Super-Connectivity of the Folded Locally Twisted Cube

Abstract

The hypercube $Q_n$ is one of the most popular interconnection networks with high symmetry. To reduce the diameter of $Q_n$, many variants of $Q_n$ have been proposed, such as the n-dimensional locally twisted cube $LT{Q}_n$. To further optimize the diameter of $LT{Q}_n$, the n-dimensional folded locally twisted cube $FLT{Q}_n$ is proposed, which is built based on $LT{Q}_n$ by adding $2^{n-1}$ complementary edges. Connectivity is an important indicator to measure the fault tolerance and reliability of a network. However, the connectivity has an obvious shortcoming, in that it assumes all the adjacent vertices of a vertex will fail at the same time. Super-connectivity is a more refined index to judge the fault tolerance of a network, which ensures that each vertex has at least one neighbor. In this paper, we show that the super-connectivity ${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)=2n$ for any integer $n\ge 6$, which is about twice $\kappa \left(FLT{Q}_n\right)$.

1. Introduction

High-performance computers can be widely used in many fields thanks to the development of high performance computing technology. The topological properties of interconnection networks are very important for high-performance computers. One typically uses an undirected graph $G=\left(V\right(G),E(G\left)\right)$ to model the topology of a multiprocessor system H, where the processor set of H is represented by $V\left(G\right)$ and the link set of H is represented by $E\left(G\right)$.

Interconnection networks have many important properties, one of which is the connectivity denoted by $\kappa \left(G\right)$. A graph’s connectivity is the minimum number of vertices whose removal makes the graph disconnected or trivial [1]. Connectivity is an important indicator to measure the fault tolerance and reliability of a network. In a large interconnection network, each vertex has a large number of neighbors. This property has an obvious deficiency, in that it assumes that all the adjacent vertices of a vertex will fail at the same time. However, this situation does not happen frequently in real networks. To address this deficiency, Esfahanian et al. [2] introduced the concept of restricted connectivity by imposing additionally restricted conditions on a network. Super-connectivity is a special case of restricted connectivity. When determing the super-connectivity of a network, one needs to ensure that each vertex has at least one neighbor. Hence, super-connectivity is a more refined index to judge the fault tolerance of a network.

Let K be a subset of $V\left(G\right)$. $G\setminus K$ (or $G-K$) denotes a graph obtained by removing all the vertices in K and edges incident to at least one vertex in K from G. If $G\setminus K$ is disconnected and each component of $G\setminus K$ has at least two vertices, then K is called a super vertex cut. Let S be a subset of $E\left(G\right)$. If $G\setminus S$ is disconnected and each component of $G\setminus S$ has at least two vertices, then S is called a super edge cut. The super-connectivity of G (or, respectively, the super edge connectivity), denoted by ${\kappa }^{\left(1\right)}\left(G\right)$ (or ${\lambda }^{\left(1\right)}\left(G\right)$), is the minimum cardinality of all super vertex cuts (or super edge cuts) in G, if any exist. Many relevant results have been obtained regarding super-connectivity and super edge connectivity [3,4,5,6,7,8,9,10,11,12,13,14,15,16].

The hypercube $Q_n$ has become one of the most popular interconnection networks, because of its many attractive properties, such as its regularity and symmetry. $Q_n$ is a Cayley graph and hence vertex-transitive and edge-transitive. However, the diameter of $Q_n$ is not optimal. In order to enhance the hypercube, researchers have proposed many variants, such as crossed cubes [17], locally twisted cubes [18], and spined cubes [19]. The n-dimensional locally twisted cube $LT{Q}_n$ was proposed by Yang et al. [18], whose diameter was only about half that of $Q_n$. Many research results have been published on the properties of $LT{Q}_n$ [20,21,22,23,24,25]. $LT{Q}_n$ is vertex-transitive if and only if $n\le 3$, and it is edge-transitive if and only if $n=2$ [25]. To further enhance the hypercube, inspired by the folded cube [26], Peng et al. [27] proposed a new network topology called the folded locally twisted cube $FLT{Q}_n$. So far there, no work has been reported on the super-connectivity of $FLT{Q}_n$. In this work, we studied the super-connectivity of $FLT{Q}_n$ and obtained the result that the super-connectivity ${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)$ is $2n$ for $n\ge 6$, which is about twice $\kappa \left(FLT{Q}_n\right)$.

