The structure connectivity of Data Center Networks

Lina Ba and Heping Zhang Corresponding author.

(School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000, P.R. China
E-mails: baln19@lzu.edu.cn, zhanghp@lzu.edu.cn)

Abstract

Last decade, numerous giant data center networks are built to provide increasingly fashionable web applications. For two integers $m\geq 0$ and $n\geq 2$ , the $m$ -dimensional DCell network with $n$ -port switches $D_{m,n}$ and $n$ -dimensional BCDC network $B_{n}$ have been proposed. Connectivity is a basic parameter to measure fault-tolerance of networks. As generalizations of connectivity, structure (substructure) connectivity was recently proposed. Let $G$ and $H$ be two connected graphs. Let $\mathcal{F}$ be a set whose elements are subgraphs of $G$ , and every member of $\mathcal{F}$ is isomorphic to $H$ (resp. a connected subgraph of $H$ ). Then $H$ -structure connectivity $\kappa(G;H)$ (resp. $H$ -substructure connectivity $\kappa^{s}(G;H)$ ) of $G$ is the size of a smallest set of $\mathcal{F}$ such that the rest of $G$ is disconnected or the singleton when removing $\mathcal{F}$ . Then it is meaningful to calculate the structure connectivity of data center networks on some common structures, such as star $K_{1,t}$ , path $P_{k}$ , cycle $C_{k}$ , complete graph $K_{s}$ and so on. In this paper, we obtain that $\kappa(D_{m,n};K_{1,t})=\kappa^{s}(D_{m,n};K_{1,t})=\lceil\frac{n-1}{1+t}\rceil+m$ for $1\leq t\leq m+n-2$ and $\kappa(D_{m,n};K_{s})=\lceil\frac{n-1}{s}\rceil+m$ for $3\leq s\leq n-1$ by analyzing the structural properties of $D_{m,n}$ . We also compute $\kappa(B_{n};H)$ and $\kappa^{s}(B_{n};H)$ for $H\in\{K_{1,t},P_{k},C_{k}|1\leq t\leq 2n-3,6\leq k\leq 2n-1\}$ and $n\geq 5$ by using $g$ -extra connectivity of $B_{n}$ .

Keywords: Fault-tolerance; Structure connectivity; DCell; BCDC.

1 Introduction

Networks are often modeled by connected graphs. A node replaces with a vertex and a link replaces with an edge. Let $G$ be a graph. We set $V(G)$ as vertex set of $G$ and $E(G)$ as edge set of $G$ . Connectivity is a basic parameter in measuring fault tolerance of networks. In order to more accurately measure fault-tolerance of large-scale parallel processing systems, $\rho$ -conditional connectivity for some property $\rho$ proposed by Harary [9] referred to the minimum cardinality of a vertex subset $S$ of $G$ such that $G-S$ is disconnect and every component has property $\rho$ . Thus, $h$ -extra connectivity $\kappa_{h}(G)$ is introduced by Fàbrega and Fiol [5], and the definition emphasizes that the every component left behind has at least $h+1$ vertices; $h$ -restricted connectivity $\kappa^{h}(G)$ is came up with by Esfahanian and Hakimi [3], and $\kappa^{h}(G)$ requires that any one vertex of every component left behind has at least $h$ neighborts. In the Network-on-Chip technology, if there is a fault node on the chip, then we assume that the whole chip is fault. Because of the facts, structure (substructure) connectivity is defined by Lin et al. [13]. Let $\mathcal{F}$ be a set of subgraphs of $G$ . If $\mathcal{F}$ disconnects $G$ or $G-\mathcal{F}$ is a trivial graph, then $\mathcal{F}$ is called a subgraph cut of $G$ . For a connected subgraph $H$ and a subgraph cut $\mathcal{F}$ of $G$ , if $\mathcal{F}$ ’s every member is isomorphic to $H$ (resp. a connected subgraph of $H$ ), then $\mathcal{F}$ is called an $H$ -structure cut (resp. $H$ -substructure cut). The size of the smallest $H$ -structure cuts (resp. $H$ -substructure cuts) of $G$ is $H$ -structure connectivity of $G$ , $\kappa(G;H)$ (resp. $H$ -substructure connectivity of $G$ , $\kappa^{s}(G;H)$ ). Thus, $\kappa^{s}(G;H)\leqslant\kappa(G;H)$ . Especially, $K_{1}$ -structure connectivity is equal to vertex connectivity. For the past few years, there are many studies on structure connectivity for some well-known networks, such as $HL$ -networks [12], alternating group graphs [15], star graph [14], wheel network [6] and so on.

Data center network, DCN, is a networking infrastructure inside a data center, which connects many servers by links and switches [7]. Due to the the advances of science and technology, DCN runs thousands of servers. For instance, Google has over 450,000 servers operating in 30 data centers by 2006 [1, 10], and Microsoft and Yahoo! have hundreds of millions of servers running at the same time in their data centers [2, 18]. Then how to efficiently connect numerous servers become a fundamental challenge in DCN. Therefore, DCell and BCDC were proposed by Guo et al. [7] and Wang et al. [19], respectively. DCell defines by a recursive structure. A server connects to distinct levels of DCells by multiple links (for details see Definition 1). According to its structure, with the increase of node degree, DCell shows a doubly exponential growth. Then, without the use of costly core-switches or core-routers, a DCell with a degree less than 4 can operate over thousands of servers [7]. An $n$ -dimensional BCDC, $B_{n}$ , consists of an independent set and two $(n-1)$ -dimensional BCDC’s (for details see Definition 4). Then the quantity of servers in BCDC add up fast as the dimension of BCDC grows. Such as, with the use of 16-port switches, $B_{16}$ operates 524 288 servers at once [19]. It is easy to find that DCell and BCDC provide two ways to connect plentiful servers efficiently. For these two networks, the best application scenarios are large data centers, which contain end-user applications (e.g. Web search and IM) and distributed system operations (e.g. MapReduce and GFS) [7]. Because DCell and BCDC based on complete graphs and crossed cubes, respectively, they still have lots of good properties, such as high-capacity, minor diameter and high fault-tolerance. There are a few primary characters of graphs $D_{m,n}$ and $B_{n}$ , such as connectivity, diameter, symmetry, broadcasting, have been studied recently [7, 19]. As extensions of connectivity, $h$ -restricted connectivity [20, 17] and $h$ -extra connectivity [16, 17] have been studied, yet. Then, we compute the structure connectivities of $D_{m,n}$ and $B_{n}$ on some common structures, star, complete graph, path and cycle in this paper.

The paper has the following sections. Section 2 defines a few notations. Sections 3 presents the star-structure (substructure) connectivity and complete graph-structure connectivitives of DCell. Sections 4 shows the star-structure (substructure) connectivity, path-structure (substructure) connectivity and cycle-structure (substructure) connectivity of BCDC. Section 5 summarizes the content of this paper and indicates some unsolved questions.

2 Notations

In this paper, we only consider finite and simple graphs. For a graph $H$ that satisfies the conditions $V(H)\subseteq V(G)$ and $E(H)\subseteq E(G)$ , we call $H$ a subgraph of $G$ . For a vertex subset $S$ of $G$ , $G-S$ is a graph obtained by deleting the vertex set $S$ and all edges incident to them from $G$ . If $H$ is a subgraph of $G$ , then we set $G-H=G-V(H)$ . Let $\mathcal{F}$ be a set whose members are subgraphs of $G$ . Then $G-\mathcal{F}=G-V(\mathcal{F})$ , where $V(\mathcal{F})$ is the union of the vertex set of members of $\mathcal{F}$ . For a vertex subset $V^{\prime}$ of $G$ , we set $G[V^{\prime}]$ as the induced subgraph by $V^{\prime}$ of $G$ , where $V(G[V^{\prime}])=V^{\prime}$ and $E(G[V^{\prime}])=\{(u,u^{\prime})\in E(G)|u,u^{\prime}\in V^{\prime}\}.$ For any one vertex $x$ of $G$ , $N_{G}(x)=\{y|xy\in E(G)\}$ . For a vertex subset $A$ of $G$ , $N_{G}(A)=\cup_{x\in A}N_{G}(x)\setminus A$ . Let $P_{k}=\langle v_{1},v_{2},\ldots,v_{k}\rangle$ be a path. We set $P^{-1}_{k}=\langle v_{k},v_{k-1},\ldots,v_{1}\rangle$ and $P_{k}-v_{i}\cup w=\langle v_{1},\ldots,v_{i-1},w,v_{i+1},\ldots,v_{k}\rangle$ . If $v_{1}=v_{k}$ in $P_{k}$ , $k\geq 3$ , then we call it cycle $C_{k}$ . Let $\{x;x_{i}|1\leq i\leq t\}$ be a star $K_{1,t}$ . Then we set $x$ and $x_{i}$ , $1\leq i\leq t$ , are center and leaves, respectively.

3 The structure connectivity of DCell

Given an integer $s$ , we set $\langle s\rangle=\{0,1,\ldots,s\}$ and $[s]=\{1,2,\ldots,s\}$ . Define $I_{0,n}=\langle n-1\rangle$ and $I_{i,n}=\langle t_{i-1,n}\rangle$ for $i\in[m]$ . An $m$ -dimensional DCell with $n$ -port switches is denoted by $D_{m,n}$ for $m\geq 0$ and $n\geq 2$ . Let $t_{m,n}$ be the number of servers in $D_{m,n}$ , where $t_{0,n}=n$ and $t_{m,n}=t_{m-1,n}\cdot(t_{m-1,n}+1)$ for $m\geq 1$ . Servers of $D_{m,n}$ can be labeled by $\{x_{m}x_{m-1}\ldots x_{1}x_{0}|x_{i}\in I_{i,n},i\in\langle m\rangle\}$ .

