Constant-Cost Communication is not Reducible to k-Hamming Distance

Yuting Fang, Mika Göös, Nathaniel Harms, Pooya Hatami
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Abstract

Every known communication problem whose randomized communication cost is constant (independent of the input size) can be reduced to $k$-Hamming Distance, that is, solved with a constant number of deterministic queries to some $k$-Hamming Distance oracle. We exhibit the first examples of constant-cost problems which cannot be reduced to $k$-Hamming Distance. To prove this separation, we relate it to a natural coding-theoretic question. For $f : \{2, 4, 6\} \to \mathbb{N}$, we say an encoding function $E : \{0, 1\}^n \to \{0, 1\}^m$ is an $f$-code if it transforms Hamming distances according to $\mathrm{dist}(E(x), E(y)) = f(\mathrm{dist}(x, y))$ whenever $f$ is defined. We prove that, if there exist $f$-codes for infinitely many $n$, then $f$ must be affine: $f(4) = (f(2) + f(6))/2$.
恒定成本通信与 k-Hamming 距离无关
每一个随机通信成本恒定(与输入大小无关)的已知通信问题,都可以简化为 $k$-汉明距离,也就是说,只需对 $k$-Hamming Distance 甲骨文进行恒定次数的确定性查询即可求解。我们首次举例说明了不能简化为 $k$-Hamming 距离的实体成本问题。为了证明这种分离,我们把它与一个自然的编码理论问题联系起来。对于 $f :\{2, 4, 6\}\我们说一个编码函数 $E:\{0, 1\}^n \to \{0, 1\}^m$ 是一个 $f$ 代码,如果它在 $f$ 被定义的情况下按照 $\mathrm{dist}(E(x), E(y)) = f(\mathrm{dist}(x, y))$ 转换汉明距离的话。我们证明,如果存在无限多 $n$ 的 $f$ 代码,那么 $f$ 一定是仿射的:$f(4) = (f(2) + f(6))/2$。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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