2. Preliminaries

In this paper, we use the terms vertex and node interchangeably. We also use $(x,y)$ to denote an edge between vertices x and y. For any vertex $x\in V\left(G\right)$, the neighboring set of x is denoted by $N_G\left(x\right)=\left\{y\right|(x,y)\in E\left(G\right)\}$ (or $N\left(x\right)$ for short). Let $S\subset V\left(G\right)$. The neighboring set of S is defined as $N_G\left(S\right)=\left({\bigcup }_{x\in S}N\left(x\right)\right)\setminus S$ (or $N\left(S\right)$ for short). We define $N_G\left[S\right]={\bigcup }_{x\in S}N\left(x\right)$ and $N_G\left[x\right]=N_G\left(x\right)\cup \left\{x\right\}$. We use $x_n{x}_{n-1}\cdots x_2{x}_1$ to represent a binary string $\mu$ of length n, where $x_i\in \{0,1\}$ for $1\le i\le n$ is a part of $\mu$. $x_1$ is the first part of $\mu$, and $x_n$ is the nth part of $\mu$. The symbol ${\overline{x}}_i$ is used to represent the complement of $x_i$. As a variant of $Q_n$, $LT{Q}_n$ has the same number of vertices as $Q_n$. Each vertex of $LT{Q}_n$ is denoted by a unique binary string of length n. The definition of $LT{Q}_n$ is given below.

Definition 1

([18]). For $n\ge 2$, an n-dimensional locally twisted cube, $LT{Q}_n$, is defined recursively as follows:

(1) $LT{Q}_2$ is a graph consisting of four nodes labeled with 00, 01, 10, and 11, which are connected by four edges, (00, 01), (00, 10), (01, 11), and (10, 11).

(2) For $n\ge 3$, $LT{Q}_n$ is built from two disjointed copies of $LT{Q}_{n-1}$ named $LT{Q}_{n-1}^0$ and $LT{Q}_{n-1}^1$. Let $LT{Q}_{n-1}^0$ (or, respectively, $LT{Q}_{n-1}^1$) be the graph obtained by prefixing the label of each node of one copy of $LT{Q}_{n-1}$ with 0 (or with 1); each node $x=0x_{n-1}{x}_{n-2}\cdots x_2{x}_1$ of $LT{Q}_{n-1}^0$ is connected to the node $1(x_{n-1}+x_1)x_{n-2}\cdots x_2{x}_1$ of $LT{Q}_{n-1}^1$ with an edge, where $‘+’$ represents modulo 2 addition.

$LT{Q}_3$ and $LT{Q}_4$ are demonstrated in Figure 1. Each node in $LT{Q}_{n-1}^0$ has only one adjacent node in $LT{Q}_{n-1}^1$. The set of edges between $LT{Q}_{n-1}^0$ and $LT{Q}_{n-1}^1$ is called a perfect matching M of $LT{Q}_n$. Hence, we can write $LT{Q}_n=G(LT{Q}_{n-1}^0,LT{Q}_{n-1}^1,M)$. In [18], Yang et al. also provided a non-recursive definition of $LT{Q}_n$.

Definition 2

([18]). Let $\mu =x_n{x}_{n-1}\cdots x_1$ and $\nu =y_n{y}_{n-1}\cdots y_1$ be any two distinct vertices of $LT{Q}_n$ for $n\ge 2$. μ and ν are connected if and only if one of the following conditions is satisfied:

1. There is an integer $3\le k\le n$ such that

(a) $x_k={\overline{y}}_k$;

(b) $x_{k-1}=y_{k-1}+x_1$ ($‘+’$ represents modulo 2 addition);

(c) all the remaining bits of μ and ν are the same.

2. There is an integer $1\le k\le 2$ such that μ and ν only differ in the kth bit.

Let $\mu =x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_2{x}_1$ be any vertex of $LT{Q}_n$. By Definition 2, all the n neighbors of $\mu$ are listed as follows:

${\mu }_1=x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_2{\overline{x}}_1$;

${\mu }_2=x_n{x}_{n-1}{x}_{n-2}\dots x_3{\overline{x}}_2{x}_1$;

${\mu }_3=x_n{x}_{n-1}{x}_{n-2}\dots {\overline{x}}_3(x_2+x_1)x_1$;

…

${\mu }_{n-1}=x_n{\overline{x}}_{n-1}(x_{n-2}+x_1)x_{n-3}\dots x_2{x}_1$;

${\mu }_n={\overline{x}}_n(x_{n-1}+x_1)x_{n-2}\dots x_3{x}_2{x}_1$.

We call ${\mu }_i$ the ith dimensional neighbor of $\mu$ for $1\le i\le n$.

Definition 3

([27]). For any integer $n\ge 2$, an n-dimensional folded locally twisted cube, denoted by $FLT{Q}_n$, is a graph constructed based on $LT{Q}_n$ by adding all complementary edges. Each vertex $x=x_n{x}_{n-1}\dots x_1$ in $LT{Q}_n$ is incident to another vertex $\overline{x}={\overline{x}}_n{\overline{x}}_{n-1}\dots {\overline{x}}_1$ through a complementary edge, where ${\overline{x}}_i=1-x_i$.

