# Fault-Tolerant Resolvability and Extremal Structures of Graphs
## 2 Preliminaries
### Anti-Prism Graphs  
The cross product of a cycle \(C_{n}\) and \(P_{2}\) is actually called a prism, usually denoted by \(D(n)\). In [11], it was shown that  
\[\beta(P_{m}\Box C_{n})=\left\{ \begin{array}{ll}2,&\text{if }n\text{ is odd;}\\ 3,&\text{if }n\text{ is even.}\end{array} \right.\]  
This implies that  
\[\beta(D(n))=\left\{ \begin{array}{ll}2,&\text{if }n\text{ is odd;}\\ 3,&\text{if }n\text{ is even.}\end{array} \right.\]  
By applying Equation (1) and Theorem 4 to the prism graph \(D(n)\), we find the following result. Figure 3: (**a**) The extended Petersen graph \(P(5)\), (**b**) The extended Petersen graph \(P(6)\). **Proposition 3**. : _The prism graph \(D(n)\) has a constant fault-tolerant metric dimension._  
We investigate fault-tolerant resolvability in the anti-prism graphs. The anti-prism \(A(n)\)[35] is a 4-regular graph. It is the octahedron for \(n\geq 3\). For \(n\geq 3\), the anti-prism \(A(n)\) consists of an inner cycle \(y_{1},y_{2},\ldots,y_{n}\), an outer cycle \(z_{1},z_{2},\ldots,z_{n}\), and a set of \(n\) spokes \(y_{k}z_{k}\) and \(y_{k+1}z_{k}\), \(k=1,2,\ldots,n\), with indices taken as modulo \(n\). Thus, \(|V(A(n))|=2n\) and \(|E(A(n))|=4n\). [PERSON] et al. [34] showed the following result. **Proposition 4**. : [34] _Let \(\Gamma\) be the anti-prism graph \(A(n)\) with \(n\geq 3\); then, \(\beta(\Gamma)=3\)._  
They also showed the following:  
**Proposition 5**. : [34] _The anti-prism graph \(A(n)\) has a constant metric dimension._  
In this section, we present the main results, and, for the anti-prism graph \(A(n)\), the upper and lower bounds on the fault-tolerant metric dimension are proved. Note that Claim 1 in the following result was essentially shown in Proposition 4. **Theorem 10**. : _Let \(\Gamma\) be the anti-prism graph \(A(n)\), with \(n\geq 10\); then, \(4\leq\beta^{\prime}(\Gamma)\leq 14\)._  
Proof. : Let \(n=2\ell\) or \(n=2\ell+1\) for even or odd \(n\), respectively. **Claim 1:** A resolving set \(R\) of order 3 exists in \(\Gamma\). Based on the location of basis elements in \(G\), we divide this in two cases. **Case 1:** When \(n\) is even, \(n=2\ell\), with \(\ell\geq 3\). For \(n\geq 6\), there exists a resolving set \(R\) of cardinality 3. \(R=\{z_{1},z_{3},z_{\ell+1}\}\) is a resolving set. Representation of the vertices in the outer cycle with respect to \(\{z_{1},z_{3}\}\) is as follows. As we can see, \(C_{R\setminus\{z_{\ell+1}\}}(z_{2})=(1,1)\); in general, the representations of the vertices in the outer cycle are  
\[C_{R\setminus\{z_{\ell+1}\}}(z_{k})=\left\{ \begin{array}{ll}(k-1,k-3),&4\leq k \leq\ell+1;\\ (n-k+1,k-3),&k=\ell+2,k=\ell+3;\\ (n-k+1,n-k+3),&\ell+4\leq k\leq n.\end{array} \right.\]  
Representations of the vertices in the inner cycle are \(C_{R\setminus\{z_{\ell+1}\}}(y_{1})=(1,3)\), \(C_{R\setminus\{z_{\ell+1}\}}(y_{2})=(1,2)\), \(C_{R\setminus\{z_{\ell+1}\}}(y_{3})=(2,1)\). In general,  
\[C_{R\setminus\{z_{\ell+1}\}}(y_{k})=\left\{ \begin{array}{ll}(k-1,k-3),&4\leq k \leq\ell+1;\\ (\ell,\ell-1),&k=\ell+2;\\ (\ell-1,\ell),&k=\ell+3;\\ (n-k+2,n-k+4),&\ell+4\leq k\leq n.\end{array} \right.\]  
**Case 2:** For odd \(n\), \(n=2\ell+1\) with \(\ell\geq 3\). Then,  
\[C_{R\setminus\{z_{\ell+1}\}}(y_{k})=\left\{ \begin{array}{ll}(k-1,k-3),&4\leq k \leq\ell+2;\\ (\ell,\ell),&k=\ell+3;\\ (n-k+2,n-k+4),&\ell+4\leq k\leq n.\end{array} \right.\]  
From the above discussion, we can see there are few vertices with the same representation \(y_{k}\),\(z_{k}\), with \(4\leq k\leq\ell+1\); for even and odd \(n\), \(y_{1}\),\(z_{n}\) and \(y_{s+1}\),\(z_{s}\) with \(\ell+3\leq s\leq n-1\) and \(\ell+4\leq s\leq n-1\)respectively. In order to distinguish the pairs with the same vertices, we take \(y_{\ell+1}\) in the outer and inner cycle. Representation in the outer cycle is  
\[d(y_{\ell+1},z_{s})=\left\{ \begin{array}{ll}\ell-s+1,&1\leq s\leq\ell;\\ s-\ell-1,&\ell+2\leq s\leq n.\end{array} \right.\]  
Now, representation in the inner cycle is  
\[d(y_{\ell+1},y_{k})=\left\{