For 1≤l,r≤m, p≥3, s≥2p≡1mod2 and s≡0mod2. Then,(a)
Rl=Ral1,al+12=Rr=Rar1,ar−12=m and ∪l=1mRl∪∪r=1mRr=2m.(b)
Forp≥3, s≥2, Rr<R¯1=Rar1,ar+s1=R¯2=Rar2,ar+s2=R¯3=Rar1,ar+p2=R¯4=Rar1,ar−p2=2m−2 and R¯u∩∪r=1mRr=2m, where 1≤u≤4.(c)
Forp≥3, s≥0, Rr<R¯5=Rar1,ar+p1=R¯6=Rar2,ar+p2=R¯7=Rar1,ar+s2=R¯8=Rar1,ar−s2=2m and R¯u∩∪r=1mRr=2m, where 5≤u≤8.Proof.
(a)
The RNs ofar1,ar+12 and ar1,ar−12 are given by Rl=Ral1,al+12=Vℚm1−ah1|h≡l+1,l+2,…,l+m/2modm∪ah2|h≡l,l−1,…,l−m/2+1modm and Rr=Rar1,ar−12=Vℚm1−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2|h≡r,r−1,…,r+m/2−1modm, respectively. We note that ∪l=1mRl=Vℚm1, ∪r=1mRr=Vℚm1 and ∪l=1mRl∪∪r=1mRr=2m.(b)
The RNs ofR¯u for 1≤u≤4 are as follows: R¯1=Rar1,ar+s1=Vℚm1−ah1|h≡r−s/2,r+s/2modm∪ah2|h≡r−s/2,r+s/2modm=R¯2=Rar2,ar+s2, R¯3=Rar1,ar+p2=Vℚm1−ah1|h≡r+p+1/2,r+p+m−1/2modm∪ah2|h≡r+p−1/2,r+m−p+1/2modm and R¯4=Rar1,ar−p2=Vℚm1−ah1|h≡r+p,r−p+1/2modm∪ah2|h≡r+p+1/2,r−p−1/2modm respectively. Clearly, R¯u=2m−2. Since Rr=2m−2<R¯u, then R¯u∩∪r=1mRr=2m−2≥Rr, it can be easily seen that ∪r=1mRr=Vℚm1 and ∪r=1mRr=2m.(c)
The RNs ofar1,ar+p1, ar2,ar+p2, ar1,ar+s2 and ar1,ar−p1 are R¯5=Rar1,ar+p1=R¯6=Rar2,ar+p2=R¯7=Rar1,ar+s2=R¯8=Rar1,ar−p1=VG. We can see that R¯u=2m and R¯u∩∪r=1mRr=2n≥Rr.Theorem 2.
Ifℵ≅ℚm1 with m≥6 and m≡0mod2. Then, dimfℚm1≤2.Proof.
Case 1. Whenm=6.The RNs are given as follows.We can see that Tables3 and 4 show the RNs with maximum cardinalities, that is, 12 and 8, respectively, whereas, Table 5 shows the RNs with minimum cardinality, that is, 6, and ∪r=112Rr=Vℚ61 which implies ∪r=112Rr=12 and Ru¯∩∪r=112Rr≥Rr, where 1≤u≤36.Here, we define a functionϕ:Vℚ61⟶0,1 such that κar1=κar2=1/6. Also Rr for 1≤r≤12 of ℚ61 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℚ61 such that κ¯≤κ. As a result, dimfℚ61≤∑r=1121/6=2.Case 2. Whenm≥6.We have seen from Lemma2 that the RNs with minimum cardinality of m are Rar1,ar2 and Rar1,ar−12, respectively.Letη=∪r=1mRr=2m and λ=Rr=2m. Then, we define a function κ:Vℚm1⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈Vℚm1−∪r=1mRr. We see that κ is a resolving function for ℚm1 because κRa,b≥1∀a,b⊂Vℵ. Now, suppose on contrary that there is another resolving function ρ, such that ρu≤ϕu, for at least one u∈Vℚm1ρu≠κu. As a result, ρRa,b<1, where Ra,b is an RN of ℚm1 having minimum cardinality λ. This shows that ρ is not a resolving function which is a contradiction to our supposition. Therefore, κ is a minimal resolving function that achieves minimum κ for ℚm1. Since all the RNs of ℚm1 are equal, hence ∃ another minimal resolving function κ¯ of ℚm1 such that κ¯≤κ. Assigning 1/λ to all the vertices of ℚm1 and calculating their sum, we get: dimfℚm1≤∑r=1η1/m=2.Lemma 3.
Letℵ≅ℚm2 be the 3-faced QCL with m≥6 and m≡0mod2. For 1≤l,r≤m, p≥3, s≥2p≡1mod2 and s≡0mod2. Then(a)
Rl=Ral1,al−12=Rr=Rar1,ar2=m with a02=am2, ∪l=1mRl=2m, ∪r=1mRr=2m and ∪l=1mRl∪∪r=1mRr=2m.(b)
Rr<R¯1=Rar1,ar+p1=R¯2=Rar1,ar+s1=R¯3=Rar2,ar+s2=R¯4=Rar1,ar+p2=R¯5=Rar1,ar+s2=R¯6=Rar1,ar−p2=R¯7=Rar1,ar−s2 and R¯u∩∪r=1mRr≥Rr with am+11=a11 and 1≤u≤8.Proof.
(a)
The RNs ofar1,ar−12 and ar1,ar2 are Rar1,ar−12=Vℚm2−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2|h≡r,r+1,…,r+m/2−1modm and Rar1,ar2=Vℚm2−ah1|h≡r+1,r+2,…,r+m/2modm∪ah2|h≡r−1,r−2,…,r−m/2modm respectively.We note that∪r=1mRr=Vℚm2 and ∪r=1mRr=3m.(b)
The RNs ofar1,ar+p1,ar1,ar+s1,ar2,ar+s2,ar1,ar+p2,ar1,ar+s2,ar1,ar−p1 and ar1,ar−s2 are given by: R¯1=Rar1,ar+p1=Vℚm2−ah2|h≡r+p−1/2,r+m+p−1/2modm, R¯2=Rar1,ar+s1=Vℚm2−ah1|h≡r+s/2,r+m+s/2modm, R¯3=Rar2,ar+s2=Vℚm2−ah2|h≡r+s/2,r+m+s/2modm, R¯4=Rar1,ar+p2=Vℚm2−ah1|h≡r+p+1/2modm∪ah2|h≡r−p−1/2modm, R¯5=Rar1,ar+s2=Vℚm2−ah1|h≡r+m+s/2modm∪ah2|h≡r+m+s/2modm, R¯6=Rar1,ar−p2=Vℚm2−ah1|h≡r−p+1/2modm∪ah2|h≡r+m+s/2modm, R¯7=Rar1,ar−s2=Vℚm2−ah1|h≡r−s/2modm∪ah2|h≡r−s/2modm, R¯8=Rar2,ar+p2=Vℚm2−ah1|h≡r+p+1/2,r+m+p+1/2modm.Clearly,R¯u=2m−1. Since Rar1ar−12=Rar1ar2=n<Rar1ar+11 then Rar1ar+11∩∪r=1mRr≥Rr.Theorem 3.
Letℵ≅ℚm2 with m≥6 and m≡0mod2. Then, dimfℚm2≤2.Proof.
Case 1. Whenm=6.The RNs are given as follows.Tables6 and 7 given above shows the RNs with cardinalities of 10 and 6 respectively. Clearly, the table shows the RNs bearing the minimum cardinality of 6 and ∪r=112Rr=Vℚ62 which implies ∪r=112Rr=12 and Ru¯∩∪r=112Rr≥Rr, where 1≤u≤52.Now, we define a functionκ:Vℚ62⟶0,1 such that κar1=κar2=1/6. Moreover, Rr for 1≤r≤12 of ℚ62 are pairwise overlapping; thus, ∃ is another minimal resolving function κ¯ of ℚ62 such that κ¯≤κ. As a result, dimfℚ62≤∑r=1121/6=2.Case 2.m≥8It is clear from Lemma3 that the RNs bearing the minimum cardinality of m in ℚm2 are Rl=Ral1,al2 and Rr=Rar1,ar−12 respectively, with R¯u∩∪r=1mRr=2n≥Rr.Letη=∪r=1mRr=2m and λ=Rr=m. Then we define a mapping κ:Vℚm2⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈Vℚm2−∪r=1mRr. We see that κ is a resolving function for ℚm1 because κRa,b≥1∀a,b⊂Vℵ. Now suppose on contrary that there is another resolving function ρ, such that ρu≤ϕu, for at least one u∈Vℚm1ρu≠κu. As a result, ρRa,b<1, where Ra,b is a RN of ℚm1 having minimum cardinality λ. This shows that ρ is not a resolving function which is a contradiction to our supposition. Therefore, κ is a minimal resolving function that achieves minimum κ for ℚm1. Since all the RNs of ℚm2 are equal, hence ∃ another minimal resolving function κ¯ of ℚm2 such that κ¯≤κ\enleadertwodots Assigning 1/λ to all the vertices of ℚm2 and calculating their sum, we get dimfℚm2≤∑r=1η1/m=2.Corollary 1.
Letℵ≅ℚm3 with m≥6 and m≡0mod2. Then, dimfℚm3≤2.Proof.
The result is evident asℚm3≅ℚm2.


## 5.3. Pentagonal Circular Ladders

Lemma 4.
Letℵ≅ℙm1 be the 2-faced QCL with m≥10 and m≡0mod2. For 1≤l,r≤m, p≥3, s≥2p≡1mod2 and s≡0mod2. Then,(a)
Rl=Ral1,al−23=Rr=Rar1,ar+13=5m/2−2 with am+13=a13, ∪l=1mRl=∪r=1mRr=3m and ∪l=1mRl∪∪r=1mRr=3m.(b)
Rr<R¯1=Rar2,ar3=R¯2=Rar2,ar−13=5m/2 with a03=am3 and Ru∩∪r=1mRr≥Rr.(c)
Rr<R¯3=Rar1,ar+11=R¯4=Rar1,ar+p1=R¯5=Rar2,ar+12=R¯6=Rar2,ar+p2=R¯7=Rar3,ar+s3=R¯8=Rar1,ar+s2=3m−2 and R¯u∩∪r=1mRr≥Rr with am+11=a11.(d)
Rr<R¯9=Rar1,ar2=R¯10=Rar1,ar+s1=R¯11=Rar3,ar+13=R¯12=Rar3,ar+p3=R¯13=Rar2,ar+s2=R¯14=Rar1,ar3 and R¯u∩∪r=1mRr≥Rr.(e)
Rr<R¯15=Rar1,ar+12=R¯16=Rar1,ar−12 and R¯u∩∪r=1mRr≥Rr.Proof.
