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Article

# Resistance Distance in H-Join of Graphs G1,G2,…,Gk

by 1, 1,* and
1
School of Mathematics and Physics, Anhui Jianzhu University, Hefei 230601, China
2
Department of Mathematics, Loyola College, Chennai 600034, India
*
Author to whom correspondence should be addressed.
Mathematics 2018, 6(12), 283; https://doi.org/10.3390/math6120283
Received: 11 October 2018 / Revised: 18 November 2018 / Accepted: 21 November 2018 / Published: 26 November 2018
(This article belongs to the Special Issue Discrete Optimization: Theory, Algorithms, and Applications)

## Abstract

:
In view of the wide application of resistance distance, the computation of resistance distance in various graphs becomes one of the main topics. In this paper, we aim to compute resistance distance in H-join of graphs $G 1 , G 2 , … , G k$. Recall that H is an arbitrary graph with $V ( H ) = { 1 , 2 , … , k }$, and $G 1 , G 2 , … , G k$ are disjoint graphs. Then, the H-join of graphs $G 1 , G 2 , … , G k$, denoted by $⋁ H { G 1 , G 2 , … , G k }$, is a graph formed by taking $G 1 , G 2 , … , G k$ and joining every vertex of $G i$ to every vertex of $G j$ whenever i is adjacent to j in H. Here, we first give the Laplacian matrix of $⋁ H { G 1 , G 2 , … , G k }$, and then give a ${ 1 }$-inverse $L ( ⋁ H { G 1 , G 2 , … , G k } ) { 1 }$ or group inverse $L ( ⋁ H { G 1 , G 2 , … , G k } ) #$ of $L ( ⋁ H { G 1 , G 2 , … , G k } )$. It is well know that, there exists a relationship between resistance distance and entries of ${ 1 }$-inverse or group inverse. Therefore, we can easily obtain resistance distance in $⋁ H { G 1 , G 2 , … , G k }$. In addition, some applications are presented in this paper.

## 1. Introduction

Throughout this paper, “G is a graph” always means that “G is a simple and undirected graph”. Moreover, we denote a graph G by $G = ( V ( G ) , E ( G ) )$, where $V ( G ) = { v 1 , v 2 , … , v n }$ is the vertex set and $E ( G ) = { e 1 , e 2 , … , e m }$ is the edge set of G. Associated with a graph G, some matrices characterize the structure of G, such as the adjacency matrix $A ( G )$, which is an $n × n$ matrix with entry $a i j = 1$ if $v i$ and $v j$ are adjacent in G, and $a i j = 0$ otherwise, the diagonal matrix $D ( G )$ with diagonal entries $d G ( v 1 ) , d G ( v 2 ) , … , d G ( v n )$ and the Laplacian matrix $L ( G )$, which is $D ( G ) − A ( G )$. Let $I n$ denote the unit matrix of order n, 1n be the all-one column vector of dimension n and $J n × m$ be the all-one $n × m$-matrix. For more detail, one can refer to [1,2] for the definitions and notions in the paper.
It is rather clear that, from some given graphs, a big graph arises by the help of graph operations, such as the Cartesian product, the Kronecker product, the corona graph, the neighborhood corona graph and subdivision-vertex join and subdivision-edge join of graphs (see [3,4,5,6,7]). Furthermore, following [8], from an arbitrary graph H of order k and graphs $G 1 , G 2 , … , G k$, we obtain a new graph called H-join of graphs $G 1 , G 2 , … , G k$, which is denoted by $⋁ H { G 1 , G 2 , … , G k }$, for detail:
Definition 1.
Let H be an arbitrary graph with $V ( H ) = { 1 , 2 , … , k }$, and $G 1 , G 2 , … , G k$ be disjoint graphs of orders $n 1 , n 2 , … , n k$. The H-join of graphs $G 1 , G 2 , … , G k$, which is denoted by $⋁ H { G 1 , G 2 , … , G k }$, is a graph formed by taking $G 1 , G 2 , … , G k$ and joining every vertex of $G i$ to every vertex of $G j$ whenever i is adjacent to j in H. Particularly, $⋁ H { G 1 , G 1 , … , G 1 }$ is denoted by $H ⨀ G 1$.
Example 1.
Let $P n$ and $C n$ be a path and a cycle with n vertices. Then, $⋁ P 3 { P 3 , P 1 , P 2 }$, $P 3 ⨀ P 2$ and $C 3 ⨀ P 3$ are as follows (Figure 1 and Figure 2).
As we know, the length of a shortest path between vertices $v i$ and $v j$, which is denoted by $d i j$, is the conventional distance. However, it does not apply to some practical situations, such as electrical network. Thus, based on electrical network theory, Klein and Randić introduced a new distance called resistance distance ([9]). The resistance distance between vertices $v i$ and $v j$ is denoted by $r i j$, and, in fact, $r i j$ is the effective electrical resistance between $v i$ and $v j$ if every edge of G is replaced by a unit resistor. In view of its practical application, resistance distance was widely explored by many authors. One of the main topics in the study of resistance distance is to determine it in various graphs. For example, from [10], one would know that how $r i j$ can be computed from the Laplacian matrix of the graph; in [11], authors gave the resistance distance between any two vertices of a wheel and a fan; in [3], authors obtained formulae for resistance distance in subdivision-vertex join and subdivision-edge join of graphs; recently, in [12], authors gave the resistance distance in corona and the neighborhood corona graphs of two disjoint graphs. Except for the above, one can refer to [13,14,15,16,17,18,19,20] for more information.
Motivated by the study of resistance distance and graph operations, a natural question arises: what is the resistance distance in $⋁ H { G 1 , G 2 , … , G k }$? In fact, this paper focuses on this question, gives resistance distance in H-join of graphs $G 1 , G 2 , … , G k$ and finally presents some applications.

