## 5 Fault-Free Routing Algorithm of BCCC under the Restricted Connectivity
### Analysis of Fault-Free Unicast Algorithm  
First, the time complexity of \(\mathit{PATHSEQ}(u,F,H_{1},H_{2})\) and \(\mathit{BCTP}(u,v,\mathit{BCC},F)\) functions is analyzed. \(u\) in \(H_{1}\) exists in the \(\mathit{PATHSEQ}(u,F,H_{1},H_{2})\) function. This depends on whether \(u\) is connected to other subgraphs, including \(u_{1}\) and \(u_{2}\). The path from \(u_{1}\) to the subgraph \(H_{2}\) is \((u,x)\), \((u,y_{1},y_{2},y_{3},y_{4})\). The time complexity of \(\mathit{PATHSEQ}\) in different situations is as follows. First, the path is \((u,x)\), because \((N_{G}(u)\cap V(H_{2}))\backslash F\
eq\emptyset\) finds a vertex \(x\) in \(H_{2}\), and returns a path \((u,x)\). The time complexity is \(O(1)\) in this case. Second, the path is \((u,y_{1},y_{2},y_{3},y_{4})\). Each vertex in the set is \(y_{1}\) and the time complexity is \(O(1)\) after traversing \((N_{G}(u)\cap V(H_{1}))\backslash F\). Each vertex in the set is \(y_{2}\) and the time complexity is less than \(O(\kappa(G))\) after traversing \((N_{G}(y_{1})\cap V(H_{1}))\backslash(F\cup u)\). Each vertex in the set is \(y_{3}\) and the time complexity is less than \(O(\kappa(G))\) after we go through \((N_{G}(y_{2})\cap V(H_{1}))\backslash(F\cup y_{1})\). Each vertex in the set is \(y_{4}\) and the time complexity is \(O(1)\). After traversing \((N_{G}(y_{3})\cap V(H_{2}))\backslash F\), paths \((u,y_{1},y_{2},y_{3},y_{4}))\) are obtained. The time complexity is less than \(O(\kappa(G)^{2})\) in this case. The path from \(u_{2}\) to subgraph \(H_{2}\) includes \((u,x_{1},x_{2})\), \((u,y_{1},y_{2},y_{3})\), and \((u,z_{1},z_{2},z_{3},z_{4})\). First, the path is \((u,x_{1},x_{2})\). Each vertex in the set is \(x\) and the time complexity is \(O(1)\) after the function traverses \((N_{G}(u)\cap V(H_{1}))\backslash F\). Next, vertex \(x_{2}\) is found from \((N_{G}(x_{1})\cap V(H_{2}))\backslash F\). The time complexity in this case is \(O(k)<O(k(G))\). Second, the path is \((u,y_{1},y_{2},y_{3})\). The function traverses \((N_{G}(u)\cap V(H_{1}))\backslash F\). The node is set on the same type \(A\) switch as \(u\) to \(y_{1}\) and the time complexity to \(O(1)\). Then, traversing \((N_{G}(y1)\cap V(H_{1}))\backslash(F\cup u)\) sets the point connected with other subgraphs as \(y_{2}\), and the time complexity is less than \(O(\kappa(G))\). Finally, \((N_{G}(y_{2})\cap V(H_{2}))\backslash F\) is traversed and node \(y_{3}\) selected. In this case, the time complexity is less than \(O(\kappa(G))\). Third, the path of \(u_{2}\)\((u,z_{1},z_{2},z_{3},z_{4})\) is the same as the path of \(u_{1}\)\((u,y_{1},y_{2},y_{3},y_{4})\). In summary, the time complexity of the \(\mathit{PATHSEQ}(u,F,H_{1},H_{2})\) function is \(O(\kappa(G)^{2})\) combined with \(u_{1}\) and \(u_{2}\). Next, we analyze \(\mathit{BCCFTP}(u,v,\alpha,\beta,F,\mathit{BCC}(u,k))\). \(\mathit{BCCFTP}\) is divided into two cases, including \(\alpha\
eq\beta\) and \(\alpha=\beta\). In the case of \(\alpha\
eq\beta\), if \(|F_{\alpha}|\geq k\), Algorithm \(\mathit{BCCFTP}\) will first obtain a fault-free path \(P_{1}\) from \(u\) to a vertex \(s\) in \(G-\mathit{BCC}_{n,k}^{\alpha}\) by \(\mathit{PATHSEQ}(u,F,\mathit{BCC}_{n,k}^{\alpha},G-\mathit{BCC}_{n,k}^{\alpha})\). The time complexity of this part is \(O(\kappa(G)^{2})\). Next, Algorithm \(\mathit{BCCFTP}\) will return \(P_{1}\) if \(s=v\); then, the time to obtain the path from \(u\) to \(v\) in \(G-F\) is \(O(\kappa(G)^{2})\). Otherwise, Algorithm \(\mathit{BCCFTP}\) will return \((P_{1}\),\(\mathit{BCTP}(s,v,G-\mathit{BCC}_{n,k}^{\alpha},F\backslash F_{\beta}))\); then, the time to run \(\mathit{BCTP}(s,v,G-\mathit{BCC}_{n,k}^{\alpha},F\backslash F_{\beta})\) is \(O(\kappa n\kappa^{1}(G))\) and the time to obtain a path \((P_{1}\),\(\mathit{BCTP}(s,v,\mathit{G-BCC}_{n,k}^{\alpha},F\backslash F_{\beta}))\) by connecting path \(P_{1}\) and \(\mathit{BCTP}\) is \(O(1)\). Thus, the time to obtain the path from \(u\) to \(v\) in \(G-F\) is \(O(\kappa(G)^{2}+k\kappa^{1}(G)+1)=O(kn\kappa^{1}(G))\).