Minimum fundamental cycle basis of some bipartite graphs

Abstract

We determine the number of cycles of different lengths in the minimum fundamental cycle basis of some bipartite graphs with centralized spanning trees. As applications, we characterize the minimum fundamental cycle basis of some $(n-k)$-regular balanced graphs with order $2n$.

Keywords:

• Cycle basis of graph
• Minimum cycle basis
• Fundamental cycle basis
• Diameter

1 Introduction

All graphs in this paper are undirected and simple. Let $G=(V,E)$ be a graph. Any subgraph $H$ of $G$ can be represented by an incidence vector $\mathbf{x}\in {\mathbb{Z}}_2^{\left|E\right|}$ such that $e$ is an edge of $H$ if and only if ${\mathbf{x}}_e=1$. A cycle of $G$ is a connected subgraph with all vertices of degree 2. The cycle space of $G$ is the vector space spanned by the incidence vectors of the cycles in $G$ over ${\mathbb{Z}}_2$. A cycle basis of $G$ consists of some cycles which form a basis of the cycle space of $G$. If $G$ is a connected graph with $n$ vertices and $m$ edges, then the cycle matrix corresponding to a cycle basis $\mathcal{B}$ is an $m$ by $m-n+1$ matrix whose rows are indexed by the edges of $G$ and columns by the elements of $\mathcal{B}$. The length of a cycle basis is the sum of the lengths of the cycles in the basis. A minimum cycle basis (or MCB for short) of a graph is a cycle basis with minimum length. One can verify that any MCB of $G$ consists of $m-n+1$ cycles, and the dimension of the cycle space of $G$ is $m-n+1$.

In [1] five different classes of cycle basis are defined, namely: fundamental, weakly fundamental, totally unimodular, integral and undirected cycle basis. In this article, we just consider the fundamental cycle basis of some bipartite graphs.

Definition 1.1

A cycle basis $\mathcal{B}=\{C_1,C_2,\dots ,C_s\}$ of a graph $G$ is called fundamental (or FCB for short), if there exists some spanning tree $T$ of $G$ such that $\mathcal{B}=\left\{C_T\left(e\right)\right|e\in E\left(G\right)\setminus E\left(T\right)\}$, where $C_T\left(e\right)$ denotes the unique cycle in $T\cup \left\{e\right\}$. A minimum fundamental cycle basis of a graph is a fundamental cycle basis with minimum length (or MFCB for short).

Although a minimum cycle basis of a graph can be computed in polynomial time [2], but to give an explicit description of such a basis of some families of graphs seems hard. This problem is solved for some families of graphs, for example see [3–6].

The direct product of graphs $G$ and $H$ is the graph $G\times H$ whose vertex set is the Cartesian product $V\left(G\right)\times V\left(H\right)$ and whose edges are $(w,x)(y,z)$ where $wy\in E\left(G\right)$ and $xz\in E\left(H\right)$. In [6], the authors considered the direct product of a complete graph $K_n$ with $K_2$ ( which is an $(n-1)$-regular balanced bipartite graph with order $2n$), constructed a minimum fundamental cycle basis of the graph.

Motivated by their work, we consider the minimum fundamental cycle basis of some special bipartite graphs, determine the number of cycles of different lengths in their minimum fundamental cycle basis. As applications, we characterize the minimum fundamental cycle basis of some $(n-k)$-regular balanced graphs with order $2n$, which is a generalization of Theorem 3 in [6].

For a subset $S$ of $E\left(G\right)$, we denote by $G-S$ the subgraph of $G$ obtained by deleting the edges of $S$. The neighborhood of a vertex $v$ in $G$ is the set $N_G\left(v\right)=\{w\in V\left(G\right)|vw\in E\left(G\right)\}$. The degree of a vertex $v$ in $G$, denoted by $d_G\left(v\right)$, is $\left|N_G\left(v\right)\right|$. If $G$ is a bipartite graph with two parts $X$ and $Y$, for any $v\in X$ (or $v\in Y$), we define $N_G^{\prime }\left(v\right)=Y\setminus N_G\left(v\right)$ (or $N_G^{\prime }\left(v\right)=X\setminus N_G\left(v\right)$ ), and use $d_G^{\prime }\left(v\right)$ to denote the number of vertices in $N_G^{\prime }\left(v\right)$. If $uv$ is an edge of the bipartite graph $G$, we denote $m_{uv}^{\prime }$ the number of edges between $N_G^{\prime }\left(u\right)$ and $N_G^{\prime }\left(v\right)$.

