# The Application of Fault-Tolerant Partition Resolvability in Cycle-Related Graphs
## 1 Introduction and Preliminaries
Now, we consider the subsequent cases by assuming that \(|\aleph_{1}|\geq 2\) and \(v_{1}\in\aleph_{1}\):  
**Case 1:**: If \(N(v_{1})\cap\aleph_{1}=\{v_{2},v_{l},v_{l+t+1}\}\), then \(r(v_{1}|\aleph)=(0,a_{0},b_{0})\), \(r(v_{2}|\aleph)=(0,a_{1},b_{1})\), \(r(v_{l}|\aleph)=(0,a_{2},b_{2})\) and \(r(v_{l+t+1}|\aleph)=(0,a_{3},b_{3})\). As \(a_{0}-1\leq a_{1},a_{2},a_{3}\leq a_{0}+1\), two vertices will have two identical coordinates in their identification, thus creating a contradiction. **Case 2:**: If \(N(v_{1})\cap\aleph_{1}=\{v_{2},v_{l}\}\), and one vertex \(v_{l+t+1}\in\aleph_{2}\), then \(r(v_{1}|\aleph)=(0,1,b_{0})\), \(r(v_{2}|\aleph)=(0,a_{1},b_{1})\), \(r(v_{l}|\aleph)=(0,a_{2},b_{2})\) and \(r(v_{l+t+1}|\aleph)=(1,0,b_{3})\). Since \(1\leq a_{1},a_{2}\leq 2\), two vertices will have two identical coordinates, thus creating a contradiction. **Case 3:**: If \(N(v_{1})\cap\aleph_{1}=\{v_{2}\}\), and two vertices \(v_{l},v_{l+t+1}\in\aleph_{2}\), then \(r(v_{1}|\aleph)=(0,1,b_{0})\), \(r(v_{2}|\aleph)=(0,a_{1},b_{1})\), \(r(v_{l}|\aleph)=(1,0,b_{2})\) and \(r(v_{l+t+1}|\aleph)=(1,0,b_{3})\). Again, \(r(v_{l}|\aleph)\) and \(r(v_{l+t+1}|\aleph)\) are identical in two places, thus creating a contradiction. **Case 4:**: If \(N(v_{1})\cap\aleph_{1}=\{v_{2}\}\), \(v_{l}\in\aleph_{2}\) and \(v_{l+t+1}\in\aleph_{3}\), then \(r(v_{1}|\aleph)=(0,1,1)\). Hence, we have the following four cases:  
**Case 4(a):**: At least one of \(l\) or \(t\) is \(3\), so without loss of generality, we have \(l=3\). As \(r(v_{2}|\aleph)=(0,1,2)\), having the first two coordinates being the same as \(r(v_{1}|\aleph)\), therefore, it is against our assumption. **Case 4(b):**: At least one of \(l\) or \(t\) is \(4\), so without loss of generality, we have \(l=4\). Here, \(N(v_{2})=\{v_{3},v_{1}\}\), and let \(v_{3}\in\aleph_{1}\). It can be seen that \(r(v_{2}|\aleph)=(0,2,2)\) and \(r(v_{3}|\aleph)=(0,1,3)\). As \(r(v_{1}|\aleph)\) and \(r(v_{3}|\aleph)\) have the same first two coordinates, this negates our claim. **Case 4(c):**: If at least one of \(l\) or \(t\) is \(5\), then without loss of generality, we have \(l=5\). Here, \(N(v_{2})=\{v_{3},v_{1}\}\), \(N(v_{5})=\{v_{4},v_{1}\}\) and \(N(v_{l+t+1})=\{v_{l+t+2},v_{1}\}\). Let \(v_{3}\in\aleph_{1}\) and \(v_{4},v_{l+t+2},\in\aleph_{2}\cup\aleph_{3}\). Without loss of generality, we suppose that \(v_{4},v_{l+t+2}\) belong to \(\aleph_{2}\). Here, \(r(v_{2}|\aleph)=(0,2,2)\) and \(r(v_{3}|\aleph)=(0,1,d_{1})\). As \(r(v_{1}|\aleph)\) and \(r(v_{3}|\aleph)\) have the same first two coordinates, this negates our claim. Now, if \(v_{4}\in\aleph_{2}\) and \(v_{l+t+2}\in\aleph_{3}\), then \(r(v_{3}|\aleph)=(0,1,d_{2})\). As \(r(v_{1}|\aleph)\) and \(r(v_{3}|\aleph)\) have their first two coordinates as identical, this opposes our claim. Now, if \(v_{l+t+2}\in\aleph_{2}\) and \(v_{4}\in\aleph_{3}\), then \(r(v_{l}|\aleph)=(1,0,1)\) and \(r(v_{l+t+2}|\aleph)=(d_{3},0,1)\). As \(r(v_{1}|\aleph)\) and \(r(v_{l+t+2}|\aleph)\) have the same last two coordinates, which negates our claim. **Case 4(d):**: Now, for \(l\geq 6\) and \(t\geq 6\), consider \(N(v_{2})=\{v_{3},v_{1}\}\), \(N(v_{l})=\{v_{l-1},v_{1}\}\) and \(N(v_{l+t+1})=\{v_{l+t+2},v_{1}\}\). Let \(v_{3}\in\aleph_{1}\) and \(v_{l-1},v_{l+t+2}\in\aleph_{2}\cup\aleph_{3}\). Without loss of generality, we assume that \(v_{l-1},v_{l+t+2}\in\aleph_{2}\). Here, \(r(v_{l}|\aleph)=(1,0,2)\), \(r(v_{l-1}|\aleph)=(a_{0},0,c_{1})\) and \(r(v_{l+t+2}|\aleph)=(a_{1},0,1)\). Since \(1\leq a_{0},a_{1}\leq 2\), then two vertices will have two identical coordinates, which contradicts the claim. Now, if \(v_{l+t+2}\in\aleph_{2}\) and \(v_{l-1}\in\aleph_{3}\), then \(r(v_{l}|\aleph)=(1,0,1)\) and \(r(v_{l+t+2}|\aleph)=(a_{2},0,1)\). As \(r(v_{l}|\aleph)\) and \(r(v_{l+t+2}|\aleph)\) have the same last two coordinates, this negates our claim. If \(v_{l-1}\in\aleph_{2}\) and \(v_{l+t+2}\in\aleph_{3}\), then \(N(v_{l-1})=\{v_{l},v_{l-2}\}\). Consider \(v_{l-2}\in\aleph_{1}\) such that \(r(v_{l}|\aleph)=(1,0,2)\) and \(r(v_{l-1}|\aleph)=(1,0,a_{3})\).