## 2. Preliminaries

Recall that, for a matrix M, a ${ 1 }$-inverse of M, which is always denoted by $M { 1 }$, is a matrix X such that $M X M = M$. For a square matrix M, the group inverse of M, which is denoted by $M #$, is the unique matrix X such that the following hold: $( 1 ) M X M = M ; ( 2 ) X M X = X ; ( 3 ) M X = X M$. It is well-known that $M #$ exists if and only if rank(M)=rank($M 2$). Therefore, $A #$ exists and it is a ${ 1 }$-inverse of A, whenever A is a real symmetric. In fact, assume that A is a real symmetric matrix and U is an orthogonal matrix (i.e., $U U T = U T U = I$), such that $A = U T d i a g { λ 1 , λ 2 , ⋯ , λ n } U$, where $λ 1 , λ 2 , ⋯ , λ n$ are eigenvalues of A. Then, $A # = U T d i a g { f ( λ 1 ) , f ( λ 2 ) , ⋯ , f ( λ n ) } U$, where
$f ( λ i ) = 1 / λ i , if λ i ≠ 0 , 0 , if λ i = 0 .$
Note that the Laplacian matrix $L ( G )$ of a graph G is real symmetric. Thus, $L ( G ) #$ exists. For more detail about the group inverse of the Laplacian matrix of a graph, see [21].
Lemma 1
([3,22]). Let $L = L 1 L 2 L 2 T L 3$ 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,
(1)
$L 1 − 1 + L 1 − 1 L 2 S # L 2 T L 1 − 1 − L 1 − 1 L 2 S # − S # L 2 T L 1 − 1 S #$ is a symmetric ${ 1 }$-inverse of L.
(2)
If each column vector of $L 2$ is $1$ or a zero vector, then $L 1 − 1 0 0 S #$ is a symmetric ${ 1 }$-inverse of L.
In order to compute the inverse of a matrix, the next lemma is useful.
Lemma 2
([3]). Let $M = A B C D$ be a nonsingular matrix. If A and D are nonsingular, then
$M − 1 = A − 1 + A − 1 B S − 1 C A − 1 − A − 1 B S − 1 − S − 1 C A − 1 S − 1 ,$
where $S = D − C A − 1 B$ is the Schur complement of A in M.
One of the important applications of group inverse $L ( G ) #$ or ${ 1 }$-inverse $L ( G ) { 1 }$ is based on the following fact, which gives the formulae for resistance distance.
Lemma 3
([3]). Let G be a connected graph and $( L ( G ) ) i j$ be the $( i , j )$-entry of the Laplacian matrix $L ( G )$. Then,
$r i j ( G ) = ( L ( G ) { 1 } ) i i + ( L ( G ) { 1 } ) j j − ( L ( G ) { 1 } ) i j − ( L ( G ) { 1 } ) j i = ( L ( G ) # ) i i + ( L ( G ) # ) j j − 2 ( L ( G ) # ) i j .$

