# Fault-Tolerant Resolvability and Extremal Structures of Graphs
## 2 Preliminaries
### Extended Petersen Graphs
This shows that \(R\) resolves vertices of \(\Gamma\), which means \(\beta(\Gamma)\leq 3\) when \(n\equiv 0(\mod 4)\). **Subcase 1.2:** When \(n\equiv 2(\mod 4)\). It can be written as \(n=4\ell+2\), \(2\leq\ell\in\mathbb{Z}^{+}\). In this case, again, \(R=\{y_{1},y_{2},y_{3}\}\) resolves \(V(\Gamma)\). In order to show that \(R\) resolves the vertices of \(V(P(n))\), we first represent the vertices in \(\Gamma\) with respect to \(R\setminus\{y_{3}\}\). Again, it is clear that the vertices \(y_{1}\) and \(y_{2}\) distinguish the inner and outer cycle vertices. Note that for the outer cycle, we have \(C_{R\setminus\{y_{3}\}}(z_{1})=(1,2)\), \(C_{R\setminus\{y_{3}\}}(z_{2})=(2,1)\),  
\[C_{R\setminus\{y_{3}\}}(z_{2k})=\left\{ \begin{array}{ll}(k+1,k),&2\leq k \leq\ell+1;\\ (2\ell-k+3,2\ell-k+3),&\ell+2\leq k\leq 2\ell+1.\end{array} \right.\]  
and  
\[C_{R\setminus\{y_{3}\}}(z_{2k+1})=\left\{ \begin{array}{ll}(k+1,k+1),&1\leq k \leq\ell;\\ (2\ell-k+2,2\ell-k+3),&\ell+1\leq k\leq 2\ell.\end{array} \right.\]  
In the inner cycle,  
\[C_{R\setminus\{y_{3}\}}(y_{2k})=\left\{ \begin{array}{ll}(k+2,k-1),&2\leq k \leq\ell+1;\\ (2\ell-k+4,2\ell-k+2),&\ell+2\leq k\leq 2\ell+1.\end{array} \right.\]  
and  
\[C_{R\setminus\{y_{3}\}}(y_{2k+1})=\left\{ \begin{array}{ll}(k,k+2),&1\leq k \leq\ell;\\ (2\ell-k+1,2\ell-k+4),&\ell+1\leq k\leq 2\ell.\end{array} \right.\]  
Again, in this case, it is clear for the inner cycle that there are no two vertices with the same representation. However, for the outer cycle, \(C_{R\setminus\{y_{3}\}}(z_{3+k})=C_{R\setminus\{y_{3}\}}(z_{n-k})\) for \(k=0,2,\ldots,2\ell-2\). Note that the pairs with the same representations are distinguished by \(y_{3}\) since \(d(y_{3},z_{3+k})=\lfloor\frac{3+k}{2}\rfloor\
eq d(y_{3},z_{n-k})=\lfloor\frac {3+k}{2}\rfloor+2\) for \(k=0,2,\ldots,2\ell-2\). This shows that \(R\) resolves the vertices of \(\Gamma\), which means \(\beta(\Gamma)\leq 3\), when \(n\equiv 2(\mod 4)\). **Claim 2:** When \(n\geq 8\), the cardinality of the fault-tolerant resolving set in \(\Gamma\) is 10. We can write \(n=4\ell\), \(\ell\geq 2\), \(\ell\in\mathbb{Z}^{+}\). Note that, for this, \(R=\{y_{1},y_{2},y_{3}\}\) is a resolving set of \(\Gamma\). We show that \(\Gamma\) has a fault-tolerant resolving set of cardinality \(10\). As seen from Figure 3, it can be observed that \(N[y_{1}]=\{y_{1},y_{3},y_{n-1},z_{1}\}\), \(N[y_{2}]=\{y_{2},y_{4},y_{n},z_{2}\}\), and \(N[y_{3}]=\{y_{1},y_{3},y_{5},z_{3}\}\). Moreover, we find that \(\lambda(N_{\Gamma}(y_{1}))=\lambda(N_{\Gamma}(y_{2}))=\lambda(N_{\Gamma}(y_{3} ))=\emptyset\). Thus, by using Lemma 3, we find that \(R^{\prime}=\{y_{1},y_{2},y_{3},y_{4},y_{5},y_{n-1},y_{n},z_{1},z_{2},z_{3}\}\) is a fault-tolerant resolving set of \(\Gamma\). Thus, a fault-tolerant resolving set of \(\Gamma\) with cardinality \(10\) exists when \(n\geq 8\). **Case 2:** When \(n\equiv 1(\mod 2)\) with \(n\geq 11\). Based on the location of basis elements in \(\Gamma\), we further divide this case into two subcases. **Subcase 2.1:** When \(n\equiv 1(\mod 4)\). **Claim 1:**\(\Gamma\) has a resolving set \(R\) of order \(3\). In this case, we can write \(n=4\ell+1\), \(1\leq\ell\in\mathbb{Z}^{+}\). It can be seen that \(\{y_{1},y_{2},z_{3}\}\) is a resolving set for the standard Petersen graph \(P(5)\). For \(P(9)\), we see that \(W=\{y_{1},y_{2},z_{4}\}\) is a resolving set. Now, we show that, for \(n\geq 9\), \(R=\{y_{1},y_{2},z_{2\ell-1}\}\) resolves the vertices of \(\Gamma\), where \(n\equiv 1(\mod 4)\). In order to show this, first we present representations of the vertices with respect to \(R\setminus\{z_{2\ell-1}\}\). The representations of the vertices in the outer cycle are \(C_{R\setminus\{z_{2\ell-1}\}}(z_{1})=(1,2)\), \(C_{R\setminus\{z_{2\ell-1}\}}(z_{2})=(2,1)\),  
\[C_{R\setminus\{z_{2\ell-1}\}}(z_{2k})=\left\{ \begin{array}{ll}(k+1,k),&2 \leq k\leq\ell;\\ (k,k),&k=\ell+1;\\ (2\ell-k+2,2\ell-k-3),&\ell+2\leq k\leq 2\ell.\end{array} \right.\]  
and  
\[C_{R\setminus\{z_{2\ell-1}\}}(z_{2k+1})=\left\{