arXivDaily arXiv每日学术速递 周一至周五更新
arXiv周末暂无论文更新,休息一下吧,周末愉快~~
arXiv 2609.28691cs.CC

确定性通信复杂度中的无损硬度凝聚

Lossless Hardness Condensation in Deterministic Communication Complexity

发表机构新南威尔士大学
查看机构详情
  • University of New South Wales (UNSW)(新南威尔士大学)

机构由 AI 辅助整理,请以论文原文为准。

Simon Mackenzie

首次发表
浏览论文内容

中文总结 AI 辅助

本文证明确定性通信复杂度为 c 的矩阵存在接近最大难度的子矩阵,肯定回答无损凝聚问题,并给出构造性算法。

中文摘要 AI 辅助

一个通信问题可能拥有比其通信代价所暗示的多得多的可能输入。其难度是否必然已经存在于一个更小的输入集合上?我们证明,每一个确定性通信复杂度为 $c\ge4$ 的有限全布尔矩阵,都包含一个子矩阵,该子矩阵由原始行中的 $2^k$ 行和原始列中的 $2^k$ 列组成,其中 $k=\Theta_\varepsilon(c)$,且其复杂度至少为 $(1-\varepsilon)(k+1)$,对于每个固定的 $0<\varepsilon<1$ 成立。由于 $k+1$ 是这种正方形上的最大可能代价,所保留的问题可以任意接近最大难度。这肯定地回答了 Hamed Hatami 的无损凝聚问题;Göös、Newman、Riazanov 和 Sokolov(STOC 2024)将其记录为开放问题 2,并猜想答案为否定。Hrubeš 此前保证了输入长度为 $\Omega(\sqrt c)$。同样的论证给出了一个原始的 $2^{c-2}$ 乘 $2^{c-2}$ 的正方形,保留了至少 $c/3-O(\log c)$ 比特的通信复杂度。该证明建立在 Hrubeš 的计数和覆盖论证之上。我们计算配备短通信协议的子矩阵:一个玩家命名一个覆盖子矩阵,然后玩家运行其协议。这避免了将矩形划分转换为协议时的损失。对矩形的递归使论证具有构造性。对于固定的有理数 $\varepsilon$ 和任何目标深度 $d\ge4$,一个确定性算法要么返回深度低于 $d$ 的协议,要么返回一个输入长度为 $\Theta_\varepsilon(d)$ 的原始输入正方形,并具有相同的近最大保证。其运行时间为 $2^{O(2^d)}$ 乘以表格大小的多项式。如果原始复杂度至少为 $d$,则该算法必然返回该正方形。一个扩展给出了任意固定数量的 number-in-hand 玩家的常数因子凝聚,其界限与有限输出字母表无关。

英文摘要

A communication problem can have far more possible inputs than its communication cost would suggest. Must its difficulty already be present on a much smaller set of inputs? We prove that every finite total Boolean matrix of deterministic communication complexity $c\ge4$ has a submatrix on $2^k$ of its original rows and $2^k$ of its original columns, with $k=Θ_\varepsilon(c)$ and complexity at least $(1-\varepsilon)(k+1)$, for every fixed $0<\varepsilon<1$. Since $k+1$ is the maximum possible cost on such a square, the retained problem can be arbitrarily close to maximally hard. This answers affirmatively the lossless condensation question of Hamed Hatami; Göös, Newman, Riazanov, and Sokolov (STOC 2024), who recorded it as Open Problem 2, conjectured a negative answer. Hrubeš previously guaranteed input length $Ω(\sqrt c)$. The same argument gives an original $2^{c-2}$-by-$2^{c-2}$ square retaining at least $c/3-O(\log c)$ bits of communication complexity. The proof builds on Hrubeš's counting and covering argument. We count submatrices equipped with short communication protocols: a player names a covering submatrix, then the players run its protocol. This avoids the loss from converting rectangle partitions into protocols. A recursion on rectangles makes the argument constructive. For fixed rational $\varepsilon$ and any target depth $d\ge4$, a deterministic algorithm returns either a protocol of depth below $d$, or a square of original inputs at input length $Θ_\varepsilon(d)$ with the same near-maximal guarantee. Its running time is $2^{O(2^d)}$ times a polynomial in the table size. If the original complexity is at least $d$, the algorithm necessarily returns the square. An extension gives constant-factor condensation for any fixed number of number-in-hand players, with bounds independent of the finite output alphabet.

补充信息

↑