Wiener polarity and Wiener index of double generalized Petersen graph

Tanveer Iqbal; Syed Ahtsham Ul Haq Bokhary; Ghulam Abbas; Jamel Baili; Hijaz Ahmad; Hafsah Tabassum; Saqib Murtaza; Zubair Ahmad; Xiao-Zhong Zhang

doi:10.1016/j.jksus.2023.102680

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Original article

07 2023

:35;

102680

doi:

10.1016/j.jksus.2023.102680

Wiener polarity and Wiener index of double generalized Petersen graph

Tanveer Iqbal^{^a}, Syed Ahtsham Ul Haq Bokhary^{^a}, Ghulam Abbas^{^a}, Jamel Baili^{^b,^c}, Hijaz Ahmad^{^d,^e}, Hafsah Tabassum^{^f}, Saqib Murtaza^{^f}, Zubair Ahmad^{^g}, Xiao-Zhong Zhang^{^h,⁎}

a Centre for Advanced Studies in Pure and Applied Mathematics, Bahauddin Zakariya University, Multan, Pakistan

b Department of Computer Engineering, College of Computer Science, King Khalid University, Abha 61413, Saudi Arabia

c Higher Institute of Applied Science and Technology of Sousse (ISSATS), Cité Taffala (Ibn Khaldoun) 4003 Sousse, University of Souse, Tunisia

d Near East University, Operational Research Center in Healthcare, Near East Boulevard, PC: 99138 Nicosia/Mersin 10, Turkey

e Section of Mathematics, International Telematic University Uninettuno, Corso Vittorio Emanuele II, 39, 00186 Roma, Italy

f Department of Mathematics, Faculty of Science, King Mongkut’s University of Technology Thonburi (KMUTT), 126 Pracha Uthit Rd., Bang Mod, Thung Khru, Bangkok 10140, Thailand

g Department of Mathematics and Physics, University of Campania “Luigi Vanvitelli”, Caserta 81100, Italy

h School of Mathematics and Information Science, Henan Polytechnic University, Jiaozuo 454000, China

⁎Corresponding author. zhangxiaozhong2000@163.com (Xiao-Zhong Zhang)

Received: 2022-5-27, Accepted: 2023-4-5,

Disclaimer:
This article was originally published by Elsevier and was migrated to Scientific Scholar after the change of Publisher.

Peer review under responsibility of King Saud University.

Abstract

A topological index is a numerical parameter of a graph which characterizes some of the topological properties of the graph. The concepts of Wiener polarity index and Wiener index were established in chemical graph theory by means of the distances. The double generalized Petersen graph denoted by $DP (n, k)$ is obtained by attaching the vertices of outer pendent vertices to inner pendent vertices lying at distance $k$ . The length of the outer and inner cycle is $n$ , thus the number of vertices are $4 n$ and the number of edges in the $DP (n, k)$ are $6 n$ . In this paper, the Wiener polarity index of $DP (n, k$ for $3 \leq n \leq 6$ and for $n \geq 6 k + 1$ is computed. Further, the Wiener index of $DP (n, k)$ , for $k = {1, 2}$ is determined.

Keywords

Double Generalized Petersen Graph

Wiener Polarity Index

Wiener Index

05C22

05C12

1 Introduction

Let $G$ be a simple, connected, undirected graph with a vertex set $V (G)$ and an edge set $E (G)$ . The distance between two vertices $v_{i}$ and $v_{j}$ , denoted by $d (v_{i}, v_{j})$ is the length of the shortest path between the vertices $v_{i}$ and $v_{j}$ in $G$ . The diameter $diam (G)$ of a connected graph $G$ is the length of any longest geodesic. The degree of a vertex $v_{i}$ in $G$ is the number of edges incident to $v_{i}$ and is denoted by $d_{i} = d e g (v_{i})$ (Adnan et al., 2021; Buckley and Harary, 1990).

In the modern age, network structures have great significance in the field of chemistry, information technology, communication, and physical structures. A topological index of graph G is a numerical quantity that describes the topology of the graph. It reflects the theoretical properties of chemical compounds when applied to the molecular structure of the chemical compounds. Several topological indices have been proposed so far by various researchers.The Wiener index was the first and most studied topological index (see for details in (Wiener, 1947)). Wiener demonstrates that the Wiener index number is strongly related to the boiling points of alkane molecules. In chemistry, it was the first molecular topological index used. Since then, many indices that relate topological indices to different physical properties have been introduced in (Bokhary et al., 2021; Bokhary and Adnan, 2021; Ul Haq Bokhary et al., 2021 ).

Generalized Petersen Graphs and Double Generalized Petersen graphs are extensively studied graph networks. Many graph properties like metric dimension, partition dimension, different kinds of graph labelings, etc of this family are already been explored. In (Liu et al., 2017), Hamiltonian cycles is studied.

The Wiener polarity index of $G$ is denoted by $W_{p} (G)$ and defined as $W_{p} (G) = | {{u, v} | d (u, v) = 3, u, v \in V (G)} | .$

The name “Wiener polarity index” is introduced by Harold Wiener in $1947$ (Ul Haq Bokhary et al., 2021).

The Wiener index of a graph $G$ is denoted by $W (G)$ and is defined as: $W (G) = \sum_{u, v \in V (G)} d (x, y) = \sum_{k \geq 1} k γ (G, k) .$

The popularity of these indices is due to numerous of their chemical applications and mathematical properties reported in (Bokhary et al., 2021; Bokhary and Adnan, 2021; Du et al., 2008; Graovac and Pisanski, 1991; Harary, n.d.; Ul Haq Bokhary et al., 2021 )

Consider a graph consisting of two cycles; one is called an outer and the other is an inner cycle. The vertex set of outer and inner cycles is denoted by $X$ and $Y$ , respectively. Let, each of the vertices of the cycles are attached to a pendant vertex. The pendant vertices attached to the outer cycle are called outer pendent vertices whereas the inner pendent vertices are the vertices are attached to the inner cycle. The vertex set of outer and inner pendent vertices is denoted by $U$ and $V$ , respectively. The double generalized Petersen graph denoted by $DP (n, k)$ is a graph obtained by attaching outer pendent vertices to inner pendent vertices lying at distance $k$ . The vertices of $DP (n, k)$ are defined as follows: $V (D P (n, k)) = {X \cup Y \cup U \cup V}$

$X = {x_{i}}$ , $Y = {y_{i}}$ , $U = {u_{i}}$ , $V = {V_{i}}$ for $1 \leq i \leq n .$

The edge set is defined as follows: $E (D P (n, k)) = {x_{i} x_{i + 1}, x_{i} u_{i}, u_{i} v_{i + k}, y_{i} y_{i + 1}, y_{i} u_{i} : 1 \leq i \leq n}$

The construction follows that the order of the graph $DP (n, k)$ is $6 n$ . The graph of $DP (6, 1)$ is depicted in Fig. 1. (See Fig 2 Fig 3)

The double generalized Petersen graph DP ( 6 , 1 )

Graphical representation of W ( q , 1 ) for different values of q

Graphical representation of W ( q , 2 ) for different values of q

In this paper, the Wiener polarity and the Wiener index of dthe graphs $DP (n, k)$ , where $n$ and $k$ be the nonnegative integers are investigated. Throughout this paper, the Wiener polarity index and the Wiener index of double generalized Petersen graph $DP (n, k)$ will be denoted by $W_{p} (q, k)$ and $W (q, k)$ , respectively. All the indices that follow henceforth are taken under modulo $n$ .

2 The Wiener polarity index of Double Generalized Petersen graphs

In this section, the $W_{p} (n, 1)$ of $DP (n, k))$ is computed for different values of $n$ and $k$ .

Theorem 2.1

For $n \geq 3$ , the Wiener polarity index $W_{p} (n, 1)$ is.

W_{p} (n, 1) = \{\begin{matrix} 15 & forn = 3 \\ 40 & forn = 4 \\ 80 & forn = 5 \\ 96 & forn = 6 \\ 18 n & forn \geq 7 \end{matrix}

Proof. For $3 \leq n \leq 6$ , it is easy to verify the results. For $n \geq 7 a n d 1 \leq l \leq n$ , let $a$ be an arbitrary vertex of $DP (n, k)$ . Then, the vertex $a$ can belong to either of the set $X, Y, U$ or $V$ :

Case 1

Let $a \in X$ , then $a$ be one of the outer vertex $x_{l}$ . The vertices $x_{l + 3}$ , $u_{l + 2}$ , $u_{l - 2}$ , $v_{l + 2}$ , $v_{l - 2}$ , $v_{l}$ , $y_{l + 1}$ , $y_{l - 1}$ are the vertices which have a distance of three from $x_{l}$ . There are $8 n$ vertices that have a distance of three from the vertices of the set $X$ .

