# IFDA: Intermittent Fault Diagnosis Algorithm for Augmented Cubes Under the PMC Model
## 3 Properties of Augmented Cubes
: _Let \(x,y,z,t\) be four nodes in \(AQ_{n}\), where \(n\geq 4\). Then, \(\left|N_{AQ_{n}}(\{x,y,z,t\})\right|\geq 8n-28\) holds._  
**Lemma 7** ([28]). : _Let \(x,y,s,t,u\) be five nodes in \(AQ_{n}\), where \(n\geq 4\). If \(AQ_{n}\) contains two complement pairs among \(x,y,s,t,u\), then \(\left|N_{AQ_{n}}(\{x,y,s,t,u\})\right|\geq 10n-35\) holds._  
**Lemma 8** ([28]). : _Let \(x,y,s,t,u\) be five nodes in \(AQ_{n}\), where \(n\geq 4\). If \(AQ_{n}\) contains at least one complement pair among \(x,y,s,t,u\), then \(\left|N_{AQ_{n}}(\{x,y,s,t,u\})\right|\geq 10n-39\) holds._  
According to Lemma 4 to Lemma 8, the following lemma can be obtained:  
**Lemma 9**. : _Let \(x,y,s,t,u\) be five nodes in \(AQ_{n}\), where \(n\geq 5\). Then, \(\left|N_{AQ_{n}}(\{x,y,s,t,u\})\right|\geq 10n-41\) holds._  
Proof of Lemma 9.: The complete proof of Lemma 9 is presented in Appendix A.1. **Lemma 10** ([28]). : _Let \(x,y,s,t,u,v\) be six nodes in \(AQ_{n}\), where \(n\geq 4\). If \(AQ_{n}\) contains three complement pairs among \(x,y,s,t,u,v\), then \(\left|N_{AQ_{n}}(\{x,y,s,t,u,v\})\right|\geq 12n-46\) holds._  
**Lemma 11**. : _Let \(x,y,s,t,u,v\) be six nodes in \(AQ_{n}\), where \(n\geq 4\). If \(AQ_{n}\) contains two complement pairs among \(x,y,s,t,u,v\), then \(\left|N_{AQ_{n}}(\{x,y,s,t,u,v\})\right|\geq 12n-48\) holds._  
Proof of Lemma 11.: Let \((x,y)\) and \((s,t)\) be the two complement pairs of \(AQ_{n}=L\oplus R\), where \(L\cong AQ_{n-1}^{0}\) and \(R\cong AQ_{n-1}^{1}\). We assume that \(x,s\in V_{L}\) and \(y,t\in V_{R}\). For the distribution of \(u\) and \(v\), we consider the following cases:  
**Case 1. ** Both \(u\) and \(v\) are in the same subcube. We assume that \(u,v\in V_{L}\) (see Figure 2a). By Lemma 6, we have \(\left|N_{L}(\{x,s,u,v\})\right|\geq 8(n-1)-28=8n-36\). By Lemma 4, we have \(\left|N_{R}(\{y,t\})\right|\geq 4(n-1)-8=4n-12\). Hence, \(\left|N_{AQ_{n}}(\{x,y,s,t,u,v\})\right|\geq\left|N_{L}(\{x,s,u,v\})\right|+ \left|N_{R}(\{y,t\})\right|\geq(8n-36)+(4n-12)=12n-48\) holds. **Case 2. ** One of \(u\) and \(v\) is in one subcube, and the remaining one is in the other subcube. We assume that \(u\in V_{L}\) and \(v\in V_{R}\) (see Figure 2b). By Lemma 5, we have \(\left|N_{L}(\{x,s,u\})\right|\geq 6(n-1)-17=6n-23\). \(\left|N_{R}(\{y,t,v\})\right|\geq 6(n-1)-17=6n-23\). Hence, \(\left|N_{AQ_{n}}(\{x,y,s,t,u,v\})\right|\geq\left|N_{L}(\{x,s,u\})\right|+ \left|N_{R}(\{y,t,v\})\right|\geq(6n-23)+(6n-23)=12n-46\geq 12n-48\) holds. By combining the above two cases, we complete the proof. **Lemma 12**. : _Let \(x,y,s,t,u,v\) be six nodes in \(AQ_{n}\), where \(n\geq 5\). If \(AQ_{n}\) contains at least one complement pair among \(x,y,s,t,u,v\), then \(\big{|}N_{AQ_{n}}(\{x,y,s,t,u,v\})\big{|}\geq 12n-54\) holds._  
Proof of Lemma 12.: Let \((x,y)\) be the complement pair of \(AQ_{n}=L\oplus R\), where \(L\cong AQ_{n-1}^{0}\) and \(R\cong AQ_{n-1}^{1}\). We assume that \(x\in V_{L}\) and \(y\in V_{R}\). For the distribution of \(\{s,t,u,v\}\), we consider the following cases:  
**Case 1. ** All of \(\{s,t,u,v\}\) are in the same subcube. We assume that \(s,t,u,v\in V_{L}\) (see Figure 2c). By Lemma 9, we have \(|N_{L}(\{x,s,t,u,v\})|\geq 10(n-1)-41=10n-51\). By the fact that \(|N_{R}(\{y\})|=2(n-1)-1=2n-3\), hence \(\big{|}N_{AQ_{n}}(\{x,y,s,t,u,v\})\big{|}\geq|N_{L}(\{x,s,t,u,v\})|+|N_{R}(\{ y\})|\geq(10n-51)+(2n-3)=12n-54\) holds. **Case 2. ** Three of \(\{s,t,u,v\}\) are in the same subcube, and the remaining one is in the other subcube. We assume that \(s,t,u\in V_{L}\) and \(v\in V_{R}\) (see Figure 2d). By Lemma 6, we have \(|N_{L}(\{x,s,t,u\})|\geq 8(n-1)-28=8n-36\). By Lemma 4, we have \(|N_{R}(\{y,v\})|\geq 4(n-1)-8=4n-12\). Hence, \(\big{|}N_{AQ_{n}}(\{x,y,s,t,u,v\})\big{|}\geq|N_{L}(\{x,s,t,u\})|+|N_{R}(\{y, v\})|\geq(8n-36)+(4n-12)=12n-48\geq 12n-54\) holds. **Case 3. ** Two of \(\{s,t,u,v\}\) are in the same subcube, and the remaining two are in the other subcube. We assume that \(s,t\in V_{L}\) and \(u,v\in V_{R}\) (see Figure 2e).