Definition 1.

[7] The $m$ -dimensional DCell with $n$ -port switches $D_{m,n}$ is defined recursively as follows.

(1)

$D_{0,n}$ is a complete graph consisting of $n$ servers.
(2)
For $m\geq 1$ , $D_{m,n}$ is obtained from $t_{m-1,n}+1$ disjoint copies $D_{m-1,n}$ by the following steps.
1. (i)
  
  Let $D^{i}_{m-1,n}$ be a copy of $D_{m-1,n}$ by prefixing label of each server with $i$ for $i\in I_{m,n}.$
2. (ii)
  
  Server $u=u_{m}u_{m-1},\ldots,u_{0}$ in $D^{u_{m}}_{m-1,n}$ is adjacent to server $v=v_{m}v_{m-1},\ldots,v_{0}$ in $D^{v_{m}}_{m-1,n}$ if and only if $u_{m}=v_{0}+\sum^{m-1}_{j=1}(v_{j}\times t_{j-1,n})$ and $v_{m}=u_{0}+\sum^{m-1}_{j=1}(u_{j}\times t_{j-1,n})+1$ for any $u_{m},v_{m}\in I_{m,n}$ and $u_{m}<v_{m}$ .

If $u$ has a neighbor $u^{\prime}\notin D^{i}_{m-1,n}$ for $u\in D^{i}_{m-1,n}$ and $i\in I_{m,n}$ , then $u^{\prime}$ is called the outside neighbor of $u$ . Figure 1(a) shows 1-dimensional DCell with 4-port switches. Since switches are transparent in networks, the graph structure of $D_{1,4}$ is shown in Figure 1(b).

Lemma 3.1.

[8] For $m\geq 0$ and $n\geq 2$ , the following statements hold:

(i)

$D_{m,n}$ is $(m+n-1)$ -regular and $|V_{m,n}|=t_{m,n}$ ,
(ii)

for $m\geq 1$ , $D_{m,n}$ consists of $t_{m-1,n}+1$ copies of $D_{m-1,n}$ , denoted by $D^{i}_{m-1,n}$ , for each $i\in I_{m,n}$ . There is only one edge between $D^{i}_{m-1,n}$ and $D^{j}_{m-1,n}$ for any $i,j\in I_{m,n}$ and $i\neq j$ , which implies that the outside neighbors of vertices in $D^{i}_{m-1,n}$ belong to different copies of $D^{j}_{m-1,n}$ for $i,j\in I_{m,n}$ and $i\neq j$ .

For convenience, let

D_{m,n}=D^{0}_{m-1,n}\otimes D^{1}_{m-1,n}\otimes D^{2}_{m-1,n}\otimes\ldots\otimes D^{t_{m-1,n}}_{m-1,n}.

(3.1)

Lemma 3.2.

[7] For $m\geq 0$ and $n\geq 2$ , $\kappa(D_{m,n})=n+m-1$ and $t_{m,n}\geq(n+\frac{1}{2})^{2^{m}}-\frac{1}{2}.$

Lemma 3.3.

$\kappa^{s}(D_{m,n};K_{1,t})\geq\lceil\frac{n-1}{1+t}\rceil+m$ for $1\leq t\leq m+n-1.$

Proof.

Let’s explain this by mathematical induction on $m$ . For $m=0$ , $D_{0,n}\cong K_{n}$ . Assume $\kappa^{s}(D_{0,n};K_{1,t})\leq\lceil\frac{n-1}{1+t}\rceil-1$ for $1\leq t\leq n-1.$ Then $D_{0,n}$ has a $K_{1,t}$ -substructure cut $\mathcal{F^{\prime}}$ with $|\mathcal{F^{\prime}}|\leq\lceil\frac{n-1}{1+t}\rceil-1$ . We have

|V(\mathcal{F^{\prime}})|\leq(1+t)(\lceil\frac{n-1}{1+t}\rceil-1)\leq(1+t)(\frac{n+t-1}{1+t}-1)=n-2.

Since $\kappa(D_{0,n})=n-1$ , $|V(\mathcal{F^{\prime}})|<\kappa(D_{0,n})$ , a contradiction. Then it is true for $m=0$ .

Suppose $\kappa^{s}(D_{m-1,n};K_{1,t})\geq\lceil\frac{n-1}{1+t}\rceil+m-1$ for $1\leq t\leq m+n-2.$ Now, we consider $D_{m,n}$ . Let $\mathcal{F}=\{T_{i}|0\leq i\leq\lceil\frac{n-1}{1+t}\rceil+m-2\}$ be a set of connected subgraphs of $K_{1,t}$ for $1\leq t\leq n+m-1$ . We need to show that $D_{m,n}-\mathcal{F}$ is connected. Let $\mathcal{F}^{i}=\mathcal{F}\cap D^{i}_{m-1,n}$ , $\mathcal{F}^{M}=\cup_{i\in M}\mathcal{F}^{i}$ and $D_{m-1,n}^{M}=\cup_{i\in M}D^{i}_{m-1,n}$ by Eq. (3.1). We know that each star is in at most two $D^{i}_{m-1,n}$ ’s by Definition 1. Then there exist at most $2(\lceil\frac{n-1}{1+t}\rceil+m-1)$ $D^{i}_{m-1,n}$ ’s such that $\mathcal{F}^{i}\neq\emptyset.$ Let $S=\{0,1,2,\ldots,s\}$ be a set such that $\mathcal{F}^{i}\neq\emptyset$ for $i\in S$ . Then $|S|\leq 2(\lceil\frac{n-1}{1+t}\rceil+m-1)$ . By Lemma 3.2, for $m\geq 1$ and $n\geq 2$ ,

		$\displaystyle t_{m-1,n}+1-\|S\|\geq t_{m-1,n}+1-2(\lceil\frac{n-1}{1+t}\rceil+m-1)\geq t_{m-1,n}+1-2(\frac{n+t-1}{1+t}+m-1)$
		$\displaystyle\geq t_{m-1,n}+1-2(\frac{n}{2}+m-1)\geq(n+\frac{1}{2})^{2^{m-1}}-n-2m+\frac{5}{2}\geq(n+\frac{1}{2})-n-2+\frac{5}{2}=1,$

which implies that there exists $D^{i}_{m-1,n}$ satisfied $\mathcal{F}^{i}=\emptyset$ for $i\in I_{m,n}\setminus S$ . Then $\mathcal{F}^{I_{m,n}\setminus S}=\emptyset$ and $D_{m-1,n}^{I_{m,n}\setminus S}-\mathcal{F}^{I_{m,n}\setminus S}=D_{m-1,n}^{I_{m,n}\setminus S}$ is connected. Let $S_{1}=\{0,1,2,\ldots,s_{1}\}$ be a subset of $S$ satisfied that there exist star-centers in $\mathcal{F}^{i}$ for $i\in S_{1}$ . Then there only leaves of stars in $\mathcal{F}^{i}$ for $i\in S\setminus S_{1}$ .

If there exist $D^{i}_{m-1,n}$ and $D^{j}_{m-1,n}$ such that both $D^{i}_{m-1,n}-\mathcal{F}^{i}$ and $D^{j}_{m-1,n}-\mathcal{F}^{j}$ are disconnected for $i\in S$ , $j\in S$ and $i\neq j$ , then $|\mathcal{F}^{i}|\geq\lceil\frac{n-1}{1+t}\rceil+m-1$ and $|\mathcal{F}^{j}|\geq\lceil\frac{n-1}{1+t}\rceil+m-1$ by induction hypothesis. Since $|\mathcal{F}|=\lceil\frac{n-1}{1+t}\rceil+m-1$ , $|\mathcal{F}^{i}|\geq\lceil\frac{n-1}{1+t}\rceil+m-1$ and $|\mathcal{F}^{j}|\geq\lceil\frac{n-1}{1+t}\rceil+m-1$ are impossible by Definition 1. Thus there is at most one $D^{i}_{m-1,n}$ satisfied that $D^{i}_{m-1,n}-\mathcal{F}^{i}$ is disconnected for $i\in S$ .

Case 1. Each $D^{i}_{m-1,n}-\mathcal{F}^{i}$ is connected for $i\in S$ .