Case 2

Let $a \in U$ . In this case, $a$ be one of the inner vertex $u_{l}$ . The vertices $x_{l + 2}$ , $x_{l - 2}$ , $u_{l + 1}$ , $v_{l + 3}$ , $v_{l - 3}$ , $y_{l + 2}$ , $y_{l - 2}$ , $y_{l}$ are the vertices at distance three from each $u_{l}$ . So, $6 n$ vertices from $8 n$ that are yet to be counted.

Case 3

Let $a \in V$ , then $a$ is one of the inner vertex $v_{l} .$ The vertices $x_{l + 2}$ , $x_{l - 2}$ , $x_{l}$ , $u_{l + 3}$ , $u_{l - 3}$ , $v_{l + 1}$ , $y_{l + 2}$ , $y_{l - 2}$ are the vertices at distance of three from each vertex of $V$ . So, 3 $n$ vertices from $8 n$ which are yet to be counted.

Case 4

Let $a \in Y$ , then $a$ be one of the inner vertex $y_{i} .$ The vertices $x_{l + 1}$ , $x_{l - 1}$ , $u_{l + 2}$ , $u_{l - 2}$ , $v_{l + 2}$ , $u_{l}$ , $v_{l - 2}$ , $y_{l + 3}$ are the vertices at distance of three. So, $n$ vertices from $8 n$ which are yet to be counted.The above cases imply that.

W_{p} (n, 1) = 8 n + 6 n + 3 n + n = 18 n .

Theorem 2.2

For $n \geq 6 k + 1$ , $W_{p} (n, k) = 24 n,$ where $k \in N$ .

Proof. For $1 \leq l \leq n$ , let $a$ be an arbitrary vertex of $DP (n, k)$ . Then, the vertex $a$ can belong to either of the set $X, Y, U$ or $V$ :

If $a \in X$ , then $a$ is one of the outer vertex $x_{l} .$ The vertices $x_{l + 3}$ , $u_{l + 2}$ , $u_{l - 2}$ , $u_{l + 2 k}$ , $u_{l - 2 k}$ , $v_{l + 1 + k}$ , $v_{l + 1 - k}$ , $v_{l + 1}$ , $v_{l - k - 1} y_{l + k}$ , and $y_{l - k}$ are the vertices at distance three from each $x_{l}$ . It is clear that $u_{l + 2 k}$ and $u_{l - 2 k}$ are the vertices that have the largest gap from the vertex $x_{l}$ . Thus, the maximum value of $n$ for which the vertices $u_{l + 2 k}$ and $u_{l - 2 k}$ can be equal is $n = 4 k + 1$ but $n \geq 6 k + 1$ . Therefore, $u_{l + 2 k}$ and $u_{l - 2 k}$ are distinct vertices and hence, all the vertices mentioned above are distinct. Since, $1 \leq i \leq n$ , therefore there are total $11 n$ vertices that have a distance of three from the set $X$ .

If the vertex $a$ belongs to set $U$ or $V$ . Without loss of generality, one can suppose that $a \in U$ , then $a$ is one of the inner vertex $u_{l}$ . The vertices $x_{l + 2}$ , $x_{l - 2}$ , $u_{l + 1}$ , $v_{l + 3 k}$ , $v_{l - 3 k}$ , $y_{l + 1 + k}$ , $y_{l + 1 - k}$ , $y_{l + 3}$ , $y_{l - 3}$ are the vertices at distance three. The vertices $v_{i + 3 k}$ and $v_{i - 3 k}$ have the largest gap from the vertex $u_{l}$ . Thus, the maximum value of $n$ for which the vertex $v_{l + 3 k}$ and $v_{l - 3 k}$ are equal is $n = 6 k$ but $n \geq 6 k + 1$ . Therefore, $v_{l + 3 k}$ and $v_{l - 3 k}$ are distinct vertices and hence all the vertices mentioned above are distinct. Since $4$ of these vertices belong to a set $X$ and are already counted, therefore there are total $7 n$ vertices that are yet to be counted and have a distance of three from the vertices of the set $U$ . Similarly, for each $a \in V$ , there are $11$ vertices with distance $3$ and out of them $4$ belong to set $U$ and $2$ belong to a set $X$ . Therefore there are $5 n$ vertices that are yet to be counted and have a distance of three from $V$ .

Finally, if $a \in Y$ then for any arbitrary vertex $y_{l}$ , the vertex $y_{l + 3}$ is the vertex with distance three and is not counted before. Thus, there are $n$ new vertices having a distance of three from a set $Y$ . Therefore, $W_{p} (n, k) = 11 n + 7 n + 5 n + n = 24 n .$

2.1

2.1 The Wiener index of Double Generalized Petersen Graphs $D P (n, k)$

In this section, the $W (q, k)$ for $k = 1, 2$ is computed, where $q$ and $k$ are positive integers. Let $d^{*}$ (x, y) and $d^{* *} (x, y)$ be the minimum and maximum distances between any pair of vertices $x$ and $y .$

Theorem 2.3

For $q \geq 3$ ,

W (q, 1) = \{\begin{matrix} 2 q^{3} + 8 q^{2} + 8 q & ifqiseven, \\ 2 q^{3} + 8 q^{2} + 4 q & ifqisodd, \end{matrix}

Proof. The equation for calculating the Wiener index is

(1)

W (G) = \sum_{p_{1}, p_{2} \in V (G)} d (p_{1}, p_{2})

The proof is divided into two cases.

Case 1

$q$ is even.

If $p_{1}, p_{2} \in X o r Y$ , then $d^{* *} (p_{1}, p_{2}) = \frac{q}{2} .$ There are $q$ and $\frac{q}{2}$ pairs of vertices when $1 \leq d (x, y) \leq \frac{q - 2}{2}$ and $d (p_{1}, p_{2}) = \frac{q}{2}$ respectively. Therefore,

(2)

\sum_{p_{1}, p_{2} \in X} d (p_{1}, p_{2}) = q (1 + 2 + 3 + \dots + \frac{q - 2}{2}) + \frac{q}{2} (\frac{q}{2}) = \frac{q^{3}}{8}

If $p_{1} \in X$ and $p_{2} \in U$ or $p \in Y$ and $y \in V$ , then $d^{* *} (p_{1}, p_{2}) = \frac{q + 2}{2} .$ There are $2 q$ and $q$ pairs of vertices when $2 \leq d (p_{1}, p_{2}) \leq \frac{q}{2}$ and $d (p_{1}, p_{2}) = 1 o r \frac{q + 2}{2}$ respectively. Thus,

(3)

\sum_{p_{1} \in X, p_{2} \in U} d (p_{1}, p_{2}) = 2 q (2 + \dots + \frac{q}{2}) + q (1 + \frac{q + 2}{2}) = \frac{q^{3} + 4 q^{2}}{4}

If $p_{1} \in X$ and $p_{2} \in V$ or $p_{1} \in Y$ and $p_{2} \in U$ . In this case, $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 2}{2}$ . There are $2 q$ and $q$ pairs of vertices when $2 \leq d (p_{1}, p_{2}) \leq \frac{q}{2}$ and $d (p_{1}, p_{2}) = 3 o r \frac{q + 2}{2}$ respectively. Thus,

(4)

\sum_{p_{1} \in X, p_{2} \in V} d (p_{1}, p_{2}) = 2 q (2 + \dots + \frac{q}{2}) + q (3 + \frac{q + 2}{2}) = \frac{q^{3} + 4 q^{2} + 8 q}{4}

If $p_{1} \in X$ and $p_{2} \in Y$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 3 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 4}{2} .$ There are $2 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 2}{2}$ and $q$ if $d (p_{1}, p_{2}) = \frac{q}{2}$ or $\frac{q + 4}{2}$ . Thus,

(5)

\sum_{p_{1} \in Y, p_{2} \in X} d (p_{1}, p_{2}) = 2 q (3 + 4 + \dots + \frac{q + 2}{2}) + q ((\frac{q}{2}) + (\frac{q + 4}{2})) = \frac{q^{3} + 8 q^{2} + 8 q}{4}

If $p_{1}, p_{2} \in U$ or $p_{1}, p_{2} \in V$ . In both these cases, $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 2}{2} .$ There are $q$ pairs of vertices when $2 \leq d (p_{1}, p_{2}) \leq \frac{q + 2}{2}$ and $\frac{q}{2}$ if $d (p_{1}, p_{2}) = \frac{q}{2}$ . Therefore,

