text:
\section{Introduction}
For an interconnection network, one major concern is the fault tolerance. An interconnection network is usually modelled by a connected graph $G=(V, E)$, where nodes represent processors and edges represent communication links between processors. The connectivity is one of the important parameters to evaluate the reliability and fault tolerance of a network. {\it The connectivity $\kappa (G)$} of a graph $G$ is defined as the minimum number of vertices whose deletion results in a disconnected graph. Whitney~\cite{w} provides another definition of connectivity. For any subset $S=\{u, v\}\subseteq V(G)$, let $\kappa_{G}(S)$ denote the maximum number of internally disjoint paths between $u$ and $v$ in $G$. Then $\kappa(G)=min\{\kappa_{G}(S)|S\subseteq V(G)$ and $|S|=2\}$. As generalizations of the traditional connectivity, the {\it generalized $k$-connectivity} and the $r$-component connectivity were introduced by Chartrand $et$ $al.$~\cite{c} in $1984$.

Let $G$ be a non-complete graph. A $r$-component cut of $G$ is a set of vertices whose deletion results in a graph with at least $r$ components. The $r$-component connectivity $c\kappa_{r}(G)$ of a graph $G$ is the size of the smallest $r$-component cut of $G$. By the definition of $c\kappa_{r}(G)$, it can be seen that $ c\kappa_{r+1}(G)\geq c\kappa_{r}(G)$ for every positive integer $r$. The $r$-component connectivity is an extension of the usual connectivity $c\kappa_{2}(G)$. In \cite{Hsu}, Hsu $et$ $al.$ determined the $r$-component connectivity of the hypercube $Q_{n}$ for $r=2,3,\cdots, n+1$. In \cite{Z}, Zhao $et$ $al.$ determine the $r$-component connectivity of the hypercube $Q_{n}$ for $r=n+2, n+3, \cdots, 2n-4$.

Let $S\subseteq V(G)$ and $\kappa_{G}(S)$ denote the maximum number $k$ of edge-disjoint trees $T_{1}, T_{2}, \cdots, \\T_{k}$ in $G$ such that $V(T_{i})\bigcap V(T_{j})=S$ for any $i, j \in \{1, 2, \cdots, k\}$ and $i\neq j$. For an integer $r$ with $2\leq r\leq n$, the {\em generalized $r$-connectivity} of a graph $G$ is defined as $\kappa_{r}(G)= min\{\kappa_{G}(S)|S\subseteq V(G)$ and $|S|=r\}$. It is a parameter that can measure the reliability of a network $G$ to connect any $k$ vertices in $G$. If $\kappa_{r}(G)=m$, then there are $m$ internally disjoint trees in $G$ and each of them connecting the vertices of $S$, where $|S|=r$ be any $S$ of $V(G)$. The generalized $2$-connectivity is exactly the traditional connectivity. In~\cite{l4}, Li $et$ $al.$ derived that it is NP-complete for a general  graph $G$ to decide whether there are $k$ internally disjoint trees connecting $S$, where $k$ is a fixed integer and $S\subseteq V(G).$ There are some results~\cite{l2,l5} about the upper and lower bounds of the generalized connectivity. In addition, there are some results of the generalized $k$-connectivity for some classes of graphs and most of them are about $k=3$. For example, Chartrand {\em et al.}~\cite{ch} studied the generalized connectivity of complete graphs; Li $et$ $al.$~\cite{l1} first studied the generalized $3$-connectivity of Cartesian product graphs, then Li $et$ $al.$ ~\cite{l8} studied the generalized $3$-connectivity of graph products; Li $et$ $al.$~\cite{l3} studied the generalized connectivity of the complete bipartite graphs and Lin $et$ $al.$~\cite{L} studied the generalized $4$-connectivity of hypercubes. As the Cayley graph has some attractive properties to design interconnection networks, Li $et$ $al.$~\cite{l6} studied the generalized $3$-connectivity of star graphs and bubble-sort graphs and Li $et$ $al.$~\cite{l7} studied the generalized $3$-connectivity of the Cayley graph generated by trees and cycles. Except hypercubes and complete bipartite graphs, almost all known exact value of the generalized $k$-connectivity are about $k=3$.

In this paper, we first show that $\kappa_{4}(D_{n})=\kappa_{3}(D_{n})=n-1$ for $n\geq 4$, then we show that $c\kappa_{r+1}(D_{n})=rn-\frac{r(r+1)}{2}+1$ for $n\geq 2$ and $1\leq r \leq n-1$.

The paper is organized as follows. In section 2, some notations and definitions are given. In section 3, the generalized $4$-connectivity and the generalized $3$-connectivity of dual cubes are determined. In section 4, the $r$-component connectivity of the dual cube for $n\geq 2$ and $2\leq r \leq n-1$ is determined. In section 5, the paper is concluded.

\section{Preliminary}

Let $G=(V, E)$ be a simple, undirected graph. Let $|V(G)|$ and $|E(G)|$ denote the order and size of graph $G$, respectively. Let $V^{\prime}\subseteq V(G)$, then $G[V^{\prime}]$ is the subgraph of $G$ whose vertex set is $V^{\prime}$ and whose edge set consists of all edges of $G$ which have both ends in $V^{\prime}$. For a vertex $v\in V(G)$, the set of neighbours of a vertex $v$ in a graph $G$ is denoted by $N_{G}(v)$. For a vertex set $U \subseteq V(G)$, the {\it neighborhood} of $U$ in $G$ is defined as $N_G(U)=\bigcup\limits_{v\in U}N_{G}(v)-U$. Let $d_{G}(v)$ denote the number of edges incident with $v$ and $\delta(G)$ denote the {\em minimum degree} of the graph $G$. A graph is said to be {\em $k$-regular} if for any vertex $v$ of $G$, $d_{G}(v)=k$. Two $xy$- paths $P$ and $Q$ in $G$ are {\em internally disjoint} if they have no common internal vertices, that is $V(P)\bigcap V(Q)=\{x, y\}$. Let $Y\subseteq V(G)$ and $X\subset V(G)\setminus Y$, the $(X, Y)$-paths is a family of internally disjoint paths starting at a vertex $x\in X$, ending at a vertex $y\in Y$ and whose internal vertices belong neither to $X$ nor $Y$. If $X=\{x\}$, then the $(X, Y)$-paths is a family of internal disjoint paths whose starting vertex is $x$ and the terminal vertices are distinct in $Y$, which is referred to as a {\em $k$-fan} from $x$ to $Y$. For terminologies and notations not defined here we follow ~\cite{B}.

The hypercube is one of the most fundamental interconnection network. An $n$-dimensi\\onal hypercube is an undirected graph $Q_{n}=(V,E)$ with $|V|=2^{n}$ and $|E|=n2^{n-1}.$ Each vertex can be represented by an $n$-bit binary string. There is an edge between two vertices whenever their binary string representation differs in only one bit position. The dual-cube was introduced by Li and Peng in \cite{L}. As an invariant of the hypercube, it not only keeps numerous desirable properties of the hypercube, but also reduces the interconnection complexity.  The $n$-dimensional dual cube, denoted by $D_{n}$, has $2^{2n-1}$ vertices, each labeled by a $(2n-1)$-bit binary string $u_{1}u_{2}\cdots u_{2n-1}$, and $u_{i}\in \{0,1\}$ for $i=1,2\cdots, 2n-1.$ Two vertices $u=u_{1}u_{2}\cdots u_{2n-1}$ and $v=v_{1}v_{2}\cdots v_{2n-1}$ are adjacent if and only if the following conditions are satisfied:

$(1)$ $u$ and $v$ differ in exactly one bit position $i$.

$(2)$ If $1\leq i\leq n-1$, then $u_{2n-1}=v_{2n-1}=0$.

$(3)$ If $n\leq i\leq 2n-2$, then $u_{2n-1}=v_{2n-1}=1$.

By the definition of the $n$-dimensional dual cube, we can see that the set of vertices with fixed bits in positions $n,\cdots, 2n-2$ and last bit $0$ form an $(n-1)$-dimensional hypercube, which is called a cluster of class $0$. And the set of vertices with fixed bits in positions $1,\cdots, n-1$ and last bit $1$ also form an $(n-1)$-dimensional hypercube, which is called a cluster of class $1$. Edges connecting two vertices in different clusters of different class are called cross edges. Let $v$ be a vertex of cluster $0 (1)$, then we call the neighbour of the vertex $v$ in cluster $1 (0)$ the outside neighbour of $v$ in $D_{n}$. The $n$-dimensional dual cube has $2^{n-1}$ clusters of class $0$ and $2^{n-1}$ clusters of class $1$. For convenience, we denote the clusters of class $0$ by $D_{01}, D_{02},\cdots $ and $D_{02^{n-1}}$, respectively. And we denote the clusters of class $1$ by $D_{11}, D_{12},\cdots$ and $D_{12^{n-1}}$, respectively. The $3$-dimensional dual cube is shown as figure $1$.

\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.6]{a.eps}
\end{center}
\vskip0.5cm
\caption{The $3-$dimensional dual cube}\label{F1}
\end{figure}

The $n$-dimensional dual cube $D_{n}$ is an $n$-regular graph. It consists of $2^{n}$ clusters, half of them are class $0$, and the other half of them is class $1$. There are many research results about dual cubes, one can refer \cite{Ar, Yang2, Ya} etc. for the detail.

\section{The generalized $4$-connectivity  of dual cubes}

In this section, we will study the generalized $4$-connectivity  of dual cubes. The following results are useful to the main result.

\begin{lem}\label{lem1} Let $D_{n}$ be an $n$-dimensional dual cube for $n\geq 2$. Then the following results hold.
\begin{enumerate}
\item [{\rm (1)}] Any two different vertices in the same cluster of $D_{n}$ have different outside neighbours and they belong to different clusters of $D_{n}$.

\item [{\rm (2)}] There is exactly one cross edge between two clusters of different class in $D_{n}$.
\end{enumerate}
\end{lem}
\f {\bf Proof.} (1) Let $D_{01}, D_{02},\cdots, D_{02^{n-1}}$ be the clusters of class $0$ and $D_{11}, D_{12},\cdots, D_{12^{n-1}}$ be the clusters of class $1$ of $D_{n}$. Without loss of generality, let $u,v\in V(D_{01})$ and $u\neq v$. Let $u=u_{1}u_{2}\cdots u_{n-1}u_{n}\cdots u_{2n-2}0$ and $v=v_{1}v_{2}\cdots v_{n-1}v_{n}\cdots v_{2n-2}0$. As $u\neq v$, then $u_{1}u_{2}\cdots u_{n-1}$ and $v_{1}v_{2}\cdots v_{n-1}$ have at least one different digit. The outside neighbour $u^{\prime}$ of $u$ is $u^{\prime}=u_{1}u_{2}\cdots u_{n-1}u_{n}\cdots u_{2n-2}1$ and
the outside neighbour $v^{\prime}$ of $v$ is $v^{\prime}=v_{1}v_{2}\cdots v_{n-1}v_{n}\cdots v_{2n-2}1$. If $u^{\prime}$ and $v^{\prime}$ belong to the same cluster, then $u_{1}u_{2}\cdots u_{n-1}=v_{1}v_{2}\cdots v_{n-1}$ by the definition of $D_{n}$, which is a contradiction.

(2) By $(1)$, the result holds easily.
\hfill\qed
\begin{lem}{\rm(\cite{B})}\label{lem2}
 $\kappa(Q_{n})=n$ for $n\geq 2$.
\end{lem}

\begin{theorem}{\rm(\cite{L})}\label{lem3}
 The generalized $4$-connectivity of hypercube $Q_{n}$ is $n-1$, where $n\geq 2$.
\end{theorem}

\begin{lem}{\rm(\cite{L})}\label{lem4}
Let $G$ be an $r$-regular graph. If $\kappa_{k}(G)=r-1$, then $\kappa_{k-1}(G)=r-1$, where $k\geq 4$.
\end{lem}

\begin{lem}{\rm(\cite{l5})}\label{lem5}
If there are two adjacent vertices of degree $\delta(G)$, then $\kappa_{k}(G)\leq \delta-1$ for $3\leq k \leq |V(G)|$.
\end{lem}

\begin{lem}{\rm(\cite{B})}\label{lem6}
Let $G=(V, E)$ be a $k$-connected graph, let $x$ be a vertex of $G$, and let $Y\subseteq V\setminus \{x\}$ be a set of at least $k$ vertices of $G$. Then there exists a $k$-fan in $G$ from $x$ to $Y$. That is, there exists a family of $k$ internally vertex-disjoint $(x, Y)$-paths whose terminal vertices are distinct in $Y$.
\end{lem}

\begin{theorem}\label{lem8}
Let $H=D_{n}[(\bigcup_{j=1}^{k}V(D_{0i_{j}}))\bigcup(\bigcup_{i=1}^{l}V(D_{1j_{i}}))]$, where $k\geq 1, l\geq 1,$ $D_{0i_{j}}$ is a cluster of class $0$ of $D_{n}$ and $D_{1j_{i}}$ is a cluster of class $1$ of $D_{n}$. Then $H$ is connected for $n\geq 3$.
\end{theorem}
\f {\bf Proof.} Without loss of generality, let $H=D_{n}[(\bigcup_{j=1}^{k}V(D_{0j}))\bigcup(\bigcup_{i=1}^{l}V(D_{1i}))]$, where $k\geq 1, l\geq 1,$ $D_{0j}$ is a cluster of class $0$ of $D_{n}$ and $D_{1i}$ is a cluster of class $1$ of $D_{n}$. To prove the result, we just need to show for any two distinct
vertices $x$ and $y$ of $H$, there is a path between $x$ and $y$ in $H$.

Case $1$. $x$ and $y$ belong to the same cluster.

Without loss of generality, let $x, y\in V(D_{01})$. As $D_{01}$ is connected. Then there is a path $P$ between $x$ and $y$ in $D_{01}$. As $D_{01}$ is a subgraph of $H$, then $P$ is a path between $x$ and $y$ in $H$.

Case $2$. $x$ and $y$ belong to two clusters of the same class.