(a)
The RNs ofar1,ar+13 and al1,al−22 are Rl=Rar1,ar+13=Vℙm1−ah1|h≡r+2,r+3,…,r+m/2modm∪ar+42∪ah3|h≡r,r+4modm and R2=Ral1,al−23=Vℙm1−Vℙm1−ah1|h≡r+2,r+3,…,r+m/2modm∪ar−42∪ah3|h≡r,r−5modm, respectively. We can see that Rr=5m/2−2 and ∪r=1mRr=∪l=1mRl=Vℙm1 and ∪l=1mRl∪∪r=1mRr=3m.(b)
The RNs ofar2,ar−13 and ar2,ar3 are R¯1=Rar2,ar−13=Vℙm1−ah1|h≡r−1,r−2,…,r−m/2modm and R¯1=Rar2,ar3=Vℵ−ah1|h≡r+1,r+2,…,r+m/2modm mod m}, respectively.Since for1≤u≤2R¯u>Rr=5m/2−1, thus R¯u∩∪r=1mRr=3m.(c)
The RNs ofar1,ar+11,ar1,ar+p1,ar2,ar+12,ar2,ar+p2,ar3,ar+s3 and ar1,ar+s2 are R¯3=Rar1,ar+11=Vℙm1−ah3|h≡r,r+m/2modm=R¯5=Rar2,ar+12, R¯4=Rar1,ar+p1=Vℙm1−ah3|h≡r+p−1/2,r+p+m−1/2modm=R¯6=Rar2,ar+p2, R¯7=Rar3,ar+s3=Vℙm1−ah3|h≡r+s/2,r+m+s/2modm and R¯8=Rar1,ar+s2=Vℙm1−ah1|h≡r+s/2modm∪ah3|h≡r+m+s−2/2modm, respectively.Clearly, for3≤u≤8,R¯u=3m−2. Since Rar2,ar3|=5m/2<Rar1ar+11, then R¯u∩∪r=1mRr=3m−2≥Rr.(d)
The RN ofar1,ar2 is R¯9=Rar1,ar2=Vℙm1−ah2|h≡r−1,r+1modm∪ah3|h≡r−2,r+1modm. R¯10=Rar1,ar+s1=Vℙm1−ah1||h≡r+s/2r+m+s/2modm∪ah2||h≡r+s/2,r+m+s/2modm,R¯11=Rar3,ar+13=ar1,ar+21, R¯12=Rar3,ar+p3=R¯13=Rar2,ar+s2=Rar1,ar+s1, R¯14=Rar1,ar3=Vℙm1−ah2|h≡r,r+2modm∪ah3|h≡r−1,r+2modm.Clearly, for9≤u≤14,Rr=5m/2<R¯u and R¯u∩∪r=1mRr≥Rr.(e)
The RNs ofar1,ar+12 and ar1,ar−12 are R¯15=Rar1,ar+12=Vℙm1−ah1|h≡r+1,r+2,…,r+m/2mod5∪ah2|h≡r+m/2modm and R¯16=Rar1,ar−12=Vℙm1−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2h≡r−m/2modm, respectively. We can see that Rr=5m/2 and R¯u∪r=1mRr≥Rr.Theorem 4.
Ifℵ≅ℙm1 with m≥8 and m≡0mod2. Then, dimfℙm1≤6m/5m−2for6≤m≤8,6m/5m−4form≥10.Proof.
Case 1. (i) Whenm=6.The RNs are given as follows.We can see that Tables8 and 9 show the RNs with maximum cardinality of 16 and 15, respectively, whereas both Tables 10 and 11 show the RNs with minimum cardinality of 14. Moreover, ∪r=145Rr=Vℙ61 which implies ∪r=145Rr=18 and Rr¯∩∪r=124Rr≥Rr.Now, we define a mappingκ:Vℙ61⟶0,1 such that κar1=κar2=κar3=1/14. It is observed that Rr for 1≤r≤45 of ℙ61 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙ61 such that κ¯≤κ. As a result, dimfℙ61≤∑r=1161/14=18/14=9/7.(ii)
Whenm=8.The RNs are given as follows.We can see that Tables12–14 shows the RNs with maximum cardinalities that is 22 and 20, respectively, whereas Table 15 shows the RNs with minimum cardinality of 19. Moreover, ∪r=116Rr=Vℙ81 which implies ∪r=116Rr=19 and Rr¯∩∪r=124Rr≥Rr.Now, we define a mappingκ:Vℙ81⟶0,1 such that κar1=κar2=κar3=1/19. It is observed that Rr for 1≤r≤16 of ℙ81 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙ81 such that κ¯≤κ. As a result, dimfℙ81≤∑r=1161/19=24/19.(iii)
For6≤m≤8.We have seen thatdimfℙ61≤9/7 whereas dimfℙ81≤24/19. In generic form, we can write it as dimf≤6m/5m−2.Case 2. Whenm=10.The RNs are given as follows.We can see that Tables16–18 show the RNs with maximum cardinalities, that is, 28 and 26 and 25, respectively, whereas Table 19 shows the RNs with minimum cardinality of 23. Moreover, ∪r=116Rr=Vℙ101 which implies ∪r=116Rr=23 and Rr¯∩∪r=124Rr≥Rr.Now, we define a mappingκ:Vℙ101⟶0,1 such that κar1=κar2=κar3=1/23. It is observed that Rr for 1≤r≤16 of ℙ101 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙ101 such that κ¯≤κ. As a result, dimfℙ101≤∑r=1231/19≤30/23.Case 3.m≥12.From Lemma4, we can see that the RNs with minimum cardinality of 5m/2−2 are Rar1,ar+13 and Rar1,ar−23, respectively. Moreover, for 1≤u≤15,Ru∩∪r=1mRr≥Rr.Letη=∪r=1mRr=3m and λ=Rr=5m/2−2. Then, we define a mapping κ:Vℙm1⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈Vℙm1−∪r=1mRr. We note that κ is a resolving function for ℙm1 because κRa,b≥1∀a,b∈Vℙm1. Now, assume that there is another resolving function ρ, such that ρu≤κu, for at least one u∈Vℙm1ρu≠κu. As a result, ρRa,b<1, where Ra,b is a RN of ℙm1 bearing minimum cardinality λ. This shows that ρ is not a resolving function a contradiction. Therefore, κ is a minimal resolving function for ℙm1 that achieves minimum κ. As a matter of fact Rr for 1≤t≤2m of ℙm1 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙm1 such that κ¯≤κ. Therefore, assigning 1/λ to all the vertices of ℙm1 and calculating their sum, we get: dimfℙm1≤∑r=1η1/λ=3m/5m/2−2=6m/5m−4.Lemma 5.
Letℵ≅ℙm2 be the 3-faced QCL with m≥6 and m≡0mod2. For 1≤l,r≤m, p≥3, s≥2p≡1mod2 and s≡0mod2. Then,(a)
Rl=Ral2,al3=Rr=Rar2,ar−13=3m/2+1, ∪l=1mRl, ∪r=1mRr=3m and ∪l=1mRl∪∪r=1mRr=3m.(b)
Rr<R¯1=Rar1,ar+13=R¯2=Rar1,ar−13 and R¯u∩∪r=1mRr≥Rr.(c)
Rr<R¯3=Rar1,ar+11=R¯4=Rar1,ar+p1=R¯5=Rar2,ar+12=R¯6=Rar2,ar+p2=R¯7=Rar3,ar+s3=R¯8=Rar1,ar+s2=R¯9=Rar1,ar+s3=R¯10=Rar1,ar+p3 and R¯u∩∪r=1mRr≥Rr with am+11=a11 and am+12=a12.(d)
Rr<R¯11=Rar2,ar+p3=R¯12=Rar2,ar+s3 and R¯u∩∪r=1mRr≥Rr.(e)
Rr<R¯13=Rar1,ar2 and R¯u∩∪r=1mRr≥Rr.Proof.
(a)
The RNs ofar2ar3 and ar2ar−13 are Rar2ar3=Vℙm2−ah1|h≡r+1,r+2,…,r+m/2modm∪ah2|h≡r+1,r+2,…,r+m/2modm∪ah3|h≡r+1,r+2,…,r+m/2−1modm and Rar2ar−13=Vℙm2−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2|h≡r−1,r−2,…,r−m/2modm∪ah3|h≡r−2,r−3,…,r−m/2+1modm. We note that ∪r=1mRr=Vℙm2 and ∪r=1mRr=3m.(b)
The RNs ofar1,ar−13 and ar1,ar−13 are R¯1=Rar1,ar+13=Vℙm2−ah1|h≡r+2,r+3,…,r+m/2modm∪ah3|h≡r,r+m/2modm and R¯2=Rar1,ar−13=Vℙm2−ah1|h≡r−2,r−3,…,r−m/2modm∪ah3|h≡rmodm respectively. It is clear from the above that Rr<R¯u and R¯u∩∪r=1mRr≥Rr.(c)
The RNs ofar1,ar+11,ar1,ar+p1,ar2,ar+12,ar2,ar+p2,ar3,ar+s3,ar1,ar+s2,ar1,ar+s3 and ar1,ar+p3 are R¯3=Rar1,ar+11=Vℙm2−ah3|h≡r,r+m/2modm=R¯5=Rar2,ar+12, R¯4=Rar1,ar+p1=Vℙm2−ah3|h≡r+p−1/2,r+p+m−1/2modm=R¯6=Rar2,ar+p2, R¯7=Rar3,ar+s3=Vℙm2−ah3|h≡r+s/2,r+m+s/2modm, R¯8=Rar1,ar+s2=Vℙm2−ah3|h≡r+s/2,r+s+m/2modm, R¯9=Rar1,ar+s3=Vℙm2−ah1|h≡r+s+2/2modm∪ah3|h≡r+s+m/2modm, R¯10=Rar1,ar+p3=Vℙm2−ah1|h≡r+p+m+1/2modm∪ah3|h≡r+p−1/2modm, Clearly, R¯u=3m−2. Since Rr=3m/2+1<R¯u, then R¯u∩∪r=1mRr=3m−2≥Rr.(d)
The RNs ofar2,ar+p3 and ar2,ar+s3 are R¯11=Rar2,ar+p3=Vℙm2−ah1|h≡r+p+1/2modm∪ah2|h≡r+p+1/2modm∪ah3|h≡r+p+m+1/2modm and R¯12=Rar2,ar+s3=Vℙm2−ah1|h≡r+m+s/2modm∪ah2|h≡r+m+s/2modm∪ah3|h≡r+s/2modmClearly,R¯u=3m−1. Since Rr=3m/2+1<R¯u, then R¯u∩∪r=1mRr≥Rr.(e)
The RN ofar1,ar2 is R¯13=Rar1,ar2=Vℙm2.Clearly,Rar2ar3=3m/2+1<Rar1ar2 and Rar1ar2∩∪r=1mRr=3n≥Rr.Theorem 5.
Ifℵ≅ℙm2 with m≥6 and m≡0mod2. Then, dimfℙm2≤6m/3m+2.Proof.
