## 2 Preliminaries
### Determination of the Fuzzification Coefficients for Supervised Elements  
In the proposed sSMC-FCM algorithm, when an element \(i\) is to be supervised, \(U_{ij}\) will be calculated according to Equations (17)-(20) in Step 2 of the algorithm. In this sub-section, we will discuss these formulas in more detail, as well as how to determine the exponential parameter \(M^{\prime}\) for the supervised elements. **Proposition 1**. : \(U_{ij}\) _calculated from Equations (17)-(20) satisfies Equation (16)._  
Proof of Proposition 1.: From Equation (20), taking the sum of \(U_{ij}\), we can see that it satisfies the condition of \(\sum_{j=1}^{C}U_{ij}=1\) in Equation (16). \(\square\)  
Next, we calculate \(M^{\prime}U_{ij}{}^{M^{\prime}-1}D_{ij}{}^{2}\) by gradually replacing from the formulas (20), (19), (18), (17), as follows:  
\[ \begin{split} M^{\prime}U_{ik}{}^{M^{\prime}-1}D_{ik}{}^{2}& =M^{\prime}\bigg{(}\frac{\mu_{ik}}{\sum_{i=1}^{C}\mu_{i}}\bigg{)}^{M^{ \prime}-1}(d_{ik}d_{min})^{2}\\ =M^{\prime}\Bigg{(}\frac{\left(\frac{1}{M^{\prime}d_{ik}{}^{2}} \right)^{\frac{1}{M^{\prime}-1}}\left(\mu_{ik}+\sum_{j=1,j,\mu_{j}}^{C}\mu_{ij }\right)^{\frac{M^{\prime}-M}{M^{\prime}-1}}\bigg{)}^{M^{\prime}-1}}{\sum_{i=1 }^{C}\mu_{i}}\\ =M^{\prime}\Bigg{(}\frac{\left(\frac{1}{M^{\prime}d_{ik}{}^{2}} \right)^{\frac{1}{M^{\prime}-1}}\left(\sum_{j=1}^{C}\mu_{ij}\right)^{\frac{M^ {\prime}-M}{M^{\prime}-1}}}{\sum_{i=1}^{C}\mu_{i}}\Bigg{)}^{M^{\prime}-1}{d_{ ik}}^{2}d_{min}{}^{2}\\ =M^{\prime}\frac{\frac{1}{M^{\prime}d_{ik}{}^{2}}\left(\sum_{j=1 }^{C}\mu_{ij}\right)^{M^{\prime}-M}}{\left(\sum_{j=1}^{C}\mu_{i}\right)^{M^{ \prime}-1}}{d_{ik}}^{2}d_{min}{}^{2}=\frac{d_{min}{}^{2}}{\left(\sum_{j=1}^{C} \mu_{i}\right)^{M^{\prime}-1}}\end{split}  \tag{21}\]  
Similarly, calculate \(MU_{ij}{}^{M-1}D_{ij}{}^{2}\) for all \(j\
eq k\)  
\[ \begin{split} MU_{ij}{}^{M-1}D_{ij}{}^{2}&=M\bigg{(} \frac{\mu_{ij}}{\sum_{i=1}^{C}\mu_{i}}\bigg{)}^{M-1}(d_{ij}d_{min})^{2}\\ =M\Bigg{(}\frac{\left(\frac{1}{Md_{ij}{}^{2}}\right)^{\frac{1}{ M-1}}}{\sum_{i=1}^{C}\mu_{i}}\Bigg{)}^{M-1}{d_{ij}}^{2}d_{min}{}^{2}=\frac{d_{ min}{}^{2}}{\left(\sum_{j=1}^{C}\mu_{j}\right)^{M^{\prime}-1}}\end{split}  \tag{22}\]  
From (21) and (22), it can be seen that \(M^{\prime}U_{ik}{}^{M^{\prime}-1}D_{ik}{}^{2}=MU_{i1}{}^{M-1}D_{i1}{}^{2}=\ldots =MU_{ij}{}^{M-1}D_{ij}{}^{2}=\ldots=MU_{ic}{}^{M-1}D_{ic}{}^{2}\) for all \(j\
eq k\). Therefore, combining with \(\sum_{j=1}^{C}U_{ij}=1\), Equation (16) is satisfied, and Proposition 1 is proven. We can hence apply Equations (17)-(20) in Step 2 of the proposed algorithm. Next, we investigate how to determine the exponential parameter \(M^{\prime}\). We will utilize Proposition 2 after proving it. **Proposition 2**. : _The function \(f(x)=\left(\frac{1}{xx}\right)^{\frac{1}{x-1}}\) with \(a\geq 1\) is an increasing function when \(x>1\)._  
**Proof of Proposition 2. ** Consider the function \(f(x)=\left(\frac{1}{ax}\right)^{\frac{1}{x-1}}=e^{\ln((\frac{1}{ax})^{\frac{1}{x-1 }})}=e^{\frac{1}{x-1}ln(\frac{1}{ax})}\), Calculate its derivative:  
\[f^{\prime}(x)=e^{\frac{1}{x-1}ln(\frac{1}{ax})}.\left(\left(\frac{ 1}{x-1}\right)^{\prime}ln\left(\frac{1}{ax}\right)+\left(\frac{1}{x-1}\right) ln\left(\frac{1}{ax}\right)^{\prime}\right)\] \[=e^{\frac{1}{x-1}ln(\frac{1}{ax})}.\left(\frac{1}{(x-1)^{2}}(-1) ln(ax)+\left(\frac{1}{x-1}\right)\frac{1}{ax}\left(\frac{1}{ax}\right)^{ \prime}\right)\] \[=e^{\frac{1}{x-1}ln(\frac{1}{ax})}.\left(\frac{ln(ax)}{(x-1)^{2}} -\frac{1}{x(x-1)}\right)=e^{\frac{1}{x-1}ln(\frac{1}{ax})},\frac{\lambda ln(ax )-x+1}{x(x-1)^{2}}\]  
For \(x>1\), \(a\geq 1\), in \(f^{\prime}(x)\), we have \(e^{\frac{1}{x-1}ln(\frac{1}{ax})}>0\), \(xln(ax)-x+1\geq xlnx-x+1\), and \(x(x-1)^{2}>0\). For \(g(x)=xlnx-x+1\), we have \(g^{\prime}(x)=lnx+\frac{x}{x}-1=lnx>0\) for all \(x>1\). Therefore, \(g(x)\) is an increasing function in the range \([1,+\infty)\), and thus, \(g(x)>g(1)=0\), and hence, \(xln(ax)-x+1\geq xlnx-x+1>0\). Overall, all the components of \(f^{\prime}(x)\) are greater than 0.