Counting Spanning Trees in Various Products of Two Complete Bipartite Graphs

Abstract

Calculating the number of spanning trees in a graph is a crucial problem in combinatorics and physics that has been thoroughly researched for many years by mathematicians and physicists. The higher the quality and perfection of the network, the greater the number of trees spanning it, leading to greater possibilities for connection between two vertices and ensuring good rigidity and resistance. In this work, we use matrix theory and linear algebra techniques to derive simple explicit formulas for calculating the complexity of various products of two complete bipartite graphs, including the Cartesian product, tensor product, normal product, composition product, symmetric product, disjunction, and strong sum.

1. Introduction

A spanning tree of an undirected graph $G$ is a subgraph that is a tree which contains all of the vertices of $G$. In sectors including computer networking, telecommunications, and transportation, spanning trees are essential for establishing the bare minimum of connections required to link every part of a network or system. This is necessary for network stability, redundancy, and efficiency. By allowing only one active data stream and offering backup paths for fault tolerance, they stop loops and broadcast storms in networks [1,2,3]. Additionally, the number of spanning trees is counted because it is a basic problem in graph theory and combinatorics with applications in physics and engineering, aids in the design of efficient network infrastructure and routing algorithms, and can provide a sparse subgraph that reflects graph properties [4,5].

In 1847, the physicist Kirchhoff [6] proposed the matrix tree theorem, which marked the beginning of the history of calculating the number of spanning trees, $\tau \left(G\right)$, in a graph $G$. This theorem was based on the determinants of a particular matrix obtained from the Laplacian matrix, $\mathrm{L}$, which is defined as the difference between the adjacency matrix, $A\left(G\right)$, and the degree matrix, $D\left(G\right)$, which is a diagonal matrix, $D\left(G\right)=dig(d_1,d_2,\dots ,d_n)$, corresponding to a graph $G$ with $n$ vertices and a $1$ or $0$ in position $(u_i,u_j)$, depending on whether or not they are adjacent. That is,

$$L_{ij}=\left\{\begin{array}{l}{d}_{i } \mathrm{i}\mathrm{f} i=j\\ -1 \mathrm{i}\mathrm{f} i\ne j \mathrm{a}\mathrm{n}\mathrm{d} i \mathrm{i}\mathrm{s} \mathrm{a}\mathrm{d}\mathrm{j}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t} \mathrm{t}\mathrm{o} j\\ 0 \mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}\end{array}\right.,$$

where the degree of the vertex $u_i$ is indicated by $d_i$. The matrix tree theorem states that the complexity $\tau \left(G\right)$ is equal to the common value of all the cofactors of $L$.

Consequently, Cayley’s formula [7] states that the number of spanning trees in the complete graph with $n$ vertices as $n^{n-2}$. According to Fieldler and Sedlacek [8], the number of spanning trees of the complete bipartite graph on $r$ and $s$ vertices is equal to $r^{s-1}{s}^{r-1}$.

Lastly, Onodera [9] provides the number of spanning trees of a $k$-partite graph on the vertex set $V_1+V_2+\dots +V_k$, where $\left|V_i\right|=p_i$ for $i=\mathrm{1,2},\dots ,k$, and “+” signifies disjoint union. This number is expressed as $n^{k-2}{\prod }_{i=1}^k(n-p_i{)}^{p_i-1}$, $n=\sum p_i$.

Daoud recently derived the exact formulas for calculating the complexity of several products of complete and complete bipartite graphs [10].

There are numerous methods for calculating $\tau \left(G\right)$. Laplacian eigenvalues are one such method. Assume that the eigenvalues of the Laplacian matrix $L$ of a graph $G$ with $k$ vertices are represented by ${\mu }_1\ge {\mu }_2\ge \dots \ge {\mu }_k=0$. As stated by “Kelmans” and “Chelnokov” [11],

$$\tau (G)={\displaystyle \frac{1}{k}}{\prod }_{i=1}^{k-1}{\mu }_k.$$

(1)

Following that, Temperley [12] has proven that

$$\tau (G)={\displaystyle \frac{1}{k^2}}\mathrm{d}\mathrm{e}\mathrm{t}(L+J)$$

(2)

where $J$ is the $k\times k$ matrix, each element of which is unity.

One popular way for determining complexity is the deletion-contraction method. It is feasible to recursively determine a graph’s complexity in this way: if $e$ is any edge in a graph $G$, its complexity $\tau \left(G\right)$ equals $\tau \left(G\right)=\tau (G-e)+\tau (G/e)$, where $G/e$ is the contraction of $e$ in $G$, and $G-e$ is the deletion of $e$ from $G$ [13,14].

It is also possible to determine the number of spanning trees iteratively by tracking changes in the number of weighted spanning trees and simplifying the graph. The electrically equivalent transformation strategy uses graph transformations such as switching between triangle ($∆$) and star ($Y$) configurations, combining serial edges, and merging parallel edges, using the connection between the counting spanning tree problem and electrical networks, where conductance is represented by edge weights [15,16].

An efficient method for counting spanning trees in graphs is modular decomposition, which breaks a graph into its constituent modules bottom-up. For some graph types, this allows for a linear-time solution. The total spanning tree count is computed by multiplying the values associated with the vertices after the modular decomposition tree is shrunk until only one node is left [17].

Unfortunately, counting the number of spanning trees in large graphs by directly evaluating the determinant of Kirchhoff’s matrix is challenging. As a result, strategies to overcome the difficulties have been developed, and there has been much interest in obtaining explicit and possibly simple formulas for specific types of graphs. We now present the following lemma, which explains how to use an extension of Kirchhoff’s formula to get the number of spanning trees.

The following Lemma is easily proved from Temperley’s Equation (2):

Lemma 1.

Assume that the graph $G$ has $n$ vertices.

$$\tau (G)={\displaystyle \frac{1}{n^2}}\mathit{det}(nI-\overline{D}+\overline{A})$$

where $I$is the $n\times n$ identity matrix, and $\overline{A}, \overline{D }$ are the adjacency and degree matrices of $\overline{G}$, the complement of $G$, respectively.

This formula has a big advantage because it does not need cofactors or eigenvalues, instead representing $\tau \left(G\right)$ directly as a determinant.

Lemma 2.

Ref. [10] assumes that $A_n\left(t\right)$ is an $n\times n$ matrix with $t\ge 2$such that

$$A_n(t)=\left(\begin{array}{cccccc}t& 1& \cdots & \cdots & \cdots & 1\\ 1& t& \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & t& 1\\ 1& \cdots & \cdots & \cdots & 1& t\end{array}\right)$$

Then,

$$\mathit{det}(A_n\left(t\right))=(t+n-1\left)\right(t-1{)}^{n-1}.$$

Proof. 

We obtain the following from the circulant determinants’ definition:

$$\begin{array}{l}\mathit{det}(A_n(t))=\mathit{det}\left(\begin{array}{cccccc}t& 1& \cdots & \cdots & \cdots & 1\\ 1& t& \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & t& 1\\ 1& \cdots & \cdots & \cdots & 1& t\end{array}\right)={\displaystyle \prod _{j=1}^n}(t+{\omega }_j+{\omega }_j^2+{\omega }_j^3+\dots \dots \dots \dots +{\omega }_j^{n-1})\\ =(t+1+1+\dots \dots +1)\times {\displaystyle \prod _{j=1,{\omega }_j\ne 1}^n}(t+\underset{=-1}{\underbrace{{\omega }_j+{\omega }_j^2+{\omega }_j^3+\dots \dots \dots +{\omega }_j^{n-1})}}\\ =(t+n-1)\times (t-1{)}^{n-1}.\end{array}$$

□

We can generalize the previous Lemma as follows:

Lemma 3.

Let$P,Q\in F^{n\times n}$and $\mathbb{F}\in F^{kn\times kn}$ be such that

$$\mathbb{F}=\left(\begin{array}{cccccc}P& Q& \cdots & \cdots & \cdots & Q\\ Q& P& \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & P& Q\\ Q& \cdots & \cdots & \cdots & Q& P\end{array}\right)$$

Then,

$$\mathit{det}(\mathbb{F})=[\mathit{det}(P-Q){]}^{k-1}\mathit{det}[P+(k-1\left)Q\right].$$

Lemma 4.

Let$A\in F^{n\times n}$, $B\in F^{n\times m}$, $C\in F^{m\times n}$, and $D\in F^{m\times m}$ [18]. Assume that $A$ and $D$ are nonsingular matrices. Then,

$$\mathit{det}\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)=\mathit{det}(A-B{D}^{-1}C)\mathit{det}(D)=\mathit{det}(A)\mathit{det}(D-C{A}^{-1}B).$$

The formulas in Lemmas 2–4 provide some symmetry in certain matrices, making it easier to calculate determinants.

2. Complexity of Cartesian Product of Two Complete Bipartite Graphs

The Cartesian product of two graphs $G$ and $H$, denoted by $G \square H$, is a graph where the vertices are ordered pairs $(g,h)$ with $g\in V\left(G\right)$) and $h\in V\left(H\right)$. Two vertices $(g_1,h_1)$ and $(g_2,h_2)$ are adjacent if $g_1=g_2$ and $h_1,h_2$ are adjacent in graph $H$ or if $h_1=h_2$ and $g_1,g_2$ are adjacent in graph $G$ [19].

One way to view the resulting graph $G\square H$ is to replace each vertex in $G$ with a copy of $H$, connecting the relevant vertices between copies if they were adjacent in $G$. On the other hand, it can be thought of as swapping out each vertex in $H$ for a copy of $G$. This operation is commutative up to isomorphism, meaning $G\square H\cong H\square G$.

Theorem 1.

For $n,r$ and $s$ are all greater than or equal to 1, we get

$$\begin{gathered} \tau (K_{n,n}\square K_{r,s})=r^{s-1}\times s^{r-1}\times n^{2\left(n-1\right)}\times {\left(n+r\right)}^{2\left(s-1\right)\left(n-1\right)}\times {\left(n+s\right)}^{2\left(r-1\right)\left(n-1\right)}\times {\left(n+r+s\right)}^{2\left(n-1\right)} \\ \times {\left(2n+r\right)}^{\left(s-1\right)}\times {\left(2n+s\right)}^{\left(r-1\right)}\times \left(2n+r+s\right) \end{gathered}$$

Proof. 

By using Lemma 1, we have

$$\tau (K_{n,n}\square K_{r,s})={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times \mathit{det}\left(\begin{array}{cc}A& B\\ B& A\end{array}\right),$$

where $A$ and $B$ are block matrices of order $n\times n$ made up of matrices of order $(r+s)\times (r+s)$ that are described as follows:

$$\begin{gathered} A=\left(\begin{array}{ccccccc}P& J& \cdots & \cdots & \cdots & \cdots & J\\ J& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & J\\ J& \cdots & \cdots & \cdots & \cdots & J& P\end{array}\right),B=\left(\begin{array}{ccccccc}(J-I)& \cdots & \cdots & \cdots & \cdots & \cdots & (J-I)\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ (J-I)& \cdots & \cdots & \cdots & \cdots & \cdots & (J-I)\end{array}\right), \\ P=\left(\begin{array}{cccccccc}(n+s+1)& 1& \cdots & 1& 0& \cdots & \cdots & 0\\ 1& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & (n+s+1)& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& (n+r+1)& 1& \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 1\\ 0& \cdots & \cdots & 0& 1& \cdots & 1& (n+r+1)\end{array}\right). \end{gathered}$$

Thus, we have: $\tau (K_{n,n}\square K_{r,s})={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times \mathit{det}(A+B)\times \mathit{det}(A-B).$ Such that

$$\begin{gathered} A+B=\left(\begin{array}{ccccccc}X& Y& \cdots & \cdots & \cdots & \cdots & Y\\ Y& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & Y\\ Y& \cdots & \cdots & \cdots & \cdots & Y& X\end{array}\right),A-B=\left(\begin{array}{ccccccc}Z& W& \cdots & \cdots & \cdots & \cdots & W\\ W& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & W\\ W& \cdots & \cdots & \cdots & \cdots & W& Z\end{array}\right), \\ X=\left(\begin{array}{cccccccc}(n+s+1)& 2& \cdots & 2& 1& \cdots & \cdots & 1\\ 2& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 2& ⋮& \ddots & \ddots & ⋮\\ 2& \cdots & 2& (n+s+1)& 1& \cdots & \cdots & 1\\ 1& \cdots & \cdots & 1& (n+r+1)& 2& \cdots & 2\\ ⋮& \ddots & \ddots & ⋮& 2& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 2\\ 1& \cdots & \cdots & 1& 2& \cdots & 2& (n+r+1)\end{array}\right), \\ Y=\left(\begin{array}{cccccccc}1& 2& \cdots & 2& 2& \cdots & \cdots & 2\\ 2& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 2& ⋮& \ddots & \ddots & ⋮\\ 2& \cdots & 2& 1& 2& \cdots & \cdots & 2\\ 2& \cdots & \cdots & 2& 1& 2& \cdots & 2\\ ⋮& \ddots & \ddots & ⋮& 2& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 2\\ 2& \cdots & \cdots & 2& 2& \cdots & 2& 1\end{array}\right),W=\left(\begin{array}{cccccccc}1& 0& \cdots & 0& 0& \cdots & \cdots & 0\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & 0& 1& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& 1& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ 0& \cdots & \cdots & 0& 0& \cdots & 0& 1\end{array}\right) \mathrm{and} \\ Z=\left(\begin{array}{cccccccc}(n+s+1)& 0& \cdots & 0& -1& \cdots & \cdots & -1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & 0& (n+s+1)& -1& \cdots & \cdots & -1\\ -1& \cdots & \cdots & -1& (n+r+1)& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ -1& \cdots & \cdots & -1& 0& \cdots & 0& (n+r+1)\end{array}\right). \end{gathered}$$

Applying Lemma 3, we obtain

$$\begin{gathered} \tau (K_{n,n}\square K_{r,s})={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times (\mathit{det}\left(\begin{array}{cccccccc}n+s& 0& \cdots & 0& -1& \cdots & \cdots & -1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & n+s& -1& \cdots & \cdots & -1\\ -1& \cdots & \cdots & -1& n+r& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ -1& \cdots & \cdots & -1& 0& \cdots & 0& n+r\end{array}\right){)}^{n-1}\times \\ \mathit{det}\left(\begin{array}{cccccccc}2n+s& 2n& \cdots & 2n& 2n-1& \cdots & \cdots & 2n-1\\ 2n& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 2n& ⋮& \ddots & \ddots & ⋮\\ 2n& \cdots & 2n& 2n+s& 2n-1& \cdots & \cdots & 2n-1\\ 2n-1& \cdots & \cdots & 2n-1& 2n+r& 2n& \cdots & 2n\\ ⋮& \ddots & \ddots & ⋮& 2n& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 2n\\ 2n-1& \cdots & \cdots & 2n-1& 2n& \cdots & 2n& 2n+r\end{array}\right) \\ \times (\mathit{det}\left(\begin{array}{cccccccc}n+s& 0& \cdots & 0& -1& \cdots & \cdots & -1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & n+s& -1& \cdots & \cdots & -1\\ -1& \cdots & \cdots & -1& n+r& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ -1& \cdots & \cdots & -1& 0& \cdots & 0& n+r\end{array}\right){)}^{n-1}\times \\ \mathit{det}\left(\begin{array}{cccccccc}2n+s& 0& \cdots & 0& -1& \cdots & \cdots & -1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & 0& 2n+s& -1& \cdots & \cdots & -1\\ -1& \cdots & \cdots & -1& 2n+r& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ -1& \cdots & \cdots & -1& 0& \cdots & 0& 2n+r\end{array}\right). \end{gathered}$$

