# The Application of Fault-Tolerant Partition Resolvability in Cycle-Related Graphs
## 1 Introduction and Preliminaries
As \(r(v_{l}|\aleph)\) and \(r(v_{l-1}|\aleph)\) have their first two coordinatesof the flower graph, respectively. The degree of vertices of the center and of all other vertices are four and two, respectively. Consider a class of a flower graph for \(q=3\), denoted by \(f_{\alpha\times 3}\). The vertices of the center of \(f_{\alpha\times 3}\) are denoted by \(v_{1},v_{2},\ldots,v_{\alpha}\), and the vertices of the petals are denoted by \(v_{\alpha+1},v_{\alpha+2},\ldots,v_{2\alpha}\). The set \(\{v_{1},v_{2},\ldots,v_{2\alpha}\}\) is a vertex set, and the set \(\{v_{i}v_{i+1},1\leq i\leq\alpha-1\}\cup\{v_{1}v_{\alpha}\}\cup\{v_{1}v_{2 \alpha}\}\cup\{v_{i}v_{\alpha+i},v_{i+1}v_{\alpha+i},1\leq i\leq\alpha-1\} \cup\{v_{\alpha},v_{2\alpha}\}\) is an edge set of the flower graph \(f_{\alpha\times 3}\). Flower graph \(f_{\beta\times 3}\) is given in Figure 2.  
The \(\beta(f_{\alpha\times 3})\) was been computed in [12]:  
Lemma 4 ([12]): _Let \(f_{\alpha\times 3}\) be a flower graph. Then, we have_  
\[\beta(f_{\alpha\times 3})=\left\{ \begin{array}{ll}2,&\text{if $\alpha$ is even;}\\ 3,&\text{otherwise.}\end{array} \right.\]  
_for every \(\alpha\geq 6\)._  
Lemma 5 ([16]): _Let \(f_{\alpha\times 3}\) be a flower graph when \(\alpha\) is odd. Then, \(pd(f_{\alpha\times 3})\leq 4\)._  
The following theorem computes the exact partition dimension of \(f_{\alpha\times 3}\):  
Theorem 4.1: _Let \(f_{\alpha\times 3}\) be a flower graph with \(\alpha\geq 3\). Then, \(pd(f_{\alpha\times 3})=3\)._  
Proof: Suppose that \(\aleph=\{\aleph_{i}:1\leq i\leq 3\}\) is a partition set of \(V(f_{\alpha\times 3})\). We consider the following cases:  
**Case (i): When \(\alpha\) is even for \(\alpha\geq 4\)**. It follows from Proposition 1(b) that \(pd(f_{\alpha\times 3})\geq 3\). Now, from Lemma 4 and Proposition 1(a), \(pd(f_{\alpha\times 3})\leq 3\). Thus, \(pd(f_{\alpha\times 3})=3\). **Case (ii): When \(\alpha\) is odd for \(\alpha\geq 3\)**. **Case ii(a): When \(\alpha=3\)**. Consider \(\aleph_{1}=\{v_{i}:1\leq i\leq 4\},\aleph_{2}=\{v_{5}\}\) and \(\aleph_{3}=\{v_{6}\}\). It can easily be checked that \(\aleph\) is the resolving partition of \(f_{\alpha\times 3}\). **Case ii(b): When \(\alpha=2e+1\) for \(e\geq 2\)**. Consider \(\aleph_{1}=\{v_{i}:1\leq i\leq\alpha+e\}\cup\{v_{i}:\alpha+e+2\leq i\leq 2 \alpha-2\},\aleph_{2}=\{v_{\alpha+e+1}\}\) and \(\aleph_{3}=\{v_{2\alpha-1},v_{2\alpha}\}\). The distance codes of the vertices of \(f_{\alpha\times 3}\) with respect to \(\aleph\) are given in Table 7.  
Figure 2: Flower graph \(f_{\beta\times 3}\). \begin{tabular}{c c c c} \hline \hline
**Distance of Vertex** & **From**\(\aleph_{1}\) & **From**\(\aleph_{2}\) & **From**\(\aleph_{3}\) \\ \hline \(v_{r}(1\leq r\leq\epsilon)\) & 0 & \(\epsilon-r+2\) & \(r\) \\ \hline \(v_{\epsilon+1}\) & 0 & 1 & \(\epsilon\) \\ \hline \(v_{\epsilon+r+1}(1\leq r\leq\epsilon-1)\) & 0 & \(r\) & \(\epsilon-r\) \\ \hline \(v_{2\epsilon+1}\) & 0 & \(\epsilon\) & \(1\) \\ \hline \(v_{2\epsilon+r}(2\leq r\leq\epsilon+1)\) & 0 & \(\epsilon-r+3\) & \(r\) \\ \hline \(v_{3\epsilon+2}\) & 1 & 0 & \(\epsilon\) \\ \hline \(v_{3\epsilon+r+1}(2\leq r\leq\epsilon-1)\) & 0 & \(r\) & \(\epsilon-r+1\) \\ \hline \(v_{4\epsilon+r}(1\leq r\leq 2)\) & 1 & \(\epsilon+r-1\) & 0 \\ \hline \hline \end{tabular}   
It is obvious from Table 7 that \(\aleph\) is a resolving partition of \(f_{\alpha\times 3}\), and hence \(pd(f_{\alpha\times 3})\leq 3\). Now, by Proposition 1(b), \(f_{\alpha\times 3}\) is not a path graph, so it is not possible that the PD of \(f_{\alpha\times 3}\) is two. Therefore, \(pd(f_{\alpha\times 3})\geq 3\), and from both acquired inequalities, \(pd(f_{\alpha\times 3})=3\), which concludes the proof. In the following theorem, we compute the \(\mathcal{P}(f_{\alpha\times 3})\):  
**Theorem 3**. : _Let \(f_{\alpha\times 3}\) be a flower graph with \(\alpha\geq 3\). Then, \(\mathcal{P}(f_{\alpha\times 3})=4\)._  
Proof. : To prove that \(\mathcal{P}(f_{\alpha\times 3})=4\), we first show that \(\mathcal{P}(f_{\alpha\times 3})\leq 4\).