# Metric Dimension on Path-Related Graphs

Authors: Saqib Nazeer; Muhammad Hussain; Fatimah Abdulrahman Alrawajeh; Sultan Almotairi
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

Graph theory has a large number of applications in the fields of computer networking, robotics, Loran or sonar models, medical networks, electrical networking, facility location problems, navigation problems etc. It also plays an important role in studying the properties of chemical structures. In the field of telecommunication networks such as CCTV cameras, fiber optics, and cable networking, the metric dimension has a vital role. Metric dimension can help us in minimizing cost, labour, and time in the above discussed networks and in making them more efficient. Resolvability also has applications in tricky games, processing of maps or images, pattern recognitions, and robot navigation. We defined some new graphs and named thems−middle graphs, s-total graphs, symmetrical planar pyramid graph, reflection symmetrical planar pyramid graph, middle tower path graph, and reflection middle tower path graph. In the recent study, metric dimension of these path-related graphs is computed.


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## Body

## 1. Introduction and Preliminary Results

Term metric dimension was firstly revealed by mathematician named Slater [1]. He explained the study of sonar or Loran model station by this concept. Harary, another mathematician, with the help of Melter, independently described the same concept of metric dimension [2].ConsiderG is a simple, undirected, and finite graph and T=s1,s2,…,sf is ordered subset of vertex set of G, so rs|T is a representation of the vertex s with respect to T which is f−tupledl,s1,dl,s2,dl,s3,…,dl,sf. If each element of vertex set possesses unique representation with T, then we conclude that T is resolving set of S. If T is the smallest resolving set for G, then its cardinality becomes metric dimension of G and can be symbolized as dimG.Path graph is the only graph with metric dimension 1 [3]. Metric dimension of cycle graph is 2. Resolvability also has applications in tricky games, processing of maps or images, pattern recognitions, and networking robot navigation [4]. More details of applications of metric dimension can be seen in [5–10]. It is suggested to study the following references for better comprehension and details of metric dimension of graphs [11–16].A lot of work has been carried out by many authors on metric dimension of path-related graphs. In 2012, Ali et al. computed constant metric dimension of middle path graph and power of path graph [10]. In 2017, Alholi et al. also put contribution in this field and found constant metric dimension [17]. Shahida and Sunitha found constant metric dimension of join of any two graphs [18]. In 2016, Peng et al. calculated metric dimension of Kenser graphs [19].Ali et al., in 2012, worked on metric dimension of total graph of path graph and found it to be constant [10]. In 2019, Pan et al. calculated the unbounded metric dimension of splitting graphs of path graph and cycle graph [20]. In 2021, Nawaz et al. proved that metric dimension of path power three and four is unbounded; also, they proved some results on edges of power of path and power of total graph [21].Our research is in continuation with their work, and we computed the metric dimension ofs-middle path graphs and s-total path graphs. These graphs have bounded metric dimension. In the paper under study, we also calculated metric dimension of other path-related graphs such as middle tower path graph, reflection middle tower path graph, symmetrical planar pyramid path graph, and reflection symmetrical planar pyramid path graph.


