Local Enumeration: The Not-All-Equal Case
Mohit Gurumukhani, Ramamohan Paturi, Michael Saks, Navid Talebanfard
TL;DR
This work reframes SSETH-breaking potential by focusing on the Not-All-Equal local enumeration problem $NAE$-Enum$(k,t)$. It refines the transversal-tree based TreeSearch framework to show that solving $NAE$-Enum$(3,n/2)$ optimally suffices to break SSETH, achieving an expected time of $poly(n)\cdot 6^{n/4}$. The results tie algorithmic bounds to hypergraph Turán problems and depth-3 majority complexity, providing tight combinatorial and probabilistic analyses and interpreting the bound as a potential route to stronger circuit lower bounds. The work also outlines a path for extending the approach to larger $k$ and discusses the broader implications for exact exponential-time algorithms for SAT-like problems.
Abstract
Gurumukhani et al. (CCC'24) proposed the local enumeration problem Enum(k, t) as an approach to break the Super Strong Exponential Time Hypothesis (SSETH): for a natural number $k$ and a parameter $t$, given an $n$-variate $k$-CNF with no satisfying assignment of Hamming weight less than $t(n)$, enumerate all satisfying assignments of Hamming weight exactly $t(n)$. Furthermore, they gave a randomized algorithm for Enum(k, t) and employed new ideas to analyze the first non-trivial case, namely $k = 3$. In particular, they solved Enum(3, n/2) in expected $1.598^n$ time. A simple construction shows a lower bound of $6^{\frac{n}{4}} \approx 1.565^n$. In this paper, we show that to break SSETH, it is sufficient to consider a simpler local enumeration problem NAE-Enum(k, t): for a natural number $k$ and a parameter $t$, given an $n$-variate $k$-CNF with no satisfying assignment of Hamming weight less than $t(n)$, enumerate all Not-All-Equal (NAE) solutions of Hamming weight exactly $t(n)$, i.e., those that satisfy and falsify some literal in every clause. We refine the algorithm of Gurumukhani et al. and show that it optimally solves NAE-Enum(3, n/2), namely, in expected time $poly(n) \cdot 6^{\frac{n}{4}}$.