Using Lemma 4, we get

$$\begin{array}{l}\tau (K_{n,n}\square K_{r,s})={\displaystyle \frac{1}{4n^2(r+s{)}^2}}({\mathit{det}\left(\begin{array}{cccc}n+s& 0& \cdots & 0\\ 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0\\ 0& \cdots & \cdots & n+s\end{array}\right)}_{r\times r}{)}^{2(n-1)}\times \\ ({\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{\left(n+r\right)\left(n+s\right)-r}{n+s}}& {\displaystyle \frac{-r}{n+s}}& \cdots & {\displaystyle \frac{-r}{n+s}}\\ {\displaystyle \frac{-r}{n+s}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{-r}{n+s}}\\ {\displaystyle \frac{-r}{n+s}}& \cdots & {\displaystyle \frac{-r}{n+s}}& {\displaystyle \frac{\left(n+r\right)\left(n+s\right)-r}{n+s}}\end{array}\right)}_{s\times s}{)}^{2\left(n-1\right)}{\times \mathit{det}\left(\begin{array}{cccc}2n+s& 2n& \cdots & 2n\\ 2n& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 2n\\ 2n& \cdots & 2n& 2n+s\end{array}\right)}_{r\times r}\times \\ {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{2n\left(r^2+2r+s\right)+r\left(s-1\right)}{2nr+s}}& {\displaystyle \frac{2n\left(2r+s\right)-r}{2nr+s}}& \cdots & {\displaystyle \frac{2n\left(2r+s\right)-r}{2nr+s}}\\ {\displaystyle \frac{2n\left(2r+s\right)-r}{2nr+s}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{2n\left(2r+s\right)-r}{2nr+s}}\\ {\displaystyle \frac{2n\left(2r+s\right)-r}{2nr+s}}& \cdots & {\displaystyle \frac{2n\left(2r+s\right)-r}{2nr+s}}& {\displaystyle \frac{2n\left(r^2+2r+s\right)+r\left(s-1\right)}{2nr+s}}\end{array}\right)}_{s\times s}\times \\ {{\mathit{det}\left(\begin{array}{cccc}2n+s& 0& \cdots & 0\\ 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0\\ 0& \cdots & \cdots & 2n+s\end{array}\right)}_{r\times r}\times \mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{\left(2n+r\right)\left(2n+s\right)-r}{2n+s}}& {\displaystyle \frac{-r}{2n+s}}& \cdots & {\displaystyle \frac{-r}{2n+s}}\\ {\displaystyle \frac{-r}{2n+s}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{-r}{2n+s}}\\ {\displaystyle \frac{-r}{2n+s}}& \cdots & {\displaystyle \frac{-r}{2n+s}}& {\displaystyle \frac{\left(2n+r\right)\left(2n+s\right)-r}{2n+s}}\end{array}\right).}_{s\times s}\\ ={\displaystyle \frac{1}{4n^2(r+s{)}^2}}(n+s{)}^{2r(n-1)}\times ({\displaystyle \frac{-r}{n+s}}{)}^{2s(n-1)}\times (2n{)}^r\times \\ ({\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{(n+r)(n+s)-r}{-r}}& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{(n+r)(n+s)-r}{-r}}\end{array}\right)}_{s\times s}{)}^{2(n-1)}\times {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{2n+s}{2n}}& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& 2n+s\end{array}\right)}_{r\times r}\times \\ ({\displaystyle \frac{2n(2r+s)-r}{2nr+s}}{)}^s{\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{2n(r^2+2r+s)+r(s-1)}{2n(2r+s)-r}}& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{2n(r^2+2r+s)+r(s-1)}{2n(2r+s)-r}}\end{array}\right)}_{s\times s}\times (2n+s{)}^r\times \\ ({\displaystyle \frac{-r}{2n+s}}{)}^r{\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{\left(2n+r\right)\left(2n+s\right)-r}{-r}}& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{\left(2n+r\right)\left(2n+s\right)-r}{-r}}\end{array}\right)}_{s\times s}.\end{array}$$

Applying Lemma 2, we have

$$\begin{array}{l}\tau (K_{n,n}\square K_{r,s})={\displaystyle \frac{1}{4n^2{\left(r+s\right)}^2}}\times {\left(n+s\right)}^{2r\left(n-1\right)}\times {\left({\displaystyle \frac{r}{n+s}}\right)}^{2s\left(n-1\right)}\times {\left({\displaystyle \frac{\left(n+r\right)\left(n+s\right)-r}{-r}}+s-1\right)}^{2\left(n-1\right)}\times {\left({\displaystyle \frac{\left(n+r\right)\left(n+s\right)-r}{-r}}-1\right)}^{2\left(n-1\right)\left(s-1\right)}\\ \times {\left(2n\right)}^r\times \left({\displaystyle \frac{2n+s}{2n}}+r-1\right)\times {\left({\displaystyle \frac{2n+s}{2n}}-1\right)}^{r-1}\times {\left({\displaystyle \frac{r-\left(2n\right)\left(2r+s\right)}{2nr+s}}\right)}^s\times \left({\displaystyle \frac{2n\left(r^2+2r+s\right)+r\left(s-1\right)}{2n\left(2r+s\right)-r}}+s-1\right)\\ \times {\left({\displaystyle \frac{\left(n+r\right)\left(n+s\right)-r}{-r}}-1\right)}^{\left(n-1\right)\left(s-1\right)}\times {\left(2n+s\right)}^r\times {\left({\displaystyle \frac{r}{2n+s}}\right)}^s\left({\displaystyle \frac{\left(r+2n\right)\left(s+2n\right)-r}{-r}}+s-1\right)\\ \times {\left({\displaystyle \frac{\left(r+2n\right)\left(s+2n\right)-r}{-r}}-1\right)}^{s-1}\\ ={\displaystyle \frac{1}{4n^2{\left(r+s\right)}^2}}\times {\left(n+s\right)}^{\left(r-2s\right)\left(n-1\right)}\times {\left(\left(n+r\right)\left(n+s\right)-rs\right)}^{n-1}\times {\left(\left(n+r\right)\left(n+s\right)\right)}^{2\left(n-1\right)\left(s-1\right)}\times {\left(s\right)}^{r-1}\times {\left(r\right)}^{s-1}\times \\ \left(2n\left(r^2+2r+s\right)+r\left(s-1\right)+\left(s-1\right)\left(2n\left(2r+s\right)-r\right)\right)\times {\left(n+s\right)}^{r\left(n-1\right)}\times {\left(\left(n+r\right)\left(n+s\right)-rs\right)}^{n-1}\times \\ {\left(2n+s\right)}^r\left(\left(r+2n\right)\left(s+2n\right)-r-r\left(s-1\right)\right)\times {\left(s+2n\right)}^{-1}\times {\left(r+2n\right)}^{s-1}\\ =r^{s-1}\times s^{r-1}\times n^{2\left(n-1\right)}\times {\left(n+r\right)}^{2\left(s-1\right)\left(n-1\right)}\times {\left(n+s\right)}^{2\left(r-1\right)\left(n-1\right)}\times {\left(n+r+s\right)}^{2\left(n-1\right)}\times {\left(2n+r\right)}^{\left(s-1\right)}\times \\ {\left(2n+s\right)}^{\left(r-1\right)}\times \left(2n+r+s\right).\end{array}$$

□

3. Complexity of Tensor Product of Two Complete Bipartite Graphs

The tensor product, or the Kronecker product, $G\times H$ of two graphs $G$ and $H$ is the simple graph with $V(G\times H)=V\left(G\right)\times V\left(H\right)$, where $(g_1,h_1)$ and $(g_2,h_2)$ are adjacent in $G\times H$ if and only if $g_1$ is adjacent to $g_2$ in $G$ and $h_1$ is adjacent to $h_2$ in, $H$ [19]. This product is commutative $G\times H\cong H\times G$ and associative, often creating disconnected components even if $G$ and $H$ are connected.

Lemma 5.

For $n,r$ and $s$ are all greater than or equal to 1, we get $(K_{n,n}\times K_{r,s})=0$.

Proof. 

Tensor products of two connected bipartite graphs, such as complete bipartite graphs, are disconnected because they split into exactly two connected components. This occurs because bipartite graphs lack odd cycles, making it impossible for the two distinct sets of vertex pairs $(g,h)$ classified by the bipartite partitions of the original graphs to create paths. □

4. Complexity of Normal Product of Two Complete Bipartite Graphs

The strong (normal) product of two graphs $G$ and $H$, denoted $G⊠H$, is a graph with vertex set $V\left(G\right)\times V\left(H\right)$ where two distinct vertices $(g_1,h_1)$ and $(g_2,h_2)$ are adjacent if $g_1$ is adjacent or equal $g_2$ in $G$ and $h_1$ is adjacent or equal $h_2$ in $H$ [19].

The normal product is commutative $G⊠H\cong H⊠G$ and associative, acting as the union of the Cartesian product $G\square H$ and the tensor product $G\times H$.

Theorem 2.

For $n,r$ and $s$ are all greater than or equal to 1, we get

$$\begin{array}{ll}\tau \left(K_{n,n}⊠K_{r,s}\right)=& n^{2\left(n-1\right)}\times {\left(n+1\right)}^{r+s-1}\times r^{s-1}\times s^{r-1}\times {\left(nr+n+r\right)}^{2\left(n-1\right)\left(s-1\right)}\times {\left(ns+n+s\right)}^{2\left(n-1\right)\left(r-1\right)}\\ & \times {\left(nr+2n+r\right)}^{\left(s-1\right)}\times {\left(ns+2n+s\right)}^{\left(r-1\right)}\times {\left(r+s+2rs+n\left(1+r\right)\left(1+s\right)\right)}^{2\left(n-1\right)}\\ & \times \left(r+s+2rs+n\left(2+r+s\right)\right)\end{array}$$

Proof. 

Applying Lemma 1, we have

$$\tau \left(K_{n,n}⊠K_{r,s}\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\mathit{det}\left(\begin{array}{cc}A& B\\ B& A\end{array}\right),$$

where $A$ and $B$ are block matrices of order $n\times n$ made up of matrices of order $(r+s)\times (r+s)$ that are described as follows:

$$\begin{gathered} A=\left(\begin{array}{ccccccc}P& J& \cdots & \cdots & \cdots & \cdots & J\\ J& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & J\\ J& \cdots & \cdots & \cdots & \cdots & J& P\end{array}\right),B=\left(\begin{array}{ccccccc}Q& \cdots & \cdots & \cdots & \cdots & \cdots & Q\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ Q& \cdots & \cdots & \cdots & \cdots & \cdots & Q\end{array}\right), \\ P=\left(\begin{array}{cccccccc}(n+1)(s+1)& 1& \cdots & 1& 0& \cdots & \cdots & 0\\ 1& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & (n+1)(s+1)& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& (n+1)(r+1)& 1& \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 1\\ 0& \cdots & \cdots & 0& 1& \cdots & 1& (n+1)(r+1)\end{array}\right), \\ Q=\left(\begin{array}{cccccccc}0& 1& \cdots & 1& 0& \cdots & \cdots & 0\\ 1& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & 0& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& 0& 1& \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 1\\ 0& \cdots & \cdots & 0& 1& \cdots & 1& 0\end{array}\right) \end{gathered}$$

Thus, we have: $\tau \left(K_{n,n}⊠K_{r,s}\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times \mathit{det}(A+B)\times \mathit{det}(A-B).$ Such that

$$\begin{gathered} A+B=\left(\begin{array}{ccccccc}X& Y& \cdots & \cdots & \cdots & \cdots & Y\\ Y& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & Y\\ Y& \cdots & \cdots & \cdots & \cdots & Y& X\end{array}\right),A-B=\left(\begin{array}{ccccccc}Z& W& \cdots & \cdots & \cdots & \cdots & W\\ W& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & W\\ W& \cdots & \cdots & \cdots & \cdots & W& Z\end{array}\right), \\ X=\left(\begin{array}{cccccccc}\left(n+1\right)\left(s+1\right)& 2& \cdots & 2& 0& \cdots & \cdots & 0\\ 2& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 2& ⋮& \ddots & \ddots & ⋮\\ 2& \cdots & 2& \left(n+1\right)\left(s+1\right)& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& \left(n+1\right)\left(r+1\right)& 2& \cdots & 2\\ ⋮& \ddots & \ddots & ⋮& 2& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 2\\ 0& \cdots & \cdots & 0& 2& \cdots & 2& \left(n+1\right)\left(r+1\right)\end{array}\right), \\ Y=\left(\begin{array}{cccccccc}1& 2& \cdots & 2& 1& \cdots & \cdots & 1\\ 2& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 2& ⋮& \ddots & \ddots & ⋮\\ 2& \cdots & 2& 1& 1& \cdots & \cdots & 1\\ 1& \cdots & \cdots & 1& 1& 2& \cdots & 2\\ ⋮& \ddots & \ddots & ⋮& 2& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 2\\ 1& \cdots & \cdots & 1& 2& \cdots & 2& 1\end{array}\right),W=\left(\begin{array}{cccccccc}1& 0& \cdots & 0& 1& \cdots & \cdots & 1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & 0& 1& 1& \cdots & \cdots & 1\\ 1& \cdots & \cdots & 1& 1& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ 1& \cdots & \cdots & 1& 0& \cdots & 0& 1\end{array}\right) \mathrm{and} \\ Z=\left(\begin{array}{cccccccc}(n+1)(s+1)& 0& \cdots & 0& 0& \cdots & \cdots & 0\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & 0& (n+1)(s+1)& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& (n+1)(r+1)& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ 0& \cdots & \cdots & 0& 0& \cdots & 0& (n+1)(r+1)\end{array}\right). \end{gathered}$$