### 1.1. Metric Dimension ofs-Middle Graph

We generalized the concept of middle of a graph and extended this idea tos vertices. We took two particular cases for path graph and with the help of these results and finally computed the metric dimension of s-middle graph of the path graph.Definition 1.
For undirected simpleplanar or nonplanar graph G with VG and EG as vertex and edge set, respectively, VG=v1,v2,…,vl and EG=e1,e2,…,eu. We associate for every edge in edge set s vertices, forming another graph possessing vertex set represented by Vs−MidG which is given by Vs−MidG=VG∪∪u=1sVu, and if there are l vertices in G and number of edges is u, then s−MidG contains l+su vertices. Each vg in Vs−MidG is adjacent to vh in Vs−MidG:(1)
vg∼vgu if vg is incident to eg in G(2)
vgu∼vfu if eg is incident to ef in GExample 1.
For a particular case ofs-Mid (G), we dicussed two-middle path graph in which exactly two vertices were assigned against each edge. We take G=Pq (path graph containing q vertices). We associate hl and sl so that V∗=h1,h2,h3,…,hq−1,V∗∗=s1,s2,s3,…,sq−1.
Thus, the vertex set ofTwo−MidPq is VTwo−MidPq=VPq∪V∗Pq∪V∗∗Pq. Each vl∼vm in Two−MidPq:(1)
vl∼si and vl∼hl are incident to el in Pq(2)
sl∼sm and hl∼hm if and only if ei is adjacent to em in Pq
Two-middle path graph and three-middle path graph are explained in Figures 1 and 2.Figure 1
Two-middle path graph.Figure 2
Three-middle path graph.Theorem 1.
dimTwo−Midpq is 2, where q≥3.Proof.
Two-Mid Pq is not path when q≥3, so its metric dimension must be greater than 1. The vertex set for Two-Mid Pq is VTwo−MidPq and is given by VTwo−MidPq=v1,v2,v3,…,vq∪h1,h2,h3,…,hq−1∪s1,s2,s3,…,sq−1.
AssumeW=s1,sq−1 as a resolving set for Two-Mid Pq so that(1)rvl|W=l,q−1,l=1,l−1,q−l,l=2,3,4,…,q−1,l−1,1,l=q,rsj|W=j−1,q−j−1,∀j=1,2,3,…,q−1,rhj|W=2,q−1,j=1,j,q−j,j=2,3,4,…,q−2,j,2,j=q−1.
Whenl≠g and vl and vg have the same representation for W, contradiction arises, i.e.,(2)rvl|W=rvg|W⇒l−1,q−l=g−1,q−g⇒l=gcontradiction.
Similarly, whenl≠g,(3)rsl|W=rsg|W⇒l−1,q−l−1=g−1,q−g−1⇒1=gcontradiction.
Whenl≠g,(4)rhl|W=rhq|W⇒l,q−l=g−1,q−g−1⇒1=g−1and1=g+1contradiction.
Whenl≠g,(5)rvl|S=rsg|S⇒l,q−l=g,g−l⇒1=gcontradiction.
Whenl≠g,(6)rhl|W=rsg|W⇒l,q−l=g,q−1⇒1=gcontradiction.
Hence,Two-Mid Pq possesses W as its smallest resolving set, so its metric dimension is 2.Theorem 2.
dimThree−MidPq is 3, where q≥3.Proof.
The vertex set forThree−MidPq is VThree−MidPq and is given by(7)VThree−MidPq=v1,v2,v3,…,vq∪u1,u2,u3,…,uq−1∪w1,w2,w3,…,wq−1∪m1,m2,m3,…,mq−1.
AsThree−MidPq is not a path graph, so it possesses metric dimension greater than 1.
AssumeW=u1,v1,m1 as a resolving set for Three−MidPq so that(8)rvg|W=1,0,1,g=1,g−1,g,g−1,g=2,3,4,…,q,rug|W=0,1,2,g=1,g−1,g,g,g=2,3,4,…,q−1,rwg|W=2,1,2,g=1,g,g,g,g=2,3,4,…,q−1,rmg|W=2,1,0,g=1,g,g,g−1,g=2,3,4,…,q−1.
Whenl≠g and vl and vg have the same representation with respect to w, a contradiction arises, i.e.,(9)rvl|W=rvg|W⇒l−1,l,l−1=g−1,q,g−1⇒l=gcontradiction.
Similarly, whenl≠g,(10)rul|W=rug|W⇒l−1,l,l=g−1,g,g⇒l=gcontradiction.
Whenl≠g,(11)rwl|W=rwg|W⇒l,l,l=g,g,g⇒l=gcontradiction.
Whenl≠g,(12)rml|W=rmg|W⇒l,l,l−l=g,g,g−1⇒l=gcontradiction.
Whenl≠g,(13)rvl|W=rug|W⇒l−1,l,l−1=g−1,g,g⇒l=gcontradiction.
Whenl≠g,(14)rvl|W=rwg|W⇒l−1,l,l−l=g,g,g⇒l=gcontradiction.
Whenl≠g,(15)rul|W=rwg|W⇒l−1,l,l−1=g,g,g⇒l=gcontradiction.
Whenl≠g,(16)rul|W=rwg|W⇒l−1,l,l=g,g,g⇒l=gcontradiction.
Whenl≠g,(17)rul|W=rmg|W⇒l−1,l,l=g,g,g−1⇒l=gcontradiction.
Whenl≠g,(18)rwl|W=rmg|W⇒l,l,l=g,g,g−1⇒l=gcontradiction.
Hence,W is the resolving set for Three−MidPq. Now, we shall show that no resolving set having cardinality 2 exists.LetT1=vl,vg be metric basis of Three−MidPq; then, ruj|T1=rmj|T1=rwj|T1, so uniqueness is violated.LetT2=ul,ug be metric bases of Three−MidPq; then, rwj|T2=rmj|T2, so uniqueness is violated.LetT3=wl,wg be metric bases of Three−MidPq; then, ruj|T3=rmj|T3, so uniqueness is violated.LetT4=ml,mg be metric bases of Three−MidPq; then, rwj|T4=ruj|T4, so uniqueness is violated.LetT5=vl,wg be metric bases of Three−MidPq; then, ruj|T5=rmj|T5, so uniqueness is violated.LetT6=vl,ug be metric bases of Three−MidPq; then, rwj|T6=rmj|T6, so uniqueness is violated.LetT7=vl,mg be metric bases of Three−MidPq; then, rwj|T7=ruj|T7, so uniqueness is violated.LetT8=ul,wg be metric bases of Three−MidPq; then, rvj|T8=rmj−1|T8, so uniqueness is violated.LetT9=ul,mg be metric bases of Three−MidPq; then, rvj|T9=rmj−1|T9, so uniqueness is violated.LetT10=wl,mg be metric bases of Three−MidPq; then, rvj|T10=ruj−1|T10, so uniqueness is violated.
So, there are no metric bases of order 2. Hence,W is the resolving set of order 3 and metric dimension of Three−MidPq is 3.Theorem 3.
Metric dimension ofs-Middle Pq is s.Proof.
The vertex set ofs-Middle Pq is v1,v2,v3,…,vp∪wgm, where g=1,2,3,…,p−1 and m=1,2,3,…,s. Let W=w11,w12,w13,…,w1s be the resolving set of the s-Middle Pq; then, rvg|W=g,g−1,g−1,…,g−1s−tuple, where g=1,2,3,…,p:(19)rwhg|W=g−1,g,g,…,gs−tuple,g=1,2,3,…,p−1;h=1,g,g−1,g,…,gs−tuple,g=1,2,3,…,p−1;h=2,g,g,g−1,…,gs−tuple,g=1,2,3,…,p−1;h=3,⋅⋅⋅g,g,g,…,g−1s−tuple,g=1,2,3,…,p−1;h=s,
Whenl≠m and vl and vm possess the same representation for W, a contradiction arises, i.e.,(20)rvl|W=rvm|W⇒l,l−1,l−1,…,l−1s−tuple=m,m−1,m−1,…,m−1s−tuple⇒l=mcontradiction.
Similarly, whenl≠m,(21)rwlh|W=rwmh|W⇒l,l,l,…,l−1hthposition,l,…,ls−tuple=m,m,m,…,m−1hthposition,m,…,ms−tuple⇒l=mcontradiction.
Whenl≠m,(22)rwgl|W=rwgm|W⇒g,g,g,…,g−1lthposition,g,…,gs−tuple=g,g,g,…,g−1mthposition,g,…,gs−tuple⇒g−1=gcontradiction.
Whenl≠m,(23)u≠vrwlu|W=rwmv|W⇒l,l,l,…,l−1uthposition,l…,ls−tuple=m,m,m,…,m−1vthposition,m,…,ms−tuple⇒l−1=mandl=m−1contradiction.
Now, we shall show that there is no resolving set withs−1 elements that exists. So, on the contrary, let a set with s−1 elements be a resolving set for s-Middle Pq. Then, particularly, for s=3, there exists a resolving set of 3−1=2 elements. So, metric dimension of Three-Middle Pq is not 2, and it leads to contradiction.
So, there is no resolving set withs−1 elements, and hence, s-Middle Pq has metric dimension s.