For each $i\in S\setminus S_{1}$ , $\mathcal{F}^{i}$ consists of the outside neighbors of partical star-centers which are in $D^{S_{1}}_{m-1,n}$ . Then $D^{i}_{m-1,n}-\mathcal{F}^{i}$ has an outside neighbor of a vertex which is in $D^{I_{m,n}\setminus S}_{m-1,n}$ for $i\in S\setminus S_{1}$ . Since $D^{i}_{m-1,n}-\mathcal{F}^{i}$ is connected, $D_{m-1,n}^{I_{m,n}\setminus S_{1}}-\mathcal{F}^{I_{m,n}\setminus S_{1}}$ is connected. Let $p_{i}$ be the number of star-centers in $D^{i}_{m-1,n}$ for $i\in S_{1}$ . For $p_{i}=\lceil\frac{n-1}{1+t}\rceil+m-1$ , we find $S_{1}=\{i\}$ . Then each vertex of $D^{i}_{m-1,n}-\mathcal{F}^{i}$ has an outside neighbor in $D_{m-1,n}^{I_{m,n}\setminus S}$ . Since $D^{I_{m,n}\setminus\{i\}}_{m-1,n}-\mathcal{F}^{I_{m,n}\setminus\{i\}}$ is connected, $D_{m,n}-\mathcal{F}$ is connected (see Figure 2(a)). For $p_{i}\leq\lceil\frac{n-1}{1+t}\rceil+m-2$ , by Lemma 3.2, we have for $n\geq 2$ and $m\geq 2$ ,

		$\displaystyle\|V(D^{i}_{m-1,n})\|-(1+t)p_{i}-(\lceil\frac{n-1}{1+t}\rceil+m-1-p_{i})=t_{m-1,n}-tp_{i}-(\lceil\frac{n-1}{1+t}\rceil+m-1)\geq$
		$\displaystyle t_{m-1,n}-t(\lceil\frac{n-1}{1+t}\rceil+m-2)-(\lceil\frac{n-1}{1+t}\rceil+m-1)=t_{m-1,n}-(1+t)(\lceil\frac{n-1}{1+t}\rceil+m-1)+t$
		$\displaystyle\geq t_{m-1,n}-(1+t)(\frac{n+t-1}{1+t}+m-1)+t\geq(n+\frac{1}{2})^{2^{m-1}}-n-m-(m-1)t+\frac{3}{2}$
		$\displaystyle\geq(n+\frac{1}{2})^{2^{m-1}}-n-m-(m-1)(n+m-1)+\frac{3}{2}=(n+\frac{1}{2})^{2^{m-1}}-mn-m^{2}+m+\frac{1}{2}$
		$\displaystyle\geq(2+\frac{1}{2})^{2^{m-1}}-2m-m^{2}+m+\frac{1}{2}\geq(2+\frac{1}{2})^{2}-4-4+2+\frac{1}{2}>0,$

and for $m=1$ ,

\displaystyle|V(D^{i}_{0,n})|-(1+t)p_{i}-(\lceil\frac{n-1}{1+t}\rceil-p_{i})\geq t_{0,n}-(1+t)(\frac{n+t-1}{1+t}+1-1)+t>0,

which implies that $D^{i}_{m-1,n}-\mathcal{F}^{i}$ has a vertex that is connected to $D_{m-1,n}^{I_{m,n}\setminus S_{1}}-\mathcal{F}^{I_{m,n}\setminus S_{1}}$ for each $i\in S_{1}$ (see Figure 2(b)). Thus $D_{m,n}-\mathcal{F}$ is connected.

Case 2. There is exactly one $D^{i}_{m-1,n}$ such that $D^{i}_{m-1,n}-\mathcal{F}^{i}$ is disconnected for $i\in S$ .

Since $D^{j}_{m-1,n}-\mathcal{F}^{j}$ is connected for $j\in S$ and $j\neq i$ , $D^{I_{m,n}\setminus\{i\}}_{m-1,n}-\mathcal{F}^{I_{m,n}\setminus\{i\}}$ is connected by the similar argument as Case 1. By induction hypothesis, $|\mathcal{F}^{i}|\geq\lceil\frac{n-1}{1+t}\rceil+m-1$ . If $i\in S_{1}$ , then each $\mathcal{F}^{j}$ has exactly one star-center and exactly one of leaves of that star in $\mathcal{F}^{i}$ for $j\in S_{1}\setminus\{i\}$ , and each $\mathcal{F}^{j}$ has exactly one leaf of star and its center in $\mathcal{F}^{i}$ for $j\in S\setminus S_{1}$ (see Figure 2(c)). By Lemma 3.1 $(ii)$ , each vertex of $D^{i}_{m-1,n}-\mathcal{F}^{i}$ has an outside neighbor in $D^{I_{m,n}\setminus S}_{m-1,n}$ . Thus $D_{m,n}-\mathcal{F}$ is connected. If $i\in S\setminus S_{1}$ , then $S\setminus S_{1}=\{i\}$ and $|S_{1}|=\lceil\frac{n-1}{1+t}\rceil+m-1$ . Otherwise, $|\mathcal{F}^{i}|<\lceil\frac{n-1}{1+t}\rceil+m-1$ , a contradiction. For $j\in S_{1}$ , we have that each $\mathcal{F}^{j}$ has exactly one star-center and one leaf of the star is in $\mathcal{F}^{i}$ (see Figure 2(d)). Thus $\mathcal{F}^{i}$ has all outside neighbors of vertices which from different $D^{j}_{m-1,n}$ s, $j\in S\setminus\{i\}$ . By Lemma 3.1 $(ii)$ , each vertex of $D^{i}_{m-1,n}-\mathcal{F}^{i}$ has an outside neighbor in $D^{I_{m,n}\setminus S}_{m-1,n}$ . Thus $D_{m,n}-\mathcal{F}$ is connected. ∎

Lemma 3.4.

$\kappa(D_{m,n};K_{1,t})\leq\lceil\frac{n-1}{1+t}\rceil+m$ for $1\leq t\leq m+n-2.$

Proof.

Given two vertices $u=00\ldots 00$ and $v=11\ldots 11$ in $D_{m,n}$ . Then $N_{D_{m,n}}(u)=\{00\ldots 0i|i\in[n-1]\}\cup\{10\ldots 00,01\ldots 00,\ldots,00\ldots 10\}$ . Since $D_{m,n}[\{00\ldots 0i|i\in[n-1]\}]$ is a complete graph $K_{n-1}$ by Definition 1, we need $\lceil\frac{n-1}{1+t}\rceil$ $K_{1,t}$ ’s to cover all vertices of $D_{m,n}[\{00\ldots 0i|i\in[n-1]\}]$ . Then we set

S_{i}=\left\{\begin{aligned} &\{00\ldots 0[(i-1)t+i];00\ldots 0[(i-1)t+i+k]|1\leq k\leq t\},~{}~{}for~{}~{}1\leq i\leq\lfloor\frac{n-1}{1+t}\rfloor;\\ &\{00\ldots 0(n-1);00\ldots 0k|n-t-1\leq k\leq n-2\},~{}~{}for~{}~{}i=\lceil\frac{n-1}{1+t}\rceil.\\ \end{aligned}\right.

Next, we will construct several stars $K_{1,t}$ ’s to cover the vertex subset $\{10\ldots 00,01\ldots 00,\ldots,00\ldots 10\}$ . For $1\leq j\leq m$ , let $S_{\lceil\frac{n-1}{1+t}\rceil+j}$ be a star with center $0\ldots 010\ldots 0$ ( $u_{j}=1$ ) and its leaves are the neighbors of $0\ldots 010\ldots 0$ except $u$ (see Figure 3(a)).

Let $\mathcal{F}=\{S_{i}|i\in[\lceil\frac{n-1}{1+t}\rceil+m]\}$ . Since $N_{D_{m,n}}(u)\subseteq V(\mathcal{F})$ and $v\in V(D_{m,n}-\mathcal{F})$ , $D_{m,n}-\mathcal{F}$ has two components. Then $u$ is a singleton and $v$ belongs to the other component. Thus $\kappa(D_{m,n};K_{1,t})\leq|\mathcal{F}|=\lceil\frac{n-1}{1+t}\rceil+m$ .

∎

Lemmas 3.3 and 3.4 imply that $\lceil\frac{n-1}{1+t}\rceil+m\leq\kappa^{s}(D_{m,n};K_{1,t})\leq\kappa(D_{m,n};K_{1,t})\leq\lceil\frac{n-1}{1+t}\rceil+m$ for $1\leq t\leq m+n-2$ . Then following result holds.

Theorem 3.5.

$\kappa(D_{m,n};K_{1,t})=\kappa^{s}(D_{m,n};K_{1,t})=\lceil\frac{n-1}{1+t}\rceil+m$ for $1\leq t\leq m+n-2$ .

Lemma 3.6.

$\kappa(D_{m,n};K_{s})\geq\lceil\frac{n-1}{s}\rceil+m$ for $3\leq s\leq n-1$ .

Proof.

Induction on $m$ . For $m=0$ , $D_{0,n}\cong K_{n}$ . Assume $\kappa(D_{0,n};K_{s})\leq\lceil\frac{n-1}{s}\rceil-1$ for $3\leq s\leq n-1$ . Then there exists a $K_{s}$ -structure cut $\mathcal{F^{\prime}}$ of $D_{0,n}$ , $|\mathcal{F^{\prime}}|\leq\lceil\frac{n-1}{s}\rceil-1.$ We have

|V(\mathcal{F^{\prime}})|\leq s\cdot(\lceil\frac{n-1}{s}\rceil-1)\leq s\cdot(\frac{n+s-2}{s}-1)=n-2<n-1.

Since $\kappa(D_{0,n})=n-1$ , $|V(\mathcal{F^{\prime}})|<\kappa(D_{0,n})$ , a contradiction. Then $\kappa(D_{0,n};K_{s})\geq\lceil\frac{n-1}{s}\rceil$ for $3\leq s\leq n-1$ .

Assume $\kappa(D_{m-1,n};K_{s})\geq\lceil\frac{n-1}{s}\rceil+m-1$ for $3\leq s\leq n-1$ . Now, we consider $D_{m,n}$ . Let $\mathcal{F}=\{T_{i}|1\leq i\leq\lceil\frac{n-1}{s}\rceil+m-1\}$ be a set such that $T_{i}\cong K_{s}$ for $1\leq i\leq\lceil\frac{n-1}{s}\rceil+m-1$ . We need to show that $D_{m,n}-\mathcal{F}$ is connected. Let $\mathcal{F}^{i}=\mathcal{F}\cap D^{i}_{m-1,n}$ , $\mathcal{F}^{T}=\cup_{i\in T}\mathcal{F}^{i}$ and $D_{m-1,n}^{T}=\cup_{i\in T}D^{i}_{m-1,n}$ by Eq. (3.1). We know that a complete graph $K_{s}$ for each $3\leq s\leq n-1$ is in exactly one $D^{i}_{m-1,n}$ by Definition 1. Let $p_{i}$ be the number of $K_{s}$ ’s in $D^{i}_{m-1,n}$ . If $p_{i}=\lceil\frac{n-1}{s}\rceil+m-1$ for some $i\in I_{m,n}$ , then $\mathcal{F}^{I_{m,n}\setminus\{i\}}=\emptyset$ . Thus $D^{I_{m,n}\setminus\{i\}}_{m-1,n}-\mathcal{F}^{I_{m,n}\setminus\{i\}}$ is connected. By Definition 1, each vertex of $D^{i}_{m-1,n}-\mathcal{F}^{i}$ has an outside neighbor in $D^{I_{m,n}\setminus\{i\}}_{m-1,n}$ . Then $D_{m,n}-\mathcal{F}$ is connected.