## 3. Main Results

Now, we turn to compute resistance distance in H-join of graphs $G 1 , G 2 , … , G k$. Denote $G = ⋁ H { G 1 , G 2 , … , G k }$. Keeping Lemma 3 in mind, we only need to compute the group inverse $L ( G ) #$ or a ${ 1 }$-inverse $L ( G ) { 1 }$.
First, we give the Laplacian matrix $L ( G )$ of G.
Theorem 1.
Let H be an arbitrary graph with $V ( H ) = { 1 , 2 , … , k }$, and $G i$ be the disjoint graph of order $n i ( i = 1 , 2 , … , k )$. Assume that the adjacency matrix of H is $A ( H ) = ( a i j ) k$ and
$A ( H ) ( n 1 , n 2 , … , n k ) T = ( m 1 , m 2 , … , m k ) T .$
Denote $G = ⋁ H { G 1 , G 2 , … , G k }$, and label the $n i$ vertices of $G i$ with
$V ( G i ) = { v i n 1 + ⋯ + n i − 1 + 1 , v i n 1 + ⋯ + n i − 1 + 2 , … , v i n 1 + ⋯ + n i − 1 + n i } .$
Then, $V ( G ) = { v 1 1 , … , v 1 n 1 , … , v i n 1 + ⋯ + n i − 1 + 1 , … , v i n 1 + ⋯ + n i − 1 + n i , … , v k n 1 + ⋯ + n k − 1 + 1 , … , v k n 1 + ⋯ + n k − 1 + n k }$, and the Laplacian matrix $L ( G )$ of G is
$L ( G 1 ) + m 1 I n 1 0 ⋯ 0 0 L ( G 2 ) + m 2 I n 2 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ L ( G k ) + m k I n k − a 11 J n 1 × n 1 a 12 J n 1 × n 2 ⋯ a 1 k J n 1 × n k a 21 J n 2 × n 1 a 22 J n 2 × n 2 ⋯ a 2 k J n 2 × n k ⋮ ⋮ ⋮ a k 1 J n k × n 1 a k 2 J n k × n 2 ⋯ a k k J n k × n k .$
Proof.
Clearly, all of the diagonal matrix $D ( G )$, the adjacency matrix $A ( G )$ and the Laplacian matrix $L ( G )$ are partitioned $k × k$-matrixes, whose $( i j )$-entry is a $n i × n j$-matrix. We proceed via the following steps:
(1) The diagonal matrix $D ( G )$ of G.
Obviously, the degree increment of $V ( G i )$ depends on the i-th line $( a i 1 a i 2 ⋯ a i k )$ of $A ( H )$. For detail, if $a i j = 1 , j = 1 , 2 , ⋯ , k$, then every vertex of $G j$ is joined to every vertex of $G i$, that is, the increment of each vertex in $V ( G i )$ is $a i j n j$. Otherwise, that is $a i j = 0$, the increment is zero, which can also be written by $a i j n j$. In general, the degree increment of each vertex of $V ( G i )$ is
$a i 1 n 1 + a i 2 n 2 + ⋯ + a i k n k = m i$
.
Consequently, the diagonal matrix of G is
$D ( G ) = D ( G 1 ) + m 1 I n 1 0 ⋯ 0 0 D ( G 2 ) + m 2 I n 2 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ D ( G k ) + m k I n k .$
(2) The adjacency matrix $A ( G )$ of G.