(6)

\sum_{p_{1}, p_{2} \in U} d (p_{1}, p_{2}) = q (2 + 3 + \dots + \frac{q - 2}{2} + \frac{q + 2}{2}) + \frac{q}{2} (\frac{q}{2}) = \frac{q^{3} + 4 q^{2}}{8}

If $p_{1} \in U$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 4}{2} .$ It is important to note that there is no pair which have a distance of three. There are $2 q$ pairs of vertices when $d (p_{1}, p_{2}) = 1$ or $3 \leq d (p_{1}, p_{2}) \leq \frac{q}{2}$ and $q$ if the $d (p_{1}, p_{2}) = \frac{q}{2} o r \frac{q + 4}{2}$ . So,

(7)

\sum_{p_{1} \in V, p_{2} \in U} d (p_{1}, p_{2}) = 2 q (1 + 3 + \dots + \frac{q}{2}) + q (\frac{q}{2} + \frac{q + 4}{2}) = \frac{q^{3} + 4 q^{2} + 8 q}{4}

By adding Eqs. (2), (3), (4), (5), (6) and (7), we get $W (q, 1) = 2 q^{3} + 8 q^{2} + 8 q .$

Case 2

$q$ is odd.

If $p_{1}, p_{2} \in X o r Y$ . In both these cases, $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q - 1}{2} .$ There are $q$ pairs of vertices when $1 \leq d (p_{1}, p_{2}) \leq \frac{q - 1}{2}$ . Thus,

(8)

\sum_{p_{1}, p_{2} \in X} d (p_{1}, p_{2}) = q (1 + 2 + 3 + \dots + \frac{q - 1}{2}) = \frac{q^{3} - q}{8}

If $p_{1} \in X$ and $p_{2} \in Y$ or $p_{1} \in Y$ and $p_{2} \in V$ . In both these cases, $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 1}{2} .$ There are $q$ pairs of vertices when $d (p_{1}, p_{2}) = 1$ and $2 q$ if the $2 \leq d (p_{1}, p_{2}) \leq \frac{q + 1}{2}$ . Therefore,

(9)

\sum_{p_{1} \in X, p_{2} \in U} d (p_{1}, p_{2}) = 2 q (2 + 3 + \dots + \frac{q + 1}{2}) + 1 (q) = \frac{q^{3} + 4 q^{2} - q}{4}

If $p_{1} \in X$ and $p_{2} \in V$ or $p_{1} \in Y$ and $p_{2} \in U$ . Then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 1}{2} .$ There are $2 q$ pairs of vertices when $2 \leq d (p_{1}, p_{2}) \leq \frac{q + 1}{2}$ and $q$ if the $d (p_{1}, p_{2}) = 3$ . Thus,

(10)

\sum_{p_{1} \in X, p_{2} \in V} d (p_{1}, p_{2}) = = 2 q (2 + 3 + \dots + \frac{q + 1}{2}) + 3 q = \frac{q^{3} + 4 q^{2} + 7 q}{4}

If $p_{1} \in X$ and $p_{2} \in Y$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 3 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 3}{2} .$ There are $2 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 3}{2}$ and $q$ if the $d (p_{1}, p_{2}) = 4$ . Therefore,

(11)

\sum_{p_{1} \in X, p_{2} \in Y} d (p_{1}, p_{2}) = 2 q (3 + 4 + \dots + \frac{q + 3}{2}) + 4 q = \frac{q^{3} + 8 q^{2} + 7 q}{4}

If $p_{1}, p_{2} \in U o r V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 1}{2}$ . There are $q$ pairs of vertices when $2 \leq d (p_{1}, p_{2}) \leq \frac{q + 1}{2}$ . Thus,

(12)

\sum_{p_{1}, p_{2} \in U} d (p_{1}, p_{2}) = q (2 + 3 + . . . + \frac{q + 1}{2}) = \frac{q^{3} + 4 q^{2} - 5 q}{8}

If $p_{1} \in U$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 1}{2} .$ There are $2 q$ pairs of vertices when $1 o r 3 \leq d (p_{1}, p_{2}) \leq \frac{q + 1}{2}$ and $q$ if the $d (p_{1}, p_{2}) = \frac{q - 1}{2}$ . So,

(13)

\sum_{p_{1} \in U, p_{2} \in V} d (p_{1}, p_{2}) = 2 q (1 + 3 + 4 . . . + \frac{q + 1}{2}) + q (\frac{q - 1}{2}) = \frac{q^{3} + 4 q^{2} + 3 q}{4}

By adding Eqs. (8), (9), (10), (11), (12) and (13), we get $W (q, 1) = 2 q^{3} + 8 q^{2} + 4 q .$

Theorem 2.4

For $7 \leq q \leq 11$ ,

W ((q, 2)) = \{\begin{matrix} 2 q^{3} + 12 q^{2} - 30 q & ifqiseven, \\ 2 q^{3} + 11 q^{2} - 37 q & ifqisodd, \end{matrix}

Proof. The proof is divided into two cases.

Case 1

$q$ is even.

If $p_{1}, p_{2} \in Y$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q}{2}$ . There are $q$ pairs of vertices when $1 \leq d (p_{1}, p_{2}) \leq \frac{q - 2}{2}$ and $\frac{q}{2}$ if the $d (p_{1}, p_{2}) = \frac{q}{2}$ . Therefore,

(14)

\sum_{p_{1}, p_{2} \in Y} d (p_{1}, p_{2}) = q (1 + 2 + 3 + \dots + \frac{q - 2}{2}) + \frac{q}{2} (\frac{q}{2}) = \frac{q^{3}}{8}

If $p_{1} \in Y$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q}{2}$ . There are $2 q$ pairs of vertices when $2 \leq d (p_{1}, p_{2}) \leq \frac{q}{2}$ and $q$ if the $d (p_{1}, p_{2}) = 1 0 r \frac{q - 2}{2}$ . Thus,

(15)

\sum_{p_{1} \in Y, p_{2} \in V} d (p_{1}, p_{2}) = 2 q (2 + . . . + \frac{q}{2}) + q (1 + \frac{q - 2}{2}) = \frac{q^{3} + 4 q^{2} - 8 q}{4}

If $p_{1} \in Y$ and $p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q}{2} .$ There are $2 q$ pairs of vertices when $2 \leq d (p_{1}, p_{2}) \leq \frac{q}{2}$ and $2 q$ if the $d (p_{1}, p_{2}) = \frac{q - 2}{2}$ . Therefore,

(16)

\sum_{p_{1} \in y, p_{2} \in U} d (p_{1}, p_{2}) = 2 q (2 + 3 + . . . + \frac{q}{2}) + 2 q (\frac{q - 2}{2}) = \frac{q^{3} + 6 q^{2} - 16 q}{4}

If $p_{1} \in Y$ and $p_{2} \in X$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 3 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 2}{2} .$ There are $2 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 2}{2}$ and $2 q$ if the $d (p_{1}, p_{2}) = \frac{q}{2}$ . Therefore,

(17)

\sum_{p_{1} \in Y, p_{2} \in X} d (p_{1}, p_{2}) = 2 q (3 + 4 + . . . + \frac{q + 2}{2}) + 2 q (\frac{q}{2}) = \frac{q^{3} + 10 q^{2} - 16 q}{4}

If $p_{1}, p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 3 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 2}{2} .$ There are $q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 2}{2}$ and $\frac{q}{2}$ if the $d (p_{1}, p_{2}) = \frac{q - 4}{2}$ . Thus,

(18)

\sum_{p_{1}, p_{2} \in V} d (p_{1}, p_{2}) = q (3 + 4 + . . . + \frac{q + 2}{2}) + \frac{q}{2} (\frac{q - 4}{2}) = \frac{q^{3} + 8 q^{2} - 24 q}{8}

If $p_{1} \in V$ and $p_{2} \in U$ , then then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 2}{2} .$ There are $2 q$ pairs of vertices when $1 \leq d (p_{1}, p_{2}) \leq \frac{q + 2}{2}$ and $2 q$ if the $d (p_{1}, p_{2}) = \frac{q}{2}$ . Thus,

(19)

\sum_{p_{1} \in V, p_{2} \in U} d (p_{1}, p_{2}) = 2 q (1 + . . . + \frac{q + 2}{2}) + 2 q (\frac{q}{2}) = \frac{q^{3} + 10 q^{2} - 32 q}{4}

By adding Eqs. (14), (15), (16), (17), (18) and (19), we get $W (D P (q, 2)) = 2 q^{3} + 12 q^{2} - 30 q .$

Case 2

$q$ is odd.

