Resistance Distance in the Double Corona Based on R-Graph

Abstract

Let $G_0$ be a connected graph on n vertices and m edges. The R-graph $R\left(G_0\right)$ of $G_0$ is a graph obtained from $G_0$ by adding a new vertex corresponding to each edge of $G_0$ and by joining each new vertex to the end points of the edge corresponding to it. Let $G_1$ and $G_2$ be graphs on $n_1$ and $n_2$ vertices, respectively. The R-graph double corona $G_0^{\left(R\right)}\circ \{G_1,G_2\}$ of $G_0$, $G_1$ and $G_2$, is the graph obtained by taking one copy of $R\left(G_0\right)$, n copies of $G_1$ and m copies of $G_2$ and then by joining the i-th old-vertex of $R\left(G_0\right)$ to every vertex of the i-th copy of $G_1$ and the j-th new vertex of $R\left(G_0\right)$ to every vertex of the j-th copy of $G_2$. In this paper, we consider resistance distance in $G_0^{\left(R\right)}\circ \{G_1,G_2\}$. Moreover, we give an example to illustrate the correction and efficiency of the proposed method.

1. Introduction

All graphs considered in this paper are simple and undirected. A graph G whose vertex set is $V\left(G\right)=\{v_1,v_2,\dots ,v_n\}$ and edge set is $E\left(G\right)=\{e_1,e_2,\dots ,e_m\}$, is denoted by $\left(V\right(G),E(G\left)\right)$. As we know, the conventional distance $d_{ij}$ is the length of a shortest path between vertices $v_i$ and $v_j$. Connected with practical applications, such as electrical network, Klein and Randić introduced resistance distance [1], which is the effective electrical resistance between two vertices if every edge is replaced by a unit resistor, and is denoted by $r_{ij}$ for resistance distance between $v_i$ and $v_j$. Some results on resistance distance can be found in [2,3,4].

One of the main topics about resistance distance is to determine it in various graphs. Now one can easily obtain resistance distance in wheels and fans [5], in subdivision-vertex join and subdivision-edge join of graphs [6], in corona and the neighborhood corona graphs of two disjoint graphs [7], in H-Join of Graphs $G_1,G_2,\dots ,G_k$ [8]. Please turn to [9,10,11,12,13,14,15] for more detail.

Motivated by the above works, we consider resistance distance in double corona based on R-graph. The R-graph of G, which is denoted by $R\left(G\right)$, appeared in [16]. Moreover, $R\left(G\right)$ is defined as the graph obtained from G by adding a new vertex corresponding to each edge of G and by joining each new vertex to the end points of the edge corresponding to it. Recently, in 2017, Barik and Sahoo introduced the R-graph double corona of $G_0$, $G_1$ and $G_2$ [17].

Definition 1 ([17]).

Let $G_0$ be a connected graph on n vertices and m edges. Let $G_1$ and $G_2$ be graphs on $n_1$ and $n_2$ vertices, respectively. The R-graph double corona of $G_0$, $G_1$ and $G_2$, denoted by $G_0^{\left(R\right)}\circ \{G_1,G_2\}$, is the graph obtained by taking one copy of $R\left(G_0\right)$, n copies of $G_1$ and m copies of $G_2$ and then by joining the i-th old-vertex of $R\left(G_0\right)$ to every vertex of the i-th copy of $G_1$ and the j-th new vertex of $R\left(G_0\right)$ to every vertex of the j-th copy of $G_2$.

Example 1.

Please refer to Example 3 of [17] for $C_4^{\left(R\right)}\circ \{P_3,P_2\}$, where $P_n$ and $C_n$ are a path and a cycle with n vertices. Moreover, one can refer to Example 2 for $P_2^{\left(R\right)}\circ \{P_2,P_2\}$.

This paper will compute resistance distance in $G_0^{\left(R\right)}\circ \{G_1,G_2\}$. However, first of all, we turn the readers’ attention to some matrices associated with a graph G. The adjacency matrix $A_G$, which is a $n\times n$-matrix with entry $a_{ij}=1$ if $v_i$ and $v_j$ are adjacent in G and $a_{ij}=0$ otherwise, the diagonal matrix $D_G$ with diagonal entries $d_G\left(v_1\right),d_G\left(v_2\right),\dots ,d_G\left(v_n\right)$ and the incidence matrix $M_G$ which is a $n\times m$-matrix with $m_{ij}=1$ (or 0) if vertex $v_i$ is (not) incident with $e_j$. Moreover, the Laplacian matrix $L_G$ of G is $D_G-A_G$. For more detail, please refer to [18,19].

Here we list some symbols. Let $I_n$ denote the unit matrix of order n, 1${}_n$ be the all-one column vector of dimension n and $J_{n\times m}$ be the all-one $n\times m$-matrix. Recall that the Kronecker product $A\otimes B$ [20] of two matrices $A={\left(a_{ij}\right)}_{m\times n}$ and $B={\left(b_{ij}\right)}_{p\times q}$ is an $mp\times nq$-matrix obtained from A by replacing every element $a_{ij}$ by $a_{ij}B$. Moreover, $(A\otimes B)(C\otimes D)=AC\otimes BD$, whenever the products $AC$ and $BD$ exist, which implies that ${(A\otimes B)}^{-1}=A^{-1}\otimes B^{-1}$.

2. Preliminaries

Let M be a matrix. If X is a matrix such that $MXM=M$, then X is a $\left\{1\right\}$-inverse of M, and X is always denoted by $M^{\left\{1\right\}}$. Further assume that M is a square matrix. If X is the matrix satisfying $\left(1\right)MXM=M;\left(2\right)XMX=X;\left(3\right)MX=XM$, then $X=M^{\#}$ is the group inverse of M. It is well-known that $M^{\#}$ exists if and only if rank(M)=rank($M^2$), and $M^{\#}$ is unique.

Let A be a real symmetric matrix. Obviously, $A^{\#}$ exists and it is a $\left\{1\right\}$-inverse of A. In fact, assume that U is an orthogonal matrix (i.e., $U{U}^T=U^{T}U=I$) such that $A=Udiag\{{\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n\}{U}^T$, where ${\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n$ are eigenvalues of A. Then $A^{\#}=Udiag\{f\left({\lambda }_1\right),f\left({\lambda }_2\right),\cdots ,f\left({\lambda }_n\right)\}{U}^T$, where $f\left({\lambda }_i\right)=\left\{\begin{array}{cc}1/{\lambda }_i,& \mathrm{if}{\lambda }_i\ne 0,\\ 0,& \mathrm{if}{\lambda }_i=0.\end{array}\right.$ Moreover, in [21], the existence and the representation of the group inverse for the block matrices with an invertible subblock were given.