Similarly, the i-th line of the partitioned matrixes $A ( G )$ also relies on $( a i 1 a i 2 ⋯ a i k )$. Assume that $a i j = 1$. Then, every vertex of $G j$ is joined to every vertex of $G i$. Thus, the $( i j )$-entry of $A ( G )$ is $J n i × n j$, which is $a i j J n i × n j$. If $a i j = 0$, then there is no edge between $V ( G i )$ and $V ( G j )$, that is, the $( i j )$-entry of $A ( G )$ is zero. However, in this case, we can also denote it by $a i j J n i × n j$. Note that the above holds for $i = j$. Therefore, the adjacency matrix of G is
$A ( G ) = A ( G 1 ) 0 ⋯ 0 0 A ( G 2 ) ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ A ( G k ) + a 11 J n 1 × n 1 a 12 J n 1 × n 2 ⋯ a i k J n 1 × n k a 21 J n 2 × n 1 a 22 J n 1 × n 2 ⋯ a 2 k J n 2 × n k ⋮ ⋮ ⋮ a k 1 J n k × n 1 a k 2 J n k × n 2 ⋯ a 2 k J n k × n k .$
(3) The Laplacian matrix $L ( G )$ of G.
With respect to the above results, the Laplacian matrix $L ( G )$ of G is the Theorem 1.  ☐
According to Theorem 1 and Lemma 1, we finally obtain a symmetric ${ 1 }$-inverse of $L ( G )$.
Theorem 2.
Let H be an arbitrary connected graph with $V ( H ) = { 1 , 2 , … , k }$, and $G i$ be disjoint connected graph of order $n i$$( i = 1 , 2 , … , k )$. Assume that $A ( H ) = ( a i j ) k$ and $A ( H ) ( n 1 , n 2 , … , n k ) T = ( m 1 , m 2 , … , m k ) T$. Denote $G = ⋁ H { G 1 , G 2 , … , G k }$. Then, the following matrix
$L 1 − 1 + L 1 − 1 L 2 S # L 2 T L 1 − 1 − L 1 − 1 L 2 S # − S # L 2 T L 1 − 1 S #$
is a symmetric ${ 1 }$-inverse of $L ( G )$, where
$L 1 = L ( G 1 ) + m 1 I n 1 ; L 2 = − ( a 12 J n 1 × n 2 a 13 J n 1 × n 3 ⋯ a 1 k J n 1 × n k ) ; L 3 = d i a g { L ( G 2 ) + m 2 I n 2 , … , L ( G k ) + m k I n k } − ( a i j J n i × n j ) i , j = 2 , 3 , … , k ; S = L 3 − L 2 T L 1 − 1 L 2 = d i a g { L ( G 2 ) + m 2 I n 2 , … , L ( G k ) + m k I n k } − ( ( a i j + a i 1 a 1 j s ) J n i × n j ) i , j = 2 , 3 , … , k = L 3 − ( ( a i 1 a 1 j s ) J n i × n j ) i , j = 2 , 3 , … , k = L 3 − s B B T .$
Here, $s = 1 n 1 T L 1 − 1 1 n 1$ and $B T = a 12 1 n 2 T a 13 1 n 3 T ⋯ a 1 k 1 n k T$.
Proof.
Note that all of H and $G 1 , G 2 , … , G k$ are connected. Thus, it is easy to show that G is connected. By Theorem 1, we have the Laplacian matrix $L ( G )$ of G. In order to give a ${ 1 }$-inverse of $L ( G )$ with the help of Lemma 1, we further divide $L ( G )$ into blocks $L ( G ) = L 1 L 2 L 2 T L 3$, where