If $p_{1}, p_{2} \in Y$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q - 1}{2} .$ There are $q$ pairs of vertices when $1 \leq d (p_{1}, p_{2}) \leq \frac{q - 1}{2}$ . Therefore,

(20)

\sum_{p_{1}, p_{2} \in Y} d (p_{1}, p_{2}) = q (1 + 2 + 3 + \dots + \frac{q - 1}{2}) = \frac{q^{3} - q}{8}

If $p_{1} \in Y$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q - 1}{2} .$ There are $2 q$ pairs of vertices when $2 \leq d (p_{1}, p_{2}) \leq \frac{q - 1}{2}$ and $q$ if the $d (p_{1}, p_{2}) = 1$ and $2 q$ if the $\frac{q - 3}{2}$ . Therefore,

(21)

\sum_{p_{1} \in Y, p_{2} \in V} d (p_{1}, p_{2}) = q (1) + 2 q (2 + . . . + \frac{q - 1}{2}) + 2 q (\frac{q - 3}{2}) = \frac{q^{3} + 4 q^{2} - 17 q}{4}

If $p_{1} \in Y$ and $p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q - 1}{2} .$ There are $2 q$ pairs of vertices when $2 \leq d (p_{1}, p_{2}) \leq \frac{p_{1} - 1}{2}$ and $q$ if the $d (p_{1}, p_{2}) = \frac{q - 1}{2}$ and $2 q$ if the $\frac{q - 3}{2}$ . Thus,

(22)

\sum_{p_{1} \in Y, p_{2} \in U} d (p_{1}, p_{2}) = 2 q (2 + 3 + . . . + \frac{q - 1}{2}) + q (\frac{q - 1}{2}) + 2 q (\frac{q - 3}{2}) = \frac{q^{3} + 6 q^{2} - 23 q}{4}

If $p_{1} \in Y$ and $p_{2} \in X$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 3 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 1}{2} .$ There are $2 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 1}{2}$ and $q$ if the $d (p_{1}, p_{2}) = \frac{q + 1}{2}$ and $2 q$ if the $\frac{q - 1}{2}$ . Therefore,

(23)

\sum_{p_{1} \in Y, p_{2} \in X} d (p_{1}, p_{2}) = 2 q (3 + 4 + . . . + \frac{q + 1}{2}) + 2 q (\frac{q - 1}{2}) + q (\frac{q + 1}{2}) = \frac{q^{3} + 10 q^{2} - 23 q}{4}

If $p_{1}, p_{2} \in V$ , then then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 1}{2} .$ There are $q$ pairs of vertices when $2 \leq d (p_{1}, p_{2}) \leq \frac{q + 1}{2}$ . Therefore,

(24)

\sum_{p_{1}, p_{2} \in V} d (p_{1}, p_{2}) = q (2 + 3 + . . . + \frac{q + 1}{2}) = \frac{q^{3} + 4 q^{2} - 5 q}{8}

If $p_{1} \in V$ and $p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 1}{2}$ . There are $2 q$ pairs of vertices when $1 o r 4 \leq d (p_{1}, p_{2}) \leq \frac{q + 1}{2}$ and $q$ if the $d (p_{1}, p_{2}) = \frac{q + 1}{2}$ and $2 q$ if the $d (p_{1}, p_{2}) = \frac{q - 1}{2}$ . Thus,

(25)

\sum_{p_{1} \in V, p_{2} \in U} d (p_{1}, p_{2}) = q (1 + 4 . . . + \frac{q + 1}{2}) + 2 q (\frac{q - 1}{2}) + q (\frac{q + 1}{2}) = \frac{q^{3} + 10 q^{2} - 39 q}{4}

By adding Eqs. (20), (21), (22), (23), (24) and (25), we get

q $W (D P (q, 2)) = 2 q^{3} + 11 q^{2} - 37 q .$

Theorem 2.5

For $q \geq 12$ ,

W ((q, 2)) = \{\begin{matrix} q^{3} + 17 q^{2} - 6 q & ifq \equiv 0 (m o d 8), \\ q^{3} + 17 q^{2} - 12 q & ifq \equiv 1 (m o d 8), \\ q^{3} + 17 q^{2} - 10 q & ifq \equiv 2, 5, 6 (m o d 8), \\ q^{3} + 17 q^{2} - 18 q & ifq \equiv 3 (m o d 8), \\ \frac{4 q^{3} + 69 q^{2} - 44 q}{4} & ifq \equiv 4 (m o d 8), \\ q^{3} + 17 q^{2} - 17 q & ifq \equiv 7 (m o d 8), \end{matrix}

Proof. The proof is divided into two cases.

Case 1

$q \equiv 0 (m o d 8)$

If $p_{1}, p_{2} \in X$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = ⌊ \frac{q}{4} ⌋ + 3 .$ There are $q$ pairs of vertices when $1 \leq d (p_{1}, p_{2}) \leq 5$ and $\frac{q + 12}{4}$ and $2 q$ pairs of vertices if the $\frac{q}{4} \leq d (p_{1}, p_{2}) \leq \frac{q + 4}{4}$ and also $\frac{3 q}{2}$ pairs of vertices when $d (p_{1}, p_{2}) = \frac{q + 8}{4}$ . Thus,

\begin{matrix} \sum_{p_{1}, p_{2} \in X} d (p_{1}, p_{2}) = q (1 + 2 + 3 + \dots + 5) + q (\frac{q + 8}{4} + \frac{q + 12}{4}) + \frac{q}{2} (\frac{q + 8}{4}) \\ + 2 q (6 + 7 + \dots + \frac{q + 4}{4}) \end{matrix}

(26)

= \frac{q^{3} + 22 q^{2} - 112 q}{16}

If $p_{1} \in X$ and $p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 8}{4} .$ There are $4 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q}{4}$ and $3 q$ if the $d (p_{1}, p_{2}) = \frac{q + 4}{4}$ and $2 q$ if the $d (p_{1}, p_{2}) = 2 0 r \frac{q + 8}{4}$ and also $q$ if the $d (p_{1}, p_{2}) = 1$ . Therefore,

(27)

\sum_{p_{1} \in X, p_{2} \in U} d (p_{1}, p_{2}) = q (1 + \frac{q + 4}{4}) + 2 q (2 + \frac{q + 4}{4} + \frac{q + 8}{4}) + 4 q (3 + 4 + \dots + \frac{q}{4}) = \frac{q^{3} + 14 q^{2}}{8}

If $p_{1} \in X$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 8}{4} .$ There are $4 q$ pairs of vertices when $3 o r 5 \leq d (p_{1}, p_{2}) \leq \frac{q + 4}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 4$ and $2 q$ if the $d (p_{1}, p_{2}) = 2$ and $q$ if the $d (p_{1}, p_{2}) = \frac{q + 8}{4}$ . Thus,

(28)

\sum_{p_{1} \in X, p_{2} \in V} d (p_{1}, p_{2}) = q (4 + \frac{q + 8}{4}) + 2 q (2) + 4 q (3 + 4 + \dots + \frac{q + 4}{4}) = \frac{q^{3} + 14 q^{2} + 16 q}{8}

If $p_{1} \in X$ and $p_{2} \in Y$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 3 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 12}{4} .$ There are $4 q$ pairs of vertices when $4 o r 6 \leq d (p_{1}, p_{2}) \leq \frac{q + 8}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $2 q$ if the $d (p_{1}, p_{2}) = 2$ and $q$ if the $d (p_{1}, p_{2}) = \frac{q + 12}{4}$ . Therefore,

(29)

\sum_{p_{1} \in X, p_{2} \in Y} d (p_{1}, p_{2}) = q (5 + \frac{q + 12}{4}) + 2 q (3) + 4 q (4 + 5 + \dots + \frac{q + 8}{4}) = \frac{q^{3} + 22 q^{2} + 16 q}{8}

If $p_{1}, p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 12}{4} .$ There are $2 q$ pairs of vertices when $4 \leq d (p_{1}, p_{2}) \leq 5 o r \frac{q + 4}{4}$ and $q$ if the $d (p_{1}, p_{2}) = 2, 3, \frac{q + 8}{4}, \frac{q + 12}{4}$ and $\frac{3 q}{2}$ if the $d (p_{1}, p_{2}) = \frac{q}{4}$ . So,

(30)