Case 1. Whenm=6.The RNs are given as follows.Tables20–22 show the RNs with maximum cardinality of 18, 16, and 14, respectively. On the other hand, Table 23 bears RNs with the minimum cardinality of 10. Also, it is observed that ∪r=112Rr=Vℙ62 which implies ∪r=112Rr=18 and Rr¯∩∪r=16Rr≥Rr.Now, we define a functionκ:Vℙ62⟶0,1 such that κar1=κar2=κar3=1/10. As Rr for 1≤t≤12 of ℙ62 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙ62 such that κ¯≤κ. As a result, dimfℙ62≤∑r=1181/10=9/5.Case 2.m≥8.As Lemma5 clears the fact that among the RNs of ℙm2, those that have the minimum cardinality of 3m/2+1 are Rar2,ar3 and Rar2,ar−13, respectively. Also, R¯u∩∪r=1mRr≤Rr, where 1≤u≤17.Letη=∪r=1mRr=3m and λ=Rr=3m/2+1. Then, we define a function κ:Vℙm2⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈Vℙm2−∪r=1mRr. It is seen that κ is a resolving function for ℙm2 because κRa,b≥1∀a,b⊂Vℙm2. Now, suppose on contrary that there is another resolving function ρ, such that ρu≤κu, for at least one u∈Vℙm2ρu≠κu. As a result, ρRa,b<1, where Ra,b is an RN of ℙm2 having minimum cardinality λ. This shows that ρ is not a resolving function, a contradiction. Thus, κ is a minimal resolving function for ℙm2 which attains minimum κ. Also, Rr for 1≤t≤2m of ℙm2 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙm2 such that κ¯≤κ. Therefore, assigning 1/λ to all the vertices of ℙm2 and calculating their sum, we get dimfℙm2≤∑r=1η1/λ=3m/3m+2/2=6m/3m+2.Lemma 6.
Letℵ≅ℙm3 be the 3-faced QCL with m≥6 and m≡0mod2. For 1≤r≤m, p≥3, s≥2p≡1mod2 and s≡0mod2. Then,(a)
Rl=Rar1,ar2=Rr=2m−1, ∪r=1mRl=3m and ∪r=1mRr=3m.(b)
Rr<R¯1=Rar1,ar+11=R¯2=Rar1,ar+p1=R¯3=Rar2,ar+12=R¯4=Rar2,ar+p2=R¯5=Rar3,ar+13=R¯6=Rar3,ar+p3=R¯7=Rar1,ar+s1=R¯8=Rar1,ar+13=R¯9=Rar1,ar−13=R¯10=Rar1,ar3 and R¯u∩∪r=1mRr≥Rr with am+11=a11 and am+12=a12.(c)
Rr<R¯11=Rar2,ar+s2=R¯12=Rar3,ar+s3=R¯13=Rar1,ar+p3=R¯14=Rar1,ar+p3R¯u∩∪r=1mRr≥Rr.(d)
Rr<R¯15=Rar2,ar+s2=R¯16=Rar3,ar+s3=R¯17=Rar1,ar+p3=R¯18=Rar1,ar+p3R¯u∩∪r=1mRr≥Rr.(e)
Rr<R¯19=Rar1,ar−12 and Rar1,ar−12∩∪r=1mRr≥Rr.(f)
Rr<R¯20=Rar2,ar−13 and Rar2,ar−13∩∪r=1mRr≥Rr.(g)
Rr<R¯21=Rar2,ar3 and Rar2,ar3∩∪r=1mRr≥Rr.Proof.
(a)
The RN ofar1,ar2 is Rar1ar2=Vℙm3−ah1|h≡r+1,r+2,…,r+m/2modm∪ah2|h≡r+1,r+2,…,r+m/2−1modm∪ah3|h≡r+1,r+2modm.We note that∪r=1mRr=Vℙm3 and ∪r=1mRr=3m.(b)
The RNs ofar1,ar+11,ar1,ar+p1,ar2,ar+12,ar2,ar+p2,ar3,ar+13,ar3,ar+p3,ar1,ar+s1,ar1,ar+13, ar1,ar−13, and ar1,ar3 are R¯1=Rar1,ar+11=Vℙm3−ah2|h≡r,r+m/2modm∪ah3|h≡r,r+m/2modm, R¯2=Rar1,ar+p1=Vℙm3−ah2|h≡r+p−1/2,r+p+m−1/2modm∪ah3|h≡r+p−1/2,r+p+m−1/2modm, R¯3=Rar2,ar+12=Vℙm3−ah1|h≡r+1,r+m+2/2modm∪ah3|h≡r,r+m/2modm, R¯4=Rar2,ar+p2=Vℙm3−ah1|h≡r+p+1/2,r+p+m+1/2modm∪ah3|h≡r+p+1/2,r+p+m+1/2modm, R¯5=Rar3,ar+13=Vℙm3−ah1|h≡r+1,r+m+2/2modm∪ah2|h≡r+1,r+m+2/2modm, R¯6=Rar3,ar+p3=Vℙm3−ah1|h≡r+p+1/2,r+p+m+1/2modm∪ah2|h≡r+p+1/2,r+p+m+1/2modm, R¯7=Rar1,ar+s1=Vℙm3−ah1|h≡r+s/2,r+s+m/2modm∪ah3|h≡r+s−2/2,r+s+m−2/2modm=R¯8=Rar1,ar+13, R¯9=Rar1,ar−13=Vℙm3−ah1|h≡r−s/2,r−s+m/2modm∪ah3|h≡r−s−2/2,r−s+m−2/2modm, R¯10=Rar1,ar3=Vℙm3−ah2|h≡r,r+2modm∪ah3|h≡r−1,r+2modm, Clearly, R¯u=3m−4. Since Rr=3m/2+1<R¯u, then R¯u∩∪r=1mRr=3m−2≥Rr.(c)
The RN ofar1,ar+p2, ar1,ar+s2, ar1,ar−p2, and ar1,ar−s2 are R¯11=Rar1,ar+p2=VG−ah1|h≡r+p+1/2modm∪ah2|h≡r+p+m−1/2modm∪ah3|h≡r+p−1/2modm, R¯12=Rar1,ar+s2=VG−ah1|h≡r+m+s/2modm∪ah2|h≡r+s/2modm∪ah3|h≡r+m+s−2/2modm, R¯13=Rar1,ar−p2=VG−ah1|h≡r−p+1/2modm∪ah2|h≡r−p+m−1/2modm∪ah3|h≡r−p−1/2modm, R¯14=Rar1,ar+s2=VG−ah1|h≡r−m+s/2modm∪ah2|h≡r−s/2modm∪ah3|h≡r−m+s−2/2modm, Clearly, R¯u=3m−1. Since Rr=3m/2+1<R¯u, then R¯u∩∪r=1mRr=3m−2≥Rr.(d)
The RNs ofar2,ar+s2, ar3,ar+s3, ar1,ar+p3, and ar1,ar+p3 are R¯15=ar2,ar+s2=VG−ah2|r≡r+s/2,r+s+m−2/2modm, R¯16=ar3,ar+s3=VG−ah3|r≡r+s/2,r+s+m−2/2modm, R¯17=ar1,ar+s3=VG−ah1|r≡r+s+2/2,r+s+m+2/2modm, R¯18=ar1,ar+p3=VG−ah2|r≡r+p+1/2modm∪ah1|r≡r+p+m+1/2,modm, Clearly, R¯u=3m−2. Since Rr=3m/2+1<R¯u, then R¯u∩∪r=1mRr=3m−2≥Rr.(e)
The RNs ofar1,ar+12 and ar1,ar−12 are R¯19=Rar1ar−12=Vℙm3−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2|h≡r−2modm∪ah3|h≡r−3modm.Clearly,Rar1ar2=2m−1<Rar1ar−12 and Rar1ar−12∩∪r=1mRr=5m−4/2≥Rr.(f)
The RN ofar2ar−13 is given by R¯20=Rar2ar−13=Vℙm3−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2|h≡r−1modm.Clearly,Rar1ar2=2m−1<Rar2ar−13 and Rar2ar−13∩∪r=1mRr=5m−2/2≥Rr.(g)
The RN ofar2ar3 is given by R¯21=Rar2,ar3=Vℙm3−ah2|h≡r−1,r−2,…,r−m/2+1modm.Clearly,Rar1ar2=2m−1<Rar2,ar3 and Rar2ar−13∩∪r=1mRr=5m+2/2≥Rr.Theorem 6.
Ifℵ≅ℙm3 with m≥6 and m≡0mod2. Then, dimfℙm3≤3m/2m−1.Proof.
Case 1. Whenm=6The RNs are given as follows.We note that the cardinalities of RNs as shown in Tables24–26 are 16, 14, and 13, respectively. Similarly, the cardinality of RNs given in Table 27 is 11 which is less than the least among all the RNs. Moreover, ∪r=16Rr=Vℙ63 which implies ∪r=16Rr=18 and Rr¯∩∪r=16Rr≥Rr.Now, we define a functionκ:Vℙ63⟶0,1 such that κar1=κar2=κar3=1/11. Furthermore, Rr for 1≤t≤6 of ℙ63 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙ63 such that κ¯≤κ. As a result, dimfℙ63≤∑r=1181/11=18/11.Case 2.m≥8We have seen from Lemma6Rar1,ar2=2m−1≤Ra,b for all a,b⊂Vℙm3 andRa,b∩∪r=1mRr≥Rr.Letη=∪r=1mRr=3m and λ=Rr=2m−1. Then, we define a function κ:Vℙm3⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈Vℙm3−∪r=1mRr. We see that κ is a resolving function for ℙm3 because κRa,b≥1∀uv∈Eℙm3. Now, suppose on contrary that there is another resolving function ρ, such that ρu≤κu, for at least one u∈Vℙm3ρu≠κu. As a result, ρRa,b<1, where Ra,b is an RN of ℙm3 having the minimum cardinality of λ . This shows that ρ is not a resolving function, a contradiction. Thus, κ is a minimal resolving function for ℙm3 that achieves minimum κ. Moreover, Rr for 1≤r≤m of ℙm3 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙm3 such that κ¯≤κ. Therefore, assigning 1/λ to all the vertices of ℙm3 and calculating their sum, we get dimfℙm3≤∑r=1η1/λ=3m/2m−1=3m/2m−1.Corollary 2.
Ifℵ≅ℙm4 with m≥6 and m≡0mod2. Then, dimfℙm4≤3m/2m−1.Proof.
The above assertion is valid asℙm4≅ℙm3.Lemma 7.
Letℵ≅ℙm5 be the 4-faced QCL with m≥6 and m≡0mod2. For 1≤l,r≤m, p≥3, s≥2p≡1mod2 and s≡0mod2. Then,(a)
Rl=Ral1,al2=Rr=Rar2,ar−13=3m/2+1 with a03=am3,a02=am2, ∪l=1mRl=3m, ∪r=1mRr=3m and ∪l=1mRl∪∪r=1mRr=3m,(b)
Rr<R¯1=Rar1,ar3=R¯2=Rar2,ar+13=R¯3=Rar2,ar−13 and R¯u∩∪r=1mRr≥Rr.(c)
Rr<R¯4=Rar1,ar+s1=R¯5=Rar2,ar+12=R¯6=Rar2,ar+p2=R¯7=Rar1,ar+s3 with am+11=a11 and R¯u∩∪r=1mRr≥Rr.(d)
Rr<R¯8=Rar1,ar+11=R¯9=Rar1,ar+p1=R¯10=Rar2,ar+p3=R¯11=Rar2,ar−p3 and R¯u∩∪r=1mRr≥Rr.(e)
Rr<R¯12=Rar2,ar+s2=R¯13=Rar1,ar+p2=R¯14=Rar1,ar+s2=R¯15=Rar1,ar+p3 and R¯u∩∪r=1mRr≥Rr.(f)
Rr<R¯16=Rar2,ar3 and Rar2,ar3∩∪r=1mRr≥Rr.Proof.