Without loss of generality, let $x\in V(D_{01})$ and $y\in V(D_{02})$. By Lemma~\ref{lem1}(1), there is a vertex $u$ of $D_{01}$ such that $u^{1}$, the outside neighbour of $u$, belongs to $D_{11}$. Similarly, there is a vertex $v$ of $D_{02}$ such that $v^{1}$, the outside neighbour of $v$, belongs to $D_{11}$. As $D_{01}, D_{02}$ and $D_{11}$ are connected. Then there is a path $P$ between $x$ and $u$ in $D_{01}$, a path $Q$ between $u^{1}$ and $v^{1}$ in $D_{11}$ and a path $R$ between $y$ and $v$ in $D_{02}$. If $u=x$, then the path $P$ is the single vertex $x$ and if $v=y$, then the path $R$ is the single vertex $y$. Let $\widehat{P}=xPuu^{1}Qv^{1}vRy$, then it is a path between $x$ and $y$ in $H$.

Case $3$. $x$ and $y$ belong to two clusters of different class.

Without loss of generality, let $x\in V(D_{01})$ and $y\in V(D_{11})$. By Lemma~\ref{lem1}(1), there is a vertex $u$ of $D_{01}$ such that $u^{1}$, the outside neighbour of $u$, belongs to $D_{11}$. As $D_{01}$ and $D_{11}$ are connected. Then there is a path $P$ between $x$ and $u$ in $D_{01}$ and a path $Q$ between $u^{1}$ and $y$ in $D_{11}$. If $u=x$, then the path $P$ is the single vertex $x$ and if $u^{1}=y$, then the path $Q$ is the single vertex $y$. Let $\widehat{P}=xPuu^{1}Qy$, then it is a path between $x$ and $y$ in $H$.
\hfill\qed

To prove the generalized $4$-connectivity of dual cubes, the following results are useful.

\begin{theorem}\label{Lem81}
Let $D_{n}$ be an $n$-dimensional dual cube and $S=\{x, y, z, w\}$, where $x, y, z$ and $w$ are any four distinct vertices of $D_{n}$ for $n\geq 4$. If the vertices in $S$ belong to two clusters of $D_{n}$, then there are $n-1$ internally disjoint trees connecting $S$ in $D_{n}$.
\end{theorem}
\f {\bf Proof.} Let $D_{01}, D_{02},\cdots, D_{02^{n-1}}$ be the clusters of class $0$ and $D_{11}, D_{12},\cdots$ and $D_{12^{n-1}}$ be the clusters of class $1$ of $D_{n}$. Let $S=\{x, y, z, w\}$, where $x, y, z$ and $w$ are any four distinct vertices of $D_{n}$ for $n\geq 4$. Let the vertices in $S$ belong to two clusters of $D_{n}$. By the symmetry of $D_{n}$, the following cases are considered.

Case $1$. $|S\bigcap V(D_{0i})|=3$ and $|S\bigcap V(D_{0j})|=1$ for distinct $i,j\in\{1,2,\cdots, 2^{n-1}\}$.

Without loss of generality, let $|S\bigcap V(D_{01})|=3$ and $|S\bigcap V(D_{02})|=1$. Let $\{x,y,z\}\subseteq V(D_{01})$ and $w\in V(D_{02})$. See Figure $2$. As $x\neq y$, without loss of generality, we may assume that the $(n-1)$-th digit of $x$ and $y$ are different. Hence, we may assume that $x=x_{1}x_{2}\cdots x_{n-2}0H0$ and $y=y_{1}y_{2}\cdots y_{n-2}1H0$, where $H=x_{n}x_{n+1}\cdots x_{2n-2}$. Assume the $(n-1)$-th digit of $z$ is $0$ and we divide $D_{01}$ along the $(n-1)$th dimension into two copies of $Q_{n-2}$, denoted by $Q^{n-1}[0]$ and $Q^{n-1}[1]$, respectively. Then $x, z\in V(Q^{n-1}[0])$ and $y\in V(Q^{n-1}[1])$. By Lemma~\ref{lem2}, $\kappa(Q_{n-2})=n-2$. Then there are $n-2$ internally disjoint paths $P_{1}, P_{2},\cdots, P_{n-2}$ between $x$ and $z$ in $Q^{n-1}[0]$. Let $x_{i}\in V(P_{i})$ such that $y_{i}\in V(Q^{n-1}[1])$ and $y_{i}\neq y$, where $y_{i}$ is the neighbour of $x_{i}$ in $Q^{n-1}[1]$. This can be done as any two different vertices in $Q^{n-1}[0]$ have different neighbours in $Q^{n-1}[1]$ and $P_{i}$s are internally disjoint for $1\leq i\leq n-2$. Let $Y=\{y_{1}, y_{2}, \ldots, y_{n-2}\}$. By Lemma~\ref{lem2}, $\kappa(Q^{n-1}[1])=n-2$. By Lemma~\ref{lem6}, there are $n-2$ internally disjoint paths $P_{1}^{\prime}, P_{2}^{\prime},\cdots, P_{n-2}^{\prime}$ from $y$ to $Y$. Let $\widehat{T}_{i}=P_{i}\bigcup x_{i}y_{i}\bigcup P_{i}^{\prime}$ for each $i\in\{1,2,\cdots, n-2\}$, then $n-2$ internally disjoint trees $\widehat{T}_{i}$s for $1\leq i\leq n-2$ that connecting $x, y$ and $z$ are obtained in $D_{01}$.

\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.7]{1.eps}
\end{center}
\vskip0.1cm
\caption{The illustration of Case $1$}\label{F2}
\end{figure}

Let $y_{i}^{1}$ be the outside neighbour of $y_{i}$ in $D_{n}$ for each $i\in\{1, 2, \cdots, n-2\}$. Without loss of generality, let $y_{i}^{1}\in V(D_{1i})$ for $1\leq i\leq n-2$. Then choose $n-1$ vertices $w_{1}, w_{2},\cdots, w_{n-1}$ from $D_{02}$ such that $w_{i}^{1}$, the outside neighbour of $w_{i}$ in $D_{n}$, belongs to $D_{1i}$ for each $i\in\{1, 2, \cdots, n-1\}$. For convenience, let $W=\{w_{1}, w_{2},\cdots, w_{n-1}\}$. By Lemma~\ref{lem2}, $\kappa(D_{02})=n-1$ as $D_{02}$ is a copy of $Q_{n-1}$. By Lemma~\ref{lem6}, there are $n-1$ internally disjoint paths $Q_{1}, Q_{2}, \cdots, Q_{n-1}$ from $w$ to $W$ and let the terminal vertex of $Q_{i}$ is $w_{i}$ for each $i\in\{1, 2, \cdots, n-2\}$. As $D_{1i}$ is connected for $1\leq i\leq n-2$, then there is a path $R_{i}$ between $y_{i}^{1}$ and $w_{i}^{1}$ in $D_{1i}$ for each $i\in\{1, 2, \cdots, n-2\}$.

Let $T_{i}=\widehat{T}_{i}\bigcup Q_{i}\bigcup R_{i}\bigcup y_{i}y_{i}^{1}\bigcup w_{i}w_{i}^{1}$ for each $i\in\{1, 2, \cdots, n-2\}$. Then $n-2$ internally disjoint trees $T_{i}$s for $1\leq i\leq n-2$ that connecting $x, y, z$ and $w$ are obtained. Let $x^{1}, y^{1}$ and $z^{1}$ be the outside neighbours of $x, y$ and $z$ in $D_{n}$, respectively. By Lemma~\ref{lem1}(2), $\{x^{1}, y^{1}, z^{1}, w_{n-1}^{1}\}\subset V(D_{n}-\bigcup_{i=1}^{2}D_{0i}-\bigcup_{i=1}^{n-2}D_{1i})$. As $2^{n-1}> n-2$ for $n\geq 4$. By Theorem~\ref{lem8}, $D_{n}-\bigcup_{i=1}^{2}D_{0i}-\bigcup_{i=1}^{n-2}D_{1i}$ is connected. Then there is a tree $\widetilde{T}_{n-1}$ connecting $x^{1}, y^{1}, z^{1}$ and $w_{n-1}^{1}$ in $D_{n}-\bigcup_{i=1}^{2}D_{0i}-\bigcup_{i=1}^{n-2}D_{1i}$. Let $T_{n-1}=\widetilde{T}_{n-1}\bigcup Q_{n-1}\bigcup xx^{1}\bigcup yy^{1}\bigcup zz^{1}\bigcup w_{n-1}w_{n-1}^{1}$, then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

Case $2$. $|S\bigcap V(D_{0i})|=3$ and $|S\bigcap V(D_{1j})|=1$ for $i,j\in\{1,2,\cdots, 2^{n-1}\}$.

Without loss of generality, let $|S\bigcap V(D_{01})|=3$ and $|S\bigcap V(D_{11})|=1$. Let $\{x,y,z\}\subseteq V(D_{01})$ and $w\in V(D_{11})$. See Figure $3$. Let $Y$ and $\widehat{T}_{i}$s for $1\leq i\leq n-2$ be the same as case $1$. Note that $Y=\{y_{1}, y_{2}, \ldots, y_{n-2}\}$ and $\widehat{T}_{1}, \widehat{T}_{2}, \cdots, \widehat{T}_{n-2}$ are $n-2$ internally disjoint trees connecting $x, y$ and $z$ in $D_{01}$. Let $Y^{1}=\{y_{1}^{1}, y_{2}^{1},\ldots, y_{n-2}^{1}\}$, where $y_{i}^{1}$ is the outside neighbour of $y_{i}$ in $D_{n}$ for each $y_{i}\in Y$.

Subcase $2.1$. If no element in $Y^{1}$ belongs to $D_{11}$. Without loss of generality, let $y_{i}^{1}\in V(D_{1(i+1)})$ for each $i\in\{1,2,\cdots, n-2\}$. Let $N_{D_{11}}(w)=\{w_{1}, w_{2},\cdots,w_{n-2}, w_{n-1}\}$. Then choose $n-2$ vertices from $N_{D_{11}}(w)$ such that none of them with outside neighbour in $D_{01}$, say $w_{1}, w_{2},\cdots,w_{n-2}$ and let $W=\{w_{1}, w_{2},\cdots,w_{n-2}\}$. Without loss of generality, let $w_{i}^{0}\in V(D_{0(i+1)})$, where $w_{i}^{0}$ is the outside neighbour  of $w_{i}$ in $D_{n}$. By Theorem~\ref{lem8}, $D_{n}[V(D_{0(i+1)})\bigcup V(D_{1(i+1)})]$ is connected. Then there is a path $Q_{i}$ between $w_{i}^{0}$ and $y_{i}^{1}$ in $D_{n}[V(D_{0(i+1)})\bigcup V(D_{1(i+1)})]$ for each $i\in\{1,2,\cdots, n-2\}$. Let $T_{i}=\widehat{T}_{i}\bigcup Q_{i}\bigcup y_{i}y_{i}^{1}\bigcup w_{i}w_{i}^{0}\bigcup \\w_{i}w$ for each $i\in\{1,2,\cdots, n-2\}$. Then $n-2$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-2$ that connecting $x, y, z$ and $w$ are obtained. Let $x^{1}, y^{1}, z^{1}, w^{0}$ and $w_{n-1}^{0}$ be the outside neighbours of $x, y, z, w$ and $w_{n-1}$ in $D_{n}$, respectively.

Subcase $2.1.1$. If one of $x^{1}, y^{1}$ and $z^{1}$ belongs to $D_{11}$, say $x^{1}$. By Lemma~\ref{lem2}, $\kappa(D_{11})=n-1$. Then $D_{11}\setminus W$ is connected. Then there is a path $Q$ between $x^{1}$ and $w$ in $D_{11}\setminus W$.

If $x^{1}\neq w$. By Lemma~\ref{lem1}(2), $\{y^{1}, z^{1}, w^{0}\}\subseteq V(D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i})$. As $2^{n-1}> n-1$ for $n\geq 4$. By Theorem~\ref{lem8}, $D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i}$ is connected. Then there is a tree $\widetilde{T}_{n-1}$ connecting $y^{1}, z^{1}$ and $w^{0}$ in $D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i}$. Let $T_{n-1}= xx^{1}\bigcup Q\bigcup ww^{0}\bigcup yy^{1}\bigcup zz^{1}\bigcup \widetilde{T}_{n-1}$, then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

If $x^{1}=w$. By Lemma~\ref{lem1}(2), $\{w_{n-1}^{0}, y^{1}, z^{1}\}\subseteq V(D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i})$. As $2^{n-1}> n-1$ for $n\geq 4$. By Theorem~\ref{lem8}, $D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i}$ is connected. Then there is a tree $\widetilde{T}_{n-1}$ connecting $y^{1}, z^{1}$ and $w_{n-1}^{0}$ in $D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i}$. Let $T_{n-1}= xw\bigcup ww_{n-1}\bigcup w_{n-1}w_{n-1}^{0}\bigcup \widetilde{T}_{n-1}\bigcup yy^{1}\bigcup zz^{1}$. Then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

Subcase $2.1.2$. If none of $x^{1}, y^{1}$ and $z^{1}$ belongs to $D_{11}$. Then $\{x^{1}, y^{1}, z^{1}\}\subseteq V(D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i})$. By Lemma~\ref{lem1}(2), at least one of $w^{0}$ and $w_{n-1}^{0}$ does not belong to $D_{01}$.

