带噪声的k-means++并非过于嘈杂
Noisy k-means++ is Not too Noisy
中文总结 AI 辅助
该研究解决了带噪声k-means++的近似保证问题，证明其可达到与经典k-means++相差1+O(ε)的近似比，同时揭示了逐点乘性控制的必要性及噪声线性依赖的必然性。
中文摘要 AI 辅助
Arthur和Vassilvitskii提出的经典k-means++算法（SODA 2007）采用D²采样技术，对经典k-means问题实现了O(log k)的期望近似比，该技术现已在聚类算法设计中普遍应用。Bhattacharya等人（ESA 2020）引入了ε-噪声k-means++，其采样概率可能存在对抗性乘性误差(1±ε)，但仅获得了O(log² k)的保证。Grunau等人（ESA 2023）恢复了渐近O(log k)的保证，但其分析中即使ε→0，也损失了约147638的常数因子，这留下了一个问题：k-means++是否对少量噪声高度敏感，是否可能得到与经典保证相差1+O(ε)范围内的界？我们正面解决了该问题，证明了期望近似保证为8(ln k+2)((1+ε)/(1-ε))⁴=(1+O(ε))8(ln k+2)。我们通过两个分离结果补充上界：第一，Arthur和Vassilvitskii下界实例的带噪声版本，相比精确k-means++会产生1+Ω(ε)的损失，因此对噪声的线性依赖是必要的；第二，逐点乘性控制在本质上是必不可少的：即使k=2，将其替换为每轮总变差接近度也无法获得有限近似保证。
英文摘要
The celebrated $k$-means++ algorithm of Arthur and Vassilvitskii (SODA 2007) achieves an $O(\log k)$ expected approximation for the classical $k$-means problem using $D^2$-sampling, a technique now ubiquitous in clustering algorithm design. Bhattacharya et al. (ESA 2020) introduced $\varepsilon$-noisy $k$-means++, where sampling probabilities may incur an adversarial multiplicative error of $(1\pm\varepsilon)$, but obtained only an $O(\log^2 k)$ guarantee. Grunau et al. (ESA 2023) recovered the asymptotic $O(\log k)$ guarantee, but their analysis loses a constant factor of roughly $147{,}638$ even as $\varepsilon\to0$, leaving open whether $k$-means++ is highly sensitive to even a small amount of noise. They asked whether a bound within $1+O(\varepsilon)$ of the classical guarantee is possible. We resolve this affirmatively, proving an expected approximation guarantee of $8(\ln k+2)\left(\frac{1+\varepsilon}{1-\varepsilon}\right)^4 = (1+O(\varepsilon))\,8(\ln k+2)$. We complement the upper bound with two separations. First, a noisy version of the Arthur and Vassilvitskii lower-bound instance incurs a $1+Ω(\varepsilon)$ loss over exact $k$-means++, so linear dependence on the noise is necessary. Second, pointwise multiplicative control is qualitatively essential: replacing it with per-round total variation closeness admits no finite approximation guarantee, even for $k=2$.