(a)
The RNs ofal1,al2 and ar2ar−13 are Ral1,al2=Vℙm5−ah1|h≡l+1,l+2,…,l+m/2modm∪ah2|h≡l−1,l−2,…,l−m/2modm∪ah3|h≡l−1,l−2,…,l−m/2+1modm and Rar2ar−13=Vℙm5−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2|h≡r−1,r−2,…,r−m/2modm∪ah3|h≡r−1,r−2,…,r−m/2+1modm, respectively. We note that ∪r=1mRr=Vℙm5 and ∪r=1mRr=3m.(b)
The RNs ofar1,ar3,ar2,ar+13 and ar2,ar−13 are R¯1=Rar1,ar3=Vℙm5−ah1|h≡r+2,r+3,…,r+m/2modm∪ah2|h≡rmodm∪ah3|h≡r−1modm, R¯2=Rar2,ar+13=Vℙm5−ah1|h≡r+2,r+3,…,r+m+2/2modm∪ah2|h≡rmodm∪ah3|h≡r+m/2modm and R¯3=Rar2,ar−13=Vℙm5−ah1|h≡r−2,r−3,…,r−m+2/2modm∪ah2|h≡rmodm∪ah3|h≡r−m/2modm, respectively. We can see that R¯u=5m/2+2. Since Rar1ar2=Rar2ar−13=3m/2+1<R¯u, then R¯u∩∪r=1mRr=3m−4≥Rr, where 1≤u≤3.(c)
The RNs ofar1,ar+s1,ar2,ar+p2 and ar1,ar+s3 are R¯4=Rar1,ar+s1=Vℙm5−ah2|h≡r+s/2,r+m+s/2modm∪ah3|h≡r+s−2/2,r+s+m−2/2modm,R¯5=Rar2,ar+12=Rar1,ar+21, for s≥4, R¯6=Rar2,ar+p2=Rar1,ar+s1, R¯7=Rar1,ar+s3=Vℙm5−ah1|h≡r+s+2/2,r+m+s/2modm∪ah2|h≡r+s/2,r+m+s/2modm, Clearly, R¯u=3m−4. Since Rar1,ar2=Rar2,ar−13=3m/2+1<R¯u, then R¯u∩∪r=1mRr=3m−4≥Rr.(d)
The RNs ofar1,ar+p1,ar2,ar+p3 and ar2,ar−p3 are R¯8=Rar1,ar+11=Vℙm5−ah2|h≡r,r+m/2modm∪ah3|h≡rmodm, R¯9=Rar1,ar+p1=Vℙm5−ah2|h≡r+p−1/2,r+p+m−1/2modm∪ah3|h≡r+p−1/2modm, R¯10=Rar2,ar+p3=Vℙm5−ah1|h≡r−1modm∪ah2|h≡r+p+1/2modm∪ah3|h≡r+p+m−1/2modm and R¯11=Rar2,ar−p3=Vℙm5−ah1|h≡r−1modm∪ah2|h≡r−p+1/2modm∪ah3|h≡r−p+m+1/2modm, respectively. We can see from these sets that R¯u=3m−1>Rr and R¯u∩∪r=1mRr=3m−4≥Rr.(e)
The RNs ofar2,ar+s2,ar1,ar+p2,ar1,ar+s2,ar1,ar+p3 and Rar1,ar+s3| are R¯12=Rar2,ar+s2=Vℙm5−ah2|h≡r+s/2,r+s+m/2modm, R¯13=Rar1,ar+p2=Vℙm5−ah2|h≡r+p−1/2modm∪ah3|h≡r+p+m−1/2modm, R¯14=Rar1,ar+s2=Vℙm5−ah1|h≡r+m+s/2modm∪ah3|h≡r+s−2/2modm, R¯15=Rar1,ar+p3=Vℙm5−ah1|h≡r+p+1/2modm∪ah3|h≡r+p−1/2modm, respectively. We note that R¯u=3m−2>Rr and R¯u∩∪r=1mRr=3m−2≥Rr.(f)
The RN ofar2,ar3 is R¯16=Rar2,ar3=Vℙm5−ah2|h≡r+1,r+2,…,r+m/2modm∪ah3|h≡r+1,r+2,…,r+m/2−1modm.Clearly,Rar1ar2=Rar2ar−13=3m/2+1<Rar2ar3 and Rar2ar3∩∪r=1mRr=2m+1≥Rr.Theorem 7.
Ifℵ≅ℙm5 with m≥6 and m≡0mod2. Then, dimfℙm5≤6m/3m+2.Proof.
Case 1. Whenm=6.The RNs are given as follows.In Tables28–31, the cardinalities of RNs are 16, 15, 14, and 13, respectively. In the same manner, Table 32 is having RNs withholding the cardinality of 10. Moreover, ∪r=112Rr=Vℙ65 which implies ∪r=112Rr=18 and Rr¯∩∪r=112Rr≥Rr.Now, we define a functionκ:Vℙ65⟶0,1 such that κar1=κar2=κar3=1/10. We have seen that Rr for 1≤t≤12 of ℙ65 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙ65 such that κ¯≤κ. As a result, dimfℙ65≤∑r=1181/10=9/5.Case 2.m≥8.It can be seen from Lemma7 that among all the RNs of ℙm5, Rar1ar2=Rar2ar−13=3m/2+1≤Ra,b for all a,b⊂Vℙm5 andRa,b∩∪r=1mRr=3m/2+1≤Rr.Letη=∪r=1mRr=3m and λ=Rr=3m/2+1. Then, we define a function κ:Vℙm5⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈Vℙm5−∪r=1mRr. It is found that κ is a resolving function for ℙm5 because κRa,b≥1∀a,b⊂Vℙm5. On contrary, assume that there is another resolving function ρ, such that ρu≤κu, for at least one u∈Vℙm5ρu≠κu. As a result, ρRa,b<1, where Ra,b is an RN of ℙm5 having the minimum cardinality of λ . This shows that ρ is not a resolving function, a contradiction. Thus, κ is a minimal resolving function for ℙm5 that achieves minimum κ. Moreover, Rr for 1≤t≤2m of ℙm5 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙm5 such that κ¯≤κ. Therefore, assigning 1/λ to all the vertices of ℙm5 and calculating their sum, we get dimfℙm5≤∑r=1η1/λ=3m/3m+2/2=6m/3m+2.Corollary 3.
Ifℵ≅ℙm6 with m≥6 and m≡0mod2. Then, dimfℙm6≤6m/3m+2.Proof.
This can be derived easily asℙm6≅ℙm5.Lemma 8.
Letℵ≅ℙm7 be the 4-faced QCL with m≥6 and m≡0mod2. For 1≤l,r≤m, p≥3, s≥2p≡1mod2 and s≡0mod2. Then,(a)
Rl=Ral1,al3=Rr=Rar1,ar−13, ∪l=1mRl=3m, ∪r=1mRr=3m and ∪l=1mRl∪∪r=1mRr=3m.(b)
Rr<R¯1=Rar1,ar+11=R¯2=Rar1,ar+p1=R¯3=Rar2,ar+12=R¯4=Rar2,ar+p2=R¯5=Rar1,ar+s2=R¯6=Rar1,ar+p2=R¯7=Rar2,ar+s3 with am+11=a11 and R¯u∩∪r=1mRr≥Rr.(c)
Rr<R¯8=Rar1,ar2=R¯9=Rar1,ar+s1=R¯10=Rar2,ar+s2=R¯11=Rar3,ar+13R¯12=Rar3,ar+p3 and R¯u∩∪r=1mRr≥Rr.(d)
Rr<R¯13=Rar1,ar+p3=R¯14=Rar1,ar+s3 and R¯u∩∪r=1mRr≥Rr.(e)
Rr<R¯15=Rar2,ar3 and Rar2ar3∩∪r=1mRr≥Rr.(f)
Rr<R¯16=Rar2,ar−13 and Rar2,ar−13∩∪r=1mRr≥Rr.(g)
Rr<R¯17=Rar2,ar+p3 and R¯17∩∪r=1mRr≥Rr.(h)
Rr<R¯18=Rar1,ar+12 and R¯18∩∪r=1mRr≥Rr.Proof.
(a)
The RNs ofal1al3 and ar1ar−13 are Rl=Ral1,al3=Vℙm7−ah1|h≡l+1,l+2,…,l+m/2modm∪ah2|h≡l,l+2,…,r+m/2modm∪ah3|t≡l+1,l+2,…,l+m/2modm and Rr=Rar1ar−13=Vℙm7−ah1|h≡r−1,r−2,…,r−m/2modm∪ah2|h≡r,r−2,…,r−m/2modm∪ah3|t≡r−1,r−2,…,r−m/2modm, respectively. Clearly, Rr=3/2m+1 for 1≤r≤m.We note that∪r=1mRr=Vℙm7 and ∪r=1mRr=3m.(b)
The RNs ofar1,ar+11,ar1,ar+p1,ar2,ar+p2,ar2,ar+p2,ar1,ar+s2,ar1,ar+p2, and ar2,ar+s3| are R¯1=Rar1,ar+11=VG−ah3|h≡r,r+m/2modm=R¯3=Rar2,ar+12, R¯2=Rar1,ar+p1=VG−ah3|h≡r+p−1/2,r+p+m−1/2modm=R¯4=Rar2,ar+p2=R¯5=Rar1,ar+s2=VG−ah2|h≡r+s/2modm∪ah3|h≡r+m+s−2/2modm, R¯6=Rar1,ar+p2=VG−ah1|h≡r+p+1/2,r+p+m+1/2modm and R¯7=Rar2,ar+s3=VG−ah1|h≡r+s/2modm∪ah3|h≡r+m+s/2modm, respectively. We can see that Rr<R¯u=3m−2 and R¯u∩∪r=1mRr≥Rr.(c)
The RNs ofar1,ar2,ar1,ar+s1,ar2,ar+s2,ar3,ar+13and ar3,ar+p3 are R¯8=Rar1,ar2=Vℙm7−ah2|h≡r−1,r+1modm∪ah3|h≡r,r−1modm, R¯9=Rar1,ar+s1=Vℙm7−ah1|h≡r+s/2,r+s+m/2modm∪ah2|h≡r+s/2,r+s+m/2modm=R¯10=ar2,ar+s2, R¯11=Rar3,ar+13=Rar2,ar+22 and R¯12=Rar3,ar+p3=Rar2,ar+s2 where s≥4, respectively.Clearly,Rr=3m−4<Rar1,ar2 and Rar1,ar2∩∪r=1mRr=3m−4≥Rr.(d)
The RNs ofar1,ar+p3 and ar1,ar+s3 are R¯13=Rar1,ar+p3=VG−ah1|h≡r+p+1/2modm∪ah2|h≡r+p+1/2modm∪ah3|h≡r+p+m−1/2modm and R¯14=Rar1,ar+s3=VG−ah1|h≡r+m+s/2modm∪ah2|h≡r+m+s/2modm∪ah3|h≡r+s/2modm, respectively. Clearly, Rr=3m−1<Rar1ar2 and Rar1ar2∩∪r=1mRr=3m−4≥Rr.(e)
The RN ofar2,ar3 is R¯15=Rar2,ar3=Vℙm7−ah1|h≡r,r−1,…,r−m/2+1modm.Clearly,Rr=5m/2<Rar2ar3 and Rar2ar3∩∪r=1mRr=5m+2/2≥Rr.(f)
The RN ofar2,ar−13 is R¯16=Rar2,ar3=Vℙm7−ah1|h≡r,r+1,…,r+m/2−1modm.Clearly,Rr=5m/2<Rar2,ar3 and Rar2,ar3∩∪r=1mRr=5m+2/2≥Rr.(g)
The RN ofar2,ar+p3 is R¯17=Rar2,ar+p3=VG−ah1|h≡r+p+m−1/2modm. Clearly, Rr<R¯15=3m−1 and Rar2,ar+p3∩∪r=1mRr=5m+2/2≥Rr.(h)
The RN ofar1,ar+12 is R¯18=Rar1,ar+12=VG−ah1|h≡r+1,r+2,…,r+m/2modm∪ah2|h≡r+3modm∪ah3|h≡r,r+2modm. Clearly, Rr<R¯18=3m−1 and Rar2,ar+p3∩∪r=1mRr=5m+2/2≥Rr.Theorem 8.