\sum_{p_{1}, p_{2} \in U} d (p_{1}, p_{2}) = q (2 + 3 + \frac{q}{4} + \frac{q + 8}{4} + \frac{q + 12}{4}) + \frac{q}{2} (\frac{q}{4}) + 2 q (\frac{q + 4}{4}) + 2 q (4 + 5 + \dots + \frac{q - 4}{4}) = \frac{q^{3} + 18 q^{2}}{16}

If $p_{1} \in U$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 12}{4} .$ There are $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $4 q$ if the $d (p_{1}, p_{2}) = 4 o r \frac{q}{4} o r \frac{q + 8}{4}$ and $2 q$ if the $d (p_{1}, p_{2}) = 1 o r 3 o r \frac{q + 4}{4}$ and $q$ if the $d (p_{1}, p_{2}) = \frac{q + 12}{4}$ . Thus,

(31)

\sum_{p_{1} \in U, p_{2} \in V} d (p_{1}, p_{2}) = q (5 + \frac{q + 12}{4}) + 2 q (1 + 3 + \frac{q + 4}{4}) + 4 q (\frac{q + 8}{4}) + 4 q (4 + 5 + \dots + \frac{q}{4}) \frac{q^{3} + 18 q^{2} + 16 q}{8}

By adding Eqs. (26), (27), (28), (29), (30) and (31), we get $W (D P (q, 2)) = q^{3} + 17 q^{2} - 6 q .$

Case 2

$q \equiv 1 (m o d 8)$

If $p_{1}, p_{2} \in X$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 11}{4} .$ There are $q$ pairs of vertices when $1 \leq d (p_{1}, p_{2}) \leq 5 o r \frac{q + 11}{4}$ and $2 q$ if the $6 \leq d (p_{1}, p_{2}) \leq \frac{q + 7}{4}$ . Thus,

\sum_{p_{1}, p_{2} \in X} d (p_{1}, p_{2}) = q (1 + 2 + 3 + \dots + 5) + q (\frac{q + 11}{4}) + 2 q (6 + \dots + \frac{q + 7}{4})

(32)

= \frac{q^{3} + 22 q^{2} - 119 q}{16}

If $p_{1} \in X$ and $p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 7}{4} .$ There are $4 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 3}{4}$ and $2 q$ if the $d (p_{1}, p_{2}) = 2 o r \frac{q + 7}{4}$ and also $q$ if the $d (p_{1}, p_{2}) = 1$ . Therefore,

(33)

\sum_{p_{1} \in X, p_{2} \in U} d (p_{1}, p_{2}) = q (1) + 2 q (2 + \frac{q + 7}{4}) + 4 q (3 + 4 + \dots + \frac{q + 3}{4}) = \frac{q^{3} + 14 q^{2} - 7 q}{8}

If $p_{1} \in X$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 7}{4} .$ There are $4 q$ pairs of vertices when $3 o r 5 \leq d (p_{1}, p_{2}) \leq \frac{q + 3}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 4$ and $2 q$ if the $d (p_{1}, p_{2}) = 2 o r \frac{q + 7}{4}$ . Thus,

(34)

\sum_{p_{1} \in X, p_{2} \in V} d (p_{1}, p_{2}) = 2 q (2 + \frac{q + 7}{4}) + q (4) + 4 q (3 + 4 + \dots + \frac{q + 3}{4}) = \frac{q^{3} + 14 q^{2} + 17 q}{8}

If $p_{1} \in X$ and $p_{2} \in Y$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 3 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 11}{4} .$ There are $4 q$ pairs of vertices when $4 o r 6 \leq d (p_{1}, p_{2}) \leq \frac{q + 7}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $2 q$ if the $d (p_{1}, p_{2}) = 3 o r \frac{q + 11}{4}$ . Thus,

(35)

\sum_{p_{1} \in X, p_{2} \in Y} d (p_{1}, p_{2}) = q (5) + 2 q (3 + \frac{q + 11}{4}) + 4 q (4 + 5 + \dots + \frac{q + 7}{4}) = \frac{q^{3} + 22 q^{2} + 17 q}{8}

If $p_{1}, p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 11}{4} .$ There are $2 q$ pairs of vertices when $4 \leq d (p_{1}, p_{2}) \frac{q + 3}{4}$ and $q$ if the $d (p_{1}, p_{2}) = 2, 3, \frac{q + 7}{4}, \frac{q + 11}{4}$ . Therefore,

(36)

\sum_{p_{1}, p_{2} \in U} d (p_{1}, p_{2}) = q (2 + 3 + \frac{q + 7}{4} + \frac{q + 11}{4}) + 2 q (4 + 5 + \dots + \frac{q + 3}{4}) = \frac{q^{3} + 18 q^{2} - 19 q}{16}

If $p_{1} \in U$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 11}{4} .$ There are $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $4 q$ if the $4 \leq d (p_{1}, p_{2}) \leq \frac{q + 3}{4}$ and $2 q$ if the $d (p_{1}, p_{2}) = 1 o r 3 o r \frac{q + 7}{4} o r \frac{q + 11}{4}$ . Thus,

(37)

\sum_{p_{1} \in U, p_{2} \in V} d (p_{1}, p_{2}) = 2 q (1 + 3 + \frac{q + 7}{4} + \frac{q + 11}{4}) + q (5) + 4 q (4 + \dots + \frac{q + 3}{4}) = \frac{q^{3} + 18 q^{2} + 5 q}{8}

By adding Eqs. (32), (33), (34), (35), (36) and (37), we get $W (D P (q, 2)) = q^{3} + 17 q^{2} - 12 q .$

Case 3

$q \equiv 2 (m o d 8)$

If $p_{1}, p_{2} \in X$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 10}{4} .$ There are $q$ pairs of vertices when $1 \leq d (p_{1}, p_{2}) \leq 5$ and $2 q$ if the $6 \leq d (p_{1}, p_{2}) \leq \frac{q + 6}{4}$ and $\frac{3 q}{2}$ if the $d (p_{1}, p_{2}) = \frac{q + 10}{4}$ . Thus,

\sum_{p_{1}, p_{2} \in X} d (p_{1}, p_{2}) = q (1 + 2 + 3 + \dots + 5) + 2 q (6 + \dots + \frac{q + 6}{4}) + \frac{3 q}{2} (\frac{q + 10}{4})

(38)

= \frac{q^{3} + 22 q^{2} - 120 q}{16}

If $p_{1} \in X$ and $p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) =$ $\frac{q + 6}{4} .$ There are $4 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 2}{4}$ and $3 q$ if the $d (p_{1}, p_{2}) = \frac{q + 6}{4}$ and $2 q$ if the $d (p_{1}, p_{2}) = 2$ and also $q$ if the $d (p_{1}, p_{2}) = 1$ . Thus,

(39)

\sum_{p_{1} \in X, p_{2} \in U} d (p_{1}, p_{2}) = q (1 + \frac{q + 6}{4}) + 2 q (2 + \frac{q + 6}{4}) + 4 q (3 + 4 + \dots + \frac{q + 2}{4}) = \frac{q^{3} + 14 q^{2} - 8 q}{8}

If $p_{1} \in X$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 10}{4} .$ There are $4 q$ pairs of vertices when $3 o r 5 \leq d (p_{1}, p_{2}) \leq \frac{q + 2}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 4$ and $2 q$ if the $d (p_{1}, p_{2}) = 2 o r \frac{q + 6}{4}$ and $n$ if the $d (p_{1}, p_{2}) = \frac{q + 10}{4}$ . Therefore,

(40)

\sum_{p_{1} \in X, p_{2} \in V} d (p_{1}, p_{2}) = q (4 + \frac{q + 10}{4}) + 2 q (2 + \frac{q + 6}{4}) + 4 q (3 + 4 + \dots + \frac{q + 2}{4}) = \frac{q^{3} + 14 q^{2} + 24 q}{8}

If $p_{1} \in X$ and $p_{2} \in Y$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 3 and $d^{* *} (p_{1}, p_{2}) =$ $\frac{q + 14}{4} .$ There are $4 q$ pairs of vertices when $4 o r 6 \leq d (p_{1}, p_{2}) \leq \frac{q + 6}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $2 q$ if the $d (p_{1}, p_{2}) = 3 o r \frac{q + 10}{4}$ and also $q$ if the $d (p_{1}, p_{2}) = \frac{q + 14}{4}$ . So,

(41)

\sum_{p_{1} \in X, p_{2} \in Y} d (p_{1}, p_{2}) = q (5 + \frac{q + 14}{4}) + 2 q (3 + \frac{q + 10}{4}) + 4 q (4 + 5 + \dots + \frac{q + 6}{4}) = \frac{q^{3} + 22 q^{2} + 24 q}{8}

