基于对抗序的无替代k-means聚类
No-substitution k-means Clustering with Adversarial Order
摘要：
我们调查$k$ -均值聚类在线无替换设置时，输入到达\emph{arbitrary}顺序。在这种设置下，点一个接一个地到达，算法需要在观察下一个点之前立即决定是否以当前点为中心。决定是不可撤销的。目标是最小化中心数量和$k$均值成本。以前的作品在此设置假设输入的顺序是随机的，或者输入的长宽比是有界的。众所周知，如果阶数是任意的，并且对输入没有任何假设，那么任何算法都必须以所有点为中心。此外，假设有界纵横比限制太多--它不包括从混合模型生成的自然输入。我们引入了一个新的复杂性度量，量化了以任意顺序到达的数据集聚类的难度。我们设计了一个新的随机算法，并证明了，如果应用于数据的复杂性$d$，该算法采取$O(d\log(n) k\log(k))$中心，是一个$O(k^3)$ -近似。我们还证明，如果数据是从一个“自然”的分布，如混合$k$高斯采样，那么新的复杂性措施是等于$O(k^2\log(n))$。这意味着，对于从这些分布生成的数据，我们的新算法只需要$\text{poly}(k\log(n))$中心，是一个$\text{poly}(k)$ -近似。在否定的结果，我们证明，需要达到$α$ -近似的中心数至少是$Ω\left(\frac{d}{k\log(nα)}\right)$。
英文摘要：
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 $α$-approximation is at least $Ω\left(\frac{d}{k\log(nα)}\right)$.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
