Faster Algorithms for the Constrained k-means Problem

April 10, 2015 Β· Declared Dead Β· πŸ› Theory of Computing Systems

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Authors Anup Bhattacharya, Ragesh Jaiswal, Amit Kumar arXiv ID 1504.02564 Category cs.DS: Data Structures & Algorithms Citations 56 Venue Theory of Computing Systems Last Checked 3 months ago
Abstract
The classical center based clustering problems such as $k$-means/median/center assume that the optimal clusters satisfy the locality property that the points in the same cluster are close to each other. A number of clustering problems arise in machine learning where the optimal clusters do not follow such a locality property. Consider a variant of the $k$-means problem that may be regarded as a general version of such problems. Here, the optimal clusters $O_1, ..., O_k$ are an arbitrary partition of the dataset and the goal is to output $k$-centers $c_1, ..., c_k$ such that the objective function $\sum_{i=1}^{k} \sum_{x \in O_{i}} ||x - c_{i}||^2$ is minimized. It is not difficult to argue that any algorithm (without knowing the optimal clusters) that outputs a single set of $k$ centers, will not behave well as far as optimizing the above objective function is concerned. However, this does not rule out the existence of algorithms that output a list of such $k$ centers such that at least one of these $k$ centers behaves well. Given an error parameter $\varepsilon > 0$, let $\ell$ denote the size of the smallest list of $k$-centers such that at least one of the $k$-centers gives a $(1+\varepsilon)$ approximation w.r.t. the objective function above. In this paper, we show an upper bound on $\ell$ by giving a randomized algorithm that outputs a list of $2^{\tilde{O}(k/\varepsilon)}$ $k$-centers. We also give a closely matching lower bound of $2^{\tildeΞ©(k/\sqrt{\varepsilon})}$. Moreover, our algorithm runs in time $O \left(n d \cdot 2^{\tilde{O}(k/\varepsilon)} \right)$. This is a significant improvement over the previous result of Ding and Xu who gave an algorithm with running time $O \left(n d \cdot (\log{n})^{k} \cdot 2^{poly(k/\varepsilon)} \right)$ and output a list of size $O \left((\log{n})^k \cdot 2^{poly(k/\varepsilon)} \right)$.
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