### 1.2.s-Total Graph of a Path

We discuss here more generalized concept of total graph of some graphs. We shall apply the very concept fors=2,3 on the path graph and use these results to find metric dimension of s-total graph of the path graph.Definition 2.
For undirected simple (planar or nonplanar) graphG with vertex set VG and EG as the edge set, VG=v1,v2,…,vs and EG=e1,e2,…,eu. We associate for every edge in edge set s vertices, forming another graph possessing vertex set represented by Vs−TG which is given by Vs−TG=VG∪∪u=1sVu, and if the total number of vertices in G is r and number of edges is u, then s-Total G contains r+su vertices. Each vg in Vs−TG is adjacent to vh in Vs−TG:(1)
vg∼vf if vg∼vf in G(2)
vg∼vgu if vg is incident to eg in G(3)
vgu∼vfu if eg is incident to ef in GTwo-total graph of a path and Three-total graph of a path are explained in Figures 3 and 4.Figure 3
Two-total graph of the path graph.Figure 4
Three-total graph of the path graph.Theorem 4.
dimTwo−Tpq is 2, where q≥2.Proof.
Two−TPq is not the path when q≥2, so its metric dimension must be greater than 1. The vertex set for Two−TPq is VTwo−TPq and is given by VTwo−TPq=v1,v2,v3,…,vq∪h1,h2,h3,…,hq−1∪s1,s2,s3,…,sq−1.
AssumeW=h1,hq−1 as a resolving set for Two−TPq so that(24)rvl|W=l,q−1,j=1,l−1,q−l,j=2,3,4,…,q−1,l−1,1,l=q,rsh|W=l−1,q−l−1∀j=1,…,q−1,rhj|W=2,q−1l=1,l,q−l,q=2,3,4,…,q−2,l,2,l=q−1.
Whenu≠r and hu and hr possess the same representation for W, a contradiction arises, i.e.,(25)rhu|W=rhr|W⇒u−1,q−u−1=r−1,q−r⇒l=gcontradiction.
Similarly, whenu≠r,(26)rsu|W=rsr|W⇒u−1,q−u=r−1,q−r⇒u=rcontradiction.
Whenu≠r,(27)rhu|W=rsr|W⇒u,q−u=r,q−r⇒u=rcontradiction.
Whenu≠r,(28)rvu|W=rvr|W⇒u−l,q−u=r−1,q−r⇒u=rcontradiction.
Hence,Two−TPq possesses W as its smallest resolving set, so its metric dimension is 2.Theorem 5.
dimThree−Tpq is 3, where q≥3.Proof.
The vertex set forThree−TPq is VThree−TPq and is given by VThree−TPq=v1,v2,v3,…,vq∪u1,u2,u3,…,uq−1∪w1,w2,w3,…,wq−1∪m1,m2,m3,…,mq−1. As Three−TPq is not a path graph, so it possesses metric dimension greater than 1.
AssumeW=u1,v1,m1 as a resolving set for Three−TPq so that(29)rvg|W=1,0,1,g=1,g−1,g−1,g−1,g=2,3,4,…,q−1,rug|W=0,1,2,g=1,g−1,g,g,g=2,3,4,…,q−1,rwg|W=2,1,2,g=1,g,g,g,g=2,3,4,…,q−1,rmg|W=2,1,0,g=1,g,g,g−1,g=2,3,4,…,q−1.
Whenl≠g and vl and vg possess the same representation for W, a contradiction arises, i.e.,(30)rvl|W=rvg|W⇒l−1,l,l−1=g−1,g−1,g−1⇒l=gcontradiction.
Similarly, wheng≠l,(31)rug|W=rul|W⇒g−1,g,g=l−1,l,l⇒g=lcontradiction.
Whenl≠g,(32)rwl|W=rwg|W⇒l,l,l=g,g,g⇒l=gcontradiction.
Whenl≠g,(33)rml|W=rmg|W⇒l,l,l−l=g,g,g−1⇒l=gcontradiction.
Whenl≠g,(34)rvl|W=rug|W⇒l−1,l−1,l−1=g−1,g,g⇒l=gcontradiction.
Whenl≠g,(35)rul|W=rwg|W⇒l−1,l,l=g,g,g⇒l=gcontradiction.
Whenl≠g,(36)rul|W=rmg|W⇒l−1,l,l=g,g,g−1⇒l=gcontradiction.
Whenl≠g,(37)rwl|W=rmg|W⇒l,l,l=g,g,g−1⇒l=gcontradiction.
Hence,W is resolving set for Three-Total Pq. Now, we shall show that no resolving set having cardinality 2 exists.LetT1=vl,vg be a resolving set of Three−TPq; then, ruj|T1=rmj|T1=rwj|T1, so uniqueness is violated.LetT2=ul,ug be metric bases of Three−TPq; then, rwj|T2=rmj|T2, so uniqueness is violated.LetT3=wl,wg be metric bases of Three−TPq; then, ruj|T3=rmj|T3, so uniqueness is violated.LetT4=ml,mg be metric bases of Three−TPq; then, rwj|T4=ruj|T4, so uniqueness is violated.LetT5=vl,wg be metric bases of Three−TPq; then, ruj|T5=rmj|T5, so uniqueness is violated.LetT6=vl,ug be metric bases of Three−TPq; then, rwj|T6=rmj|T6, so uniqueness is violated.LetT7=vl,mg be metric bases of Three−TPq; then, rwj|T7=ruj|T7, so uniqueness is violated.LetT8=ul,wg be metric bases of Three−TPq; then, rvj|T8=rmj−1|T8, so uniqueness is violated.LetT9=ul,mg be metric bases of Three−TPq; then, rvj|T9=rwj−1|T9, so uniqueness is violated.LetT10=wl,mg be metric bases of Three−TPq; then, rvj|T10=ruj−1|T10, so uniqueness is violated.