If $p_{i}\leq\lceil\frac{n-1}{s}\rceil+m-2$ for some $i\in I_{m,n}$ , then $D^{i}_{m-1,n}-\mathcal{F}^{i}$ is connected by induction hypothesis. Since $|\mathcal{F}|=\lceil\frac{n-1}{s}\rceil+m-1$ , there are at most $(\lceil\frac{n-1}{s}\rceil+m-1)$ $D^{i}_{m-1,n}$ ’s such that $\mathcal{F}^{i}\neq\emptyset$ . Let $X=\{0,1,\ldots,x\}$ be a set such that $\mathcal{F}^{i}\neq\emptyset$ for $i\in X$ . Then $|X|\leq\lceil\frac{n-1}{s}\rceil+m-1$ . By Lemma 3.2,

		$\displaystyle t_{m-1,n}+1-\|X\|\geq t_{m-1,n}+1-(\lceil\frac{n-1}{s}\rceil+m-1)\geq t_{m-1,n}+1-(\frac{n+s-2}{s}+m-1)$
		$\displaystyle\geq(n+\frac{1}{2})^{2^{m-1}}-\frac{1}{2}+1-(\frac{n+1}{3}+m-1)\geq n+1-\frac{n+1}{3}\geq 2,$

which implies that there exist $D^{i}_{m-1,n}$ ’s such that $\mathcal{F}^{i}=\emptyset$ for $i\in I_{m,n}\setminus X$ . Then $\mathcal{F}^{I_{m,n}\setminus X}=\emptyset$ and $D_{m-1,n}^{I_{m,n}\setminus X}-\mathcal{F}^{I_{m,n}\setminus X}$ is connected. By Lemma 3.2, we find, for $i\in X$ , $m\geq 1$ and $n\geq 2$ ,

		$\displaystyle\|V(D^{i}_{m-1,n})\|-sp_{i}-(\lceil\frac{n-1}{s}\rceil+m-1-p_{i})=t_{m-1,n}-(s-1)p_{i}-(\lceil\frac{n-1}{s}\rceil+m-1)$
		$\displaystyle\geq t_{m-1,n}-(s-1)(\lceil\frac{n-1}{s}\rceil+m-2)-(\lceil\frac{n-1}{s}\rceil+m-1)=t_{m-1,n}-s(\lceil\frac{n-1}{s}\rceil+m-2)-1$
		$\displaystyle\geq t_{m-1,n}-s(\frac{n+s-2}{s}+m-2)-1\geq(n+\frac{1}{2})^{2^{m-1}}-\frac{1}{2}-n-(m-1)s+1$
		$\displaystyle\geq(n+\frac{1}{2})^{2^{m-1}}-n-(m-1)(n-1)+\frac{1}{2}=(n+\frac{1}{2})-n+\frac{1}{2}=1>0,$

which implies that $D^{i}_{m-1,n}-\mathcal{F}^{i}$ has a vertex that is connected to $D_{m-1,n}^{I_{m,n}\setminus X}$ for each $i\in X$ . Since $D^{i}_{m-1,n}-\mathcal{F}^{i}$ is connected for $i\in X$ , $D_{m,n}-\mathcal{F}$ is connected. ∎

Lemma 3.7.

$\kappa(D_{m,n};K_{s})\leq\lceil\frac{n-1}{s}\rceil+m$ for $3\leq s\leq n-1$ .

Proof.

Given two vertices $u=00\ldots 00$ and $v=11\ldots 11$ in $D_{m,n}$ . Then $N_{D_{m,n}}(u)=\{00\ldots 0i|i\in[n-1]\}\cup\{10\ldots 00,01\ldots 00,\ldots,00\ldots 10\}$ . Let $S_{i}=\{x_{i}|1\leq i\leq s\}$ be a complete graph $K_{s}$ . Since $D_{m,n}[\{00\ldots 0i|i\in[n-1]\}]$ is a complete graph $K_{n-1}$ by Definition 1, we need $\lceil\frac{n-1}{s}\rceil$ $K_{s}$ ’s to cover all vertices of $D_{m,n}[\{00\ldots 0i|i\in[n-1]\}]$ . Then we set

S_{i}=\left\{\begin{aligned} &\{00\ldots 0[(i-1)s+k+1]|0\leq k\leq s-1\},~{}~{}for~{}~{}1\leq i\leq\lfloor\frac{n-1}{s}\rfloor;\\ &\{00\ldots 0k|n-s\leq k\leq n-1\},~{}~{}for~{}~{}i=\lceil\frac{n-1}{s}\rceil.\\ \end{aligned}\right.

Next, we will construct several complete graphs $K_{s}$ ’s to cover the vertex subset $\{10\ldots 00,01\ldots 00,\ldots,00\ldots 10\}$ . For each $j\in[m]$ , let $S_{\lceil\frac{n-1}{s}\rceil+j}$ be a $K_{s}$ with vertices $0\ldots 010\ldots 0k$ ( $u_{j}=1$ and $u_{0}=k$ ) for $0\leq k\leq s-1$ . Precisely, $S_{\lceil\frac{n-1}{s}\rceil+1}=\{00\ldots 01k|0\leq k\leq s-1\}$ , $S_{\lceil\frac{n-1}{s}\rceil+2}=\{00\ldots 10k|0\leq k\leq s-1\}$ , analogously, $S_{\lceil\frac{n-1}{s}\rceil+m}=\{10\ldots 00k|0\leq k\leq s-1\}$ (see Figure 3(b)).

Let $\mathcal{F}=\{S_{i}|i\in[\lceil\frac{n-1}{s}\rceil+m]\}$ . Since $N_{D_{m,n}}(u)\subseteq V(\mathcal{F})$ and $v\in V(D_{m,n}-\mathcal{F})$ , $D_{m,n}-\mathcal{F}$ has two components. Then $u$ is a singleton and $v$ belongs to the other component. Thus $\kappa(D_{m,n};K_{s})\leq|\mathcal{F}|=\lceil\frac{n-1}{s}\rceil+m$ . ∎

On the basis of 3.6 and 3.7, the following result holds.

Theorem 3.8.

$\kappa(D_{m,n};K_{s})=\lceil\frac{n-1}{s}\rceil+m$ for $3\leq s\leq n-1$ .

4 The structure connectivity of BCDC

Since $n$ -dimensional BCDC consists of two $n-1$ -dimensional crossed cubes and an independent set, we introduce $n$ -dimensional crossed cube first.

Definition 2.

[4] Two binary strings $x=x_{1}x_{0}$ and $y=y_{1}y_{0}$ are called pair related (denoted by $x\sim y$ ) if and only if $(x,y)\in\{(00,00),(10,10),(01,11),(11,01)\}$ .

Definition 3.

[4] We set $CQ_{1}$ as $K_{2}$ with vertices $0$ and $1$ . $CQ_{n}$ consists of $CQ^{0}_{n-1}$ and $CQ^{1}_{n-1}$ . Two vertices $u=0u_{n-2}\ldots u_{0}\in V(CQ^{0}_{n-1})$ and $v=1v_{n-2}\ldots v_{0}\in V(CQ^{1}_{n-1})$ are joined by an edge in $CQ_{n}$ if and only if

(i)

$u_{n-2}=v_{n-2}$ if $n$ is even, and
(ii)

$u_{2i+1}u_{2i}\sim v_{2i+1}v_{2i}$ , for $0\leq i<\lfloor\frac{n-1}{2}\rfloor$ .

For any two vertices $u=u_{n-1}u_{n-2}\ldots u_{0}$ and $v=v_{n-1}v_{n-2}\ldots v_{0}$ , $u$ is adjacent to $v$ if and only if there is $0\leq d\leq n-1$ satisfying the following conditions: (1) $u_{n-1}\ldots u_{d+1}=v_{n-1}\ldots v_{d+1}$ , (2) $u_{d}\neq v_{d}$ , (3) $u_{d-1}=v_{d-1}$ if $d$ is odd, (4) $u_{2i+1}u_{2i}\sim v_{2i+1}v_{2i}$ , for all $0\leq i\leq\lfloor\frac{d}{2}\rfloor-1$ . Then we set $v$ as $d$ -dimensional neighbor of $u$ , denoted by $u^{d}$ .

BCDC’s switches are vertices of $CQ_{n}$ and servers are edges of $CQ_{n}$ . Each switch is an $n$ -bit binary string $x=x_{n-1}x_{n-2}\ldots x_{0}$ and each server is a pair $[x,y]$ . Then we get the original graph $A_{n}$ of the $n$ -dimensional BCDC network, $B_{n}$ . Consider that switches are transparent, the graph structure of BCDC network as follows:

Definition 4.