If $w^{0}\notin V(D_{01})$, then $w^{0}\in V(D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i})$. By Theorem~\ref{lem8}, $D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i}$ is connected. Then there is a tree $\widetilde{T}_{n-1}$ connecting $x^{1}, y^{1}$ and $z^{1}$ and $w^{0}$ in $D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i}$. Let $T_{n-1}= \widetilde{T}_{n-1}\bigcup xx^{1}\bigcup yy^{1}\bigcup zz^{1}\bigcup ww^{0}$, then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

If $w_{n-1}^{0}\notin V(D_{01})$, then $w_{n-1}^{0}\in V(D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i})$. By Theorem~\ref{lem8}, $D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i}$ is connected. Then there is a tree $\widetilde{T}_{n-1}$ connecting $x^{1}, y^{1}$ and $z^{1}$ and $w_{n-1}^{0}$ in $D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i}$. Let $T_{n-1}= \widetilde{T}_{n-1}\bigcup xx^{1}\bigcup yy^{1}\bigcup zz^{1}\bigcup w_{n-1}w_{n-1}^{0} \bigcup ww_{n-1}$, then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

Subcase $2.2$. If there is some element in $Y^{1}$ belongs to $D_{11}$. Without loss of generality, let $y_{1}^{1}\in V(D_{11})$ and $y_{i}^{1}\in V(D_{1i})$ for $2\leq i\leq n-2$. Define $W$ same as Subcase $2.1$. By Lemma~\ref{lem2}, $\kappa(D_{11})=n-1$. Then $D_{11}\setminus W$ is connected. Then there is a path $Q$ between $y_{1}^{1}$ and $w$ in $D_{11}\setminus W$. Then let $T_{1}=\widehat{T}_{1}\bigcup y_{1}y_{1}^{1}\bigcup Q$. By Theorem~\ref{lem8}, $D_{n}[V(D_{0(i+1)})\bigcup V(D_{1i})]$ is connected. Then there is a path $Q_{i}$ between $w_{i}^{0}$ and $y_{i}^{1}$ in $D_{n}[V(D_{0(i+1)})\bigcup V(D_{1i})]$ for each $i\in\{2,\cdots, n-2\}$. Let $T_{i}=\widehat{T}_{i}\bigcup Q_{i}\bigcup y_{i}y_{i}^{1}\bigcup w_{i}w_{i}^{0}\bigcup w_{i}w$ for each $2\leq i\leq n-2$. Combining with $T_{1}$, then $n-2$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-2$ that connecting $x, y, z$ and $w$ are obtained.

Subcase $2.2.1$. If $y_{1}^{1}\neq w$. By Lemma~\ref{lem1}(2), $\{x^{1}, y^{1}, z^{1}, w^{1} \}\subseteq V(D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i})$. As $2^{n-1}> n-1$ for $n\geq 4$. By Theorem~\ref{lem8}, $D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i}$ is connected. Then there is a tree $\widetilde{T}_{n-1}$ connecting $x^{1}, y^{1}$ and $z^{1}$ and $w^{1}$ in $D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i}$. Let $T_{n-1}= \widetilde{T}_{n-1}\bigcup xx^{1}\bigcup yy^{1}\bigcup zz^{1}\bigcup ww^{1}$, then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

Subcase $2.2.2$. If $y_{1}^{1}= w$. Then $T_{1}=\widehat{T}_{1}\bigcup y_{1}y_{1}^{1}$. By Lemma~\ref{lem1}(2), $\{w_{n-1}^{0}, x^{1}, y^{1}, z^{1}\}\subseteq V(D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i})$. As $2^{n-1}> n-1$ for $n\geq 4$. By Theorem~\ref{lem8}, $D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i}$ is connected. Then there is a tree $\widetilde{T}_{n-1}$ connecting $w_{n-1}^{0}, x^{1}, y^{1}$ and $z^{1}$ in $D_{n}-\bigcup_{i=1}^{n-1}D_{0i}-\bigcup_{i=1}^{n-1}D_{1i}$. Let $T_{n-1}= \widetilde{T}_{n-1}\bigcup xx^{1}\bigcup yy^{1}\bigcup zz^{1}\bigcup ww_{n-1}\bigcup w_{n-1}w_{n-1}^{0}$, then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.7]{2.eps}
\end{center}
\vskip0.1cm
\caption{The illustration of Subcase $2.1$}\label{F3
\end{figure}

Case $3$. $|S\bigcap V(D_{0i})|=2$ and $|S\bigcap V(D_{0j})|=2$ for $i\neq j$ and $i,j\in\{1,2,\cdots, 2^{n-1}\}$.

Without loss of generality, let $|S\bigcap V(D_{01})|=2$ and $|S\bigcap V(D_{02})|=2$. Let $\{x,y\}\subseteq V(D_{01})$ and $\{z, w\}\subseteq V(D_{02})$. Recall that $D_{ij}$ is a copy of $Q_{n-1}$. By Lemma~\ref{lem2}, $\kappa(D_{01})=\kappa(D_{02})=n-1$. Then there are $n-1$ internally disjoint paths $P_{1}, P_{2}, \cdots, P_{n-1} $ between $x$ and $y$ in $D_{01}$ and $n-1$ internally disjoint paths $P_{1}^{\prime}, P_{2}^{\prime}, \cdots, P_{n-1}^{\prime} $ between $z$ and $w$ in $D_{02}$. Let $x_{i}\in V(P_{i})$ and $y_{i}\in V(P_{i}^{\prime})$ for each $i\in\{1,2,\cdots, n-1\}$. Let $x_{i}^{1}$ and $y_{i}^{1}$ be the outside neighbour of $x_{i}$ and $y_{i}$ in $D_{n}$ for each $i\in\{1,2,\cdots, n-1\}$, respectively. Without loss of generality, let $x_{i}^{1}\in V(D_{1i})$ for each $i\in\{1,2,\cdots, n-1\}$. Let $X^{\prime}=\{x_{1}^{1}, x_{2}^{1},\cdots, x_{n-1}^{1}\}$ and $Y^{\prime}=\{y_{1}^{1}, y_{2}^{1},\cdots, y_{n-1}^{1}\}$, then $X^{\prime}\bigcap Y^{\prime}=\emptyset$.

Subcase $3.1$. If there is no vertex in $X^{\prime}$ belongs to the same cluster with the vertices in $Y^{\prime}$. Then without loss of generality, let $y_{i}^{1}\in V(D_{1(i+n-1)})$ for each $i\in\{1,2,\cdots, n-1\}$. This can be done as $2^{n-1}> 2(n-1)$ for $n\geq 4$. See Figure $4$. By Theorem~\ref{lem8}, $D_{n}[V(D_{1i})\bigcup V(D_{1(i+n-1)})\bigcup V(D_{0(i+n-1)})]$ is connected for each $i\in\{1,2,\cdots, n-1\}$. Then there is a path $Q_{i}$ between $x_{i}^{1}$ and $y_{i}^{1}$ in $D_{n}[V(D_{1i})\bigcup V(D_{1(i+n-1)})\bigcup V(D_{0(i+n-1)})]$ for each $i\in\{1,2,\cdots, n-1\}$. Let $T_{i}=P_{i}\bigcup P_{i}^{\prime}\bigcup Q_{i}\bigcup x_{i}x_{i}^{1}\bigcup y_{i}y_{i}^{1}$ for each $i\in\{1,2,\cdots, n-1\}$. Then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

Subcase $3.2$. If there are some vertices in $X^{\prime}$ belong to the same cluster with some vertices in $Y^{\prime}$.
Then without loss of generality, let $x_{i}^{1}$ and $y_{i}^{1}$ belong to the same cluster for $t\leq i\leq n-1$ and let $y_{i}^{1}\in V(D_{1(i+n-1)})$ for $1\leq i\leq t-1$, where $1\leq t\leq n-1$. By Theorem~\ref{lem8}, $D_{n}[V(D_{1i})\bigcup V(D_{1(i+n-1)})\bigcup V(D_{0(i+n-1)})]$ is connected for $1\leq i\leq t-1$. Then there is a path $Q_{i}$ between $x_{i}^{1}$ and $y_{i}^{1}$ in $D_{n}[V(D_{1i})\bigcup V(D_{1(i+n-1)})\bigcup V(D_{0(i+n-1)})]$ for each $i\in\{1,2,\cdots, t-1\}$. As $x_{i}, y_{i}\in V(D_{1i})$ and $D_{1i}$ is connected for each $i\in\{t, t+1,\cdots, n-1\}$, then there is a path $R_{i}$ between $x_{i}$ and $y_{i}$ in $D_{1i}$ for $t\leq i \leq n-1$. Let $T_{i}=P_{i}\bigcup P_{i}^{\prime}\bigcup Q_{i}\bigcup x_{i}x_{i}^{1}\bigcup y_{i}y_{i}^{1}$ for $1\leq i \leq t-1$ and $T_{i}=P_{i}\bigcup P_{i}^{\prime}\bigcup R_{i}\bigcup x_{i}x_{i}^{1}\bigcup y_{i}y_{i}^{1}$ for $t\leq i \leq n-1$. Then $n-1$ internally disjoint trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.
\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.7]{3.eps}
\end{center}
\vskip0.1cm
\caption{The illustration of Subcase $3.1$}\label{F4
\end{figure}

Case $4$. $|S\bigcap V(D_{0i})|=2$ and $|S\bigcap V(D_{1j})|=2$ for $i,j\in\{1,2,\cdots, 2^{n-1}\}$.

Without loss of generality, let $|S\bigcap V(D_{01})|=2$ and $|S\bigcap V(D_{11})|=2$. Let $\{x,y\}\subseteq V(D_{01})$ and $\{z, w\}\subseteq V(D_{11})$. See Figure $5$. Recall that $D_{ij}$ is a copy of $Q_{n-1}$. By Lemma~\ref{lem2}, $\kappa(D_{01})=\kappa(D_{11})=n-1$. Then there are $n-1$ internally disjoint paths $P_{1}, P_{2}, \cdots, P_{n-1} $ between $x$ and $y$ in $D_{01}$ and $n-1$ internally disjoint paths $P_{1}^{\prime}, P_{2}^{\prime}, \cdots, P_{n-1}^{\prime} $ between $z$ and $w$ in $D_{11}$. Let $x_{i}\in V(P_{i})$ such that $x_{i}^{1}$, the outside neighbour of $x_{i}$, does not belong to $D_{11}$. Without loss of generality, let $x_{i}^{1}\in V(D_{1(i+1)})$. Similarly, let $y_{i}\in V(P_{i}^{\prime})$ such that $y_{i}^{0}$, the outside neighbour of $y_{i}$, does not belong to $D_{01}$. Without loss of generality, let $y_{i}^{0}\in V(D_{0(i+1)})$. This can be done as $2^{n-1}> n$ for $n\geq 4$. By Theorem~\ref{lem8}, $D_{n}[V(D_{0(i+1)})\bigcup V(D_{1(i+1)})]$ is connected. Then there is a path $Q_{i}$ between $x_{i}^{1}$ and $y_{i}^{0}$ in $D_{n}[V(D_{0(i+1)})\bigcup V(D_{1(i+1)})]$ for each $i\in\{1,2,\cdots, n-1\}$. Let $T_{i}=P_{i}\bigcup P_{i}^{\prime}\bigcup Q_{i}\bigcup x_{i}x_{i}^{1}\bigcup y_{i}y_{i}^{0}$ for each $i\in\{1,2,\cdots, n-1\}$. Then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.7]{4.eps}
\end{center}
\vskip0.5cm
\caption{The illustration of Case $4$}\label{F5}
\end{figure}
\hfill\qed

\begin{theorem}\label{Lem82}
Let $D_{n}$ be an $n$-dimensional dual cube and $S=\{x, y, z, w\}$, where $x, y, z$ and $w$ are any four distinct vertices of $D_{n}$ for $n\geq 4$. If the vertices in $S$ belong to three clusters of $D_{n}$, then there are $n-1$ internally disjoint trees connecting $S$ in $D_{n}$.
\end{theorem}
\f {\bf Proof.} Let $D_{01}, D_{02},\cdots, D_{02^{n-1}}$ be the clusters of class $0$ and $D_{11}, D_{12},\cdots$ and $D_{12^{n-1}}$ be the clusters of class $1$ of $D_{n}$. Let $S=\{x, y, z, w\}$, where $x, y, z$ and $w$ are any four distinct vertices of $D_{n}$ for $n\geq 4$. Let the vertices in $S$ belong to three clusters of $D_{n}$. By the symmetry of $D_{n}$, the following cases are considered.

Case $1$. $|S\bigcap V(D_{0i})|=2$, $|S\bigcap V(D_{0j})|=1$ and $|S\bigcap V(D_{0k})|=1$ for distinct $i,j,k\in\{1,2,\cdots, 2^{n-1}\}$.

Without loss of generality, let $|S\bigcap V(D_{01})|=2, |S\bigcap V(D_{02})|=1$ and $|S\bigcap V(D_{03})|=1$. Let $x, y\in V(D_{01}), z\in V(D_{02})$ and $w\in V(D_{03})$. See Figure $6$. Recall that $D_{01}$ is a copy of $Q_{n-1}$. By Lemma~\ref{lem2}, $\kappa(D_{01})=n-1$. Then there are $n-1$ internally disjoint paths $P_{1}, P_{2}, \cdots, P_{n-1} $ between $x$ and $y$ in $D_{01}$. Let $x_{i}\in V(P_{i})$ and $x_{i}^{1}$ be the outside neighbour of $x_{i}$ in $D_{n}$ for $1\leq i\leq n-1$. Without loss of generality, let $x_{i}^{1}\in V(D_{1i})$. Then choose $n-1$ vertices $z_{1}, z_{2},\cdots, z_{n-1}$ from $D_{02}$ such that $z_{i}^{1}$, the outside neighbour  of $z_{i}$, belongs to $D_{1i}$. Similarly, choose $n-1$ vertices $w_{1}, w_{2},\cdots, w_{n-1}$ from $D_{03}$ such that $w_{i}^{1}$, the outside neighbour  of $w_{i}$, belongs to $D_{1i}$ for each $i\in\{1,2,\cdots, n-1\}$. Let $Z=\{z_{1}, z_{2},\cdots, z_{n-1}\}$ and $W=\{w_{1}, w_{2},\cdots, w_{n-1}\}$. By Lemma~\ref{lem2}, $\kappa(D_{02})=\kappa(D_{03})=n-1$. By Lemma~\ref{lem6}, there are $n-1$ internally disjoint paths $Q_{1}, Q_{2},\cdots, Q_{n-1}$ from $z$ to $Z$ in $D_{02}$ and $n-1$ internally disjoint paths $R_{1}, R_{2},\cdots, R_{n-1}$ from $w$ to $W$ in $D_{03}$. It is possible that one of the paths $Q_{i}s (R_{i}s)$ is a single vertex. Let $X^{\prime}=\{x_{i}^{1}, z_{i}^{1}, w_{i}^{1}\}$, then $X^{\prime}\subset V(D_{1i})$. As $D_{1i}$ is connected, then there is a $X^{\prime}$-tree $\widehat{T}_{i}$ in $D_{1i}$ for each $i\in \{1, 2, \cdots, n-1\}$. Let $T_{i}=P_{i}\bigcup Q_{i}\bigcup R_{i}\bigcup \widehat{T}_{i} \bigcup x_{i}x_{i}^{1}\bigcup z_{i}z_{i}^{1}\bigcup w_{i}w_{i}^{1}$ for each $i\in \{1, 2, \cdots, n-1\}$. Then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.7]{5.eps}
\end{center}
\vskip0.1cm
\caption{The illustration of Case $1$}\label{F6}
\end{figure}

Case $2$. $|S\bigcap V(D_{0i})|=2$, $|S\bigcap V(D_{0j})|=1$ and $|S\bigcap V(D_{1k})|=1$ for $i\neq j$ and $i,j,k\in\{1,2,\cdots, 2^{n-1}\}$.