Ifℵ≅ℙm7 with m≥6 and m≡0mod2. Then, dimfℙm7≤6m/3m+2.Proof.
Case 1. Whenm=6.The RNs are given as follows.Tables33–37 show the RNs of ℙ67 with cardinalities 16, 15, 14, 13, and 10, respectively. Moreover, in Table 37, ∪r=112Rr=Vℙ67 which implies ∪r=112Rr=18 and Ru¯∩∪r=112Rr≥Rr = , where 1≤u≤81.Now, we define a functionκ:Vℙ67⟶0,1 such that κar1=κar2=κar3=1/10. Furthermore, Rr for 1≤r≤12 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙ67 such that κ¯≤κ. As a result, dimfℙ67≤∑r=1181/10=9/5.Case 2.m≥8.As Lemma8 clears the fact that Rar2,ar3=Rar2,ar−13=3/2m+1≤Ra,bs for all a,b⊂Vℙm7 andRa,b∩∪r=1mRr≤Rr.Letη=∪r=1mRr=3m and λ=Rr=3/2m+1. Then, we define a function κ:Vℙm7⟶0,1 such that κc=1/λforc∈∪r=1mRr,0forc∈Vℙm7−∪r=1mRr. We find that κ is a resolving function for ℙm7 because κRa,b≥1∀a,b⊂Vℙm7. Assume on contrary that there is another resolving function ρ, such that ρu≤κu, for at least one u∈Vℙm7. As a result, ρRa,b<1, where Ra,b is an RN of ℙm7 having the minimum cardinality of λ. This shows that ρ is not a resolving function, a contradiction. Thus, κ is a minimal resolving function for ℙm7 that achieves minimum κ. Also Rr for 1≤t≤2m of ℙm7 are pairwise overlapping; hence, ∃ another minimal resolving function κ¯ of ℙm7 such that κ¯≤κ. Therefore, assigning 1/λ to all the vertices of ℙm7 and calculating their sum, we get dimfℙm7≤∑r=1η1/λ=3m/3/2m+1=6m/3m+2.


## 6. Conclusion

In this article,(i)
we have found the upper bounds of FMD of symmetric networks called by TCL (Tm), QCLs (ℚm1andℚm2), and PCLs (ℙm1,ℙm2,ℙm3,ℙm5, and ℙm7).(ii)
Table38 shows the summary of main results, and Table 39 gives the values of FMDs as they tend to ∞.(iii)
the networks having the maximum FMD of 2 areℚm1 and ℚm2.(iv)
in contrast,Tm is the network bearing the minimum FMD of 2m/m+1.(v)
the obtained results are the generalization of [23].(vi)
to find the extremal values of FMD of asymmetric networks is still an open problem.(vii)
moreover, the evaluation of the FMD and LFMD of connected networks such as convex polytopes is an open problem as well.Table 1
R¯u for 1≤u≤49.
RNsElementsEqualityRa11,a21VT6−a12,a42Ra41,a51,Ra12,a22Ra21,a31VT6−a22,a52Ra51,a61,Ra22,a32Ra31,a41VT6−a32,a62Ra11,a61,Ra32,a42Ra11,a31VT6−a21,a51Ra12,a32Ra21,a41VT6−a31,a61Ra22,a42Ra31,a51VT6−a11,a41Ra32,a52Ra41,a61VT6−a21,a51Ra42,a62Ra11,a51VT6−a31,a61Ra12,a52Ra11,a41VT6−a22,a52Ra12,a42Ra21,a51VT6−a32,a62Ra22,a52Ra31,a61VT6−a12,a42Ra32,a62Ra21,a61VT6−a41,a61Ra22,a62Ra11,a22VT6−a21∪a42Ra21,a32VT6−a31∪a52Ra11,a42Ra31,a42VT6−a41∪a62Ra21,a52Ra41,a52VT6−a51∪a12Ra32,a62Ra51,a62VT6−a61∪a22Ra22,a52Ra61,a12VT6−a11∪a32Ra32,a62Ra11,a32VT6−a51∪a22Ra21,a42VT6−a61∪a32Ra11,a42Ra31,a52VT6−a11∪a42Ra21,a52Ra41,a62VT6−a21∪a52Ra32,a62Ra11,a52VT6−a31∪a62Ra22,a52Ra21,a62VT6−a41∪a12Ra32,a62Table 2
RNs having minimum cardinality.
RNsElementsR1=Ra11,a12VT6−a21,a31,a41∪a22,a32R3=Ra31,a32VT6−a41,a51,a61∪a42,a52R5=Ra51,a52VT6−a11,a21,a61∪a12,a62R7=Ra21,a12VT6−a11,a51,a61∪a52,a62R9=Ra41,a32VT6−a11,a21,a31∪a12,a22R11=Ra61,a52VT6−a31,a41,a51∪a32,a42R2=Ra21,a22VT6−a31,a41,a51∪a32,a42R4=Ra41,a42VT6−a11,a51,a61∪a52,a62R6=Ra61,a62VT6−a11,a21,a31∪a12,a22R8=Ra31,a22VT6−a11,a21,a61∪a12,a62R10=Ra51,a42VT6−a21,a31,a41∪a22,a32R12=Ra11,a62VT6−a41,a51,a61∪a42,a52Table 3
R¯u for 1≤u≤21.
Resolving neighbourhoodsSet/elementsRa12,a22Ra42,a52Ra12,a42Vℚ61Ra22,a32Ra42,a52Ra11,a21Vℚ61Ra32,a42Ra12,a62Ra22,a52Vℚ61Ra32,a62Ra31,a41Ra21,a31Vℚ61Ra11,a32Ra41,a51Ra51,a61Vℚ61Ra11,a61Ra31,a52Ra21,a42Vℚ61Ra21,a62Ra41,a62Ra11,a52Vℚ61Table 4
R¯ufor21≤u≤36.
RNsElementsEqualityRa12,a32Vℚ61−a21,a51∪a22,a52Ra11,a31,Ra42,a62,Ra41,a61Ra22,a42Vℚ61−a31,a61∪a32,a62Ra21,a51,Ra12,a52,Ra11,a51Ra32,a52Vℚ61−a11,a41∪a12,a42Ra31,a51,Ra22,a62,Ra21,a61Ra11,a42Vℚ61−a31,a51∪a22,a62Ra21,a52Vℚ61−a41,a61∪a12,a32Ra31,a62Vℚ61−a11,a51∪a22,a42Table 5
RNs with minimum cardinality.
RNsElementsR1=Ra11,a22Vℚ61−a21,a31,a41∪a12,a52,a62R3=Ra31,a42Vℚ61−a41,a51,a61∪a12,a22,a32R5=Ra51,a62Vℚ61−a11,a21,a61∪a32,a42,a52R7=Ra21,a12Vℚ61−a11,a51,a61∪a22,a32,a42R9=Ra41,a32Vℚ61−a11,a21,a31∪a42,a52,a62R11=Ra61,a52Vℚ61−a31,a41,a51∪a12,a22,a62R2=Ra21,a32Vℚ61−a31,a41,a51∪a12,a22,a62R4=Ra41,a52Vℚ61−a11,a51,a61∪a22,a32,a42R6=Ra61,a12Vℚ61−a11,a21,a31∪a42,a52,a62R8=Ra31,a22Vℚ61−a11,a21,a61∪a32,a42,a52R10=Ra51,a42Vℚ61−a21,a31,a41∪a12,a52,a62R12=Ra11,a62Vℚ61−a41,a51,a61∪a12,a22,a32Table 6
R¯u for 1≤u≤52.
RNsElementsEqualityRa12,a22Vℚ62−a21,a51Ra42,a52Ra22,a32Vℚ62−a31,a61Ra42,a52Ra32,a42Vℚ62−a11,a41Ra12,a62Ra12,a32Vℚ62−a22,a52Ra42,a62Ra22,a42Vℚ62−a32,a62Ra12,a52Ra32,a52Vℚ62−a12,a42Ra22,a62Ra12,a42Vℚ62−a31,a61Ra22,a52Vℚ62−a11,a41Ra32,a62Vℚ62−a21,a51Ra11,a21Vℚ62−a12,a42Ra41,a51Ra21,a31Vℚ62−a22,a52Ra51,a61Ra31,a41Vℚ62−a32,a62Ra11,a61Ra11,a31Vℚ62−a21,a51∪a12,a52Ra12,a32Ra21,a41Vℚ62−a31,a61∪a22,a62Ra22,a42Ra31,a51Vℚ62−a11,a41∪a12,a42Ra32,a52Ra41,a61Vℚ62−a21,a51∪a22,a52Ra42,a62Ra11,a51Vℚ62−a31,a61∪a32,a62Ra12,a52Ra21,a61Vℚ62−a11,a41∪a12,a42Ra22,a62Ra11,a41Vℚ62−a22,a52Ra21,a51Vℚ62−a32,a62Ra31,a61Vℚ62−a12,a52Ra11,a22Vℚ62−a21∪a12Ra21,a32Vℚ62−a31∪a22Ra31,a42Vℚ62−a41∪a32Ra41,a52Vℚ62−a51∪a42Ra51,a62Vℚ62−a61∪a52Ra61,a12Vℚ62−a11∪a62Ra11,a32Vℚ62−a51∪a52Ra21,a42Vℚ62−a61∪a62Ra31,a52Vℚ62−a11∪a12Ra41,a62Vℚ62−a21∪a22Ra11,a52Vℚ62−a31∪a32Ra21,a62Vℚ62−a41∪a42Ra11,a42Vℚ62−a31∪a22Ra21,a52Vℚ62−a41∪a32Ra31,a62Vℚ62−a51∪a42Table 7
Rr with minimum cardinalities for 1≤r≤12.
