The multiplayer promise set disjointness is one of the most widely used problems from communication complexity in applications. In this problem there are $k$ players with subsets $S^1, \ldots, S^k$, each drawn from $\{1, 2, \ldots, n\}$, and we are promised that either the sets are (1) pairwise disjoint, or (2) there is a unique element $j$ occurring in all the sets, which are otherwise pairwise disjoint. The total communication of solving this problem with constant probability in the blackboard model is $\Omega(n/k)$. We observe for most applications, it instead suffices to look at what we call the ``mostly'' set disjointness problem, which changes case (2) to say there is a unique element $j$ occurring in at least half of the sets, and the sets are otherwise disjoint. This change gives us a much simpler proof of an $\Omega(n/k)$ randomized total communication lower bound, avoiding Hellinger distance and Poincare inequalities. Using this we show several new results for data streams: \begin{itemize} \item for $\ell_2$-Heavy Hitters, any $O(1)$-pass streaming algorithm in the insertion-only model for detecting if an $\eps$-$\ell_2$-heavy hitter exists requires $\min(\frac{1}{\eps^2}\log \frac{\eps^2n}{\delta}, \frac{1}{\eps}n^{1/2})$ bits of memory, which is optimal up to a $\log n$ factor. For deterministic algorithms and constant $\eps$, this gives an $\Omega(n^{1/2})$ lower bound, improving the prior $\Omega(\log n)$ lower bound. We also obtain lower bounds for Zipfian distributions. \item for $\ell_p$-Estimation, $p > 2$, we show an $O(1)$-pass $\Omega(n^{1-2/p} \log(1/\delta))$ bit lower bound for outputting an $O(1)$-approximation with probability $1-\delta$, in the insertion-only model. This is optimal, and the best previous lower bound was $\Omega(n^{1-2/p} + \log(1/\delta))$. \end{itemize}
翻译:多个玩家承诺的脱节是应用中最常用的通信复杂度问题之一。 在这个问题中,有美元玩家的子集$S%1,\ldots,Sk$,每个子集来自$1,2,\ldots,n ⁇,我们得到的保证是,这些组合是(1)双向脱节,或者(2)在所有组合中都出现一个独特的要素$j美元,否则就是对齐的脱节。在黑板模型中,用恒定的概率来解决这个问题的沟通总量是$(Omega(n/k)美元。我们在大多数应用程序中观察的是美元玩家$“最接近的脱节问题 ”,这样就足够看看我们所谓的“最接近的脱节问题 ” ($2,\ldolototos) 美元。这个游戏让我们更简单地证明$(n/nf) 美元的总交流的亮度, 避免地狱距离和波尔的不平等性。如果我们在数据流中显示一些新的结果:O_xx