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Symmetric Perceptrons, Number Partitioning and Lattices (2501.16517v1)

Published 27 Jan 2025 in math.ST, cs.CC, math-ph, math.MP, math.PR, and stat.TH

Abstract: The symmetric binary perceptron ($\mathrm{SBP}{\kappa}$) problem with parameter $\kappa : \mathbb{R}{\geq1} \to [0,1]$ is an average-case search problem defined as follows: given a random Gaussian matrix $\mathbf{A} \sim \mathcal{N}(0,1){n \times m}$ as input where $m \geq n$, output a vector $\mathbf{x} \in {-1,1}m$ such that $$|| \mathbf{A} \mathbf{x} ||{\infty} \leq \kappa(m/n) \cdot \sqrt{m}~.$$ The number partitioning problem ($\mathrm{NPP}{\kappa}$) corresponds to the special case of setting $n=1$. There is considerable evidence that both problems exhibit large computational-statistical gaps. In this work, we show (nearly) tight average-case hardness for these problems, assuming the worst-case hardness of standard approximate shortest vector problems on lattices. For $\mathrm{SBP}$, for large $n$, the best that efficient algorithms have been able to achieve is $\kappa(x) = \Theta(1/\sqrt{x})$ (Bansal and Spencer, Random Structures and Algorithms 2020), which is a far cry from the statistical bound. The problem has been extensively studied in the TCS and statistics communities, and Gamarnik, Kizildag, Perkins and Xu (FOCS 2022) conjecture that Bansal-Spencer is tight: namely, $\kappa(x) = \widetilde{\Theta}(1/\sqrt{x})$ is the optimal value achieved by computationally efficient algorithms. We prove their conjecture assuming the worst-case hardness of approximating the shortest vector problem on lattices. For $\mathrm{NPP}$, Karmarkar and Karp's classical differencing algorithm achieves $\kappa(m) = 2{-O(\log2 m)}~.$ We prove that Karmarkar-Karp is nearly tight: namely, no polynomial-time algorithm can achieve $\kappa(m) = 2{-\Omega(\log3 m)}$, once again assuming the worst-case subexponential hardness of approximating the shortest vector problem on lattices to within a subexponential factor.

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