# Computation of Metric Dimension of Certain Subdivided Convex Polytopes

Authors: S. Imran; Z. Ali; N. Nigar; Syed Ajaz K. Kirmani; M. K. Siddiqui; S. A. Fufa
Journal: Journal of Mathematics
 (2022)

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## Abstract

The distancedz1,z2 from vertex z1∈VG to z2∈VG is minimum length of z1,z2-path in a given connected graph G having E(G) and V(G) edges and vertices’/nodes’ sets, respectively. Suppose Z=z1,z2,z3,…,zm⊆VG is an order set and c∈VG, and the code of c with reference to Z is the m-tuple {d(c, z1), d(c, z2), d(c, z13), …, d(c, zk)}. Then, Z is named as the locating set or resolving set if each node of G has unique code. A locating set of least cardinality is described as a basis set for the graph G, and its cardinal number is referred to as metric dimension symbolized by dimG. Metric dimension of certain subdivided convex polytopes STn has been computed, and it is concluded that just four vertices are sufficient for unique coding of all nodes belonging to this family of convex polytopes.


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## Body

## 1. Introduction

In the discipline of computer science and mathematics, graph theory [1] is the survey of graphs that considers the link between edges and vertices. This is the most celebrated discipline these days that has applications [2] in computer science, information technology, biosciences, mathematics, social sciences, physics, chemistry, and linguistics. To illustrate pairwise relationship of objects, graph theory analysis is very important [3, 4].Formally, a graph is the collection of vertices and edges. Among several types of different graphs, we will analyze a particular class of graph known as convex polytopes [5]. Convex polytopes are the principal geometric structures which are under investigation since antiquity. The charm of this concept is nowadays complemented by their significance for various mathematical fields, extending from algebraic geometry, linear programming, integration, and combinatorial optimization. Convex polytope is the simplest kind of polytopes [6] which satisfies the property of convex set in k-dimensional Euclidean space Rk. Convex polytopes play a vital part in enormous areas of mathematics as well as in applied disciplines, but its role in linear programming is most influential [7, 8].Moreover, subdividing is a process in which we add an extra vertex on each edge of the graph in such a way such that each will be splitted into two edges, and the resulting graph is called subdivided graph of the original graphG. Since a couple of years, the variables associated with distances in graphs have enchanted the focus of various researchers, but in the recent years, the phenomenon that has centered certain surveys is termed as metric dimension [9]. The distance dz1,z2 from vertex z1∈VG to z2∈VG is minimum length of z1,z2-path in a given connected graph G having EG and VG edges and vertices’/nodes’ sets, respectively. Suppose Z=z1,z2,z3,…,zm⊆VG is an order set and c∈VG; the code of c with reference to Z is the m-tuple {d(c, z1), d(c, z2), d(c, z13), …, d(c, zk)}. Then, Z is named as the locating set or resolving set if each node of G has a unique code. A locating set of least cardinality is described as a basis set for the graph G, and its cardinal number is referred as metric dimension symbolized by dimG. Moreover, if every member of a family of graphs possess the same metric dimension, then it is known as family with constant metric dimension [10, 11].To find intruder’s location in a network, Slater, in [12, 13], described metric dimension. Harary and Melter [14] are also the founder of metric dimension. The concept of constant metric dimension of certain class of convex polytopes is studied in [15, 16]. Generalization of Petersen graphs with its Bounden metric dimension has been studied in [3]. It is given in [17] that if we remove the join of the nodes bi and bi+1, the resulting class of convex polytopes is Tn, as given in Figure 1, where both the categories of convex polytopes have identical metric dimension 3.Figure 1
Convex polytopeTn.Metric dimension of a specific class of convex polytope has been studied in this paper. The scope of metric dimension is broadened, the generalized family of convex polytope undergoes subdivision ofTn, and the metric dimension of subdivided convex polytope STn has been determined. The set of vertices that belongs to the graph of subdivided convex polytope STn consists of fi;n≥i≥1andn=2k∪gi;n≥i≥1andn=2k∪hi;1≤i≤k∪li;1≤i≤k∪mi;1≤i≤kandn=2k. The family of vertices on inner cycle is fi;n≥i≥1andn=2k and collection of vertices of outer cycle is mi;1≤i≤kandn=2k. The collection of vertices assigned to central cycles is hi;1≤i≤k∪li;1≤i≤k. This study comprises of three main theorems which define its scope. Minimal cardinal number for the locating set of subdivided convex polytope STn is greater than two and is the topic of first theorem. In the second theorem, the work is extended to prove that minimal cardinality of the resolving set of convex polytope is greater than 3. Finally, it is demonstrated in the third theorem that the subdivided convex polytope STn has metric dimension equal to 4, and the formulation for the subdivided convex polytope STn is presented, as given in Figure 2. Lemma 1 and Theorem 1 presented in [15] were useful in finding the results of this paper.Figure 2
Subdivided convex polytopeSTn.Lemma 1.