If $G$ is connected, the distance between vertices $u$ and $v$, denoted by $d_G(u,v)$, is the length of a shortest path in $G$ connecting them. The diameter of $G$, denoted by $d\left(G\right)$, is the largest distance of $G$. For a cycle basis $\mathcal{B}$ of $G$, we use $m_6\left(\mathcal{B}\right)$ to denote the number of cycles of length 6 in $\mathcal{B}$.

2 MFCB of some bipartite graphs with centralized spanning trees

We begin this section with defining a special spanning tree of connected graph.

Definition 2.1

A spanning tree $T$ of $G$ is centralized, if there exists an edge $v_1{v}_2$ of $T$ which satisfies the following two conditions:

1.	$N_T\left(v_i\right)=N_G\left(v_i\right)$, where $i=1,2$;
2.	the distance between $v_i$ and any other vertex (if exists) in the same part of $T$ is 2, where $i=1,2$.

$v_1{v}_2$ is called a key edge of $T$.

If $T$ is a centralized spanning tree, the diameter of $T$ is no more than 5. A centralized spanning tree with diameter 5 has exactly one key edge, but, if the diameter is less than 5, the key edge maybe not unique. We use $\mathcal{G}$ to denote the set of all bipartite graphs with centralized spanning trees, $\mathcal{T}\left(G\right)$ to denote the set of all centralized spanning trees of $G$, and $\mathcal{K}\left(T\right)$ to denote the set of all key edges of the centralized spanning tree $T$. For convenience, for any $G\in \mathcal{G}$, we define $$N^{\prime }\left(G\right)=\underset{\genfrac{}{}{0ex}{}{uv\in \mathcal{K}\left(T\right)}{T\in \mathcal{T}\left(G\right)}}{min}{m}_{uv}^{\prime }.$$If $G$ has a centralized spanning tree $T$ such that $d\left(T\right)\le 4$, it is not difficult to see that for any key edge $uv$, $N_G^{\prime }\left(u\right)$ or $N_G^{\prime }\left(v\right)$ is empty, so there is no edge between $N_G^{\prime }\left(u\right)$ and $N_G^{\prime }\left(v\right)$, and then $m_{uv}^{\prime }=0$, which implies that $N^{\prime }\left(G\right)=0$. So for a bipartite graph $G\in \mathcal{G}$, $$N^{\prime }\left(G\right)=\left\{\begin{array}{cc}\hfill 0,\hfill & \hfill \text{if}G\text{has a centralized spanning tree}T\text{with}d\left(T\right)\le 4;\hfill \\ \hfill \underset{T\in \mathcal{T}\left(G\right)}{min}{m}_{uv}^{\prime },\hfill & \hfill \text{otherwise}.\hfill \end{array}\right.$$

Let $G$ be a connected bipartite graph and let $v\in G$. If $N^{\prime }\left(G\right)=0̸$, then we say that $v$ is saturated. Otherwise $v$ is unsaturated.

Since $G$ is bipartite, the lengths of its cycles are at least 4. In the following lemma we will show that for any FCB of $G\in \mathcal{G}$ produced by a centralized spanning tree does not contain any cycle of length more than 6, and will determine the number of cycles of length 6 in the FCB.

Lemma 2.1

Let $G\in \mathcal{G}$, and$\mathcal{B}$ be a fundamental cycle basis of $G$ corresponding to a centralized spanning tree $T$, then

1.	$\mathcal{B}$ does not contain any cycle of length more than 6,
2.	$m_6\left(\mathcal{B}\right)=m_{uv}^{\prime }$, where $uv$ is a key edge of $T$.

Proof

1.

Note that $d\left(T\right)\le 5$, so there is no cycle with length more than 6 in the corresponding FCB produced by $T$.

2.

