## 4 Restricted Connectivity of BCCC
: According to the construction rules of BCCC, \(U\) is set as the point adjacent to \(BCC_{n,k}^{i}\) in \(BCC_{n,k}^{R}\) to reach the conclusion that the point set of \(U\) is located in the same type \(A\) switch of subgraph \(BCC_{n,k}^{i}\) and has no neighbors. Therefore, the point in the set of \(U\) points is the one with the fewest neighbors in \(BCC_{n,k}^{i}\). These vertices have \(k\) vertex disjoint paths in the \(BCC_{n,k}^{i}\) subgraph. This achieves \(\kappa(BCC_{n,k}^{i})=k(k\geq 2)\). Moreover, for any vertex \(x\) in \(BCC_{n,k}^{R}\) to connect to \(BCC_{n,k}^{i}\), one of \(x\)'s neighbors in the same type \(A\) switch is not in the subgraph. Therefore, \(\kappa(BCC_{n,k}^{R})=\kappa(G)-1\). Because \(|F\cap V(G-V(BCC_{n,k}^{i}))|\leq\kappa(G)-2\) and \(\kappa(BCC_{n,k}^{R})=\kappa(G)-1\), \(BCC_{n,k}^{R}\) are connected, and the Lemma holds. According to the position of \(u\) and \(v\) vertices in BCCC, we have the following cases. Case 1. \(\alpha\
eq\beta\). Subcase 1.1. \(|F_{\alpha}|\geq\kappa(BCC_{n,k}^{a})or|F_{\beta}|\geq\kappa(BCC_{n,k}^{\beta})\). Without losing generality, suppose \(|F_{\alpha}|\geq\kappa(BCC_{n,k}^{a})\). Apparently, \(|F_{R}|=|F\backslash F_{\alpha}|\leq 2\kappa(G)-n-1-(\kappa(G)-n+1)\leq \kappa(G)-2\). According to Lemma 2, \(BCC_{n,k}^{R}-F_{R}\) is connected. Next, we discuss the following cases. Subcase 1.1.1. \(u\) is \(u_{1}\). Subcase 1.1.1.1. \(N_{G}(u_{1})\cap V(BCC_{n,k}^{R})\
ot\subseteq F\). Selecting a vertex \(x\), \(x\subseteq(N_{G}(u_{1})\cap V(BCC_{n,k}^{R}))\), there is a path \(P_{1}\) from \(u_{1}\) to \(x\) in \(BCC_{n,k}^{a}-F_{\alpha}\). Because \(BCC_{n,k}^{R}-F_{R}\) is connected, there exists a fault-free path \(P_{2}\) from \(x\) to \(v\) in \(BCC_{n,k}^{R}\). Subcase 1.1.1.2. \(N_{G}(u_{1})\cap V(BCC_{n,k}^{R})\subseteq F\). Because the neighbors of \(u_{1}\) are all fault vertices in subgraph \(BCC_{n,k}^{R}\), there must be a fault-free neighbor in subgraph \(BCC_{n,k}^{a}\). Now, we consider how many fault-free paths from \(u_{1}\) to \(BCC_{n,k}^{R}\) exist in \(BCC_{n,k}^{a}\). Furthermore, \(N_{G}(u_{1})\cap V(BCC_{n,k}^{R})=n-1\). Let \(x_{1},x_{2},\ldots,x_{n-1}\) be all neighbors of \(u_{1}\) in \(BCC_{n,k}^{R}\). Let \(y_{1},y_{2},\ldots,y_{k}\) be all neighbors of \(u_{1}\) in \(BCC_{n,k}^{a}\); \(y_{1},y_{2},\ldots,y_{k^{\prime}}\) be the same type \(A\) switch neighbors of \(y_{1},y_{2},\ldots,y_{k}\); \(y_{1},y_{2},\ldots,y_{k^{\prime}}\), \(\ldots,y_{k^{\prime}}\) be all neighbors of \(y_{1},y_{2},\ldots,y_{k^{\prime}}\) in \(BCC_{n,k}^{a}\) and \(N_{G}(y_{1}^{\prime\prime})\cap v(BCC_{n,k}^{R})\
eq\emptyset,i\in\{1,2,\ldots,k\}\). Thus, there are \((n-1)(k+1)\) paths from \(u_{1}\) to \(BCC_{n,k}^{R}\) as follows (refer to Figure 3). The specific paths are listed below:\[ \begin{cases}X_{i}=(u_{1},x_{i}),1\leq i\leq n-1\\ Y_{i}=(u_{1},y_{i},y_{i^{\prime}},y_{i^{\prime\prime}}),1\leq i\leq k,i^{\prime}=k (n-1)\end{cases} \]  
Because \(N_{G}(u_{1})\cap V(BCC^{R}_{n,k})\subseteq F,F\cap V(BCC^{R}_{n,k})\geq n-1,F \cap V(BCC^{a}_{n,k})\leq\kappa^{1}(G)-1-n+1=\kappa^{1}(G)-n\). Whether \(\kappa^{1}(G)=2\kappa(G)-n\) or \(2\kappa(G)-k-1\), \((n-1)(k+1)-k^{1}(G)-n\geq 1\). Therefore, there is at least one fault-free path \(P_{1}\) from \(u_{1}\) to \(BCC^{R}_{n,k}\). Assume that \(t\) is the last vertex in the path \(P_{1}\). If \(t=v\), then \(P_{1}\) is a path from \(u_{1}\) to \(v\) in \(G-F\). Otherwise, since \(BCC^{R}_{n,k}-F_{R}\) is connected, there is a fault-free path \(P_{2}\) from \(t\) to \(v\) in \(BCC^{R}_{n,k}\). Then, the path \((P_{1},P_{2})\) is a fault-free path from \(u_{1}\) to \(v\) in \(G-F\). Subcase 1.1.2. \(u\) is \(u_{2}\). It can be obtained that \(u_{2}\) is a vertex in \(BCC^{a}_{n,k}\) and its neighbors are all in this subgraph. According to the definition of restricted connectivity, there is a fault-free neighbor in \(BCC^{a}_{n,k}\). Now, we consider how many fault-free paths there are from \(u_{2}\) to \(BCC^{R}_{n,k}\) in \(BCC^{a}_{n,k}\).