We call the added complementary edges c-links. $FLT{Q}_n$ has $2^{n-1}$c-links, and each vertex $\mu =x_n{x}_{n-1}\dots x_1$ is connected to a complementary vertex ${\mu }_c={\overline{x}}_n{\overline{x}}_{n-1}\dots {\overline{x}}_1$ by a c-link. The set of complementary edges between $LT{Q}_{n-1}^0$ and $LT{Q}_{n-1}^1$ is a perfect matching C of $FLT{Q}_n$. Hence, we can write $FLT{Q}_n=G(LT{Q}_{n-1}^0,LT{Q}_{n-1}^1,M,C)$ or $G(LT{Q}_n,C)$. Each node $\mu \in V\left(FLT{Q}_n\right)$ in $LT{Q}_{n-1}^0$ (or, respectively, $LT{Q}_{n-1}^1$) has two neighbors, ${\mu }_n$ and ${\mu }_c$, in $LT{Q}_{n-1}^1$ (or $LT{Q}_{n-1}^0$) for $n\ge 3$. Compared with $LT{Q}_n$, each vertex in $FLT{Q}_n$ has one more neighbor. Then, the node degree of $FLT{Q}_n$ is $n+1$ and $\kappa \left(FLT{Q}_n\right)=n+1$ [27]. Figure 2 demonstrates $FLT{Q}_3$ and $FLT{Q}_4$, respectively, and Figure 3 demonstrates $FLT{Q}_5$.

3. Super Connectivity of ${\mathbf{FLTQ}}_{\mathbf{n}}$

In this section, we study the super connectivity of $FLT{Q}_n$ for any integer $n\ge 6$. Since $FLT{Q}_n$ is composed of $LT{Q}_n$ and the complementary edge set C, we can use some properties of $LT{Q}_n$ to prove the super-connectivity property of $FLT{Q}_n$.

Lemma 1

([18]). For $n\ge 2$, $\kappa \left(LT{Q}_n\right)=\lambda \left(LT{Q}_n\right)=n$.

Lemma 2

([28]). For any two vertices μ, $\nu \in V\left(LT{Q}_n\right)$($n\ge 2$), we have $|N_{LT{Q}_n}\left(\mu \right)\cap N_{LT{Q}_n}\left(\nu \right)|\le 2$.

Lemma 3

([28]). Let μ and ν be any two distinct vertices in $LT{Q}_n(n\ge 4)$ such that $|N_{LT{Q}_n}\left(\mu \right)\cap N_{LT{Q}_n}\left(\nu \right)|=2$.

(1) If $\mu \in V\left(LT{Q}_{n-1}^0\right)$ and $\nu \in V\left(LT{Q}_{n-1}^1\right)$, then the one common neighbor is in $LT{Q}_{n-1}^0$, and the other one is in $LT{Q}_{n-1}^1$.

(2) If μ, $\nu \in V\left(LT{Q}_{n-1}^0\right)$ or $V\left(LT{Q}_{n-1}^1\right)$, then the two common neighbors are in $LT{Q}_{n-1}^0$ or $LT{Q}_{n-1}^1$.

Lemma 4.

Let μ and ν be any two distinct vertices in the same $LT{Q}_{n-1}^i$ for $0\le i\le 1$ and $n\ge 6$. If ${\mu }_n={\nu }_c$ or ${\mu }_c={\nu }_n$, then $|N_{FLT{Q}_n}\left(\mu \right)\cap N_{FLT{Q}_n}\left(\nu \right)|=1$.

Proof. 

Without loss of generality, we suppose that $\mu$, $\nu \in V\left(LT{Q}_{n-1}^0\right)$, and ${\mu }_n={\nu }_c$. Then, ${\mu }_n$ is the common neighbor for $\mu$ and $\nu$. Let $\mu =x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_2{x}_1$ and $X=FLT{Q}_n\setminus \left\{{\mu }_n\right\}$. Next, we consider the neighbors of $\mu$ and $\nu$ in X according to different values of the first part $x_1$ of $\mu$.

Case 1. $x_1=0$.

${\mu }_n={\overline{x}}_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}0={\nu }_c$ and $\nu =x_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$. We list $N_X\left(\mu \right)$ and $N_X\left(\nu \right)$ separately in

Table 1. The neighbors of $\mu$ and $\nu$ in X, where $x_1=0$.

${\mathit{N}}_{\mathit{X}}\mathbf{\left(}\mathit{\mu }\mathbf{\right)}$	${\mathit{N}}_{\mathit{X}}\mathbf{\left(}\mathit{\nu }\mathbf{\right)}$
${\mu }_1=x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}1$	${\nu }_1=x_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$
${\mu }_2=x_n{x}_{n-1}{x}_{n-2}\dots x_3{\overline{x}}_{2}0$	${\nu }_2=x_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{x}_{2}1$
${\mu }_3=x_n{x}_{n-1}{x}_{n-2}\dots {\overline{x}}_3{x}_{2}0$	${\nu }_3=x_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots x_3{x}_{2}1$
…	…
${\mu }_{n-1}=x_n{\overline{x}}_{n-1}{x}_{n-2}\dots x_3{x}_{2}0$	${\nu }_{n-1}=x_n{x}_{n-1}{x}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$
${\mu }_c={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$	${\nu }_n={\overline{x}}_n{x}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$

.

