No-substitution k-means Clustering with Adversarial Order

• URL: http://arxiv.org/abs/2012.14512v1
• Date: Mon, 28 Dec 2020 22:35:44 GMT
• Title: No-substitution k-means Clustering with Adversarial Order
• Authors: Robi Bhattacharjee and Michal Moshkovitz
• Abstract summary: We introduce a new complexity measure that quantifies the difficulty of clustering a dataset arriving in arbitrary order. Our new algorithm takes only $textpoly(klog(n))$ centers and is a $textpoly(k)$-approximation.
• Score: 8.706058694927613
• License: http://creativecommons.org/licenses/by/4.0/
• Abstract: We investigate $k$-means clustering in the online no-substitution setting when the input arrives in \emph{arbitrary} order. In this setting, points arrive one after another, and the algorithm is required to instantly decide whether to take the current point as a center before observing the next point. Decisions are irrevocable. The goal is to minimize both the number of centers and the $k$-means cost. Previous works in this setting assume that the input's order is random, or that the input's aspect ratio is bounded. It is known that if the order is arbitrary and there is no assumption on the input, then any algorithm must take all points as centers. Moreover, assuming a bounded aspect ratio is too restrictive -- it does not include natural input generated from mixture models. We introduce a new complexity measure that quantifies the difficulty of clustering a dataset arriving in arbitrary order. We design a new random algorithm and prove that if applied on data with complexity $d$, the algorithm takes $O(d\log(n) k\log(k))$ centers and is an $O(k^3)$-approximation. We also prove that if the data is sampled from a ``natural" distribution, such as a mixture of $k$ Gaussians, then the new complexity measure is equal to $O(k^2\log(n))$. This implies that for data generated from those distributions, our new algorithm takes only $\text{poly}(k\log(n))$ centers and is a $\text{poly}(k)$-approximation. In terms of negative results, we prove that the number of centers needed to achieve an $\alpha$-approximation is at least $\Omega\left(\frac{d}{k\log(n\alpha)}\right)$.

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