So, there are no metric bases of order 2. Hence,W is the resolving set of order 3 and metric dimension of Three−TPq is 3.Theorem 6.
Metric dimension ofs−TotalPq is s.Proof.
The vertex set ofs-Total Pq is v1,v2,v3,…,vp∪wgm, where g=1,2,3,…,p−1 and m=1,2,3,…,s. Let W=w11,w12,w13,…,w1s be the resolving set of the s-Total Pq; then, rvg|W=g,g−1,g−1,…,g−1s−tuple, where g=1,2,3,…,p,(38)rwgh|W=g−1,g,g,…,gs−tuple,g=1,2,3,…,p−1;h=1,g,g−1,g,…,gs−tuple,g=1,2,3,…,p−1;h=2,g,g,g−1,…,gs−tuple,g=1,2,3,…,p−1;h=3,⋅⋅⋅g,g,g,…,g−1s−tuple,g=1,2,3,…,p−1;h=s.
Wheng≠m and vg and vm possess the same representation for W, a contradiction arises, i.e.,(39)rvg|W=rvm|W⇒g,g−1,g−1,…,g−1s−tuple=m,m−1,m−1,…,m−1s−tuple⇒g=mcontradiction.
Similarly, whenl≠m,(40)rwlh|W=rwmh|W⇒l,l,l,…,l−1hthposition,l,…,ls−tuple=m,m,m,…,m−1hthposition,m,…,ms−tuple⇒l=mcontradiction.
Whenl≠m,(41)rwgl|W=rwgm|W⇒g,g,g,…,g−1lthposition,g…,gs−tuple=g,g,g,…,g−1mthposition,g…,gs−tuple⇒g−1=gcontradiction.
Whenl≠m and u≠v,(42)rwlu|W=rwmv|W⇒l,l,l,…,l−1uthposition,l…,ls−tuple=m,m,m,…,m−1vthposition,m,…,ms−tuple⇒l−1=mandl=m−1contradiction.
Now, we shall show that there is no resolving set withs−1 elements exists. So, on the contrary, let a set with s−1 elements be a resolving set for s-Total Pq. Then, particularly, for s=3, there exists a resolving set of 3−1=2 elements. So, metric dimension of Three-Total Pq is not 2, and it leads to contradiction.
So, there is no resolving set withs−1 elements, and hence, s-Total Pq has metric dimension s.


### 1.3. Middle Tower Path Graph

Middle tower path graph is a special graph represented by MiddleTowers, which is related to the path graph, and later, we extended its concept to its reflection.Definition 3.
An undirected, simple, and planar graph possessing vertex setV=Vgt|1≤t≤s,1≤g≤t, where t=1,2,3,…,s. Each vertex of Middle Towers is adjacent in the following manner.(1)
vgt∼vgt+1 when g=1,2,3,…,t and t=1,2,3,…,s−1(2)
vgt∼vg+1t+1 when g=1,2,3,…,t and t=1,2,3,…,s−1(3)
vgt∼vg+1t when g=1,2,3,…,t and t=1,2,3,…,s−1Middle tower path graph is explained in Figure5.Figure 5
Middle tower path graph.Theorem 7.
Metric dimension of MiddleTowers is 1, for s=2.Proof.
As MiddleTower2 is a path graph (by definition) and every path graph has metric dimension 1, so metric dimension of this graph is 1.Theorem 8.
Metric dimension of MiddleTowers is 2, for s≥3.Proof.
As MiddleTowers is not path graph; hence, it possesses metric dimension ≥2. Vertex set of Middle Towers is denoted by V (Middle Towers) and is given by VMiddle TowerPs=∪p=1sh1p,h2p,h3p,…,hqp, where q=1,2,3,…,s−p+1. We take W=h11,h1s resolving set of Middle Towers; then, representation rhqm|W=q+m−2,s−m,q=1,2,3,…,s−m+1 and m=1,2,3,…,s.
Wheng≠r and vg and vr possess the same representation for W, a contradiction arises, i.e.,(43)rhqm|W=rhrm|W⇒q+m−2,s−m=r+m−2,s−m⇒q=rcontradiction.
Similarly, whenm≠n,(44)rhqm|W=rhqn|W⇒q+m−2,s−m=q+n−2,s−n⇒m=ncontradiction.
Whenm≠n and q≠d,(45)rhqm|W=rhdn|W⇒q+m−2,s−m=d+n−2,s−n⇒m=nq=dcontradiction.
So,W is the smallest resolving set for Middle TowerPs and its metric dimension is 2.