LetZ be a locating set for a connected graph G and z1,z2∈VG. If dz1,z≠dz2,z,∀z∈VG∖z1,z2, then z1,z2∩Z≠ϕ.Theorem 1.
LetG be a graph with metric dimension 2, and let the basis set in G be z1,z2⊂VG. Then, the following are true:(a)
There is a unique path betweenz1 and z2(b)
The degree of eachz1 and z2 is at most 3


## 2. Main Results

Theorem 2.
Prove that minimal cardinality for the locating set of subdivided convex polytopeSTn is greater than 2.Proof.
To prove that minimal cardinality for the locating set of subdivided convex polytopeSTn is greater than 2, the following cases exist.Case I: both locating nodes have a place in the inner cycle. Without loss of generality, we can assume that the first locating node isf1. Assume that the other locating node is ftk+1≥t≥2. Thus, Z=f1,ft is the possible locating set. If t varies from 2 to k, then rfn/f1,ft=1,t=rgn/f1,ft and rg1/f1,ft=rgn/f1,ft=1,t, for t = k + 1, which contradict unique representation.Case II: one vertex has a place on the inner cycle, and the other has a place with the family of points whose distance from the inward cycle is 1. With no loss of universality, we can assume that the first locating node isf1. Assume the other node is gtk+1≥t≥1. Thus, Z=f1,gt is the possible locating set. If t varies from 1 to k − 1 when t is odd, we get rgn−2/f1,gt=rgn−1/f1,gt=3,t+3, and if 2≤t≤k, t is even, we get rgn−2/f1,gt=rgn−1/f1,gt=3,t+4 and rgn−2/f1,gt=rgn−1/f1,gt=3,t−1 for t = k + 1, which contradict unique representation.Case III: when one vertex has a place on the inner cycle and the other has a place with the family of points whose distance from the inner cycle is 2. Without loss of universality, we can assume that the first locating node isf1 and the other locating node is htk/2+1≥t≥1. Thus, Z=f1,ht is the possible locating set. If 1≤t≤k/2, where t is odd, then we get rgn−2/f1,ht=rgn−1/f1,ht=3,t+4, and if 2≤t≤k/2+1, where t is even, we get rg1/f1,ht=rgn/f1,ht=1,t+3, which contradict unique representation.Case IV: when one vertex has a place on the inner cycle and the other has a place with the family of points whose distance from the inner cycle is 3. Without loss of universality, we are able to assume that the first locating node isf1 and the other locating node is ltk/2+1≥t≥1. Thus, Z=f1,lt is the possible locating set. If t varies from 1 to k/2 when t is odd, we get rgn−2/f1,lt=rgn−1/f1,lt=3,t+5, and if 2≤t≤k/2+1, where t is even, then we get rg1/f1,lt=rgn/f1,lt=1,t+4, which contradict unique representation.Case V: one vertex has a place on the inner cycle and the other has a place with the family of points whose distance from the inner cycle is 4. Without loss of universality, we are able to assume that the first locating node isf1 and the other locating node is mtk+1≥t≥1. Thus, Z=f1,mt is the possible locating set. If t varies from 1 to k − 1, we get rfn/f1,mt=1,t+4=rgn/f1,mt and rg1/f1,mt=1,t+2=rgn/f1,mt when t = k, k + 1 which contradict unique representation.Case VI: both vertices have a place with the family of points whose distance from the inner cycle is 1. Without loss of universality, we are able to assume that the first locating node isg1 and the other node is gtk+1≥t≥2. Thus, Z=g1,gt is the possible locating set. If t varies from 2 to k when t is even, we get rgn−2/g1,gt=rgn−1/g1,gt=4,t+4, and if k/2≤t≤k−1, where t is odd, we get rgn−2/g1,gt=rgn−1/g1,gt=4,t+3, and if t = k + 1, then rgn−2/g1,gt=rgn−1/g1,gt=4,t−1, which contradict unique representation.Case VII: family of vertices whose distance from the inward cycle is 1 and 2, respectively. Without loss of universality, we are able to assume that the first locating node isg1 and the other node is ht1≤t≤k/2+1. Thus, Z=g1,ht is the possible locating set. If t varies from 1 to k/2 