$L 1 = L ( G 1 ) + m 1 I n 1 − a 11 J n 1 × n 1 = L ( G 1 ) + m 1 I n 1 ; L 2 = − ( a 12 J n 1 × n 2 a 13 J n 1 × n 3 ⋯ a 1 k J n 1 × n k ) ; L 3 = L ( G 2 ) + m 2 I n 2 ⋯ 0 ⋮ ⋱ ⋮ 0 ⋯ L ( G k ) + m k I n k − a 22 J n 2 × n 2 ⋯ a 2 k J n 2 × n k ⋮ ⋮ a k 2 J n k × n 2 ⋯ a k k J n k × n k .$
Note that $L 2 T = − a 12 J n 2 × n 1 − a 13 J n 3 × n 1 ⋮ − a 1 k J n k × n 1$. Thus, we have
$L 2 T L 1 − 1 L 2 = a 12 J n 2 × n 1 a 13 J n 3 × n 1 ⋱ a 1 k J n k × n 1 L 1 − 1 ( a 12 J n 1 × n 2 a 13 J n 1 × n 3 ⋯ a 1 k J n 1 × n k ) = a 12 a 12 J n 2 × n 1 L 1 − 1 J n 1 × n 2 ⋯ a 12 a 1 k J n 2 × n 1 L 1 − 1 J n 1 × n k a 13 a 12 J n 3 × n 1 L 1 − 1 J n 1 × n 2 ⋯ a 13 a 1 k J n 3 × n 1 L 1 − 1 J n 1 × n k ⋮ ⋮ a 1 k a 12 J n k × n 1 L 1 − 1 J n 1 × n 2 ⋯ a 1 k a 1 k J n k × n 1 L 1 − 1 J n 1 × n k .$
Since $J n i × n 1 L 1 − 1 J n 1 × n j = s J n i × n j$, where $s = 1 n 1 T L 1 − 1 1 n 1 ∈ R$, we have
$L 2 T L 1 − 1 L 2 = s a 12 a 12 J n 2 × n 2 ⋯ a 12 a 1 k J n 2 × n k a 13 a 12 J n 3 × n 2 ⋯ a 13 a 1 k J n 3 × n k ⋮ ⋮ a 1 k a 12 J n k × n 2 ⋯ a 1 k a 1 k J n k × n k = s a 12 1 n 2 a 13 1 n 3 ⋮ a 1 k 1 n k ( a 12 1 n 2 T a 13 1 n 3 T … a 1 k 1 n k T ) .$
Assume that B is a column vector of dimension $n 2 + n 3 + ⋯ + n k$ satisfying
$B T = a 12 1 n 2 T a 13 1 n 3 T … a 1 k 1 n k T .$
Therefore, $S = L 3 − L 2 T L 1 − 1 L 2$ has three forms:
$S = d i a g { L ( G 2 ) + m 2 I n 2 , … , L ( G k ) + m k I n k } − ( ( a i j + a i 1 a 1 j s ) J n i × n j ) i , j = 2 , 3 , … , k = L 3 − s ( a i 1 a 1 j J n i × n j ) i , j = 2 , 3 , … , k = L 3 − s B B T .$
By Lemma 1, we know that Theorem 2 holds.  ☐
Recall that the Kronecker product $A ⊗ B$ ([23]) of two matrices $A = ( a i j ) m × n$ and $B = ( b i j ) p × q$ is an $m p × n q$-matrix obtained from A by replacing every element $a i j$ by $a i j B$. As an application of Theorem 2, we easily obtain a symmetric ${ 1 }$-inverse of $L ( H ⨀ G )$.
Corollary 1.
Let H be an arbitrary connected graph with k vertices and G be a connected graph with n vertices. Assume that $A ( H ) = 0 1 × 1 H 2 H 2 T H 3$ and $n A ( H ) 1 n = n D ( H ) 1 n = ( m 1 , m 2 , … , m k ) T$. Then, the following matrix
$L 1 − 1 + L 1 − 1 L 2 S # L 2 T L 1 − 1 − L 1 − 1 L 2 S # − S # L 2 T L 1 − 1 S #$
is a symmetric ${ 1 }$-inverse of $L ( H ⨀ G )$, where
$L 1 = L ( G ) + m 1 I n ; L 2 = − H 2 ⊗ J n × n ; L 3 = I k − 1 ⊗ L ( G ) + d i a g { m 2 , … , m k } ⊗ I n − H 3 ⊗ J n × n ; S = L 3 − L 2 T L 1 − 1 L 2 = L 3 − s ( H 2 T ⊗ 1 n ) ( H 2 ⊗ 1 n T ) = L 3 − s ( H 2 T H 2 ) ⊗ J n × n .$
Here, $s = 1 n T L 1 − 1 1 n$.