### 1.4. Reflection Middle Tower Path Graph

Mirror image of MiddleTowers is named as Reflection Middle TowerPs and is denoted by rlTowers.Definition 4.
Simple undirected graph possessingVrlTowers=V1gt|1≤t≤s,1≤g≤t∪V2gt|1≤t≤s,1≤g≤t, where t=1,2,3,…,s. Each vertex of rlTowers is adjacent in the following manner.(1)
vgit∼vgit+1 when i=1,2,g=1,2,3,…,t, and t=1,2,3,…,s−1(2)
vgit∼vg+1it+1 when i=1,2,g=1,2,3,…,t, and t=1,2,3,…,s−1(3)
vgit∼vg+1it when i=1,2,g=1,2,3,…,t, and t=1,2,3,…,s−1(4)
vg1t∼vg+12t−1 when g=1,2,3,…,t−1 and t=s(5)
vg+11t∼vg2t−1 when g=1,2,3,…,t−1 and t=sReflection middle tower path graph is explained in Figure6.Figure 6
rlTowers graph.Theorem 9.
Metric dimension ofrlTowers is 2, for s≥3.Proof.
The vertex set ofrlTowers is VrlTowers=∪g=2s−1l11g,l21g,l31g,…,lj1g∪∪g=2s−2l12g,l22g,l33g,…,lj2g, where j=1,2,3,…,s−g+1.
AssumeW=l11g,l21g as the resolving set; then,(46)rl11j|W=s−1,s−j+1,j=1,2,3,…,s−1,s−1,0,j=s.
rlj|W1q=s−q,s−j, where j=1,2,3,…,s−1 and q=2,3,4,…,s.
rlj|W2q=s−q−2,s−j, where j=1,2,3,…,q and q=2,3,4,…,s.
Whenm≠n and lm and ln possess the same representation for W, a contradiction arises, i.e.,(47)rlm11|W=rln11|W⇒s−1,s−m+1=s−1,s−n+1⇒m=ncontradiction.
Similarly, whenm≠n,(48)rlm1h|W=rln1h|W⇒s−h,s−m=s−h,s−n⇒m=ncontradiction.
Whenm≠n,(49)rlp1m|W=rlp1n|W⇒s−m,s−1=s−h,s−1⇒m=ncontradiction.
Whenm≠n and u≠v,(50)rlu1m|W=rlv1n|W⇒s−m,s−l=s−n,s−v⇒m=nandu=vcontradiction.
Whenm≠n,(51)rlm2g|W=rln2g|W⇒s+g−2,s−m=s+g−2,s−n⇒m=ncontradiction.
Whenm≠n and u≠v,(52)rlu2m|W=rlv2n|W⇒s+m−2,s−u=s+n−2,s−v⇒m=nandu=vcontradiction.
Whenm≠n,(53)rlc2m|W=rlc2n|W⇒s+m−2,s−c=s+n−2,s−c⇒m=ncontradiction.
Hence,W is the smallest resolving set and metric dimension of rlTowers is 2.


### 1.5. Symmetrical Planar Pyramid Graph

A graph is obtained by joining the neighbouring vertices of MiddleTowerPs in its base and is represented by SPPs. If there are s vertices in the base of SPPs, then its vertex set contains ss−1/2 vertices.Definition 5.
We defineSPPs with s vertices in base possesses vertex set ∪t=1sVpt, where t=1,2,3,…,s and p≤t.
Each vertex oftowers is adjacent in the following manner.(1)
vpt∼vpt+1 when p=1,2,3,…,s−1 and t=1,2,3,…,s−1 and p≤t(2)
vpt∼vp+1t+1 when p=1,2,3,…,s−1 and t=1,2,3,…,s−1(3)
vpt∼vp+1t when p=1,2,3,…,s and t=1,2,3,…,s−1Symmetrical planar pyramid graph is explained in Figure7.Figure 7
Symmetrical planar pyramid graph.Theorem 10.
Metric dimension ofSPPs is 2.Proof.
TheSPP has vertex set VSPPs=∪t=1s, where p=1,2,3,…s and t=1,2,3,…,s. Also, p≤t.
We takeW=v11,vss resolving set of Middle SPPs and then representation rvpt|W=p+j−1,s−t+jp=1,2,3,…,s and t=1,2,3,…,s. Also, p≤t.
Whenl≠g and vl and vg possess the same representation for W, a contradiction arises, i.e.,(54)rvlt|W=rvgt|W⇒l+j−1,s−t+j=g+j−1,s−t+j⇒l=gcontradiction.
Similarly, whenl≠g,(55)rvpl|W=rvpg|W⇒p+j−1,s−l+j=p+j−1,s−g+j⇒l=gcontradiction.
Whenl≠g and u≠v,(56)rvul|W=rvvg|W⇒l+j−1,s−u+j=g+j−1,s−v+j⇒l=gu=vcontradiction.
So,W is the smallest resolving set for SPPs and its metric dimension is 2.


### 1.6. Reflection Symmetrical Planar Pyramid Graph

Mirror image ofSPPs is named as reflection symmetrical planar pyramid graph and is denoted by rlSPPs.Definition 6.
Simple undirected graph possessing vertex setVrlSPPs=∪t=1sVp1t∪∪t=2s−1Vp2t, where p=1,2,3,…,s and t=1,2,3,…,s; also, s≤t. Each vertex of rlSPPs is adjacent in the following manner.(1)
vpit∼vpit+1 when p=1,2,3,…,s, t=1,2,3,…,s−1, p≤t, and i=1,2(2)
vpit∼vp+1it+1 when p=1,2,3,…,s, t=1,2,3,…,s−1, p≤t, and i=1,2(3)
vpit∼vpit when p=1,2,3,…,s, t=1,2,3,…,s−1, p≤t, and i=1,2(4)
vp1t∼vp2t−1 when t=s and g=2,3,4,…,s(5)
vp+11t∼vp2t−1 when t=s and p=1,2,3,…,s−1Reflection symmetrical planar pyramid graph is explained in Figure8.Figure 8
Reflection symmetrical planar pyramid graph.Theorem 11.
Metric dimension ofrlSPPs.Proof.
The vertex set ofrlSPPs is VrlSPPs=∪t=2s∪∪t=1s−1Vp2t, where p=1,2,3,…,s and t=1,2,3,…,s; also, s≤t.
AssumeW=v11,vss as the resolving set; then,(57)rv1tp|W=p−j−1,s−t+1J,p=1,2,3,…,st=1,2,3,…,sp≤tandj=t−p,rv2tp|W=s−t+p−1,2s−t−1,j=1,2,3,…,s−1.
Whenl≠g and vl and vg possess the same representation for W, a contradiction arises, i.e.,(58)rvl1t|W=rvg1t|W⇒l+j−1,s−t+j=g+j−1,s−g+j⇒l=gcontradiction.
Similarly, whenl≠g,(59)rvp1l|W=rvp1g|W⇒p+j−1,s−l+j=p−j−1,s−g+j⇒l=gcontradiction.
Whenl≠g,(60)rvl1u|W=rvg1p|W⇒l+j−1,s−u+j=g+j−1,s−p+j⇒l=gu=pcontradiction.
Whenl≠g,(61)rvl2t|W=rvg2t|W⇒s−t+l−1,2s−t−l=s−t+g,2s−t−1⇒l=gcontradiction.
Whenl≠g,(62)rvp2t|W=rvp2g|W⇒s−l+p−1,2s−t−1=s−g+p−1,2s−t−1⇒l=gcontradiction.
Whenl≠gandu≠p,(63)rvl2u|W=rvg2p|W⇒s−u+l−1,2s−u−1=s−v+g−1,2s−p−1⇒l=gandu=pcontradiction.
Hence,W is the smallest resolving set and metric dimension of rlSPPs is 2.