when t is odd, we get rgn−2/g1,ht=rgn−1/g1,ht=4,t+4, and if 2≤t≤k/2+1, where t is even, then we get rmn−8/g1,ht=rgn−7/g1,ht=6,t+1, which contradict unique representation.Case VIII: family of vertices whose distance from the inner cycle is 1 and 3, respectively. Without loss of universality, we are able to assume that the first locating node isg1 and the other locating node is ltk/2+1≥t≥1. Thus, Z=g1,lt is the possible locating set. If t varies from 1 to k/2 when t is odd, we get rgn−2/g1,lt=rgn−1/g1,lt=4,t+5, and if 2≤t≤k/2+1, where t is even, then we get rfn−8/g1,lt=rmn/g1,lt=4,t+2, which contradict unique representation.Case IX: family of vertices whose distance from the inner cycle is 1 and 4, respectively. Without loss of universality, we are able to assume that the first locating node isg1 and the other node is mtk+1≥t≥1. Thus, Z=g1,mt is the possible locating set. If t varies from 1 to k − 1, we get rfn/g1,mt=2,t+4=rgn/g1,mt and rf2/g1,mt=rgn/g1,mt=2,t+2 when t = k, k + 1, which contradict unique representation.Case X: both vertices have a place with the family of points whose distance from the inner cycle is 2. Without loss of universality, we are able to assume that the first locating node ish1 and the other node is htk/2+1≥t≥2. Thus, Z=h1,ht is the possible locating set. If t varies from 2 to k/2 when t is odd, we get rh14/h1,ht=rmn−5/h1,ht=8,t+9, and if 2≤t≤k/2, where t is even, we get rg6/h1,ht=rg7/h1,ht=7,t+3, and if t = k/2 + 1, we get rh14/h1,ht=rmn−5/h1,ht=8,t+3, which contradict unique representation.Case XI: family of vertices whose distance from the inner cycle is 2 and 3, respectively. Without loss of universality, we are able to assume that the first locating node ish1 and the other node is lt1≤t≤k/2+1. Thus, Z=h1,lt is the possible locating set. If t varies from 1 to k/2 when t is odd, we get rf2/h1,lt=rg3/h1,lt=3,t+3, and if k/2−1≤t≤k/2+1, where t is even, then we get rfn−8/h1,lt=rmn/h1,lt=3,t+2, which contradict unique representation.Case XII: family of vertices whose distance from the inner cycle is 2 and 4, respectively. Without loss of universality, we are able to assume that the first locating node ish1 and the other node is mtk+1≥t≥1. Thus, Z=h1,mt is the possible locating set. If t varies from 1 to k + 1, we get rg1/h1,mt=1,t+2=rg2/h1,mt, which contradict unique representation.Case XIII: both the vertices have a place with the family of points whose distance from the inner cycle is 3. Without loss of universality, we are able to assume that the first locating node isl1 and the other node is ltk/2+1≥t≥2. Thus, Z=l1,lt is the possible locating set. If t varies from 2 to k/2, where t is odd, we get rfn−4/l1,lt=rl14/l1,lt=8,t+9, and if 2 2≤t≤k/2, where t is even, we get rg6/l1,lt=rg7/l1,lt=8,t+4, and if t = k/2 + 1, then we get rfn−4/l1,lt=rl14/l1,lt=8,t+3, which contradict unique representation.Case XIV: family of vertices whose distance from the inner cycle is 3 and 4 respectively. Without loss of universality, we are able to assume that the first locating node isl1 and the other node is mtk+1≥t≥1. Thus, Z=l1,mt is the possible locating set. If t varies from 1 to k + 1, we get rg1/l1,mt=2,t+2=rg2/l1,mt, which contradict unique representation.Case XV: both vertices have a place on the outer cycle. Without loss of universality, we are able to assume that the first locating node ism1 and the other node is mtk+1≥t≥2. Thus, Z=m1,mt is the possible locating set. If t varies from 2 to k, we get rl1/m1,mt=1,t=rmn/m1,mt and rm2/m1,mt=rmn/m1,mt=1,t−2 when t = k + 1, which contradict unique representation. From all the above cases, it is concluded that, for unique representation of each node, cardinality of the locating set is greater than two. Hence, the result is obtained.Theorem 3.