It is obvious that $|N_X\left(\mu \right)\cap N_X\left(\nu \right)|=0$.

Case 2. $x_1=1$.

${\mu }_n={\overline{x}}_n{\overline{x}}_{n-1}{x}_{n-2}\dots x_3{x}_{2}1={\nu }_c$ and $\nu =x_n{x}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$. We list $N_X\left(\mu \right)$ and $N_X\left(\nu \right)$ separately in

Table 2. The neighbors of $\mu$ and $\nu$ in X, where $x_1=1$.

${\mathit{N}}_{\mathit{X}}\mathbf{\left(}\mathit{\mu }\mathbf{\right)}$	${\mathit{N}}_{\mathit{X}}\mathbf{\left(}\mathit{\nu }\mathbf{\right)}$
${\mu }_1=x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}0$	${\nu }_1=x_n{x}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$
${\mu }_2=x_n{x}_{n-1}{x}_{n-2}\dots x_3{\overline{x}}_{2}1$	${\nu }_2=x_n{x}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{x}_{2}0$
${\mu }_3=x_n{x}_{n-1}{x}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$	${\nu }_3=x_n{x}_{n-1}{\overline{x}}_{n-2}\dots x_3{\overline{x}}_{2}0$
…	…
${\mu }_{n-1}=x_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots x_3{x}_{2}1$	${\nu }_{n-1}=x_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$
${\mu }_c={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$	${\nu }_n={\overline{x}}_n{x}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$

.

It is obvious that $|N_X\left(\mu \right)\cap N_X\left(\nu \right)|=0$.

Hence, $\mu$ and $\nu$ have only one common neighbor in $FLT{Q}_n$ and $|N_{FLT{Q}_n}\left(\mu \right)\cap N_{FLT{Q}_n}\left(\nu \right)|=1$. □

Lemma 5.

Let μ be any node in $FLT{Q}_n$, where $n\ge 6$ and $X=FLT{Q}_n\setminus \left\{\mu \right\}$. Then, $|N_X\left({\mu }_n\right)\cap N_X\left({\mu }_c\right)|=0$.

Proof. 

Let $\mu =x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_2{x}_1$. We consider the different values of the first part $x_1$ of $\mu$.

Case 1. $x_1=0$.

Let $\alpha ={\mu }_n={\overline{x}}_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}0$ and $\beta ={\mu }_c={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$. All the neighbors of $\alpha$ and $\beta$ in X are listed separately in

Table 3. The neighbors of $\alpha$ and $\beta$ in X, where $x_1=0$.

${\mathit{N}}_{\mathit{X}}\mathbf{\left(}\mathit{\alpha }\mathbf{\right)}$	${\mathit{N}}_{\mathit{X}}\mathbf{\left(}\mathit{\beta }\mathbf{\right)}$
${\alpha }_1={\overline{x}}_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}1$	${\beta }_1={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$
${\alpha }_2={\overline{x}}_n{x}_{n-1}{x}_{n-2}\dots x_3{\overline{x}}_{2}0$	${\beta }_2={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{x}_{2}1$
${\alpha }_3={\overline{x}}_n{x}_{n-1}{x}_{n-2}\dots {\overline{x}}_3{x}_{2}0$	${\beta }_3={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots x_3{x}_{2}1$
…	…
${\alpha }_{n-1}={\overline{x}}_n{\overline{x}}_{n-1}{x}_{n-2}\dots x_3{x}_{2}0$	${\beta }_{n-1}={\overline{x}}_n{x}_{n-1}{x}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$
${\alpha }_c=x_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$	${\beta }_n=x_n{x}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$

.

It is obvious that $N_X\left(\alpha \right)\cap N_X\left(\beta \right)=\varnothing$.

Case 2. $x_1=1$.

Let $\alpha ={\mu }_n={\overline{x}}_n{\overline{x}}_{n-1}{x}_{n-2}\dots x_3{x}_{2}1$ and $\beta ={\mu }_c={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$. All the neighbors of $\alpha$ and $\beta$ in X are listed separately in

Table 4. The neighbors of $\alpha$ and $\beta$ in X, where $x_1=1$.

${\mathit{N}}_{\mathit{X}}\left(\mathit{\alpha }\right)$	${\mathit{N}}_{\mathit{X}}\left(\mathit{\beta }\right)$
${\alpha }_1={\overline{x}}_n{\overline{x}}_{n-1}{x}_{n-2}\dots x_3{x}_{2}0$	${\beta }_1={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$
${\alpha }_2={\overline{x}}_n{\overline{x}}_{n-1}{x}_{n-2}\dots x_3{\overline{x}}_{2}1$	${\beta }_2={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{x}_{2}0$
${\alpha }_3={\overline{x}}_n{\overline{x}}_{n-1}{x}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$	${\beta }_3={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots x_3{\overline{x}}_{2}0$
…	…
${\alpha }_{n-1}={\overline{x}}_n{x}_{n-1}{\overline{x}}_{n-2}\dots x_3{x}_{2}1$	${\beta }_{n-1}={\overline{x}}_n{x}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$
${\alpha }_c=x_n{x}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$	${\beta }_n=x_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$

.