[19] We set $B_{2}$ as a $4$ -cycle with vertices $[00,01],[00,10],[01,11]$ and $[10,11].$ For $n\geq 3$ , $B^{0}_{n-1}$ (resp. $B^{1}_{n-1}$ ) is obtained by prefixing each vertex $[x,y]$ of $B_{n-1}$ with $0$ (resp. $1$ ). $B_{n}$ consists of $B^{0}_{n-1}$ , $B^{1}_{n-1}$ and a vertex subset $S_{n}=\{[a,b]|a\in V(CQ^{0}_{n-1}),b\in V(CQ^{1}_{n-1})$ and $(a,b)\in E(CQn)\}$ . For three vertices $u=[a,b]\in V(B^{0}_{n-1}),v=[c,d]\in S_{n}$ and $w=[e,f]\in V(B^{1}_{n-1})$ ,

(i)

$(u,v)\in E(B_{n})$ if and only if $a=c$ or $b=c$ ;
(ii)

$(v,w)\in E(B_{n})$ if and only if $e=d$ or $f=d$ .

Lemma 4.1.

[11] $CQ_{n}$ is $n$ -regular and triangle-free.

Lemma 4.2.

[17] For $B_{n}$ , we have the statements below:

(i)

$B_{n}$ is $(2n-2)$ -regular and has $n2^{n-1}$ vertices, $n(n-1)2^{n-1}$ edges.
(ii)

$\kappa(B_{n})=2n-2$ .
(iii)

$B_{n}$ is the line graph of $CQ_{n}$ .

Lemma 4.3.

[17] For $n\geq 3$ and $0\leq g\leq n-3$ ,

\kappa_{g}(B_{n})=\left\{\begin{aligned} 4,~{}~{}if~{}n=3;\\ 2n+g(n-2)-2,~{}~{}if~{}n\geq 4.\\ \end{aligned}\right.

Lemma 4.4.

$B_{n}[N_{B_{n}}(u)]$ is two independent complete graphs $K_{n-1}$ .

Proof.

Given a vertex $u=[v,w]$ in $B_{n}$ , where $vw$ is an edge of $CQ_{n}$ . Since $CQ_{n}$ is $n$ -regular, we set $N_{CQ_{n}}(v)=\{w\}\cup\{v^{i}|0\leq i\leq n-2\}$ and $N_{CQ_{n}}(w)=\{v\}\cup\{w^{i}|0\leq i\leq n-2\}$ . Then $N_{B_{n}}(u)=\{[v,v^{i}],[w,w^{i}]|0\leq i\leq n-2\}$ . According to the Definition 4, $B_{n}[\{[v,v^{i}]|0\leq i\leq n-2\}]\cong K_{n-1}$ and $B_{n}[\{[w,w^{i}]|0\leq i\leq n-2\}]\cong K_{n-1}$ . Suppose $[v,v^{i}]$ is adjacent to $[w,w^{j}]$ in $B_{n}$ . Then $v^{i}=w^{j}$ , which implies that $v,w$ and $v^{i}$ (or $w^{j}$ ) induce a triangle in $CQ_{n}$ , a contradiction by Lemma 4.1. Thus $B_{n}[N_{B_{n}}(u)]$ is two independent complete graphs $K_{n-1}.$ ∎

Theorem 4.5.

\kappa(B_{n};K_{1,1})=\kappa^{s}(B_{n};K_{1,1})=\left\{\begin{aligned} n-1,~{}~{}for~{}odd~{}n\geq 5;\\ n,~{}~{}for~{}even~{}n\geq 4.\\ \end{aligned}\right.

Proof.

Assume $\kappa^{s}(B_{n};K_{1,1})\leq n-2$ ( $\leq n-1$ ) for odd (even) $n$ . Then $B_{n}$ has a $K_{1,1}$ -substructure cut $\mathcal{F}$ . Let $C$ be a smallest component of $B_{n}-\mathcal{F}$ . For odd $n\geq 5$ , $|V(\mathcal{F})|\leq 2(n-2)<\kappa(B_{n})$ by Lemma 4.2, a contradiction. Thus $\kappa^{s}(B_{n};K_{1,1})\geq n-1$ for odd $n\geq 5$ . For even $n\geq 4$ , $|V(\mathcal{F})|\leq 2(n-1)<\kappa_{1}(B_{n})$ by Lemma 4.3. Then $|V(C)|=1$ , say $u$ . We have $N_{B_{n}}(u)\subseteq V(\mathcal{F})$ . By Lemma 4.4, $n-1$ subgraphs of $K_{1,1}$ can not cover all vertices of $N_{B_{n}}(u)$ , a contradiction. Thus $\kappa^{s}(B_{n};K_{1,1})\geq n$ for even $n\geq 4$ .

We give a $K_{1,1}$ -structure cut of $B_{n}$ for $n\geq 4$ . Let $u=[v,w]$ be a vertex of $B_{n}$ with $v=00\ldots 00$ and $w=10\ldots 00$ . Then $N_{B_{n}}(u)=\{[v,v^{i}],[w,w^{i}]|0\leq i\leq n-2\}$ . For $0\leq j\leq\lfloor\frac{n-2}{2}\rfloor-1$ , we set $S_{j}=\{[v,v^{2j}];[v,v^{2j+1}]\}$ and $S^{\prime}_{j}=\{[w,w^{2j}];[w,w^{2j+1}]\}$ . For odd $n$ , we set

\displaystyle S_{\lfloor\frac{n-2}{2}\rfloor}=\{[v,v^{n-3}];[v,v^{n-2}]\},~{}~{}S^{\prime}_{\lfloor\frac{n-2}{2}\rfloor}=\{[w,w^{n-3}];[w,w^{n-2}]\}.

For even $n$ , we set

\displaystyle S_{\lfloor\frac{n-2}{2}\rfloor}=\{[v,v^{n-2}];[v^{n-2},v^{n-2,n-1}]\},~{}~{}S^{\prime}_{\lfloor\frac{n-2}{2}\rfloor}=\{[w,w^{n-2}];[w^{n-2},w^{n-2,n-1}]\}.

Let $\mathcal{F}=\{S_{j},S^{\prime}_{j}|0\leq j\leq\lfloor\frac{n-2}{2}\rfloor\}$ . We find $N_{B_{n}}(u)\subseteq V(\mathcal{F})$ and $[11\ldots 11,11\ldots 10]\notin V(B_{n}-\mathcal{F})$ , which implies that $B_{n}-\mathcal{F}$ has two components. Then $u$ is a singleton and $[11\ldots 11,11\ldots 10]$ belongs to the other component. Thus $\kappa(B_{n};K_{1,1})\leq n-1$ for odd $n\geq 5$ and $\kappa(B_{n};K_{1,1})\leq n$ for even $n\geq 4$ . ∎

Lemma 4.6.

For $n\geq 4$ and $2\leq t\leq 2n-3$ , let $r$ be remainder of $n-1$ divided by $1+t$ . Then

\kappa^{s}(B_{n};K_{1,t})\geq\left\{\begin{aligned} &\frac{2n-4}{1+t}+1,~{}~{}if~{}2\leq t\leq n-3~{}and~{}r=1,\\ &2\lceil\frac{n-1}{1+t}\rceil,~{}~{}otherwise.\\ \end{aligned}\right.

Proof.

Suppose to the contrary that $\kappa^{s}(B_{n};K_{1,t})\leq\frac{2n-4}{1+t}$ for $2\leq t\leq n-3$ and $r=1$ , and $\kappa^{s}(B_{n};K_{1,t})\leq 2\lceil\frac{n-1}{1+t}\rceil-1$ for otherwise. Then $B_{n}$ has a $K_{1,t}$ -substructure cut $\mathcal{F}$ . Let $C$ be a smallest component of $B_{n}-\mathcal{F}$ . Then $|V(\mathcal{F})|\geq\kappa(B_{n})=2n-2$ by Lemma 4.2. For $2\leq t\leq n-3$ , if $r=0$ , then $|V(\mathcal{F})|\leq(1+t)(2\lceil\frac{n-1}{1+t}\rceil-1)<2n-2$ , a contradiction. If $r=1$ , then $|V(\mathcal{F})|\leq(1+t)\frac{2n-4}{1+t}=2n-4<2n-2$ , a contradiction. Similarly, for $n-2\leq t\leq 2n-4$ , $|V(\mathcal{F})|\leq 1+t\leq 2n-3<2n-2$ , a contradiction. Then we consider the cases $t=2n-3$ , or $2\leq t\leq n-3$ and $r\geq 2$ . We find $|V(\mathcal{F})|\leq(1+t)(2\lceil\frac{n-1}{1+t}\rceil-1)=1+t\leq 2n-2$ for $t=2n-3$ , and $|V(\mathcal{F})|\leq(1+t)(2\lceil\frac{n-1}{1+t}\rceil-1)\leq(1+t)(2\frac{n-1+t}{1+t}-1)=2n-3+t\leq 3n-6$ for $2\leq t\leq n-3$ and $r\geq 2$ . By Lemma 4.3, $|V(\mathcal{F})|<\kappa_{1}(B_{n})$ . Then $|V(C)|=1$ , say $u=[v,w]$ . Thus $N_{B_{n}}(u)\subseteq V(\mathcal{F})$ , which implies that $B_{n}[N_{B_{n}}(u)]$ has a star $K_{1,2n-3}$ for $t=2n-3$ , contradicting Lemma 4.4. For $2\leq t\leq n-3$ and $r\geq 2$ , $|\mathcal{F}|\leq 2\lceil\frac{n-1}{1+t}\rceil-1$ . If all star-centers are in $N_{B_{n}}(u)$ , then $2\lceil\frac{n-1}{1+t}\rceil-1$ stars can not cover all vertices of $N_{B_{n}}(u)$ by Lemma 4.4. Thus there exists a star whose center $x$ is not in $N_{B_{n}}(u)\cup\{u\}$ . We know that $x$ has at most two neighbors in $N_{B_{n}}(u)$ since $[v,v^{i}]$ and $[v,v^{j}]$ ( $[w,w^{i}]$ and $[w,w^{j}]$ ) have no common neighbor in $V(B_{n})-N_{B_{n}}(u)-\{u\}$ . Then $V(\mathcal{F})\leq(1+t)\cdot 2\cdot\lfloor\frac{n-1}{1+t}\rfloor+2=(1+t)\cdot 2\cdot\frac{n-1-r}{1+t}+2=2n-2r<2n-2$ for $r\geq 2$ . It is a contradiction. ∎

Lemma 4.7.