Prove that minimal cardinality for the locating set of subdivided convex polytopeSTn is greater than 3.Proof.
To prove that minimal cardinality for the locating set of subdivided convex polytopeSTn is greater than 3, the following cases exist.Case I: all the three vertices have a place on the inner cycle. Without loss of universality, we are able to assume that the first resolving vertex isf1. Let the second resolving vertex is fr2≤r≤k. Suppose the third resolving vertex is ftr+1≤t≤k+1. Hence, Z=f1,fr,ft. If 2≤r≤k−1 and r+1≤t≤k−1, we get rgn−2/f1,fr,ft=rgn−1/f1,fr,ft=3,r+2,t+2, and if 2≤r≤k−1, t = k, we get rgn−2/f1,fr,ft=rgn−1/f1,fr,ft=3,r+2,t, and if r = k, t = k + 1, we get rgn−2/f1,fr,ft=rgn−1/f1,fr,ft=3,r,t−2, which contradict unique representation.Case II: one vertex has a place on the inner cycle and the other two have a place with the family of points whose distance from the inner cycle is 1. Without loss of universality, we are able to assume that the first resolving vertex isf1. Let the second resolving vertex is gr1≤r≤k. Suppose the third resolving vertex is gtr+1≤t≤k+1. Hence, Z=f1,gr,gt. If 1≤r≤k−1, where r is even and r+1≤t≤k−1, where t is odd, we get rgn−2/f1,gr,gt=rgn−1/f1,gr,gt=3,r+4,t+3; if 1≤r≤k−1, where r is odd and r+1≤t≤k−1, where t is even, we get rgn−2/f1,gr,gt=rgn−1/f1,gr,gt=3,r+3,t+4; if 1≤r≤k−1, where r is even and t = k, we get rgn−2/f1,gr,gt=rgn−1/f1,gr,gt=3,r+4,t; if 1≤r≤k−1, where r is odd and t = k, we get rgn−2/f1,gr,gt=rgn−1/f1,gr,gt=3,r+3,t; if 1≤r≤k−1, where r is even and t = k + 1, we get rgn−2/f1,gr,gt=rgn−1/f1,gr,gt=3,r+4,t−1; if 1≤r≤k−1, r is odd and t = k + 1, we get rgn−2/f1,gr,gt=rgn−1/f1,gr,gt=3,r+3,t−1; if r = k and t = k + 1, we get rgn−2/f1,gr,gt=rgn−1/f1,gr,gt=3,r,t−1, which contradict unique representation.Case III: one vertex has a place on the inner cycle and the other two have a place with the family of points whose distance from the inner cycle is 2. Without loss of universality, we are able to assume that the first resolving vertex isf1. Let the second resolving vertex be hr1≤r≤k/2. Suppose the third resolving vertex is htr+1≤t≤k/2+1. Hence, Z=f1,hr,ht. If 1≤r≤k/2, where r is odd and r+1≤t≤k/2−1, we get rgn−2/f1,hr,ht=rgn−1/f1,hr,ht=3,r+4,t+5; if 1≤r≤k/2, where r is even and t = k/2, we get (3, r + 5, t + 4); if 1≤r≤k/2, where r is odd and t = k/2, we get (3, r + 4, t + 4); if 1≤r≤k/2, where r is odd and t = k/2 + 1, we get rgn−2/f1,hr,ht=rgn−1/f1,hr,ht=3,r+4,t+1; if 1≤r≤k/2, where r is even and t = k/2 + 1, we get rgn−2/f1,hr,ht=rgn−1/f1,hr,ht=3,r+5,t+1, which contradict unique representation.Case IV: one vertex has a place on the inner cycle and the other two have a place with the family of points whose distance from the inner cycle is 3. Without loss of universality, we are able to assume that the first resolving vertex