RNsElementsR1=Ra11a12Vℚ62−a21,a31,a41∪a42,a52,a62R3=Ra31a32Vℚ62−a41,a51,a61∪a12,a22,a62R5=Ra51a52Vℚ62−a11,a21,a61∪a22,a32,a42R7=Ra21a12Vℚ62−a11,a51,a61∪a22,a32,a42R9=Ra41a32Vℚ62−a11,a21,a31∪a42,a52,a62R11=Ra61a52Vℚ62−a31,a41,a51∪a12,a22,a62R2=Ra21a22Vℚ62−a31,a41,a51∪a12,a52,a62R4=Ra41a42Vℚ62−a11,a51,a61∪a12,a22,a32R6=Ra61a62Vℚ62−a11,a21,a31∪a32,a42,a52R8=Ra31a22Vℚ62−a11,a21,a61∪a32,a42,a52R10=Ra51a42Vℚ62−a21,a31,a41∪a12,a52,a62R12=Ra11a62Vℚ62−a41,a51,a61∪a12,as22,a32Table 8
R¯u for 1≤u≤23.
RNsElementsEqualityRa11,a21Vℙ61−a13,a43Ra41,a51,Ra12,a22, Ra42,a52,Ra31,a61,Ra32,a62Ra21,a31Vℙ61−a23,a53Ra51,a61,Ra22,a32, Ra51,a61,Ra11,a41, Ra12,a42,Ra11,a41,Ra12,a42, Ra13,a33,Ra52,a82Ra31,a41Vℙ61−a33,a63Ra11,a61,Ra32,a42,Ra12,a62, Ra21,a51,Ra22,a52,Ra11,a61Table 9
R¯u for 24≤u≤35.
RNsElementsRa12,a13Vℙ61−a21,a31,a41Ra22,a23Vℙ61−a31,a41,a51Ra32,a33Vℙ61−a41,a51,a61Ra42,a43Vℙ61−a11,a51,a61Ra52,a53Vℙ61−a11,a21,a61Ra62,a63Vℙ61−a11,a21,a31Ra22,a13Vℙ61−a11,a51,a61Ra32,a23Vℙ61−a11,a21,a61Ra42,a33Vℙ61−a11,a21,a31Ra52,a43Vℙ61−a21,a31,a41Ra62,a53Vℙ61−a31,a41,a51Ra12,a63Vℙ61−a41,a51,a61Table 10
Rr for 1≤r≤27.
RNsElementsEqualityRa11,a31Vℙ61−a21,a51∪a22,a52Ra41,a61,Ra12,a32,Ra42,a62, Ra13,a23, Ra43,a53,Ra33,a63Ra21,a41Vℙ61−a31,a61∪a32,a62Ra11,a51,Ra22,a42,Ra12,a52, Ra11,a51,Ra12,a52,Ra23,a33, Ra53,a63,Ra13,a43,Ra13,a63Ra31,a51Vℙ61−a11,a41∪a12,a42Ra21,a61,Ra32,a52,Ra22,a62, Ra21,a61,Ra22,a62,Ra33,a43, Ra13,a63,Ra23,a53,Ra23,a53Table 11
Rr for 28≤r≤45.
RNsElementsRa11,a12Vℙ61−a22,a62∪a23,a53Ra31,a32Vℙ61−a22,a42∪a13,a43Ra51,a52Vℙ61−a42,a62∪a33,a63Ra11,a23Vℙ61−a31,a41∪a13,a43Ra31,a43Vℙ61−a51,a61∪a33,a63Ra51,a63Vℙ61−a11,a21∪a23,a53Ra21,a63Vℙ61−a51,a61∪a13,a43Ra41,a23Vℙ61−a11,a21∪a33,a63Ra61,a43Vℙ61−a31,a41∪a23,a53Ra21,a22Vℙ61−a12,a32∪a33,a63Ra41,a42Vℙ61−a32,a52∪a23,a53Ra61,a62Vℙ61−a12,a52∪a13,a43Ra21,a33Vℙ61−a41,a51∪a23,a53Ra41,a53Vℙ61−a11,a61∪a13,a43Ra11,a63Vℙ61−a21,a31∪a33,a63Ra31,a13Vℙ61−a11,a61∪a23,a53Ra51,a33Vℙ61−a21,a31∪a13,a43Ra11,a53Vℙ61−a41,a51∪a33,a63Table 12
R¯u for 1≤u≤30.
RNsElementsEqualityRa11,a21Vℙ81−a13,a53Ra51,a61,Ra12,a22, Ra52,a62,Ra41,a71,Ra42,a72Ra21,a31Vℙ81−a23,a63Ra61,a71,Ra22,a32, Ra51,a81,Ra52,a82, Ra62,a72,Ra11,a41,Ra12,a42, Ra13,a33,Ra52,a82Ra31,a41Vℙ81−a33,a73Ra71,a81,Ra32,a42,Ra72,a82, Ra21,a51,Ra22,a52,Ra11,a61, Ra12,a62Ra41,a51Vℙ81−a43,a83Ra11,a81,Ra42,a52,Ra12,a82, Ra21,a61,Ra32,a62Table 13
R¯u for 28≤u≤38.
RNsElementsEqualityRa11,a31Vℙ81−a21,a51∪a22,a52Ra51,a71,Ra12,a32,Ra52,a72, Ra41,a81,Ra42,a82,Ra13,a23, Ra43,a53,Ra53,a83Ra21,a41Vℙ81−a31,a61∪a32,a62Ra61,a81,Ra22,a42,Ra62,a82, Ra11,a51,Ra12,a52,Ra23,a33, Ra53,a63,Ra13,a43,Ra13,a63Ra31,a51Vℙ81−a41,a71∪a42,a72Ra11,a71,Ra32,a52,Ra12,a72, Ra21,a61,Ra22,a62,Ra33,a43, Ra63,a73,Ra23,a53,Ra23,a73Ra41,a61Vℙ81−a51,a81∪a52,a82Ra21,a81,Ra32,a52,Ra22,a82, Ra31,a71,Ra32,a72,Ra43,a53, Ra13,a83,Ra33,a73, Ra33,a83Table 14
R¯u for 39≤u≤54.
RNsElementsRa11a12Vℙ81−a22,a82∪a23,a73Ra31a32Vℙ81−a22,a42∪a13,a43Ra51a52Vℙ81−a42,a62∪a33,a63Ra71a72Vℙ81−a62,a72∪a53,a83Ra12a13Vℙ81−a21,a31,a41,a51Ra32a33Vℙ81−a41,a51,a61,a71Ra52a53Vℙ81−a11,a61,a71,a81Ra72a73Vℙ81−a11,a21,a31,a81Ra22a13Vℙ81−a11,a61,a71,a81Ra42a33Vℙ81−a11,a21,a31,a81Ra62a53Vℙ81−a21,a31,a41,a51Ra82a73Vℙ81−a41,a51,a61,a71Ra21a22Vℙ81−a12,a32∪a33,a83Ra41a42Vℙ81−a32,a52∪a23,a53Ra61a62Vℙ81−a52,a72∪a43,a73Ra81a82Vℙ81−a12,a72∪a13,a63Ra22a23Vℙ81−a31,a41,a51,a61Ra42a43Vℙ81−a51,a61,a71,a81Ra62a63Vℙ81−a11,a21,a71,a81Ra82a83Vℙ81−a11,a21,a31,a41Ra32a23Vℙ81−a11,a21,a71,a81Ra52a43Vℙ81−a11,a21,a31,a41Ra72a63Vℙ81−a31,a41,a51,a61Ra12a83Vℙ81−a51,a61,a71,a81Table 15
Rr with minimum cardinalities for 1≤r≤16.
RNsElementsR1=a11,a23Vℙ81−a21,a31,a41∪a13,a53R3=a31,a43Vℙ81−a41,a51,a61∪a33,a73R5=a51,a63Vℙ81−a11,a21,a61∪a13,a53R7=a71,a83Vℙ81−a21,a31,a41∪a33,a73R9=a21,a83Vℙ81−a61,a71,a81∪a13,a53R11=a41,a23Vℙ81−a11,a21,a81∪a33,a73R13=a61,a43Vℙ81−a21,a31,a41∪a13,a53R15=a81,a63Vℙ81−a41,a51,a61∪a33,a73R2=a21,a33Vℙ81−a31,a41,a51∪a23,a63R4=a41,a53Vℙ81−a11,a51,a61∪a43,a83R6=a61,a73Vℙ81−a11,a21,a31∪a23,a63R8=a81,a13Vℙ81−a31,a41,a51∪a43,a83R10=a31,a13Vℙ81−a11,a71,a81∪a23,a63R12=a51,a33Vℙ81−a11,a21,a31∪a43,a83R14=a71,a53Vℙ81−a31,a41,a51∪a23,a63R16=a11,a73Vℙ81−a51,a61,a71∪a43,a83Table 16
R¯u for 1≤u≤57.
RNsElementsEqualityRa11,a21Vℙ101−a13,a63Ra61,a71,Ra12,a22, Ra62,a72,Ra51,a81,Ra52,a82, Ra31,a101,Ra32,a102, Ra41,a91,Ra42,a92, Ra53,a73Ra21,a31Vℙ101−a23,a73Ra71,a81,Ra22,a32, Ra71,a81,Ra11,a41, Ra62,a72,Ra11,a41,Ra12,a42, Ra61,a91,Ra62,a92, Ra13,a33,Ra63,a83, Ra61,a91,Ra62,a92Ra31,a41Vℙ101−a33,a83Ra81,a91,Ra32,a42,Ra82,a92, Ra21,a51,Ra22,a52,Ra21,a51, Ra71,a101,Ra72,a102, Ra23,a43,Ra73,a93, Ra12,a62Ra41,a51Vℙ101−a43,a93Ra91,a101,Ra42,a52,Ra92,a102, Ra31,a61,Ra32,a62, Ra11,a81,Ra12,a82, Ra33,a53,Ra83,a103Ra41,a51Vℙ101−a53,a103Ra11,a101,Ra42,a52,Ra12,a102, Ra41,a71,Ra32,a62, Ra21,a91,Ra22,a92, Ra43,a63,Ra13,a93Table 17
R¯u for 58≤u≤122.