Lemma 1 ([21]).

Let $M=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$ be a $m\times m$ matrix, where A is an invertible $n\times n$ matrix, $S=D-C{A}^{-1}B$. If $S^{\#}$ exists, then

$\left(1\right)$$M^{\#}$ exists if and only if R is invertible, where $R=A^2+B{S}^{\pi }C$ and $S^{\pi }=I_{m-n}-S{S}^{\#}$;

$\left(2\right)$ if $M^{\#}$ exists, then $M^{\#}=\left(\begin{array}{cc}X& Y\\ Z& W\end{array}\right),$ where

$$\begin{array}{cc}\hfill X=& A{R}^{-1}(A+B{S}^{\#}C)R^{-1}A,\hfill \\ \hfill Y=& A{R}^{-1}(A+B{S}^{\#}C)R^{-1}B{S}^{\pi }-A{R}^{-1}B{S}^{\#},\hfill \\ \hfill Z=& S^{\pi }C{R}^{-1}(A+B{S}^{\#}C)R^{-1}A-S^{\#}C{R}^{-1}A,\hfill \\ \hfill W=& S^{\pi }C{R}^{-1}(A+B{S}^{\#}C)R^{-1}B{S}^{\pi }-S^{\#}C{R}^{-1}B{S}^{\pi }-S^{\pi }C{R}^{-1}B{S}^{\#}+S^{\#}.\hfill \end{array}$$

Please note that, the Laplacian matrix $L\left(G\right)$ of a graph G is real symmetric. So $L{\left(G\right)}^{\#}$ exists and consequently, $L{\left(G\right)}^{\left\{1\right\}}$ exists. The representations of $L{\left(G\right)}^{\left\{1\right\}}$ were investigated in [6,11,22] under different conditions. We list two in the next lemma.

Lemma 2 ([6,11,22]).

Let $L=\left(\begin{array}{cc}{L}_1& L_2\\ L_2^T& L_3\end{array}\right)$ be the Laplacian matrix of a connected graph. Assume that $L_1$ is nonsingular. Denote $S=L_3-L_2^T{L}_1^{-1}{L}_2$. Then

$\left(1\right)$$\left(\begin{array}{cc}{L}_1^{-1}+L_1^{-1}{L}_2{S}^{\#}{L}_2^T{L}_1^{-1}& -L_1^{-1}{L}_2{S}^{\#}\\ -S^{\#}{L}_2^T{L}_1^{-1}& S^{\#}\end{array}\right)$ is a symmetric $\left\{1\right\}$-inverse of L;

$\left(2\right)$ If each column vector of $L_2$ is $\mathbf{1}$ or a zero vector, then $\left(\begin{array}{cc}{L}_1^{-1}& 0\\ 0& S^{\#}\end{array}\right)$ is a symmetric $\left\{1\right\}$-inverse of L.

To compute the inverse of a matrix, the next lemma is useful.

Lemma 3 ([6]).

Let $M=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$ be a nonsingular matrix. If A and D are nonsingular, then

$$M^{-1}=\left(\begin{array}{cc}{A}^{-1}+A^{-1}B{S}^{-1}C{A}^{-1}& -A^{-1}B{S}^{-1}\\ -S^{-1}C{A}^{-1}& S^{-1}\end{array}\right),$$

where $S=D-C{A}^{-1}B$ is the Schur complement of A in M.

This paper is devoted to the compute of resistance distance in $G_0^{\left(R\right)}\circ \{G_1,G_2\}$. In [6], authors obtained the formulae for resistance distance by elements of group inverse $L{\left(G\right)}^{\#}$ or $\left\{1\right\}$-inverse $L{\left(G\right)}^{\left\{1\right\}}$ of $L\left(G\right)$, where $G=G_0^{\left(R\right)}\circ \{G_1,G_2\}$.

Lemma 4 ([6]).

Let G be a connected graph and ${\left(A\right)}_{ij}$ be the $(i,j)$-entry of a matrix A. Then

$$\begin{array}{cc}\hfill r_{ij}\left(G\right)& ={\left(L{\left(G\right)}^{\left\{1\right\}}\right)}_{ii}+{\left(L{\left(G\right)}^{\left\{1\right\}}\right)}_{jj}-{\left(L{\left(G\right)}^{\left\{1\right\}}\right)}_{ij}-{\left(L{\left(G\right)}^{\left\{1\right\}}\right)}_{ji}\hfill \\ & ={\left(L{\left(G\right)}^{\#}\right)}_{ii}+{\left(L{\left(G\right)}^{\#}\right)}_{jj}-2{\left(L{\left(G\right)}^{\#}\right)}_{ij}.\hfill \end{array}$$

Keep Lemma 4 in mind, we only need to compute $L{\left(G\right)}^{\#}$ or $L{\left(G\right)}^{\left\{1\right\}}$. Before calculating, we list more preliminaries below.

Lemma 5 ([11]).

For any graph G, $L{\left(G\right)}^{\#}\mathbf{1}=0$.

Lemma 6 ([23]).

Let G be a simple connected graph. Then its adjacency matrix $A\left(G\right)$, diagonal matrix $D\left(G\right)$ and incidence matrix $M\left(G\right)$ satisfy $M\left(G\right)M{\left(G\right)}^T=A\left(G\right)+D\left(G\right)$.

Lemma 7.

Assume that A is symmetric and 0 is a simple eigenvalue. Let u be the unitary 0-eigenvector of A. Then the group inverse $A^{\#}$ is characterized as the unique singular matrix satisfying

$$A{A}^{\#}=A^{\#}A=I-u{u}^T.$$

Proof. 

Assume that ${\lambda }_i$ is a non-zero eigenvalue of A and $u_i$ is the unitary ${\lambda }_i$-eigenvector of A, for $i=1,2,\cdots ,n-1$. Let $U=(u_1{u}_2\cdots u_{n-1}u)$. Then

$$A=Udiag\{{\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_{n-1},0\}{U}^T,A^{\#}=Udiag\{\frac{1}{{\lambda }_1},\frac{1}{{\lambda }_2},\cdots ,\frac{1}{{\lambda }_{n-1}},0\}{U}^T.$$

Clearly, $I-A{A}^{\#}=I-A^{\#}A$ and

$$I-A^{\#}A=U\left(\begin{array}{cccc}0& & & \\ & \ddots & & \\ & & 0& \\ & & & 1\end{array}\right)U^T=U\left(\begin{array}{c}0\\ ⋮\\ 0\\ 1\end{array}\right)\left(\begin{array}{cccc}0& \cdots & 0& 1\end{array}\right)U^T=u{u}^T.$$

 □

Lemma 8.

