The CAP theorem of Clustering: Why Every Algorithm Must Sacrifice Something

Re:The CAP theorem of Clustering: Why Every Algorithm Must S... by keithalewis on Friday December 27, 2024 @04:24AM (#42520945)

Let S = {1, 2}. Every distance function is determined by d(1,2) = a, a >= 0. Define f(d) = {{1,2}} if a = 0 and f(d) = {{1},{2}} otherwise. Isn't this a clustering algorithm that is scale invariant, rich, and consistent?

[ Parent | on HN ]

Re:The CAP theorem of Clustering: Why Every Algorithm Must S... by bubblyworld on Friday December 27, 2024 @04:33AM (#42520986)

Looks that way to me, yeah, though this is obviously a super simple case. It's clearly scale invariant and there are only two partitions, which your algorithm hits, so it's rich. Completeness is trivially satisfied in both cases too.

[ Parent | on HN ]

Re:The CAP theorem of Clustering: Why Every Algorithm Must S... by n0bra1n on Friday December 27, 2024 @05:36AM (#42521224)

i think i found the issue: the paper says distance function is 0 IFF elements are equal. so for this example, you can not define d(1,2) as equal to 0. so it is not rich, as this is the only way to get the partition {{1,2}}.

[ Parent | on HN ]

Re:The CAP theorem of Clustering: Why Every Algorithm Must S... by bubblyworld on Friday December 27, 2024 @05:50AM (#42521275)

Oh, I see, it's not a true metric. That's fair enough, though I wonder if the result depends critically on that assumption.

(you can pass to equivalence classes to recover a true metric, and I didn't see anything obviously incompatible with that in the paper, but I admit I didn't look very deeply)

[ Parent | on HN ]