If $p_{1}, p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 10}{4} .$ There are $2 q$ pairs of vertices when $4 \leq d (p_{1}, p_{2}) \leq \frac{q + 2}{4}$ and $q$ if the $d (p_{1}, p_{2}) = 2, 3, \frac{q + 6}{4}$ and also $\frac{3 q}{2}$ if the $d (p_{1}, p_{2}) = \frac{q + 10}{4}$ . Thus,

(42)

\sum_{p_{1}, p_{2} \in U} d (p_{1}, p_{2}) = q (2 + 3 + \frac{q + 6}{4} + \frac{q + 10}{4}) + 2 q (4 + 5 + \dots + \frac{q + 2}{4}) + \frac{q}{2} (\frac{q + 10}{4}) = \frac{q^{3} + 18 q^{2} - 16 q}{16}

If $p_{1} \in U$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 14}{4} .$ There are $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $4 q$ if the $4 \leq d (p_{1}, p_{2}) \leq \frac{q + 6}{4}$ and $2 q$ if the $d (p_{1}, p_{2}) = 1 o r 3$ and also $q$ if the $d (p_{1}, p_{2}) = \frac{q + 14}{4}$ . Therefore,

(43)

\sum_{p_{1} \in U, p_{2} \in V} d (p_{1}, p_{2}) = 2 q (1 + 3) + q (5 + \frac{q + 14}{4}) + 4 q (4 + \dots + \frac{q + 6}{4}) = \frac{q^{3} + 18 q^{2}}{8}

By adding Eqs. (38), (39), (40), (41), (42) and (43), we get $W (D P (q, 2)) = q^{3} + 17 q^{2} - 10 q .$

Case 4

$q \equiv 3 (m o d 8)$

If $p_{1}, p_{2} \in X$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 9}{4} .$ There are $q$ pairs of vertices when $1 \leq d (p_{1}, p_{2}) \leq 5$ and $2 q$ if the $6 \leq d (p_{1}, p_{2}) \leq \frac{q + 9}{4}$ . So,

(44)

\sum_{p_{1}, p_{2} \in X} d (p_{1}, p_{2}) = q (1 + 2 + 3 + \dots + 5) + 2 q (6 + \dots + \frac{q + 9}{4})) = \frac{q^{3} + 22 q^{2} - 123 q}{16}

If $p_{1} \in X$ and $p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 5}{4} .$ There are $4 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 5}{4}$ and $2 q$ if the $d (p_{1}, p_{2}) = 2$ and also $q$ if the $d (p_{1}, p_{2}) = 1$ . Therefore,

(45)

\sum_{p_{1} \in X, p_{2} \in U} d (p_{1}, p_{2}) = q (1) + 2 q (2) + 4 q (3 + 4 + \dots + \frac{q + 5}{4}) = \frac{q^{3} + 14 q^{2} - 11 q}{8}

If $p_{1} \in X$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 5}{4} .$ There are $4 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 5}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 4$ and $2 q$ if the $d (p_{1}, p_{2}) = 2$ . Therefore,

(46)

\sum_{p_{1} \in X, p_{2} \in V} d (p_{1}, p_{2}) = 2 q (2) + 4 (q) + 4 q (3 + 4 + \dots + \frac{q + 5}{4}) = \frac{q^{3} + 14 q^{2} + 13 q}{8}

If $p_{1} \in X$ and $p_{2} \in Y$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 3 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 9}{4} .$ There are $4 q$ pairs of vertices when $4 o r 6 \leq d (p_{1}, p_{2}) \leq \frac{q + 9}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $2 q$ if the $d (p_{1}, p_{2}) = 3$ . Therefore,

(47)

\sum_{p_{1} \in X, p_{2} \in Y} d (p_{1}, p_{2}) = q (5) + 2 q (3) + 4 q (4 + 5 + \dots + \frac{q + 9}{4}) = \frac{q^{3} + 22 q^{2} + 13 q}{8}

If $p_{1}, p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 9}{4} .$ There are $2 q$ pairs of vertices when $4 \leq d (p_{1}, p_{2}) \leq \frac{q + 5}{4}$ and $q$ if the $d (p_{1}, p_{2}) = 2, 3, \frac{q + 9}{4}$ . Therefore,

(48)

\sum_{p_{1}, p_{2} \in U} d (p_{1}, p_{2}) = q (2 + 3 + \frac{q + 9}{4}) + 2 q (4 + 5 + \dots + \frac{q + 5}{4}) = \frac{q^{3} + 18 q^{2} - 31 q}{16}

If $p_{1} \in U$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 9}{4} .$ There are $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $4 q$ if the $4 \leq d (p_{1}, p_{2}) \leq \frac{q + 5}{4}$ and $2 q$ if the $d (p_{1}, p_{2}) = 1 o r 3 o r \frac{q + 9}{4}$ . So, we have

(49)

\sum_{p_{1} \in U, p_{2} \in V} d (p_{1}, p_{2}) = 2 q (1 + 3 + \frac{q + 9}{4}) + q (5) + 4 q (4 + \dots + \frac{q + 5}{4}) = \frac{q^{3} + 18 q^{2} - 7 q}{8}

By adding Eqs. (44), (45), (46), (47), (48) and (49) we get $W (D P (q, 2)) = q^{3} + 17 q^{2} - 18 q .$

Case 5

$q \equiv 4 (m o d 8)$

If $p_{1}, p_{2} \in X$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) =$ $\frac{q + 12}{4} .$ There are $q$ pairs of vertices when $1 \leq d (p_{1}, p_{2}) \leq 5$ and $2 q$ if the $6 \leq d (p_{1}, p_{2}) \leq \frac{q + 8}{4}$ and $\frac{q}{2}$ if the $d (p_{1}, p_{2}) = \frac{q + 12}{8}$ . Thus,

\sum_{p_{1}, p_{2} \in X} d (p_{1}, p_{2}) = q (1 + 2 + 3 + \dots + 5) + 2 q (6 + \dots + \frac{q + 8}{4}) + \frac{q}{2} (\frac{q + 12}{4})

(50)

= \frac{q^{3} + 22 q^{2} - 120 q}{16}

If $p_{1} \in X$ and $p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) =$ $\frac{q + 8}{4} .$ There are $4 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 4}{4}$ and $2 q$ if the $d (p_{1}, p_{2}) = 2$ and also $q$ if the $d (p_{1}, p_{2}) = 1 o r \frac{q + 8}{4}$ . Therefore,

(51)

\sum_{p_{1} \in X, p_{2} \in U} d (p_{1}, p_{2}) = q (1 + \frac{q + 8}{4}) + 2 q (2) + 4 q (3 + 4 + \dots + \frac{q + 4}{4}) = \frac{q^{3} + 14 q^{2} - 8 q}{8}

If $p_{1} \in X$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 8}{4} .$ There are $4 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 4$ and $2 q$ if the $d (p_{1}, p_{2}) = 2 o r \frac{q + 8}{4}$ and $3 q$ if the $d (p_{1}, p_{2}) = \frac{q + 4}{4}$ . Therefore,

(52)

\sum_{p_{1} \in X, p_{2} \in V} d (p_{1}, p_{2}) = 2 q (2 + \frac{q + 4}{4} + \frac{q + 8}{4}) + q (4 + \frac{q + 4}{4}) + 4 q (3 + 4 + \dots + \frac{q}{4}) = \frac{q^{3} + 14 q^{2} + 24 q}{8}

If $p_{1} \in X$ and $p_{2} \in Y$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 3 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 12}{4} .$ There are $4 q$ pairs of vertices when $4 \leq d (p_{1}, p_{2}) \leq \frac{q + 4}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $2 q$ if the $d (p_{1}, p_{2}) = 3 o r \frac{q + 12}{4}$ and also $3 q$ if the $d (p_{1}, p_{2}) = \frac{q + 8}{4}$ . So,

(53)

\sum_{p_{1} \in X, p_{2} \in Y} d (p_{1}, p_{2}) = q (5 + \frac{q + 8}{4}) + 2 q (3 + \frac{q + 8}{4} + \frac{q + 12}{4}) + 4 q (4 + 5 + \dots + \frac{q + 4}{4}) = \frac{+ 22 q^{2} + 24 q}{8}

If $p_{1}, p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 12}{4} .$ There are $2 q$ pairs of vertices when $4 \leq d (p_{1}, p_{2}) \frac{q}{4} o r \frac{q + 8}{4}$ and $q$ if the $d (p_{1}, p_{2}) = 2, 3, \frac{q + 4}{4}$ and $\frac{q}{2}$ if the $d (p_{1}, p_{2}) = \frac{q + 12}{4}$ . Therefore,