Without loss of generality, let $|S\bigcap V(D_{01})|=2, |S\bigcap V(D_{02})|=1$ and $|S\bigcap V(D_{11})|=1$. Let $\{x,y\}\subseteq V(D_{01}), z\in V(D_{02})$ and $w\in V(D_{11})$. Recall that $D_{01}$ is a copy of $Q_{n-1}$. By Lemma~\ref{lem2}, $\kappa(D_{01})=n-1$. Then there are $n-1$ internally disjoint paths $P_{1}, P_{2}, \cdots, P_{n-1} $ between $x$ and $y$ in $D_{01}$. See Figure $7$. Let $x_{i}\in V(P_{i})$ and $x_{i}^{1}$ be the outside neighbour of $x_{i}$ in $D_{n}$ for each $i\in \{1,2,\cdots, n-1\}$. Without loss of generality, let $x_{i}^{1}\in V(D_{1(i+1)})$. Choose $n-1$ vertices $z_{1}, z_{2},\cdots, z_{n-1}$ from $D_{02}$ such that $z_{i}^{1}$, the outside neighbour of $z_{i}$, belongs to $D_{1(i+1)}$ for each $i\in \{1,2,\cdots, n-1\}$.
Then choose $n-1$ vertices $w_{1}, w_{2},\cdots, w_{n-1}$ from $D_{11}$ such that $w_{i}^{0}$, the outside neighbour of $w$, belongs to neither $D_{01}$ nor $D_{02}$. Without loss of generality, let $w_{i}^{0}\in V(D_{0(i+2)})$. This can be done as $2^{n-1}> n+1$ for $n\geq 4$. Let $W=\{w_{1}, w_{2},\cdots, w_{n-1}\}$ and $Z=\{z_{1}, z_{2},\cdots, z_{n-1}\}$.
By Lemma~\ref{lem6}, there are $n-1$ internally disjoint paths $Q_{1}, Q_{2},\cdots, Q_{n-1}$ from $w$ to $W$ in $D_{11}$ and $n-1$ internally disjoint paths $R_{1}, R_{2},\cdots, R_{n-1}$ from $z$ to $Z$ in $D_{02}$. It is possible that one of the paths $Q_{i}s (Q_{i}s)$ is a single vertex. By Theorem~\ref{lem8}, $D_{n}[V(D_{1(i+1)})\bigcup V(D_{0(i+2)})]$ is connected. Then there is a tree $\widehat{T}_{i}$ connecting  $x_{i}^{1}, z_{i}^{1}$ and $w_{i}^{0}$ in $D_{n}[V(D_{1(i+1)})\bigcup V(D_{0(i+2)})]$. Let $T_{i}=P_{i}\bigcup Q_{i}\bigcup R_{i} \bigcup\widehat{T}_{i}\bigcup x_{i}x_{i}^{1}\bigcup z_{i}z_{i}^{1}\bigcup w_{i}w_{i}^{0}$ for each $i\in \{1, 2, \cdots, n-1\}$. Then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.7]{6.eps}
\end{center}
\vskip0.1cm
\caption{The illustration of Case $2$}\label{F7}
\end{figure}

Case $3$. $|S\bigcap V(D_{0i})|=2$, $|S\bigcap V(D_{1j})|=1$ and $|S\bigcap V(D_{1k})|=1$ for $j\neq k$ and $i,j,k\in\{1,2,\cdots, 2^{n-1}\}$.

Without loss of generality, let $|S\bigcap V(D_{01})|=2, |S\bigcap V(D_{11})|=1$ and $|S\bigcap V(D_{12})|=1$. Let $\{x,y\}\subseteq V(D_{01}), z\in V(D_{11})$ and $w\in V(D_{12})$. Recall that $D_{01}$ is isomorphic to $Q_{n-1}$, then there are $n-1$ internally disjoint paths $P_{1}, P_{2}, \cdots, P_{n-1} $ between $x$ and $y$ in $D_{01}$. See Figure $8$.

Subcase $3.1$. If there exists $x_{i}\in V(P_{i})$ such that $x_{i}^{1}$, the outside neighbour of $x_{i}$, belongs to neither $D_{11}$ nor $D_{12}$ for each path $P_{i}$ with $i\in\{1,2,\cdots, n-1\}$. Without loss of generality, let $x_{i}^{1}\in (V(D_{1(i+2)})$. This can be done as $2^{n-1}> n+1$ for $n\geq 4$. Now, choose $n-1$ vertices $z_{1}, z_{2},\cdots, z_{n-1}$ from $D_{11}\setminus \{z\}$ such that $z_{i}^{0}$, the outside neighbour of $z_{i}$, does not belong to $D_{01}$. Without loss of generality, let $z_{i}^{0}\in V(D_{0(i+1)})$. Then choose $n-1$ vertices $w_{1}, w_{2},\cdots, w_{n-1}$ from $D_{12}\setminus \{w\}$ such that $w_{i}^{0}$, the outside neighbour of $w_{i}$, belongs to $D_{0(i+1)}$ for each $i\in\{1,2,\cdots, n-1\}$. Let $Z=\{z_{1}, z_{2},\cdots, z_{n-1}\}$ and $W=\{w_{1}, w_{2},\cdots, w_{n-1}\}$. By Lemma~\ref{lem2}, $\kappa(D_{11})=\kappa(D_{12})=n-1$. By Lemma~\ref{lem6}, there are $n-1$ internally disjoint paths $Q_{1}, Q_{2},\cdots, Q_{n-1}$ from $z$ to $Z$ in $D_{11}$ and $n-1$ internally disjoint paths $R_{1}, R_{2},\cdots, R_{n-1}$ from $w$ to $W$ in $D_{12}$. By Theorem~\ref{lem8}, $D_{n}[V(D_{0(i+1)})\bigcup V(D_{1(i+2)})]$ is connected for each $i\in\{1,2,\cdots,n-1\}$. Then there is a tree $\widehat{T}_{i}$ connecting $z_{i}^{0}, w_{i}^{0}$ and $x_{i}^{1}$ in $D_{n}[V(D_{0(i+1)})\bigcup V(D_{1(i+2)})]$. Let $T_{i}=P_{i}\bigcup Q_{i}\bigcup R_{i}\bigcup \widehat{T}_{i} \bigcup x_{i}x_{i}^{1}\bigcup z_{i}z_{i}^{0}\bigcup w_{i}w_{i}^{0}$ for each $i\in \{1, 2,\cdots, n-1\}$. Then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.7]{7.eps}
\end{center}
\vskip0.5cm
\caption{The illustration of Subcase $3.1$}\label{F8}
\end{figure}

Subcase $3.2$. If not, then there is a path $P_{i}=xy$ with length one such that $x^{1}\in V(D_{11}), y^{1}\in V(D_{12})$ or $x^{1}\in V(D_{12}), y^{1}\in V(D_{11})$, where $x^{1}$ and $y^{1}$ are the outside neighbour of $x$ and $y$, respectively.
Without loss of generality, let $P_{n-1}=xy, x^{1}\in V(D_{11})$ and $y^{1}\in V(D_{12})$. Let $x_{i}\in V(P_{i})$ and $x_{i}^{1}$ be the outside neighbour of $x_{i}$ for $1\leq i\leq n-2$. Without loss of generality, let $x_{i}^{1}\in (V(D_{1(i+2)})$ for $1\leq i\leq n-2$.
Let $N_{D_{11}}(z)=\{z_{1}, z_{2},\cdots, z_{n-2}, z_{n-1}\}$, $N_{D_{12}}(w)=\{w_{1}, w_{2},\cdots, w_{n-2}, w_{n-1}\}$, $Z=\{z_{1}, z_{2},\cdots, z_{n-2}\}$ and $W=\{w_{1}, w_{2},\cdots, w_{n-2}\}$. By Lemma~\ref{lem2}, $\kappa(D_{11})=\kappa(D_{12})=n-1$. Then $D_{11}\setminus Z$ and $D_{12}\setminus W$ are both connected. Then there is a path $Q$ between $x^{1}$ and $z$ in $D_{11}\setminus Z$ and a path $R$ between $y^{1}$ and $w$ in $D_{12}\setminus W$. Let $T_{1}=P_{n-1}\bigcup xx^{1}\bigcup yy^{1}\bigcup Q\bigcup R$. Let $z_{i}^{0}$ and $w_{i}^{0}$ be the outside neighbour of $z_{i}$ and $w_{i}$, respectively. Without loss of generality, let $z_{i}^{0}\in V(D_{0(i+1)})$. Let $Z^{0}=\{z_{1}^{0}, z_{2}^{0},\cdots, z_{n-2}^{0}\}$ and $W^{0}=\{w_{1}^{0}, w_{2}^{0},\cdots, w_{n-2}^{0}\}$. Then $Z^{0}\bigcap W^{0}=\emptyset$.

If there is no vertex in $Z^{0}$ belongs to the same cluster with the vertices in $W^{0}$. Then without loss of generality, let $w_{i}^{0}\in V(D_{0(i+n-1)})$ for each $i\in\{1,2,\cdots, n-2\}$. This can be done as $2^{n-1}> 2(n-1)$ for $n\geq 4$. By Theorem~\ref{lem8}, $D_{n}[V(D_{1(i+2)})\bigcup V(D_{0(i+1)}\bigcup V(D_{0(i+n-1)}]$ is connected for each $i\in\{1,2,\cdots, n-2\}$. Then there is a tree $\widehat{T}_{i}$ connecting $z_{i}^{0}, w_{i}^{0}$ and $x_{i}^{1}$ in $D_{n}[V(D_{1(i+2)})\bigcup V(D_{0(i+1)}\bigcup V(D_{0(i+n-1)}]$ for each $i\in\{1,2,\cdots, n-2\}$. Let $T_{i}=P_{i}\bigcup \widehat{T}_{i}\bigcup x_{i}x_{i}^{1}\bigcup z_{i}z_{i}^{0}\bigcup w_{i}w_{i}^{0}\bigcup zz_{i}\bigcup ww_{i}$ for each $i\in\{1,2,\cdots, n-2\}$. Combining with $T_{1}$, then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

If there are some vertices in $Z^{0}$ belong to the same cluster with some vertices in $W^{0}$.
Then without loss of generality, let $z_{i}^{0}$ and $w_{i}^{0}$ belong to the same cluster for $t\leq i\leq n-2$ and let $w_{i}^{0}\in V(D_{0(i+n-1)})$ for $1\leq i\leq t-1$, where $1\leq t\leq n-2$. By Theorem~\ref{lem8}, $D_{n}[V(D_{1(i+2)})\bigcup V(D_{0(i+1)})\bigcup V(D_{0(i+n-1)})]$ is connected for $1\leq i\leq t-1$. Then there is a tree $\widehat{T}_{i}$ connecting $z_{i}^{0}, w_{i}^{0}$ and $x_{i}^{1}$ in $D_{n}[V(D_{1(i+2)})\bigcup V(D_{0(i+1)})\bigcup V(D_{0(i+n-1)})]$ for each $i\in\{1,2,\cdots, t-1\}$. As $z_{i}^{0}, w_{i}^{0}\in V(D_{0(i+1)})$ and $D_{n}[V(D_{1(i+2)})\bigcup V(D_{0(i+1)})]$ is connected for each $i\in\{t, t+1,\cdots, n-1\}$, then there is a tree $\widehat{T}_{i}$ connecting $z_{i}^{0}, w_{i}^{0}$ and $x_{i}^{1}$ in $D_{n}[V(D_{1(i+2)})\bigcup V(D_{0(i+1)})]$ for each $t\leq i\leq n-2$. Let $T_{i}=P_{i}\bigcup \widehat{T}_{i}\bigcup x_{i}x_{i}^{1}\bigcup z_{i}z_{i}^{0}\bigcup zz_{i}\bigcup \\ww_{i}\bigcup w_{i}w_{i}^{0}$ for each $i\in\{1,2,\cdots, n-2\}$. Combining with $T_{1}$, then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.
\hfill\qed

\begin{theorem}\label{Lem83}
Let $D_{n}$ be an $n$-dimensional dual cube and $S=\{x, y, z, w\}$, where $x, y, z$ and $w$ are any four distinct vertices of $D_{n}$ for $n\geq 4$. If the vertices in $S$ belong to four clusters of $D_{n}$, then there are $n-1$ internally disjoint trees connecting $S$ in $D_{n}$.
\end{theorem}
\f {\bf Proof.} Let $D_{01}, D_{02},\cdots, D_{02^{n-1}}$ be the clusters of class $0$ and $D_{11}, D_{12},\cdots$ and $D_{12^{n-1}}$ be the clusters of class $1$ of $D_{n}$. Let $S=\{x, y, z, w\}$, where $x, y, z$ and $w$ are any four distinct vertices of $D_{n}$ for $n\geq 4$. Let the vertices in $S$ belong to four clusters of $D_{n}$. By the symmetry of $D_{n}$, the following cases are considered.