Let $V^{\prime }=N_G^{\prime }\left(u\right)\cup N_G^{\prime }\left(v\right)$, where $uv\in \mathcal{K}\left(T\right)$. We consider two cases. Case 1 $u$ or $v$ is saturated. Then the induced subgraph of $V^{\prime }$ is empty (without edges), and the diameter of $T$ is no more than 4. Hence, there is no any cycle of length 6 in $\mathcal{B}$, i.e., $m_6\left(\mathcal{B}\right)=0$, and thus the equality holds. Case 2 Neither $u$ nor $v$ is saturated. Let $V_1$, $V_2$ be the two parts of $G$. Without loss of generality, we suppose $u\in V_1$, $v\in V_2$. Then $V_1$ can be divided into three disjoint parts $\left\{u\right\},N_G\left(v\right)\setminus \left\{u\right\}$ and $N_G^{\prime }\left(v\right)$. Similarly, $V_2$ can be divided into three disjoint parts $\left\{v\right\},N_G\left(u\right)\setminus \left\{v\right\}$ and $N_G^{\prime }\left(u\right)$. It is not difficult to check that only the distances between the vertices in $N_G^{\prime }\left(u\right)$ and the vertices in $N_G^{\prime }\left(v\right)$ are 5. If we add one edge of $G$ between $N_G^{\prime }\left(u\right)$ and $N_G^{\prime }\left(v\right)$ to $T$, we will obtain a cycle of length 6. Hence, $m_6\left(\mathcal{B}\right)=m_{uv}^{\prime }$. __

Lemma 2.2

Let $G\in \mathcal{G}$, and $T$ be the corresponding spanning tree of an MFCB. If for any two vertices$s$,$t$ in the same part of $G$, (1) $$\mid N_G\left(s\right)\cap N_G\left(t\right)\mid >max\{2,N^{\prime }\left(G\right)\}+1$$(1) holds, then there does not exist any path of length 6 in$T$.

Proof

Let $V_1$, $V_2$ be the two parts of $G$, and ${\mathcal{B}}^{\ast }$ be the MFCB produced by $T$. Suppose that $T$ contains a path $P$ of length 6. With no loss of generality, let $P=a_1{b}_1{a}_2{b}_2{a}_3{b}_3{a}_4$, and $\{a_1,a_2,a_3,a_4\}\subseteq V_1$, $\{b_1,b_2,b_3\}\subseteq V_2$. Obviously, $T-E\left(P\right)$ has 7 components. For $u\in V\left(P\right)$, we use $T_u$ to denote the component containing $u$ of $T-E\left(P\right)$. Let $W=(N_G\left(a_1\right)\cap N_G\left(a_4\right))\setminus \{b_1,b_2,b_3\}$, from the inequality (1), we have $$\mid W\mid \ge |N_G\left(a_1\right)\cap N_G\left(a_4\right)|-3>max\{2,N^{\prime }\left(G\right)\}-2\ge 0.$$ So $W$ is non-empty. To prove the result, we show the following claim first.

Claim There are at least $\mid W\mid$ cycles of length at least 6 in ${\mathcal{B}}^{\ast }$.

Since $W$ is non-empty, there are at least one component of $T-E\left(P\right)$, say $T_x$ (where $x\in V\left(P\right)$), shares common vertices with $W$.

Let $\beta \in V\left(T_x\right)\cap W$. Note that $6=d_T(a_1,a_4)=d_T(a_1,x)+d_T(x,a_4)$. If $d_T(a_1,x)\ge d_T(x,a_4)$, i.e., $d_T(a_1,x)\ge 3$ and $x\in \{b_2,b_3,a_3,a_4\}$. From the definition of $W$, we know $\beta a_1$ is an edge of $G$.

We consider the case $x\in \{b_2,b_3\}$ first. It is not difficult to find that $d_T(\beta ,a_1)=d_T(\beta ,x)+d_T(x,a_1)\ge 5$ since both $\beta$ and $x$ are in $V_2$ which yields $d_T(\beta ,x)=d_{T_x}(\beta ,x)\ge 2$. Now adding the edge $\beta a_1$ to $T$, it will produce a cycle of length at least 6 in ${\mathcal{B}}^{\ast }$. And then, $m_6\left({\mathcal{B}}^{\ast }\right)\ge \mid W\mid$.

If $x\in \{a_3,a_4\}$. Observe that $d_T(a_1,x)\ne 3$ for $x$ and $a_1$ are in the same part of bipartite graph, so we have $d_T(a_1,x)\ge 4$. Then $d_T(\beta ,a_1)=d_T(\beta ,x)+d_T(x,a_1)\ge 5$. If adding the edge $\beta a_1$ to $T$, then it will form a cycle of length at least 6 in ${\mathcal{B}}^{\ast }$. In this case, we also have $m_6\left({\mathcal{B}}^{\ast }\right)\ge \mid W\mid$.

