## 3 Derivation of the Algorithm  
We consider the objective function (4) that should be minimised under the following constraints:  
\[\sum_{i=1}^{k}u_{ij}\ =\ 1\quad\quad\text{for all }j\in\{1,\ldots,n\} \tag{5}\]  
and  
\[u_{ij}\ \geq\ 0\quad\quad\text{for all }i\in\{1,\ldots,k\}\text{ and for all }j\in\{1,\ldots,n\} \tag{6}\]  
where \(u_{ij}\) indicates how much data point \(x_{j}\) is assigned to cluster \(i\). \(d_{ij}=\parallel x_{j}-v_{i}\parallel^{2}\) is the squared Euclidean distance between data point \(x_{j}\) and cluster centre \(v_{i}\). The second term in Equation (4) should reward non-crisp membership degrees, i.e., values greater than zero but smaller than one. For this purpose, we require that the function \(g:[0,1]\rightarrow\mathbb{R}\) be twice differentiable and that \(g^{\prime\prime}(x)>0\) holds. The parameter \(\alpha\) controls the extent to which membership degrees deviating from zero and one are rewarded. The corresponding Lagrange function for this optimisation problem is as follows:  
\[f_{\text{Lagrange}}\ =\ \sum_{i=1}^{k}\sum_{j=1}^{n}u_{ij}d_{ij}\ +\ \alpha\sum_{i=1}^{k}\sum_{j=1}^{n}g(u_{ij})\ +\ \sum_{j=1}^{n}\lambda_{j}\Bigg{(}1-\sum_{i=1}^{k}u_{ij} \Bigg{)}. \tag{7}\]  
The partial derivative with respect to \(u_{ij}\) is then  
\[\frac{\partial f_{\text{Lagrange}}}{\partial u_{ij}}\ =\ d_{ij}\ +\ \alpha\cdot g^{\prime}(u_{ij})\ -\ \lambda_{j}\ =\ 0 \tag{8}\]  
for all \(i\in\{1,\ldots,k\}\) and for all \(j\in\{1,\ldots,n\}\). This implies  
\[\lambda_{j}\ =\ d_{ij}\ +\ \alpha\cdot g^{\prime}(u_{ij})\ =\ d_{rj}\ +\ \alpha\cdot g^{\prime}(u_{rj}) \tag{9}\]  
for \(i,r\in\{1,\ldots,k\}\), \(i\
eq r\). For the special case of \(g^{\prime}(x)=x\), i.e., \(g(x)=\frac{x^{2}}{2}\), we obtain the following:  
\[u_{rj}\ =\ \frac{d_{ij}-d_{rj}}{\alpha}+u_{ij} \tag{10}\]  
implying  
\[1\ =\ \sum_{r=1}^{k}u_{rj}\ =\ \frac{k\cdot d_{ij}-\sum_{r=1}^{k}d_{rj}}{\alpha}+k \cdot u_{ij} \tag{11}\]  
leading to  
\[u_{ij}\ =\ \frac{1}{k}+\frac{\frac{1}{k}\Big{(}\sum_{r=1}^{k}d_{rj}\Big{)}-d_{ ij}}{\alpha}. \tag{12}\]  
Equation (12) can lead to negative values for \(u_{ij}\), violating constraint (6). One can either choose \(\alpha\) sufficiently large so that \(u_{ij}\geq 0\) is always guaranteed or one has to remove, step by step, the largest negative \(u_{ij}\) and simultaneously reduce the number of clusters, \(k\), in Equation (12), in a similar manner as in [26]. We now restrict the problem to one single cluster and a noise cluster, meaning that we have only two membership degrees for each data point, i.e., the membership degrees to the single cluster and the noise cluster. Such an approach can be used to identify a main cluster as in Section 5 or to apply a subtractive clustering strategy removing single clusters step by step [27; 28; 29]. Assuming a sufficiently large \(\alpha\), Equation (12) simplifies to the following:  
\[u_{j}\ =\ \frac{1}{2}+\frac{\delta-d_{j}}{2\alpha} \tag{13}\]  
where \(u_{j}\) is the membership degree of data point \(x_{j}\) to the single cluster and \(\delta\) is the (squared) noise distance. The membership degree of data point \(x_{j}\) to the noise cluster is \(1-u_{j}\). In the one-dimensional case, Equation (4) is as follows:\[f(x) = \sum_{j=1}^{n}\biggl{(}u_{j}(x_{j}-x)^{2}\ +\ (1-u_{j})\delta\ +\ \alpha \left(\frac{u_{j}^{2}}{2}+\frac{(1-u_{j})^{2}}{2}\right)\biggr{)} \tag{14}\] \[u_{ij}\ \geq\ 0\quad\quad\text{for all }i\in\{1,\ldots,k\}\text{ and for all }j\in\{1,\ldots,n\} \tag{6}\]0 \[u_{ij}\ \geq\ 0\quad\quad\text{for all }i\in\{1,\ldots,k\}\text{ and for all }j\in\{1,\ldots,n\} \tag{6}\]1 \[u_{ij}\ \geq\ 0\quad\quad\text{for all }i\in\{1,\ldots,k\}\text{ and for all }j\in\{1,\ldots,n\} \tag{6}\]2 \[u_{ij}\ \geq\ 0\quad\quad\text{for all }i\in\{1,\ldots,k\}\text{ and for all }j\in\{1,\ldots,n\} \tag{6}\]3  
i.e., it is a fourth-degree equation in the cluster centre, \(x\), so that the derivative is a cubic equation in \(x\).