## 4. Some Applications

Now, we give a specific application of formation mentioned in the Section 2. Let A be a real symmetric such that $λ 1 , λ 2 , ⋯ , λ n − 1 , 0$ are eigenvalues of A and 0 is a simple eigenvalue. Assume that A is a real symmetric and U is an orthogonal matrix such that $A = U T d i a g { λ 1 , λ 2 , ⋯ , λ n − 1 , 0 } U$. Then, $A # = U T d i a g { 1 λ 1 , 1 λ 2 , ⋯ , 1 λ n − 1 , 0 } U$.
Example 2.
Compute resistance distance in $G = ⋁ P 3 { P 3 , P 1 , P 2 }$ (see Figure 1).
Step 1. We label the vertices $P 3 = { v 1 1 , v 1 2 , v 1 3 } , P 1 = { v 2 4 } , P 2 = { v 3 5 , v 3 6 }$. Then,
$V ( G ) = { v 1 1 , v 1 2 , v 1 3 , v 2 4 , v 3 5 , v 3 6 } .$
Note that $A ( P 3 ) 3 1 2 = 0 1 0 1 0 1 0 1 0 3 1 2 = 1 5 1 .$ Thus, the Laplacian matrix of G is
$L ( G ) = L ( P 3 ) + I 3 0 0 0 L ( P 1 ) + 5 I 1 0 0 0 L ( P 2 ) + I 2 − 0 3 × 3 J 3 × 1 0 3 × 2 J 1 × 3 0 1 × 1 J 1 × 2 0 2 × 3 J 2 × 1 0 2 × 2 = L 1 L 2 L 2 T L 3 ,$
where $L 1 = L ( P 3 ) + I 3 = 2 − 1 0 − 1 3 − 1 0 − 1 2$, $L 2 = − ( J 3 × 1 0 3 × 2 ) = − 1 0 0 − 1 0 0 − 1 0 0$ and
$L 3 = L ( P 1 ) + 5 I 1 − J 1 × 2 − J 2 × 1 L ( P 2 ) + I 2 = 5 − 1 − 1 − 1 2 − 1 − 1 − 1 2 .$
Step 2.$L 1 − 1 = 1 8 5 2 1 2 4 2 1 2 5$ and so $s = 1 3 T L 1 − 1 1 3 = 3$. By Theorem 2, $B = 1 0 0$ and $S = 2 − 1 − 1 − 1 2 − 1 − 1 − 1 2 .$ By the formula at the beginning of this section, $S # = 1 9 2 − 1 − 1 − 1 2 − 1 − 1 − 1 2 .$ Furthermore, $− L 1 − 1 L 2 S # = 1 9 2 − 1 − 1 2 − 1 − 1 2 − 1 − 1$ and $L 1 − 1 L 2 S # L 2 T L 1 − 1 = 2 9 J 3 × 3$.
Step 3. By Lemma 1 or Theorem 2, $1 8 5 2 1 2 4 2 1 2 5 + 2 9 J 3 × 3 1 9 2 − 1 − 1 2 − 1 − 1 2 − 1 − 1 1 9 2 2 2 − 1 − 1 − 1 − 1 − 1 − 1 1 9 2 − 1 − 1 − 1 2 − 1 − 1 − 1 2$ is a ${ 1 }$-inverse of $L ( G )$.
Step 4. In view of Lemma 3, the matrix whose $( i , j )$-entry is the resistance distance $r i j$ between vertices $v i$ and $v j$ is
$0 5 8 1 5 8 31 24 31 24 5 8 0 5 8 1 2 7 6 7 6 1 5 8 0 5 8 31 24 31 24 5 8 1 2 5 8 0 2 3 2 3 31 24 7 6 31 24 2 3 0 2 3 31 24 7 6 31 24 2 3 2 3 0 .$
Example 3.
Assume that $G = P 3 ⨀ P 2$ (see Figure 1). Then, the Laplacian matrix of G is
$L ( G ) = L ( P 2 ) + 2 I 2 0 0 0 L ( P 2 ) + 4 I 2 0 0 0 L ( P 2 ) + 2 I 2 − 0 2 × 2 J 2 × 2 0 2 × 2 J 2 × 2 0 2 × 2 J 2 × 2 0 2 × 2 J 2 × 2 0 2 × 2 .$
From Theorem 2, we have that the matrix $1 16 7 3 3 7 1 16 1 1 − 1 − 1 1 1 − 1 − 1 1 16 1 1 1 1 − 1 − 1 − 1 − 1 1 48 7 − 1 − 3 − 3 − 1 7 − 3 − 3 − 3 − 3 9 − 3 − 3 − 3 − 3 9$ is a ${ 1 }$-inverse of $L ( G )$.
Thus, the matrix whose $( i , j )$-entry is $r i j$ is
$0 1 2 11 24 11 24 3 4 3 4 1 2 0 11 24 11 24 3 4 3 4 11 24 11 24 0 1 3 11 24 11 24 11 24 11 24 1 3 0 11 24 11 24 3 4 3 4 11 24 11 24 0 1 2 3 4 3 4 11 24 11 24 1 2 0 .$
Example 4.
Assume that $G = C 3 ⨀ P 3$ (see Figure 2). Then, the Laplacian matrix of G is
$L ( G ) = L ( P 3 ) + 6 I 3 0 0 0 L ( P 3 ) + 6 I 3 0 0 0 L ( P 3 ) + 6 I 3 − 0 3 × 3 J 3 × 3 J 3 × 3 J 3 × 3 0 3 × 3 J 3 × 3 J 3 × 3 J 3 × 3 0 3 × 3 .$
Based on Theorem 2, the matrix $A 1 27 J 3 × 3 0 3 × 3 1 27 J 3 × 3 B 0 3 × 3 0 3 × 3 0 3 × 3 S #$ is a ${ 1 }$-inverse of $L ( G )$, where
$A = B = 31 189 1 27 4 189 1 27 4 27 1 27 4 189 1 27 31 189 , S # = 17 189 − 1 27 − 10 189 − 1 27 2 27 − 1 27 − 10 189 − 1 27 17 189 .$
Thus, the matrix whose $( i , j )$-entry is $r i j$ is
$0 5 21 2 7 16 63 5 21 16 63 16 63 5 21 16 63 5 21 0 5 21 5 21 2 9 5 21 5 21 2 9 5 21 2 7 5 21 0 16 63 5 21 16 63 16 63 5 21 16 63 16 63 5 21 16 63 0 5 21 2 7 16 63 5 21 16 63 5 21 2 9 5 21 5 21 0 5 21 5 21 2 9 5 21 16 63 5 21 16 63 2 7 5 21 0 16 63 5 21 16 63 16 63 5 21 16 63 16 63 5 21 16 63 0 5 21 2 7 5 21 2 9 5 21 5 21 2 9 5 21 5 21 0 5 21 16 63 5 21 16 63 16 63 5 21 16 63 2 7 5 21 0 .$