Case $1$. $|S\bigcap V(D_{0i})|=1, |S\bigcap V(D_{0j})|=1, |S\bigcap V(D_{0k})|=1$ and $|S\bigcap V(D_{0l})|=1$ for distinct $i,j, k, l\in\{1,2,\cdots, 2^{n-1}\}$.

Without loss of generality, let $|S\bigcap V(D_{01})|=1, |S\bigcap V(D_{02})|=1, |S\bigcap V(D_{03})|=1$ and $|S\bigcap V(D_{04})|=1$. Let $x\in V(D_{01}), y\in V(D_{02}), z\in V(D_{03})$ and $w\in V(D_{04})$. See Figure $9$. Choose $n-1$ vertices $x_{1}, x_{2},\cdots, x_{n-1}$ from $D_{01}\setminus\{x\}$. Without loss of generality, let $x_{i}^{1}\in V(D_{1i})$, where $x_{i}^{1}$ is the outside neighbour of $x_{i}$ in $D_{n}$. Then choose $n-1$ vertices $y_{1}, y_{2},\cdots, y_{n-1}$ from $D_{02}$, $n-1$ vertices $z_{1}, z_{2},\cdots, z_{n-1}$ from $D_{03}$ and $n-1$ vertices $w_{1}, w_{2},\cdots, w_{n-1}$ from $D_{04}$ such that $y_{i}^{1}, z_{i}^{1}$ and $w_{i}^{1}$ all belong to $D_{1i}$, where $y_{i}^{1}, z_{i}^{1}$ and $w_{i}^{1}$ are the outside neighbour $y_{i}, z_{i}$ and $w_{i}$, respectively. Let $X=\{x_{1}, x_{2},\cdots, x_{n-1}\}, Y=\{y_{1}, y_{2},\cdots, y_{n-1}\}, Z=\{z_{1}, z_{2},\cdots, z_{n-1}\}$ and $W=\{w_{1}, w_{2},\cdots, w_{n-1}\}$. By Lemma~\ref{lem6}, there are $n-1$ internally disjoint paths $P_{1}, P_{2}, \cdots, P_{n-1} $ from $x$ to $X$ in $D_{01}$, $n-1$ internally disjoint paths $Q_{1}, Q_{2}, \cdots, Q_{n-1} $ from $y$ to $Y$ in $D_{02}$, $n-1$ internally disjoint paths $R_{1}, R_{2}, \cdots, R_{n-1} $ from $z$ to $Z$ in $D_{03}$ and $n-1$ internally disjoint paths $S_{1}, S_{2}, \cdots, S_{n-1} $ from $w$ to $W$ in $D_{04}$, respectively. It is possible that one of the paths $Q_{i}s(R_{i}s, S_{i}s)$ is a single vertex. As $D_{1i}$ is connected, there is tree $\widehat{T}_{i}$ connecting $x_{i}^{1}, y_{i}^{1}$, $z_{i}^{1}$ and $w_{i}^{1}$ in $D_{1i}$ for each $i\in\{1,2,\cdots, n-1\}$. Let $T_{i}=P_{i}\bigcup Q_{i}\bigcup R_{i}\bigcup S_{i}\bigcup \widehat{T}_{i} \bigcup x_{i}x_{i}^{1}\bigcup y_{i}y_{i}^{1}\bigcup z_{i}z_{i}^{1}\bigcup w_{i}w_{i}^{1}$ for each $i\in \{1, 2, \cdots, n-1\}$. Then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.
\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.7]{8.eps}
\end{center}
\vskip0.5cm
\caption{The illustration of Case $1$}\label{F9}
\end{figure}

Case $2$. $|S\bigcap V(D_{0i})|=1, |S\bigcap V(D_{0j})|=1, |S\bigcap V(D_{0k})|=1$ and $|S\bigcap V(D_{1l})|=1$ for distinct $i,j, k$ and $i,j, k ,l \in\{1,2,\cdots, 2^{n-1}\}$.

Without loss of generality, let $|S\bigcap V(D_{01})|=1, |S\bigcap V(D_{02})|=1, |S\bigcap V(D_{03})|=1$ and $|S\bigcap V(D_{11})|=1$. Let $x\in V(D_{01}), y\in V(D_{02}), z\in V(D_{03})$ and $w\in V(D_{11})$. See Figure $10$. Choose $n-1$ vertices $x_{1}, x_{2},\cdots, x_{n-1}$ from $D_{01}$. Without loss of generality, let $x_{i}^{1}\in V(D_{1(i+3)})$, where $x_{i}^{1}$ is the outside neighbour of $x_{i}$. Then choose $n-1$ vertices $y_{1}, y_{2},\cdots, y_{n-1}$ from $D_{02}$ and $n-1$ vertices $z_{1}, z_{2},\cdots, z_{n-1}$ from $D_{03}$ such that $y_{i}^{1}, z_{i}^{1}\in V(D_{1(i+3)})$, where $y_{i}^{1}$ and $z_{i}^{1}$ are the outside neighbour $y_{i}$ and $z_{i}$ in $D_{n}$, respectively. Then choose $n-1$ vertices $w_{1}, w_{2},\cdots, w_{n-1}$ from $D_{11}$ such that $w_{i}^{0}$, the outside neighbour of $w_{i}$, does not belong to $D_{01}, D_{02}$ and $D_{03}$. Without loss od generality, let $w_{i}^{0}\in V(D_{0(i+3)})$. This can be done as $2^{n-1}> n+2$ for $n\geq 4$. Let $X=\{x_{1}, x_{2},\cdots, x_{n-1}\}, Y=\{y_{1}, y_{2},\cdots, y_{n-1}\}, Z=\{z_{1}, z_{2},\cdots, z_{n-1}\}$ and $W=\{w_{1}, w_{2},\cdots, w_{n-1}\}$. By Lemma~\ref{lem6}, there are $n-1$ internally disjoint paths $P_{1}, P_{2}, \cdots, P_{n-1} $ from $x$ to $X$ in $D_{01}$, $n-1$ internally disjoint paths $Q_{1}, Q_{2}, \cdots, Q_{n-1} $ from $y$ to $Y$ in $D_{02}$, $n-1$ internally disjoint paths $R_{1}, R_{2}, \cdots, R_{n-1} $ from $z$ to $Z$ in $D_{03}$ and $n-1$ internally disjoint paths $S_{1}, S_{2}, \cdots, S_{n-1} $ from $w$ to $W$ in $D_{11}$, respectively. It is possible that one of the paths $P_{i}s(Q_{i}s, R_{i}s, S_{i}s)$ is a single vertex. By Theorem~\ref{lem8}, $D_{n}[V(D_{1(i+3)})\bigcup V(D_{0(i+3)})]$ is connected. Then there is tree $\widehat{T}_{i}$ connecting $x_{i}^{1}, y_{i}^{1}$, $z_{i}^{1}$ and $w_{i}^{0}$ in $D_{n}[V(D_{1(i+3)})\bigcup V(D_{0(i+3)})]$. Let $T_{i}=P_{i}\bigcup Q_{i}\bigcup R_{i}\bigcup S_{i}\bigcup \widehat{T}_{i}\\\bigcup x_{i}x_{i}^{1}\bigcup y_{i}y_{i}^{1}\bigcup z_{i}z_{i}^{1}\bigcup w_{i}w_{i}^{0}$ for each $i\in \{1, 2, \cdots, n-1\}$. Then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.7]{9.eps}
\end{center}
\vskip0.5cm
\caption{The illustration of Case $2$}\label{F10}
\end{figure}

Case $3$. $|S\bigcap V(D_{0i})|=1, |S\bigcap V(D_{0j})|=1, |S\bigcap V(D_{1k})|=1$ and $|S\bigcap V(D_{1l})|=1$ for $i\neq j$, $k\neq l$ and $i,j, k, l\in\{1,2,\cdots, 2^{n-1}\}$.

Without loss of generality, let $|S\bigcap V(D_{01})|=1, |S\bigcap V(D_{02})|=1, |S\bigcap V(D_{11})|=1$ and $|S\bigcap V(D_{12})|=1$. Let $x\in V(D_{01}), y\in V(D_{02}), z\in V(D_{11})$ and $w\in V(D_{12})$. See Figure $11$. Choose $n-1$ vertices $x_{1}, x_{2},\cdots, x_{n-1}$ from $D_{01}$. Without loss of generality, let $x_{i}^{1}\in V(D_{1(i+2)})$, where $x_{i}^{1}$ is the outside neighbour of $x_{i}$ for $1\leq i\leq n-1$. Then choose $n-1$ vertices $y_{1}, y_{2},\cdots, y_{n-1}$ from $D_{02}$ such that $y_{i}^{1}\in V(D_{1(i+2)})$, where $y_{i}^{1}$ is the outside neighbour of $y_{i}$. Similarly, choose $n-1$ vertices $z_{1}, z_{2},\cdots, z_{n-1}$ from $D_{11}$. Without loss of generality, let $z_{i}^{0}\in V(D_{0(i+2)})$, where $z_{i}^{0}$ is the outside neighbour of $z_{i}$ for $1\leq i\leq n-1$. Then choose $n-1$ vertices $w_{1}, w_{2},\cdots, w_{n-1}$ from $D_{12}$ such that $w_{i}^{0}\in V(D_{0(i+2)})$, where $w_{i}^{0}$ is the outside neighbour of $w_{i}$. Let $X=\{x_{1}, x_{2},\cdots, x_{n-1}\}, Y=\{y_{1}, y_{2},\cdots, y_{n-1}\}, Z=\{z_{1}, z_{2},\cdots, z_{n-1}\}$ and $W=\{w_{1}, w_{2},\cdots, w_{n-1}\}$. By Lemma~\ref{lem6}, there are $n-1$ internally disjoint paths $P_{1}, P_{2}, \cdots, P_{n-1} $ from $x$ to $X$ in $D_{01}$, $n-1$ internally disjoint paths $Q_{1}, Q_{2}, \cdots, Q_{n-1} $ from $y$ to $Y$ in $D_{02}$, $n-1$ internally disjoint paths $R_{1}, R_{2}, \cdots, R_{n-1} $ from $z$ to $Z$ in $D_{11}$ and $n-1$ internally disjoint paths $S_{1}, S_{2}, \cdots, S_{n-1} $ from $w$ to $W$ in $D_{12}$, respectively. It is possible that one of the paths $P_{i}s(Q_{i}s, R_{i}s, S_{i}s)$ is a single vertex. By Theorem~\ref{lem8}, $D_{n}[V(D_{0(i+2)})\bigcup V(D_{1(i+2)})]$ is connected for each $i\in\{1,2,\cdots, n-1\}$. Then there is tree $\widehat{T}_{i}$ connecting $x_{i}^{1}, y_{i}^{1}$, $z_{i}^{0}$ and $w_{i}^{0}$ in $D_{n}[V(D_{0(i+2)})\bigcup V(D_{1(i+2)})]$. Let $T_{i}=P_{i}\bigcup Q_{i}\bigcup R_{i}\bigcup S_{i}\bigcup \widehat{T}_{i} \bigcup x_{i}x_{i}^{1}\bigcup y_{i}y_{i}^{1}\bigcup z_{i}z_{i}^{0}\bigcup w_{i}w_{i}^{0}$ for each $i\in \{1, 2, \cdots, n-1\}$. Then $n-1$ internally disjoint $S$-trees $T_{i}$s for $1\leq i\leq n-1$ are obtained in $D_{n}$.

\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=0.7]{10.eps}
\end{center}
\vskip0.5cm
\caption{The illustration of Case $3$}\label{F3}
\end{figure}

\hfill\qed

\begin{theorem}\label{thm1}
Let $D_{n}$ be an $n$-dimensional dual cube, then $\kappa_{4}(D_{n})=n-1$ for $n\geq 4$.
\end{theorem}

\f {\bf Proof.} As $D_{n}$ is $n$ regular, by Lemma~\ref{lem5}, $\kappa_{4}(D_{n})\leq \delta-1= n-1.$ To prove the result, we just need to show that $\kappa_{4}(D_{n})\geq n-1.$

Let $D_{01}, D_{02},\cdots, D_{02^{n-1}}$ be the clusters of class $0$ and $D_{11}, D_{12},\cdots$ and $D_{12^{n-1}}$ be the clusters of class $1$ of $D_{n}$. Let $x, y, z$ and $w$ be any four distinct vertices of $D_{n}$. For convenience, let $S=\{x, y, z, w\}.$ By the symmetry of $D_{n}$, we prove the result by considering the following cases.

Case 1. $x, y, z$ and $w$ belong the same cluster of $D_{n}$.

Without loss of generality, let $S\subseteq V(D_{01})$. Recall that $D_{01}$ is a copy of $Q_{n-1}$. By Theorem~\ref{lem3}, $\kappa_{4}(Q_{n-1})=n-2$. Then there are $(n-2)$-internally disjoint $S$-trees $T_{1}, T_{2},\cdots, T_{n-2}$ in $D_{01}$. Let $x^{1}, y^{1}, z^{1}$ and $w^{1}$ be the outside neighbours of $x, y, z$ and $w$ in $D_{n}$. By Lemma~\ref{lem1}, $\{x^{1}, y^{1}, z^{1}, w^{1}\}\subseteq V(D_{n}\setminus D_{01})$. By Theorem~\ref{lem8}, $D_{n}\setminus D_{01}$ is connected. Thus, there is a tree $\widehat{T}_{n-1}$ connecting $x^{1}, y^{1}, z^{1}$ and $w^{1}$ in $D_{n}\setminus D_{01}$. Let $T_{n-1}=\widehat{T}_{n-1}\bigcup xx^{1}\bigcup yy^{1}\bigcup zz^{1}\bigcup ww^{1}$, then $T_{1}, T_{2},\cdots, T_{n-1}$ are $(n-1)$-internally disjoint $S$-trees in $D_{n}$ and the result is desired.

Case 2. $x, y, z$ and $w$ belong to two different clusters of $D_{n}$.