If $d_T(a_1,x)<d_T(x,a_4)$, by adding the edge $\beta a_4$ to $T$, we can get the claim similarly.

Now we continue the proof of the lemma. It is not difficult to see that, we can obtain a cycle of length 6 by adding one edge (if there exists in $G$) in $\{a_1{b}_3,a_4{b}_1\}$ to $T$. Let $\mid \{b_1,b_3\}\cap (N_G^{\prime }\left(a_1\right)\cup N_G^{\prime }\left(a_4\right))\mid =k$ (where $k$ is 0, 1 or 2), then $\mid N_G\left(a_1\right)\cap N_G\left(a_4\right)\cap \{b_1,b_3\}\mid =2-k$. Combining these with the above claim, we can conclude that ${\mathcal{B}}^{\ast }$ contains at least $\left|W\right|+2-k$ cycles of length at least 6. By the definition of $W$, we have $$\mid W\mid -k+2=\mid N_G\left(a_1\right)\cap N_G\left(a_4\right)\setminus \{b_1,b_2,b_3\}\mid -k+2=\mid N_G\left(a_1\right)\cap N_G\left(a_4\right)\mid -\mid N_G\left(a_1\right)\cap N_G\left(a_4\right)\cap \{b_1,b_2,b_3\}\mid -k+2\ge \mid N_G\left(a_1\right)\cap N_G\left(a_4\right)\mid -\mid N_G\left(a_1\right)\cap N_G\left(a_4\right)\cap \{b_1,b_3\}\mid -k+1=\mid N_G\left(a_1\right)\cap N_G\left(a_4\right)\mid -1>max\{2,N^{\prime }\left(G\right)\}.$$ From the above discussion, we know that MFCB contains more than $N^{\prime }\left(G\right)$ cycles of length no less than 6. If the dimension of the cycle space of $G$ is $\ell$, then the length of ${\mathcal{B}}^{\ast }$ is more than $4\ell +2N^{\prime }\left(G\right)$.

Note that $G$ is a finite graph, by Lemma 2.1, there exists an FCB, say $\mathcal{B}$, which contains no cycle of length more than 6, so the length of $\mathcal{B}$ is $4\ell +2N^{\prime }\left(G\right)$, which contradicts to the fact that MFCB is an FCB with minimum length. __

Theorem 2.1

Let $G\in \mathcal{G}$. If for any two vertices $s$,$t$ in the same part of $G$, (2) $$\mid N_G\left(s\right)\cap N_G\left(t\right)\mid >max\{2,N^{\prime }\left(G\right)\}+1$$(2) holds, then $G$ has an MFCB whose corresponding spanning tree is centralized.

Proof

We consider two cases.

Case 1 There exists some vertex $v$ of $G$ is saturated.

If one neighbor of $v$ is also saturated, then $G$ has a spanning tree with diameter 3 which is centralized. By adding one edge of $G$ to the spanning tree, we will obtain a cycle of length no more than 4. For $G$ is bipartite, the length of the cycle is 4, and thus the FCB produced by this spanning tree contains only cycles of length 4, so it is an MFCB.

If any neighbor of $v$ is unsaturated, then $G$ has a spanning tree with diameter 4 which is also centralized. If we add one edge of $G$ to the spanning tree, then we will obtain an FCB which contains only cycles of length 4. Since $G$ is bipartite, we deduce that the obtained FCB is consequently an MFCB.

Case 2 All vertices of $G$ are unsaturated.

Let ${\mathcal{B}}^{\ast }$ be an MFCB of $G$, and $F$ be the corresponding spanning tree. By Lemma 2.2, the diameter of $F$ is no more than $5$. We first prove that the diameter is exactly 5.

It is not difficult to see that the diameter of $F$ is not 3. Now suppose that the diameter is 4, and $P=abcde$ is a longest path of $F$. By the definition of unsaturated vertex, there are some vertices of $N_G^{\prime }\left(c\right)$ not in $F$, which contradicts to the fact that $F$ is a spanning tree. So the diameter of $F$ is 5.