RNsElementsEqualityRa11,a31Vℙ101−a21,a71∪a22,a72Ra61,a81,Ra12,a32,Ra62,a82, Ra51,a91,Ra52,a92,Ra13,a23, Ra63,a73,Ra53,a83,Ra33,a103Ra21,a41Vℙ101−a31,a81∪a32,a82Ra71,a91,Ra22,a42,Ra72,a92, Ra11,a51,Ra12,a52, Ra61,a101,Ra62,a102, Ra23,a33,Ra73,a83, Ra13,a43,Ra63,a93Ra31,a51Vℙ101−a41,a91∪a42,a92Ra81,a101,Ra32,a52,Ra82,a102, Ra21,a61,Ra22,a62, Ra11,a71,Ra12,a72,Ra33,a43, Ra83,a93,Ra23,a53, Ra73,a103Ra41,a61Vℙ101−a51,a101∪a52,a102Ra11,a91,Ra42,a62,Ra12,a92, Ra31,a71,Ra32,a72, Ra21,a81,Ra22,a82, Ra43,a53,Ra93,a103, Ra13,a83Ra51,a71Vℙ101−a11,a61∪a12,a62Ra21,a101,Ra52,a72,Ra22,a102, Ra41,a81,Ra42,a82,Ra53,a63, Ra13,a103,Ra43,a73, Ra23,a93Ra11,a12Vℙ101−a22,a102∪a23,a93Ra21,a22Vℙ101−a12,a32∪a33,a103Ra31,a32Vℙ101−a22,a42∪a13,a43Ra41,a42Vℙ101−a32,a52∪a23,a53Ra51,a52Vℙ101−a42,a52∪a33,a63Ra61,a62Vℙ101−a52,a72∪a43,a73Ra71,a72Vℙ101−a62,a82∪a53,a83Ra81,a82Vℙ101−a72,a82∪a63,a93Ra91,a92Vℙ101−a82,a102∪a73,a103Ra101,a102Vℙ101−a12,a92∪a13,a83Table 18
R¯u for 123≤u≤142.
RNsElementsRa12,a13Vℙ101−a21,a31,a41,a51,a61Ra32,a33Vℙ101−a41,a51,a61,a71,a81Ra52,a53Vℙ101−a61,a71,a81,a91,a101Ra72,a73Vℙ101−a11,a21,a81,a91,a101Ra92,a93Vℙ101−a11,a21,a31,a41,a101Ra22,a13Vℙ101−a11,a71,a81,a91,a101Ra42,a33Vℙ101−a11,a21,a31,a91,a101Ra62,a53Vℙ101−a11,a21,a31,a41,a51Ra82,a73Vℙ101−a31,a41,a51,a61,a71Ra102,a93Vℙ101−a51,a61,a71,a81,a91Ra22,a23Vℙ101−a31,a41,a51,a61,a71Ra42,a43Vℙ101−a51,a61,a71,a81,a91Ra62,a63Vℙ101−a11,a71,a81,a91,a101Ra82,a83Vℙ101−a11,a21,a31,a91,a101Ra102,a103Vℙ101−a11,a21,a31,a41,a51Ra32,a23Vℙ101−a11,a21,a81,a91,a101Ra52,a43Vℙ101−a11,a21,a31,a41,a101Ra72,a63Vℙ101−a21,a31,a41,a51,a61Ra92,a83Vℙ101−a41,a51,a61,a71,a81Ra12,a103Vℙ101−a61,a71,a81,a91,a101Table 19
Rr with minimum cardinalities for 1≤r≤20.
RNsElementsR1=a11,a23Vℙ101−a31,a41,a51,a61∪a52∪a13,a53R3=a31,a43Vℙ101−a51,a61,a71,a81∪a72∪a33,a73R5=a51,a63Vℙ101−a71,a81,a91,a101∪a92∪a53,a93R7=a71,a83Vℙ101−a11,a21,a91,a101∪a12∪a13,a73R9=a91,a103Vℙ101−a11,a21,a31,a41∪a32∪a33,a93R11=a21,a103Vℙ101−a71,a81,a91,a101∪a82∪a13,a73R13=a41,a23Vℙ101−a11,a21,a91,a101∪a102∪a33,a93R15=a61,a43Vℙ101−a11,a21,a31,a41∪a22∪a13,a53R17=a81,a63Vℙ101−a31,a41,a51,a61∪a42∪a33,a73R19=a101,a83Vℙ101−a51,a61,a71,a81∪a62∪a53,a93R2=a21,a33Vℙ101−a41,a51,a61,a71∪a62∪a23,a63R4=a41,a53Vℙ101−a61,a71,a81,a91∪a82∪a43,a83R6=a61,a73Vℙ101−a11,a81,a91,a101∪a102∪a63,a103R8=a81,a93Vℙ101−a11,a21,a31,a101∪a22∪a23,a83R10=a11,a103Vℙ101−a21,a31,a41,a51∪a42∪a43,a103R12=a31,a13Vℙ101−a11,a81,a91,a101∪a92∪a23,a83R14=a51,a33Vℙ101−a11,a21,a31,a101∪a12∪a43,a103R16=a71,a53Vℙ101−a21,a31,a41,a51∪a32∪a23,a63R18=a91,a73Vℙ101−a41,a51,a61,a71∪a52∪a43,a83R20=a11,a93Vℙ101−a61,a71,a81,a91∪a72∪a63,a103Table 20
R¯u for 1≤u≤6.
Resolving neighbourhoodsSet/elementsRa11,a12Ra21,a22Ra31,a32Vℙ62Ra41,a42Ra51,a52Ra61,a62Vℙ62Table 21
R¯u for 7≤u≤24.
RNsElementsEqualityRa11,a21Vℙ62−a13,a53Ra41,a51,Ra12,a22,Ra42,a52, Ra41,a71,Ra42,a72Ra21,a31Vℙ62−a23,a63Ra11,a61,Ra22,a32,Ra62,a72, Ra11,a41,Ra12,a42,Ra51,a81, Ra52,a82,Ra13,a33,Ra52,a82Ra31,a41Vℙ62−a33,a73Ra71,a81,Ra32,a42,Ra72,a82, Ra21,a51,Ra22,a52,Ra11,a61, Ra12,a62Table 22
R¯u for 25≤u≤29.
RNsElementsEqualityRa11,a31Vℙ62−a21,a51∪a22,a52Ra51,a71,Ra12,a32,Ra52,a72, Ra41,a81,Ra42,a82,Ra13,a23, Ra43,a53,Ra53,a83Ra21,a41Vℙ62−a31,a61∪a32,a62Ra61,a81,Ra22,a42,Ra62,a82, Ra11,a51,Ra12,a52,Ra23,a33, Ra53,a63,Ra13,a43,Ra13,a63Ra31,a51Vℙ62−a41,a71∪a42,a72Ra11,a71,Ra32,a52,Ra12,a72, Ra22,a62,Ra33,a43,Ra63,a73, Ra23,a53,Ra23,a73,Ra21,a61Table 23
Rr with minimum cardinalities for 1≤r≤12.
RNsElementsR1=Ra12,a13Vℙ62−a21,a31,a41∪a22,a32,a42∪a23,a33R2=Ra22,a23Vℙ62−a31,a41,a51∪a32,a42,a52∪a33,a43R3=Ra32,a33Vℙ62−a41,a51,a61∪a42,a52,a62∪a43,a53R4=Ra42,a43Vℙ62−a11,a51,a61∪a12,a52,a62∪a53,a63R5=Ra52,a53Vℙ62−a11,a21,a61∪a12,a22,a62∪a13,a63R6=Ra62,a63Vℙ62−a11,a21,a31∪a12,a22,a32∪a13,a23R7=Ra22,a13Vℙ62−a11,a51,a61∪a12,a52,a62∪a53,a63R8=Ra32,a23Vℙ62−a11,a21,a61∪a12,a22,a62∪a13,a63R9=Ra42,a33Vℙ62−a11,a21,a31∪a12,a22,a32∪a13,a23R10=Ra52,a43Vℙ62−a21,a31,a41∪a22,a32,a42∪a23,a33R11=Ra62,a53Vℙ62−a31,a41,a51∪a32,a42,a52∪a33,a43R12=Ra12,a63Vℙ62−a41,a51,a61∪a42,a52,a62∪a43,a53Table 24
R¯u for 1≤u≤30.
RNsElementsEqualityRa12,a32Vℙ63−a22,a52Ra42,a62Ra22,a42Vℙ63−a32,a62Ra51,a61,Ra11,a51Ra32,a52Vℙ63−a12,a42Ra22,a62Ra13,a33Vℙ63−a23,a53Ra43,a63Ra23,a43Vℙ63−a33,a63Ra53,a63,Ra13,a53Ra33,a53Vℙ63−a13,a43Ra23,a63Ra33,a53Vℙ63−a13,a43Ra23,a63Ra11,a33Vℙ63−a31,a51Ra51,a13Ra21,a43Vℙ63−a41,a61Ra61,a23Ra31,a53Vℙ63−a11,a51Ra11,a43Vℙ63−a51∪a32Ra21,a53Vℙ63−a61∪a42Ra31,a63Vℙ63−a11∪a52Ra22,a23Vℙ63−a42,a52Ra32,a33Vℙ63−a52,a62Ra42,a43Vℙ63−a12,a62Ra52,a53Vℙ63−a12,a22Ra62,a63Vℙ63−a22,a32Ra12,a13Vℙ63−a32,a42Table 25
R¯u for 31≤u≤78.
RNsElementsEqualityRa22,a13Vℙ63−a11,a51,a61∪a62Ra32,a23Vℙ63−a11,a21,a61∪a12Ra42,a33Vℙ63−a11,a21,a31∪a22Ra52,a43Vℙ63−a21,a31,a41∪a32Ra62,a53Vℙ63−a31,a41,a51∪a42Ra12,a63Vℙ63−a41,a51,a61∪a52Ra11,a21Vℙ83−a12,a42∪a13,a43Ra41,a51,Ra31,a61,Ra32,a62Ra21,a31Vℙ83−a22,a52∪a23,a53Ra51,a61,Ra11,a41,Ra12,a42,Ra51,a81,Ra52,a82,Ra13,a33Ra31,a41Vℙ83−a32,a62∪a33,a63Ra11,a61,Ra21,a51,Ra11,a61Ra11,a31Vℙ83−a21,a51∪a13,a43Ra41,a61,Ra12,a22,Ra31,a61,Ra42,a52,Ra32,a62Ra21,a41Vℙ83−a31,a61∪a23,a53Ra11,a51,Ra22,a32,Ra11,a41,Ra51,a81,Ra52,a62,Ra12,a42Ra31,a51Vℙ83−a11,a41∪a33,a63Ra21,a51,Ra11,a61,Ra32,a42,Ra12,a62,Ra22,a52Ra13,a23Vℙ83−a21,a51∪a22,a52Ra43,a53,Ra13,a43Ra23,a33Vℙ83−a31,a61∪a32,a62Ra53,a63,Ra23,a53Ra33,a43Vℙ83−a11,a41∪a12,a42Ra13,a63,Ra33,a63Table 26
R¯u for 79≤u≤84.
RNsElementsRa21a12Vℙ63−a11,a51,a61∪a62∪a53Ra31a22Vℙ63−a11,a21,a61∪a12∪a63Ra41a32Vℙ63−a11,a21,a31∪a22∪a13Ra51a42Vℙ63−a21,a31,a41∪a32∪a23Ra61a52Vℙ63−a31,a41,a51∪a42∪a33Ra11a62Vℙ63−a41,a51,a61∪a52∪a43Table 27
Rr with minimum cardinalities for 1≤r≤6.
RNsElementsR1=Ra11a12Vℙ63−a21,a31,a41∪a22,a32∪a23,a33R2=Ra21a22Vℙ63−a31,a41,a51∪a32,a42∪a33,a43R3=Ra31a32Vℙ63−a41,a51,a61∪a42,a52∪a43,a53R4=Ra41a42Vℙ63−a11,a51,a61∪a52,a62∪a53,a63R5=Ra51a52Vℙ63−a11,a21,a61∪a12,a62∪a13,a63R6=Ra61a62Vℙ63−a11,a21,a31∪a12,a22∪a13,a23Table 28
R¯u for 1≤u≤31.