Applying Lemma 3, we obtain

$$\begin{gathered} \tau \left(K_{n,n}⊠K_{r,s}\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times (\mathit{det}\left(\begin{array}{cccccccc}n(s+1)+s& 0& \cdots & 0& -1& \cdots & \cdots & -1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & n(s+1)+s& -1& \cdots & \cdots & -1\\ -1& \cdots & \cdots & -1& n(r+1)+r& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ -1& \cdots & \cdots & -1& 0& \cdots & 0& n(r+1)+r\end{array}\right){)}^{n-1}\times \\ \mathit{det}\left(\begin{array}{cccccccc}n(s+2)+s& 2n& \cdots & 2n& n-1& \cdots & \cdots & n-1\\ 2n& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 2n& ⋮& \ddots & \ddots & ⋮\\ 2n& \cdots & 2n& n(s+2)+s& n-1& \cdots & \cdots & n-1\\ n-1& \cdots & \cdots & n-1& n(r+2)+r& 2n& \cdots & 2n\\ ⋮& \ddots & \ddots & ⋮& 2n& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 2n\\ n-1& \cdots & \cdots & n-1& 2n& \cdots & 2n& n(r+2)+r\end{array}\right) \\ \times (\mathit{det}\left(\begin{array}{cccccccc}n(s+1)+s& 0& \cdots & 0& -1& \cdots & \cdots & -1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & n(s+1)+s& -1& \cdots & \cdots & -1\\ -1& \cdots & \cdots & -1& n(r+1)+r& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ -1& \cdots & \cdots & -1& 0& \cdots & 0& n(r+1)+r\end{array}\right){)}^{n-1}\times \\ \mathit{det}\left(\begin{array}{cccccccc}n(s+2)+s& 0& \cdots & 0& n-1& \cdots & \cdots & n-1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & 0& n(s+2)+s& n-1& \cdots & \cdots & n-1\\ n-1& \cdots & \cdots & n-1& n(r+2)+r& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ n-1& \cdots & \cdots & n-1& 0& \cdots & 0& n(r+2)+r\end{array}\right). \end{gathered}$$

Using Lemma 4, we get

$$\begin{array}{l} \tau \left(K_{n,n}⊠K_{r,s}\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}({\mathit{det}\left(\begin{array}{cccc}n(s+1)+s& 0& \cdots & 0\\ 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0\\ 0& \cdots & 0& n(s+1)+s\end{array}\right)}_{r\times r}{)}^{2(n-1)}\\ \times {\mathit{det}\left(\begin{array}{cccc}n(s+2)+s& 2n& \cdots & 2n\\ 2n& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 2n\\ 2n& \cdots & 2n& n(s+2)+s\end{array}\right)}_{r\times r}\\ \times {\mathit{det}\left(\begin{array}{cccc}n(s+2)+s& 0& \cdots & 0\\ 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0\\ 0& \cdots & 0& n(s+2)+s\end{array}\right)}_{r\times r}\\ ({\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{(nr+n+r)(ns+n+s)-r}{ns+n+s}}& {\displaystyle \frac{-r}{ns+n+s}}& \cdots & {\displaystyle \frac{-r}{ns+n+s}}\\ {\displaystyle \frac{-r}{ns+n+s}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{-r}{ns+n+s}}\\ {\displaystyle \frac{-r}{ns+n+s}}& \cdots & {\displaystyle \frac{-r}{ns+n+s}}& {\displaystyle \frac{(nr+n+r)(ns+n+s)-r}{ns+n+s}}\end{array}\right)}_{s\times s}{)}^{2(n-1)}\times \\ {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{(n+1)(-r+rs+n((3r+2s)+r(2r+s)\left)\right)}{n(2r+s)+s}}& {\displaystyle \frac{(n+1)(-r+n(3r+2s\left)\right)}{n(2r+s)+s}}& \cdots & {\displaystyle \frac{(n+1)(-r+n(3r+2s\left)\right)}{n(2r+s)+s}}\\ {\displaystyle \frac{(n+1)(-r+n(3r+2s\left)\right)}{n(2r+s)+s}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{(n+1)(-r+n(3r+2s\left)\right)}{n(2r+s)+s}}\\ {\displaystyle \frac{(n+1)(-r+n(3r+2s\left)\right)}{n(2r+s)+s}}& \cdots & {\displaystyle \frac{(n+1)(-r+n(3r+2s\left)\right)}{n(2r+s)+s}}& {\displaystyle \frac{(n+1)(-r+rs+n((3r+2s)+r(2r+s)\left)\right)}{n(2r+s)+s}}\end{array}\right)}_{s\times s}\times \\ {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{\left(n\right(r+2)+r)\left(n\right(s+2)+s)-r(n-1{)}^2}{n(s+2)+s}}& {\displaystyle \frac{-r(n-1{)}^2}{n(s+2)+s}}& \cdots & {\displaystyle \frac{-r(n-1{)}^2}{n(s+2)+s}}\\ {\displaystyle \frac{-r(n-1{)}^2}{n(s+2)+s}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{-r(n-1{)}^2}{n(s+2)+s}}\\ {\displaystyle \frac{-r(n-1{)}^2}{n(s+2)+s}}& \cdots & {\displaystyle \frac{-r(n-1{)}^2}{n(s+2)+s}}& {\displaystyle \frac{\left(n\right(r+2)+r)\left(n\right(s+2)+s)-r(n-1{)}^2}{n(s+2)+s}}\end{array}\right)}_{s\times s}\\ ={\displaystyle \frac{1}{4n^2(r+s{)}^2}}(n(s+1)+s{)}^{2r(n-1)}\times ({\displaystyle \frac{-r}{n(s+1)+s}}{)}^{2s(n-1)}\times (2n{)}^r\times \\ ({\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{\left(n\right(r+1)+r)\left(n\right(s+1)+s)-r}{-r}}& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{\left(n\right(r+1)+r)\left(n\right(s+1)+s)-r}{-r}}\end{array}\right)}_{s\times s}{)}^{2(n-1)}\times (2n{)}^r\times \\ ({\displaystyle \frac{n(s+2)+s}{2n}}-1{)}^{r-1}\times ({\displaystyle \frac{n(s+2)+s}{2n}}+r-1)\times (n+1{)}^s\times ({\displaystyle \frac{n(3r+s)-r}{n(2r+s)+s}}{)}^s\times \\ {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{n\left(\right(3r+2s)+r(2r+s\left)\right)+rs-r}{n(3r+2s)-r}}& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{n\left(\right(3r+2s)+r(2r+s\left)\right)+rs-r}{n(3r+2s)-r}}\end{array}\right)}_{s\times s}\times (n(s+2)+s{)}^r\times \\ ({\displaystyle \frac{-r(n-1{)}^2}{n\left(s+2\right)+s}}{)}^s{\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{\left(n\right(r+2)+r)\left(n\right(s+2)+s)-r(n-1{)}^2}{-r(n-1{)}^2}}& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{\left(n\right(r+2)+r)\left(n\right(s+2)+s)-r(n-1{)}^2}{-r(n-1{)}^2}}\end{array}\right)}_{s\times s}.\end{array}$$

Applying Lemma 2, we have:

$$\begin{array}{l} \tau \left(K_{n,n}⊠K_{r,s}\right)={\displaystyle \frac{1}{4n^2{\left(r+s\right)}^2}}\times {\left(ns+n+s\right)}^{2r\left(n-1\right)}\times {\left({\displaystyle \frac{-r}{ns+n+s}}\right)}^{2s\left(n-1\right)}\\ \times {\left({\displaystyle \frac{\left(nr+n+r\right)\left(ns+n+s\right)-r}{-r}}-1\right)}^{2\left(s-1\right)\left(n-1\right)}\times {\left({\displaystyle \frac{\left(nr+n+r\right)\left(ns+n+s\right)-r}{-r}}+s-1\right)}^{2\left(n-1\right)}\\ \times {\left(2n\right)}^r{\left({\displaystyle \frac{sn+s}{2n}}\right)}^{r-1}\left({\displaystyle \frac{ns+s+2nr}{2n}}\right)\times {\left(n+1\right)}^s\times {\left({\displaystyle \frac{3nr+2ns-r}{ns+s+2nr}}\right)}^s\\ \times \left({\displaystyle \frac{n\left(\left(3r+2s\right)+r\left(2r+s\right)\right)+rs-r}{3nr+2ns-r}}+s-1\right)\times {\left({\displaystyle \frac{n\left(\left(3r+2s\right)+r\left(2r+s\right)\right)+rs-r}{3nr+2ns-r}}-1\right)}^{s-1}\\ \times {\left(n\left(s+2\right)+s\right)}^r\times {\left({\displaystyle \frac{1}{n\left(s+2\right)+s}}\right)}^s\times \left(\left(n\left(r+2\right)+r\right)\left(n\left(s+2\right)+s\right)-rs{\left(n-1\right)}^2\right)\\ \times {\left(\left(n\left(r+2\right)+r\right)\left(n\left(s+2\right)+s\right)\right)}^{s-1}\\ ={\displaystyle \frac{1}{4n^2{\left(r+s\right)}^2}}\times {\left(ns+n+s\right)}^{2\left(r-1\right)\left(n-1\right)}\times {\left(nr+n+r\right)}^{2\left(s-1\right)\left(n-1\right)}\times {\left(\left(nr+n+r\right)\left(ns+n+s\right)-rs\right)}^{2\left(n-1\right)}\\ \times {\left(sn+s\right)}^{r-1}\times {\left(n+1\right)}^s\times \left(-r+rs+n\left(\left(3r+2s\right)+r\left(2r+s\right)\right)+\left(s-1\right)\left(3nr+2ns-r\right)\right)\\ \times {\left(r\right)}^{s-1}\times {\left(n\left(s+2\right)+s\right)}^{r-s}\times \left(\left(n\left(r+2\right)+r\right)\left(n\left(s+2\right)+s\right)-rs{\left(n-1\right)}^2\right)\\ \times {\left(\left(n\left(r+2\right)+r\right)\left(n\left(s+2\right)+s\right)\right)}^{s-1}\\ =n^{2\left(n-1\right)}\times {\left(n+1\right)}^{r+s-1}\times r^{s-1}\times s^{r-1}\times {\left(nr+n+r\right)}^{2\left(n-1\right)\left(s-1\right)}\times {\left(ns+n+s\right)}^{2\left(n-1\right)\left(r-1\right)}\times {\left(nr+2n+r\right)}^{\left(s-1\right)}\times \\ {\left(ns+2n+s\right)}^{\left(r-1\right)}\times {\left(r+s+2rs+n\left(1+r\right)\left(1+s\right)\right)}^{2\left(n-1\right)}\left(r+s+2rs+n\left(2+r+s\right)\right)\end{array}$$

□

Complexity of Symmetric Difference in Two Complete Bipartite Graphs

The symmetric difference,$G\oplus H$, of two graphs $G$ and $H$ is the simple graph with $V(G\oplus H)=V\left(G\right)\times V\left(H\right)$, where $(g_1,h_1)$ and $(g_2,h_2)$ are adjacent in $G\oplus H$ if either $g_1$ is adjacent to $g_2$ in $G$ and $h_1$ is not adjacent to $h_2$ in $H$, or $g_1$ is not adjacent to $g_2$ in $G$ and $h_1$ is adjacent to $h_2$ in $H$ [20]. The symmetric difference is commutative, $G\oplus H\cong H\oplus G$ and associative.

Theorem 3.

For $n,r$ and $s$ are all greater than or equal to 1, we get

$$\tau (K_{n,n}\oplus K_{r,s})={\left(n\left(r+s\right)\right)}^{2n \left(\mathrm{r}+\mathrm{s}\right)-2}$$

Proof. 

Applying Lemma 1, we have

$$\tau (K_{n,n}\oplus K_{r,s})={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\mathit{det}\left(\begin{array}{cc}A& B\\ B& A\end{array}\right),$$

where $A$ and $B$ are block matrices of order $n\times n$ made up of matrices of order $(r+s)\times (r+s)$ that are described as follows:

$$\begin{gathered} A=\left(\begin{array}{ccccccc}P& T& \cdots & \cdots & \cdots & \cdots & T\\ T& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & T\\ T& \cdots & \cdots & \cdots & \cdots & T& P\end{array}\right),B=\left(\begin{array}{ccccccc}Q& \cdots & \cdots & \cdots & \cdots & \cdots & Q\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ Q& \cdots & \cdots & \cdots & \cdots & \cdots & Q\end{array}\right), \\ P=\left(\begin{array}{cccccccc}n\left(r+s\right)+1& 1& \cdots & 1& 0& \cdots & \cdots & 0\\ 1& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & n\left(r+s\right)+1& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& n\left(r+s\right)+1& 1& \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 1\\ 0& \cdots & \cdots & 0& 1& \cdots & 1& n\left(r+s\right)+1\end{array}\right), \\ T=\left(\begin{array}{cccccccc}1& \cdots & \cdots & 1& 0& \cdots & \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & 1& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& 1& \cdots & \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & 0& 1& \cdots & \cdots & 1\end{array}\right),Q=\left(\begin{array}{cccccccc}0& \cdots & \cdots & 0& 1& \cdots & \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & 0& 1& \cdots & \cdots & 1\\ 1& \cdots & \cdots & 1& 0& \cdots & \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & 1& 0& \cdots & \cdots & 0\end{array}\right). \end{gathered}$$

Thus, we obtain $\tau (K_{n,n}\oplus K_{r,s})={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times \mathit{det}(A+B)\times \mathit{det}(A-B),$ such that

$$\begin{gathered} A+B=\left(\begin{array}{ccccccc}X& J& \cdots & \cdots & \cdots & \cdots & J\\ J& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & J\\ J& \cdots & \cdots & \cdots & \cdots & J& X\end{array}\right),A-B=\left(\begin{array}{ccccccc}Y& Z& \cdots & \cdots & \cdots & \cdots & Z\\ Z& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & Z\\ Z& \cdots & \cdots & \cdots & \cdots & Z& Y\end{array}\right), \\ X=\left(\begin{array}{cccccccc}n(r+s)+1& 1& \cdots & 1& 1& \cdots & \cdots & 1\\ 1& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & 1& n(r+s)+1& 1& \cdots & \cdots & 1\\ 1& \cdots & \cdots & 1& n(r+s)+1& 1& \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 1\\ 1& \cdots & \cdots & 1& 1& \cdots & 1& n(r+s)+1\end{array}\right) \\ Y=\left(\begin{array}{cccccccc}n(r+s+1)& 1& \cdots & 1& -1& \cdots & \cdots & -1\\ 1& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & 1& n(r+s+1)& -1& \cdots & \cdots & -1\\ -1& \cdots & \cdots & -1& n(r+s+1)& 1& \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 1\\ -1& \cdots & \cdots & -1& 1& \cdots & 1& n(r+s+1)\end{array}\right) \\ \mathrm{and} Z=\left(\begin{array}{cccccccc}1& \cdots & \cdots & 1& -1& \cdots & \cdots & -1\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & 1& -1& \cdots & \cdots & -1\\ -1& \cdots & \cdots & -1& 1& \cdots & \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ -1& \cdots & \cdots & -1& 1& \cdots & \cdots & 1\end{array}\right). \end{gathered}$$