Let $P=a_1{b}_1{a}_2{b}_2{a}_3{b}_3$ be a longest path of $F$. Then $m_6\left({\mathcal{B}}^{\ast }\right)$ equals the number of edges in $G$ between $N_F^{\prime }\left(a_2\right)$ and $N_F^{\prime }\left(b_2\right)$. Note that $N_F\left(a_2\right)\subseteq N_G\left(a_2\right)$ and $N_F\left(b_2\right)\subseteq N_G\left(b_2\right)$, we find that $N_G^{\prime }\left(a_2\right)\subseteq N_F^{\prime }\left(a_2\right)$ and $N_G^{\prime }\left(b_2\right)\subseteq N_F^{\prime }\left(b_2\right)$, which yield $m_6\left({\mathcal{B}}^{\ast }\right)\ge m_{a_2{b}_2}^{\prime }$. In light of Lemma 2.1, ${\mathcal{B}}^{\ast }$ contains only cycles of length 4 and 6. Since the dimension of the cycle space is fixed, and ${\mathcal{B}}^{\ast }$ is an MFCB, so we have $m_6\left({\mathcal{B}}^{\ast }\right)=m_{a_2{b}_2}^{\prime }$, which yields that $N_F\left(a_2\right)=N_G\left(a_2\right)$ and $N_F\left(b_2\right)=N_G\left(b_2\right)$. Hence $F$ is a centralized spanning tree. __

Theorem 2.2

Let $G\in \mathcal{G}$. If for any two vertices $s$,$t$ in the same part of $G$, (3) $$\mid N_G\left(s\right)\cap N_G\left(t\right)\mid >max\{2,N^{\prime }\left(G\right)\}+1$$(3) holds, then $G$ has an MFCB which consists of $N^{\prime }\left(G\right)$ cycles of length 6 and the rest of length 4.

Proof

From the proof of Theorem 2.1, if there exists some vertex $v$ of $G$ is saturated, then $G$ has an MFCB whose corresponding spanning tree with diameter less than 5, so $N^{\prime }\left(G\right)=0$, and the desired result holds. If all vertices of $G$ are unsaturated, $G$ has an MFCB with diameter 5, and $m_6\left({\mathcal{B}}^{\ast }\right)=m_{a_2{b}_2}^{\prime }\ge N^{\prime }\left(G\right)$. By Lemma 2.1, $m_6\left({\mathcal{B}}^{\ast }\right)\le N^{\prime }\left(G\right)$, and thus, $m_6\left({\mathcal{B}}^{\ast }\right)=N^{\prime }\left(G\right)$. _

3 MFCB of regular balanced bipartite graph

In this section, we consider the MFCB of regular balanced bipartite graph with centralized spanning trees. A balanced bipartite graph is a bipartite graph whose two parts have equal cardinality.

Theorem 3.1

Let $G\in \mathcal{G}$ be an$(n-k)$-regular balanced bipartite graph with order $2n$. If for any two vertices $s$,$t$ in the same part of $G$ satisfy $$\mid N_G\left(s\right)\cap N_G\left(t\right)\mid >max\{2,k^2\}+1,$$then any minimum fundamental cycle basis ${\mathcal{B}}^{\ast }$ of $G$ consists of$m_6\left({\mathcal{B}}^{\ast }\right)$ cycles of length 6 and the rest of length 4, and $k\le m_6\left({\mathcal{B}}^{\ast }\right)\le k^2$.

Proof

It suffices to prove $k\le m_6\left({\mathcal{B}}^{\ast }\right)\le k^2$.

Since $G$ is a balanced $(n-k)$-regular bipartite graph, for any edge $uv$, both $N_G^{\prime }\left(u\right)$ and $N_G^{\prime }\left(v\right)$ have $k$ vertices, so $m_{uv}^{\prime }\le k^2$, and thus $N^{\prime }\left(G\right)={min}_{\genfrac{}{}{0ex}{}{uv\in \mathcal{K}\left(T\right)}{T\in \mathcal{T}\left(G\right)}}{m}_{uv}^{\prime }\le k^2$. Hence, $\mid N_G\left(s\right)\cap N_G\left(t\right)\mid >max\{2,N^{\prime }\left(G\right)\}+1$ follows from $\mid N_G\left(s\right)\cap N_G\left(t\right)\mid >max\{2,k^2\}+1$.