By Theorem~\ref{Lem81}, $(n-1)$-internally disjoint $S$-trees $T_{1}, T_{2},\cdots, T_{n-1}$ can be obtained in $D_{n}$.

Case 3. $x, y, z$ and $w$ belong to three different clusters of $D_{n}$.

By Theorem~\ref{Lem82}, $(n-1)$-internally disjoint $S$-trees $T_{1}, T_{2},\cdots, T_{n-1}$ can be obtained in $D_{n}$.

Case 4. $x, y, z$ and $w$ belong to four different clusters of $D_{n}$.

By Theorem~\ref{Lem83}, $(n-1)$-internally disjoint $S$-trees $T_{1}, T_{2},\cdots, T_{n-1}$ can be obtained in $D_{n}$.

Thus, $\kappa_{4}(D_{n})\geq n-1$ and the result is desired.
\hfill\qed

By Theorem~\ref{thm1} and Lemma~\ref{lem4}, we have the following corollary.

\begin{cor}\label{cor1}
Let $D_{n}$ be an $n$-dimensional dual cube. Then $\kappa_{3}(D_{n})=n-1$ for $n\geq 4$.
\end{cor}

\section{The $r$-component connectivity of dual cubes}

In this section, we prove the $r$-component connectivity $c\kappa_{r}(D_{n})$ of $D_{n}$. To prove the result, the following result is useful.

\begin{lem}{\rm(\cite{Zh})}\label{lem9}
Any two vertices in $V(Q_{n})$ have exactly two common neighbours for $n\geq 3$ if they have any.
\end{lem}

\begin{theorem}{\rm(\cite{Ar})}\label{lem10}
Let $n$ and $k$ be positive integers such that $n\geq 3$ and $1\leq k \leq n-1$. Let $D_{n}$ be a dual cube network of order $n$ and $T\subseteq V(D_{n})$. If $|T|\leq kn-\frac{k(k+1)}{2}$, then $D_{n}-T$ is either connected or it has a large connected component and small components with at most $k-1$ vertices in total. Moreover, there is a set of vertices $T$ in $D_{n}$ such that $|T|= kn-\frac{k(k+1)}{2}+1$ and $D_{n}-T$ has a component containing exactly $k$ vertices.
\end{theorem}

Following, we show the $r$-component connectivity of the dual cube $D_{n}$.

\begin{theorem}\label{thm2}
Let $D_{n}$ be the $n$-dimensional dual cube, then $c\kappa_{r+1}(D_{n})=rn-\frac{r(r+1)}{2}+1$ for $n\geq 2$ and $1\leq r \leq n-1$.
\end{theorem}

\f {\bf Proof.} By Theorem~\ref{lem10}, $c\kappa_{r+1}(D_{n})\geq rn-\frac{r(r+1)}{2}+1$. Following, we show that $c\kappa_{r+1}(D_{n})\leq rn-\frac{r(r+1)}{2}+1$. Let $u\in V(D_{01})$, then $u$ has $n-1$ neighbours $u_{1}, u_{2},\cdots,$ and $u_{n-1}$ in $D_{01}$ as $D_{01}$ is a copy of $Q_{n-1}$. As $D_{n}$ is bipartite, then $u_{i}$ and $u_{j}$ are non-adjacent for $i\neq j$ and $i,j\in\{1,2,\cdots,n-1\}$. Let $1\leq r\leq n-1$ and $S=N_{D_{n}}(\{u_{1},u_{2},\cdots,u_{r}\})$. Then $u\in S$. By Lemma~\ref{lem9}, $|S|=r(n-1)-\frac{r(r+1)}{2}+1+r=rn-\frac{r(r+1)}{2}+1$. It is obvious that $D_{n}-S$ contains at least $r+1$ components, where at least $r$ of them are singletons. Hence, $c\kappa_{r+1}(D_{n})\leq rn-\frac{r(r+1)}{2}+1$. Combining with $c\kappa_{r+1}(D_{n})\geq rn-\frac{r(r+1)}{2}+1$, we have that $c\kappa_{r+1}(D_{n})=rn-\frac{r(r+1)}{2}+1$ for $n\geq 2$ and $1\leq r \leq n-1$.
\hfill\qed

\section{Concluding remarks}
The dual cube has some attractive properties to design interconnection networks. In this paper, we focus on the dual cube, which is an invariant of the hypercube and denoted by $D_{n}$. We first show that $\kappa_{4}(D_{n})=n-1$ for $n\geq 4$. That is, for any four vertices in $D_{n}$, there are $n-1$ internally disjoint trees connecting them in $D_{n}$.
As a corollary, we obtain $\kappa_{3}(D_{n})=n-1$ for $n\geq 4$. Furthermore, we show that $c\kappa_{r+1}(D_{n})=rn-\frac{r(r+1)}{2}+1$ for $n\geq 2$ and $1\leq r \leq n-1$. So far, the results about the generalized $k$-connectivity of networks are almost about $k=3$ and there are few results about larger $k$. We are interested in this topic and we would like to study the generalized $k$-connectivity of the dual cube for $k\geq 5$ in the future. In addition, we conjecture that $\kappa_{r}(D_{n})=n-1$ for $5\leq r \leq n $ and $n\geq 2$. We would like to prove the result for the dual cube and show the corresponding results for other classical networks such as arrangement graphs, balanced graphs, $(n, k)$-star networks and so on.

\section*{Acknowledgments}
This work was supported by the National Natural Science Foundation of China (Nos.11371052, 11731002, 11571035), the Fundamental Research Funds for the Central Universities (No. 2016JBM071, 2016JBZ012) and the $111$ Project of China (B16002).

redpajama_set_name: RedPajamaArXiv


Bio: Jeremy Lichstein is an assistant professor of Biology at the University of Florida. Professor Lichstein got his Ph. D. from Princeton University and was a postdoctoral research fellow at Princeton's department of Ecology and Evolutionary Biology. He was the recipient of the University of Florida Excellence Award for Assistant Professor, and was named a Florida Climate Institute Fellow for 2016-2017. His research interests are forest dynamics, biodiversity, carbon cycle and climate change.
Terrestrial ecosystems currently offset roughly 25% of global annual anthropogenic fossil fuel emissions. However, the fate of this carbon sink is highly uncertain, in large part because global models diverge in their predictions of ecosystem responses to climate change, drought, and other perturbations. Although there is little agreement on how terrestrial ecosystems will respond to global change on decadal and longer time-scales, there is wide consensus that current global models are overly simplistic in their representation of important ecological processes. I will discuss our current understanding of how tree functional diversity is maintained in forests, the consequences of including more realistic levels of functional diversity in global models, and the computational challenges that need to be overcome in order to introduce ecological realism into the Earth System Models that the scientific and policy communities rely on for climate projections. A key result that is emerging from empirical and theoretical studies is that shifts in species composition across time or space (beta diversity) have different (and sometimes opposite) effects on ecosystem stability as local (alpha) diversity.


The obstacle of the "not invented here" was faced by Chester Carlson, inventor of xerography. Of more recent vintage is the story of Copaxone.
Arnon: "Not really. It wasn't as if other companies were waiting by my door. At the time, I had ties with scientists at several companies, but they weren't happy bringing outside products to their management. What, they'd reveal to their superiors that in-house R&D wasn't good enough? At Teva, Michael had a direct relationship with the company's management, and it didn't have an in-house innovative R&D department anyway.
**Separately, one notes that the discovery of Copaxone was not part of a coordinated drug discovery effort.
Prof. Michael Sela and Prof. Ruth Arnon from the Weizmann Institute of Science discovered Copaxone for the treatment of multiple sclerosis while seeking to develop a product to induce the disease. The serendipitous discovery became the drug with the biggest sales by an Israeli pharmaceutical company. It would be no exaggeration to say that the discovery has even affected the Israeli economy.
Teva Pharmaceutical Industries Ltd. (NYSE: TEVA; TASE: TEVA) is expected to report $3.7-3.9 billion in Copaxone sales in 2013, after steady growth from $2.3 billion in 2008 to $4 billion in 2012.
Arnon and Sela have repeatedly said that they did not embark on their study to develop a drug. Sela told "Globes", "For me, curiosity is the determining factor; curiosity and truth. I am allergic to lies, but I can wake up at 2 am because something interests me. What interested me during the discovery of Copaxone was immunology - how the immune system works."
Arnon, who was Sela's doctoral student when Copaxone's development was initiated, has stressed that she was not seeking to develop a drug, but to complete a detective story to create understanding where previously there was only ignorance.
The year was 1967, and Arnon and Sela attempted to induce multiple sclerosis in animals to discover the causes of the disease and to create a model for understanding it.


text:
anti-morality / cultural Marxism / egalitarianism / feminism / http://schemas.google.com/blogger/2008/kind#post / libertarianism / morality / pathological tolerance / social decay
Of bad girls and bad societies
November 29, 2015 bonnyblue16072 Comments
RamZPaul discusses whether or not feminism has anything to do with the 'bad girl' epidemic of today, and how, if at all, it affects the "AltRight." I offer my thoughts here rather than in his comment box because for some reason I cannot always get a comment to post on Blogger. The issue requires more than a couple of sentences in response, anyway.
First, I will agree with RamZPaul that 'bad girls' have always existed, because human nature has always contained the potential for bad behaviors. However, I disagree that things were 'no better in the old days.' This is an often-heard sentiment these days, though it's most often to be heard from liberals, who believe the old days to be worthy only of condemnation. After all, the 'old days' is where all the sexists, racists, homophobes, transphobes, bigots, prudes, and hypocrites resided, and therefore the old days have to be spat on at every turn. Popular culture constantly depicts the old days in a sneering fashion, or didactic tones. One example: the movie 'Pleasantville', which was apparently a critical success but was nonetheless a blatant piece of propaganda, meant to deride sexual restraint and morality as 'repression', something that robbed life of color and fulfillment. The answer was for everybody to learn to break the rules with gusto, and really live. The past of Pleasantville was a heavy-handed caricature of what real 1950s America was, but then when most movie viewers of today have no first-hand knowledge of life before cultural Marxism destroyed everything, few question this negative view of the past.
Feminism is, in fact, in large part responsible for much of the sluttishness which is prevalent among women today. It isn't just the younger women; it's across age groups. Even 'Christian' girls and women of my acquaintance are almost as casual about sex and as lacking in modesty as are the non-believing ones. The Church (by which I mean all organized Christianity) has failed to teach young people about sexual restraint or modesty — but then, even if they tried, the siren call of the media and of course the wide-open public square 'educate' young people about hedonism and so-called 'sexual freedom' at a very young age. Children in grade school are more sexually savvy than most high-school students were in the 1950s.
Does that mean that there was no bad behavior and no 'bad girls' back in that pre-feminist era? Of course not; I concede that. But as with any generalizations, it's a question of percentages or proportions. The 'bad girls' in the 1930s, for example (RamZPaul uses an image of female outlaw Bonnie Parker as an illustration of the presence of bad girls in that era) were a smaller subset of girls then, and what is more important, there was a strong social stigma attached to the behavior of such girls and women. Bonnie Parker no doubt was already considered beyond the pale when she took up with her partner-in-crime Clyde Barrow. Women and girls knew then that when they adopted a certain kind of persona and lifestyle, they forfeited social respect. Nowadays, no stigma attaches to the worst behaviors among women and young girls. Look at the female celebrities who are most popular. Sluttish behavior is described as "empowering", and women who flaunt such attitudes are admired and rewarded. This is all the fruit of feminism, and feminists have made being a 'slut' a good thing.
The act of admiring and rewarding any behavior will reinforce it and produce more of it.
Young girls today don't see anything demeaning about, say, becoming a stripper. A young 'Christian' woman I know said that she saw nothing wrong or immoral about stripping, and that it's "just another way of making a living, and it pays better than most jobs." Even prostitution is now labeled euphemistically as "the sex industry." How is that for removing the element of morality from the equation?
Back in the old days (or the 'not-so-good-old-days', depending on your point of view) most people would have agreed with what I am writing here, where morality is concerned. Now, thanks to leftist social engineering and propaganda, of which feminism was a big part, views like those I'm expressing are judged as 'prudish' or 'old-fashioned.' Almost everybody now believes that it's better to be without any restraints at all than to be "repressed", as they insist people of the past were.
Society is decidedly worse today by all objective measures, such as prevalence of STDs, infertility resulting directly from those STDs, marriage and divorce rates, low birth rates amongst our folk, teen pregnancy and abortion, and 'mental illness' rates. Just count the number of people you know who are on some kind of psychoactive drug. None of this reflects improvement in society over the decades; quite the opposite.
In the past, too, there was no such mass phenomenon as 'self-hating' Whites, wiggerized teens and young people, or out-of-control immigration, tolerated by a passive population. This doesn't indicate a healthier society, in comparison with the past. With all the talk about 'personal self-esteem', why don't we as a people think more of ourselves? What's needed is not phony self-esteem, but self-respect, which is a different thing. Being worthy of respect leads to respecting oneself and one's own. And women should be at the heart of raising children with the right attitudes to life and themselves. I wonder, too, if young people raised without any personal boundaries — for example, girls who are promiscuous, who let their private space be invaded so easily, grow up to be people who think there should be no boundaries in the wider world — that we should tolerate anything and anyone, in the name of 'openness.' Isn't promiscuity just a personal and unhealthy kind of 'openness'? It reflects our society.
I think the left has done a spectacular job of getting most Americans, especially those too young to remember, to believe the past was a dystopian place, not fit for today's "enlightened" people to live in. Above all, they don't want anyone to respect the past or the people of past eras. Tearing down the past, comparing it unfavorably to today's chaotic world, is part of their plan. If we have nothing left worth reviving from the past, much less preserving what remnants are left, then the social engineers can go on remaking the world in their ugly image.
If the AltRight considers it desirable to incorporate a lot of the left's social attitudes in order to succeed amongst the younger generations, then they've already compromised themselves to the extent that they cannot provide a healthy alternative to this unacceptable mess we find ourselves in.
Useless non-'reporting' from Vancouver
The 'right' enemies
2 thoughts on "Of bad girls and bad societies"
AnalogMan says:
Well said. Good point about promiscuity being the exact same thing that is destroying the West through immigration. Being able to say No is basic to morality and survival, both.
"If the AltRight considers it desirable to incorporate a lot of the left's social attitudes in order to succeed amongst the younger generations, then they've already compromised themselves to the extent that they cannot provide a healthy alternative to this unacceptable mess we find ourselves in."
I agree with this one hundred percent. It's always a catastrophic mistake to compromise on what should be absolutely unshakeable principles.

redpajama_set_name: RedPajamaCommonCrawl


The infamous Chinese hopping zombie, known for their stiff gait. Whispered in folklore, the Jiangshi are all too real. They do not rely on physical power; instead their merest touch will drain the life force from hapless victims. Their formal burial robes make a horrific contrast to their decaying bodies, terrifying villagers that knew them when they were among the living. Beware village cemeteries at night!This boxed set adds terrifying Undead to enemies faced by our heroic Pilgrims, together with the scenery, cards and rules. Three completely new scenarios can be downloaded from the Marrow Production website.
Not a standalone product. Journey: Wrath of Demons™ is required to play.


