# IFDA: Intermittent Fault Diagnosis Algorithm for Augmented Cubes Under the PMC Model
## 3 Properties of Augmented Cubes
By Lemma 5, we have \(|N_{L}(\{x,s,t\})|\geq 6(n-1)-17=6n-23\). \(|N_{R}(\{y,u,v\})|\geq 6(n-1)-17=6n-23\). Hence, \(|N_{AQ_{n}}(\{x,y,s,t,u,v\})|\geq|N_{L}(\{x,s,t\})|+|N_{R}(\{y,u,v\})|\geq(6 n-23)+(6n-23)=12n-46\geq 12n-54\) holds. By combining the above three cases, we complete the proof. **Lemma 13**. : _Let \(x,y,s,t,u,v\) be six nodes in \(AQ_{n}\), where \(n\geq 5\). Then, \(\big{|}N_{AQ_{n}}(\{x,y,s,t,u,v\})\big{|}\geq 12n-56\) holds._  
Proof of Lemma 13.: We use mathematical induction to conduct the proofs. The base case for \(AQ_{5}\) can be verified according to its structure diagram, i.e., let \(x,y,s,t,u,v\) be six nodes in \(AQ_{5}\); then, \(\big{|}N_{AQ_{5}}(\{x,y,s,t,u,v\})\big{|}\geq 12\times 5-56=4\) holds. Assume that the lemma holds for \(AQ_{n-1}\), where \(n\geq 6\). Now, we consider \(AQ_{n}=L\oplus R\), where \(L\cong AQ_{n-1}^{0}\) and \(R\cong AQ_{n-1}^{0}\). For the distribution of \(x,y,s,t,u,v\), we consider the following four cases:  
Figure 2: Illustration of the proofs of Lemma 11 and Lemma 12. **Case 1. ** All of \(\{x,y,s,t,u,v\}\) are in the same subcube. We assume that \(x,y,s,t,u,v\in V_{L}\). According to the definition of the augmented cube, \(L\) contains at most three complement pairs among \(x,y,s,t,u,v\). Hence, there are the following subcases:  
**Case 1.1. **\(L\) contains three complement pairs among \(x,y,s,t,u,v\). We assume that \((x,y)\), \((s,t)\), and \((u,v)\) are the three complement pairs of \(L\) (see Figure 3a). By Lemma 10, we have that \(|N_{L}(\{x,y,s,t,u,v\})|\geq 12(n-1)-46=12n-58\). By Property 2, \(|N_{R}(x,y)|=|N_{R}(s,t)|=|N_{R}(u,v)|=2\). Hence, \(|N_{R}(\{x,y,s,t,u,v\})|=|N_{R}(x,y)|+|N_{R}(s,t)|+|N_{R}(u,v)|=2+2+2=6\). Therefore, \(|N_{AQ_{n}}(\{x,y,s,t,u,v\})|=|N_{L}(\{x,y,s,t,u,v\})|+|N_{R}(\{x,y,s,t,u,v\})| \geq(12n-58)+6=12n-52\) holds. **Case 1.2. **\(L\) contains two complement pairs among \(x,y,s,t,u,v\). We assume that \((x,y)\) and \((s,t)\) are the two complement pairs of \(L\) (see Figure 3b). By Lemma 11, we have that \(|N_{L}(\{x,y,s,t,u,v\})|\geq 12(n-1)-48=12n-60\). By Property 2, \(|N_{R}(x,y)|=|N_{R}(s,t)|=2\). Because \(|N_{R}(u)|=|N_{R}(v)|=2\), \(|N_{R}(\{x,y,s,t,u,v\})|=|N_{R}(x,y)|+|N_{R}(s,t)|+|N_{R}(u)|+|N_{R}(v)|=2+2+2 +2=8\). Therefore, \(|N_{AQ_{n}}(\{x,y,s,t,u,v\})|=|N_{L}(\{x,y,s,t,u,v\})|+|N_{R}(\{x,y,s,t,u,v\})| \geq(12n-60)+8=12n-52\) holds. **Case 1.3. **\(L\) contains only one complement pair among \(x,y,s,t,u,v\). We assume that \((x,y)\) is the complement pair of \(L\) (see Figure 3c). By Lemma 12, we have that \(|N_{L}(\{x,y,s,t,u,v\})|\geq 12(n-1)-54=12n-66\). By Property 2, \(|N_{R}(x,y)|=2\). Because \(|N_{R}(s)|=|N_{R}(t)|=|N_{R}(u)|=|N_{R}(v)|=2\), \(|N_{R}(\{x,y,s,t,u,v\})|=|N_{R}(x,y)|+|N_{R}(s)|+|N_{R}(t)|+|N_{R}(u)|+|N_{R}( v)|=2+2+2+2=10\). Therefore, \(|N_{AQ_{n}}(\{x,y,s,t,u,v\})|=|N_{L}(\{x,y,s,t,u,v\})|+|N_{R}(\{x,y,s,t,u,v\})| \geq(12n-66)+10=12n-56\) holds. **Case 1.4. **\(L\) contains no complement pair among \(x,y,s,t,u\) (see Figure 3d). By the induction hypothesis, we have that \(|N_{L}(\{x,y,s,t,u,v\})|\geq 12(n-1)-56=12n-68\). Because \(|N_{R}(x)|=|N_{R}(y)|=|N_{R}(s)|=|N_{R}(t)|=|N_{R}(u)|=|N_{R}(v)|=2\), hence \(|N_{R}(\{x,y,s,t,u,v\})|=|N_{R}(x)|+|N_{R}(y)|+|N_{R}(s)|+|N_{R}(t)|+|N_{R}(u)| +|N_{R}(v)|=2+2+2+2+2=12\). Therefore, \(|N_{AQ_{n}}(\{x,y,s,t,u,v\})|=|N_{L}(\{x,y,s,t,u,v\})|+|N_{R}(\{x,y,s,t,u,v\})| \geq(12n-68)+12=12n-56\) holds. **Case 2. ** Five of \(\{x,y,s,t,u,v\}\) are in one subcube, and the remaining one is in the other subcube. We assume that \(x,y,s,t,u\in V_{L}\) and \(v\in V_{R}\) (see Figure 3e). By Lemma 9, we have \(|N_{L}(\{x,y,s,t,u\})|\geq 10(n-1)-41=10n-51\). By Property 1, we have \(|N_{R}(v)|=2(n-1)-1=2n-3\). Therefore, \(|N_{AQ_{n}}(\{x,y,s,t,u,v\})|\geq|N_{L}(\{x,y,s,t,u\})|+|N_{R}(v)|\geq(10n-51) +(2n-3)=12n-54\geq 12n-56\) holds. Figure 3. Illustration of the proof of Lemma 13. **Case 3. ** Four of \(\{x,y,s,t,u,v\}\) are in one subcube, and the remaining two are in the other subcube.