## 4 Restricted Connectivity of BCCC  
In this section, we prove the restricted connectivity of BCCC. We refer to the entire graph of \(BCCC\) as \(G\). Theorem 4.1: _The restricted connectivity of \(BCCC\) is_  
\(\kappa^{1}(G)\leq 2n+k-3=2\kappa(G)-k-1\)_, if \(n<k+1\),_  
\(\kappa^{1}(G)\leq n+2k-2=2\kappa(G)-n\)_, if \(n\geq k+1\), when \(n\geq 3\) and \(k\geq 2\)._  
Proof: Let \(u\) and \(v\) be two vertices of the same type \(A\) switch (\(v\in N_{G}(u)\)). Therefore, \(|N_{G}(u)\cap N_{G}(v)|=n-2\). If \(|N_{G}(u)|=|N_{G}(v)|=\kappa(G)\), according to the definition,  
\[ \begin{split}|N_{G}(\{u,v\})|&=|N_{G}(u)\cup N_{G} (v)\backslash(N_{G}(u)\cap N_{G}(v))|\\ &=|N_{G}(u)-1|+|N_{G}(v)-1|-(n-2)\\ &=2(\kappa(G)-1)-(n-2)\\ &=2\kappa(G)-n\end{split}  \tag{1}\]  
Let \(F=N_{G}(\{u,v\})\), so \(G-F\) is divided into two disconnected subgraphs: one is \(G(u,v)\) and the other is \(G-(N_{G}(\{u,v\})\cap\{u,v\})\). Furthermore, \(u\) has a fault-free neighbor \(v\). Therefore, each vertex of subgraph \(G(u,v)\) has at least one fault-free neighbor. Let \(H=N_{G}(\{u,v\})\backslash\{u,v\}\), for \(BCCC(n,k)\), when \(u\) and \(v\) are located in the same type \(A\) switch, each node in \(N_{G}(H)\) has only one neighbor that is a faulty node. According to Lemma 1 (2), \(\kappa(G)=n+k-1\). Therefore, \(\kappa(G)\) is constantly greater than \(2\) when \(n\geq 3\) and \(k\geq 2\). It can be ensured that each node in \(G-(N_{G}(\{u,v\})\cap\{u,v\})\) has at least one fault-free neighbor. In summary, \(G-F\) is disconnected and each vertex in both subgraphs contains at least one fault-free neighbor. Hence, \(\kappa^{1}(G)\leq 2n+k-2=2\kappa(G)-n\). Therefore, the theorem holds. Let \(x\) and \(y\) be two vertices of the same type \(B\) switch (\(v\in N_{G}(u)\)) (\(y\in N_{G}(x)\)). Therefore, \(|N_{G}(x)\cap N_{G}(y)|=k-1\). If \(|N_{G}(x)|=|N_{G}(y)|=\kappa(G)\), according to the definition,  
\[ \begin{split}|N_{G}(\{x,y\})|&=|N_{G}(x)\cup N_{G} (y)\backslash(N_{G}(x)\cap N_{G}(y))|\\ &=|N_{G}(x)-1|+|N_{G}(y)-1|-(k-1)\\ &=2(\kappa(G)-1)-(k-1)\\ &=2\kappa(G)-k-1\end{split}  \tag{2}\]  
Proving that the two vertices are in the same type \(B\) switch is the same as proving the two vertices are in the same type \(A\) switch. Let \(F=N_{G}(\{x,y\})\), so \(G-F\) is disconnected and each vertex in both subgraphs contains at least one fault-free neighbor. Above all, \(\kappa^{1}(G)\leq 2k+n-3=2\kappa(G)-k-1\). Therefore, the theorem holds. Theorem 4.2: _The restricted connectivity of \(BCCC\) is_\(\kappa^{1}(G)\geq 2n+k-3=2\kappa(G)-k-1\), if \(n<k+1\), \(\kappa^{1}(G)\geq n+2k-2=2\kappa(G)-n\), if \(n\geq k+1\), when \(n\geq 3\) and \(k\geq 2\)._  
Proof. : Let \(F\) be the set of faulty vertices in \(G\). Proving this theorem is equivalent to proving, for any two vertices \(u\) and \(v\) in \(G-F\), there exists a fault-free path from \(u\) to \(v\). Furthermore, any vertex in \(G-F\) has at least one fault-free neighbor. If \(u\) and \(v\) are adjacent, the theorem holds. Hence, we discuss the case that \(u\) and \(v\) are not adjacent. According to the structure of BCCC, BCCC can be divided into \(n\) subgraphs. Let \(BCC_{n,k}^{\delta}\) for \(\kappa\in\langle n\rangle\backslash\{0\}\) be the subgraph of \(G\). Furthermore, \(u\in BCC_{n,k}^{a}\), \(v\in BCC_{n,k}^{\beta}\) for \(\alpha,\beta\in\langle n\rangle\backslash\{0\}\). Let \(F_{\alpha}=V(BCC_{n,k}^{a})\cap F\) and \(F_{\beta}=V(BCC_{n,k}^{\beta})\cap F\). Let \(R=\langle n\rangle\backslash\{0,\alpha\}\), \(BCC_{n,k}^{R}\) be a subgraph other than \(BCC_{n,k}^{a}\) in \(G\) and \(F_{R}=V(BCC_{n,k}^{R})\cap F\). If \(N_{G}(u)\cap V(BCC_{n,k}^{R})\
eq\emptyset\), \(u\) is named as \(u_{1}\). Otherwise, \(u\) is named as \(u_{2}\). Lemma 2: _If \(F\) is a set of vertices in \(BCCC(n,k)\), and for any subgraph \(i\) in \(BCCC(n,k)\), \(BCC_{n,k}^{R}\) is connected if \(|F\cap V(G-V(BCC_{n,k}^{i}))|\leq\kappa(G)-2\)._  
Proof.