Applying Lemma 3, we obtain

$$\begin{gathered} \tau (K_{n,n}\oplus K_{r,s})={\displaystyle \frac{1}{4n^2(r+s{)}^2}}(\mathit{det}\left(\begin{array}{cccccccc}n(r+s)& 0& \cdots & 0& 0& \cdots & \cdots & 0\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & n(r+s)& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& n(r+s)& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ 0& \cdots & \cdots & 0& 0& \cdots & 0& n(r+s)\end{array}\right){)}^{n-1}\times \\ \mathit{det}\left(\begin{array}{cccccccc}n\left(r+s+1\right)& n& \cdots & n& n& \cdots & \cdots & n\\ n& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & n& ⋮& \ddots & \ddots & ⋮\\ n& \cdots & n& n\left(r+s+1\right)& n& \cdots & \cdots & n\\ n& \cdots & \cdots & n& n\left(r+s+1\right)& n& \cdots & n\\ ⋮& \ddots & \ddots & ⋮& n& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & n\\ n& \cdots & \cdots & n& n& \cdots & n& n\left(r+s+1\right)\end{array}\right) \\ \times (\mathit{det}\left(\begin{array}{cccccccc}n(r+s)& 0& \cdots & 0& 0& \cdots & \cdots & 0\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & n(r+s)& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& n(r+s)& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ 0& \cdots & \cdots & 0& 0& \cdots & 0& n(r+s)\end{array}\right){)}^{n-1}\times \\ \mathit{det}\left(\begin{array}{cccccccc}n(r+s+1)& n& \cdots & n& -n& \cdots & \cdots & -n\\ n& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & n& ⋮& \ddots & \ddots & ⋮\\ n& \cdots & n& n(r+s+1)& -n& \cdots & \cdots & -n\\ -n& \cdots & \cdots & -n& n(r+s+1)& n& \cdots & n\\ ⋮& \ddots & \ddots & ⋮& n& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & n\\ -n& \cdots & \cdots & -n& n& \cdots & 0& n(r+s+1)\end{array}\right). \end{gathered}$$

Applying Lemma 4, we obtain

$$\begin{gathered} \tau (K_{n,n}\oplus K_{r,s})={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times (n(r+s){)}^{2(n-1)(r+s)}\times n^{(r+s)}\times (r+s{)}^{(r+s)-1}\times (2r+2s)\times n^{(r+s)}\times \\ \mathit{det}={\left(\begin{array}{ccccc}r+s+1& 1& \cdots & \cdots & 1\\ 1& \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & 1\\ 1& \cdots & \cdots & 1& r+s+1\end{array}\right)}_{r\times r}\times ({\displaystyle \frac{r+s}{2r+s}}{)}^s\times \\ {\mathit{det}\left(\begin{array}{ccccc}{\displaystyle \frac{r^2+r(r+2s+1)+s(r+s+1)}{r+s}}& 1& \cdots & \cdots & 1\\ 1& \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & 1\\ 1& \cdots & \cdots & 1& {\displaystyle \frac{r^2+r(r+2s+1)+s(r+s+1)}{r+s}}\end{array}\right)}_{s\times s}, \end{gathered}$$

Using Lemma 2, we have

$$\begin{array}{l}\tau (K_{n,n}\oplus K_{r,s})={\displaystyle \frac{1}{4n^2{\left(r+s\right)}^2}}\times {\left(n\left(r+s\right)\right)}^{2\left(r+s\right)\left(n-1\right)}\times n^{r+s}\times 2{\left(r+s\right)}^{r+s}\times n^{r+s}\times {\left(r+s\right)}^{r-1}\times \left(2r+s\right)\times {\left({\displaystyle \frac{r+s}{2r+s}}\right)}^s\times \\ {\left({\displaystyle \frac{r^2+r\left(r+2s+1\right)+s\left(r+s+1\right)}{r+s}}-1\right)}^{s-1}\times \left({\displaystyle \frac{r^2+r\left(r+2s+1\right)+s\left(r+s+1\right)}{r+s}}+s-1\right){=\left(n\left(r+s\right)\right)}^{2 \left(nr+ns-1\right)}.\end{array}$$

□

5. Complexity of Composition Product of Two Complete Bipartite Graphs

The composition, or lexicographic product of two graphs $G$ and $H$ denoted, $G\left[H\right]$ is the simple graph with $V\left(G\right)\times V\left(H\right)$ as the vertex set in which the vertices $(a,b)$ and $(c,d)$ are adjacent if either $a$ is adjacent to $c$ or $a=c$ and $b$ is adjacent to $d$ in $H$ [19]. This operation is non-commutative, $G\left[H\right]\ncong H\left[G\right]$.

One way to view the resulting graph $G\left[H\right]$ is to replace each vertex in $G$ with a copy of $H$ and each edge in $G$ with a complete bipartite graph connecting the corresponding copies of $H.$

Theorem 4.

For $n,r$ and $s$ are all greater than or equal to 1, we get

$$\tau \left(K_{n,n}\left[K_{r,s}\right]\right)=n^{2\left(n-1\right)}\times {\left(n+1\right)}^{2n}\times {\left(r+s\right)}^{2\left(2n-1\right)}\times {\left(n\left(r+s\right)+r\right)}^{2n\left(s-1\right)}\times {\left(n\left(r+s\right)+s\right)}^{2n\left(r-1\right)}$$

Proof. 

Applying Lemma 1, we have

$$\tau \left(K_{n,n}\left[K_{r,s}\right]\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\mathit{det}\left(\begin{array}{cc}A& B\\ B& A\end{array}\right),$$

where $A$ and $B$ are block matrices of order $n\times n$ made up of matrices of order $(r+s)\times (r+s)$ that are described as follows:

$$\begin{gathered} A=\left(\begin{array}{ccccccc}P& J& \cdots & \cdots & \cdots & \cdots & J\\ J& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & J\\ J& \cdots & \cdots & \cdots & \cdots & J& P\end{array}\right),B=\left(\begin{array}{ccccccc}0& \cdots & \cdots & \cdots & \cdots & \cdots & 0\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & \cdots & \cdots & \cdots & 0\end{array}\right), \\ P=\left(\begin{array}{cccccccc}n(r+s)+s+1& 1& \cdots & 1& 0& \cdots & \cdots & 0\\ 1& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & n(r+s)+s+1& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& n(r+s)+r+1& 1& \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 1\\ 0& \cdots & \cdots & 0& 1& \cdots & 1& n(r+s)+r+1\end{array}\right) \end{gathered}$$

Thus, we obtain: $\tau \left(K_{n,n}\left[K_{r,s}\right]\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}(\mathit{det}(A){)}^2$

Applying Lemma 3, we have

$$\begin{gathered} \tau \left(K_{n,n}\left[K_{r,s}\right]\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times (det\left(\begin{array}{cccccccc}n(r+s)+s& 0& \cdots & 0& -1& \cdots & \cdots & -1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & n(r+s)+s& -1& \cdots & \cdots & -1\\ -1& \cdots & \cdots & -1& n(r+s)+r& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ -1& \cdots & \cdots & -1& 0& \cdots & 0& n(r+s)+r\end{array}\right){)}^{2(n-1)}\times \\ (\mathit{det}\left(\begin{array}{cccccccc}n(r+s+1)+s& n& \cdots & n& n-1& \cdots & \cdots & n-1\\ n& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & n& ⋮& \ddots & \ddots & ⋮\\ n& \cdots & n& n(r+s+1)+s& n-1& \cdots & \cdots & n-1\\ n-1& \cdots & \cdots & n-1& n(r+s+1)+r& n& \cdots & n\\ ⋮& \ddots & \ddots & ⋮& n& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & n\\ n-1& \cdots & \cdots & n-1& n& \cdots & n& n(r+s+1)+r\end{array}\right){)}^2. \end{gathered}$$

Using Lemma 4, we obtain

$$\begin{array}{l}\tau \left(K_{n,n}\left[K_{r,s}\right]\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times ({\mathit{det}\left(\begin{array}{cccc}n(r+s)+s& 0& \cdots & 0\\ 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0\\ 0& \cdots & \cdots & n(r+s)+s\end{array}\right)}_{r\times r}{)}^{2(n-1)}\times \\ ({\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{\left(n\right(r+s)+r)\left(n\right(r+s)+s)-r}{n(r+s)+s}}& {\displaystyle \frac{-r}{n(r+s)+s}}& \cdots & {\displaystyle \frac{-r}{n(r+s)+s}}\\ {\displaystyle \frac{-r}{n(r+s)+s}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{-r}{n(r+s)+s}}\\ {\displaystyle \frac{-r}{n(r+s)+s}}& \cdots & {\displaystyle \frac{-r}{n(r+s)+s}}& {\displaystyle \frac{\left(n\right(r+s)+r)\left(n\right(r+s)+s)-r}{n(r+s)+s}}\end{array}\right)}_{s\times s}{)}^{2(n-1)}\times \\ \times (\mathit{det}\left(\begin{array}{cccc}n\left(r+s+1\right)+s& n& \cdots & n\\ n& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & n\\ n& \cdots & n& n\left(r+s+1\right)+s\end{array}\right){)}^2\times \\ (\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{(n+r+n(r+s\left)\right)(2nr+s(n+1\left)\right)-r(n-1{)}^2}{n(2r+s)+s}}& {\displaystyle \frac{n^2(r+s)+n(2r+s)-r}{n(2r+s)+s}}& \cdots & {\displaystyle \frac{n^2(r+s)+n(2r+s)-r}{n(2r+s)+s}}\\ {\displaystyle \frac{n^2(r+s)+n(2r+s)-r}{n(2r+s)+s}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{n^2(r+s)+n(2r+s)-r}{n(2r+s)+s}}\\ {\displaystyle \frac{n^2(r+s)+n(2r+s)-r}{n(2r+s)+s}}& \cdots & {\displaystyle \frac{n^2(r+s)+n(2r+s)-r}{n(2r+s)+s}}& {\displaystyle \frac{(n+r+n(r+s\left)\right)(2nr+s(n+1\left)\right)-r(n-1{)}^2}{n(2r+s)+s}}\end{array}\right){)}^2\times \\ ={\displaystyle \frac{1}{4n^2(r+s{)}^2}}(n(r+s)+s{)}^{2r(n-1)}\times ({\displaystyle \frac{-r}{n(r+s)+s}}{)}^{2s(n-1)}\times \\ ({\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{\left(n\right(r+s)+r)\left(n\right(r+s)+s)-r}{-r}}& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{\left(n\right(r+s)+r)\left(n\right(r+s)+s)-r}{-r}}\end{array}\right)}_{s\times s}{)}^{2(n-1)}\times \\ n^{2r}\times ({\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{n(r+s+1)+s}{n}}& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{n(r+s+1)+s}{n}}\end{array}\right)}_{r\times r}{)}^2\times ({\displaystyle \frac{n^2(r+s)+n(2r+s)-r}{n(2r+s)+s}}{)}^{2s}\\ \times {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{(n+r+n(r+s\left)\right(2nr+s(n+1))-r(n-1{)}^2}{n^2(r+s)+n(2r+s)-r}}& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{(n+r+n(r+s\left)\right(2nr+s(n+1))-r(n-1{)}^2}{n^2(r+s)+n(2r+s)-r}}\end{array}\right)}_{s\times s}\end{array}$$

Applying Lemma 2, we get

$$\begin{array}{l} \tau \left(K_{n,n}\left[K_{r,s}\right]\right)={\displaystyle \frac{1}{4n^2{\left(r+s\right)}^2}}\times {\left(n\left(r+s\right)+s\right)}^{2r\left(n-1\right)}\times {\left(-{\displaystyle \frac{r}{n\left(r+s\right)+s}}\right)}^{2s\left(n-1\right)}\\ \times {\left({\displaystyle \frac{\left(n\left(r+s\right)+r\right)\left(n\left(r+s\right)+s\right)-r}{-r}}-1\right)}^{2\left(n-1\right)\left(s-1\right)}\\ \times {\left({\displaystyle \frac{\left(n\left(r+s\right)+r\right)\left(n\left(r+s\right)+s\right)-r}{-r}}+s-1\right)}^{2\left(n-1\right)}\times n^{2r}\times {\left({\displaystyle \frac{n\left(r+s+1\right)+s}{n}}-1\right)}^{2\left(r-1\right)}\\ \times {\left({\displaystyle \frac{n\left(r+s+1\right)+s}{n}}+r-1\right)}^2\times {\left({\displaystyle \frac{n\left(2r+s\right)+n^2\left(r+s\right)-r}{s+n\left(r+r+s\right)}}\right)}^{2s}\\ \times {\left({\displaystyle \frac{\left(n+r+n\left(r+s\right)\right)\left(2nr+s\left(1+n\right)\right)-r{\left(n-1\right)}^2}{n\left(2r+s\right)+n^2\left(r+s\right)-r}}-1\right)}^{2\left(s-1\right)}\\ \times {\left({\displaystyle \frac{\left(n+r+n\left(r+s\right)\right)\left(2nr+s\left(1+n\right)\right)-r{\left(n-1\right)}^2}{n\left(2r+s\right)+n^2\left(r+s\right)-r}}+s-1\right)}^2\\ ={\displaystyle \frac{1}{4n^2{\left(r+s\right)}^2}}\times {\left(n\left(r+s\right)+s\right)}^{2\left(r-s\right)\left(n-1\right)}\times {\left(\left(n\left(r+s\right)+r\right)\left(n\left(r+s\right)+s\right)\right)}^{2\left(n-1\right)\left(s-1\right)}\times ((n\left(r+s\right)+r)\left(n\left(r+s\right)+s\right)-\\ {rs)}^{2\left(n-1\right)}\times {\left(n\left(r+s\right)+s\right)}^{2\left(r-1\right)}\times {\left(2nr+ns+s\right)}^{2-2s}\times ((n+r+n\left(r+s\right))\left(2nr+s\left(1+n\right)\right)-r{\left(n-1\right)}^2-\\ {\left(n\left(2r+s\right)+n^2\left(r+s\right)-r\right)}^{2\left(s-1\right)}\times ((n+r+n\left(r+s\right)\left(2nr+s\left(1+n\right)\right)-r{\left(n-1\right)}^2+\left(s-1\right)(n\left(2r+s\right)+\\ {n^2\left(r+s\right)-r))}^2\\ =n^{2\left(n-1\right)}\times {\left(n+1\right)}^{2n}\times {\left(r+s\right)}^{2\left(2n-1\right)}\times {\left(n\left(r+s\right)+r\right)}^{2n\left(s-1\right)}\times {\left(n\left(r+s\right)+s\right)}^{2n\left(r-1\right)}.\end{array}$$

□

Theorem 5.