Let M be a matrix and X be a $\left\{1\right\}$-inverse of M. If $X_0$ is a matrix satisfying $M{X}_{0}M=0$, then $X-X_0$ is also a $\left\{1\right\}$-inverse of M.

Proof. 

Please note that $M(X-X_0)M=MXM-M{X}_{0}M=M-0=M$. Thus, $X-X_0$ is a $\left\{1\right\}$-inverse of M. □

3. Main Results

In this section, $G=G_0^{\left(R\right)}\circ \{G_1,G_2\}$, where $G_0$ is a connected r-regular graph on n vertices $V\left(G_0\right)=\{v_1,v_2,\dots ,v_n\}$ and m edges $E\left(G_0\right)=\{e_1,e_2,\dots ,e_m\}$, and $G_1,G_2$ are graphs on $n_1$ vertices $V\left(G_1\right)=\{u_1,u_2,\dots ,u_{n_1}\}$ and $n_2$ vertices $V\left(G_2\right)=\{w_1,w_2,\dots ,w_{n_2}\}$. We give a $\left\{1\right\}$-inverse of $L\left(G\right)$ in Theorem 1. However, before Theorem 1, we show the labeling rule of vertices of G.

(1) For $i=1,2,\dots ,m$, label the $n_2$ vertices of the i-th copy of $G_2$ with

$$V{\left(G_2\right)}_i=\{w_1^{(i-1)n_2+1},w_2^{(i-1)n_2+2},\dots ,w_{n_2}^{i{n}_2}\}.$$

(2) For $j=1,2,\dots ,n$, label the $n_1$ vertices of the j-th copy of $G_1$ with

$$V{\left(G_1\right)}_j=\{u_1^{m{n}_2+(j-1)n_1+1},u_2^{m{n}_2+(j-1)n_1+2},\dots ,u_{n_1}^{m{n}_2+j{n}_1}\}.$$

(3) Label the m new-vertices of $R\left(G_0\right)$ corresponding to edges of $E\left(G_0\right)$ with

$$\{e_1^{m{n}_2+n{n}_1+1},e_2^{m{n}_2+n{n}_1+2},\dots ,e_m^{m(n_2+1)+n{n}_1}\}.$$

(4) Label the n old-vertices $V\left(G_0\right)=\{v_1,v_2,\dots ,v_n\}$ of $R\left(G_0\right)$ with

$$\{v_1^{m(n_2+1)+n{n}_1+1},v_2^{m(n_2+1)+n{n}_1+2},\dots ,v_n^{m(n_2+1)+n(n_1+1)}\}.$$

Thus,

$$\begin{array}{cc}\hfill V\left(G\right)=& V{\left(G_2\right)}_1\cup \cdots \cup V{\left(G_2\right)}_m\cup V{\left(G_1\right)}_1\cup \cdots \cup V{\left(G_1\right)}_n\hfill \\ & \cup \{e_1^{m{n}_2+n{n}_1+1},\dots ,e_m^{m(n_2+1)+n{n}_1}\}\cup \{v_1^{m(n_2+1)+n{n}_1+1},\dots ,v_n^{m(n_2+1)+n(n_1+1)}\}.\hfill \end{array}$$

Theorem 1.

The following matrix is a $\left\{1\right\}$-inverse of $L\left(G\right)$,

$$\begin{array}{c}\left(\begin{array}{cccc}{I}_m\otimes \left({(L_{G_2}+I_{n_2})}^{-1}+\frac{1}{2}{J}_{n_2\times n_2}\right)& 0& \frac{1}{2}{I}_m\otimes {\mathbf{1}}_{n_2}& 0\\ 0& I_n\otimes {(L_{G_1}+I_{n_1})}^{-1}& 0& 0\\ \frac{1}{2}{I}_m\otimes {\mathbf{1}}_{n_2}^T& 0& \frac{1}{2}{I}_m& 0\\ 0& 0& 0& 0\end{array}\right)\hfill \\ +\left(\begin{array}{c}\frac{1}{2}{M}_{G_0}^T\otimes {\mathbf{1}}_{n_2}\\ I_n\otimes {\mathbf{1}}_{n_1}\\ \frac{1}{2}{M}_{G_0}^T\\ I_n\end{array}\right)\frac{2}{3}{L}_{G_0}^{\#}\left(\frac{1}{2}{M}_{G_0}\otimes {\mathbf{1}}_{n_2}^T{I}_n\otimes {\mathbf{1}}_{n_1}^T\frac{1}{2}{M}_{G_0}{I}_n\right).\hfill \end{array}$$

Moreover, denote $\left(\begin{array}{c}\frac{1}{2}{M}_{G_0}^T\\ I_n\end{array}\right)\frac{2}{3}{L}_{G_0}^{\#}\left(\frac{1}{2}{M}_{G_0}{I}_n\right)=\left(\begin{array}{cc}\frac{1}{6}{M}_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}& \frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}\\ \frac{1}{3}{L}_{G_0}^{\#}{M}_{G_0}& \frac{2}{3}{L}_{G_0}^{\#}\end{array}\right)$ by $\left(\begin{array}{cc}{X}_{11}& X_{12}\\ X_{12}^T& X_{22}\end{array}\right)$. Then the second term in the above matrix is

$$\left(\begin{array}{cccc}{X}_{11}\otimes J_{n_2\times n_2}& X_{12}\otimes J_{n_2\times n_1}& X_{11}\otimes {\mathbf{1}}_{n_2}& X_{12}\otimes {\mathbf{1}}_{n_2}\\ X_{12}^T\otimes J_{n_1\times n_2}& X_{22}\otimes J_{n_1\times n_1}& X_{12}^T\otimes {\mathbf{1}}_{n_1}& X_{22}\otimes {\mathbf{1}}_{n_1}\\ X_{11}\otimes {\mathbf{1}}_{n_2}^T& X_{12}\otimes {\mathbf{1}}_{n_1}^T& X_{11}& X_{12}\\ X_{12}^T\otimes {\mathbf{1}}_{n_2}^T& X_{22}\otimes {\mathbf{1}}_{n_1}^T& X_{12}^T& X_{22}\end{array}\right).$$

Proof. 