Pam Rhodes experiences an Edwardian Christmas in rural Shropshire.
Pam Rhodes visits Holmfirth, the town made famous on Last of the Summer Wine.
Pam Rhodes joins pupils in Milton Keynes for popular school hymns.


text:
Wednesday 17th of July 2019 - Last Update: 13:27
WINE AND AWARDS
University of Sannio awards Riccardo Cotarella an Honorary degree in Economics and Management
An academic award, after the one in Agriculture from the University of Tuscia, that acknowledges his social commitment, and devotion to the South
Benevento, 08 February 2019, 17:06
Riccardo Cotarella, president of Assoenologi
Two honorary degrees for Riccardo Cotarella, one of the most famous Italian winemakers, president of the Italian sommeliers association, Assoenologi and of the Union Internationale des Enologues. The University of Tuscia (where he teaches Viticulture and Oenology) awarded him a degree in Agriculture and now, on March 1st, the University of Sannio will award him an Honorary Degree in Economics and Management.
"The University of Sannio awards this Honorary Degree to an innovative researcher and enlightened entrepreneur, untiring in the development of a completely new managerial and global approach towards the production of wine and the culture and economy of the territories", explained the University.
A further recognition linked not only to his profession of winemaker, but also to his constant social commitment, "accompanying the youth of the community of San Patrignano in the process of recovery and agro-wine training, participating in the "Wine for Life" project of the Community of Sant Egidio and following the Cremisan winery in Bethlehem (Palestine), one of the most important communities of Salesian priests. In all of these experiences, Cotarella has shown that the multidimensional value of wine can become a "means" of redemption, rebirth, inclusion and renewal of the community".
Further, the particular attention he has shown to the viticulture of Southern Italy, emphasized the University of Sannio, and "Cotarella's strong commitment to the protection of the territory-wine connection, showing that great wines can be created not only from suitable territories but there is a less known Italy which deserves to be valued as territories of great wines. His merit is having discovered and enhanced less fortunate areas as wine territories, revitalizing entire local economies. He has always maintained a privileged relationship with the lands of the South, showing special attention and care to the territories and vines of Southern Italy. In recent years he has given, and is still giving, a great contribution to the viticulture of Campania, collaborating with important regional companies and, especially with "La Guardiense", one of the most important winegrowing cooperatives in Southern Italy operating in the Sannio territory. Cotarella's presence in the Sannio has marked a turning point in the evolutionary path of viticulture. In just a few years it has lead to prestigious awards that have brought the wines of the Sannio area to the attention of the vast audience of national and international experts, opening up new perspectives for the economic-social development of the local territory".
TAG: BENEVENTO, RICCARDO COTARELLA, UNIVERSITÀ DEL SANNIO
Wine, a family affair? Only 70% of family businesses are handed down to the second generation
"André-The Voice of Wine", the story of the enologist, who has revolutionized the Californian wine
Value over time, efficiency, brand: the keys to a successful wine business
In Canada, sparkling consumptions are increasing, with Italy leader, according to Wine Intelligence
The wine world in mourning for Lucio Caputo
Edoardo Valentini, symbol of Italian wine, to be immortalized in the heart of Rome, the Eternal City
Sandro Boscaini is vice president of Federalimentare (also president of Federvini and Masi Agricola)
In 2020 Italy will enter the Standing Committee on Wine with France and Spain
Antinori in Chianti Classico, the only Italian winery in the "Top 50 - The World's Best Vineyards"
The "Toscana" brand will be on the label of Nobile di Montepulciano (Rosso and Vinsanto)
The "Prosecco Hills of Conegliano and Valdobbiadene" UNESCO World Heritage,
Ian D'Agata: the name is not enough, Italian wine must focus on quality on mature markets
Russia, the ban on Georgian wine could open the doors to growth of Italian exports
Stranger Things, in the second episode of the new season, the special cameo of Chianti
Italians, food and wine, according to Beppe Severgnini: a fundamental part of our identity
With Ernesto Abbona, Stefano Accorsi's passion for wine
Mauro Corona: civil disobedience to guarantee a future for villages and agricultural territories
Prosecco Docg, Frascati Superiore and Moscato d'Asti for the dinner between Putin and Conte
Wine, peace and brotherhood, "Vigneto del Mediterraneo" created in the heart of Gibellina in Sicily
In 2018 the e-commerce of wine and spirits in the world registered 3.6%


fre:ac Portable can run from a cloud folder, external drive, or local folder without installing into Windows. It's even better with the PortableApps.com Platform for easy installs and automatic updates.
fre:ac is a free/open source audio extracter, encoder, and converter. You can use fre:ac to extract audio files from CDs, encode and reencode audio files to a lower bitrate, and convert audio files to and from various formats, including: MP3, Ogg Vorbis, MP4/AAC, FLAC, and more. Additional plugins are available as well.


text:
Introduction: Scientific manuscript writing requires certain art and skills and is challenging even to the best of the researchers. Scientific manuscripts have specific format and language and must adhere to certain guidelines so research must be thoroughly done and manuscript drafted as per the guidelines to be successfully published in a journal.
Objective of the Training: To introduce participants with the format of scientific manuscripts, describe what to write various parts of manuscript and acquaint them with the process of publishing the manuscript.
Details of the Course: The Training on Manuscript writing for Authors 2021- I was a One Day training conducted on Wednesday, 10 Feb 2021 (2077 Magh 28). The training was jointly organized by Nepalese Army Institute of Health Sciences (NAIHS) and Medical Journal of Shree Birendra Hospital (MJSBH). The master of ceremony, Dr Barsha Bajracharya, welcomed everyone which was followed by the opening remarks by Col Dr Srijan Malla, Editor in chief of MJSBH. The opening ceremony was then inaugurated by Brig Gen Dr Arun K Neopane. The course co-ordinator and Executive editor of MJSBH, Lt Col Dr Bikash Shrestha then gave an introduction to manuscript writing and publication following which the training was conducted as per the scientific program. The training was closed by Maj. Gen. Prof. Dr. Nagendra Bahadur K.C. (Retd.), Executive Director of NAIHS. The participants were awarded with certificates after completion of the course.
Participants: 22 Participants including MBBS, Specialists (MD/MS), Nurses and faculties from College of Allied Health Sciences with at least bachelor degree (demonstrator level) were included in the training. 12 participants were affiliated to College of Medicine, 2 from College of Nursing and 8 from College of Medical Polytechnic. The resource persons were as per Table 1.
Outcome of the Training: Participants from various colleges of NAIHS learnt what to include in their manuscript. Due to participation of all faculties, it is hoped that it will facilitate in better teaching and learning as well as assist on writing manuscript and submitting it in journals for the scientific community to disseminate research work done in NAIHS.
Glimpses of the Training:


text:
Home>World>Sit down before you see how Negro Arija's daughters flaunt their beauty
Sit down before you see how Negro Arija's daughters flaunt their beauty
nwcadminNovember 20, 2022Last Updated: November 20, 2022
One of the most beloved conductors of Mexican TV Raul ArizaAlthough few people know that it has two daughtersWhose beauty And the personality has given a lot to talk about on social networks and of which the actor is also very proud.
driver ofon-air memberThe man who started his media career at the age of eight claims good relations with his offspring on social networks, whom he openly supports in all their decisions, as in the case of his daughter Camilla, who recently declared herself pansexual,
Camila Ariza
Camila is the eldest daughter of Arija, this young woman is dedicated to modeling and photography. Last June, he openly declared himself to be part of LGBT+ community and introduced his girlfriend, the model camila solorzano aroesti,
Through her Instagram account, Camila published a message in which she appears next to her partner and where she explains that it is difficult for her to pigeonhole, specify and justify herself, because it is more than one nameMore than age, gender, body.
Raul supports Camilla
,I owe everything to me because it is self-love and that is my greatest mission", Camila Ariza wrote on her Instagram account, where she shared the message with more than 100,000 subscribers, on the day pride march in mexico,
Camilla has stood outside to resign stereotypes of beauty, because a few months ago she decided to stay baldbecause she's been diagnosed since she was three alopecia areataA condition that caused her to lose her hair from a very young age.
Photo: Exclusive
Roberta Ariza
"Negro" Ariza, on the other hand, stands out for being also the youngest daughter of Roberta. influencer on social networks, although since a few months, with a more discreet profile instagram account It got personal.
This young woman has a passion for fashion, art, and pets, and has posed for a few magazine covers with her dad. Roberta's beauty is reflected in some of her pictures, where she shows off her more sensuous,
She is Roberta, daughter of Raul "El Negro" Ariza who succeeds as a professional chef.
Raul "El Negro" Ariza turns 57 and celebrates at Hoy event
Camila Ariza raul ariza Roberta Ariza


….I know that I couldn't be the advocate I need to be for my children without the non-judgmental support of these parents. I've even gotten to know some of the people I first met online in person. Having these men and women and these groups is the most valuable tool in my fight to keep my personal sanity on this wild ride. Without them I'd spend a lot more time in the fetal position, and I wouldn't have the strength to pull myself up by my bootstraps the morning after my pity party and be the mom they need me to be….
Read the entire post at Support For Special Needs.
Thank you for sharing with us and I'm so grateful you will be doing it monthly for us! So, so grateful (and if you know me, I don't just throw that word around).


It is with regret that we notify you, after 30 years in business we are having to close Knight Plumbers Ltd. Over the last 5 years it has been a constant struggle to get and retain competent plumbers & not being able to do this in the numbers that we need, is the only reason for our closure. Our last day of trade will be Friday 31 August 2018.
Our phone lines will not be active after this date but our email admin@knightplumbers.co.nz will be checked periodically & we will reply to emails are received. Please not - there will be an auto response on our email but we will check and respond also.
We can look back over & our time in business & know that we have always tried to be the best we can be, we have had a lot of great times & met a lot of good people.
We thank you very much for letting us be your plumbers for this time & wish you the best in the future.


text:
Q: changing pointers(increasing , decreasing)? the program below shows *q=46 but q points to an address that is not defined because it points to : p-1 so its not pointing to p or m but another address(because of -1)
then how can it be *q=46
i think it must have a garbage number based on what ive learnt
i hope you get what im trying to say because im not native speaker
int m = 44;
int* p = &m;
int& r = m;
int n = (*p++);
int* q = p – 1;
r = *(--p) + 1;
++*q;

the exercise wants the amount of each variable and pointer and reference above once the program is run
i tested it with a program
output:
p=0022fa00
*p=46
&p=0022f9f4
&n=0022f9dc
n=44
m=46
q=0022fa00  <<<<<<<
r=46
*q=46  <<<<<<<
&m=0022fa00

i thought maybe it's because of int n=(*p++) but i think that shouldnt be the reason because that statement must increment the number in p not the address

A: int m = 44;//m=44
int* p = &m;//p has address of m *p=44
int& r = m;// r is allias(nickname) of m (*p m r is same)=44
int n = (*p++);//n=44 *p=garbage m=44 r=44 p is m's address+1
// *p++ means give *p then increase p(not *p)
int* q = p - 1;//q shows m's address *q=44
r = *(--p) + 1;//*(--p) means m r= m+1=45 r=45 m=45 because r is allias of m *q=45
++*q;//increase r=46 m=46 (*q)=46

It is what happens at there.I hope this will help.

A:
i thought maybe it's because of int n=(*p++) but i think that shouldnt be the reason because that statement must increment the number in p not the address

Nope. *p++ is *(p++), not (*p)++.

redpajama_set_name: RedPajamaStackExchange


Guide for Web Developers→How FusionCharts? works?
//We keep flags to check whether the charts have loaded successfully.
* against the DOM id and update charts or loading flags.
*	@param	DOMId	Dom Id of the chart that was succesfully loaded.
* updateChart method is invoked when the user clicks on a pie slice.
* update the Column Charts.
//Sometimes, the above URL and XML data gets cached by the browser.
//the browser and forces it to re-load the XML data every time.
//Get reference to chart object using Dom ID "EmployeeDetails"
* dataURL for effective non-caching.