For $\mathrm{n},\mathrm{r}$ and $\mathrm{s}$ are all greater than or equal to 1, we get

$$\tau \left(K_{r,s}\left[K_{n,n}\right]\right)=4^{r+s-1}\times n^{2\left(n r+n s-1\right)}\times r^{s-1}\times s^{r-1}\times {\left(r+1\right)}^s\times {\left(s+1\right)}^r\times {\left(2r+1\right)}^{2s \left(n-1\right)}\times {\left(2s+1\right)}^{2r \left(n-1\right)}$$

Proof. 

Applying Lemma 1, we have

$$\tau \left(K_{r,s}\left[K_{n,n}\right]\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\mathit{det}\left(\begin{array}{cc}A& O\\ O& B\end{array}\right),$$

where $A$ and $B$ are block matrices of order $r\times r$ and $s\times s$, respectively, made up of matrices of order $\left(2n\right)\times \left(2n\right)$ that are described as follows:

$$\begin{gathered} A=\left(\begin{array}{ccccccc}P& J& \cdots & \cdots & \cdots & \cdots & J\\ J& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & J\\ J& \cdots & \cdots & \cdots & \cdots & J& P\end{array}\right),B=\left(\begin{array}{ccccccc}Q& J& \cdots & \cdots & \cdots & \cdots & J\\ J& \ddots & \ddots & \ddots & \ddots & \ddots & J\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & J\\ J& \cdots & \cdots & \cdots & \cdots & J& Q\end{array}\right), \\ P=\left(\begin{array}{cccccccc}n(2s+1)+1& 1& \cdots & 1& 0& \cdots & \cdots & 0\\ 1& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & n(2s+1)+1& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& n(2s+1)+1& 1& \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 1\\ 0& \cdots & \cdots & 0& 1& \cdots & 1& n(2s+1)+1\end{array}\right) \\ Q=\left(\begin{array}{cccccccc}n(2r+1)+1& 1& \cdots & 1& 0& \cdots & \cdots & 0\\ 1& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & n(2r+1)+1& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& n(2r+1)+1& 1& \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 1\\ 0& \cdots & \cdots & 0& 1& \cdots & 1& n(2r+1)+1\end{array}\right) \end{gathered}$$

Thus, we get $\tau \left(K_{r,s}\left[K_{n,n}\right]\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times \mathit{det}(A)\times \mathit{det}(B)$.

Thus, applying Lemma 3, we have

$$\begin{gathered} \tau \left(K_{r,s}\left[K_{n,n}\right]\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times (\mathit{det}\left(\begin{array}{cccccccc}n(2s+1)& 0& \cdots & 0& -1& \cdots & \cdots & -1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & n(2s+1)& -1& \cdots & \cdots & -1\\ -1& \cdots & \cdots & -1& n(2s+1)& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ -1& \cdots & \cdots & -1& 0& \cdots & 0& n(2s+1)\end{array}\right){)}^{(r-1)}\times \\ \mathit{det}\left(\begin{array}{cccccccc}n(2s+1)+r& r& \cdots & r& r-1& \cdots & \cdots & r-1\\ r& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & r& ⋮& \ddots & \ddots & ⋮\\ r& \cdots & r& n(2s+1)+r& r-1& \cdots & \cdots & r-1\\ r-1& \cdots & \cdots & r-1& n(2s+1)+r& r& \cdots & r\\ ⋮& \ddots & \ddots & ⋮& r& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & r\\ r-1& \cdots & \cdots & r-1& r& \cdots & r& n(2s+1)+r\end{array}\right)\times \\ (\mathit{det}\left(\begin{array}{cccccccc}n(2r+1)& 0& \cdots & 0& -1& \cdots & \cdots & -1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & n(2r+1)& -1& \cdots & \cdots & -1\\ -1& \cdots & \cdots & -1& n(2r+1)& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ -1& \cdots & \cdots & -1& 0& \cdots & 0& n(2r+1)\end{array}\right){)}^{(s-1)}\times \\ \mathit{det}\left(\begin{array}{cccccccc}n(2r+1)+s& s& \cdots & s& s-1& \cdots & \cdots & s-1\\ s& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & s& ⋮& \ddots & \ddots & ⋮\\ s& \cdots & s& n(2r+1)+s& s-1& \cdots & \cdots & s-1\\ s-1& \cdots & \cdots & s-1& n(2r+1)+s& s& \cdots & s\\ ⋮& \ddots & \ddots & ⋮& s& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & s\\ s-1& \cdots & \cdots & s-1& s& \cdots & s& n(2r+1)+s\end{array}\right). \end{gathered}$$

Using Lemma 4, we obtain

$$\begin{gathered} \tau \left(K_{r,s}\left[K_{n,n}\right]\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times (n(2s+1){)}^{n\left(r-1\right)}\times \\ (\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{n(2s+1{)}^2-1}{2s+1}}& {\displaystyle \frac{-1}{2s+1}}& \cdots & {\displaystyle \frac{-1}{2s+1}}\\ {\displaystyle \frac{-1}{2s+1}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{-1}{2s+1}}\\ {\displaystyle \frac{-1}{2s+1}}& \cdots & {\displaystyle \frac{-1}{2s+1}}& {\displaystyle \frac{n(2s+1{)}^2-1}{2s+1}}\end{array}\right){)}^{(r-1)}\times \\ r^n\times \mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{n(2s+1)+r}{r}}& 1& \cdots & 1\\ n& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{n(2s+1)+r}{r}}\end{array}\right)\times \\ ({\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{(2ns+n+r)(2s+r+1)-(r-1{)}^2}{2s+r+1}}& {\displaystyle \frac{r(2s+3)-1}{2s+r+1}}& \cdots & {\displaystyle \frac{r(2s+3)-1}{2s+r+1}}\\ {\displaystyle \frac{r(2s+3)-1}{2s+r+1}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{r(2s+3)-1}{2s+r+1}}\\ {\displaystyle \frac{r(2s+3)-1}{2s+r+1}}& \cdots & {\displaystyle \frac{r(2s+3)-1}{2s+r+1}}& {\displaystyle \frac{(2ns+n+r)(2s+r+1)-(r-1{)}^2}{2s+r+1}}\end{array}\right)}_{n\times n}{)}^2\times \\ (n(2r+1){)}^{n(s-1)}\times {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{n(2r+1{)}^2-1}{2r+1}}& {\displaystyle \frac{-1}{2r+1}}& \cdots & {\displaystyle \frac{-1}{2r+1}}\\ {\displaystyle \frac{-1}{2r+1}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{-1}{2r+1}}\\ {\displaystyle \frac{-1}{2r+1}}& \cdots & {\displaystyle \frac{-1}{2r+1}}& {\displaystyle \frac{n(2r+1{)}^2-1}{2r+1}}\end{array}\right)}_{n\times n}{)}^{(s-1)}\times s^n\times \\ \times \mathit{det}{\left(\begin{array}{cccc}{\displaystyle \frac{n(2r+1)+s}{s}}& 1& \cdots & 1\\ n& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{n(2r+1)+s}{s}}\end{array}\right)}_{n\times n}\times \\ {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{(2nr+n+s)(2r+s+1))-(s-1{)}^2}{2r+s+1}}& {\displaystyle \frac{s\left(2r+3\right)-1}{2r+s+1}}& \cdots & {\displaystyle \frac{s\left(2r+3\right)-1}{2r+s+1}}\\ {\displaystyle \frac{s\left(2r+3\right)-1}{2r+s+1}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{s\left(2r+3\right)-1}{2r+s+1}}\\ {\displaystyle \frac{s\left(2r+3\right)-1}{2r+s+1}}& \cdots & {\displaystyle \frac{s\left(2r+3\right)-1}{2r+s+1}}& {\displaystyle \frac{(2nr+n+s)(2r+s+1))-(s-1{)}^2}{2r+s+1}}\end{array}\right)}_{n\times n} \end{gathered}$$

Applying Lemma 2, we obtain

$$\begin{gathered} \begin{array}{l}\tau \left(K_{r,s}\left[K_{n,n}\right]\right)={\displaystyle \frac{1}{4n^2{\left(r+s\right)}^2}}\times {\left(n\left(2s+1\right)\right)}^{n\left(r-1\right)}\times {\left(-{\displaystyle \frac{1}{2s+1}}\right)}^{n\left(r-1\right)}\times {\left({\displaystyle \frac{n{\left(2s+1\right)}^2-1}{-1}}-1\right)}^{\left(n-1\right)\left(r-1\right)}\\ \times {\left({\displaystyle \frac{n{\left(2s+1\right)}^2-1}{-1}}+n-1\right)}^{\left(r-1\right)}\times r^n\times {\left({\displaystyle \frac{n\left(2s+1\right)+r}{r}}-1\right)}^{\left(n-1\right)}\times \left({\displaystyle \frac{n\left(2s+1\right)+r}{r}}+n-1\right)\\ \times {\left({\displaystyle \frac{r\left(2s+3\right)-1}{2s+r+1}}\right)}^n\times {\left({\displaystyle \frac{\left(2ns+n+r\right)\left(2s+r+1\right)-{\left(r-1\right)}^2}{r\left(2s+3\right)-1}}-1\right)}^{\left(n-1\right)}\\ \times \left({\displaystyle \frac{\left(2ns+n+r\right)\left(2s+r+1\right)-{\left(r-1\right)}^2}{r\left(2s+3\right)-1}}+n-1\right)\times {\left(n\left(2r+1\right)\right)}^{n\left(s-1\right)}\times {\left(-{\displaystyle \frac{1}{2r+1}}\right)}^{n\left(s-1\right)}\\ \times {\left({\displaystyle \frac{n{\left(2r+1\right)}^2-1}{-1}}-1\right)}^{\left(n-1\right)\left(s-1\right)}\times {\left({\displaystyle \frac{n{\left(2r+1\right)}^2-1}{-1}}+n-1\right)}^{\left(s-1\right)}\times s^n\times {\left({\displaystyle \frac{n\left(2r+1\right)+s}{s}}-1\right)}^{\left(n-1\right)}\\ \times \left({\displaystyle \frac{n\left(2r+1\right)+s}{s}}+n-1\right)\times {\left({\displaystyle \frac{s\left(2r+3\right)-1}{2r+s+1}}\right)}^n\times {\left({\displaystyle \frac{\left(2nr+n+s\right)\left(2r+s+1\right)-{\left(s-1\right)}^2}{s\left(2r+3\right)-1}}-1\right)}^{\left(n-1\right)}\\ \times \left({\displaystyle \frac{\left(2nr+n+s\right)\left(2r+s+1\right)-{\left(s-1\right)}^2}{s\left(2r+3\right)-1}}+n-1\right)\end{array} \\ ={\displaystyle \frac{1}{4{\left(r+s\right)}^2}}\times {\left(-n\right)}^{n\left(r-1\right)}\times {\left(-n{\left(2s+1\right)}^2\right)}^{\left(n-1\right)\left(r-1\right)}\times {\left(n-n{\left(2s+1\right)}^2\right)}^{\left(r-1\right)}\times {\left(n\left(2s+1\right)\right)}^{\left(n-1\right)}\times \\ {\left(n\left(1+2s\right)\right)}^{\left(n-1\right)}\times \left(\left(2ns+n+r\right)\left(2s+r+1\right)-{\left(r-1\right)}^2+\left(n-1\right)\left(r\left(2s+3\right)-1\right)\right)\times {\left(-n\right)}^{n\left(s-1\right)}\times \\ {\left(-n{\left(2r+1\right)}^2\right)}^{\left(n-1\right)\left(s-1\right)}\times {\left(n-n{\left(2r+1\right)}^2\right)}^{\left(s-1\right)}\times {\left(n\left(2r+1\right)\right)}^{\left(n-1\right)}\times {\left(n\left(1+2r\right)\right)}^{\left(n-1\right)}\times \\ \left(\left(2nr+n+s\right)\left(2r+s+1\right)-{\left(s-1\right)}^2+\left(n-1\right)\left(s\left(2r+3\right)-1\right)\right) \\ {=4}^{r+s-1}\times n^{2\left(n r+n s-1\right)}\times r^{s-1}\times s^{r-1}\times {\left(r+1\right)}^s\times {\left(s+1\right)}^r\times {\left(2r+1\right)}^{2s\left(n-1\right)}\times {\left(2s+1\right)}^{2r\left(n-1\right)} \end{gathered}$$

□

6. Complexity of Disjunction of Two Complete Bipartite Graphs

The disjunction $G\vee H$ of two graphs $G$ and $H$ is the simple graph with $V(G\vee H)=V\left(G\right)\times V\left(H\right)$ where $(g_1,h_1)$ and $(g_2,h_2)$ are adjacent in $G\vee H$ if $g_1$ is adjacent to $g_2$ in $G$ or $h_1$ is adjacent to $h_2$ in $H$ [20]. This operation is commutative, $G\vee H\cong H\vee G$.

Theorem 6.

For $n,r$ and $s$ are all greater than or equal to 1, we get

$$\tau \left(K_{n,n}\vee K_{r,s}\right)=4(r+s{)}^2\times n^{2n\left(r+s\right)-2}\times {\left(2s+r\right)}^{2\left(nr-1\right)}\times {\left(2r+s\right)}^{2\left(ns-1\right)}$$

Proof. 