For $n\geq 4$ and $2\leq t\leq 2n-3$ , let $r$ be remainder of $n-1$ divided by $1+t$ . Then

\kappa(B_{n};K_{1,t})\leq\left\{\begin{aligned} &\frac{2n-4}{1+t}+1,~{}~{}if~{}2\leq t\leq n-3~{}and~{}r=1,\\ &2\lceil\frac{n-1}{1+t}\rceil,~{}~{}otherwise.\end{aligned}\right.

Proof.

Let $u=[v,w]$ be a vertex of $B_{n}$ with $v=00\ldots 00$ and $w=10\ldots 00$ . Then $N_{B_{n}}(u)=\{[v,v^{i}],[w,w^{i}]|0\leq i\leq n-2\}$ .

For $n-2\leq t\leq 2n-3$ , let $S_{1}$ ( $S_{2}$ ) be a star with center $[v,v^{0}]$ ( $[w,w^{0}]$ ) and the leaves are neighbors of $[v,v^{0}]$ ( $[w,w^{0}]$ ). Precisely,

\left\{\begin{aligned} &S_{1}=\{[v,v^{0}];[v,v^{i}],t_{j}|1\leq i\leq n-2,n-1\leq j\leq t\},\\ &S_{2}=\{[w,w^{0}];[w,w^{i}],t^{\prime}_{j}|1\leq i\leq n-2,n-1\leq j\leq t\},\\ \end{aligned}\right.

where $\{t_{j}|n-1\leq j\leq t\}\subseteq N_{B_{n}}([v,v^{0}])\setminus\{u,[v,v^{i}]|1\leq i\leq n-2\}$ and $\{t^{\prime}_{j}|n-1\leq j\leq t\}\subseteq N_{B_{n}}([w,w^{0}])\setminus\{u,[w,w^{i}]|1\leq i\leq n-2\}$ . Let $\mathcal{F}=\{S_{1},S_{2}\}$ . Since $N_{B_{n}}(u)\subseteq V(\mathcal{F})$ and $[11\ldots 11,11\ldots 10]\notin V(B_{n}-\mathcal{F})$ , $B_{n}-\mathcal{F}$ has two components. Then $u$ is a singleton and $[11\ldots 11,11\ldots 10]$ belongs to the other component.

For $2\leq t\leq n-3$ and $1\leq i\leq\lfloor\frac{n-1}{1+t}\rfloor$ , we set $S_{i}$ ( $S^{\prime}_{i}$ ) as a star with center $[v,v^{(i-1)(t+1)}]$ ( $[w,w^{(i-1)(t+1)}]$ ) and the leaves are $[v,v^{(i-1)(t+1)+k}]$ ( $[w,w^{(i-1)(t+1)+k}]$ ) for each $1\leq k\leq t$ . The more detailed expressions are

\left\{\begin{aligned} &S_{i}=\{[v,v^{(i-1)(t+1)}];[v,v^{(i-1)(t+1)+k}]|1\leq k\leq t\}\\ ~{}~{}&S^{\prime}_{i}=\{[w,w^{(i-1)(t+1)}];[w,w^{(i-1)(t+1)+k}]|1\leq k\leq t\}~{}~{}\end{aligned}\right.

If $r=1$ , then we set $S^{\prime\prime}=\{[v^{n-2},w^{n-2}];[v,v^{n-2}],[w,w^{n-2}],t_{i}|1\leq i\leq t-2\}$ where $\{t_{i}|1\leq i\leq t-2\}\subseteq N_{B_{n}}([v^{n-2},w^{n-2}])\setminus\{[v,v^{n-2}],[w,w^{n-2}]\}$ . If $r\geq 2$ , then we set

\left\{\begin{aligned} &S^{\prime\prime}_{1}=\{[v,v^{n-2}];[v,v^{i}],t_{j}|n-r-1\leq i\leq n-3,1\leq j\leq t-r+1\},\\ &S^{\prime\prime}_{2}=\{[w,w^{n-2}];[w,w^{i}],t^{\prime}_{j}|n-r-1\leq i\leq n-3,1\leq j\leq t-r+1\},\\ \end{aligned}\right.

where $\{t_{j}|1\leq j\leq t-r\}\subseteq N_{B_{n}}([v,v^{n-2}])\setminus\{u,[v,v^{i}]|0\leq i\leq n-3\}$ and $\{t^{\prime}_{j}|1\leq j\leq t-r\}\subseteq N_{B_{n}}([w,w^{n-2}])\setminus\{u,[w,w^{i}]|0\leq i\leq n-3\}$ . Let $\mathcal{F}=\{S_{i},S^{\prime}_{i}|1\leq i\leq\lfloor\frac{n-1}{1+t}\rfloor\}$ for $r=0$ , $\mathcal{F}=\{S_{i},S^{\prime}_{i},S^{\prime\prime}|1\leq i\leq\lfloor\frac{n-1}{1+t}\rfloor\}$ for $r=1$ and $\mathcal{F}=\{S_{i},S^{\prime}_{i},S^{\prime\prime}_{1},S^{\prime\prime}_{2}|1\leq i\leq\lfloor\frac{n-1}{1+t}\rfloor\}$ for $r\geq 2$ . We find $N_{B_{n}}(u)\subseteq V(\mathcal{F})$ and $[11\ldots 11,11\ldots 10]\notin V(B_{n}-\mathcal{F})$ , which implies that $B_{n}-\mathcal{F}$ has two components and $u$ is a singleton and $[11\ldots 11,11\ldots 10]$ belongs to other component. ∎

By Lemmas 4.6 and 4.7, we have the following result.

Theorem 4.8.

For $n\geq 4$ and $2\leq t\leq 2n-3$ , let $r$ be remainder of $n-1$ divided by $1+t$ . Then

\kappa(B_{n};K_{1,t})=\kappa^{s}(B_{n};K_{1,t})=\left\{\begin{aligned} &\frac{2n-4}{1+t}+1,~{}~{}if~{}2\leq t\leq n-3~{}and~{}r=1,\\ &2\lceil\frac{n-1}{1+t}\rceil,~{}~{}otherwise.\end{aligned}\right.

Lemma 4.9.

For $n\geq 4$ and $4\leq k\leq 2n-1$ , we have

\kappa^{s}(B_{n};P_{k})\geq\left\{\begin{aligned} &\frac{2n-2}{k},~{}~{}if~{}4\leq k\leq n-1~{}and~{}k\mid n-1,\\ &\lceil\frac{2n-1}{k}\rceil,~{}~{}otherwise.\\ \end{aligned}\right.

Proof.

Suppose $\kappa^{s}(B_{n};P_{k})\leq\frac{2n-2}{k}-1$ for $4\leq k\leq n-1$ and $k\mid n-1$ , and $\kappa^{s}(B_{n};P_{k})\leq\lceil\frac{2n-1}{k}\rceil-1$ for otherwise. Then there exists a $P_{k}$ -substructure cut $\mathcal{F}$ . Let $C$ be a smallest component of $B_{n}-\mathcal{F}$ . Then $|V(\mathcal{F})|\geq\kappa(B_{n})=2n-2$ by Lemma 4.2. For $4\leq k\leq n-1$ and $k\mid n-1$ , $|V(\mathcal{F})|\leq k\cdot(\frac{2n-2}{k}-1)=2n-2-k<2n-2$ , a contradiction. Similarly, for $n\leq k\leq 2n-3$ , $|V(\mathcal{F})|\leq k\cdot(\lceil\frac{2n-1}{k}\rceil-1)=k\leq 2n-3<2n-2$ , a contradiction.

We have $|V(C)|=1$ , say $C=\{u\}$ , for $k=2n-2$ or $4\leq k\leq n-1$ and $k\nmid n-1$ since $|V(\mathcal{F})|\leq k\cdot(\lceil\frac{2n-1}{k}\rceil-1)\leq k(\frac{2n-1+k-1}{k}-1)=2n-2<\kappa_{1}(B_{n})$ by Lemma 4.3. Then $N_{B_{n}}(u)\subseteq V(\mathcal{F})$ , which implies that $B_{n}[N_{B_{n}}(u)]$ has a hamiltonian path $P_{2n-2}$ and $\frac{2n-2}{k}$ vertex-disjoint $P_{k}$ ’s for $4\leq k\leq n-1$ and $k\nmid n-1$ . It is impossible by Lemma 4.4. ∎

Lemma 4.10.

For $n\geq 4$ and $4\leq k\leq 2n-1$ , we have

\kappa(B_{n};P_{k})\leq\left\{\begin{aligned} &\frac{2n-2}{k},~{}~{}if~{}4\leq k\leq n-1~{}and~{}k\mid n-1,\\ &\lceil\frac{2n-1}{k}\rceil,~{}~{}otherwise.\\ \end{aligned}\right.

