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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 cluster
by keithalewis 2y ago
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?
- bubblyworld 2y agoLooks 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.
- n0bra1n 2y agoi 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}}.
- bubblyworld 2y agoOh, 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)