It is obvious that $N_X\left(\alpha \right)\cap N_X\left(\beta \right)=\varnothing$.

Hence, $|N_X\left({\mu }_n\right)\cap N_X\left({\mu }_c\right)|=0$. □

Lemma 6.

Let μ, $\nu \in V\left(FLT{Q}_n\right)$ where $n\ge 6$. Then $|N_{FLT{Q}_n}\left(\mu \right)\cap N_{FLT{Q}_n}\left(\nu \right)|\le 2$.

Proof. 

Since $FLT{Q}_n$ is constructed from $LT{Q}_n$ by adding the complementary edge set C, we can study this lemma based on $LT{Q}_n$.

Case 1. $\mu$, $\nu$ are in the same $LT{Q}_{n-1}^i$ for $0\le i\le 1$.

Without loss of generality, we suppose that $\mu ,\nu \in V\left(LT{Q}_{n-1}^0\right)$. According to Lemmas 2 and 3, $|N_{LT{Q}_n}\left(\mu \right)\cap N_{LT{Q}_n}\left(\nu \right)|\le 2$ for $n\ge 6$, and the two common neighbors are in $LT{Q}_{n-1}^0$. According to the definition of $FLT{Q}_n$, we have $N_{LT{Q}_{n-1}^1}\left(\mu \right)=\{{\mu }_n,{\mu }_c\}$, $N_{LT{Q}_{n-1}^1}\left(\nu \right)=\{{\nu }_n,{\nu }_c\}$, ${\mu }_n\ne {\nu }_n$, and ${\mu }_c\ne {\nu }_c$. If ${\mu }_c\ne {\nu }_n$ and ${\mu }_n\ne {\nu }_c$, then $\mu$ and $\nu$ do not have the same neighbors in $LT{Q}_{n-1}^1$. Hence, $|N_{FLT{Q}_n}\left(\mu \right)\cap N_{FLT{Q}_n}\left(\nu \right)|\le 2$. According to Lemma 4, if ${\mu }_c={\nu }_n$ or ${\mu }_n={\nu }_c$, then $\mu$ and $\nu$ have only one common neighbor in $FLT{Q}_n$ and $|N_{FLT{Q}_n}\left(\mu \right)\cap N_{FLT{Q}_n}\left(\nu \right)|=1\le 2$.

Case 2. $\mu$ and $\nu$ are in a different $LT{Q}_{n-1}^i$ for $0\le i\le 1$.

Without loss of generality, we suppose that $\mu \in V\left(LT{Q}_{n-1}^0\right)$ and $\nu \in V\left(LT{Q}_{n-1}^1\right)$. According to Lemma 2, $|N_{LT{Q}_n}\left(\mu \right)\cap N_{LT{Q}_n}\left(\nu \right)|\le 2$. Based on the definition of $FLT{Q}_n$, we have $N_{LT{Q}_{n-1}^1}\left(u\right)=\{u_n,u_c\}$ and $N_{LT{Q}_{n-1}^0}\left(v\right)=\{v_n,v_c\}$. According to Lemma 5, $|N_{FLT{Q}_n\setminus \left\{\mu \right\}}\left({\mu }_n\right)\cap N_{FLT{Q}_n\setminus \left\{\mu \right\}}\left({\mu }_c\right)|=0$ and $|N_{FLT{Q}_n\setminus \left\{\nu \right\}}\left({\nu }_n\right)\cap N_{FLT{Q}_n\setminus \left\{\nu \right\}}\left({\nu }_c\right)|=0$. Hence, we cannot find a vertex ${\mu }^{\prime }\in V\left(LT{Q}_{n-1}^1\right)$, where ${\mu }^{\prime }$ and $\mu \in V\left(LT{Q}_{n-1}^0\right)$ have two common neighbors, nor can we find a vertex ${\nu }^{\prime }\in V\left(LT{Q}_{n-1}^0\right)$, where ${\nu }^{\prime }$ and $\nu \in V\left(LT{Q}_{n-1}^1\right)$ have two common neighbors. Then, u and v cannot have three or four common neighbors in $FLT{Q}_n$. Hence, $|N_{FLT{Q}_n}\left(\mu \right)\cap N_{FLT{Q}_n}\left(\nu \right)|\le 2$. □

Lemma 7

([28]). If μ and ν are two vertices of $LT{Q}_n$ and $(\mu ,\nu )\in E\left(LT{Q}_n\right)$, where $n\ge 2$, then $|N_{LT{Q}_n}\left(\mu \right)\cap N_{LT{Q}_n}\left(\nu \right)|=0$.

Lemma 8.

If μ and ν are two vertices of $FLT{Q}_n$ and $(\mu ,\nu )\in E\left(FLT{Q}_n\right)$, where $n\ge 3$, then $|N_{FLT{Q}_n}\left(\mu \right)\cap N_{FLT{Q}_n}\left(\nu \right)|=0$.