Proof.

Let $u=[v,w]$ be a vertex of $B_{n}$ with $v=00\ldots 00$ and $w=10\ldots 00$ . Then $N_{B_{n}}(u)=\{[v,v^{i}],[w,w^{i}]|0\leq i\leq n-2\}$ . For convenience, we set $P_{n-1}(w)=\langle[w,w^{0}],[w,w^{1}],\\ \ldots,[w,w^{n-2}]\rangle$ as a path with $n-1$ vertices.

For $k=2n-1$ , we set $\mathcal{F}=P_{2n-1}=\langle P_{n-1}(v),[v^{n-2},w^{n-2}],P^{-1}_{n-1}(w)\rangle$ . For $n\leq k\leq 2n-2$ , let $P^{1}_{2n-2}=\langle P_{n-1}(v),P_{n-1}(v^{n-2})-[v,v^{n-2}]\cup[v^{n-2},v^{n-2,n-1}]\rangle$ and $P^{2}_{2n-2}=\langle P_{n-1}(w),P_{n-1}(w^{n-2})-[w,w^{n-2}]\cup[w^{n-2},w^{n-2,n-1}]\rangle$ . Then we set $\mathcal{F}=\{P^{1}_{k},P^{2}_{k}\}$ , where $P^{1}_{k}$ ( $P^{2}_{k}$ ) along $P^{1}_{2n-2}$ ( $P^{2}_{2n-2}$ ) with end-vertices $[v,v^{0}]$ ( $[w,w^{0}]$ ) and the $k$ -th vertex of $P^{1}_{2n-2}$ ( $P^{2}_{2n-2}$ ).

For $4\leq k\leq n-1$ and $k\mid n-1$ , we set $\mathcal{F}=\{P^{i}_{k}|1\leq i\leq\frac{2n-2}{k}\}$ , where $P^{1}_{k}$ along $P_{n-1}(v)$ with end-vertices $[v,v^{0}]$ and $[v,v^{k-1}]$ , $P^{2}_{k}$ along $P_{n-1}(v)$ with end-vertices $[v,v^{k}]$ and $[v,v^{2k-1}]$ , by the analogous, $P^{\frac{n-1}{k}}_{k}$ along $P_{n-1}(v)$ with end-vertices $[v,v^{n-k-1}]$ and $[v,v^{n-2}]$ . Similarly, we set $P^{\frac{n-1}{k}+1}_{k}$ along $P_{n-1}(w)$ with end-vertices $[w,w^{0}]$ and $[w,w^{k-1}]$ , by the analogous, $P^{2\frac{n-1}{k}}_{k}$ along $P_{n-1}(v)$ with end-vertices $[w,w^{n-k-1}]$ and $[w,w^{n-2}]$ . For $4\leq k\leq n-1$ and $k\nmid n-1$ , let $P_{3n-2}=\langle P_{2n-1},P_{n-1}(w^{0})-[w,w^{0}]\cup[w^{0},w^{0,n-1}]\rangle$ and $\mathcal{F}=\{P^{i}_{k}|1\leq i\leq\lceil\frac{2n-1}{k}\rceil\}$ , where $P^{1}_{k}$ along $P_{3n-2}$ with end-vertices $[v,v^{0}]$ and $[v,v^{k-1}]$ , $P^{2}_{k}$ along $P_{3n-2}$ with end-vertices $[v,v^{k}]$ and $[v,v^{2k-1}]$ , by the analogous, $P^{\lceil\frac{2n-1}{k}\rceil}_{k}$ along $P_{3n-2}$ with end-vertices are the $(k(\lceil\frac{2n-1}{k}\rceil-1)+1)$ -th and $k\lceil\frac{2n-1}{k}\rceil$ -th vertices of $P_{3n-2}$ .

We find $N_{B_{n}}(u)\subseteq V(\mathcal{F})$ and $[11\ldots 11,11\ldots 10]\notin V(B_{n}-\mathcal{F})$ , which implies that $B_{n}-\mathcal{F}$ has two components. Then $u$ is a singleton and $[11\ldots 11,11\ldots 10]$ belongs to other component. ∎

Theorem 4.11.

For $n\geq 4$ and $4\leq k\leq 2n-1$ , we have

\kappa(B_{n};P_{k})=\kappa^{s}(B_{n};P_{k})=\left\{\begin{aligned} &\frac{2n-2}{k},~{}~{}if~{}4\leq k\leq n-1~{}and~{}k\mid n-1,\\ &\lceil\frac{2n-1}{k}\rceil,~{}~{}otherwise.\\ \end{aligned}\right.

We have $\kappa^{s}(B_{n};C_{k})\leq\kappa^{s}(B_{n};P_{k})$ . By the similar argument as Lemma 4.9, we obtain the following result.

Theorem 4.12.

For $n\geq 4$ and $4\leq k\leq 2n-1$ , we have

\kappa^{s}(B_{n};C_{k})=\left\{\begin{aligned} &\frac{2n-2}{k},~{}~{}if~{}4\leq k\leq n-1~{}and~{}k\mid n-1,\\ &\lceil\frac{2n-1}{k}\rceil,~{}~{}otherwise.\\ \end{aligned}\right.

Lemma 4.13.

For $n\geq 5$ and $3\leq k\leq 2n$ , let $r$ be remainder of $n-1$ divided by $k$ . Then

\kappa(B_{n};C_{k})\geq\left\{\begin{aligned} &2\lceil\frac{n-1}{k}\rceil-1,~{}~{}if~{}3\leq k\leq n-1~{}and~{}1\leq r\leq\lfloor\frac{k}{2}\rfloor-1,~{}or~{}k=2n,\\ &3,~{}~{}if~{}k=n,\\ &2\lceil\frac{n-1}{k}\rceil,~{}~{}otherwise.\\ \end{aligned}\right.

Proof.

Suppose to the contrary that $\kappa(B_{n};C_{k})\leq 2\lceil\frac{n-1}{k}\rceil-2$ for $3\leq k\leq n-1$ and $1\leq r\leq\lfloor\frac{k}{2}\rfloor-1$ , or $k=2n$ , $\kappa(B_{n};C_{k})\leq 2$ for $k=n$ and $\kappa(B_{n};C_{k})\leq 2\lceil\frac{n-1}{k}\rceil-1$ for otherwise. Then there exists a $C_{k}$ -structure cut $\mathcal{F}$ . Let $C$ be a smallest component. We have $|V(\mathcal{F})|\geq\kappa(B_{n})=2n-2$ by Lemma 4.2. For $3\leq k\leq n-1$ and $1\leq r\leq\lfloor\frac{k}{2}\rfloor-1$ , or $k=2n$ , $|V(\mathcal{F})|\leq k(2\lceil\frac{n-1}{k}\rceil-2)\leq k(\frac{2(n+k-2)}{k}-2)=2n-4<2n-2$ , a contradiction. We have a Similar contradiction for $3\leq k\leq n-1$ and $r=0$ since $|V(\mathcal{F})|<2n-2$ .

For $n\leq k\leq 2n-1$ , it is easy to find $|V(\mathcal{F})|\leq 2n$ . For $3\leq k\leq n-1$ and $\lfloor\frac{k}{2}\rfloor\leq r\leq k-1$ , we have

\displaystyle|V(\mathcal{F})|\leq k\cdot(2\lceil\frac{n-1}{k}\rceil-1)\leq k\cdot(2\frac{n-1+\lceil\frac{k}{2}\rceil}{k}-1)\leq 2(n-1+\frac{k+1}{2})-k=2n-1.

Then $|V(\mathcal{F})|<\kappa_{1}(B_{n})$ for $n\geq 5$ by Lemma 4.3. Thus $|V(C)|=1$ , say $u=[v,w]$ , which implies that $N_{B_{n}}(u)\subseteq V(\mathcal{F})$ for $n\leq k\leq 2n-1$ , or $3\leq k\leq n-1$ and $\lfloor\frac{k}{2}\rfloor\leq r\leq k-1$ . We find $|\mathcal{F}|\leq 1$ for $n+1\leq k\leq 2n-1$ , that is, a $C_{k}$ covers all vertices of $N_{B_{n}}(u)$ , contradicting Lemma 4.4. We find $|\mathcal{F}|\leq 2$ for $n=k$ , which implies that we need two $C_{n}$ ’s to cover all vertices of $N_{B_{n}}(u)$ . We know the longest cycle is $C_{n-1}$ in $B_{n}[N_{B_{n}}(u)]$ by Lemma 4.4. Then a $C_{n}$ cover $n-1$ vertices of $N_{B_{n}}(u)$ and these $n-1$ vertices are in a $K_{n-1}$ . Let $x=[x_{1},x_{2}]$ be a vertex of $C_{n}$ but $x\notin N_{B_{n}}(u)$ . Then $x$ is adjacent to both $[v,v^{i}]$ and $[v,v^{j}]$ (or $[w,w^{i}]$ and $[w,w^{j}]$ ) for $i,j\in[n]$ . Thus $x_{1}=v^{i}$ (or $w^{i}$ ) and $x_{2}=v^{j}$ (or $w^{j}$ ). It is a contradiction since $[v^{i},v^{j}]\notin V(B_{n})$ and $[w^{i},w^{j}]\notin V(B_{n})$ . For $3\leq k\leq n-1$ and $\lfloor\frac{k}{2}\rfloor\leq r\leq k-1$ , we have $2r\geq k-1$ and these $2r$ vertices induce two disjoint $P_{r}$ ’s. Then we know a $C_{k}$ can not cover two disjoint $P_{r}$ ’s by Lemma 4.4, a contradiction. ∎

Lemma 4.14.