Applying Lemma 1, we have

$$\tau \left(K_{n,n}\vee K_{r,s}\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\mathit{det}\left(\begin{array}{cc}A& O\\ O& A\end{array}\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}(\mathit{det}A{)}^2,$$

where $A$ is a block matrix of order $n\times n$ made up of matrices of order $(r+s)\times (r+s)$ that are described as follows:

$$\begin{gathered} A=\left(\begin{array}{ccccccc}P& Q& \cdots & \cdots & \cdots & \cdots & Q\\ Q& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & Q\\ Q& \cdots & \cdots & \cdots & \cdots & Q& P\end{array}\right), \\ P=\left(\begin{array}{cccccccc}n\left(r+2s\right)+1& 1& \cdots & 1& 0& \cdots & \cdots & 0\\ 1& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & n\left(r+2s\right)+1& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& n\left(2r+s\right)+1& 1& \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 1\\ 0& \cdots & \cdots & 0& 1& \cdots & 1& n\left(2r+s\right)+1\end{array}\right) \\ Q=\left(\begin{array}{cccccccc}1& \cdots & \cdots & 1& 0& \cdots & \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& \ddots & \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & 1& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& 1& \cdots & \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & 0& 1& \cdots & \cdots & 1\end{array}\right). \end{gathered}$$

Applying Lemma 3, we obtain $\tau \left(K_{n,n}\vee K_{r,s}\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}$

$$\begin{gathered} \times (\mathit{det}\left(\begin{array}{cccccccc}n(r+2s)& 0& \cdots & 0& 0& \cdots & \cdots & 0\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & n(r+2s)& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& n(2r+s)& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ 0& \cdots & \cdots & 0& 0& \cdots & 0& n(2r+s)\end{array}\right){)}^{2(n-1)}\times \\ (\mathit{det}\left(\begin{array}{cccccccc}n(r+2s+1)& n& \cdots & n& 0& \cdots & \cdots & 0\\ n& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & n& ⋮& \ddots & \ddots & ⋮\\ n& \cdots & n& n(r+2s+1)& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& n(2r+s+1)& n& \cdots & n\\ ⋮& \ddots & \ddots & ⋮& n& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & n\\ 0& \cdots & \cdots & 0& n& \cdots & n& n(2r+s+1)\end{array}\right){)}^2 \\ ={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times n^{2(r+s)(n-1)}\times (2s+r{)}^{2r(n-1)}\times (2r+s{)}^{2s(n-1)}\times n^{2r}\times \\ ({\mathit{det}\left(\begin{array}{cccc}r+2s+1& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& r+2s+1\end{array}\right)}_{r\times r}{)}^2\times n^{2s}\times ({\mathit{det}\left(\begin{array}{cccc}2r+s+1& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& 2r+s+1\end{array}\right)}_{s\times s}{)}^2. \end{gathered}$$

Applying Lemma 2, we get

$$\begin{gathered} \tau \left(K_{n,n}\vee K_{r,s}\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times n^{2(r+s)(n-1)}\times (2s+r{)}^{2r(n-1)}\times (2r+s{)}^{2s(n-1)}\times n^{2r}\times (r+2s{)}^{2(r-1)}\times (2r \\ +2s{)}^2\times n^{2s}\times (2r+s{)}^{2(s-1)}\times (2r+2s{)}^2 \\ =4(r+s{)}^2\times n^{2n\left(r+s\right)-2}\times {\left(2s+r\right)}^{2\left(nr-1\right)}\times {\left(2r+s\right)}^{2\left(ns-1\right)}. \end{gathered}$$

□

7. Complexity of Strong Sum of Two Complete Bipartite Graphs

The strong sum $G\ast H$ of two graphs $G$ and $H$ is the simple graph with $V(G\ast H)=V\left(G\right)\times V\left(H\right)$ where $(g_1,h_1)$ and $(g_2,h_2)$ are adjacent in $G\ast H$ if $h_1$ is adjacent to $h_2$ in $H$ and either $g_1$ is adjacent to $g_2$ in $G$, or $g_1=g_2$ [20]. This operation is non-commutative, $G\ast H\ncong H\ast G$.

Theorem 7.

For $n,r$ and $s$ are all greater than or equal to 1, we get

$$\tau \left(K_{n,n}\ast K_{r,s}\right)=2n^{2\left(n-1\right)}\times {\left(n+1\right)}^{2n\left(r+s-2\right)+1}\times {\left(n+2\right)}^{2\left(n-1\right)}\times r^{2ns-1}\times s^{2nr-1}.$$

Proof. 

By using Lemma 1, we have

$$\tau (K_{n,n}\ast K_{r,s})={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times \mathit{det}\left(\begin{array}{cc}A& B\\ B& A\end{array}\right),$$

where $A$ and $B$ are block matrices of order $n\times n$ made up of matrices of order $(r+s)\times (r+s)$ that are described as follows:

$$\begin{gathered} A=\left(\begin{array}{ccccccc}P& J& \cdots & \cdots & \cdots & \cdots & J\\ J& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & J\\ J& \cdots & \cdots & \cdots & \cdots & J& P\end{array}\right),B=\left(\begin{array}{ccccccc}Q& \cdots & \cdots & \cdots & \cdots & \cdots & Q\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ Q& \cdots & \cdots & \cdots & \cdots & \cdots & Q\end{array}\right), \\ P=\left(\begin{array}{cccccccc}s\left(n+1\right)+1& 1& \cdots & 1& 0& \cdots & \cdots & 0\\ 1& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & 1& s\left(n+1\right)+1& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& r\left(n+1\right)+1& 1& \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 1\\ 0& \cdots & \cdots & 0& 1& \cdots & 1& r\left(n+1\right)+1\end{array}\right), \\ Q=\left(\begin{array}{cccccccc}1& \cdots & \cdots & 1& 0& \cdots & \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & 1& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& 1& \cdots & \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & 0& 1& \cdots & \cdots & 1\end{array}\right). \end{gathered}$$

Thus, we have $\tau (K_{n,n}\ast K_{r,s})={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times \mathit{det}(A+B)\times \mathit{det}(A-B).$ Such that

$$\begin{gathered} A+B=\left(\begin{array}{ccccccc}X& Y& \cdots & \cdots & \cdots & \cdots & Y\\ Y& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & Y\\ Y& \cdots & \cdots & \cdots & \cdots & Y& X\end{array}\right),A-B=\left(\begin{array}{ccccccc}Z& W& \cdots & \cdots & \cdots & \cdots & W\\ W& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & W\\ W& \cdots & \cdots & \cdots & \cdots & W& Z\end{array}\right), \\ X=\left(\begin{array}{cccccccc}s\left(n+1\right)+2& 2& \cdots & 2& 0& \cdots & \cdots & 0\\ 2& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 2& ⋮& \ddots & \ddots & ⋮\\ 2& \cdots & 2& s\left(n+1\right)+2& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& r\left(n+1\right)+2& 2& \cdots & 2\\ ⋮& \ddots & \ddots & ⋮& 2& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 2\\ 0& \cdots & \cdots & 0& 2& \cdots & 2& r\left(n+1\right)+2\end{array}\right), \\ Y=\left(\begin{array}{cccccccc}2& \cdots & \cdots & 2& 1& \cdots & \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 2& \cdots & \cdots & 2& 1& \cdots & \cdots & 1\\ 1& \cdots & \cdots & 1& 2& \cdots & \cdots & 2\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & 1& 2& \cdots & \cdots & 2\end{array}\right),W=\left(\begin{array}{cccccccc}0& \cdots & \cdots & 0& 1& \cdots & \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & 0& 1& \cdots & \cdots & 1\\ 1& \cdots & \cdots & 1& 0& \cdots & \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & 1& 0& \cdots & \cdots & 0\end{array}\right) \end{gathered}$$

 and

$$Z=\left(\begin{array}{cccccccc}s(n+1)& 0& \cdots & 0& 0& \cdots & \cdots & 0\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & 0& s(n+1)& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& r(n+1)& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ 0& \cdots & \cdots & 0& 0& \cdots & 0& r(n+1)\end{array}\right).$$

Applying Lemma 3, we obtain $\tau \left(K_{n,n}\ast K_{r,s}\right)=$

$$\begin{gathered} {\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times (\mathit{det}\left(\begin{array}{cccccccc}s(n+1)& 0& \cdots & 0& -1& \cdots & \cdots & -1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & s(n+1)& -1& \cdots & \cdots & -1\\ -1& \cdots & \cdots & -1& r(n+1)& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ -1& \cdots & \cdots & -1& 0& \cdots & 0& r(n+1)\end{array}\right){)}^{n-1}\times \\ \mathit{det}\left(\begin{array}{cccccccc}s\left(n+1\right)+2n& 2n& \cdots & 2n& n-1& \cdots & \cdots & n-1\\ 2n& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 2n& ⋮& \ddots & \ddots & ⋮\\ 2n& \cdots & 2n& s\left(n+1\right)+2n& n-1& \cdots & \cdots & n-1\\ n-1& \cdots & \cdots & n-1& r\left(n+1\right)+2n& 2n& \cdots & 2n\\ ⋮& \ddots & \ddots & ⋮& 2n& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 2n\\ n-1& \cdots & \cdots & n-1& 2n& \cdots & 2n& r\left(n+1\right)+2n\end{array}\right) \\ \times (\mathit{det}\left(\begin{array}{cccccccc}s\left(n+1\right)& 0& \cdots & 0& -1& \cdots & \cdots & -1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & s\left(n+1\right)& -1& \cdots & \cdots & -1\\ -1& \cdots & \cdots & -1& r\left(n+1\right)& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ -1& \cdots & \cdots & -1& 0& \cdots & 0& r\left(n+1\right)\end{array}\right){)}^{n-1}\times \\ \mathit{det}\left(\begin{array}{cccccccc}s\left(n+1\right)& 0& \cdots & 0& n-1& \cdots & \cdots & n-1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & 0& s\left(n+1\right)& n-1& \cdots & \cdots & n-1\\ n-1& \cdots & \cdots & n-1& r\left(n+1\right)& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ n-1& \cdots & \cdots & n-1& 0& \cdots & 0& r\left(n+1\right)\end{array}\right). \end{gathered}$$

Using Lemma 4, we get

$$\begin{gathered} \begin{array}{l}\tau (K_{n,n}\ast K_{r,s})={\displaystyle \frac{1}{4n^2(r+s{)}^2}}({\mathit{det}\left(\begin{array}{cccc}s(n+1)& 0& \cdots & 0\\ 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0\\ 0& \cdots & 0& s(n+1)\end{array}\right)}_{r\times r}{)}^{(n-1)}\times \\ ({\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{rs{(n+1)}^2-r}{s(n+1)}}& {\displaystyle \frac{-r}{s(n+1)}}& \cdots & {\displaystyle \frac{-r}{s(n+1)}}\\ {\displaystyle \frac{-r}{s(n+1)}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{-r}{s(n+1)}}\\ {\displaystyle \frac{-r}{s(n+1)}}& \cdots & {\displaystyle \frac{-r}{s(n+1)}}& {\displaystyle \frac{rs{(n+1)}^2-r}{s(n+1)}}\end{array}\right)}_{s\times s}{)}^{(n-1)}{\times \mathit{det}\left(\begin{array}{cccc}s\left(n+1\right)+2n& 2n& \cdots & 2n\\ 2n& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 2n\\ 2n& \cdots & 2n& s\left(n+1\right)+2n\end{array}\right)}_{r\times r}\times \\ {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{\left(n+1\right)(-r+rs+n\left(3r+2s+r\left(2r+s\right))\right)}{s+n\left(2r+s\right)}}& {\displaystyle \frac{\left(n+1\right)\left(-r+n(3r+2s\right))}{s+n\left(2r+s\right)}}& \cdots & {\displaystyle \frac{\left(n+1\right)\left(-r+n(3r+2s\right))}{s+n\left(2r+s\right)}}\\ {\displaystyle \frac{\left(n+1\right)\left(-r+n(3r+2s\right))}{s+n\left(2r+s\right)}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{\left(n+1\right)\left(-r+n(3r+2s\right))}{s+n\left(2r+s\right)}}\\ {\displaystyle \frac{\left(n+1\right)\left(-r+n(3r+2s\right))}{s+n\left(2r+s\right)}}& \cdots & {\displaystyle \frac{\left(n+1\right)\left(-r+n(3r+2s\right))}{s+n\left(2r+s\right)}}& {\displaystyle \frac{\left(n+1\right)(-r+rs+n\left(3r+2s+r\left(2r+s\right))\right)}{s+n\left(2r+s\right)}}\end{array}\right)}_{s\times s}\end{array} \\ \times ({\mathit{det}\left(\begin{array}{cccc}s(n+1)& 0& \cdots & 0\\ 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0\\ 0& \cdots & 0& s(n+1)\end{array}\right)}_{r\times r}{)}^{(n-1)}\times \\ ({\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{rs{(n+1)}^2-r}{s(n+1)}}& {\displaystyle \frac{-r}{s(n+1)}}& \cdots & {\displaystyle \frac{-r}{s(n+1)}}\\ {\displaystyle \frac{-r}{s(n+1)}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{-r}{s(n+1)}}\\ {\displaystyle \frac{-r}{s(n+1)}}& \cdots & {\displaystyle \frac{-r}{s(n+1)}}& {\displaystyle \frac{rs{(n+1)}^2-r}{s(n+1)}}\end{array}\right)}_{s\times s}{)}^{(n-1)}\times \\ {\mathit{det}\left(\begin{array}{cccc}s(n+1)& 0& \cdots & 0\\ 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0\\ 0& \cdots & 0& s(n+1)\end{array}\right)}_{r\times r}\times \\ {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{rs{(n+1)}^2-r{(n-1)}^2}{s(n+1)}}& {\displaystyle \frac{-r{(n-1)}^2}{s(n+1)}}& \cdots & {\displaystyle \frac{-r{(n-1)}^2}{s(n+1)}}\\ {\displaystyle \frac{-r{(n-1)}^2}{s(n+1)}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{-r{(n-1)}^2}{s(n+1)}}\\ {\displaystyle \frac{-r{(n-1)}^2}{s(n+1)}}& \cdots & {\displaystyle \frac{-r{(n-1)}^2}{s(n+1)}}& {\displaystyle \frac{rs{(n+1)}^2-r{(n-1)}^2}{s(n+1)}}\end{array}\right)}_{s\times s} \\ \begin{array}{l}={\displaystyle \frac{1}{4n^2(r+s{)}^2}}[(s(n+1){]}^{r(n-1)}\times ({\displaystyle \frac{-r}{s(n+1)}}{)}^{s(n-1)}\times \\ ({\mathit{det}\left(\begin{array}{cccc}1-s{(n+1)}^2& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& 1-s{(n+1)}^2\end{array}\right)}_{s\times s}{)}^{(n-1)}\times {\left(2n\right)}^r\times {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{s\left(n+1\right)+2n}{2n}}& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{s\left(n+1\right)+2n}{2n}}\end{array}\right)}_{r\times r}\times \end{array} \\ ({\displaystyle \frac{(n+1)(-r+n(3r+2s\left)\right)}{s+n(2r+s)}}{)}^s\times \\ {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{-r+rs+n\left(3r+2s+r\left(2r+s\right)\right)}{-r+n(3r+2s)}}& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{-r+rs+n\left(3r+2s+r\left(2r+s\right)\right)}{-r+n(3r+2s)}}\end{array}\right)}_{s\times s}\times \\ \left(s\right(n+1){)}^{r(n-1)}\times ({\displaystyle \frac{-r}{s(n+1)}}{)}^{s(n-1)}\times {\mathit{det}\left(\begin{array}{cccc}1-s{(n+1)}^2& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& 1-s{(n+1)}^2\end{array}\right)}_{s\times s}.\times \\ \left(s\right(n+1){)}^r\times ({\displaystyle \frac{-r{(n-1)}^2}{s(n+1)}}{)}^s\times {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{{(n-1)}^2-s{(n+1)}^2}{{(n-1)}^2}}& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{{(n-1)}^2-s{(n+1)}^2}{{(n-1)}^2}}\end{array}\right)}_{s\times s} \end{gathered}$$