Proof. 

According to the position of $\mu$ and $\nu$, we consider two cases.

Case 1. $\mu$ and $\nu$ are in the same $LT{Q}_{n-1}^i$ for $0\le i\le 1$.

Without loss of generality, we assume that $\mu ,\nu \in V\left(LT{Q}_{n-1}^0\right)$. According to Lemma 7, $|N_{LT{Q}_{n-1}^0}\left(\mu \right)\cap N_{LT{Q}_{n-1}^0}\left(\nu \right)|=0$. We have $N_{LT{Q}_{n-1}^1}\left(\mu \right)=\{{\mu }_n,{\mu }_c\}$ and $N_{LT{Q}_{n-1}^1}\left(\nu \right)=\{{\nu }_n,{\nu }_c\}$. If $N_{LT{Q}_{n-1}^1}\left(\mu \right)\cap N_{LT{Q}_{n-1}^1}\left(\nu \right)=\varnothing$, then $|N_{FLT{Q}_n}\left(\mu \right)\cap N_{FLT{Q}_n}\left(\nu \right)|=0$. Otherwise, if ${\mu }_n={\nu }_c$ or ${\mu }_c={\nu }_n$, then we let $\mu =x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_2{x}_1$. All the possible values of $\mu$ and $\nu$ are listed in

Table 5. The possible values of $\mu$ and $\nu$.

	$\mu =x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}0$
${\mu }_n={\nu }_c$	${\mu }_n={\overline{x}}_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}0={\nu }_c$	$x_1=0$
	$\nu =x_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$
	$\mu =x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}1$
${\mu }_n={\nu }_c$	${\mu }_n={\overline{x}}_n{\overline{x}}_{n-1}{x}_{n-2}\dots x_3{x}_{2}1={\nu }_c$	$x_1=1$
	$\nu =x_n{x}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$
	$\mu =x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}0$
${\mu }_c={\nu }_n$	${\mu }_c={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1={\nu }_n$	$x_1=0$
	$\nu =x_n{x}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$
	$\mu =x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}1$
${\mu }_c={\nu }_n$	${\mu }_c={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0={\nu }_n$	$x_1=1$
	$\nu =x_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$

.

It is obvious that $(\mu ,\nu )\notin E\left(FLT{Q}_n\right)$; then, we reach a contradiction, and all these values of $\mu$ and $\nu$ are impossible. Hence, $|N_{FLT{Q}_n}\left(\mu \right)\cap N_{FLT{Q}_n}\left(\nu \right)|=0$.

Case 2. $\mu$ and $\nu$ are in a different $LT{Q}_{n-1}^i$ for $0\le i\le 1$.

Without loss of generality, we assume that $\mu \in V\left(LT{Q}_{n-1}^0\right)$ and $\nu \in V\left(LT{Q}_{n-1}^1\right)$. Since $(\mu ,\nu )\in E\left(FLT{Q}_n\right)$, $\nu$ should be ${\mu }_n$ or ${\mu }_c$. If ${\mu }_n=\nu$, let $K=\{\mu ,\nu ,{\mu }_c,{\nu }_c\}$. Otherwise, If ${\mu }_c=\nu$, let $K=\{\mu ,\nu ,{\mu }_n,{\nu }_n\}$. Let $\mu =x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_2{x}_1$. All the possible values of K are listed in

Table 6. The possible values of K.

	$\mu =x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}0={\nu }_n$
${\mu }_n=\nu$	${\mu }_n={\overline{x}}_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}0=\nu$	$x_1=0$
	${\nu }_c=x_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$
	${\mu }_c={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$
	$\mu =x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}1={\nu }_n$
${\mu }_n=\nu$	${\mu }_n={\overline{x}}_n{\overline{x}}_{n-1}{x}_{n-2}\dots x_3{x}_{2}1=\nu$	$x_1=1$
	${\nu }_c=x_n{x}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$
	${\mu }_c={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$
	$\mu =x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}0={\nu }_c$
${\mu }_c=\nu$	${\mu }_c={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1=\nu$	$x_1=0$
	${\nu }_n=x_n{x}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}1$
	${\mu }_n={\overline{x}}_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}0$
	$\mu =x_n{x}_{n-1}{x}_{n-2}\dots x_3{x}_{2}1={\nu }_c$
${\mu }_c=\nu$	${\mu }_c={\overline{x}}_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0=\nu$	$x_1=1$
	${\nu }_n=x_n{\overline{x}}_{n-1}{\overline{x}}_{n-2}\dots {\overline{x}}_3{\overline{x}}_{2}0$
	${\mu }_n={\overline{x}}_n{\overline{x}}_{n-1}{x}_{n-2}\dots x_3{x}_{2}1$

.