By Theorem 2.2, the number of cycles of length 6 in ${\mathcal{B}}^{\ast }$ is $N^{\prime }\left(G\right)$, i.e., $m_6\left({\mathcal{B}}^{\ast }\right)=N^{\prime }\left(G\right)$ , so it suffices to prove $N^{\prime }\left(G\right)\ge k$. Observe that for any vertex $\alpha$ of $N_G^{\prime }\left(v\right)$, there are at most $k-1$ vertices in $N_G^{\prime }\left(u\right)$ which are not adjacent to $\alpha$, that is to say, $N_G^{\prime }\left(u\right)$ contains at least one neighbor of $\alpha$. So there are at least $k$ edges between $N_G^{\prime }\left(u\right)$ and $N_G^{\prime }\left(v\right)$. From the arbitrariness of $uv$, we know that $N^{\prime }\left(G\right)\ge k$. _

Theorem 3.2

(1)	If $G=K_{n,n}$ (where $n>1$), then any MFCB of $G$ consists of only cycles of length 4.
(2)	If $G$ is an $(n-1)$-regular balanced bipartite graph with order $2n$ and $n>5$, then any MFCB of $G$ consists of one cycle of length 6 and the rest of length 4.
(3)	If $G$ is an $(n-2)$-regular balanced bipartite graph with order $2n$ and $n>9$, then any MFCB of $G$ consists of $k$ cycles of length 6 and the rest of length 4, where $k\in \{2,3,4\}$.

Proof

(1) It is easy to check that the graph obtained from two stars $K_{1,n-1}$ by adding one edge between the centers is a centralized spanning tree of $K_{n,n}$, and thus $G\in \mathcal{G}$. If $n\le 3$, the result holds obviously. If $n\ge 4$, for any two vertices in the same part have $n$ common neighbors, the desired result holds from Theorem 3.1.

(2) Let $V_1=\{u_1,\dots ,u_n\}$ and $V_2=\{v_1,\dots ,v_n\}$ be the two parts of $G$. Without loss of generality, we suppose $N_G\left(u_1\right)=\{v_1,\dots ,v_{n-1}\}$ and $N_G\left(v_1\right)=\{u_1,\dots ,u_{n-1}\}$. Since $n>2$, there exist $i,j\in \{1,\dots ,n-1\}$ such that $u_n{v}_i\in E\left(G\right)$ and $v_n{u}_j\in E\left(G\right)$. Then the tree with edge set $\{u_1{v}_1,u_1{v}_2\dots ,u_1{v}_{n-1},v_1{u}_2,\dots ,v_1{u}_{n-1},u_n{v}_i,v_n{u}_j\}$ is a centralized spanning tree of $G$, i.e. $G\in \mathcal{G}$. Note that any two vertices in the same part have at least $n-2>3=max\{2,1^2\}+1$ common neighbors, so by Theorem 3.1, we get the result.

(3) Let $V_1=\{u_1,\dots ,u_n\}$ and $V_2=\{v_1,\dots ,v_n\}$ be the two parts of $G$. Without loss of generality, we suppose $N_G\left(u_1\right)=\{v_1,\dots ,v_{n-2}\}$ and $N_G\left(v_1\right)=\{u_1,\dots ,u_{n-2}\}$. Since $n>9$, there exist $i,j,i^{\prime },j^{\prime }\in \{1,\dots ,n-2\}$ such that $\{u_{n-1}{v}_i,u_n{v}_{i^{\prime }}\}\subset E\left(G\right)$ and $\{v_{n-1}{u}_j,v_n{u}_{j^{\prime }}\}\subset E\left(G\right)$. Then the tree with edge set $\{u_1{v}_1,\dots ,u_1{v}_{n-2},v_1{u}_2,\dots ,v_1{u}_{n-2},u_{n-1}{v}_i,u_n{v}_{i^{\prime }},v_{n-1}{u}_j,v_n{u}_{j^{\prime }}\}$ is a centralized spanning tree of $G$, i.e. $G\in \mathcal{G}$. Note that any two vertices in the same part have at least $n-4>5=max\{2,2^2\}+1$ common neighbors, so by Theorem 3.1, we get the result. __

Since $K_n\times K_2$ is the $(n-1)$-regular balanced bipartite graph with order $2n$, Theorem 3 in [6] is a special case of Theorem 3.2(2).

Acknowledgment

The authors are thankful to the referee for his/her valuable comments and suggestions which improved the presentation of the paper. Research supported by the National Natural Science Foundation of China (No. 11101284, 11201303 and 11301340).