By the definition, it is easy to show that $G=G_0^{\left(R\right)}\circ \{G_1,G_2\}$ is connected. Furthermore, from the vertex-labeling rule of G, we know that all the diagonal matrix $D_G$, the adjacency matrix $A_G$ and the Laplacian matrix $L_G$ are partitioned $(m+n+2)\times (m+n+2)$-matrices. Particularly, the Laplacian matrix of G is

$$L_G=\left(\begin{array}{cccc}{I}_m\otimes (L_{G_2}+I_{n_2})& 0& -I_m\otimes {\mathbf{1}}_{n_2}& 0\\ 0& I_n\otimes (L_{G_1}+I_{n_1})& 0& -I_n\otimes {\mathbf{1}}_{n_1}\\ -I_m\otimes {\mathbf{1}}_{n_2}^T& 0& (2+n_2)I_m& -M_{G_0}^T\\ 0& -I_n\otimes {\mathbf{1}}_{n_1}^T& -M_{G_0}& 2D_{G_0}-A_{G_0}+n_1{I}_{n_1}\end{array}\right).$$

We proceed via the following steps.

Step 1. To use Lemma 2, we further divide $L_G$ into blocks $L_G=\left(\begin{array}{cc}{L}_1& L_2\\ L_2^T& L_3\end{array}\right)$, where

$$\begin{array}{cc}\hfill L_1& =\left(\begin{array}{cc}{I}_m\otimes (L_{G_2}+I_{n_2})& 0\\ 0& I_n\otimes (L_{G_1}+I_{n_1})\end{array}\right);\hfill \\ \hfill L_2& =\left(\begin{array}{cc}-I_m\otimes {\mathbf{1}}_{n_2}& 0\\ 0& -I_n\otimes {\mathbf{1}}_{n_1}\end{array}\right);L_2^T=\left(\begin{array}{cc}-I_m\otimes {\mathbf{1}}_{n_2}^T& 0\\ 0& -I_n\otimes {\mathbf{1}}_{n_1}^T\end{array}\right);\hfill \\ \hfill L_3& =\left(\begin{array}{cc}(2+n_2)I_m& -M_{G_0}^T\\ -M_{G_0}& 2D_{G_0}-A_{G_0}+n_1{I}_{n_1}\end{array}\right).\hfill \end{array}$$

Clearly, $L_1^{-1}=\left(\begin{array}{cc}{I}_m\otimes {(L_{G_2}+I_{n_2})}^{-1}& 0\\ 0& I_n\otimes {(L_{G_1}+I_{n_1})}^{-1}\end{array}\right)$.

Step 2. Please note that $L_{G_1}{\mathbf{1}}_{n_1}=0$. So $(L_{G_1}+I_{n_1}){\mathbf{1}}_{n_1}={\mathbf{1}}_{n_1}$, which shows that

$${(L_{G_1}+I_{n_1})}^{-1}{\mathbf{1}}_{n_1}={\mathbf{1}}_{n_1}.$$

Furthermore,

$${\mathbf{1}}_{n_1}^T{(L_{G_1}+I_{n_1})}^{-1}{\mathbf{1}}_{n_1}=n_1.$$

Similarly, we have ${\mathbf{1}}_{n_2}^T{(L_{G_2}+I_{n_2})}^{-1}{\mathbf{1}}_{n_2}=n_2$. Keep these in mind and recall the Kronecker product. Then

$$\begin{array}{cc}\hfill L_2^T{L}_1^{-1}{L}_2& =\left(\begin{array}{cc}{I}_m\otimes {\mathbf{1}}_{n_2}^T{(L_{G_2}+I_{n_2})}^{-1}{\mathbf{1}}_{n_2}& 0\\ 0& I_n\otimes {\mathbf{1}}_{n_1}^T{(L_{G_1}+I_{n_1})}^{-1}{\mathbf{1}}_{n_1}\end{array}\right)\hfill \\ & =\left(\begin{array}{cc}{n}_2{I}_m& 0\\ 0& n_1{I}_n\end{array}\right).\hfill \end{array}$$

From this result, it follows that

$$\begin{array}{cc}\hfill S=L_3-L_2^T{L}_1^{-1}{L}_2& =\left(\begin{array}{cc}(2+n_2)I_m& -M_{G_0}^T\\ -M_{G_0}& 2D_{G_0}-A_{G_0}+n_1{I}_{n_1}\end{array}\right)-\left(\begin{array}{cc}{n}_2{I}_m& 0\\ 0& n_1{I}_n\end{array}\right)\hfill \\ & =\left(\begin{array}{cc}2I_m& -M_{G_0}^T\\ -M_{G_0}& 2D_{G_0}-A_{G_0}\end{array}\right).\hfill \end{array}$$

Since S is real symmetric, $S^{\#}$ exists.

Step 3. Here we computer $S^{\#}$ by Lemma 1. Please note that $M_{G_0}{M}_{G_0}^T=D_{G_0}+A_{G_0}$. We denote

$$S_0=2D_{G_0}-A_{G_0}-M_{G_0}\left(\frac{1}{2}{I}_m\right)M_{G_0}^T.$$

Then $S_0=\frac{3}{2}{L}_{G_0}$. Consequently, $S_0^{\#}$ exists and in fact, $S_0^{\#}=\frac{2}{3}{L}_{G_0}^{\#}$. Denote $I_n-S_0{S}_0^{\#}$ by $S_0^{\pi }$. Combing with Lemma 7, we have

$$S_0^{\pi }=I_n-L_{G_0}{L}_{G_0}^{\#}=\frac{1}{n}{J}_{n\times n}.$$

Please note that $\left(\begin{array}{c}1\cdots 1\end{array}\right)M_{G_0}=2\left(\begin{array}{c}1\cdots 1\end{array}\right)$ and $M_{G_0}^T{\mathbf{1}}_n=2{\mathbf{1}}_n$. So, we further have

$$R=4I_m+M_{G_0}^T\left(\frac{1}{n}{J}_{n\times n}\right)M_{G_0}=4\left(I_m+\frac{1}{n}{J}_{m\times m}\right).$$

Therefore, it is easy to obtain that

$$R^{-1}=\frac{1}{4}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right).$$

