## 2 Preliminaries
### Determination of the Fuzzification Coefficients for Supervised Elements
Since \(\mu_{ik}\) is to be solved in Equation (19), \(\mu_{ik}>0\), and the above can be simplified into \(\frac{U_{ik}^{M^{\prime}-M}U_{ik}^{M-1}}{\left(1-U_{ik}\right)^{M-1}}\geq\frac{ \alpha^{M^{\prime}-1}}{\left(1-\alpha\right)^{M-1}}\) or  
\[\frac{U_{ik}^{M^{\prime}-1}}{\left(1-U_{ik}\right)^{M-1}}\geq\frac{\alpha^{M^{ \prime}-1}}{\left(1-\alpha\right)^{M-1}} \tag{25}\]  
Therefore, if Equation (23) is satisfied, then Equation (25) is also satisfied. Consider the function \(f(x)=\frac{x^{M^{\prime}-1}}{\left(1-x\right)^{M-1}}\) with \(x\in(0,1)\) we have its derivative  
\[=e^{\frac{1}{x-1}ln(\frac{1}{ax})}.\left(\frac{1}{(x-1)^{2}}(-1) ln(ax)+\left(\frac{1}{x-1}\right)\frac{1}{ax}\left(\frac{1}{ax}\right)^{ \prime}\right)\]0  
Since \(x^{M^{\prime}-2}>0\) and \(\left(1-x\right)^{M}>0\), hence, \(f^{\prime}(x)>0\), and therefore, \(f^{\prime}(x)\) is an increasing function. Equation (25) has the form of \(f(U_{ik})\geq f(\alpha)\), so then \(U_{ik}\geq\alpha\). \(\square\)  
From Proposition 3, Equation (23) demonstrates an approach to determine parameter \(M^{\prime}\). The right-hand side of (23) is a value that can be calculated knowing \(M\), \(\alpha\), and the distances, while the left-hand side of (23) is a decreasing function as \(M^{\prime}\) increases. From this, we can start at \(M^{\prime}=M\), and then increase \(M^{\prime}\) value and check simultaneously until Equation (23) is satisfied to solve for \(M^{\prime}\). Example: Given the data set from Table 1, the element \(X_{9}\) has \(U_{91}=0.189\), with \(M=2\). If we want to apply supervision such that \(U_{91}\geq 0.5\) to put \(X_{9}\) into cluster 1, then from Equation (20), we need to determine \(M^{\prime}\) such that the condition \(0.5^{M^{\prime}-1}M^{\prime}\leq 0.233\) is satisfied. For \(M^{\prime}=2\), \(0.5^{M^{\prime}-1}M^{\prime}=1\), and hence it does not satisfy the condition. We will have to increase the value of \(M^{\prime}\) to decrease the value of \(0.5^{M^{\prime}-1}M^{\prime}\) to satisfy the condition. Through certain iterations, we obtain \(M^{\prime}=5.582\), and \(0.5^{M^{\prime}-1}M^{\prime}=0.233\), which satisfies the condition, and effectively places \(X_{9}\) into cluster 1. A point that can be further discussed is that in Step 2 of the sSMC-FCM algorithm, it is necessary to solve Equation (19) to obtain \(\mu_{ik}\). In Equation (19), its right-hand side has a value in the range [0,1]. Set \(A=\sum_{j=1,j\
eq k}^{C}\mu_{ij}\), \(b=\frac{M^{\prime}-M}{M^{\prime}-1}\), with \(b<1\), the left-hand side of this equation has the form of \(f(x)=\frac{x}{\left(x+A\right)^{b}}\), which has the derivative of \(f^{\prime}(x)=\frac{\left(x+A\right)^{b}-xb(x+A)^{b-1}}{\left(x+A\right)^{2b}}= \frac{x+A-xb}{\left(x+A\right)^{b+1}}=\frac{\left(1-b\right)x+A}{\left(x+A \right)^{b+1}}>0\) for \(x>0\). From this, we can see that the left-hand side of Equation (19) is an increasing function, while the right-hand side has a value in the range [0,1]. Therefore, we can use an approximation method to obtain \(\mu_{ik}\), starting from \(\mu_{ik}=0\), and then increasing gradually to determine the solution.