On the metric-based resolving parameter of the line graph of certain structures

Abstract

Let G be a graph and R = {r₁, r₂, …, r_k} be an ordered subset of vertices of G, if every two vertices of G have different representation r (v|R) = (d (v, r₁) , d (v, r₂) , …, d (v, r_k)) with respect to R, then R is said to be a metric-based resolving parameter or resolving set of G and its minimum cardinality is called the metric dimension of graph G . Metric dimension is considered as an important applied concept of graph theory especially in the localization of a network and also in the chemical graph theoretical study of molecular compounds. Therefore, it is hot topic to study for different families of graphs as well. Convex polytopes play an important role both in various branches of mathematics and in applied areas, most notably in linear programming. In this paper, we determine the metric-based resolving parameter of line graph of a convex polytope S_n, and conclude that it has constant metric dimension but vary with the parity of n . This article presents a measurement of the line graph of a convex polytope, denoted as $(S_{n}) .$ The subsequent section provides the metric dimension of the resulting graph. There are two scenarios pertaining to the metric dimension of a selected graph with respect to the metric dimension. The metric dimension of even cycle-based convex polytopes is three, whereas for other values, the metric dimension is four.

Keywords

Convex polytope metric dimension resolving set constant metric dimension line graph of convex polytope

1 Introduction

In 1975, Slater first gave the concept of metric dimension [31], and after that, in 1976, this concept was studied independently by Harary and Melter in [7]. The metric dimension has numerous studies, for surveys on the metric dimension, see [4, 6]. On the other parameters of graph like topological indices one can see the work [1], localization is given by [36], metric and edge metric dimension of hollow coronoid is given by [3]. For the metric dimension of different convex polytopes, see [13-16]. Similarly, regarding the theoretical investigations of the resolvability parameter namely metric dimension, various results have been found and extensively studied in the literature, which has highly contributed to understanding this parameter’s mathematical properties associated with distances in graphs. Such as, product of graphs [27], strong metric dimension of power graphs [22, 24] find the metric dimension of newly designed network. Due to the vast applications of metric dimension different work has been done in the chemical graph theory in terms of this parameter, like the structure of H-Naphtalenic and VC₅C₇ nano-tubes discussed with metric concept [11], some upper bounds of cellulose network considering metric dimension as a point of discussion [29], resolving sets of silicate star are determined in [30], computing metric dimension and metric basis of 2D lattice of alpha-boron nanotubes [9, 10].

Let G be a graph, then the shortest path between two vertices of graph G is called distance and R = {r₁, r₂, …, r_k} is an ordered subset of vertices of G, if every two vertices of G have different representation r (v|R) = (d (v, r₁) , d (v, r₂) , …, d (v, r_k)) with respect to R then R is said to be a resolving set of G and its minimum cardinality is called metric dimension of graph G . The line graph L (G) of graph G is the graph whose vertices are the edges of G, and two vertices v₁ and v₂ of L (G) are connected if and only if they have a common end vertex in G . The line graphs have numerous applications not only in graph theory also in chemistry and computer science.

Computational complexity and NP-hardness of metric dimension, studied in [5 , 20]. Metric dimension has various practical applications in our daily life, which motivate the researchers, and it has been widely studied. Metric dimension is extensively applied in the various fields such as facility location problems, sonar and coastguard Loran [31], computer networks [23], combinatorial optimization [28], robot navigation [18, 19], pharmaceutical chemistry [5, 6], weighing problems [32], image processing for further detail regarding the application see [25, 26], and for the further study on this particular topic see [33 , 37]. There are many other parameters which are somehow related to the graph theory and systems with their applications, refer to see the articles [38-44].

1.1 Convex polytopes and their importance

Convex polytopes are essential entities in the field of geometry, possessing extensive utility across several academic domains. The significance and utility of these entities stem from their multifaceted characteristics and wide-ranging applications. The following are essential elements that underscore their significance:

Geometric Modeling: Convex polytopes are essential geometric structures that have the capacity to represent a diverse array of shapes and structures. The inherent simplicity and well defined geometric features of these objects render them highly valuable in the representation and analysis of intricate geometrical arrangements within both two- and three-dimensional spatial contexts. Linear Programming: Convex polytopes hold a pivotal position within the realm of linear programming, a mathematical methodology employed for the purpose of optimization. The representation of the feasible region in a linear program is commonly depicted as a convex polytope. Within this geometric framework, optimization problems that incorporate linear constraints can be effectively and efficiently handled.

Computational Geometry: Computational geometry employs techniques that pertain to convex polytopes in order to address challenges associated with geometric entities. Algorithms utilized in the determination of the convex hull, which refers to the smallest convex set encompassing a specified set of points, hold significant importance in diverse computer science applications. Combinatorial Structure: The combinatorial properties of convex polytopes are extensive and have been well investigated. The examination of the facets, vertices, and edges of convex polytopes yields valuable insights on the combinatorial characteristics of the corresponding polytope, hence making significant contributions to the fields of combinatorial geometry and combinatorics. Polyhedral Combinatorics: The study of polyhedral combinatorics involves an examination of the combinatorial characteristics of polyhedra and polytopes, with convex polytopes being particularly interconnected in this field. The discipline of mathematics in question finds practical utility in various domains, including but not limited to algebraic geometry, optimization, and discrete mathematics.

Statistical Modeling and Data Analysis: Convex polytopes are widely employed in the field of statistics for the purpose of data modeling and analysis. In the context of robust statistics, convex hulls are employed to delineate central regions that exhibit reduced sensitivity to outliers, hence facilitating a more resilient estimation of central tendencies. Operations Research: Convex polytopes find application in the field of operations research as they serve as effective tools for representing feasible regions and restrictions inside optimization issues. Linear programming, integer programming, and various other optimization approaches utilize the geometric characteristics of convex polytopes to effectively address problems. Quantum Information Theory: Convex polytopes are utilized within the field of quantum information theory to serve as a representation for the collection of quantum states. The use of this application is crucial in comprehending and delineating the conduct of quantum systems, hence bearing significant ramifications for the domains of quantum computing and communication. Topological Data Analysis: Convex hulls and other convex structures are employed in the field of topological data analysis for the purpose of examining the geometric properties and topological characteristics of data. The aforementioned phenomenon possesses significant implications in comprehending the fundamental framework of intricate datasets across diverse domains, such as biology and neuroscience. Polytope Theory in Algebraic Geometry: The study of convex polytopes is closely linked to algebraic geometry by means of toric varieties. The field of polytope theory offers a geometric framework for the characterization of algebraic varieties, hence facilitating their use in the domains of algebraic geometry and mathematical physics.

