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K-means clustering uses the sum of squared errors (SSE)

$E = \sum\limits_{i=1}^k \sum\limits_{p \in C_i} (p-m_i)^2$ (with k clusters, C the set of objects in a cluster, m the center point of a cluster)

after each iteration to check if SSE is decreasing, until reaching the local minimum/optimum. The benefit of k-medoid is "It is more robust, because it minimizes a sum of dissimilarities instead of a sum of squared Euclidean distances". Though understanding that further distance of a cluster increases the SSE, I still don't understand why it is needed for k-means but not for k-medoids.

p.s. 1. The original paper on k-means would probably explain such a matter, but how to find? Couldn't find on google scholar, or where else to search? 2. This is my first post, happy for any feedback. Should I go more into detail on k-means before a question (how other do), thought it wouldn't be necessary because if someone doesn't know what it is, a small introduction won't really help.

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  • $\begingroup$ Because mean is the locus of minimal sum of squared deviations from it. So "find mean" and "minimize SSE" are almost equivalent expressions. Minimizing absolute deviations won't give you mean (geometric centroid) and finding mean won't minimize absolute deviations. See also. $\endgroup$
    – ttnphns
    Commented Apr 22, 2015 at 14:21
  • $\begingroup$ (Thank you) After reading some threads on "mean is locus of sum of squared deviations" it makes sense. Because the sum of absolute errors will be much closer to the median then the mean. The square of SSE weights outliers much more like the mean would do, while the median ignores outliers. My question would be, using the SSE is just a better approximation? Because e.g. the sum of absolute errors (SAE) gets effected by outliers, though the actual median wouldn't get affected by outliers. $\endgroup$
    – dominic
    Commented Apr 23, 2015 at 4:54

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