Solving Random Planted CSPs below the $n^{k/2}$ Threshold

TL;DR

This paper introduces algorithms for solving random planted Boolean CSPs that operate below the $n^{k/2}$ constraint threshold, generalizing previous methods and matching known trade-offs.

Contribution

It presents a new family of algorithms that recover planted solutions in random CSPs with improved runtime and constraint bounds, extending prior work to larger runtimes.

Findings

Algorithm succeeds with high probability when constraints are above a certain threshold.

The approach combines Sum-of-Squares SDP with a novel rounding procedure.

Matches the known trade-off between constraints and runtime for planted CSPs.

Abstract

We present a family of algorithms to solve random planted instances of any $k$-ary Boolean constraint satisfaction problem (CSP). A randomly planted instance of a Boolean CSP is generated by (1) choosing an arbitrary planted assignment $x^*$, and then (2) sampling constraints from a particular "planting distribution" designed so that $x^*$ will satisfy every constraint. Given an $n$ variable instance of a $k$-ary Boolean CSP with $m$ constraints, our algorithm runs in time $n^{O(\ell)}$ for a choice of a parameter $\ell$, and succeeds in outputting a satisfying assignment if $m \geq O(n) \cdot (n/\ell)^{\frac{k}{2} - 1} \log n$. This generalizes the $\mathrm{poly}(n)$-time algorithm of [FPV15], the case of $\ell = O(1)$, to larger runtimes, and matches the constraint number vs.\ runtime trade-off established for refuting random CSPs by [RRS17]. Our algorithm is conceptually different from the recent algorithm of [GHKM23], which gave a $\mathrm{poly}(n)$-time algorithm to solve semirandom CSPs with $m \geq \tilde{O}(n^{\frac{k}{2}})$ constraints by exploiting conditions that allow a basic SDP to recover the planted assignment $x^*$ exactly. Instead, we forego certificates of uniqueness and recover $x^*$ in two steps: we first use a degree-$O(\ell)$ Sum-of-Squares SDP to find some $\hat{x}$ that is $o(1)$-close to $x^*$, and then we use a second rounding procedure to recover $x^*$ from $\hat{x}$.