Please note that G is r-regular. So $M_{G_0}{\mathbf{1}}_m=r{\mathbf{1}}_m$. Moreover, $L_{G_0}^{\#}{\mathbf{1}}_n=0$ according to Lemma 5. In general, by Lemma 1, we finally have $S^{\#}=\left(\begin{array}{cc}X& Y\\ Z& W\end{array}\right),$ where

$$\begin{array}{cc}\hfill X& =2I_m\frac{1}{4}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)2\left(I_m+\frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}\right)\frac{1}{4}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)2I_m\hfill \\ & =\frac{1}{2}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)\left(I_m+\frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}\right)\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)\hfill \\ & =\frac{1}{2}\left(I_m+\frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}-\frac{2n+m}{{(n+m)}^2}{J}_{m\times m}\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill Y=& 2I_m\frac{1}{4}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)2\left(I_m+\frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}\right)\frac{1}{4}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)(-M_{G_0}^T)\frac{1}{n}{J}_{n\times n}\hfill \\ & -2I_m\frac{1}{4}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)(-M_{G_0}^T)\frac{2}{3}{L}_{G_0}^{\#}\hfill \\ \hfill =& \frac{-1}{4n}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)\left(I_m+\frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}\right)\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)M_{G_0}^T{J}_{n\times n}\hfill \\ & +\frac{1}{3}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)M_{G_0}^T{L}_{G_0}^{\#}\hfill \\ \hfill =& \frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}-\frac{1}{4n}\left(I_m+\frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}-\frac{2n+m}{{(n+m)}^2}{J}_{m\times m}\right)M_{G_0}^T{J}_{n\times n}\hfill \\ \hfill =& \frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}-\frac{n}{2{(n+m)}^2}{J}_{m\times n},\hfill \end{array}$$

$$\begin{array}{cc}\hfill Z=& \frac{1}{n}{J}_{n\times n}(-M_{G_0})\frac{1}{4}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)2\left(I_m+\frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}\right)\frac{1}{4}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)2I_m\hfill \\ & -\frac{2}{3}{L}_{G_0}^{\#}(-M_{G_0})\frac{1}{4}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)2I_m\hfill \\ \hfill =& \frac{1}{3}L{\left(G\right)}^{\#}{M}_{G_0}-\frac{1}{4n}{J}_{n\times n}{M}_{G_0}\left(I_m+\frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}-\frac{2n+m}{{(n+m)}^2}{J}_{m\times m}\right)\hfill \\ \hfill =& \frac{1}{3}{L}_{G_0}^{\#}{M}_{G_0}-\frac{n}{2{(n+m)}^2}{J}_{n\times m},\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill W=& \frac{1}{n}{J}_{n\times n}(-M_{G_0})\frac{1}{4}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)2\left(I_m+\frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}\right)\frac{1}{4}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)(-M_{G_0}^T)\frac{1}{n}{J}_{n\times n}\hfill \\ & -\frac{2}{3}{L}_{G_0}^{\#}(-M_{G_0})\frac{1}{4}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)(-M_{G_0}^T)\frac{1}{n}{J}_{n\times n}\hfill \\ & -\frac{1}{n}{J}_{n\times n}(-M_{G_0})\frac{1}{4}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)(-M_{G_0}^T)\frac{2}{3}{L}_{G_0}^{\#}+\frac{2}{3}{L}_{G_0}^{\#}\hfill \\ \hfill =& \frac{1}{2n^2}{J}_{n\times m}\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)\left(I_m+\frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}\right)\left(I_m-\frac{1}{n+m}{J}_{m\times m}\right)J_{m\times n}+\frac{2}{3}{L}_{G_0}^{\#}\hfill \\ \hfill =& \frac{2}{3}{L}_{G_0}^{\#}+\frac{m}{2{(n+m)}^2}{J}_{n\times n}.\hfill \end{array}$$

Therefore, we have

$$\begin{array}{cc}\hfill S^{\#}=& \frac{1}{2}\left(\begin{array}{cc}{I}_m& 0\\ 0& 0\end{array}\right)+\left(\begin{array}{c}\frac{1}{2}{M}_{G_0}^T\\ I_n\end{array}\right)\frac{2}{3}{L}_{G_0}^{\#}\left(\frac{1}{2}{M}_{G_0}{I}_n\right)\hfill \\ & +\frac{1}{2(n+m)}\left(\begin{array}{cc}-J_{m\times m}& 0\\ 0& J_{n\times n}\end{array}\right)-\frac{n}{2{(n+m)}^2}{J}_{(m+n)\times (m+n)}.\hfill \end{array}$$

Step 4. In this step, we compute $-L_1^{-1}{L}_2{S}^{\#}$, $-S^{\#}{L}_2^T{L}_1^{-1}$ and $L_1^{-1}{L}_2{S}^{\#}{L}_2^T{L}_1^{-1}$.

$$\begin{array}{cc}\hfill -L_1^{-1}{L}_2{S}^{\#}=& \left(\begin{array}{cc}{I}_m\otimes {(L_{G_2}+I_{n_2})}^{-1}{\mathbf{1}}_{n_2}& 0\\ 0& I_n\otimes {(L_{G_1}+I_{n_1})}^{-1}{\mathbf{1}}_{n_1}\end{array}\right)S^{\#}\hfill \\ \hfill =& \left(\begin{array}{cc}{I}_m\otimes {\mathbf{1}}_{n_2}& 0\\ 0& I_n\otimes {\mathbf{1}}_{n_1}\end{array}\right)S^{\#}\hfill \\ \hfill =& \frac{1}{2}\left(\begin{array}{cc}{I}_m\otimes {\mathbf{1}}_{n_2}& 0\\ 0& 0\end{array}\right)+\left(\begin{array}{c}\frac{1}{2}{M}_{G_0}^T\otimes {\mathbf{1}}_{n_2}\\ I_n\otimes {\mathbf{1}}_{n_1}\end{array}\right)\frac{2}{3}{L}_{G_0}^{\#}\left(\frac{1}{2}{M}_{G_0}{I}_n\right)\hfill \\ & +\frac{1}{2(n+m)}\left(\begin{array}{cc}-J_{m{n}_2\times m}& 0\\ 0& J_{n{n}_1\times n}\end{array}\right)-\frac{n}{2{(n+m)}^2}{J}_{(m{n}_2+n{n}_1)\times (m+n)},\hfill \end{array}$$