isf1. Let the second resolving vertex be lr1≤r≤k/2. Suppose the third resolving vertex is ltr+1≤t≤k/2+1. Hence, Z=f1,lr,lt. If 1≤r≤k/2, where r is odd and r+1≤t≤k/2−1, we get rgn−2/f1,lr,lt=rgn−1/f1,lr,lt=3,r+5,t+6; if 1≤r≤k/2, where r is even and t = k/2, we get rgn−2/f1,lr,lt=rgn−1/f1,lr,lt=3,r+6,t+5; if 1≤r≤k/2, where r is odd and t = k/2, we get rgn−2/f1,lr,lt=rgn−1/f1,lr,lt=3,r+5,t+5; if 1≤r≤k/2, where r is odd and t = k/2 + 1, we get rgn−2/f1,lr,lt=rgn−1/f1,lr,lt=3,r+5,t+2; if 1≤r≤k/2, where r is even and t = k/2 + 1, then we get rgn−2/f1,lr,lt=rgn−1/f1,lr,lt=3,r+6,t+2, which contradict unique representation.Case V: one vertex has a place on the inner cycle and the other two have a place with the family of points whose distance from the inner cycle is 4. Without loss of universality, we are able to assume that the first resolving vertex isf1. Let the second resolving vertex is mr1≤r≤k. Suppose the third resolving vertex is mtr+1≤t≤k+1. Hence, Z=f1,mr,mt. If 1≤r≤k−1 and r+1≤t≤k−1, we get rfn/f1,mr,mt=rgn/f1,mr,mt=1,r+4,t+4; if 1≤r≤k−1 and t = k, we get rfn/f1,mr,mt=rgn/f1,mr,mt=1,r+4,t+2; if 1≤r≤k−1 and t = k + 1, we get rfn/f1,mr,mt=rgn/f1,mr,mt=1,r+4,t, and if r = k and t = k + 1, we get rfn/f1,mr,mt=rgn/f1,mr,mt=1,r+2,t, which contradict unique representation.Case VI: all the three vertices have a place with the family of points whose distance from the inner cycle is 1. Without loss of universality, we are able to assume that the first resolving vertex isg1. Let the second resolving vertex be gr2≤r≤k. Suppose the third resolving vertex is gtr+1≤t≤k+1. Hence, Z=g1,gr,gt. If 2≤r≤k−1, where r is even and r+1≤t≤k−1, where t is odd, we get rgn−2/g1,gr,gt=rgn−1/g1,gr,gt=4,r+4,t+3; if 2≤r≤k−1, where r is odd and r+1≤t≤k−1, where t is even, we get rgn−2/g1,gr,gt=rgn−1/g1,gr,gt=4,r+3,t+4; if 2≤r≤k−1, where r is even and t = k, we get rgn−2/g1,gr,gt=rgn−1/g1,gr,gt=4,r+4,t; if 2≤r≤k−1, where r is odd and t = k, we get rgn−2/g1,gr,gt=rgn−1/g1,gr,gt=4,r+3,t; if 2≤r≤k−1, where r is even and t = k + 1, we get rgn−2/g1,gr,gt=rgn−1/g1,gr,gt=4,r+4,t−1; if 2≤r≤k−1, where r is odd and t = k + 1, we get rgn−2/g1,gr,gt=rgn−1/g1,gr,gt=4,r+3,t−1; if r = k and t = k + 1, we get rgn−2/g1,gr,gt=rgn−1/g1,gr,gt=4,r,t−1, which contradict unique representation.Case VII: family of vertices whose distance from the inner cycle is 1 and 2, respectively. Without loss of universality, we are able to assume that the first resolving vertex isg1. Let the second resolving vertex be hr1≤r≤k/2. Suppose the third resolving vertex is htr+1≤t≤k/2+1. Hence, Z=gr,hr,ht. If 1≤r≤k/2, where r is odd and r+1≤t≤k/2−1, we get rgn−2/gr,hr,ht=rgn−1/gr,hr,ht=4,r+4,t+5; if 1≤r≤k/2, where r is even and t = k/2, we get rgn−2/gr,hr,ht=rgn−1/gr,hr,ht=4,r+5,t+4; if 1≤r≤k/2, where r is odd and t = k/2, we get (4, r + 4, t + 4); if 