在次指数时间内强反驳半随机线性系统

中文摘要 AI 辅助

本文考虑反驳具有随机右侧的 $\mathbb{F}_2$-线性方程组的问题。形式上，我们给出一个次指数 $2^{O(n/\log n)}$ 时间的随机化算法，该算法以任意 $m \times n$ 矩阵 $A$ 和均匀随机向量 $b \in \mathbb{F}_2^m$ 作为输入，并输出一个见证，表明当 $m \geq 2^{O(n/\log n)}$ 时，没有任何赋值满足超过 $\frac{1}{2}+\epsilon$ 比例的方程。上述设置是著名工作 [BKW03] 的半随机反驳变体，该工作为带噪声的学习奇偶性（LPN）问题给出了 $2^{O(n/\log n)}$ 时间的搜索算法，其中 $m \geq 2^{O(n/\log n)}$ 个方程。基于 [Lyub05] 的搜索算法，我们还给出了一个 $2^{O(n/\log \log n)}$ 时间的反驳算法，该算法仅需 $m \geq n^{1 + \gamma}$ 个方程即可成功，其中 $\gamma$ 是一个小常数。最后，我们通过证明一个度数为 $\Omega(n)$ 的平方和（sum-of-squares）下界，证明我们的算法不被平方和层级所涵盖，表明“[BKW03]式”算法能实现比平方和框架下更好的运行时间。因此，我们获得了一个自然示例，展示了一个容忍噪声的信号恢复问题，在高效算法与基于平方和层级的算法之间存在非平凡的差距。

英文摘要

In this paper, we consider the problem of refuting $\mathbb{F}_2$-linear equations with random right-hand sides. Formally, we give a sub-exponential $2^{O(n/\log n)}$-time randomized algorithm that takes as input an arbitrary $m \times n$ matrix $A$ and a uniformly random vector $b \in \mathbb{F}_2^m$, and outputs a witness showing that no assignment satisfies more than a $\frac{1}{2}+ε$ fraction of the equations provided that $m \geq 2^{O(n/\log n)}$. The setting above is the semirandom refutation variant of the famous work [BKW03] that gives a $2^{O(n/\log n)}$-time search algorithm for the learning parity with noise (LPN) problem with $m \geq 2^{O(n/\log n)}$ equations. Building on the search algorithm of [Lyub05], we also give a $2^{O(n/\log \log n)}$-time refutation algorithm that succeeds with only $m \geq n^{1 + γ}$ equations, for a small constant $γ$. Finally, we prove that our algorithm is not captured by the sum-of-squares hierarchy by proving a degree-$Ω(n)$ sum-of-squares lower bound, showing that ''[BKW03]-style'' algorithms achieve better runtime than can be done under sum-of-squares. We thus obtain a natural example of a noise-tolerant signal recovery problem that exhibits a nontrivial gap between the performance of efficient algorithms and that of those based on the sum-of-squares hierarchy.

发表机构

• Princeton University（普林斯顿大学）
• The Institute for Advanced Study（高等研究院）

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