and similarly, we would have that

$$\begin{array}{cc}\hfill -S^{\#}{L}_2^T{L}_1^{-1}=& S^{\#}\left(\begin{array}{cc}{I}_m\otimes {\mathbf{1}}_{n_2}^T{(L_{G_2}+I_{n_2})}^{-1}& 0\\ 0& I_n\otimes {\mathbf{1}}_{n_1}^T{(L_{G_1}+I_{n_1})}^{-1}\end{array}\right)\hfill \\ \hfill =& S^{\#}\left(\begin{array}{cc}{I}_m\otimes {\mathbf{1}}_{n_2}^T& 0\\ 0& I_n\otimes {\mathbf{1}}_{n_1}^T\end{array}\right)\hfill \\ \hfill =& \frac{1}{2}\left(\begin{array}{cc}{I}_m\otimes {\mathbf{1}}_{n_2}^T& 0\\ 0& 0\end{array}\right)+\left(\begin{array}{c}\frac{1}{2}{M}_{G_0}^T\\ I_n\end{array}\right)\frac{2}{3}{L}_{G_0}^{\#}\left(\frac{1}{2}{M}_{G_0}\otimes {\mathbf{1}}_{n_2}^T{I}_n\otimes {\mathbf{1}}_{n_1}^T\right)\hfill \\ & +\frac{1}{2(n+m)}\left(\begin{array}{cc}-J_{m\times m{n}_2}& 0\\ 0& J_{n\times n{n}_1}\end{array}\right)-\frac{n}{2{(n+m)}^2}{J}_{(m+n)\times (m{n}_2+n{n}_1)}.\hfill \end{array}$$

Furthermore,

$$\begin{array}{cc}\hfill L_1^{-1}{L}_2{S}^{\#}{L}_2^T{L}_1^{-1}=& \frac{1}{2}\left(\begin{array}{cc}{I}_m\otimes J_{n_2\times n_2}& 0\\ 0& 0\end{array}\right)+\left(\begin{array}{c}\frac{1}{2}{M}_{G_0}^T\otimes {\mathbf{1}}_{n_2}\\ I_n\otimes {\mathbf{1}}_{n_1}\end{array}\right)\frac{2}{3}{L}_{G_0}^{\#}\left(\frac{1}{2}{M}_{G_0}\otimes {\mathbf{1}}_{n_2}^T{I}_n\otimes {\mathbf{1}}_{n_1}^T\right)\hfill \\ & +\frac{1}{2(n+m)}\left(\begin{array}{cc}-J_{m{n}_2\times m{n}_2}& 0\\ 0& J_{n{n}_1\times n{n}_1}\end{array}\right)-\frac{n}{2{(n+m)}^2}{J}_{(m{n}_2+n{n}_1)\times (m{n}_2+n{n}_1)}.\hfill \end{array}$$

Step 5. Since $L_G$ is the Laplacian matrix of $G=G_0^{\left(R\right)}\circ \{G_1,G_2\}$, we have

$$L_G{\mathbf{1}}_{m(n_2+1)+n(n_1+1)}=0,{\mathbf{1}}_{m(n_2+1)+n(n_1+1)}^T{L}_G=0,$$

which shows that $L_G{J}_{m(n_2+1)+n(n_1+1)}{L}_G=0$. Moreover,

$$\begin{array}{c}{L}_G\left(\begin{array}{cccc}-J_{m{n}_2\times m{n}_2}& 0& -J_{m{n}_2\times m}& 0\\ 0& J_{n{n}_1\times n{n}_1}& 0& J_{n{n}_1\times n}\\ -J_{m\times m{n}_2}& 0& -J_{m\times m}& 0\\ 0& J_{n\times n{n}_1}& 0& J_{n\times n}\end{array}\right)L_G\hfill \\ =\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ -2J_{m\times m{n}_2}& -2J_{m\times n{n}_1}& -2J_{m\times m}& -2J_{m\times n}\\ r{J}_{n\times m{n}_2}& r{J}_{n\times n{n}_1}& r{J}_{n\times m}& r{J}_{n\times n}\end{array}\right)L_G\hfill \\ =0.\hfill \end{array}$$

So from Lemmas 2 and 8, we know that the following matrix is also a symmetric $\left\{1\right\}$-inverse of $L_G$,

$$\begin{array}{c}\left(\begin{array}{cccc}{I}_m\otimes {(L_{G_2}+I_{n_2})}^{-1}& 0& 0& 0\\ 0& I_n\otimes {(L_{G_1}+I_{n_1})}^{-1}& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)+\frac{1}{2}\left(\begin{array}{cccc}{I}_m\otimes J_{n_2\times n_2}& 0& I_m\otimes {\mathbf{1}}_{n_2}& 0\\ 0& 0& 0& 0\\ I_m\otimes {\mathbf{1}}_{n_2}^T& 0& I_m& 0\\ 0& 0& 0& 0\end{array}\right)\hfill \\ +\left(\begin{array}{c}\frac{1}{2}{M}_{G_0}^T\otimes {\mathbf{1}}_{n_2}\\ I_n\otimes {\mathbf{1}}_{n_1}\\ \frac{1}{2}{M}_{G_0}^T\\ I_n\end{array}\right)\frac{2}{3}{L}_{G_0}^{\#}\left(\frac{1}{2}{M}_{G_0}\otimes {\mathbf{1}}_{n_2}^T{I}_n\otimes {\mathbf{1}}_{n_1}^T\frac{1}{2}{M}_{G_0}{I}_n\right).\hfill \end{array}$$

Denote $\left(\begin{array}{c}\frac{1}{2}{M}_{G_0}^T\\ I_n\end{array}\right)\frac{2}{3}{L}_{G_0}^{\#}\left(\frac{1}{2}{M}_{G_0}{I}_n\right)=\left(\begin{array}{cc}\frac{1}{6}{M}_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}& \frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}\\ \frac{1}{3}{L}_{G_0}^{\#}{M}_{G_0}& \frac{2}{3}{L}_{G_0}^{\#}\end{array}\right)$ by $\left(\begin{array}{cc}{X}_{11}& X_{12}\\ X_{12}^T& X_{22}\end{array}\right)$. Then

$$\begin{array}{c}\left(\begin{array}{c}\frac{1}{2}{M}_{G_0}^T\otimes {\mathbf{1}}_{n_2}\\ I_n\otimes {\mathbf{1}}_{n_1}\\ \frac{1}{2}{M}_{G_0}^T\\ I_n\end{array}\right)\frac{2}{3}{L}_{G_0}^{\#}\left(\frac{1}{2}{M}_{G_0}\otimes {\mathbf{1}}_{n_2}^T{I}_n\otimes {\mathbf{1}}_{n_1}^T\frac{1}{2}{M}_{G_0}{I}_n\right)\hfill \\ =\left(\begin{array}{cccc}{X}_{11}\otimes J_{n_2\times n_2}& X_{12}\otimes J_{n_2\times n_1}& X_{11}\otimes {\mathbf{1}}_{n_2}& X_{12}\otimes {\mathbf{1}}_{n_2}\\ X_{12}^T\otimes J_{n_1\times n_2}& X_{22}\otimes J_{n_1\times n_1}& X_{12}^T\otimes {\mathbf{1}}_{n_1}& X_{22}\otimes {\mathbf{1}}_{n_1}\\ X_{11}\otimes {\mathbf{1}}_{n_2}^T& X_{12}\otimes {\mathbf{1}}_{n_1}^T& X_{11}& X_{12}\\ X_{12}^T\otimes {\mathbf{1}}_{n_2}^T& X_{22}\otimes {\mathbf{1}}_{n_1}^T& X_{12}^T& X_{22}\end{array}\right).\hfill \end{array}$$