Since $({\mu }_c,\nu ),(\mu ,{\nu }_c)\notin E\left(FLT{Q}_n\right)$ when ${\mu }_n=\nu$ and $({\mu }_n,\nu ),(\mu ,{\nu }_n)\notin E\left(FLT{Q}_n\right)$, when ${\mu }_c=\nu$, $\mu$ and $\nu$ do not have common neighbors. Hence, $|N_{FLT{Q}_n}\left(\mu \right)\cap N_{FLT{Q}_n}\left(\nu \right)|=0$. □

Lemma 9.

Let μ be any node in $LT{Q}_n$ for any integer $n\ge 3$. Then, $LT{Q}_n\setminus N_{LT{Q}_n}\left[\mu \right]$ is connected.

Proof. 

We use mathematical induction on n to prove this lemma. According to Lemma 1, we know that this lemma obviously holds when $n=3$. Suppose that this lemma holds for $n\le k(k\ge 3)$. Let $\mu$ be any node in $LT{Q}_{k+1}$. Without loss of generality, we suppose that $\mu \in V\left(LT{Q}_k^0\right)$. Then, by the induction hypothesis, $LT{Q}_k^0\setminus N_{LT{Q}_k^0}\left[\mu \right]$ is connected. Since $N_{LT{Q}_k^1}\left(\mu \right)=\left\{{\mu }_{k+1}\right\}$, according to Lemma 1, $LT{Q}_k^1\setminus \left\{{\mu }_{k+1}\right\}$ is connected. Since each node in $LT{Q}_k^0$ is connected to a node in $LT{Q}_k^1$, $LT{Q}_k^0\setminus N_{LT{Q}_k^0}\left[\mu \right]$ is connected to $LT{Q}_k^1\setminus \left\{{\mu }_{k+1}\right\}$. Then, $LT{Q}_{k+1}\setminus N_{LT{Q}_{k+1}}\left[\mu \right]$ is connected. Hence, this lemma holds. □

Since ${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)$ is the minimum cardinality of all super vertex cuts in $FLT{Q}_n$, to obtain the upper bound of ${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)$, we just need to find a super vertex cut F. Then, we have ${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)\le \left|F\right|$. If we can prove that $FLT{Q}_n$ is connected after removing $\left|F\right|-1$ vertices, then we have the lower bound ${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)\ge \left|F\right|$. With these two results, we can obtain ${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)=\left|F\right|$. In the following, we will present two important lemmas to prove the upper bound and lower bound of ${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)$.

Lemma 10.

${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)\le 2n$ for any integer $n\ge 6$.

Proof. 

Consider an edge $(x,y)\in E\left(FLT{Q}_n\right)$. Let $F=\{x,y\}$. Then, $FLT{Q}_n-N_{FLT{Q}_n}\left(F\right)$ is disconnected, and the edge $(x,y)$ is one component of $FLT{Q}_n-N_{FLT{Q}_n}\left(F\right)$. According to Lemma 8, $|N_{FLT{Q}_n}\left(F\right)|=(n+1)+(n+1)-2=2n$. Let $K=FLT{Q}_n-N_{FLT{Q}_n}\left[F\right]$. To prove that $N_{FLT{Q}_n}\left(F\right)$ is a super vertex cut, we need to show that each vertex $\alpha \in V\left(K\right)$ has at least one neighbor. According to Lemma 6, $|N_{FLT{Q}_n}\left(\alpha \right)\cap N_{FLT{Q}_n}\left(x\right)|\le 2$ and $|N_{FLT{Q}_n}\left(\alpha \right)\cap N_{FLT{Q}_n}\left(y\right)|\le 2$. Since $\kappa \left(FLT{Q}_n\right)=n+1$ and $n+1-2-2\ge 1$ for $n\ge 6$, $\alpha$ has at least one neighbor in K. Hence, $N_{FLT{Q}_n}\left(F\right)$ is a super vertex cut and ${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)\le 2n$ for $n\ge 6$. □

Lemma 11.

${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)\ge 2n$ for $n\ge 6$.

Proof. 

Suppose that F is a super vertex cut of $FLT{Q}_n$. Then, $FLT{Q}_n\setminus F$ is disconnected, and each vertex in $FLT{Q}_n\setminus F$ has at least one neighbor. To prove ${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)\ge 2n$, we will show that $FLT{Q}_n\setminus F$ is connected when $\left|F\right|\le 2n-1$. Let $F_i=F\cap LT{Q}_{n-1}^i$ for $0\le i\le 1$, $K_0=LT{Q}_{n-1}^0\setminus F_0$, and $K_1=LT{Q}_{n-1}^1\setminus F_1$. Without loss of generality, we suppose that $|F_0|\ge |F_1|$. Then, $|F_1|\le n-1$.

Case 1. $K_1$ is connected.