For $n\geq 5$ and $6\leq k\leq 2n$ , let $r$ be remainder of $n-1$ divided by $k$ . Then

\kappa(B_{n};C_{k})\leq\left\{\begin{aligned} &2\lceil\frac{n-1}{k}\rceil-1,~{}~{}if~{}6\leq k\leq n-1~{}and~{}1\leq r\leq\lfloor\frac{k}{2}\rfloor-1,~{}or~{}k=2n,\\ &3,~{}~{}if~{}k=n,\\ &2\lceil\frac{n-1}{k}\rceil,~{}~{}otherwise.\\ \end{aligned}\right.

Proof.

Let $u=[v,w]$ be a vertex of $B_{n}$ with $v=00\ldots 00$ and $w=10\ldots 00$ . Then $N_{B_{n}}(u)=\{[v,v^{i}],[w,w^{i}]|0\leq i\leq n-2\}$ . Let $P_{n-1}(v)=\langle[v,v^{1}],[v,v^{0}],[v,v^{2}],[v,v^{3}],\ldots,[v,v^{n-2}]\rangle$ be a path with $n-1$ vertices.

For $k=2n$ , we set

\displaystyle\mathcal{F}=C_{k}=\langle[v^{1},w^{1}],P_{n-1}(v),[v^{n-2},w^{n-2}],P^{-1}_{n-1}(w)\rangle.

For $n+1\leq k\leq 2n-1$ , we set $\mathcal{F}=\{C^{1}_{k},C^{2}_{k}\}$ and

\displaystyle C^{1}_{k}=\langle P_{n-1}(v),[v^{n-2},v^{n-2,1}],[v^{n-2,1},v^{1}],t_{1},\ldots,t_{k-n-1}\rangle

where $\{t_{i}|1\leq i\leq k-n-1\}\subseteq\{[v^{1},v^{1,i}]|i=0,n-1$ and $2\leq i\leq n-3\}$ ,

\displaystyle C^{2}_{k}=\langle P_{n-1}(w),[w^{n-2},w^{n-2,1}],[w^{n-2,1},w^{1}],t_{1},\ldots,t_{k-n-1}\rangle

where $\{t_{i}|1\leq i\leq k-n-1\}\subseteq\{[w^{1},w^{1,i}]|i=0,n-1$ and $2\leq i\leq n-3\}$ .

For $k=n$ , we set $\mathcal{F}=\{C^{1}_{k},C^{2}_{k},C^{3}_{k}\}$ and

\displaystyle C^{1}_{k}=\langle P_{n-1}(v)-[v,v^{n-2}],[v^{n-3},v^{n-3,1}],[v^{n-3,1},v^{1}]\rangle;

\displaystyle C^{2}_{k}=\langle P_{n-1}(w)-[w,w^{n-2}],[w^{n-3},w^{n-3,1}],[w^{n-3,1},w^{1}]\rangle;

\displaystyle C^{3}_{k}=\langle[v,v^{n-2}],[v^{n-2},w^{n-2}],[w,w^{n-2}],[w,w^{1}],[v^{1},w^{1}],[v,v^{1}],t_{1},\ldots,t_{k-6}\rangle,

where $\{t_{i}|1\leq i\leq k-6\}\subseteq\{[v,v^{i}]|2\leq i\leq n-2\}$ .

For $6\leq k\leq n-1$ and $1\leq i\leq\lfloor\frac{n-1}{k}\rfloor$ , we set

\displaystyle C^{i}_{k}=\langle[v,v^{(i-1)k}],[v,v^{(i-1)k+1}],\ldots,[v,v^{ik-1}]\rangle;

\displaystyle C^{\prime i}_{k}=\langle[w,w^{(i-1)k}],[w,w^{(i-1)k+1}],\ldots,[w,w^{ik-1}]\rangle.

If $r=0$ , then we set $\mathcal{F}=\{C^{i}_{k},C^{\prime i}_{k}|1\leq i\leq\lfloor\frac{n-1}{k}\rfloor\}$ . If $1\leq r\leq\lfloor\frac{k}{2}\rfloor-1$ , then we set $\mathcal{F}=\{C^{\lceil\frac{n-1}{k}\rceil}_{k}\}\cup\{C^{i}_{k},C^{\prime i}_{k}|1\leq i\leq\lfloor\frac{n-1}{k}\rfloor\}$ and

	$\displaystyle C^{\lceil\frac{n-1}{k}\rceil}_{k}$	$\displaystyle=\langle[w,w^{1}],[v^{1},w^{1}],[v,v^{1}],t_{1},\ldots,t_{k-2r-4},[v,v^{n-r-1}],[v,v^{n-r}],\ldots,[v,v^{n-2}],$
		$\displaystyle[v^{n-2},w^{n-2}],[w,w^{n-2}],[w,w^{n-3}],\ldots,[w,w^{n-r-1}],[w,w^{1}]\rangle;$

where $\{t_{i}|1\leq i\leq k-2r-4\}\subseteq\{[v,v^{i}]|2\leq i\leq n-r-2\}$ . If $\lfloor\frac{k}{2}\rfloor\leq r<k-1$ , then we set $\mathcal{F}=\{C^{i}_{k},C^{\prime i}_{k}|1\leq i\leq\lceil\frac{n-1}{k}\rceil\}$ and

\displaystyle C^{\lceil\frac{n-1}{k}\rceil}_{k}=\langle[v,v^{1}],[v,v^{n-r-1}],[v,v^{n-r}],\ldots,[v,v^{n-2}],t_{1},\ldots,t_{k-r-3},[v^{n-2},v^{n-2,1}],[v^{n-2,1},v^{1}]\rangle,

where $\{t_{i}|1\leq i\leq k-r-3\}\subseteq\{[v^{n-2},v^{n-2,i}]|0\leq i\leq n-3\}$ ,

	$\displaystyle C^{\prime\lceil\frac{n-1}{k}\rceil}_{k}$	$\displaystyle=\langle[w,w^{1}],[w,w^{n-r-1}],[w,w^{n-r}],\ldots,[w,w^{n-2}],t_{1},\ldots,t_{k-r-3},[w^{n-2},w^{n-2,1}],$
		$\displaystyle[w^{n-2,1},w^{1}]\rangle,$

where $\{t_{i}|1\leq i\leq k-r-3\}\subseteq\{[w^{n-2},w^{n-2,i}]|1\leq i\leq n-3\}$ . If $r=k-1$ , then we set $\mathcal{F}=\{C^{i}_{k},C^{\prime i}_{k}|1\leq i\leq\lceil\frac{n-1}{k}\rceil\}$ and

\displaystyle C^{\lceil\frac{n-1}{k}\rceil}_{k}=\langle[v,v^{0}],[v,v^{n-k}],[v,v^{n-k+1}],\ldots,[v,v^{n-2}]\rangle,

\displaystyle C^{\prime\lceil\frac{n-1}{k}\rceil}_{k}=\langle[w,w^{0}],[w,w^{n-k}],[w,w^{n-k+1}],\ldots,[w,w^{n-2}]\rangle.

By Lemmas 4.13 and 4.14, we have the following theorem.

Theorem 4.15.

For $n\geq 5$ and $6\leq k\leq 2n$ , let $r$ be remainder of $n-1$ divided by $k$ . Then

\kappa(B_{n};C_{k})=\left\{\begin{aligned} &2\lceil\frac{n-1}{k}\rceil-1,~{}~{}if~{}6\leq k\leq n-1~{}and~{}1\leq r\leq\lfloor\frac{k}{2}\rfloor-1,~{}or~{}k=2n,\\ &3,~{}~{}if~{}k=n,\\ &2\lceil\frac{n-1}{k}\rceil,~{}~{}otherwise.\\ \end{aligned}\right.

5 Conclusion

In this paper, we obtained structure connectivities of two famous data center networks DCell and BCDC on some common structures. For DCell network $D_{m,n}$ , $m\geq 0$ and $n\geq 2$ , we got that $\kappa(D_{m,n};K_{1,t})=\kappa^{s}(D_{m,n};K_{1,t})=\lceil\frac{n-1}{1+t}\rceil+m$ for $1\leq t\leq m+n-2$ and $\kappa(D_{m,n};K_{s})=\lceil\frac{n-1}{s}\rceil+m$ for $3\leq s\leq n-1$ by analyzing the structural properties of $D_{m,n}$ . For BCDC network $B_{n}$ , $n\geq 5$ , we used the existing results of $g$ -extra connectivity to obtain $\kappa(B_{n};K_{1,t})$ and $\kappa^{s}(B_{n};K_{1,t})$ for $1\leq t\leq 2n-3$ (see Theorems 4.5 and 4.8) and $\kappa(B_{n};P_{k})$ and $\kappa^{s}(B_{n};P_{k})$ for $4\leq k\leq 2n-1$ (see Theorem 4.11); $\kappa(B_{n};C_{k})$ and $\kappa^{s}(B_{n};C_{k})$ for $6\leq k\leq 2n$ (see Theorem 4.15). It is easy to find that $\kappa(D_{m,n};P_{k})$ ( $\kappa^{s}(D_{m,n};P_{k})$ ) and $\kappa(D_{m,n};C_{k})$ ( $\kappa^{s}(D_{m,n};C_{k})$ ) have no relevant conclusions. Except for $K_{1,t}$ , $P_{k}$ , $C_{k}$ and $K_{s}$ , there are still other structure connectivities of $D_{m,n}$ and $B_{n}$ have not been studied. So we will keep working on structure connectivity of data center networks.

Acknowledgement

The work is supported by NSFC (Grant No. 11871256).

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