Therefore, we complete the proof of Theorem 1. □

Please note that $\left(\begin{array}{cc}2I_m& -M_{G_0}^T\\ -M_{G_0}& 2D_{G_0}-A_{G_0}\end{array}\right)$ is just the Laplacian matrix of $G_0^{\left(R\right)}$. So, from Step 3, we obtain the group inverse of $L\left(G_0^{\left(R\right)}\right)$.

Corollary 1.

Let $G_0$ be a connected r-regular graph on n vertices and m edges, whose incidence matrix is $M_{G_0}$. Then

$$\begin{array}{cc}\hfill L{\left(G_0^{\left(R\right)}\right)}^{\#}=& \frac{1}{2}\left(\begin{array}{cc}{I}_m& 0\\ 0& 0\end{array}\right)+\left(\begin{array}{c}\frac{1}{2}{M}_{G_0}^T\\ I_n\end{array}\right)\frac{2}{3}{L}_{G_0}^{\#}\left(\frac{1}{2}{M}_{G_0}{I}_n\right)\hfill \\ & +\frac{1}{2(n+m)}\left(\begin{array}{cc}-J_{m\times m}& 0\\ 0& J_{n\times n}\end{array}\right)-\frac{n}{2{(n+m)}^2}{J}_{(m+n)\times (m+n)}.\hfill \end{array}$$

Next we consider two special situations of $G_0^{\left(R\right)}\circ \{G_1,G_2\}$. By choosing $G_2$ as a null-graph, we would reduce $G=G_0^{\left(R\right)}\circ \{G_1,G_2\}$ to R-vertex corona $G_0\odot G_1$ [24]. If $G_1$ is a null-graph, then $G=G_0^{\left(R\right)}\circ \{G_1,G_2\}$ reduces to R-edge corona $G_0\ominus G_2$ [24]. Thus, from Theorem 1, we obtain a $\left\{1\right\}$-inverse of $L(G_0\odot G_1)$ and $L(G_0\ominus G_2)$.

Corollary 2.

Denote $\left(\begin{array}{cc}\frac{1}{6}{M}_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}& \frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}\\ \frac{1}{3}{L}_{G_0}^{\#}{M}_{G_0}& \frac{2}{3}{L}_{G_0}^{\#}\end{array}\right)$ by $\left(\begin{array}{cc}{X}_{11}& X_{12}\\ X_{12}^T& X_{22}\end{array}\right)$.

$\left(1\right)$ The following matrix is a $\left\{1\right\}$-inverse of $L(G_0\odot G_1)$,

$$\left(\begin{array}{ccc}{I}_n\otimes {(L_{G_1}+I_{n_1})}^{-1}& 0& 0\\ 0& \frac{1}{2}{I}_m& 0\\ 0& 0& 0\end{array}\right)+\left(\begin{array}{ccc}{X}_{22}\otimes J_{n_1\times n_1}& X_{12}^T\otimes {\mathbf{1}}_{n_1}& X_{22}\otimes {\mathbf{1}}_{n_1}\\ X_{12}\otimes {\mathbf{1}}_{n_1}^T& X_{11}& X_{12}\\ X_{22}\otimes {\mathbf{1}}_{n_1}^T& X_{12}^T& X_{22}\end{array}\right).$$

$\left(2\right)$ The following matrix is a $\left\{1\right\}$-inverse of $L(G_0^{\left(R\right)}\ominus G_2)$,

$$\left(\begin{array}{ccc}{I}_m\otimes \left({(L_{G_2}+I_{n_2})}^{-1}+\frac{1}{2}{J}_{n_2\times n_2}\right)& \frac{1}{2}{I}_m\otimes {\mathbf{1}}_{n_2}& 0\\ \frac{1}{2}{I}_m\otimes {\mathbf{1}}_{n_2}^T& \frac{1}{2}{I}_m& 0\\ 0& 0& 0\end{array}\right)+\left(\begin{array}{ccc}{X}_{11}\otimes J_{n_2\times n_2}& X_{11}\otimes {\mathbf{1}}_{n_2}& X_{12}\otimes {\mathbf{1}}_{n_2}\\ X_{11}\otimes {\mathbf{1}}_{n_2}^T& X_{11}& X_{12}\\ X_{12}^T\otimes {\mathbf{1}}_{n_2}^T& X_{12}^T& X_{22}\end{array}\right).$$

Example 2.

Compute resistance distance in $G=P_2^{\left(R\right)}\circ \{P_2,P_2\}$, see the following Figure 1.

Step 1.$L_{G_2}+I_{n_2}=L_{G_1}+I_{n_1}=\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$. So

$${(L_{G_2}+I_{n_2})}^{-1}={(L_{G_1}+I_{n_1})}^{-1}=\frac{1}{3}\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right),$$

and $\left(\begin{array}{cccc}{I}_m\otimes {(L_{G_2}+I_{n_2})}^{-1}& 0& 0& 0\\ 0& I_n\otimes {(L_{G_1}+I_{n_1})}^{-1}& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)=\left(\begin{array}{ccc}{I}_3\otimes \frac{1}{3}\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$.

Step 2.$\frac{1}{2}\left(\begin{array}{cccc}{I}_m\otimes J_{n_2\times n_2}& 0& I_m\otimes {\mathbf{1}}_{n_2}& 0\\ 0& 0& 0& 0\\ I_m\otimes {\mathbf{1}}_{n_2}^T& 0& I_m& 0\\ 0& 0& 0& 0\end{array}\right)=\frac{1}{2}\left(\begin{array}{cccc}{J}_{2\times 2}& 0& {\mathbf{1}}_2& 0\\ 0& 0& 0& 0\\ {\mathbf{1}}_2^T& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right)$.