Applying Lemma 2, we have:

$$\begin{array}{l}\tau (K_{n,n}\ast K_{r,s})={\displaystyle \frac{1}{4n^2{\left(r+s\right)}^2}}\times {\left(s\left(n+1\right)\right)}^{r\left(n-1\right)}\times {\left({\displaystyle \frac{-r}{s\left(n+1\right)}}\right)}^{s\left(n-1\right)}\times {\left(s-s{\left(n+1\right)}^2\right)}^{\left(n-1\right)}\times {\left(-s{\left(n+1\right)}^2\right)}^{\left(n-1\right)\left(s-1\right)}\\ \times {\left(2n\right)}^r\times \left({\displaystyle \frac{s\left(n+1\right)+2n}{2n}}+r-1\right)\times {\left({\displaystyle \frac{s\left(n+1\right)+2n}{2n}}-1\right)}^{r-1}\times {\left({\displaystyle \frac{\left(n+1\right)\left(-r+n\left(3r+2s\right)\right)}{s+n\left(2r+s\right)}}\right)}^s\\ \times \left({\displaystyle \frac{-r+rs+n\left(3r+2s+r\left(2r+s\right)\right)}{n\left(3r+2s\right)-r}}+s-1\right)\times {\left({\displaystyle \frac{-r+rs+n\left(3r+2s+r\left(2r+s\right)\right)}{n\left(3r+2s\right)-r}}-1\right)}^{s-1}\\ \times {\left(s\left(n+1\right)\right)}^{r\left(n-1\right)}\times {\left({\displaystyle \frac{-r}{s\left(n+1\right)}}\right)}^{s\left(n-1\right)}\times {\left(s-s{\left(n+1\right)}^2\right)}^{\left(n-1\right)}\times {\left(-s{\left(n+1\right)}^2\right)}^{\left(n-1\right)\left(s-1\right)}\\ \times {\left(s\left(n+1\right)\right)}^r\times {\left({\displaystyle \frac{-r{\left(n-1\right)}^2}{s\left(n+1\right)}}\right)}^s\times {\left({\displaystyle \frac{{\left(n-1\right)}^2-s{\left(n+1\right)}^2}{{\left(n-1\right)}^2}}-1\right)}^{s-1}\\ \times \left({\displaystyle \frac{{\left(n-1\right)}^2-s{\left(n+1\right)}^2}{{\left(n-1\right)}^2}}+s-1\right)\\ ={\displaystyle \frac{1}{4n^2{\left(r+s\right)}^2}}\times {\left(s\left(n+1\right)\right)}^{\left(r-s\right)\left(n-1\right)}\times {\left(-r\right)}^{s\left(n-1\right)}\times {\left(s-s{\left(n+1\right)}^2\right)}^{\left(n-1\right)}\times {\left(-s{\left(n+1\right)}^2\right)}^{\left(n-1\right)\left(s-1\right)}\times {\left(s\left(n+1\right)\right)}^{r-1}\times \\ {\left(n+1\right)}^s\times \left(-r+rs+n\left(3r+2s+r\left(2r+s\right)\right)+\left(s-1\right)\left(n\left(3r+2s\right)-r\right)\right)\times {\left(r\right)}^{s-1}\times {\left(s\left(n+1\right)\right)}^{r\left(n-1\right)}\times \\ {\left(s-s{\left(n+1\right)}^2\right)}^{\left(n-1\right)}\times {\left(-s{\left(n+1\right)}^2\right)}^{\left(n-1\right)\left(s-1\right)}{\left(s\left(n+1\right)\right)}^{r-ns}\times {\left(-r\right)}^{ns}\times {\left({\left(n-1\right)}^2-s{\left(n+1\right)}^2-{\left(n-1\right)}^2\right)}^{s-1}\times \\ \left({\left(n-1\right)}^2-s{\left(n+1\right)}^2+\left(s-1\right){\left(n-1\right)}^2\right.\\ =2n^{2\left(n-1\right)}\times {\left(n+1\right)}^{2n\left(r+s-2\right)+1}\times {\left(n+2\right)}^{2\left(n-1\right)}\times r^{2ns-1}\times s^{2nr-1}.\end{array}$$

□

Theorem 8.

For $\mathrm{n},\mathrm{r}$ and $\mathrm{s}$ are all greater than or equal to 1, we get

$$\tau \left(K_{r,s}{\ast K}_{n,n}\right)=n^{2n\left(r+s\right)-2}\times r^{s-1}\times s^{r-1}\times {\left(r+1\right)}^{2s\left(n-1\right)}\times {\left(s+1\right)}^{2r\left(n-1\right)}\times {\left(r+2\right)}^{s-1}\times {\left(s+2\right)}^{r-1}\times \left(r+s+2\right)$$

Proof. 

By using Lemma 1, we have

$$\tau (K_{r,s}\ast K_{n,n})={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times \mathit{det}\left(\begin{array}{cc}A& C\\ C^T& B\end{array}\right),$$

where $A$, $B$, and $C$ are block matrices of order $r\times r$, $s\times s$, and $s\times s,$ respectively made up of matrices of order $\left(2n\right)\times \left(2n\right)$, which are described as follows:

$$\begin{gathered} A=\left(\begin{array}{ccccccc}P& J& \cdots & \cdots & \cdots & J& J\\ J& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ J& \ddots & \ddots & \ddots & \ddots & \ddots & J\\ J& J& \cdots & \cdots & \cdots & J& P\end{array}\right),B=\left(\begin{array}{ccccccc}Q& J& \cdots & \cdots & \cdots & J& J\\ J& \ddots & \ddots & \ddots & \ddots & \ddots & J\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ J& \ddots & \ddots & \ddots & \ddots & \ddots & J\\ J& J& \cdots & \cdots & \cdots & J& Q\end{array}\right), \\ \mathrm{and} C=\left(\begin{array}{ccccccc}W& \cdots & \cdots & \cdots & \cdots & \cdots & W\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ W& \cdots & \cdots & \cdots & \cdots & \cdots & W\end{array}\right), \mathrm{where} \\ P=\left(\begin{array}{cccccccc}n\left(s+1\right)+1& 1& \cdots & 1& 0& \cdots & \cdots & 0\\ 1& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & 1& n\left(s+1\right)+1& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& n\left(s+1\right)+1& 1& \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 1\\ 0& \cdots & \cdots & 0& 1& \cdots & 1& n\left(s+1\right)+1\end{array}\right), \\ Q=\left(\begin{array}{cccccccc}n\left(r+1\right)+1& \cdots & \cdots & 1& 0& \cdots & \cdots & 0\\ 1& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & n\left(r+1\right)+1& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& n\left(r+1\right)+1& \cdots & \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & 0& 1& \cdots & \cdots & n\left(r+1\right)+1\end{array}\right) \mathrm{and} \\ W=\left(\begin{array}{cccccccc}1& \cdots & \cdots & 1& 0& \cdots & \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 1& \cdots & \cdots & 1& 0& \cdots & \cdots & 0\\ 0& \cdots & \cdots & 0& 1& \cdots & \cdots & 1\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & 0& 1& \cdots & \cdots & 1\end{array}\right) \end{gathered}$$

Applying Lemma 4, we have

$$\tau \left(K_{r,s}\ast K_{n,n}\right)={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times \mathit{det}\left(A\right)\times \mathit{det}\left(D\right),$$

such that $D=B-C^T{A}^{-1}\mathrm{C}$ is a block matrix of order $s\times s$ made up of matrices of order $\left(2n\right)\times \left(2n\right)$ described as follows:

$$\begin{gathered} D=\left(\begin{array}{ccccccc}E& F& \cdots & \cdots & \cdots & \cdots & F\\ F& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & F\\ F& \cdots & \cdots & \cdots & \cdots & F& E\end{array}\right), \\ E=\left(\begin{array}{cccccccc}a& b& \cdots & b& c& \cdots & \cdots & c\\ b& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & b& ⋮& \ddots & \ddots & ⋮\\ b& \cdots & b& a& c& \cdots & \cdots & c\\ c& \cdots & \cdots & c& a& b& \cdots & b\\ ⋮& \ddots & \ddots & ⋮& b& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & b\\ c& \cdots & \cdots & c& b& \cdots & b& a\end{array}\right) \mathrm{a}\mathrm{n}\mathrm{d} F=\left(\begin{array}{cccccccc}b& \cdots & \cdots & b& d& \cdots & \cdots & d\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ b& \cdots & \cdots & b& d& \cdots & \cdots & d\\ d& \cdots & \cdots & d& b& \cdots & \cdots & b\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ d& \cdots & \cdots & d& b& \cdots & \cdots & b\end{array}\right). \end{gathered}$$

Here,

$$\begin{gathered} a={\displaystyle \frac{\left(s+2\right) \left(2 r+s\right) \left(1+n \left(r+1\right)\right)-r \left(r+s+1\right)}{\left(s+2\right) \left(2r+s\right)}},b={\displaystyle \frac{\left(s+2\right) \left(2r+s\right)-r \left(r+s+1\right)}{\left(s+2\right) \left(2r+s\right)}}, \\ c={\displaystyle \frac{r \left(r-1\right)}{\left(s+2\right) \left(2r+s\right)}} \mathrm{and} d={\displaystyle \frac{\left(r^2+r-1\right)+2r \left(1+s\right)+{\left(1+s\right)}^2}{\left(2+s\right) \left(2r+s\right)}}. \end{gathered}$$

Using Lemma 3, we obtain $\tau \left(K_{r,s}\ast K_{n,n}\right)={\displaystyle \frac{1}{4 n^2 (r+s{)}^2}}\times$

$$\begin{gathered} (\mathit{det}\left(\begin{array}{cccccccc}n(s+1)& 0& \cdots & 0& -1& \cdots & \cdots & -1\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & n(s+1)& -1& \cdots & \cdots & -1\\ -1& \cdots & \cdots & -1& n(s+1)& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ -1& \cdots & \cdots & -1& 0& \cdots & 0& n(s+1)\end{array}\right){)}^{r-1}\times \\ \mathit{det}\left(\begin{array}{cccccccc}n\left(s+1\right)+r& r& \cdots & r& r-1& \cdots & \cdots & r-1\\ r& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & r& ⋮& \ddots & \ddots & ⋮\\ r& \cdots & r& n\left(s+1\right)+r& r-1& \cdots & \cdots & r-1\\ r-1& \cdots & \cdots & r-1& n\left(s+1\right)+r& r& \cdots & r\\ ⋮& \ddots & \ddots & ⋮& r& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & r\\ r-1& \cdots & \cdots & r-1& r& \cdots & r& n\left(s+1\right)+r\end{array}\right) \\ \times (\mathit{det}\left(\begin{array}{cccccccc}a-b& 0& \cdots & 0& c-d& \cdots & \cdots & c-d\\ 0& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 0& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & \cdots & a-b& c-d& \cdots & \cdots & c-d\\ c-d& \cdots & \cdots & c-d& a-b& 0& \cdots & 0\\ ⋮& \ddots & \ddots & ⋮& 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & 0\\ c-d& \cdots & \cdots & c-d& 0& \cdots & 0& a-b\end{array}\right){)}^{s-1}\times \\ \mathit{det}\left(\begin{array}{cccccccc}a+b(s-1)& bs& \cdots & bs& c+d(s-1)& \cdots & \cdots & c+d(s-1)\\ bs& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & bs& ⋮& \ddots & \ddots & ⋮\\ bs& \cdots & bs& a+b(s-1)& c+d(s-1)& \cdots & \cdots & c+d(s-1)\\ c+d(s-1)& \cdots & \cdots & c+d(s-1)& a+b(s-1)& bs& \cdots & bs\\ ⋮& \ddots & \ddots & ⋮& bs& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮& ⋮& \ddots & \ddots & bs\\ c+d(s-1)& \cdots & \cdots & c+d(s-1)& bs& \cdots & bs& a+b(s-1)\end{array}\right). \end{gathered}$$

Using Lemma 4, we get

$$\begin{gathered} \tau (K_{r,s}\ast K_{n,n})={\displaystyle \frac{1}{4n^2(r+s{)}^2}}({\mathit{det}\left(\begin{array}{cccc}s\left(n+1\right)-1& -1& \cdots & -1\\ -1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & -1\\ -1& \cdots & -1& s\left(n+1\right)-1\end{array}\right)}_{n\times n}{)}^{(r-1)}\times \\ ({\mathit{det}\left(\begin{array}{cccc}s\left(n+1\right)+1& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& s\left(n+1\right)+1\end{array}\right)}_{n\times n}{)}^{(r-1)}\times \\ {\mathit{det}\left(\begin{array}{cccc}(n\left(s+1\right)+2r-1)& 2r-1& \cdots & 2r-1\\ 2r-1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 2r-1\\ 2r-1& \cdots & 2r-1& n(n\left(s+1\right)+2r-1)\end{array}\right)}_{n\times n} \\ {\mathit{det}\left(\begin{array}{cccc}n\left(s+1\right)+1& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& n\left(s+1\right)+1\end{array}\right)}_{n\times n}\times \\ \times ({\mathit{det}\left(\begin{array}{cccc}\left(a-b\right)+(c-d)& (c-d)& \cdots & (c-d)\\ (c-d)& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & (c-d)\\ (c-d)& \cdots & (c-d)& \left(a-b\right)+(c-d)\end{array}\right)}_{n\times n}{)}^{(s-1)}\times \\ ({\mathit{det}\left(\begin{array}{cccc}\left(a-b\right)-(c-d)& d-c& \cdots & d-c\\ d-c& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & d-c\\ d-c& \cdots & d-c& \left(a-b\right)-(c-d)\end{array}\right)}_{n\times n}{)}^{(s-1)}\times \\ {\mathit{det}\left(\begin{array}{cccc}\left(a+c\right)+(s-1)\left(b+d\right)& s\left(b+d\right)+(c-d)& \cdots & s\left(b+d\right)+(c-d)\\ s\left(b+d\right)+(c-d)& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & s\left(b+d\right)+(c-d)\\ s\left(b+d\right)+(c-d)& \cdots & s\left(b+d\right)+(c-d)& \left(a+c\right)+(s-1)\left(b+d\right)\end{array}\right)}_{n\times n}\times \\ {\mathit{det}\left(\begin{array}{cccc}\left(a-c\right)+(s-1)\left(b-d\right)& s\left(b-d\right)-(c-d)& \cdots & s\left(b-d\right)-(c-d)\\ s\left(b-d\right)-(c-d)& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & s\left(b-d\right)-(c-d)\\ s\left(b-d\right)-(c-d)& \cdots & s\left(b-d\right)-(c-d)& \left(a-c\right)+(s-1)\left(b-d\right)\end{array}\right)}_{n\times n} \\ ={\displaystyle \frac{1}{4n^2(r+s{)}^2}}\times \\ ({\mathit{det}\left(\begin{array}{cccc}n(s+1)-1& -1& \cdots & -1\\ -1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & -1\\ -1& \cdots & -1& n(s+1)-1\end{array}\right)}_{n\times n}{)}^{(r-1)}\times ({\mathit{det}\left(\begin{array}{cccc}n\left(s+1\right)+1& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& n\left(s+1\right)+1\end{array}\right)}_{n\times n}{)}^{(r-1)}\times \\ (2r-1{)}^n\times \\ {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{n\left(s+1\right)+2r-1}{2r-1}}& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& {\displaystyle \frac{n\left(s+1\right)+2r-1}{2r-1}}\end{array}\right)}_{n\times n}\times \\ {\mathit{det}\left(\begin{array}{cccc}n\left(s+1\right)+1& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& n\left(s+1\right)+1\end{array}\right)}_{n\times n}.\times \\ ({\mathit{det}\left(\begin{array}{cccc}(nr+n-1)& -1& \cdots & -1\\ -1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & -1\\ -1& \cdots & -1& (nr+n-1)\end{array}\right)}_{n\times n}{)}^{\left(s-1\right)}\times ({\mathit{det}\left(\begin{array}{cccc}(nr+n+1)& 1& \cdots & 1\\ 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & 1\\ 1& \cdots & 1& (nr+n+1)\end{array}\right)}_{n\times n}{)}^{(s-1)}\times \\ {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{\left(2r+s\right)\left(n+nr+2s-1\right)-rs}{2r+s}}& {\displaystyle \frac{s\left(3r+2s\right)-(2r+s)}{2r+s}}& \cdots & {\displaystyle \frac{s\left(3r+2s\right)-(2r+s)}{2r+s}}\\ {\displaystyle \frac{s\left(3r+2s\right)-(2r+s)}{2r+s}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{s\left(3r+2s\right)-(2r+s)}{2r+s}}\\ {\displaystyle \frac{s\left(3r+2s\right)-(2r+s)}{2r+s}}& \cdots & {\displaystyle \frac{s\left(3r+2s\right)-(2r+s)}{2r+s}}& {\displaystyle \frac{\left(2r+s\right)\left(n+nr+2s-1\right)-rs}{2r+s}}\end{array}\right)}_{n\times n} \\ {\mathit{det}\left(\begin{array}{cccc}{\displaystyle \frac{\left(s+2\right)\left(1+n+nr-r\right)+2r}{s+2}}& {\displaystyle \frac{\left(s+2\right)-rs}{s+2}}& \cdots & {\displaystyle \frac{\left(s+2\right)-rs}{s+2}}\\ {\displaystyle \frac{\left(s+2\right)-rs}{s+2}}& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & {\displaystyle \frac{\left(s+2\right)-rs}{s+2}}\\ {\displaystyle \frac{\left(s+2\right)-rs}{s+2}}& \cdots & {\displaystyle \frac{\left(s+2\right)-rs}{s+2}}& {\displaystyle \frac{\left(s+2\right)\left(1+n+nr-r\right)+2r}{s+2}}\end{array}\right)}_{n\times n} \end{gathered}$$