In essence, convex polytopes hold significant relevance across various disciplines, encompassing optimization, geometry, combinatorics, computer science, statistics, quantum information theory, and other related domains. The inherent geometric simplicity of these objects, along with their extensive combinatorial and computational features, renders them highly effective instruments for the purpose of modeling, analysis, and problem-solving across a wide range of mathematical disciplines and beyond.

1.2 Highlight and the motivation of the study

The primary objective of this work is to examine and analyse the metric-based resolving parameter in relation to the line graph of certain structures, with the aim of comprehending its complex intricacies. The decision to examine the metric dimension of these line graphs, specifically with the established values of 3 for even cycles and 4 for odd cycles, arises from the acknowledgment of their significant contribution to comprehending the inherent connection and navigability of graphs.

A comprehensive comprehension of the metric dimension of a graph holds significant importance in many domains, encompassing network architecture and information retrieval. The inclusion of even and odd cycles in this study introduces an additional level of intricacy and captivation. Cycles with an even number of vertices and a metric dimension of 3, as well as cycles with an odd number of vertices and a metric dimension of 4, demonstrate a subtle variance in the intrinsic resolving capability of these structures. The differentiation between these two concepts functions as a stimulus for delving into the various ways in which the structural properties of a graph impact its ability to resolve metrics.

Furthermore, the research is driven by a wider aspiration to make a contribution to the theoretical underpinnings of graph theory. The research seeks to enhance comprehension of graph structures and their fundamental qualities by identifying precise metric dimensions for these line graphs. This comprehension, in return, might have ramifications for various disciplines, ranging from communication networks to social network analysis, where effective navigation and information retrieval are of utmost importance.

The primary motivation for conducting this study lies in the pursuit of understanding the metric dimensions inside a certain graph framework. The objective is to gain insights into the resolving capabilities of cycles with even and odd lengths. This investigation aims to make contributions to the theoretical framework of graph theory and provide prospective real-world applications, ultimately enhancing the broader subject of network science and mathematics.

Following theorem is very useful of finding the parameter of metric dimension.

1.3 Auxiliary results

In this subsection, we will recall some important results of this topic.

Theorem 1.1. [21] Let G be a simple connected graph with dim (G) = 2 and let {a₁, a₂} ⊂ V (G) be a metric basis of G, then the degree of both a₁, a₂ is at most 3 and there exists a unique shortest path between a₁ and a₂ .

Theorem 1.2. If G be a simple, undirected connected graph, then

(1)dim (G) =1 iff G = P_n,

(2)dim (G) = n - 1 iff G = K_n,

(3)dim_f (G) =2 iff G = P_n,

(4)dim_f (G) = n iff G = K_n,

(5)dim (G) =2 if G = C_n,

(6)dim_f (G) =3 if G = C_n,

(7)Let T be a tree that is not a path and define l (T) to be the number of leaves (nodes of degree 1) in T . Further, define σ (T) as the number of exterior major vertices in T, that is vertices with degree at least 3 which are also connected to at least one leaf by an edge or a path of vertices of degree 2. Then the metric dimension of T is dim (T) = l (T) - σ (T) .

In this article, we will measured the line graph of convex polytope $(S_{n})$ and the metric dimension of resulted graph is given in the next section. There are two cases on the metric dimension of chosen graph respect to the metric dimension. For even cycle-based convex polytopes the metric dimension is 3 whhile for the other values the metric dimension is four.

In the Section 2, we have computed the main results of our work. Conclusion and Discussion is also given in the Section 3.

2 Metric dimension of line graph of convex polytope $(S_{n})$

The graph $S_{n},$ shown in Fig. 2 is the line graph of convex polytopes S_n, shown in the Fig. 1. It is study in [11]. We labeled the vertices {a_i : 1 ≤ i ≤ n} by calling inner cycle, {b_i : 1 ≤ i ≤ 2n} second inner cycle, {c_i : 1 ≤ i ≤ n} middle, {d_i : 1 ≤ i ≤ 2n} third inner cycle, {e_i : 1 ≤ i ≤ n} exterior and {f_i : 1 ≤ i ≤ n} outer cycle vertices. The order and size of $S_{n}$ is 8n and 19n respectively. Moreover, the generalize view of labeling vertices and generating line graph of convex polytope $S_{n},$ we refer to see the Fig. 2. The following theorem is the metric dimension of this structure.

Fig. 1

Convex Polytope S_n.

Fig. 2

Line graph of Convex Polytope S_n.

Theorem 2.1. Let $S_{n}$ be a line graph of convex polytope S_n, with n ≥ 6 . Then $\dim (S_{n}) = {\begin{matrix} 3, & if n = even, \\ 4, & if n = odd . \end{matrix}$

Proof. We divide the proof of theorem into three cases with respect to their metric dimension.

Case 1: For n = 6, consider the resolving set R = {b₁₁, e₄, f₁} . Following are the representations of all vertices of $S_{n}$ with respect to chosen resolving set R .

Inner cycle vertices distances representations:

The distance of a_ϵ with respect to first element of resolving set i.e. b₁₁ is following; $d (a_{ϵ}, b_{11}) = {\begin{matrix} ⌊ \frac{ϵ}{2} ⌋ + 2, & if ϵ = 1, 2, 3; \\ n - ϵ + ⌊ \frac{6}{ϵ} ⌋, & if ϵ = 4, 5, 6 . \end{matrix}$ The distance of a_ϵ with respect to second element of resolving set i.e. e₄ is following; $d (a_{ϵ}, e_{4}) = {\begin{matrix} n - 1, & if ϵ = 1, 6; \\ 4, & if ϵ = 2, 3, 4, 5 . \end{matrix}$ The distance of a_ϵ with respect to third element of resolving set i.e. f₁ is following; $d (a_{ϵ}, f_{1}) = {\begin{matrix} 5, & if ϵ = 1, 2, 5, 6; \\ 6, & if ϵ = 3, 4 . \end{matrix}$ Second Inner cycle vertices distances representations:

The distance of b_ϵ with respect to first element of resolving set i.e. b₁₁ is following, $d (b_{ϵ}, b_{11}) = {\begin{matrix} ⌊ \frac{ϵ}{2} ⌋ + 2, & if ϵ = 1, \dots, 4; \\ ⌈ \frac{2 n - ϵ}{2} ⌉, & if ϵ = 5, \dots, 10; \\ ϵ + 1 - 2 n, & if ϵ = 11, 12; . \end{matrix}$ The distance of b_ϵ with respect to second element of resolving set i.e. e₄ is following; $d (b_{ϵ}, e_{4}) = {\begin{matrix} n - ϵ + 1, & if ϵ = 1, 2; \\ n - ϵ + 2, & if ϵ = 3, 4, 5; \\ 3, & if ϵ = 6, 7, 8 \\ ⌊ \frac{ϵ}{2} ⌋, & if ϵ = 9, \dots, 12; \end{matrix}$ The distance of b_ϵ with respect to third element of resolving set i.e. f₁ is following; $d (b_{ϵ}, f_{1}) = {\begin{matrix} ⌈ \frac{n + ϵ}{2} ⌉, & if ϵ = 1, \dots, 6; \\ 5, & if ϵ = 7, 8; \\ 4, & if ϵ = 9, \dots, 12 . \end{matrix}$ Middle vertices distances representations:

The distance of c_ϵ with respect to first element of resolving set i.e. b₁₁ is following; $d (c_{ϵ}, b_{11}) = {\begin{matrix} ϵ + 2, & if ϵ = 1, 2; \\ n - ϵ + 1, & if ϵ = 3, \dots, 10 . \end{matrix}$ The distance of c_ϵ with respect to second element of resolving set i.e. e₄ is following; $d (c_{ϵ}, e_{4}) = {\begin{matrix} n - ϵ, & if ϵ = 1, 2; \\ 2, & if ϵ = 3, 4; \\ ϵ - 1, & if ϵ = 5, 6 . \end{matrix}$ The distance of c_ϵ with respect to third element of resolving set i.e. e₄ is following; $d (c_{ϵ}, f_{1}) = {\begin{matrix} ϵ + 2, & if ϵ = 1, 2, 3; \\ 4, & if ϵ = 4; \\ 3, & if ϵ = 5, 6 . \end{matrix}$ Third inner cycle vertices distances representations:

The distance of d_ϵ with respect to first element of resolving set i.e. b₁₁ is following; $d (d_{ϵ}, b_{11}) = {\begin{matrix} ϵ + 2, & if ϵ = 1, 2, 3; \\ 5, & if ϵ = 4, 5, 6 \dots, 10; \\ ⌈ \frac{2 n - ϵ + 3}{2} ⌉, & if ϵ = 7, \dots, 12 . \end{matrix}$ The distance of d_ϵ with respect to second element of resolving set i.e. e₄ is following; $d (d_{ϵ}, e_{4}) = {\begin{matrix} n - ϵ, & if ϵ = 1, 2; \\ n - ϵ + 1, & if ϵ = 3, \dots, 6; \\ ϵ - n, & if ϵ = 7, \dots, 10; \\ n - ϵ + 1, & if ϵ = 11, 12 . \end{matrix}$ The distance of d_ϵ with respect to third component of resolving set i.e. f₁ is following; $d (d_{ϵ}, f_{1}) = {\begin{matrix} ⌊ \frac{ϵ}{2} ⌋ + 2, & if ϵ = 1, \dots, 5; \\ ⌊ \frac{2 n - ϵ + 3}{2} ⌋, & if ϵ = 6, \dots, 11; \\ 2, & if ϵ = 12 . \end{matrix}$ Exterior vertices distances representations:

The distance of e_ϵ with respect to first element of resolving set i.e. b₁₁ is following; $d (e_{ϵ}, b_{11}) = {\begin{matrix} ϵ + 2, & if ϵ = 1; \\ 5, & if ϵ = 2, 3, 4; \\ n - ϵ + 3, & if ϵ = 5, 6 . \end{matrix}$ The distance of e_ϵ with respect to second element of resolving set i.e. e₄ is following; $d (e_{ϵ}, e_{4}) = {\begin{matrix} n - ϵ - 1, & if ϵ = 1, 2, 3; \\ 0, & if ϵ = 4; \\ ⌊ \frac{ϵ}{2} ⌋, & if ϵ = 5, 6 . \end{matrix}$ The distance of e_ϵ with respect to third element of resolving set i.e. f₁ is following; $d (e_{ϵ}, f_{1}) = {\begin{matrix} ϵ, & if ϵ = 1, 2, 3; \\ n - ϵ + 1, & if ϵ = 4, 5, 6 . \end{matrix}$ Outer cycle vertices distances representations:

The distance of f_ϵ with respect to first element of resolving set i.e. b₁₁ is following; $d (f_{ϵ}, b_{11}) = {\begin{matrix} ⌈ \frac{n + ϵ}{2}, & if ϵ = 1, 2, 3; \\ n - ϵ + 4, & if ϵ = 4, 5, 6 . \end{matrix}$ The distance of f_ϵ with respect to second element of resolving set i.e. e₄ is following; $d (f_{ϵ}, e_{4}) = {\begin{matrix} ⌈ \frac{n - ϵ + 1}{2} ⌉, & if ϵ = 1, 2, 3; \\ ⌊ \frac{n}{ϵ} ⌋ + ⌊ \frac{ϵ}{n} ⌋, & if ϵ = 4, 5, 6 . \end{matrix}$ The distance of f_ϵ with respect to third element of resolving set i.e. f₁ is following; $d (f_{ϵ}, f_{1}) = {\begin{matrix} ϵ - 1, & if ϵ = 1, \dots, 4; \\ n - ϵ + 1, & if ϵ = 5, 6 . \end{matrix}$ Case 2: For even values of n ≥ 8, consider the resolving set R = {b_2n-1, e₅, f₁} . Following are the representations of all vertices of $S_{n}$ with respect to chosen resolving set R .