1≤r≤k/2, where r is odd and t = k/2 + 1, we get rgn−2/gr,hr,ht=rgn−1/gr,hr,ht=4,r+4,t+1; if 1≤r≤k/2, where r is even and t = k/2 + 1, we get rgn−2/gr,hr,ht=rgn−1/gr,hr,ht=4,r+5,t+1, which contradict unique representation.Case VIII: family of vertices whose distance from the inner cycle is 1 and 3, respectively. Without loss of universality, we are able to assume that the first resolving vertex isg1. Let the second resolving vertex be lr1≤r≤k/2. Suppose the third resolving verte is ltr+1≤t≤k/2+1. Hence, Z=g1,lr,lt. If 1≤r≤k/2, where r is odd and r+1≤t≤k/2−1, we get rgn−2/g1,lr,lt=rgn−1/g1,lr,lt=4,r+5,t+6; if 1≤r≤k/2, where r is even and t = k/2, we get rgn−2/g1,lr,lt=rgn−1/g1,lr,lt=4,r+6,t+5; if 1≤r≤k/2, where r is odd and t = k/2, we get rgn−2/g1,lr,lt=rgn−1/g1,lr,lt=4,r+5,t+5; if 1≤r≤k/2, where r is odd and t = k/2 + 1, we get rgn−2/g1,lr,lt=rgn−1/g1,lr,lt=4,r+5,t+2; if 1≤r≤k/2, where r is even and t = k/2 + 1, then we get rgn−2/g1,lr,lt=rgn−1/g1,lr,lt=4,r+6,t+2, which contradict unique representation.Case IX: family of vertices whose distance from the inner cycle is 1 and 4, respectively. Without loss of universality, we are able to assume that the first resolving vertex isg1. Let the second resolving vertex be mr1≤r≤k. Suppose the third resolving vertex is mtr+1≤t≤k+1. Hence, Z=g1,mr,mt. If 1≤r≤k−1 and r+1≤t≤k−1, we get rfn/g1,mr,mt=rgn/g1,mr,mt=2,r+4,t+4; if 1≤r≤k−1 and t = k, we get rfn/g1,mr,mt=rgn/g1,mr,mt=2,r+4,t+2; if 1≤r≤k−1 and t = k + 1, we get rfn/g1,mr,mt=rgn/g1,mr,mt=2,r+4,t, and if r = k and t = k + 1, we get rfn/g1,mr,mt=rgn/g1,mr,mt=2,r+2,t, which contradict unique representation.Case X: all the three vertices have a place with the family of points whose distance from the inner cycle is 2. Without loss of universality, we are able to assume that the first resolving vertex ish1. Let the second resolving vertex is hr2≤r≤k/2. Suppose the third resolving vertex is htr+1≤t≤k/2+1. Hence, Z=h1,hr,ht. If 2≤r≤k/2, where r is even and r+1≤t≤k/2+1, where t is odd, we get rgn−2/h1,hr,ht=rgn−1/h1,hr,ht=4,r+5,t+4; if 2≤r≤k/2, where r is even and r+1≤t≤k/2+1, where t is even, we get rgn−2/h1,hr,ht=rgn−1/h1,hr,ht=4,r+5,t+1; if 2≤r≤k/2, where r is odd and r+1≤t≤k/2+1, where t is even, then we get rgn−2/h1,hr,ht=rgn−1/h1,hr,ht=4,r+4,t+1, which contradict unique representation.Case XI: family of vertices whose distance from the inner cycle is 2 and 3, respectively. Without loss of universality, we are able to assume that the first resolving vertex ish1. Let the second resolving vertex be lr1≤r≤k/2. Suppose the third resolving vertex is ltr+1≤t≤k/2+1. Hence, Z=h1,lr,lt. If 1≤r≤k/2, where r is odd and r+1≤t≤k/2−1, we get rgn−2/h1,lr,lt=rgn−1/h1,lr,lt=5,r+5,t+6; if 1≤r≤k/2, where r is even and t = k/2, we get rgn−2/h1,lr,lt=rgn−1/h1,lr,lt=5,r+6,t+5; if 1≤r≤k/2, where r is odd and t = k/2, we get rgn−2/h1,lr,lt=rgn−1/h1,lr,lt=5,r+5,t+5; if 1≤r≤k/2, where r is odd and t = k/2 + 1, we get rgn−2/h1,lr,lt=rgn−1/h1,lr,lt=5,r+5,t+2; if 1≤r≤k/2, where r is even and t = k/2 + 1, then we get rgn−2/h1,lr,lt=rgn−1/h1,lr,lt=5,r+6,t+2, which contradict unique representation.Case XII: family of vertices whose distance from the inward cycle is 2 and 4, respectively. With no loss of universality, we are able to assume that the first resolving vertex ish1. Let the second resolving vertex be mr1≤r≤k. Suppose the third resolving vertex is mtr+1≤t≤k+1. Hence, Z=h1,mr,mt. If 1≤r≤k−1 and r+1≤t≤k−1, we get rfn/h1,mr,mt=rgn/h1,mr,mt=3,r+4,t+4; if 1≤r≤k−1 and t = k, we get rfn/h1,mr,mt=rgn/h1,mr,mt=3,r+4,t+2; if 1≤r≤k−1 and t = k + 1, we get rfn/h1,mr,mt=rgn/h1,mr,mt=3,r+4,t, and if r = k and t = k + 1, we get rfn/h1,mr,mt=rgn/h1,mr,mt=3,r+2,t, which contradict unique representation.Case XIII: all the three vertices have a place with the family of points whose distance from the inner cycle is 3. Without loss of universality, we are able to assume that the first resolving vertex isl1. Let the second resolving vertex be lr2≤r≤k/2. Suppose the third resolving vertex is ltr+1≤t≤k/2+1. Hence, Z=l1,lr,lt. If 2≤r≤k/2, where r is even and r+1≤t≤k/2+1, where t is odd, we get rgn−2/l1,lr,lt=rgn−1/l1,lr,lt=6,r+6,t+5; if 2≤r≤k/2, where r is even and r+1≤t≤k/2+1, where t is even, we get rgn−2/l1,lr,lt=rgn−1/l1,lr,lt=6,r+6,t+2; if 2≤r≤k/2, where r is odd and r+1≤t≤k/2+1, where t is even, then we get rgn−2/l1,lr,lt=rgn−1/l1,lr,lt=6,r+5,t+2, which contradict unique representation.Case XIV: family of vertices whose distance from the inner cycle is 3 and 4, respectively. Without loss of universality, we are able to assume that the first resolving vertex isl1. Let the second resolving vertex be mr1≤r≤k. Suppose the third resolving vertex is mtr+1≤t≤k+1. Hence, Z=l1,mr,mt. If 1≤r≤k−1 and r+1≤t≤k−1, we get rfn/l1,mr,mt=rgn/l1,mr,mt=4,r+4,t+4; if 1≤r≤k−1 and t = k, we get rfn/l1,mr,mt=rgn/l1,mr,mt=4,r+4,t+2; if 1≤r≤k−1 and t = k + 1, we get rfn/l1,mr,mt=rgn/l1,mr,mt=4,r+4,t, and if r = k and t = k + 1, we get rfn/l1,mr,mt=rgn/l1,mr,mt=4,r+2,t, which contradict unique representation.Case XV: all the three vertices have a place on the outer cycle. Without loss of universality, we are able to assume that the first resolving vertex ism1. Let the second resolving vertex be mr2≤r≤k. Suppose the third resolving vertex is mtr+1≤t≤k+1. Hence, Z=m1,mr,mt. If 2≤r≤k−1 and r+1≤t≤k−1, we get rgn−2/m1,mr,mt=rgn−1/m1,mr,mt=5,r+4,t+4, and if 2≤r≤k−1 and t = k, we get rgn−2/m1,mr,mt=rgn−1/m1,mr,mt=5,r+4,t+2, and if r = k and t = k + 1, we get rgn−2/m1,mr,mt=rgn−1/m1,mr,mt=5,r+4,t, which contradict unique representation.Theorem 4.
Prove that metric dimension of a subdivided convex polytopeSTn denoted by dSTn is 4.Proof.
