# XOR Multiplexing Technique for Nanocomputers
## 4 Fault-Tolerant Ability Analysis  
In the last section, we analyzed the tolerant ability of the gate error probability (gate error threshold). Now let us analyze the tolerant ability of input signal error (input signal error threshold). In order to map each output probability to a logic state, we need a threshold \(z_{*}\). According to Figure 7, it is notdifficult to find out that \(z_{0}(\varepsilon_{*})\) is a good choice for \(z_{*}\). It is simple and effective. Substituting \(\varepsilon_{*}\) into Equation (8), then we have  
\[z_{0}(\varepsilon_{*})=\frac{1-\sqrt{1-4(2\varepsilon_{*}-1)(1-\varepsilon_{* })}}{2(2\varepsilon_{*}-1)}=0.6076 \tag{12}\]  
Below, we shall interpret \([0,z_{*})\) as non-stimulated state and \((z_{*},1]\) as a stimulated state. When we have fixed the input \(Y=1\) and \(Y=0\), then we can get 3-D diagrams, as shown in Figure 9. Clearly, the XOR multiplexing system has a higher fault-tolerant ability when inputs are both stimulated or both non-stimulated. Seen in Figure 9, the effectiveness of this threshold is obvious. Note that for each different fixed \(Y\), there is a different value of \(X\) (here we name it as critical point and denote it by \(x_{0}\)) that divides the output into two states when \(0\leq\varepsilon<\varepsilon_{*}\). Take \(Y=0\) as an example, in the interval \(0\leq\varepsilon<\varepsilon_{*}\), when \(X<x_{0}\), the output would be non-stimulated, and when \(X>x_{0}\), the output would be stimulated. The calculation of the critical point can help us more intuitively understand the fault tolerant ability of the system. Since when \(n\) is large enough and \(0\leq\varepsilon<\varepsilon_{*}\), the output only depends on the input condition: input \(X\) and \(Y\) have the same logic state (both stimulated or both non-stimulated) or have a different logic state (one of the inputs is stimulated and the other one is non-stimulated). Let us denote the probability that two inputs \(X\) and \(Y\) have a different logic state by \(P_{1}\), and denote the probability that two inputs \(X\) and \(Y\) have the same logic state by \(P_{2}\). Therefore, the ratio of \(P_{1}\) and \(P_{2}\) will be a key parameter to determining if the final output \(Z_{n}\) is larger than \(z_{*}\) or not. \(X\) and \(Y\) are the probabilities of inputs being stimulated, and then \(1-X\) and \(1-Y\) are the probabilities of inputs being non-stimulated. \(P_{1}\) and \(P_{2}\) are shown as follows  
\[P_{1}=X(1-Y)+Y(1-X) \tag{13}\]  
\[P_{2}=XY+(1-Y)(1-X) \tag{14}\]  
If we need the output to be stimulated, then \(P_{1}/P_{2}\) must be larger than a specific value that is greater than one. Since the output logic state is associated with the output threshold \(z_{*}\), the specific value will be a function of \(z_{*}\) and the mathematic relation between them is shown below. \[\frac{P_{1}}{P_{2}}=\frac{X(1-Y)+Y(1-X)}{XY+(1-X)(1-Y)}>\frac{z_{*}}{1-z_{*}} \tag{15}\]  
Figure 9: 3-D diagrams of XOR multiplexing for fixed input Y: (**a**) fixed \(Y=0\); (**b**) fixed \(Y=1\). Clearly, \(P_{1}+P_{2}=1\), hence Equation (15) is equivalent to  
\[P_{1}=X(1-Y)+Y(1-X)>z_{*} \tag{16}\]  
If \(P_{1}>z_{*}\), the final output would be larger than threshold (stimulated). When the inequality becomes  
\[P_{1}=X(1-Y)+Y(1-X)<z_{*} \tag{17}\]  
The final output would be smaller than threshold (non-stimulated). Hence, it is easy to obtain the critical point \(x_{0}\) for each fixed \(Y\) by solving the following equality  
\[x_{0}(1-Y)+Y(1-x_{0})=z_{*}\]  
Critical point \(x_{0}\) is a function of \(Y\); these critical points are then plotted against each \(Y\) (\(\Delta Y=0.01\)). This leads to Figure 10, which shows that the diagram has two regions and for each different \(Y\) critical point \(x_{0}\) has a different value and there is a parameter interval that makes the system no longer function, even though the system is fault-free, and the parameter interval is approximately \(0.3924<Y<0.6076\).