(54)

\sum_{p_{1}, p_{2} \in U} d (p_{1}, p_{2}) = q (2 + 3 + \frac{q + 4}{4}) + 2 q (4 + 5 + \dots + \frac{q}{4}) + 2 q (\frac{q + 8}{4}) + \frac{q}{2} (\frac{q + 12}{4}) = \frac{q^{3} + 18 q^{2} - 8 q}{16}

If $p_{1} \in U$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 12}{4} .$ There are $4 q$ if the $4 \leq d (p_{1}, p_{2}) \leq \frac{q + 4}{4}$ and $2 q$ if the $d (p_{1}, p_{2}) = 1 o r 3 o r \frac{q + 8}{4} o r \frac{q + 12}{4}$ . So,

(55)

\sum_{p_{1} \in U, p_{2} \in V} d (p_{1}, p_{2}) = 2 q (1 + 3 + \frac{q + 8}{4} + \frac{q + 12}{4}) + 4 q (4 + \dots + \frac{q + 4}{4}) = \frac{q^{3} + 20 q^{2} - 16 q}{8}

By adding Eqs. (50), (51), (52), (53), (54) and (55), we get $W (D P (q, 2)) = \frac{4 q^{3} + 69 q^{2} - 44 q}{4} .$

Case 6

$q \equiv 5 (m o d 8)$

If $p_{1}, p_{2} \in X$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 11}{4} .$ There are $q$ pairs of vertices when $1 \leq d (p_{1}, p_{2}) \leq 5 0 r \frac{q + 11}{4}$ and $2 q$ if the $6 \leq d (p_{1}, p_{2}) \leq \frac{q + 7}{4}$ . Therefore,

(56)

\sum_{p_{1}, p_{2} \in X} d (p_{1}, p_{2}) = q (1 + 2 + 3 + \dots + 5) + q (\frac{q + 11}{4}) + 2 q (6 + 7 + \dots + \frac{q + 7}{4}) = \frac{q^{3} + 22 q^{2} - 119 q}{16}

If $p_{1} \in X$ and $p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) =$ $\frac{q + 7}{4} .$ There are $4 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 3}{4}$ and $2 q$ if the $d (p_{1}, p_{2}) = 2 0 r \frac{q + 7}{2}$ and also $q$ if the $d (p_{1}, p_{2}) = 1$ . Therefore,

(57)

\sum_{p_{1} \in X, p_{2} \in U} d (p_{1}, p_{2}) = q (1) + 2 q (2 + \frac{q + 7}{4}) + 4 q (3 + 4 + \dots + \frac{q + 3}{4}) = \frac{q^{3} + 14 q^{2} - 7 q}{8}

If $p_{1} \in X$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 7}{4} .$ There are $4 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 3}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 4$ and $2 q$ if the $d (p_{1}, p_{2}) = 2 o r \frac{q + 7}{4}$ . Thus,

(58)

\sum_{p_{1} \in X, p_{2} \in V} d (p_{1}, p_{2}) = q (4) + 2 q (2 + \frac{q + 7}{4}) + 4 q (3 + 4 + \dots + \frac{q + 3}{4}) = \frac{q^{3} + 14 q^{2} + 17 q}{8}

If $p_{1} \in X$ and $p_{2} \in Y$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 3 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 11}{4}$ . There are $4 q$ pairs of vertices when $4 \leq d (p_{1}, p_{2}) \leq \frac{q + 7}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $2 q$ if the $d (p_{1}, p_{2}) = 3 o r \frac{q + 11}{4}$ . Therefore,

(59)

\sum_{p_{1} \in X, p_{2} \in Y} d (p_{1}, p_{2}) = q (5) + 2 q (3 + \frac{q + 11}{4}) + 4 q (4 + 5 + \dots + \frac{q + 7}{4}) = \frac{q^{3} + 22 q^{2} + 17 q}{8}

If $p_{1}, p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 11}{4}$ . There are $2 q$ pairs of vertices when $4 \leq d (p_{1}, p_{2}) \leq \frac{q - 1}{4} o r \frac{q + 7}{4}$ and $q$ if the $d (p_{1}, p_{2}) = 2, 3, \frac{q + 3}{4}, \frac{q + 11}{4}$ . Therefore,

(60)

\sum_{p_{1}, p_{2} \in U} d (p_{1}, p_{2}) = q (2 + 3 + \frac{q + 3}{4} + \frac{q + 11}{4}) + 2 q (4 + 5 + \dots + \frac{q - 1}{4}) + 2 q (\frac{q + 7}{4}) = \frac{q^{3} + 18 q^{2} - 3 q}{16}

If $p_{1} \in U$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 11}{4} .$ There are $4 q$ pairs of vertices if the $4 \leq d (p_{1}, p_{2}) \leq \frac{q + 3}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $2 q$ if the $d (p_{1}, p_{2}) = 1 o r 3 o r \frac{q + 7}{4} o r \frac{q + 11}{4}$ . Therefore,

(61)

\sum_{p_{1} \in U, p_{2} \in V} d (p_{1}, p_{2}) = 2 q (1 + 3 + \frac{q + 7}{4} + \frac{q + 11}{4}) + q (5) + 4 q (4 + \dots + \frac{q + 3}{4}) = \frac{q^{3} + 18 q^{2} + 5 q}{8}

By adding Eqs. (56), (57), (58), (59), (60) and (61), we get $W (D P (q, 2)) = q^{3} + 17 q^{2} - 10 q .$

Case 7

$q \equiv 6 (m o d 8)$

• If $p_{1}, p_{2} \in X$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 11}{4} .$ There are $q$ pairs of vertices when $1 \leq d (p_{1}, p_{2}) \leq 5 0 r \frac{q + 10}{4}$ and $2 q$ if the $6 \leq d (p_{1}, p_{2}) \leq \frac{q + 6}{4}$ and $\frac{q}{2}$ if the $d (p_{1}, p_{2}) = \frac{q + 14}{4}$ . Thus,

(62)

\sum_{p_{1}, p_{2} \in X} d (p_{1}, p_{2}) = q (1 + 2 + 3 + \dots + 5 + \frac{q + 10}{4}) + \frac{q}{2} (\frac{q + 14}{4}) + 2 q (6 + 7 + \dots + \frac{q + 6}{4}) = \frac{q^{3} + 22 q^{2} - 112 q}{16}

If $p_{1} \in X$ and $p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 10}{4} .$ There are $4 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 2}{4}$ and $2 q$ if the $d (p_{1}, p_{2}) = 2 0 r \frac{q + 6}{2}$ and also $q$ if the $d (p_{1}, p_{2}) = 10 r \frac{q + 10}{4}$ . Thus,

(63)

\sum_{p_{1} \in X, p_{2} \in U} d (p_{1}, p_{2}) = q (1 + \frac{q + 10}{4}) + 2 q (2 + \frac{q + 6}{4}) + 4 q (3 + 4 + \dots + \frac{q + 2}{4}) = \frac{q^{3} + 14 q^{2}}{8}

If $p_{1} \in X$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 6}{4} .$ There are $4 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 2}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 4$ and $2 q$ if the $d (p_{1}, p_{2}) = 2$ and $3 q$ if the $d (p_{1}, p_{2}) = \frac{q + 6}{4}$ . Therefore,

(64)

\sum_{p_{1} \in X, p_{2} \in V} d (p_{1}, p_{2}) = q (4 + \frac{q + 6}{4}) + 2 q (2 + \frac{q + 6}{4}) + 4 q (3 + 4 + \dots + \frac{q + 2}{4}) = \frac{q^{3} + 14 q^{2} + 16 q}{8}

If $p_{1} \in X$ and $p_{2} \in Y$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 3 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 10}{4} .$ There are $4 q$ pairs of vertices when $4 \leq d (p_{1}, p_{2}) \leq \frac{q + 6}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $2 q$ if the $d (p_{1}, p_{2}) = 3$ and $3 q$ if the $d (p_{1}, p_{2}) = \frac{q + 10}{4}$ . Thus,

(65)

\sum_{p_{1} \in X, p_{2} \in Y} d (p_{1}, p_{2}) = q (5 + \frac{q + 10}{4}) + 2 q (3 + \frac{q + 10}{4}) + 4 q (4 + 5 + \dots + \frac{q + 6}{4}) = \frac{q^{3} + 22 q^{2} + 16 q}{8}