## 5. Conclusions

This paper focuses on resistance distance in H-join of graphs $G 1 , G 2 , … , G k$. Let G be H-join of graphs $G 1 , G 2 , … , G k$. Here we first give the Laplacian matrix $L ( G )$ of G. Then we compute a symmetric ${ 1 }$-inverse of $L ( G )$. Note that there exists a relationship between resistance distance and entries of ${ 1 }$-inverse. So we can easily obtain resistance distance in G.

## Author Contributions

Funding Acquisition, L.Z. and J.-B.L.; Methodology, J.-B.L. and L.Z.; Supervision, M.A.; Writing—Original Draft, L.Z. and J.Z.; All authors read and approved the final manuscript.

## Funding

This work was supported by the Start-Up Scientific Research Foundation of Anhui Jianzhu University (2017QD20), the National Natural Science Foundation of China (11601006), the China Postdoctoral Science Foundation (2017M621579), the Postdoctoral Science Foundation of Jiangsu Province (1701081B), the Project of Anhui Jianzhu University (2016QD116 and 2017dc03) and the Anhui Province Key Laboratory of Intelligent Building and Building Energy Saving.

## Acknowledgments

The authors are grateful to the anonymous reviewers and the editor for the valuable comments and suggestions.

## Conflicts of Interest

The authors declare no conflict of interest.

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Figure 1. $⋁ P 3 { P 3 , P 1 , P 2 }$ and $P 3 ⨀ P 2$.
Figure 1. $⋁ P 3 { P 3 , P 1 , P 2 }$ and $P 3 ⨀ P 2$.
Figure 2. $C 3 ⨀ P 3$.
Figure 2. $C 3 ⨀ P 3$.

## Share and Cite

MDPI and ACS Style

Zhang, L.; Zhao, J.; Liu, J.-B.; Arockiaraj, M. Resistance Distance in H-Join of Graphs G1,G2,,Gk. Mathematics 2018, 6, 283. https://doi.org/10.3390/math6120283

AMA Style

Zhang L, Zhao J, Liu J-B, Arockiaraj M. Resistance Distance in H-Join of Graphs G1,G2,,Gk. Mathematics. 2018; 6(12):283. https://doi.org/10.3390/math6120283

Chicago/Turabian Style

Zhang, Li, Jing Zhao, Jia-Bao Liu, and Micheal Arockiaraj. 2018. "Resistance Distance in H-Join of Graphs G1,G2,,Gk" Mathematics 6, no. 12: 283. https://doi.org/10.3390/math6120283

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