## 1.1. Metric Dimension ofs-Middle Graph

We generalized the concept of middle of a graph and extended this idea tos vertices. We took two particular cases for path graph and with the help of these results and finally computed the metric dimension of s-middle graph of the path graph.Definition 1.
For undirected simpleplanar or nonplanar graph G with VG and EG as vertex and edge set, respectively, VG=v1,v2,…,vl and EG=e1,e2,…,eu. We associate for every edge in edge set s vertices, forming another graph possessing vertex set represented by Vs−MidG which is given by Vs−MidG=VG∪∪u=1sVu, and if there are l vertices in G and number of edges is u, then s−MidG contains l+su vertices. Each vg in Vs−MidG is adjacent to vh in Vs−MidG:(1)
vg∼vgu if vg is incident to eg in G(2)
vgu∼vfu if eg is incident to ef in GExample 1.
For a particular case ofs-Mid (G), we dicussed two-middle path graph in which exactly two vertices were assigned against each edge. We take G=Pq (path graph containing q vertices). We associate hl and sl so that V∗=h1,h2,h3,…,hq−1,V∗∗=s1,s2,s3,…,sq−1.
Thus, the vertex set ofTwo−MidPq is VTwo−MidPq=VPq∪V∗Pq∪V∗∗Pq. Each vl∼vm in Two−MidPq:(1)
vl∼si and vl∼hl are incident to el in Pq(2)
sl∼sm and hl∼hm if and only if ei is adjacent to em in Pq
Two-middle path graph and three-middle path graph are explained in Figures 1 and 2.Figure 1
Two-middle path graph.Figure 2
Three-middle path graph.Theorem 1.
dimTwo−Midpq is 2, where q≥3.Proof.
Two-Mid Pq is not path when q≥3, so its metric dimension must be greater than 1. The vertex set for Two-Mid Pq is VTwo−MidPq and is given by VTwo−MidPq=v1,v2,v3,…,vq∪h1,h2,h3,…,hq−1∪s1,s2,s3,…,sq−1.
AssumeW=s1,sq−1 as a resolving set for Two-Mid Pq so that(1)rvl|W=l,q−1,l=1,l−1,q−l,l=2,3,4,…,q−1,l−1,1,l=q,rsj|W=j−1,q−j−1,∀j=1,2,3,…,q−1,rhj|W=2,q−1,j=1,j,q−j,j=2,3,4,…,q−2,j,2,j=q−1.
Whenl≠g and vl and vg have the same representation for W, contradiction arises, i.e.,(2)rvl|W=rvg|W⇒l−1,q−l=g−1,q−g⇒l=gcontradiction.
Similarly, whenl≠g,(3)rsl|W=rsg|W⇒l−1,q−l−1=g−1,q−g−1⇒1=gcontradiction.
Whenl≠g,(4)rhl|W=rhq|W⇒l,q−l=g−1,q−g−1⇒1=g−1and1=g+1contradiction.
Whenl≠g,(5)rvl|S=rsg|S⇒l,q−l=g,g−l⇒1=gcontradiction.
Whenl≠g,(6)rhl|W=rsg|W⇒l,q−l=g,q−1⇒1=gcontradiction.
Hence,Two-Mid Pq possesses W as its smallest resolving set, so its metric dimension is 2.Theorem 2.
dimThree−MidPq is 3, where q≥3.Proof.
The vertex set forThree−MidPq is VThree−MidPq and is given by(7)VThree−MidPq=v1,v2,v3,…,vq∪u1,u2,u3,…,uq−1∪w1,w2,w3,…,wq−1∪m1,m2,m3,…,mq−1.
AsThree−MidPq is not a path graph, so it possesses metric dimension greater than 1.
AssumeW=u1,v1,m1 as a resolving set for Three−MidPq so that(8)rvg|W=1,0,1,g=1,g−1,g,g−1,g=2,3,4,…,q,rug|W=0,1,2,g=1,g−1,g,g,g=2,3,4,…,q−1,rwg|W=2,1,2,g=1,g,g,g,g=2,3,4,…,q−1,rmg|W=2,1,0,g=1,g,g,g−1,g=2,3,4,…,q−1.
Whenl≠g and vl and vg have the same representation with respect to w, a contradiction arises, i.e.,(9)rvl|W=rvg|W⇒l−1,l,l−1=g−1,q,g−1⇒l=gcontradiction.
Similarly, whenl≠g,(10)rul|W=rug|W⇒l−1,l,l=g−1,g,g⇒l=gcontradiction.
Whenl≠g,(11)rwl|W=rwg|W⇒l,l,l=g,g,g⇒l=gcontradiction.
Whenl≠g,(12)rml|W=rmg|W⇒l,l,l−l=g,g,g−1⇒l=gcontradiction.
Whenl≠g,(13)rvl|W=rug|W⇒l−1,l,l−1=g−1,g,g⇒l=gcontradiction.
Whenl≠g,(14)rvl|W=rwg|W⇒l−1,l,l−l=g,g,g⇒l=gcontradiction.
Whenl≠g,(15)rul|W=rwg|W⇒l−1,l,l−1=g,g,g⇒l=gcontradiction.
Whenl≠g,(16)rul|W=rwg|W⇒l−1,l,l=g,g,g⇒l=gcontradiction.
Whenl≠g,(17)rul|W=rmg|W⇒l−1,l,l=g,g,g−1⇒l=gcontradiction.
Whenl≠g,(18)rwl|W=rmg|W⇒l,l,l=g,g,g−1⇒l=gcontradiction.
Hence,W is the resolving set for Three−MidPq. Now, we shall show that no resolving set having cardinality 2 exists.LetT1=vl,vg be metric basis of Three−MidPq; then, ruj|T1=rmj|T1=rwj|T1, so uniqueness is violated.LetT2=ul,ug be metric bases of Three−MidPq; then, rwj|T2=rmj|T2, so uniqueness is violated.LetT3=wl,wg be metric bases of Three−MidPq; then, ruj|T3=rmj|T3, so uniqueness is violated.LetT4=ml,mg be metric bases of Three−MidPq; then, rwj|T4=ruj|T4, so uniqueness is violated.LetT5=vl,wg be metric bases of Three−MidPq; then, ruj|T5=rmj|T5, so uniqueness is violated.LetT6=vl,ug be metric bases of Three−MidPq; then, rwj|T6=rmj|T6, so uniqueness is violated.LetT7=vl,mg be metric bases of Three−MidPq; then, rwj|T7=ruj|T7, so uniqueness is violated.LetT8=ul,wg be metric bases of Three−MidPq; then, rvj|T8=rmj−1|T8, so uniqueness is violated.LetT9=ul,mg be metric bases of Three−MidPq; then, rvj|T9=rmj−1|T9, so uniqueness is violated.LetT10=wl,mg be metric bases of Three−MidPq; then, rvj|T10=ruj−1|T10, so uniqueness is violated.
So, there are no metric bases of order 2. Hence,W is the resolving set of order 3 and metric dimension of Three−MidPq is 3.Theorem 3.
Metric dimension ofs-Middle Pq is s.Proof.
The vertex set ofs-Middle Pq is v1,v2,v3,…,vp∪wgm, where g=1,2,3,…,p−1 and m=1,2,3,…,s. Let W=w11,w12,w13,…,w1s be the resolving set of the s-Middle Pq; then, rvg|W=g,g−1,g−1,…,g−1s−tuple, where g=1,2,3,…,p:(19)rwhg|W=g−1,g,g,…,gs−tuple,g=1,2,3,…,p−1;h=1,g,g−1,g,…,gs−tuple,g=1,2,3,…,p−1;h=2,g,g,g−1,…,gs−tuple,g=1,2,3,…,p−1;h=3,⋅⋅⋅g,g,g,…,g−1s−tuple,g=1,2,3,…,p−1;h=s,
Whenl≠m and vl and vm possess the same representation for W, a contradiction arises, i.e.,(20)rvl|W=rvm|W⇒l,l−1,l−1,…,l−1s−tuple=m,m−1,m−1,…,m−1s−tuple⇒l=mcontradiction.
Similarly, whenl≠m,(21)rwlh|W=rwmh|W⇒l,l,l,…,l−1hthposition,l,…,ls−tuple=m,m,m,…,m−1hthposition,m,…,ms−tuple⇒l=mcontradiction.
Whenl≠m,(22)rwgl|W=rwgm|W⇒g,g,g,…,g−1lthposition,g,…,gs−tuple=g,g,g,…,g−1mthposition,g,…,gs−tuple⇒g−1=gcontradiction.
Whenl≠m,(23)u≠vrwlu|W=rwmv|W⇒l,l,l,…,l−1uthposition,l…,ls−tuple=m,m,m,…,m−1vthposition,m,…,ms−tuple⇒l−1=mandl=m−1contradiction.
Now, we shall show that there is no resolving set withs−1 elements that exists. So, on the contrary, let a set with s−1 elements be a resolving set for s-Middle Pq. Then, particularly, for s=3, there exists a resolving set of 3−1=2 elements. So, metric dimension of Three-Middle Pq is not 2, and it leads to contradiction.
So, there is no resolving set withs−1 elements, and hence, s-Middle Pq has metric dimension s.