If $p_{1}, p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 14}{4} .$ There are $2 q$ pairs of vertices when $4 \leq d (p_{1}, p_{2}) \leq \frac{q + 6}{4}$ and $q$ if the $d (p_{1}, p_{2}) = 2, 3$ and $\frac{q}{2}$ if the $d (p_{1}, p_{2}) = \frac{q + 14}{2}$ . So,

(66)

\sum_{p_{1}, p_{2} \in U} d (p_{1}, p_{2}) = q (2 + 3) + \frac{q}{2} (\frac{q + 14}{4}) + 2 q (4 + 5 + \dots + \frac{q + 6}{4}) = \frac{q^{3} + 18 q^{2} - 24 q}{16}

If $p_{1} \in U$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 10}{4} .$ There are $4 q$ pairs of vertices if the $4 \leq d (p_{1}, p_{2}) \leq \frac{q + 2}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $2 q$ if the $d (p_{1}, p_{2}) = 1 o r 3 o r \frac{q + 6}{4}$ and $3 q$ if the $d (p_{1}, p_{2}) = \frac{q + 10}{4}$ . Thus,

(67)

\sum_{p_{1} \in U, p_{2} \in V} d (p_{1}, p_{2}) = 2 q (1 + 3 + \frac{q + 6}{4} + \frac{q + 10}{4}) + q (5 + \frac{q + 10}{4}) + 4 q (4 + \dots + \frac{q + 2}{4}) = \frac{q^{3} + 18 q^{2} + 8 q}{8}

By adding Eqs. (62), (63), (64), (65), (66) and (67), we get $W (D P (q, 2)) = q^{3} + 17 q^{2} - 10 q .$

Case 8

$q \equiv 7 (m o d 8)$

If $p_{1}, p_{2} \in X$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 9}{4} .$ There are $q$ pairs of vertices when $1 \leq d (p_{1}, p_{2}) \leq 5$ and $2 q$ if the $6 \leq d (p_{1}, p_{2}) \leq \frac{q + 9}{4}$ . Thus,

(68)

\sum_{p_{1}, p_{2} \in X} d (p_{1}, p_{2}) = q (1 + 2 + 3 + \dots + 5) + 2 q (6 + 7 + \dots + \frac{q + 9}{4}) = \frac{q^{3} + 22 q^{2} - 123 q}{16}

If $p_{1} \in X$ and $p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 5}{4} .$ There are $4 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 5}{4}$ and $2 q$ if the $d (p_{1}, p_{2}) = 2$ and also $q$ if the $d (p_{1}, p_{2}) = 1$ . Therefore,

(69)

\sum_{p_{1} \in X, p_{2} \in U} d (p_{1}, p_{2}) = q (1) + 2 q (2) + 4 q (3 + 4 + \dots + \frac{q + 5}{4}) = \frac{q^{3} + 14 q^{2} - 11 q}{8}

If $p_{1} \in X$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 5}{4} .$ There are $4 q$ pairs of vertices when $3 \leq d (p_{1}, p_{2}) \leq \frac{q + 5}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 4$ and $2 q$ if the $d (p_{1}, p_{2}) = 2$ . Thus,

(70)

\sum_{p_{1} \in X, p_{2} \in V} d (p_{1}, p_{2}) = q (4) + 2 q (2) + 4 q (3 + 4 + \dots + \frac{q + 5}{4}) = \frac{q^{3} + 14 q^{2} + 13 q}{8}

If $p_{1} \in X$ and $p_{2} \in Y$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 3 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 9}{4} .$ There are $4 q$ pairs of vertices when $4 \leq d (p_{1}, p_{2}) \leq \frac{q + 9}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $2 q$ if the $d (p_{1}, p_{2}) = 3$ . So,

(71)

\sum_{p_{1} \in X, p_{2} \in Y} d (p_{1}, p_{2}) = q (5) + 2 q (3) + 4 q (4 + 5 + \dots + \frac{q + 9}{4}) = \frac{q^{3} + 22 q^{2} + 13 q}{8}

If $p_{1}, p_{2} \in U$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 2 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 9}{4} .$ There are $2 q$ pairs of vertices when $4 \leq d (p_{1}, p_{2}) \frac{q + 5}{4}$ and $q$ if the $d (p_{1}, p_{2}) = 2, 3, \frac{q + 9}{4}$ . Thus,

(72)

\sum_{p_{1}, p_{2} \in U} d (p_{1}, p_{2}) = q (2 + 3 + \frac{q + 9}{4}) + 2 q (4 + 5 + \dots + \frac{q + 5}{4}) = \frac{q^{3} + 18 q^{2} - 31 q}{16}

If $p_{1} \in U$ and $p_{2} \in V$ , then $d^{*}$ ( $p_{1}, p_{2}$ ) = 1 and $d^{* *} (p_{1}, p_{2}) = \frac{q + 9}{4} .$ . There are $4 q$ pairs of vertices if the $4 \leq d (p_{1}, p_{2}) \leq \frac{q + 1}{4}$ and $5 q$ if the $d (p_{1}, p_{2}) = 5$ and $2 q$ if the $d (p_{1}, p_{2}) = 1 o r 3$ and $3 q$ if the $d (p_{1}, p_{2}) = \frac{q + 5}{4} o r \frac{q + 9}{4}$ . Thus,

(73)

\sum_{p_{1} \in U, p_{2} \in V} d (p_{1}, p_{2}) = 2 q (1 + 3 + \frac{q + 5}{4} + \frac{q + 9}{4}) + q (5 + \frac{q + 5}{4} + \frac{q + 9}{4}) + 4 q (4 + \dots + \frac{q + 1}{4}) = \frac{q^{3} + 18 q^{2} + q}{8}

By adding Eqs. (68), (69), (70), (71), (72) and (73), we get $W (D P (q, 2)) = q^{3} + 17 q^{2} - 17 q .$

3 Graphical representation of $W (q, 1)$ and $W (q, 2)$

In this section, the graphical representation of $W (q, 1)$ and $W (q, 2)$ of $DP (q, 1)$ and $D P (q, 2)$ for different valuers of $q$ are determined. These compact formulas are easy to understand and draw and can be beneficial to the people working in the area.

4 Concluding remarks

The Generalized Petersen Graph and Double Generalized Petersen graphs are extensively studied families in graph theory. We have extended the study of topological indices by finding the Wiener index and the Wiener polarity indices of the Double Generalized Petersen Graph which is constructed from the Generalized Petersen Graph. We close this section by raising the following questions.

Open Problems

Determine the Wiener Index of Double Generalized Petersen Graph $DP (q, k)$ for $k \geq 3 .$
Explore the other topological properties of $DP (q, k)$ .

Funding

This research work was supported by the Deanship of Scientific Research at King Khalid University under Grant number RGP. 1/155/42.

Acknowledgement

The Authors extend their thanks to the Deanship of Scientific Research at King Khalid University for funding this work through the large group research program under grant number RGP. 2/112/43.

Declaration of Competing Interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

References

Adnan M., Bokhary S.A.U.H., Imran M., . On Wiener polarity index and wiener index of certain triangular networks. J. Chem.. 2021;2021
[CrossRef] [Google Scholar]
Bokhary S.A.U.H., Adnan S.M.K., Cancan M., . On topological indices and QSPR analysis of drugs used for the treatment of breast cancer. Polycyclic Aromatic Compounds. 2021;42:6233-6253.
[CrossRef] [Google Scholar]
Bokhary S.A.U.H., Adnan, . On vertex PI index of certain triangular tessellation networks: Main Gr. Met. Chem.. 2021;44:203-212.
[CrossRef] [Google Scholar]
Buckley F., Harary F., . Distance in graphs. Addison-Wesley Pub Co; 1990.
Du, W., Li, X., Shi, Y., 2008. Algorithms and Extremal Problem on Wiener Polarity Index.
Graovac A., Pisanski T., . On the Wiener index of a graph. J. Math. Chem.. 1991;8:53-62.
[CrossRef] [Google Scholar]
Harary, F., n.d. Graph theory 274.
Liu Y., Zuo L., Shang C., . Chemical Indices of Generalized Petersen Graph. Int. J. Appl. Math 2017
[Google Scholar]
Ul Haq Bokhary S.A., Imran M., Akhter S., Manzoor S., . Molecular topological invariants of certain chemical networks. Main Gr. Met. Chem.. 2021;44:141-149.
[CrossRef] [Google Scholar]
Wiener H., . Structural determination of paraffin boiling points. J. Am. Chem. Soc.. 1947;69:17-20.
[CrossRef] [Google Scholar]