Inner cycle vertices distances representations:

The distance of a_ϵ with respect to first element of resolving set i.e. b_2n-1 is following; $d (a_{ϵ}, b_{2 n - 1}) = {\begin{matrix} ϵ + 1, & if k = 1, \dots, \frac{n - 2}{2}; \\ n - ϵ, & if ϵ = \frac{n}{2}, \dots, n - 1; \\ 1, & if ϵ = n . \end{matrix}$ The distance of a_ϵ with respect to second component of resolving set i.e. e₅ is following; $d (a_{ϵ}, e_{5}) = {\begin{matrix} 6 - ⌊ \frac{ϵ}{2} ⌋, & if k = 1, 2; \\ 4 & if ϵ = 3, \dots, 6; \\ ϵ - 2, & if ϵ = 7, \dots, \frac{n + 8}{2}; \\ n + ϵ - k, & if ϵ = \frac{n + 10}{2} . \end{matrix}$ The distance of a_ϵ with respect to third element of resolving set i.e. f₁ is following; $d (a_{ϵ}, f_{1}) = {\begin{matrix} 5, & if ϵ = 1; \\ ϵ + 3, & if ϵ = 2, \dots, \frac{n}{2}; \\ n - ϵ + 3, & if ϵ = \frac{n + 2}{2}, \dots, n - 3; \\ 5, & if ϵ = n - 2, \dots, n . \end{matrix}$ Second inner cycle vertices distances representations:

The distance of b_ϵ with respect to first element of resolving set i.e. b_2n-1 is following; $d (b_{ϵ}, b_{2 n - 1}) = {\begin{matrix} ⌊ \frac{ϵ}{2} ⌋ + 2, & if ϵ = 1, \dots, n - 1; \\ ⌈ \frac{2 n - k}{2} ⌉, & if ϵ = n, \dots, 2 n - 3; \\ | 2 n - ϵ - 1 |, & if ϵ = 2 n - 2, \dots, 2 n . \end{matrix}$ The distance of b_ϵ with respect to second element of resolving set i.e. e₅ is following; $d (b_{ϵ}, e_{5}) = {\begin{matrix} 7 - ⌊ \frac{ϵ}{2} ⌋, & if ϵ = 1, \dots, 5; \\ 4, & if ϵ = 6; \\ 3, & if ϵ = 7, \dots, 10; \\ ⌊ \frac{k}{2} ⌋ - 1, & if ϵ = 11, \dots, 2 n - 3; \\ 8 - 2 ⌊ \frac{15}{k} ⌋, & if ϵ = 2 n - 2, \dots, 2 n - 1; \\ 7, & if ϵ = 2 n . \end{matrix}$ The distance of b_ϵ with respect to third element of resolving set i.e. f₁ is following; $d (b_{ϵ}, f_{1}) = {\begin{matrix} ⌈ \frac{ϵ}{2} + 3, & if ϵ = 1, \dots, n - 2; \\ ⌈ \frac{2 n - ϵ + 5}{2}, & if ϵ = n - 1, \dots, 2 n - 4; \\ 4, & if ϵ = 2 n - 3, \dots, 2 n . \end{matrix}$ Middle vertices distances representations:

The distance of c_ϵ with respect to first element of resolving set i.e. b_2n-1 is following; $d (c_{ϵ}, b_{2 n - 1}) = {\begin{matrix} ϵ + 2, & if ϵ = 1, \dots, \frac{n - 2}{2}; \\ n - ϵ + 1, & if ϵ = \frac{n}{2}, \dots, n . \end{matrix}$ The distance of c_ϵ with respect to second element of resolving set i.e. e₅ is following; $d (c_{ϵ}, e_{5}) = {\begin{matrix} n - ϵ - 1, & if ϵ = 1, 2, 3; \\ 2, & if ϵ = 4, 5; \\ ϵ - 2, & if ϵ = 5, \dots, \frac{n + 8}{2}; \\ n + 7 - ϵ, & if ϵ = \frac{n + 10}{2}, \dots, n . \end{matrix}$ The distance of c_ϵ with respect to third element of resolving set i.e. f₁ is following; $d (c_{ϵ}, f_{1}) = {\begin{matrix} ϵ + 2, & if ϵ = 1, \dots, \frac{n}{2}; \\ n - ϵ + 2, & if ϵ = \frac{n + 2}{2}, \dots, n - 1; \\ 3, & if ϵ = n . \end{matrix}$ Third inner cycle vertices distances representations:

The distance of d_ϵ with respect to first element of resolving set i.e. b_2n-1 is following; $d (d_{ϵ}, b_{2 n - 1}) = {\begin{matrix} ϵ + 2, & if ϵ = 1, \dots, n - 1; \\ ⌈ \frac{ϵ}{2} ⌉ + 3, & if ϵ = n, \dots, 2 n - 3; \\ ⌊ \frac{2 n - ϵ}{2} ⌋ + 2, & if ϵ = 2 n - 2, \dots, 2 n . \end{matrix}$ The distance of d_ϵ with respect to second element of resolving set i.e. e₅ is following; $d (d_{ϵ}, e_{5}) = {\begin{matrix} 6 - ⌊ \frac{ϵ}{2} ⌋, & if ϵ = 1, \dots, 5; \\ n - ϵ + 1, & if ϵ = 6, \dots, 8; \\ ϵ - 8, & if ϵ = 9, \dots, 11; \\ ⌊ \frac{ϵ}{2} ⌋ - 2, & if ϵ = 12 \dots, 2 n - 3; \\ 7 - 2 ⌊ \frac{15}{ϵ} ⌋, & if ϵ = 2 n - 2, 2 n - 1; \\ 6, & if ϵ = 2 n \end{matrix}$ The distance of d_ϵ with respect to third element of resolving set i.e. f₁ is following; $d (d_{ϵ}, f_{1}) = {\begin{matrix} ⌊ \frac{ϵ}{2} ⌋ + 2, & if ϵ = 1, \dots, n - 1; \\ ⌈ \frac{2 n - ϵ}{2} ⌉ + 1, & if ϵ = n, \dots, 2 n - 1; \\ 2, & if ϵ = 2 n . \end{matrix}$ Exterior vertices distances representations:

The distance of e_ϵ with respect to first element of resolving set i.e. b_2n-1 is following; $d (e_{ϵ}, b_{2 n - 1}) = {\begin{matrix} 3, & if ϵ = 1; \\ ϵ + 3, & if ϵ = 2, \dots, \frac{n}{2}; \\ n - ϵ + 3, & if ϵ = \frac{n + 2}{2}, \dots, n . \end{matrix}$ The distance of e_ϵ with respect to second element of resolving set i.e. e₅ is following; $d (e_{ϵ}, e_{5}) = {\begin{matrix} 6 - ϵ, & if ϵ = 1, \dots, 4; \\ 0, & if ϵ = 5; \\ ϵ - 4, & if ϵ = 6, \dots, \frac{n + 10}{2}; \\ n - ϵ + 6, & if ϵ = \frac{n + 12}{2}, \dots, n . \end{matrix}$ The distance of e_ϵ with respect to third element of resolving set i.e. f₁ is following; $d (e_{ϵ}, f_{1}) = {\begin{matrix} ϵ, & if ϵ = 1, \dots, \frac{n + 2}{2}; \\ n - ϵ + 1, & if ϵ = \frac{n + 4}{2}, \dots, n . \end{matrix}$ Outer cycle vertices distances representations:

The distance of f_ϵ with respect to first element of resolving set i.e. b_2n-1 is following; $d (f_{ϵ}, b_{2 n - 1}) = {\begin{matrix} 4, & if ϵ = 1; \\ ϵ + 2 & if ϵ = 2, \dots, \frac{n + 2}{2} \\ n - ϵ + 4, & if ϵ = \frac{n + 4}{2}, \dots, n . \end{matrix}$ The distance of f_ϵ with respect to second element of resolving set i.e. e₅ is following; $d (f_{ϵ}, e_{4}) = {\begin{matrix} 5 - ⌊ \frac{ϵ}{8} ⌋, & if ϵ = 1; \\ n - ϵ - 2, & if ϵ = 2, \dots, 6; \\ ϵ - 5, & if ϵ = 7, \dots, n . \end{matrix}$ The distance of f_ϵ with respect to third element of resolving set i.e. f₁ is following; $d (f_{ϵ}, f_{1}) = {\begin{matrix} ϵ - 1, & if ϵ = 1, \dots, \frac{n + 2}{2}; \\ n - ϵ + 1, & if ϵ = \frac{n + 4}{2}, \dots, n . \end{matrix}$ One can easily verify that all the vertices have different representations with respect to selected resolving set R, it implies to that, $\dim (S_{n}) \leq 3 .$ (1) To prove that $\dim (S_{n}) \geq 3$ we can use contradiction method for this purpose suppose that $\dim (S_{n}) = 2$ . From Fig. 2 we see that the minimum degree of a vertex is 4, so we refer Theorem 1.1 and implies that the resolving set R with cardinality 2 is not possible and this resulted in $\dim (S_{n}) \geq 3 .$ (2) Equations (1) and (2) implies that $\dim (S_{n}) = 3 .$ Case 3: For the odd values of n ≥ 7, consider the resolving set $R = {a_{1}, a_{\frac{n - 1}{2}}, f_{2}, f_{\frac{n + 1}{2}}} .$ Following are the representations of all vertices of $S_{n}$ with respect to chosen resolving set R .

Inner cycle vertices distances representations:

The distance of a_ϵ with respect to first element of resolving set i.e. a₁ is following; $d (a_{ϵ}, a_{1}) = {\begin{matrix} ϵ - 1, & if ϵ = 1, \dots, \frac{n + 1}{2} \\ n - ϵ + 1, & if ϵ = \frac{n + 3}{2}, \dots, n; . \end{matrix}$ The distance of a_ϵ with respect to second element of resolving set i.e. $a_{\frac{n - 1}{2}}$ is following; $d (a_{ϵ}, a_{\frac{n - 1}{2}}) = {\begin{matrix} \frac{n - 2 ϵ - 1}{2}, & if ϵ = 1, \dots, \frac{n - 1}{2}; \\ \frac{2 ϵ - n + 1}{2}, & if ϵ = \frac{n + 1}{2}, \dots, n - 1; \\ \frac{2 ϵ - n - 1}{2}, & if ϵ = n . \end{matrix}$ The distance of a_ϵ with respect to third element of resolving set i.e. f₂ is following; $d (a_{ϵ}, f_{2}) = {\begin{matrix} 5, & if ϵ = 1, 2, 3; \\ ϵ + 1, & if ϵ = 4, \dots, \frac{n + 3}{2}; \\ n - ϵ + 5, & if ϵ = \frac{n + 5}{2}, \dots, n; . \end{matrix}$ The distance of a_ϵ with respect to fourth element of resolving set i.e. $f_{\frac{n + 1}{2}}$ is following; $d (a_{ϵ}, f_{\frac{n + 1}{2}}) = {\begin{matrix} \frac{n - 2 ϵ + 7}{2} - ϵ + 1, & if ϵ = 1, \dots, \frac{n - 3}{2}; \\ 5, & if ϵ = \frac{n - 1}{2}, \dots, \frac{n + 3}{2}; \\ ϵ - ⌊ \frac{n - 5}{2} ⌋, & if ϵ = \frac{n + 5}{2}, \dots, n . \end{matrix}$ Second inner cycle vertices distances representations:

The distance of b_ϵ with respect to first element of resolving set i.e. a₁ is following; $d (b_{ϵ}, a_{1}) = {\begin{matrix} 1, & if ϵ = 1, 2, 3; \\ ⌊ \frac{ϵ}{2} ⌋, & if ϵ = 4, \dots, n + 2; \\ ⌈ \frac{2 n - ϵ}{2} ⌉ + 1, & if ϵ = n + 3, \dots, 2 n - 1; \\ 1, & if ϵ = 2 n . \end{matrix}$ The distance of b_ϵ with respect to second element of resolving set i.e. $a_{\frac{n - 1}{2}}$ is following; $d (b_{ϵ}, a_{\frac{n - 1}{2}}) = {\begin{matrix} \frac{n - 1}{2}, & if ϵ = 1, 2 n; \\ ⌊ \frac{n - ϵ}{2} ⌋, & if ϵ = 2, \dots, n - 2; \\ 1 & if ϵ = n - 1, n; \\ ⌈ \frac{ϵ - n}{2} ⌉ + 1, & if ϵ = n + 1, \dots, 2 n - 1 . \end{matrix}$ The distance of b_ϵ with respect to third element of resolving set i.e. f₂ is following; $d (b_{ϵ}, f_{2}) = {\begin{matrix} 4 & if ϵ = 1, \dots, 6 \\ ⌈ \frac{ϵ}{2} ⌉ + 1, & if ϵ = 7, \dots, n + 3; \\ ⌊ \frac{2 n - ϵ}{2} ⌋ + 5, & if ϵ = n + 4, \dots, 2 n . \end{matrix}$ The distance of b_ϵ with respect to fourth element of resolving set i.e. $f_{\frac{n + 1}{2}}$ is following; $d (b_{ϵ}, f_{\frac{n + 1}{2}}) = {\begin{matrix} ⌈ \frac{n - ϵ}{2} ⌉ + 3, & if ϵ = 1, \dots, n - 1; \\ 4, & if ϵ = n, \dots, n + 3; \\ ⌊ \frac{ϵ - n}{2} ⌋ + 3, & if ϵ = n + 4, \dots, 2 n . \end{matrix}$ Middle vertices distances representations:

The distance of c_ϵ with respect to first element of resolving set i.e. a₁ is following; $d (c_{ϵ}, a_{1}) = {\begin{matrix} 2, & if ϵ = 1; \\ ϵ, & if ϵ = 2, \dots, \frac{n + 1}{2}; \\ n - ϵ + 2, & if ϵ = \frac{n + 3}{2}, \dots, n . \end{matrix}$ The distance of c_ϵ with respect to second element of resolving set i.e. $a_{\frac{n - 1}{2}}$ is following; $d (c_{ϵ}, a_{\frac{n - 1}{2}}) = {\begin{matrix} \frac{n - 2 ϵ + 1}{2}, & if ϵ = 1, \dots, \frac{n - 3}{2}; \\ 2, & if ϵ = \frac{n - 1}{2}, \dots, \frac{n + 1}{2}; \\ \frac{2 ϵ - n + 3}{2}, & if ϵ = \frac{n + 3}{2}, \dots, n - 1; \\ ⌈ \frac{ϵ}{2} ⌉, & if ϵ = n . \end{matrix}$ The distance of c_ϵ with respect to third element of resolving set i.e. f₂ is following; $d (c_{ϵ}, f_{2}) = {\begin{matrix} 3, & if ϵ = 1, 2, 3; \\ ϵ, & if ϵ = 4, \dots, \frac{n + 3}{2}; \\ n - ϵ + 4, & if ϵ = \frac{n + 5}{2}, \dots, n . \end{matrix}$ The distance of c_ϵ with respect to fourth element of resolving set i.e. $f_{\frac{n + 1}{2}}$ is following; $d (c_{ϵ}, f_{\frac{n + 1}{2}}) = {\begin{matrix} \frac{n - 2 ϵ + 5}{2}, & if ϵ = 1, \dots, \frac{n - 1}{2}; \\ 3, & if ϵ = \frac{n + 1}{2}; \\ \frac{2 ϵ - n + 3}{2}, & if ϵ = \frac{n + 3}{2}, \dots, n . \end{matrix}$ Third inner cycle vertices distances representations:

The distance of d_ϵ with respect to first element of resolving set i.e. a₁ is following; $d (d_{ϵ}, a_{1}) = {\begin{matrix} 3, & if ϵ = 1, \dots, 4; \\ ⌈ \frac{ϵ}{2} ⌉ + 1, & if ϵ = 5, \dots, n + 1; \\ ⌊ \frac{2 n - ϵ}{2} ⌋ + 3, & if ϵ = n + 2, \dots, 2 n . \end{matrix}$ The distance of d_ϵ with respect to second element of resolving set i.e. $a_{\frac{n - 1}{2}}$ is following; $d (d_{ϵ}, a_{\frac{n - 1}{2}}) = {\begin{matrix} ⌈ \frac{n - ϵ}{2} ⌉ + 1, & if ϵ = 1, \dots, n - 3; \\ 3, & if ϵ = n - 2, \dots, n + 1; \\ ⌊ \frac{ϵ - n}{2} ⌋ + 3, & if ϵ = n + 2 \dots, 2 n - 2; \\ ⌊ \frac{ϵ - n}{2} ⌋ + 2, & if ϵ = 2 n - 1, 2 n . \end{matrix}$ The distance of d_ϵ with respect to third element of resolving set i.e. f₂ is following; $d (d_{ϵ}, f_{2}) = {\begin{matrix} 3, & if ϵ = 1 \\ 2, & if ϵ = 2, \dots, 5; \\ ⌊ \frac{ϵ}{2} ⌋, & if ϵ = 6, \dots, n + 4; \\ ⌈ \frac{2 n - ϵ}{2} ⌉ + 3, & if ϵ = n + 5, \dots, 2 n . \end{matrix}$ The distance of d_ϵ with respect to fourth element of resolving set i.e. $f_{\frac{n + 1}{2}}$ is following; $d (d_{ϵ}, f_{\frac{n + 1}{2}}) = {\begin{matrix} \frac{n + 3}{2}, & if ϵ = 1; \\ ⌊ \frac{n -}{2} ⌋ + 2, & if ϵ = 2, \dots, n; \\ ⌈ \frac{ϵ - n}{2} ⌉ + 1, & if ϵ = n + 1 \dots, 2 n . \end{matrix}$ Exterior vertices distances representations:

The distance of e_ϵ with respect to first element of resolving set i.e. a₁ is following; $d (e_{ϵ}, a_{1}) = {\begin{matrix} 4, & if ϵ = 1, 2, 3; \\ ϵ + 1, & if ϵ = 4, \dots, \frac{n + 3}{2}; \\ n - ϵ + 4, & if ϵ = \frac{n + 5}{2}, \dots, n . \end{matrix}$ The distance of e_ϵ with respect to second element of resolving set i.e. $a_{\frac{n - 1}{2}}$ is following; $d (e_{ϵ}, a_{\frac{n - 1}{2}}) = {\begin{matrix} \frac{n - 2 ϵ + 5}{2}, & if ϵ = 1, \dots, \frac{n - 5}{2}; \\ 4, & if ϵ = \frac{n - 3}{2}, \dots, \frac{n + 3}{2}; \\ ⌊ \frac{2 ϵ - n + 5}{2} ⌋, & if ϵ = n . \end{matrix}$ The distance of e_ϵ with respect to third element of resolving set i.e. f₂ is following; $d (e_{ϵ}, f_{2}) = {\begin{matrix} 2 - ⌊ \frac{ϵ}{2} ⌋, & if ϵ = 1, 2; \\ ϵ - 2, & if ϵ = 3, \dots, \frac{n + 3}{2}; \\ n - ϵ + 3, & if ϵ = \frac{n + 5}{2}, \dots, n . \end{matrix}$ The distance of e_ϵ with respect to fourth element of resolving set i.e. $f_{\frac{n + 1}{2}}$ is following; $d (e_{ϵ}, f_{\frac{n + 1}{2}}) = {\begin{matrix} \frac{n - 2 ϵ + 3}{2}, & if ϵ = 1, \dots, \frac{n + 1}{2}; \\ \frac{2 ϵ - n - 1}{2}, & if ϵ = \frac{n + 3}{2}, \dots, n . \end{matrix}$ Outer cycle vertices distances representations:

The distance of f_ϵ with respect to first element of resolving set i.e. a₁ is following; $d (f_{ϵ}, a_{1}) = {\begin{matrix} 5, & if ϵ = 1, 2, 3, n - 1, n; \\ ϵ + 2, & if ϵ = 4, \dots, \frac{n + 1}{2}; \\ n - ϵ + 4, & if ϵ = \frac{n + 3}{2}, \dots, n - 2; . \end{matrix}$ The distance of f_ϵ with respect to second element of resolving set i.e. $a_{\frac{n - 1}{2}}$ is following; $d (f_{ϵ}, a_{\frac{n - 1}{2}}) = {\begin{matrix} \frac{n - 2 ϵ + 5}{2}, & if ϵ = 1, \dots, \frac{n - 5}{2}; \\ 5, & if ϵ = \frac{n - 3}{2}, \dots, \frac{n + 3}{2}; \\ \frac{2 ϵ - n + 7}{2} + ⌊ \frac{6}{ϵ} ⌋, & if ϵ = \frac{n + 5}{2}, \dots, n - 1; \\ \frac{2 ϵ - n + 5}{2}, & if ϵ = n . \end{matrix}$ The distance of f_ϵ with respect to third element of resolving set i.e. f₂ is following; $d (f_{ϵ}, f_{2}) = {\begin{matrix} | ϵ - 2 |, & if ϵ = 1, \dots, \frac{n + 3}{2}; \\ n - ϵ + 2, & if ϵ = \frac{n + 5}{2}, \dots, n . \end{matrix}$ The distance of e_ϵ with respect to fourth element of resolving set i.e. $f_{\frac{n + 1}{2}}$ is following; $d (e_{ϵ}, f_{\frac{n + 1}{2}}) = {\begin{matrix} \frac{n - 2 ϵ + 1}{2}, & if ϵ = 1, \dots, \frac{n + 1}{2}; \\ \frac{2 ϵ - n - 1}{2}, & if ϵ = \frac{n + 3}{2}, \dots, n . \end{matrix}$ One can easily verify that all the vertices have different representations with respect to selected resolving set R, it implies to that, $\dim (S_{n}) \leq 4 .$ (3) Now for $\dim (S_{n}) \geq 4,$ we are going to use contradiction method and it will resulted in $\dim (S_{n}) = 3 .$ For this purpose lets check the resolving set, R′ with cardinality 3 .

Case 3.1: All three vertices are from inner cycle. So, without loss of generality, consider the resolving set R′ = {a_p, a_q, a_s} where $1 \leq p, q, s \leq \frac{n + 1}{2},$ we have $r (b_{2 p} | R^{'}) = r (b_{p + q} | R^{'}) = (1, 1, 2) .$ Case 3.2: All three vertices are from second inner cycle. So, without loss of generality, consider the resolving set R′ = {b_p, b_q, b_s} where 1 ≤ p, q, s ≤ n, then we have $r (d_{p} | R^{'}) = r (d_{q} | R^{'}) = (2, 2, 3) .$ Case 3.3: All three vertices are from middle vertices. So, without loss of generality, consider the resolving set R′ = {c_p, c_q, c_s} where 1 ≤ p, q, s ≤ n, we have $r (b_{2 s + 2} | R^{'}) = r (b_{2 s + 3} | R^{'}) = (5, 4, 3) .$ Case 3.4: All three vertices are from third inner cycle. So, without loss of generality, consider the resolving set R′ = {d_p, d_q, d_s} where 1 ≤ p, q, s ≤ n, we have $r (b_{p} | R^{'}) = r (b_{q} | R^{'}) = (2, 2, 3) .$ Case 3.5: All three vertices are exterior cycle. So, without loss of generality, consider the resolving set R′ = {e_p, e_q, e_s} where $1 \leq p, q, s \leq \frac{n + 1}{2},$ we have $r (b_{2 s + 2} | R^{'}) = r (d_{2 s + 3} | R^{'}) = (5, 5, 4) .$ Case 3.6: All three vertices are from outer cycle. So, without loss of generality, consider the resolving set R′ = {f_p, f_q, f_s} where $1 \leq p, q, s \leq \frac{n + 1}{2},$ we have $r (d_{2 p} | R^{'}) = r (d_{s + p} | R^{'}) = (2, 2, 3) .$ Case 3.7: One vertex from inner cycle, second from second inner cycle and third from middle vertices are selected. So, without loss of generality, consider the resolving set ^′R = {a_p, b_q, c_s} where $1 \leq p, s \leq \frac{n + 1}{2}$ and 1 ≤ q ≤ n, we have $r (e_{p} | R^{'}) = r (e_{s + 1} | R^{'}) = (4, 3, 2) .$ Case 3.8: One vertex from inner cycle, second from second inner cycle and third from third inner cycle are selected. So, without loss of generality, consider the resolving set R′ = {a_p, b_q, d_s} , where $2 \leq p \leq \frac{n + 1}{2}$ and 2 ≤ q, s ≤ n, we have $r (f_{p - 1} | R^{'}) = r (f_{p} | R^{'}) = (5, 4, 2),$ and for p, q, s = 1 $r (f_{1} | R^{'}) = r (f_{n} | R^{'}) = (5, 4, 2) .$ Case 3.9: One vertex from inner cycle, second from second inner cycle and third from exterior are selected. So, without loss of generality, consider the resolving set R′ = {a_p, b_q, e_s} , where $2 \leq p, s \leq \frac{n + 1}{2}$ and 2 ≤ q ≤ n, we have $r (f_{p - 1} | R^{'}) = r (f_{p} | R^{'}) = (5, 4, 1),$ and for p, q, s = 1 $r (f_{1} | R^{'}) = r (f_{n} | R^{'}) = (5, 4, 1) .$ Case 3.10: One vertex from inner cycle, second from second inner cycle and third from outer cycle are selected. So, without loss of generality, consider the resolving set R′ = {a_p, b_q, f_s} , where $2 \leq p, s \leq \frac{n + 1}{2}$ and 2 ≤ q ≤ n, then we have $r (e_{p} | R^{'}) = r (e_{s + 1} | R^{'}) = (4, 3, 1) .$ Case 3.11: One vertex from second inner cycle, second from middle vertices and third from third inner cycle are selected. So, without loss of generality, consider the resolving set R′ = {b_p, c_q, d_s} where 2 ≤ p, s ≤ n and $2 \leq q \leq \frac{n + 1}{2},$ we have $r (f_{q - 1} | R^{'}) = r (f_{q} | R^{'}) = (4, 3, 2),$ and for p, q, s = 1 $r (f_{1} | R^{'}) = r (f_{n} | R^{'}) = (4, 3, 2) .$ Case 3.12: One vertex from second inner cycle, second from middle vertices and third from exterior vertices are selected. So, without loss of generality, consider the resolving set R′ = {b_p, c_q, e_s} , where $2 \leq q, s \leq \frac{n + 1}{2}$ and 2 ≤ p ≤ n, we have $r (f_{q - 1} | R^{'}) = r (f_{q} | R^{'}) = (4, 3, 1),$ and for p, q, s = 1 $r (f_{1} | R^{'}) = r (f_{n} | R^{'}) = (4, 3, 1) .$ Case 3.13: One vertex from second inner cycle, second from second middle and third from outer cycle are selected. So, without loss of generality, consider the resolving set R′ = {b_p, c_q, f_s} , where $1 \leq q, s \leq \frac{n + 1}{2}$ and 2 ≤ p ≤ n, we have $r (e_{q} | R^{'}) = r (e_{q + 1} | R^{'}) = (4, 3, 1) .$ Case 3.14: One vertex from middle vertices, second from third inner cycle and third from exterior cycle are selected. So, without loss of generality, consider the resolving setR′ = {c_p, d_q, e_s} , where $1 \leq p, s \leq \frac{n + 1}{2}$ and 1 ≤ q ≤ n, then we have $r (a_{p} | R^{'}) = r (a_{p + 1} | R^{'}) = (2, 3, 4) .$ Case 3.15: One vertex from middle vertices, second from third inner cycle and third from outer cycle are selected. So, without loss of generality, consider the resolving set R′ = {c_p, d_q, f_s} where $1 \leq p, s \leq \frac{n + 1}{2},$ and 1 ≤ q ≤ n we have $r (a_{p} | R^{'}) = r (a_{s + 1} | R^{'}) = (2, 3, 5) .$ Case 3.16: One vertex from third inner cycle, second from second exterior cycle and third from outer cycle are selected. So, without loss of generality, consider the resolving set R′ = {d_p, e_q, f_s} , where $1 \leq q, s \leq \frac{n + 1}{2}$ and 1 ≤ p ≤ n, we have $r (a_{q} | R^{'}) = r (a_{q + 1} | R^{'}) = (3, 4, 5) .$