## 1.2.s-Total Graph of a Path

We discuss here more generalized concept of total graph of some graphs. We shall apply the very concept fors=2,3 on the path graph and use these results to find metric dimension of s-total graph of the path graph.Definition 2.
For undirected simple (planar or nonplanar) graphG with vertex set VG and EG as the edge set, VG=v1,v2,…,vs and EG=e1,e2,…,eu. We associate for every edge in edge set s vertices, forming another graph possessing vertex set represented by Vs−TG which is given by Vs−TG=VG∪∪u=1sVu, and if the total number of vertices in G is r and number of edges is u, then s-Total G contains r+su vertices. Each vg in Vs−TG is adjacent to vh in Vs−TG:(1)
vg∼vf if vg∼vf in G(2)
vg∼vgu if vg is incident to eg in G(3)
vgu∼vfu if eg is incident to ef in GTwo-total graph of a path and Three-total graph of a path are explained in Figures 3 and 4.Figure 3
Two-total graph of the path graph.Figure 4
Three-total graph of the path graph.Theorem 4.
dimTwo−Tpq is 2, where q≥2.Proof.
Two−TPq is not the path when q≥2, so its metric dimension must be greater than 1. The vertex set for Two−TPq is VTwo−TPq and is given by VTwo−TPq=v1,v2,v3,…,vq∪h1,h2,h3,…,hq−1∪s1,s2,s3,…,sq−1.
AssumeW=h1,hq−1 as a resolving set for Two−TPq so that(24)rvl|W=l,q−1,j=1,l−1,q−l,j=2,3,4,…,q−1,l−1,1,l=q,rsh|W=l−1,q−l−1∀j=1,…,q−1,rhj|W=2,q−1l=1,l,q−l,q=2,3,4,…,q−2,l,2,l=q−1.
Whenu≠r and hu and hr possess the same representation for W, a contradiction arises, i.e.,(25)rhu|W=rhr|W⇒u−1,q−u−1=r−1,q−r⇒l=gcontradiction.
Similarly, whenu≠r,(26)rsu|W=rsr|W⇒u−1,q−u=r−1,q−r⇒u=rcontradiction.
Whenu≠r,(27)rhu|W=rsr|W⇒u,q−u=r,q−r⇒u=rcontradiction.
Whenu≠r,(28)rvu|W=rvr|W⇒u−l,q−u=r−1,q−r⇒u=rcontradiction.
Hence,Two−TPq possesses W as its smallest resolving set, so its metric dimension is 2.Theorem 5.
dimThree−Tpq is 3, where q≥3.Proof.
The vertex set forThree−TPq is VThree−TPq and is given by VThree−TPq=v1,v2,v3,…,vq∪u1,u2,u3,…,uq−1∪w1,w2,w3,…,wq−1∪m1,m2,m3,…,mq−1. As Three−TPq is not a path graph, so it possesses metric dimension greater than 1.
AssumeW=u1,v1,m1 as a resolving set for Three−TPq so that(29)rvg|W=1,0,1,g=1,g−1,g−1,g−1,g=2,3,4,…,q−1,rug|W=0,1,2,g=1,g−1,g,g,g=2,3,4,…,q−1,rwg|W=2,1,2,g=1,g,g,g,g=2,3,4,…,q−1,rmg|W=2,1,0,g=1,g,g,g−1,g=2,3,4,…,q−1.
Whenl≠g and vl and vg possess the same representation for W, a contradiction arises, i.e.,(30)rvl|W=rvg|W⇒l−1,l,l−1=g−1,g−1,g−1⇒l=gcontradiction.
Similarly, wheng≠l,(31)rug|W=rul|W⇒g−1,g,g=l−1,l,l⇒g=lcontradiction.
Whenl≠g,(32)rwl|W=rwg|W⇒l,l,l=g,g,g⇒l=gcontradiction.
Whenl≠g,(33)rml|W=rmg|W⇒l,l,l−l=g,g,g−1⇒l=gcontradiction.
Whenl≠g,(34)rvl|W=rug|W⇒l−1,l−1,l−1=g−1,g,g⇒l=gcontradiction.
Whenl≠g,(35)rul|W=rwg|W⇒l−1,l,l=g,g,g⇒l=gcontradiction.
Whenl≠g,(36)rul|W=rmg|W⇒l−1,l,l=g,g,g−1⇒l=gcontradiction.
Whenl≠g,(37)rwl|W=rmg|W⇒l,l,l=g,g,g−1⇒l=gcontradiction.
Hence,W is resolving set for Three-Total Pq. Now, we shall show that no resolving set having cardinality 2 exists.LetT1=vl,vg be a resolving set of Three−TPq; then, ruj|T1=rmj|T1=rwj|T1, so uniqueness is violated.LetT2=ul,ug be metric bases of Three−TPq; then, rwj|T2=rmj|T2, so uniqueness is violated.LetT3=wl,wg be metric bases of Three−TPq; then, ruj|T3=rmj|T3, so uniqueness is violated.LetT4=ml,mg be metric bases of Three−TPq; then, rwj|T4=ruj|T4, so uniqueness is violated.LetT5=vl,wg be metric bases of Three−TPq; then, ruj|T5=rmj|T5, so uniqueness is violated.LetT6=vl,ug be metric bases of Three−TPq; then, rwj|T6=rmj|T6, so uniqueness is violated.LetT7=vl,mg be metric bases of Three−TPq; then, rwj|T7=ruj|T7, so uniqueness is violated.LetT8=ul,wg be metric bases of Three−TPq; then, rvj|T8=rmj−1|T8, so uniqueness is violated.LetT9=ul,mg be metric bases of Three−TPq; then, rvj|T9=rwj−1|T9, so uniqueness is violated.LetT10=wl,mg be metric bases of Three−TPq; then, rvj|T10=ruj−1|T10, so uniqueness is violated.
So, there are no metric bases of order 2. Hence,W is the resolving set of order 3 and metric dimension of Three−TPq is 3.Theorem 6.
Metric dimension ofs−TotalPq is s.Proof.
The vertex set ofs-Total Pq is v1,v2,v3,…,vp∪wgm, where g=1,2,3,…,p−1 and m=1,2,3,…,s. Let W=w11,w12,w13,…,w1s be the resolving set of the s-Total Pq; then, rvg|W=g,g−1,g−1,…,g−1s−tuple, where g=1,2,3,…,p,(38)rwgh|W=g−1,g,g,…,gs−tuple,g=1,2,3,…,p−1;h=1,g,g−1,g,…,gs−tuple,g=1,2,3,…,p−1;h=2,g,g,g−1,…,gs−tuple,g=1,2,3,…,p−1;h=3,⋅⋅⋅g,g,g,…,g−1s−tuple,g=1,2,3,…,p−1;h=s.
Wheng≠m and vg and vm possess the same representation for W, a contradiction arises, i.e.,(39)rvg|W=rvm|W⇒g,g−1,g−1,…,g−1s−tuple=m,m−1,m−1,…,m−1s−tuple⇒g=mcontradiction.
Similarly, whenl≠m,(40)rwlh|W=rwmh|W⇒l,l,l,…,l−1hthposition,l,…,ls−tuple=m,m,m,…,m−1hthposition,m,…,ms−tuple⇒l=mcontradiction.
Whenl≠m,(41)rwgl|W=rwgm|W⇒g,g,g,…,g−1lthposition,g…,gs−tuple=g,g,g,…,g−1mthposition,g…,gs−tuple⇒g−1=gcontradiction.
Whenl≠m and u≠v,(42)rwlu|W=rwmv|W⇒l,l,l,…,l−1uthposition,l…,ls−tuple=m,m,m,…,m−1vthposition,m,…,ms−tuple⇒l−1=mandl=m−1contradiction.
Now, we shall show that there is no resolving set withs−1 elements exists. So, on the contrary, let a set with s−1 elements be a resolving set for s-Total Pq. Then, particularly, for s=3, there exists a resolving set of 3−1=2 elements. So, metric dimension of Three-Total Pq is not 2, and it leads to contradiction.
So, there is no resolving set withs−1 elements, and hence, s-Total Pq has metric dimension s.


## 1.3. Middle Tower Path Graph

Middle tower path graph is a special graph represented by MiddleTowers, which is related to the path graph, and later, we extended its concept to its reflection.Definition 3.
An undirected, simple, and planar graph possessing vertex setV=Vgt|1≤t≤s,1≤g≤t, where t=1,2,3,…,s. Each vertex of Middle Towers is adjacent in the following manner.(1)
vgt∼vgt+1 when g=1,2,3,…,t and t=1,2,3,…,s−1(2)
vgt∼vg+1t+1 when g=1,2,3,…,t and t=1,2,3,…,s−1(3)
vgt∼vg+1t when g=1,2,3,…,t and t=1,2,3,…,s−1Middle tower path graph is explained in Figure5.Figure 5
Middle tower path graph.Theorem 7.
Metric dimension of MiddleTowers is 1, for s=2.Proof.
As MiddleTower2 is a path graph (by definition) and every path graph has metric dimension 1, so metric dimension of this graph is 1.Theorem 8.
Metric dimension of MiddleTowers is 2, for s≥3.Proof.
As MiddleTowers is not path graph; hence, it possesses metric dimension ≥2. Vertex set of Middle Towers is denoted by V (Middle Towers) and is given by VMiddle TowerPs=∪p=1sh1p,h2p,h3p,…,hqp, where q=1,2,3,…,s−p+1. We take W=h11,h1s resolving set of Middle Towers; then, representation rhqm|W=q+m−2,s−m,q=1,2,3,…,s−m+1 and m=1,2,3,…,s.
Wheng≠r and vg and vr possess the same representation for W, a contradiction arises, i.e.,(43)rhqm|W=rhrm|W⇒q+m−2,s−m=r+m−2,s−m⇒q=rcontradiction.
Similarly, whenm≠n,(44)rhqm|W=rhqn|W⇒q+m−2,s−m=q+n−2,s−n⇒m=ncontradiction.
Whenm≠n and q≠d,(45)rhqm|W=rhdn|W⇒q+m−2,s−m=d+n−2,s−n⇒m=nq=dcontradiction.
So,W is the smallest resolving set for Middle TowerPs and its metric dimension is 2.