In this theorem, we taken = 2q, q≥3,q∈Z+. Assume that Z=f1,fq,gq+1,mq+5⊆VSTn; we will prove that Z is a locating set of STn which implies that dim STn≤4. To prove that Z is the locating set, we will just prove that representation of each node with respect to Z is unique.
First of all, we find unique representations of the collectionf1,f2,f3,…,fn:(1)rfsZ=1−s+2q,−q+s,−q+s,2+s−3q2,2q≥s≥3q2+1,1−s+2q,−q+s,−q+s,−1−s+2q,3q2≥s≥q+1,−1+s,−s+q,2−s+q,−1−s+2q,q≥s≥q2+2,−1+s,−s+q,2−s+q,−3+q+sq2+1≥r≥2.
Representations of the family of verticesg1,g2,g3,…,gn areCase I: whens is odd, then(2)rgsZ=2q−s+2,s−q+1,s−q+1,s−q−2,2q−1≥s≥q+7,2q−s+2,s−q+1,s−q+1,2q−s−2,q+5≥s≥q+3,s,q−s+1,q−s+3,2q−s−2,q+1≥s≥q2+2,s,q−s+1,q−s+3,q+s−1,q2≥s≥1.Case II: whens is even, then(3)rgsZ=1+2q−s,q,−8+q,−3+s−q,2q≥s≥q+8,1+2q−s,−8+s,−8+s,−1+2q−s,q+6≥s≥q+4,−1+q,−4+2q−s,2,−1+2q−s,s=q+2,1+s,2,2+q−s,−1+2q−s,q≥s≥q2+3,1+s,q−s,2+q−s,−3+q+s.q2+1≥s≥2.
Representations of the family of central verticesh1,h2,h3,…,hn are(4)rhsZ=2q−2s+2,2s−q+1,2s−q+1,2s−q−4,q≥s≥q2+4,2q−2s+2,2s−q+1,2s−q+1,2q−2s−2,q2+3≥s≥q2+2,q,s−3,s−5,2q−2s−2,i=q2+1,2s,3,q+3−2s,2q−2s−2,q2≥s≥q2−1,2s,q+1−2s,q+3−2s,q+2s−4,q2−2≥s≥1.
Representations of the family of verticesl1,l2,l3,…,ln are(5)rlsZ=2q−2s+3,2s−q+2,2s−q+2,2s−q−5,q≥s≥q2+4,2q−2s+3,2s−q+2,2s−q+2,2q−2s−3,q2+3≥s≥q2+2,q+1,s−2,s−4,2q−2s−3,s=q2+1,2s+1,4,q+4−2s,2q−2s−3,q2≥s≥q2−1,2s+1,q+2−2s,q+4−2s,q+2s−5,q2−2≥s≥1.
Representations of the family of verticesm1,m2,m3,…,mn on the outward cycle are(6)rmiZ=s−q−5,s−8,s−8,s−q−5,s=2q,2q−s+3,s−6,s−8,s−q−5,2q−1≥s≥q+5,2q−s+3,s−6,s−8,2q−s−5,q+4≥s≥q+2,2q−s+3,2q−s−4,q+4−s,q+5−s,q+1≥s≥q,s+3,q+4−s,q+4−s,q+5−s,q−1≥s≥q2+3,s+3,q+2−s,q+4−s,q+5−s,q2+2≥s≥q2+1,s+3,q+2−s,q+4−s,s+5,q2≥s≥1.
Thus, with the help of this formulation, we can simply demonstrate that no two such vertices are available that possess identical representations which implies thatdimTn≤4 in this case. Theorems 2 and 3 imply that neither dimSTn≠2 nor dimSTn≠3, respectively, so it can be concluded that dim dimSTn≥4. Hence, dimSTn=4.


## 3. Conclusion

Metric dimension of convex polytopeSTn after subdividing the convex polytope Tn has been studied. It is proved that the metric dimension of subdivided convex polytope STn is finite and independent of the count of vertices in all these graphs, and just four vertices which are selected properly are sufficient to locate all the vertices of this family of convex polytopes. The authors also proved that the metric dimension of subdivided convex polytope STn is neither 2 nor 3. The researcher gave the formulation of the representation of each vertex of convex polytope with respect to the resolving set and disproved lower dimensions by making different cases.Open problem: determine edge metric dimension of the convex polytopesSTn.



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