Let $\alpha$ be any node in $K_0$. We have $N_{LT{Q}_{n-1}^1}\left(\alpha \right)=\{{\alpha }_n,{\alpha }_c\}$. If $|N_{LT{Q}_{n-1}^1}\left(\alpha \right)\cap F_1|\le 1$, then $\alpha$ is connected to $K_1$. Since $K_1$ is connected, then $K_0\cup K_1$ is connected, which means that $FLT{Q}_n\setminus F$ is connected. Otherwise, since each vertex in $FLT{Q}_n\setminus F$ has at least one neighbor, there must be a vertex $\beta \in K_0$ such that $(\alpha ,\beta )\in E\left(K_0\right)$. We have $N_{LT{Q}_{n-1}^1}\left(\beta \right)=\{{\beta }_n,{\beta }_c\}$. If $|N_{LT{Q}_{n-1}^1}\left(\beta \right)\cap F_1|\le 1$, then $\alpha$ can be connected to $K_1$ through vertex $\beta$, and $FLT{Q}_n\setminus F$ is connected. Otherwise, we have $\{{\alpha }_n,{\alpha }_c,{\beta }_n,{\beta }_c\}\in F_1$, $|F_1|\ge 4$, and $|F_0|\le 2n-5$. Let $Y=N_{LT{Q}_{n-1}^0}\left(\alpha \right)\cup N_{LT{Q}_{n-1}^0}\left(\beta \right)\setminus \{\alpha ,\beta \}$. According to Lemma 8, $\left|Y\right|=(n-1)+(n-1)-2=2n-4$. Since $|F_0|\le 2n-5$, we can find at least one vertex $\gamma \in Y$ such that $\alpha$ and $\beta$ are connected to $K_1$ through $\gamma$. Hence, $FLT{Q}_n\setminus F$ is connected.

Case 2. $K_1$ is disconnected.

According to Lemma 1, we have $\kappa \left(LT{Q}_{n-1}\right)=n-1$. Since $K_1$ is disconnected, then $|F_1|=n-1$ and $|F_0|=n$. There should be an isolated vertex $\omega$ in $K_1$ and $F_1=N_{LT{Q}_{n-1}^1}\left(\omega \right)$. According to Lemma 9, $LT{Q}_{n-1}^1\setminus N_{LT{Q}_{n-1}^1}\left[\omega \right]$ is connected. For any vertex $\alpha$ in $K_0$ where $(\alpha ,\omega )\in E\left(FLT{Q}_n\right)$, based on Lemma 8, $\alpha$ and $\omega$ do not have common neighbors. Then, there exists a neighbor ${\alpha }^{\prime }$ of $\alpha$ in $LT{Q}_n^1$ such that ${\alpha }^{\prime }\notin N_{LT{Q}_{n-1}^1}\left[\omega \right]$. Hence, $\alpha$ is connected to $LT{Q}_{n-1}^1\setminus N_{LT{Q}_{n-1}^1}\left[\omega \right]$ through ${\alpha }^{\prime }$. For any vertex $\alpha$ in $K_0$ where $(\alpha ,\omega )\notin E\left(FLT{Q}_n\right)$, there must exist a neighbor $\beta$ in $K_0$. Let $Y=N_{LT{Q}_{n-1}^0}\left(\alpha \right)\cup N_{LT{Q}_{n-1}^0}\left(\beta \right)$. According to Lemma 8, $\left|Y\right|=(n-1)+(n-1)=2n-2$. Since $|F_0|=n$, we can find at least $n-2$ vertices in Y connected to $LT{Q}_{n-1}^1$. Since there exist two neighbors in $LT{Q}_{n-1}^1$ for each vertex in Y and $2n-4>n-1$ when $n\ge 6$, we can find a vertex $\gamma$ in Y such that $\alpha$ and $\beta$ are connected to $LT{Q}_{n-1}^1\setminus N_{LT{Q}_{n-1}^1}\left[\omega \right]$ through $\gamma$. Hence, $FLT{Q}_n-F$ is connected.

Thus, $FLT{Q}_n\setminus F$ is connected when $\left|F\right|\le 2n-1$ and ${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)\ge 2n$ for any integer $n\ge 6$. □

According to Lemmas 10 and 11, we obtain the following result:

Theorem 1.

${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)=2n$ for $n\ge 6$.

4. Conclusions

The folded locally twisted cube $FLT{Q}_n$ was introduced based on the locally twisted cube $LT{Q}_n$ and the folded hypercube $F{Q}_n$. In this paper, we studied the super-connectivity of folded locally twisted cubes, which is an important indicator to measure the fault tolerance and reliability of a network. The main contribution of this work was that we addressed the super-connectivity of $FLT{Q}_n$. We proved that ${\kappa }^{\left(1\right)}\left(FLT{Q}_n\right)=2n$ for any integer $n\ge 6$. Independent spanning trees and mesh embedding could be considered as future research directions. Independent spanning trees could be applied to reliable communication protocols, reliable broadcasting, and so on [29]. Meshes are fundamental guest graphs on which many algorithms, such as linear algebra algorithms and combinatorial algorithms, can be efficiently performed [30]. The results of independent spanning trees and mesh embedding for $FLT{Q}_n$ could be compared with the results of $LT{Q}_n$ [31,32].