The other day we had a question from one of our Facebook page fans asking what the problem is with paper bags. We all know there is a movement around banning single-use plastic bags, and the reasons why single-use plastic bags should be banned are pretty clear. They are made from nonrenewable resources, they do not safely breakdown, and they are a serious threat to the planet. A lot of the legislation around plastic bags includes disposable paper bags, and this leads many people to ask, "What's the problem with paper bags?".
The issues that make single-use plastic bags a clear problem aren't always applicable to paper bags. After all, paper bags are made from renewable resources and often they are made from a percentage of recycled content. Paper is biodegradable, and is not a health hazard to animals, ecosystems and the planet at large in the same way as plastic bags. So really, what's the problem? Let us break things down for you.
Many shoppers save their single-use paper bags, and they may end up being used a second or third time, but these bags are not intended or designed for long-term use and also do not offer long term marketing.
Millions of trees are cut down each year to create the billions of paper bags that are used each year by consumers; sometimes just for minutes. Single-use bags are used on average for 20 minutes, so even if they are used a few times when you tally up the amount of resources that go into creating these bags, it just doesn't measure up to the amount of time these bags are used.
Trees serve a lot of useful purposes, like being harvested to create construction materials, or converting greenhouse gases into breathable oxygen. These are very important uses of trees, and cutting trees down to create single-use paper bags just isn't the most useful application of wood or wood fibers.
There has been a very successful movement to sustainably manage forests, and this is definitely progress, but we need to be as responsible as possible when we use this resource.
When a material is biodegradable, that means it is a natural compound and that if exposed to the proper conditions, this item will break apart at a molecular level into smaller and smaller pieces. For instance, paper is a natural material typically made from processed tree fibers. When paper products are placed in just the right conditions, the carbon and other building blocks will separate and reabsorb into soil, or whatever natural elements are nearby. The key here is that the proper conditions must be met for paper bags to biodegrade.
If paper bags are tossed in a compost bin or end up as litter, they will not necessarily biodegrade. If paper bags are tossed in a municipal compost facility they will break down. Paper bags can be recycled, and they are much easier to recycle than plastic bags. When paper bags are recycled, the material can be repurposed into new paper products that can be used again. The only drawback to recycling is that paper can be recycled only a limited amount of times. Every time paper is recycled, the fibers become shorter and eventually, they will become too short to be processed.
What Material Makes the Best Shopping Bags for Retail and Marketing?
Responsible use of resources, forest management and how paper bags are handled after they have fulfilled their usefulness, all contribute to the problem with paper bags. By contrast, reusable bags can last for several years and will work to market your brand.
Our reusable bags are made from recycled or sustainable materials, which means they are easy on the planet and they can be recycled when they are no longer useful. If you want to learn more about our how reusable bags and our design, order, or production process get in touch!
We can manufacture reusable bags specifically for your project, trade show or brand so you have a unique bag that no one has seen yet, and everyone will want. Contact us to find out what we can do.


Is this face just an assembly of computer bits?
problems. Severely limiting AI research could curb that progress. But giving beneficial technologies room to grow comes at no small cost – and the potential for misuse, whether to make inaccurate "Bach-like" music or to deceive millions, is likely to grow in ways people can't yet anticipate.


text:
Tag Archives: Howard Lance
Howard Lance
in Industry News, News
Maxar to Transfer Corporate HQ to Westminster, Colorado
Maxar Technologies will relocate its corporate headquarters from San Francisco, California, to DigitalGlobe"™s existing facility in Westminster, Colorado. "This move represents an opportunity to bring together our executive leadership team in a common location, as well as to establish a base for our future shared-services organizations," Howard Lance, Maxar president and CEO, said in a statement […] More
by Jane Edwards February 15, 2018, 3:22 am
Maxar Wins "˜Deal of the Year"™ Award; Howard Lance Comments
Maxar Technologies, formerly MacDonald, Dettwiler and Associates, has received SpaceNews"™ "Deal of the Year" award in recognition of the company"™s acquisition of DigitalGlobe in an estimated $3.6 billion transaction. Maxar"™s predecessor MDA closed the deal in October in a push to offer imagery, satellite and geospatial services across the government and commercial sectors, Maxar said […] More
by Neel Mehta December 21, 2017, 12:00 am
in Executive Spotlight, General
Executive Spotlight: Rob Zitz, SVP and Chief Strategy Officer at SSL Government Systems
"I have been the 'change agent' throughout my career…understanding end users' unmet needs is an absolute priority." EM: Can you tell us a little bit about SSL Government Systems? Rob Zitz: SSL"™s Government Systems group is made up of highly experienced, cleared professionals with deep domain expertise in defense and intelligence missions. The group is […] More
by Andy Reed September 13, 2017, 9:30 am
Howard Lance: MDA Seeks to Diversify Into Earth Imagery Market Via DigitalGlobe Merger
Canada-based MacDonald, Dettwiler and Associates expects its pending merger with Earth imaging services provider DigitalGlobe to help counterbalance the fluctuating revenues and profits of its commercial satellite manufacturing business, Space News reported Tuesday. MDA agreed to buy DigitalGlobe through an estimated $3.6 billion transaction in February. Howard Lance, MDA president and CEO, told investors Tuesday the pending deal seeks […] More
by Jane Edwards May 24, 2017, 9:23 am
SSL, DARPA Finalize Agreement to Develop On-Orbit Spacecraft Servicing Tech
Space Systems Loral has finalized an agreement with the Defense Advanced Research Projects Agency to develop technologies that would help service and maintain spacecraft in geostationary orbit. SSL said Wednesday its team of engineers and technicians will perform work at a company facility in Palo Alto, California. DARPA and SSL unveiled a draft version of their public-private partnership agreement for the Robotic Servicing […] More
by Ramona Adams April 13, 2017, 10:12 am
Howard L. Lance to Speak at Jefferies Conference in New York City
Howard L. Lance, chairman, president and chief executive officer of Harris Corporation, will be a featured speaker at the Jefferies 2011 Global Industrial and A&D Conference on Aug. 11 in New York City. Lance"™s presentation is scheduled to begin 8:30 a.m. Eastern time. A live webcast and replay of the presentation will be available at […] More
by Neel Mehta July 26, 2011, 12:00 am
Harris Sets Official Opening of Virginia Cyber Integration Center for May 27
Harris Corporation (NYSE: HRS) announced it will officially open its new Harrisonburg, Va.-based Cyber Integration Center this week. The high-security 140,000 square-foot facility was designed and built with industry leading partners VCE, EMC, VMware, RSA, Cisco, HP, BMC and Lee Technologies and will offer technology and cloud-computing services for the government market. In attendance for […] More
by Neel Mehta May 24, 2011, 12:00 am
Harris' Howard Lance Joins DC Business Roundtable
Howard Lance, chairman, president and CEO of Harris Corporation, has been chosen to join Business Roundtable, a Washington, D.C.,-based association of industry and technology business leaders. "I am excited to have the opportunity to work with other CEOs to formulate positive policy recommendations designed to help spur economic growth," Lance said. "The Business Roundtable's vast […] More
by Admin October 25, 2010, 9:49 am
Harris Kicks Off Middle East Partnership
Harris Corporation will partner with Middle East technology company Atlas Telecom to provide communications intelligence services to the UAE, the companies announced this week in Abu Dhabi. The new joint venture will focus on large scale, mission-critical intelligence systems, and will take place under the name of a joint venture company, Harris Atlas. One of […] More
by Neel Mehta October 1, 2010, 12:00 am
in Cybersecurity News
New Joint Venture Harris Atlas to Deliver Intelligence Solutions in the Middle East
Harris Corporation and Middle East telecom systems and security provider Atlas Telecom yesterday launched a new joint venture, Harris Atlas, which will deliver intelligence solutions specifically designed for the United Arab Emirates. The partnership will be officially announced at a press conference at Harris Atlas' new headquarters in Abu Dhabi, which will be attended by […] More
by Admin September 30, 2010, 12:00 pm
Harris Promotes Educational Events for National Aerospace Week
Harris Corporation held events for local schools and families of employees in honor of National Aerospace Week, which took place Sept. 12-18. Activities were held at schools in Washington, D.C., Florida and Colorado, where Harris has major company locations. Students at these schools benefited from space-themed lessons and aerospace career days, while children of employees […] More
by Neel Mehta September 20, 2010, 12:00 am
Receive Daily News Briefing And Event Updates Straight to Your Inbox
ExecutiveBiz follows the executive-level business activity that drives the government contracting industry.
Through executive interviews, policy spotlights and news stories on corporate activity, ExecutiveBiz stays on top of what's shaping the industry and how GovCon leaders are positioning their businesses to win the federal business of tomorrow.
Staying in line with Executive Mosaic's dedication to covering all angles of GovCon, ExecutiveBiz also helps paint a more complete picture of the government contracting industry by featuring medium-sized business news and significant state and foreign contracting developments.
If it involves business activity that impacts contracting operations and success, ExecutiveBiz has you covered.


text:
Microsoft buys Activision Blizzard in gaming's biggest deal in history
The Great Resignation is everywhere, except in the statistics
Global eCommerce transactions to exceed $7.5 trillion by 2026, says Juniper Research
Pandemic influences digital transformation for Australian enterprise
2022: Supply Chain Pressures and Equilibrium
Partner Content Digital Nation Live 2021 2021 Sponsor Zones
Customers Growth Risk People Process Technology Resources
Focus on rapid learning to build a future-ready workforce, says MIT CISR
By Andrew Birmingham on Aug 16, 2021 5:00AM
Javascript is required to view this video.
Australian businesses are global laggards in the race to equip for the digital future of business. To catch up, they will need to invest in skills and technology – and especially in a culture of rapid learning.
Read more: Australian organisations fall badly behind in the race towards digital fitness
That's the view of MIT Sloan research scientist Kristine Dery, the keynote speaker of this year's iTnews Digital Nation series, which looks at the digital fitness of Australian organisations.
Meanwhile, Roby Sharon-Zipser, CEO of Hipages, the leading online platform for tradespeople, and Mark Cohen, the CEO of insurtech Wilber (and former CIO of Domain), stress the importance of such issues as data-led decision making and team management in a world of transnational talent and hybrid working.
Get the latest insights and analysis delivered to your inbox.
I have read and accept the privacy policy and terms and conditions and by submitting my email address I agree to receive the Digital Nation newsletter and receive special offers on behalf of Digital Nation, nextmedia and its valued partners. We will not share your details with third parties.
They make these comments in a new iTnews Digital Nation mini-documentary about the digital fitness of organisations in Australia and New Zealand (watch the video at the start of this article).
Dery says, "This requires investment in technologies, it requires realignment of resources, and rethinking where the technology investments are going to lie."
It also requires leaders to completely reimagine the types of learning in their organisations.
"I can't emphasise that enough, because in our large corporations, typically, we had skilling that relied on people acquiring skills to move from one part of the hierarchy to the next part of the hierarchy. You moved across the organisation or acquired a new job and therefore acquired new skills."
According to Dery, "In these more contemporary digital organisations, we have to be much more focused around rapid learning."
The term rapid learning is gaining traction in corporate environments. "It means we have to drive more curiosity into our enterprise. We have to be asking more questions about why it is that people are struggling. Are they struggling? We've got to be much more engaged with our people," Dery explains.
It is critical to listen and understand whether the tools employees use, or their capabilities and ways of working, are creating challenges.
"And we've seen how hard it is to move into these more agile-type ways of working, however you describe that," Dery comments.
Companies are increasingly drawing on talent wherever it resides in the world, and employees are typically in hybrid working environments. These factors create hurdles when building high-performance teams.
Mark Cohen, CEO of Wilbur, understands all too well the challenges of managing a geographically distributed team – the company has employees in Australia, the United States and Ukraine.
"Keeping people in sync is quite difficult", Cohen says. "So we try and be very, very clear. And we also try and work very clearly to define who you should be checking in with if you've got an idea that's going beyond what was expected of you, who you're talking to about it. So very structured in that regard."
Cohen sees decision-making frameworks as just as important as clear lines of communication.
"[Decision-making frameworks are] actually very useful, because you'll find situations where you have some strongly opposed views or opinions, and that you don't always want them floating up to an executive committee or to a board to make decisions on things that should really be resolved in the teams, or in between businesses.
"And that's where having some kind of a delegation, it works really well, to say 'In the event that you guys are not sure, that's the person who's going to decide'," Cohen says.
When it comes to adopting new technologies, Cohen highlights the value of shadow IT solutions. But staying on top of these new and emerging technologies isn't easy when there isn't a single person dedicated to the job.
"I guess that's a stretch goal for a lot of companies at the moment is, how do you stay aware of that? I don't have a really good answer on that one. But it's definitely a challenge," he says.
Kristine Dery says one quality that sets high performers apart from the pack is their use of analytics.
Roby Sharon-Zipser, CEO and co-founder of Hipages, says that the company's leadership team meets weekly to track real-time metrics tied to the organisation's goals.
"We do a check-in with the team, as a team, and we ask questions, we probe it. What can we do to get those weekly metrics back on track or celebrate when they're on track? Or how do we accelerate the metrics even more if we're winning?" he says.
The commitment to data-led decision making is paying off, Sharon-Zipser suggests. He says that during the pandemic, Hipages was one of the only companies in Australia that listed on the ASX with a forecast that they subsequently (favourably) revised.
decisionmakingdigital fitnesskristine deryleadershiplearningmark cohenmit cisrrapid learningroby sharonzipserskills
By Andrew Birmingham
MIT researcher's warning for Australian business leaders
Organisation shift to cross functional teams drives further innovation at hipages: CEO
Australian business leaders share how they tame prioritisation battles
Business loses billions to digital ad fraud each year. Here's how IT can help
US states allege Facebook and Google CEOs oversaw illegal deal
Pentera closes $150M Series C becoming highest valued company in its category
Leveraging digital to drive sustainability, Bain & Company
Unblocking your organisation's data flows
Infrastructure is the carbon king. Investors demand change, digital will enable it
Case Study: Baker McKenzie builds a culture of innovation
How SelfWealth manages its explosive growth in customer numbers
COVER STORY: Building sector accelerates sustainable transformation
Contact Us About Us Feedback Advertise Site Map RSS
OTHER TECH SITES: iTnews | BIT | CRN Australia | IoT Hub


text:
Q: Prove the universal class = the empty class?? I am new to set theory and am reading Pinter's Introduction to Set Theory.  The universal class is defined as the class of all elements.  The empty class is defined as the class which has no elements.  Then, in the Algebra of Classes, Theorem 1.26v states: For every class A, the universal class = the empty class. The reader is invited to prove this as an exercise.  I don't understand how this could possibly be true, much less how to prove it.  The axiom of extent says that two classes are equal iff an element belonging to one class implies that it also belongs to the other class and vice versa.  I am very confused!!

A: I have a copy of the book, and you simply misread the problem.  It's asking you to prove that ${\mathscr{U}}^{'}=\emptyset$.  That is, show that the complement of the universal class is empty. So your intuition is sound in that the problem as you had read it did not make sense.