Using Lemma 2, we have $\tau \left(K_{r,s}\ast K_{n,n}\right)={\displaystyle \frac{1}{4n^2{\left(r+s\right)}^2}}\times$

$$\begin{array}{l}{\left(-1\right)}^{r\left(n-1\right)}\times {\left(-n\left(s+1\right)\right)}^{\left(n-1\right)\left(r-1\right)}\times {\left(n-n\left(s+1\right)\right)}^{\left(r-1\right)}\times {\left(n\left(s+1\right)\right)}^{\left(n-1\right)\left(r-1\right)}\times {\left(n\left(s+1\right)+n\right)}^{\left(r-1\right)}\times {\left(2r-1\right)}^n\\ \times {\left({\displaystyle \frac{n\left(s+1\right)+2r-1}{2r-1}}-1\right)}^{n-1}\times \left({\displaystyle \frac{n\left(s+1\right)+2r-1}{2r-1}}+n-1\right)\times {\left(n\left(s+1\right)\right)}^{\left(n-1\right)}\times \left(n\left(s+1\right)+n\right)\\ \times {\left(-1\right)}^{r\left(s-1\right)}\times {\left(-n-rn\right)}^{\left(n-1\right)\left(s-1\right)}{\left(-rn\right)}^{s-1}\times {\left(n\left(r+1\right)\right)}^{\left(n-1\right)\left(s-1\right)}\times {\left(n\left(r+2\right)\right)}^{\left(s-1\right)}\times {\left({\displaystyle \frac{1}{2r+s}}\right)}^n\\ \times {\left(n\left(r+1\right)\left(2r+s\right)\right)}^{\left(n-1\right)}\times \left(2n{\left(r+s\right)}^2\right)\times {\left({\displaystyle \frac{1}{s+2}}\right)}^n\times {\left(n\left(s+2\right)\left(r+1\right)\right)}^{\left(n-1\right)}\times 2n\left(r+s+2\right)\\ =n^{2n\left(r+s\right)-2}\times r^{s-1}\times s^{r-1}\times {\left(r+1\right)}^{2s\left(n-1\right)}\times {\left(s+1\right)}^{2r\left(n-1\right)}\times {\left(r+2\right)}^{s-1}\times {\left(s+2\right)}^{r-1}\times \left(r+s+2\right.).\end{array}$$

□

8. Numerical Results

Using the formulas provided in Theorems 1–8,

Table 1. Selected numbers of spanning trees, ${\mathit{K}}_{\mathit{n},\mathit{n}}\square {\mathit{K}}_{\mathit{r},\mathit{s}}$ and ${\mathit{K}}_{\mathit{n},\mathit{n}}⊠{\mathit{K}}_{\mathit{r},\mathit{s}}.$

$\mathit{n}$	$\mathit{r}$	$\mathit{s}$	$\mathit{\tau }({\mathit{K}}_{\mathit{n},\mathit{n}}\square {\mathit{K}}_{\mathit{r},\mathit{s}})$	$\mathit{n}$	$\mathit{r}$	$\mathit{s}$	$\mathit{\tau }\left({\mathit{K}}_{\mathit{n},\mathit{n}}⊠{\mathit{K}}_{\mathit{r},\mathit{s}}\right)$
$\mathbf{1}$	$1$	$1$	$4$	$1$	$1$	$1$	$16$
$\mathbf{1}$	$1$	$2$	$15$	$1$	$1$	$2$	$192$
$\mathbf{1}$	$1$	$3$	$54$	$1$	$1$	$3$	$2048$
$\mathbf{1}$	$2$	$2$	$384$	$1$	$2$	$2$	$\mathrm{20,736}$
$\mathbf{1}$	$2$	$3$	$6720$	$1$	$2$	$3$	$\mathrm{1,327,104}$
$\mathbf{2}$	$1$	$2$	$\mathrm{31,500}$	$2$	$1$	$2$	$\mathrm{38,663,100}$
$\mathbf{2}$	$2$	$2$	$\mathrm{42,467,328}$	$2$	$2$	$2$	$\mathrm{3,822,059,520,000}$
$\mathbf{2}$	$2$	$3$	$\mathrm{34,139,750,400}$	$2$	$2$	$3$	$\mathrm{130,540,218,148,454,400}$
$\mathbf{3}$	$1$	$2$	$\mathrm{1,693,052,928}$	$3$	$1$	$2$	$\mathrm{267,411,375,000,000}$
$\mathbf{3}$	$2$	$2$	$\mathrm{194,481,000,000,000}$	$3$	$2$	$2$	$\mathrm{60,464,744,163,332,470,241,280}$

,

Table 2. Selected numbers of spanning trees, ${\mathit{K}}_{\mathit{n},\mathit{n}}\left[{\mathit{K}}_{\mathit{r},\mathit{s}}\right]$ and ${\mathit{K}}_{\mathit{r},\mathit{s}}\left[{\mathit{K}}_{\mathit{n},\mathit{n}}\right]$.

$\mathit{n}$	$\mathit{r}$	$\mathit{s}$	$\mathit{\tau }\left({\mathit{K}}_{\mathit{n},\mathit{n}}\left[{\mathit{K}}_{\mathit{r},\mathit{s}}\right]\right)$	$\mathit{n}$	$\mathit{r}$	$\mathit{s}$	$\mathit{\tau }\left({\mathit{K}}_{\mathit{r},\mathit{s}}\left[{\mathit{K}}_{\mathit{n},\mathit{n}}\right]\right)$
$\mathbf{1}$	$1$	$1$	$16$	$1$	$1$	$1$	$16$
$\mathbf{1}$	$1$	$2$	$576$	$1$	$1$	$2$	$192$
$\mathbf{1}$	$1$	$3$	$\mathrm{40,000}$	$1$	$1$	$3$	$2048$
$\mathbf{1}$	$2$	$2$	$\mathrm{82,944}$	$1$	$2$	$2$	$\mathrm{20,736}$
$\mathbf{1}$	$2$	$3$	$\mathrm{15,366,400}$	$1$	$2$	$3$	$\mathrm{1,327,104}$
$\mathbf{2}$	$1$	$2$	$\mathrm{567,106,596}$	$2$	$1$	$2$	$\mathrm{398,131,200}$
$\mathbf{2}$	$2$	$2$	$\mathrm{132,710,400,000,000}$	$2$	$2$	$2$	$\mathrm{132,710,400,000,000}$
$\mathbf{2}$	$2$	$3$	$\mathrm{62,171,080,298,496,000,000}$	$2$	$2$	$3$	$\mathrm{13,051,398,979,584,000,000}$
$\mathbf{3}$	$1$	$2$	$\mathrm{19,591,041,024,000,000}$	$3$	$1$	$2$	$\mathrm{33,891,544,377,720,000}$
$\mathbf{3}$	$2$	$2$	$\mathrm{19,723,378,462,269,065,863,888,896}$	$3$	$2$	$2$	$\mathrm{99,291,633,919,101,562,500,000,000}$

,

Table 3. Selected numbers of spanning trees, ${\mathit{K}}_{\mathit{n},\mathit{n}}\oplus {\mathit{K}}_{\mathit{r},\mathit{s}}$ and ${\mathit{K}}_{\mathit{n},\mathit{n}}\vee {\mathit{K}}_{\mathit{r},\mathit{s}}$.

$\mathit{n}$	$\mathit{r}$	$\mathit{s}$	$\mathit{\tau }({\mathit{K}}_{\mathit{n},\mathit{n}}\oplus {\mathit{K}}_{\mathit{r},\mathit{s}})$	$\mathit{n}$	$\mathit{r}$	$\mathit{s}$	$\mathit{\tau }\left({\mathit{K}}_{\mathit{n},\mathit{n}}\vee {\mathit{K}}_{\mathit{r},\mathit{s}}\right)$
$1$	$1$	$1$	$4$	$1$	$1$	$1$	$16$
$1$	$1$	$2$	$81$	$1$	$1$	$2$	$576$
$\mathbf{1}$	$1$	$3$	$4096$	$1$	$1$	$3$	$\mathrm{40,000}$
$\mathbf{1}$	$2$	$2$	$4096$	$1$	$2$	$2$	$\mathrm{82,944}$
$\mathbf{1}$	$2$	$3$	$\mathrm{390,625}$	$1$	$2$	$3$	$\mathrm{15,366,400}$
$\mathbf{2}$	$1$	$2$	$\mathrm{60,466,176}$	$2$	$1$	$2$	$\mathrm{3,774,873,600}$
$\mathbf{2}$	$2$	$2$	$\mathrm{4,398,046,511,104}$	$2$	$2$	$2$	$\mathrm{2,282,521,714,753,536}$
$\mathbf{2}$	$2$	$3$	$\mathrm{1,000,000,000,000,000,000}$	$2$	$2$	$3$	$\mathrm{1,941,155,130,215,130,726,400}$
$\mathbf{3}$	$1$	$2$	$\mathrm{1,853,020,188,851,841}$	$3$	$1$	$2$	$\mathrm{1,015,599,566,684,160,000}$
$\mathbf{3}$	$2$	$2$	$\mathrm{552,061,438,912,436,417,593,344}$	$3$	$2$	$2$	$\mathrm{7,342,984,060,644,141,028,475,928,576}$

and

Table 4. Selected numbers of spanning trees, ${\mathit{K}}_{\mathit{n},\mathit{n}}\ast {\mathit{K}}_{\mathit{r},\mathit{s}}$ and ${\mathit{K}}_{\mathit{r},\mathit{s}}\ast {\mathit{K}}_{\mathit{n},\mathit{n}}$.

$\mathit{n}$	$\mathit{r}$	$\mathit{s}$	$\mathit{\tau }({\mathit{K}}_{\mathit{n},\mathit{n}}\ast {\mathit{K}}_{\mathit{r},\mathit{s}})$	$\mathit{n}$	$\mathit{r}$	$\mathit{s}$	$\mathit{\tau }({\mathit{K}}_{\mathit{r},\mathit{s}}\ast {\mathit{K}}_{\mathit{n},\mathit{n}})$
$\mathbf{1}$	$1$	$1$	$4$	$1$	$1$	$1$	$4$
$\mathbf{1}$	$1$	$2$	$32$	$1$	$1$	$2$	$15$
$\mathbf{1}$	$1$	$3$	$192$	$1$	$1$	$3$	$54$
$\mathbf{1}$	$2$	$2$	$4096$	$1$	$2$	$2$	$384$
$\mathbf{1}$	$2$	$3$	$\mathrm{221,184}$	$1$	$2$	$3$	$6720$
$\mathbf{2}$	$1$	$2$	$\mathrm{248,832}$	$2$	$1$	$2$	$\mathrm{2,211,840}$
$\mathbf{2}$	$2$	$2$	$\mathrm{41,278,242,816}$	$2$	$2$	$2$	$\mathrm{41,278,242,816}$
$\mathbf{2}$	$2$	$3$	$\mathrm{914,039,610,015,744}$	$2$	$2$	$3$	$\mathrm{328,758,271,672,320}$
$\mathbf{3}$	$1$	$2$	$\mathrm{53,084,160,000}$	$3$	$1$	$2$	$\mathrm{13,389,252,099,840}$
$\mathbf{3}$	$2$	$2$	$\mathrm{28,499,341,391,953,920,000}$	$3$	$2$	$2$	$\mathrm{518,727,059,586,428,962,176}$

show some of the values of the number of spanning trees in the various products studied above.

9. Conclusions

Graph products are crucial in discrete mathematics because they allow the construction of large, complex networks from smaller, simpler graphs. Additionally, they facilitate network optimization, structural feature analysis, and spectral data analysis. The number of spanning trees (complexity) is measured by the complexity of graph products (such as Cartesian, tensor, and strong), which is a vital indicator of network efficiency and dependability. Breaking down vast, complex networks into smaller, simpler components enables calculation of their structural features. This is crucial for optimization in computer science, chemistry, and biology. Additionally, they enable the discovery of hidden patterns, the optimization of pathways, and the creation of complex networks from smaller ones through effective computation in social network analysis and chemistry. They facilitate the description of regular structures like grids and hypercubes and aid large-scale decomposition and graph visualization. In this work, using linear algebra techniques and block matrices, we calculated explicit formulas for the number of spanning trees of seven types of products of two complete bipartite graphs.