## 1.4. Reflection Middle Tower Path Graph

Mirror image of MiddleTowers is named as Reflection Middle TowerPs and is denoted by rlTowers.Definition 4.
Simple undirected graph possessingVrlTowers=V1gt|1≤t≤s,1≤g≤t∪V2gt|1≤t≤s,1≤g≤t, where t=1,2,3,…,s. Each vertex of rlTowers is adjacent in the following manner.(1)
vgit∼vgit+1 when i=1,2,g=1,2,3,…,t, and t=1,2,3,…,s−1(2)
vgit∼vg+1it+1 when i=1,2,g=1,2,3,…,t, and t=1,2,3,…,s−1(3)
vgit∼vg+1it when i=1,2,g=1,2,3,…,t, and t=1,2,3,…,s−1(4)
vg1t∼vg+12t−1 when g=1,2,3,…,t−1 and t=s(5)
vg+11t∼vg2t−1 when g=1,2,3,…,t−1 and t=sReflection middle tower path graph is explained in Figure6.Figure 6
rlTowers graph.Theorem 9.
Metric dimension ofrlTowers is 2, for s≥3.Proof.
The vertex set ofrlTowers is VrlTowers=∪g=2s−1l11g,l21g,l31g,…,lj1g∪∪g=2s−2l12g,l22g,l33g,…,lj2g, where j=1,2,3,…,s−g+1.
AssumeW=l11g,l21g as the resolving set; then,(46)rl11j|W=s−1,s−j+1,j=1,2,3,…,s−1,s−1,0,j=s.
rlj|W1q=s−q,s−j, where j=1,2,3,…,s−1 and q=2,3,4,…,s.
rlj|W2q=s−q−2,s−j, where j=1,2,3,…,q and q=2,3,4,…,s.
Whenm≠n and lm and ln possess the same representation for W, a contradiction arises, i.e.,(47)rlm11|W=rln11|W⇒s−1,s−m+1=s−1,s−n+1⇒m=ncontradiction.
Similarly, whenm≠n,(48)rlm1h|W=rln1h|W⇒s−h,s−m=s−h,s−n⇒m=ncontradiction.
Whenm≠n,(49)rlp1m|W=rlp1n|W⇒s−m,s−1=s−h,s−1⇒m=ncontradiction.
Whenm≠n and u≠v,(50)rlu1m|W=rlv1n|W⇒s−m,s−l=s−n,s−v⇒m=nandu=vcontradiction.
Whenm≠n,(51)rlm2g|W=rln2g|W⇒s+g−2,s−m=s+g−2,s−n⇒m=ncontradiction.
Whenm≠n and u≠v,(52)rlu2m|W=rlv2n|W⇒s+m−2,s−u=s+n−2,s−v⇒m=nandu=vcontradiction.
Whenm≠n,(53)rlc2m|W=rlc2n|W⇒s+m−2,s−c=s+n−2,s−c⇒m=ncontradiction.
Hence,W is the smallest resolving set and metric dimension of rlTowers is 2.


## 1.5. Symmetrical Planar Pyramid Graph

A graph is obtained by joining the neighbouring vertices of MiddleTowerPs in its base and is represented by SPPs. If there are s vertices in the base of SPPs, then its vertex set contains ss−1/2 vertices.Definition 5.
We defineSPPs with s vertices in base possesses vertex set ∪t=1sVpt, where t=1,2,3,…,s and p≤t.
Each vertex oftowers is adjacent in the following manner.(1)
vpt∼vpt+1 when p=1,2,3,…,s−1 and t=1,2,3,…,s−1 and p≤t(2)
vpt∼vp+1t+1 when p=1,2,3,…,s−1 and t=1,2,3,…,s−1(3)
vpt∼vp+1t when p=1,2,3,…,s and t=1,2,3,…,s−1Symmetrical planar pyramid graph is explained in Figure7.Figure 7
Symmetrical planar pyramid graph.Theorem 10.
Metric dimension ofSPPs is 2.Proof.
TheSPP has vertex set VSPPs=∪t=1s, where p=1,2,3,…s and t=1,2,3,…,s. Also, p≤t.
We takeW=v11,vss resolving set of Middle SPPs and then representation rvpt|W=p+j−1,s−t+jp=1,2,3,…,s and t=1,2,3,…,s. Also, p≤t.
Whenl≠g and vl and vg possess the same representation for W, a contradiction arises, i.e.,(54)rvlt|W=rvgt|W⇒l+j−1,s−t+j=g+j−1,s−t+j⇒l=gcontradiction.
Similarly, whenl≠g,(55)rvpl|W=rvpg|W⇒p+j−1,s−l+j=p+j−1,s−g+j⇒l=gcontradiction.
Whenl≠g and u≠v,(56)rvul|W=rvvg|W⇒l+j−1,s−u+j=g+j−1,s−v+j⇒l=gu=vcontradiction.
So,W is the smallest resolving set for SPPs and its metric dimension is 2.


## 1.6. Reflection Symmetrical Planar Pyramid Graph

Mirror image ofSPPs is named as reflection symmetrical planar pyramid graph and is denoted by rlSPPs.Definition 6.
Simple undirected graph possessing vertex setVrlSPPs=∪t=1sVp1t∪∪t=2s−1Vp2t, where p=1,2,3,…,s and t=1,2,3,…,s; also, s≤t. Each vertex of rlSPPs is adjacent in the following manner.(1)
vpit∼vpit+1 when p=1,2,3,…,s, t=1,2,3,…,s−1, p≤t, and i=1,2(2)
vpit∼vp+1it+1 when p=1,2,3,…,s, t=1,2,3,…,s−1, p≤t, and i=1,2(3)
vpit∼vpit when p=1,2,3,…,s, t=1,2,3,…,s−1, p≤t, and i=1,2(4)
vp1t∼vp2t−1 when t=s and g=2,3,4,…,s(5)
vp+11t∼vp2t−1 when t=s and p=1,2,3,…,s−1Reflection symmetrical planar pyramid graph is explained in Figure8.Figure 8
Reflection symmetrical planar pyramid graph.Theorem 11.
Metric dimension ofrlSPPs.Proof.
The vertex set ofrlSPPs is VrlSPPs=∪t=2s∪∪t=1s−1Vp2t, where p=1,2,3,…,s and t=1,2,3,…,s; also, s≤t.
AssumeW=v11,vss as the resolving set; then,(57)rv1tp|W=p−j−1,s−t+1J,p=1,2,3,…,st=1,2,3,…,sp≤tandj=t−p,rv2tp|W=s−t+p−1,2s−t−1,j=1,2,3,…,s−1.
Whenl≠g and vl and vg possess the same representation for W, a contradiction arises, i.e.,(58)rvl1t|W=rvg1t|W⇒l+j−1,s−t+j=g+j−1,s−g+j⇒l=gcontradiction.
Similarly, whenl≠g,(59)rvp1l|W=rvp1g|W⇒p+j−1,s−l+j=p−j−1,s−g+j⇒l=gcontradiction.
Whenl≠g,(60)rvl1u|W=rvg1p|W⇒l+j−1,s−u+j=g+j−1,s−p+j⇒l=gu=pcontradiction.
Whenl≠g,(61)rvl2t|W=rvg2t|W⇒s−t+l−1,2s−t−l=s−t+g,2s−t−1⇒l=gcontradiction.
Whenl≠g,(62)rvp2t|W=rvp2g|W⇒s−l+p−1,2s−t−1=s−g+p−1,2s−t−1⇒l=gcontradiction.
Whenl≠gandu≠p,(63)rvl2u|W=rvg2p|W⇒s−u+l−1,2s−u−1=s−v+g−1,2s−p−1⇒l=gandu=pcontradiction.
Hence,W is the smallest resolving set and metric dimension of rlSPPs is 2.


## 2. Conclusions

In this dissertation, some new graphs ass−MidG, s−TotalG, Middle Towers, SPPs, rlTowers, and rlSPPs are defined and metric dimension of these graphs is also computed. All these graphs possess constant and bounded metric dimension. This research work may lead to the following open problems.Open problem 1: compute metric dimension of symmetrical pyramid in space.Open problem 2: to find metric dimension ofs−MidCnOpen problem 3: to find metric dimension ofs−TCn



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