RNsElementsEqualityRa12,a32Vℙ65−a22,a52Ra42,a62Ra22,a42Vℙ65−a32,a62Ra12,a52Ra32,a62Vℙ65−a42,a62Ra22,a62Ra11,a22Vℙ65−a12∪a43Ra22,a62Ra21,a32Vℙ65−a22∪a53Ra12,a43Ra31,a42Vℙ65−a32∪a63Ra22,a53Ra41,a52Vℙ65−a42∪a13Ra32,a63Ra51,a62Vℙ65−a52∪a23Ra61,a12Vℙ65−a62∪a33Ra11,a32Vℙ65−a51∪a13Ra21,a42Vℙ65−a61∪a23Ra31,a52Vℙ65−a11∪a33Ra41,a62Vℙ65−a21∪a43Ra11,a52Vℙ65−a31∪a53Ra21,a62Vℙ65−a41∪a63Ra11,a23Vℙ65−a31∪a13Ra21,a33Vℙ65−a41∪a23Ra11,a43Ra31,a43Vℙ65−a51∪a33Ra21,a53Ra41,a53Vℙ65−a61∪a43Ra31,a63Ra51,a63Vℙ65−a11∪a53Ra61,a13Vℙ65−a21∪a63Table 29
R¯u for 32≤u≤47.
RNsElementsEqualityRa11,a21Vℙ65−a12,a42∪a13Ra41,a51,Ra31,a61,Ra32,a62Ra21,a31Vℙ65−a22,a52∪a23Ra11,a41,Ra51,a61,Ra12,a42,Ra51,a81,Ra52,a82,Ra13,a33Ra31,a41Vℙ65−a32,a62∪a33Ra21,a51,Ra11,a61,Ra11,a61Ra12,a43Vℙ65−a61∪a32∪a53Ra21,a51Vℙ65−a11∪a42∪a63Ra31,a61Vℙ65−a21∪a52∪a13Table 30
R¯u for 48≤u≤62.
RNsElementsEqualityRa11,a33Vℙ65−a31,a51∪a22,a52Ra21,a43Vℙ65−a41,a61∪a32,a62Ra31,a53Vℙ65−a11,a51∪a12,a42Ra41,a63Vℙ65−a21,a61∪a22,a52Ra11,a53Vℙ65−a31,a51∪a12,a62Ra21,a63Vℙ65−a41,a61∪a12,a32Ra13,a23Vℙ65−a21,a51∪a22,a52Ra43,a53,Ra13,a43Ra23,a33Vℙ65−a31,a61∪a32,a62Ra53,a63,Ra23,a53Ra33,a43Vℙ65−a11,a41∪a12,a42Ra13,a63,Ra33,a63Ra11,a31Vℙ65−a21,a51∪a13,a43Ra41,a61,Ra31,a61,Ra12,a22,Ra42,a52,Ra12,a42Ra21,a41Vℙ65−a31,a61∪a23,a53Ra11,a51,Ra11,a51,Ra22,a32,Ra52,a62,Ra22,a52Ra31,a51Vℙ65−a11,a41∪a33,a53Ra21,a61,Ra32,a42,Ra12,a62,Ra32,a62Table 31
R¯u for 63≤u≤68.
RNsElementsRa12,a13Vℙ65−a22,a32,a42∪a23,a33Ra22,a23Vℙ65−a32,a42,a52∪a33,a43Ra32,a33Vℙ65−a42,a52,a62∪a43,a53Ra42,a43Vℙ65−a12,a52,a62∪a53,a63Ra42,a43Vℙ65−a12,a22,a62∪a13,a63Ra42,a43Vℙ65−a12,a22,a32∪a13,a23Table 32
Rr with minimum cardinalities for 1≤r≤12.
RNsElementsR1=Ra22a13Vℙ65−a11,a51,a61∪a12,a52,a62∪a53,a63R2=Ra32a23Vℙ65−a11,a21,a61∪a12,a22,a62∪a13,a63R3=Ra42a33Vℙ65−a11,a21,a31∪a12,a22,a32∪a13,a23R4=Ra52a43Vℙ65−a21,a31,a41∪a22,a32,a42∪a23,a33R5=Ra62a53Vℙ65−a31,a41,a51∪a32,a42,a52∪a33,a43R6=Ra12a63Vℙ65−a41,a51,a61∪a42,a52,a62∪a43,a53R7=Ra11a12Vℙ65−a21,a31,a41∪a42,a52,a62∪a43,a53R8=Ra21a22Vℙ65−a31,a41,a51∪a12,a52,a62∪a53,a63R9=Ra31a32Vℙ65−a41,a51,a61∪a12,a22,a62∪a13,a63R10=Ra41a42Vℙ65−a11,a51,a61∪a12,a22,a32∪a13,a23R11=Ra51a52Vℙ65−a11,a21,a61∪a22,a32,a42∪a23,a33R12=Ra61a62Vℙ65−a11,a21,a31∪a32,a42,a52∪a33,a43Table 33
R¯u for 1≤u≤39.
RNsElementsEqualityRa11,a21Vℙ67−a13,a53Ra41,a51,Ra12,a22,Ra42,a52, Ra41,a71,Ra42,a72Ra21,a31Vℙ67−a23,a63Ra11,a61,Ra22,a32,Ra62,a72, Ra11,a41,Ra12,a42,Ra51,a81, Ra52,a82,Ra13,a33,Ra52,a82Ra31,a41Vℙ67−a33,a73Ra71,a81,Ra32,a42,Ra72,a82, Ra21,a51,Ra22,a52,Ra11,a61, Ra12,a62Ra11,a42Vℙ67−a21,a51Ra21,a52Vℙ67−a31,a61Ra31,a62Vℙ67−a11,a41Ra11,a32Vℙ67−a22∪a43Ra21,a42Vℙ67−a32∪a53Ra31,a52Vℙ67−a42∪a63Ra41,a62Vℙ67−a52∪a13Ra11,a52Vℙ67−a62∪a23Ra21,a62Vℙ67−a12∪a33Ra12,a33Vℙ67−a21∪a53Ra22,a43Vℙ67−a31∪a63Ra32,a53Vℙ67−a41∪a13Ra42,a63Vℙ67−a51∪a23Ra52,a13Vℙ67−a61∪a33Ra62,a23Vℙ67−a11∪a43Table 34
R¯u for 40≤u≤51.
RNsElementsRa11,a43Vℙ67−a31∪a32∪a53Ra21,a53Vℙ67−a41∪a42∪a63Ra31,a63Vℙ67−a51∪a52∪a13Ra11,a33Vℙ67−a51∪a52∪a23Ra11,a53Vℙ67−a61∪a62∪a23Ra21,a63Vℙ67−a11∪a12∪a33Ra11,a33Vℙ67−a51∪a52∪a23Ra21,a43Vℙ67−a61∪a62∪a33Ra31,a53Vℙ67−a11∪a12∪a43Ra41,a63Vℙ67−a21∪a22∪a53Ra11,a53Vℙ67−a31∪a32∪a63Ra21,a63Vℙ67−a41∪a42∪a13Table 35
R¯u for 52≤u≤69.
RNsElementsEqualityRa11,a12Vℙ67−a22,a62∪a13,a63Ra21,a22Vℙ67−a12,a32∪a13,a23Ra31,a32Vℙ67−a22,a42∪a23,a33Ra41,a42Vℙ67−a32,a52∪a33,a43Ra51,a52Vℙ67−a42,a62∪a43,a53Ra61,a62Vℙ67−a12,a52∪a53,a63Ra11,a31Vℙ67−a21,a51∪a22,a52Ra41,a61,Ra12,a32,Ra42,a62Ra21,a41Vℙ67−a31,a61∪a32,a62Ra11,a51,Ra22,a52, Ra12,a52Ra31,a51Vℙ67−a11,a41∪a12,a42Ra21,a61,Ra32,a52,Ra22,a62Table 36
R¯u for 70≤u≤81.
RNsElementsRa12a13Vℙ67−a11,a51,a61∪a43,a53Ra22a23Vℙ67−a11,a21,a61∪a53,a63Ra32a33Vℙ67−a11,a21,a31∪a13,a63Ra42a43Vℙ67−a21,a31,a41∪a13,a23Ra52a53Vℙ67−a31,a41,a51∪a23,a33Ra62a63Vℙ67−a41,a51,a61∪a33,a43Ra22a13Vℙ67−a21,a31,a41∪a33,a43Ra32a23Vℙ67−a31,a41,a51∪a43,a53Ra42a33Vℙ67−a41,a51,a61∪a53,a63Ra52a43Vℙ67−a11,a51,a61∪a13,a63Ra62a53Vℙ67−a11,a21,a61∪a13,a23Ra12a63Vℙ67−a11,a21,a31∪a23,a33Table 37
Rr with minimum cardinalities for 1≤r≤12.
RNsElementsR1=Ra11,a13Vℙ67−a21,a31,a41−a12,a32,a42−a23,a33R2=Ra21,a23Vℙ67−a31,a41,a51−a22,a42,a52−a33,a43R3=Ra31,a33Vℙ67−a41,a51,a61−a32,a52,a62−a43,a53R4=Ra41,a43Vℙ67−a11,a51,a61−a12,a42,a62−a53,a63R5=Ra51,a53Vℙ67−a11,a21,a61−a12,y2,a52−a13,a63R6=Ra61,a63Vℙ67−a11,a21,a31−a22,a32,a62−a13,a23R7=Ra21,a13Vℙ67−a11,a51,a61−a22,a52,a62−a53,a63R8=Ra31,a23Vℙ67−a11,a21,a61−a12,a32,a62−a13,a63R9=Ra41,a33Vℙ67−a11,a21,a31−a12,a22,a42−a13,a23R10=Ra51,a43Vℙ67−a21,a31,a41−a22,a32,a52−a23,a33R11=Ra61,a53Vℙ67−a31,a41,a51−a32,a42,a62−a33,a43R12=Ra11,a63Vℙ67−a41,a51,a61−a12,a42,a52−a43,a53Table 38
FMD of rotationally symmetric networks having different FMD form≥8.
NetworkUpper bound ofdimfCommentTm2m/m+1Boundedℚm12Bounded and constantℚm22Bounded and constantℙm16m/5m−2Boundedℙm26m/3m+2Boundedℙm33m/2m−1Boundedℙm56m/3m+2Boundedℙm76m/3m+2BoundedTable 39
Values of FMD as they tends to∞, where m≥8∧m≡0mod2.
NetworkValues of FMD as they tends to∞Tmlimm⟶∞2m/m+12ℙm2limm⟶∞6m/5m−26/5ℙm2limm⟶∞6m/3m+22ℙm3limm⟶∞3m/2m−13/2ℙm5limm⟶∞6m/3m+22ℙm7limm⟶∞6m/3m+22



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*Source: 8417127-2021-06-17.xml*