Algorithm 2. K-medoids algorithm.
Let Δ(x, y) denote the distance between two elements x, y ∈ Σ.
1.	Choose an arbitrary partition k = {Σk : k = 1,…, K} of Σ and an arbitrary set of medoids Λ = {m(Σk) ∈ Σk : k = 1,…, K}.
2.	For every Σk ∈ K, compute the elements that are wrongly classified, i.e., x ∈ Σk satisfying Δ(x, m(Σj)) < Δ(x, m(Σk)) for Σj ≠ Σk. For these elements, update the partition as follows:
	${\sum }_k^{\prime }={\sum }_k-\left\{\mathbf{\text{x}}\right\}$ and ${\sum }_j^{\prime }={\sum }_j+\left\{\mathbf{\text{x}}\right\}$. The resultant partition is ${\mathcal{\text{P}}}_K^{\prime }=\left\{{\sum }_k^{\prime }:k=1,\dots ,K\right\}$.
3.	Obtain the medoids set ${\wedge }^{\prime }=\left\{\text{m}\left({\sum }_k^{\prime }\right):{\sum }_k^{\prime }\in {\mathcal{\text{P}}}_K^{\prime }\right\}$ for the new partition ${\mathcal{\text{P}}}_K^{\prime }$ by solving the following K optimization programs:
	$m\left({\sum }_k^{\prime }\right)\in argmin\left\{max\left\{{\gamma }_k:\mathrm{\Delta }\left(\mathbf{\text{x}},\mathbf{\text{y}}\right)\le {\gamma }_k,\mathbf{\text{y}}\in {\sum }_k^{\prime }\right\}:\mathbf{\text{x}}\in {\sum }_k^{\prime }\right\},{\sum }_k^{\prime }\in {\mathcal{\text{P}}}_K^{\prime }$
4.	If the medoids set does not change, i.e., if Λ′ = Λ, then the clustering process is completed. Otherwise, do Λ = Λ′, and go to Step 2.