Step 3.$M_{G_0}=\left(\begin{array}{c}1\\ 1\end{array}\right)$ and $L_{G_0}=\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$. Hence

$$L_{G_0}^{\#}=\frac{1}{4}{L}_{G_0}=\frac{1}{4}\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right),L_{G_0}^{\#}{M}_{G_0}=\frac{1}{4}\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right).$$

Moreover,

$$M_{G_0}^T{L}_{G_0}^{\#}=\left(00\right),M_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}=0.$$

Therefore,

$$\left(\begin{array}{cc}\frac{1}{6}{M}_{G_0}^T{L}_{G_0}^{\#}{M}_{G_0}& \frac{1}{3}{M}_{G_0}^T{L}_{G_0}^{\#}\\ \frac{1}{3}{L}_{G_0}^{\#}{M}_{G_0}& \frac{2}{3}{L}_{G_0}^{\#}\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& \frac{1}{6}\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)\end{array}\right),$$

and

$$\begin{array}{c}\left(\begin{array}{c}\frac{1}{2}{M}_{G_0}^T\otimes {\mathbf{1}}_{n_2}\\ I_n\otimes {\mathbf{1}}_{n_1}\\ \frac{1}{2}{M}_{G_0}^T\\ I_n\end{array}\right)\frac{2}{3}{L}_{G_0}^{\#}\left(\frac{1}{2}{M}_{G_0}\otimes {\mathbf{1}}_{n_2}^T{I}_n\otimes {\mathbf{1}}_{n_1}^T\frac{1}{2}{M}_{G_0}{I}_n\right)\hfill \\ =\left(\begin{array}{cccc}{0}_{2\times 2}& 0_{2\times 4}& 0_{2\times 1}& 0_{2\times 2}\\ 0_{4\times 2}& \frac{1}{6}\left(\begin{array}{cccc}1& 1& -1& -1\\ 1& 1& -1& -1\\ -1& -1& 1& 1\\ -1& -1& 1& 1\end{array}\right)& 0_{4\times 1}& \frac{1}{6}\left(\begin{array}{cc}1& -1\\ 1& -1\\ -1& 1\\ -1& 1\end{array}\right)\\ 0_{1\times 2}& 0_{1\times 4}& 0& 0_{1\times 2}\\ 0_{2\times 2}& \frac{1}{6}\left(\begin{array}{cccc}1& 1& -1& -1\\ -1& -1& 1& 1\end{array}\right)& 0_{2\times 1}& \frac{1}{6}\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)\end{array}\right).\hfill \end{array}$$

Step 4. From the above and Theorem 1, we know that

$$\left(\begin{array}{ccccc}\frac{1}{3}\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)+\frac{1}{2}{J}_{2\times 2}& 0_{2\times 2}& 0_{2\times 2}& \frac{1}{2}{\mathbf{1}}_{2\times 1}& 0_{2\times 2}\\ 0_{2\times 2}& \frac{1}{3}\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)+\frac{1}{6}{J}_{2\times 2}& \frac{-1}{6}{J}_{2\times 2}& 0_{2\times 1}& \frac{1}{6}\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)\\ 0_{2\times 2}& \frac{-1}{6}{J}_{2\times 2}& \frac{1}{3}\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)+\frac{1}{6}{J}_{2\times 2}& 0_{2\times 1}& \frac{1}{6}\left(\begin{array}{cc}-1& 1\\ -1& 1\end{array}\right)\\ \frac{1}{2}{\mathbf{1}}_{2\times 1}^T& 0_{1\times 2}& 0_{1\times 2}& \frac{1}{2}& 0_{1\times 2}\\ 0_{2\times 2}& \frac{1}{6}\left(\begin{array}{cc}1& 1\\ -1& -1\end{array}\right)& \frac{1}{6}\left(\begin{array}{cc}-1& -1\\ 1& 1\end{array}\right)& 0_{2\times 1}& \frac{1}{6}\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)\end{array}\right),$$

is a $\left\{1\right\}$-inverse of $L_G$. From it, we have the matrix whose $ij$-entry is resistance distance between vertices $v^i$ and $v^j$, for detail,

$$\left(\begin{array}{ccccccccc}0& {\displaystyle \frac{2}{3}}& 2& 2& 2& 2& {\displaystyle \frac{2}{3}}& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{4}{3}}\\ {\displaystyle \frac{2}{3}}& 0& 2& 2& 2& 2& {\displaystyle \frac{2}{3}}& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{4}{3}}\\ 2& 2& 0& {\displaystyle \frac{2}{3}}& 2& 2& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{2}{3}}& {\displaystyle \frac{4}{3}}\\ 2& 2& {\displaystyle \frac{2}{3}}& 0& 2& 2& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{2}{3}}& {\displaystyle \frac{4}{3}}\\ 2& 2& 2& 2& 0& {\displaystyle \frac{2}{3}}& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{2}{3}}\\ 2& 2& 2& 2& {\displaystyle \frac{2}{3}}& 0& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{2}{3}}\\ {\displaystyle \frac{2}{3}}& {\displaystyle \frac{2}{3}}& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{4}{3}}& 0& {\displaystyle \frac{2}{3}}& {\displaystyle \frac{2}{3}}\\ {\displaystyle \frac{4}{3}}& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{2}{3}}& {\displaystyle \frac{2}{3}}& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{2}{3}}& 0& {\displaystyle \frac{2}{3}}\\ {\displaystyle \frac{4}{3}}& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{4}{3}}& {\displaystyle \frac{2}{3}}& {\displaystyle \frac{2}{3}}& {\displaystyle \frac{2}{3}}& {\displaystyle \frac{2}{3}}& 0\end{array}\right).$$

4. Conclusions

The resistance distance which is the effective electrical resistance, has wide application. For this reason, it was widely explored by so many authors. Among topics on resistance distance, its calculation plays an important role. As a continuation of this topic, in this paper, we compute resistance distance in $G=G_0^{\left(R\right)}\circ \{G_1,G_2\}$. As we known, there exists relationship between resistance distance and group inverse $L{\left(G\right)}^{\#}$ or $\left\{1\right\}$-inverse $L{\left(G\right)}^{\left\{1\right\}}$ of $L\left(G\right)$. Therefore, we aim to find a $\left\{1\right\}$-inverse $L{\left(G\right)}^{\left\{1\right\}}$ of $L\left(G\right)$, and finally give one in Theorem 1. At the end of this paper, we give an example to illustrate the correction and efficiency of the proposed method.