From above cases we can conclude that there is no resolving set with three vertices from $V (S_{n})$ implying that $\dim (S_{n}) \geq 4 .$ (4) Inequalities (3) and (4) implies that $\dim (S_{n}) = 4 .$

□

3 Conclusion and discussion

The metric dimension of line graph of convex polytope structure especially $S_{n}$ are discussed in this research work. The metric dimension of $S_{n}$ is depend on the even odd parity of n, but it is free from the order of graph and resulted in constant metric dimension in both cases of n. The final results are drawn in the following Table 1.

Table 1
Metric Dimension of Line graph of Convex Polytope $S_{n}$

G $\dim (S_{n})$

$S_{n}$ 3, (n is even);

4, (n is odd).

G	$\dim (S_{n})$
$S_{n}$	3, (n is even);
	4, (n is odd).

In summary, this research has explored the complex domain of metric-based resolution parameters, with a specific emphasis on the line graph of specific structures. Based on a rigorous examination, it has been ascertained that the metric dimension of the selected graph is 3 for even cycles and 4 for odd cycles. The aforementioned discovery not only provides significant contributions to the comprehension of metric dimension within the field of graph theory, but also emphasises the subtle differences that arise due to the structural attributes of the graph being analysed. The inclusion of these particular metric dimensions enhances the level of accuracy within the wider domain of graph theory, highlighting the need of simultaneously examining graph structures alongside metric-based resolution factors. In conclusion, the revealed metric dimensions function as reference points, encouraging additional investigation and enhancement within the domain of graph theoretic research.

3.1 Future direction

As we consider the trajectory for future research endeavours, a number of auspicious avenues arise from the discoveries made in this study pertaining to the metric-based resolving parameter of the line graph of specific structures. To begin with, conducting a thorough analysis to investigate the metric dimension of line graphs within various graph families would contribute to a more comprehensive comprehension of the potential differences and patterns that may emerge. The examination of various structural traits and their impact on the metric dimension has the potential to reveal valuable information.

Moreover, exploring the practical consequences of the established metric dimensions provides an opportunity to examine their real-world applications. Investigating potential applications of line graphs with a metric dimension of 3 or 4 could serve as a means to connect theoretical concepts with practical implementations. This exploration may have relevance in various domains, including network design, communication protocols, and optimisation challenges.

Furthermore, the examination of metric dimension fluctuations across various graph operations or transformations might offer a more comprehensive viewpoint. What is the impact on the metric dimension of a graph when edges are either added or removed, or when the graph undergoes specific transformations? Responding to these inquiries has the potential to shed light on the fluid characteristics of metric-driven resolution parameters.

Interdisciplinary investigations may be facilitated through collaborative endeavours involving academics from cognate disciplines, including computer science, optimisation, and applied mathematics. The exchange of knowledge and procedures between different academic fields has the potential to facilitate a more comprehensive comprehension of metric dimensions and their ramifications within a wider scientific framework.

In conclusion, the investigation of algorithmic elements pertaining to the computation of metric dimensions for the line graph of specific structures has the potential to provide valuable contributions towards the advancement of efficient computational tools. The development of algorithms capable of effectively processing larger or more intricate graphs has the potential to enhance the practical implementation of metric-based resolution parameters across many areas.

The study’s findings suggest that there are potential future possibilities for more investigation into metric dimensions. These directions include a call for academics to go into unexplored areas, collaborate across different disciplines, and facilitate the development of practical applications and algorithmic breakthroughs.

Footnotes

Acknowledgment

The authors extend their appreciation to Deanship of Scientific Research, Jazan University, for supporting this research work through the Research Unit Support Program Support Number: RUP2-02.

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On the metric-based resolving parameter of the line graph of certain structures

Abstract

Keywords

1 Introduction

1.1 Convex polytopes and their importance

1.2 Highlight and the motivation of the study

1.3 Auxiliary results

2 Metric dimension of line graph of convex polytope ( S n )

Table 1 Metric Dimension of Line graph of Convex Polytope S n G dim ( S n ) S n 3, (n is even); 4, (n is odd).

Footnotes

Acknowledgment

References

2 Metric dimension of line graph of convex polytope $(S_{n})$

Table 1
Metric Dimension of Line graph of Convex Polytope $S_{n}$

G $\dim (S_{n})$

$S_{n}$ 3, (n is even);

4, (n is odd).