Hardness Amplification via Group Theory

TL;DR

This paper uses group theory techniques to demonstrate that certain graph counting problems are nearly as hard in average-case scenarios as in worst-case, establishing new hardness amplification results and implications for complexity assumptions.

Contribution

It introduces novel reductions and hardness amplification techniques for graph counting problems based on group theory, extending prior work and connecting to the RETH conjecture.

Findings

Almost all corrupt answer sets enable efficient average-case reductions.

Counting Hamiltonian cycles modulo a prime is exponentially hard on most instances.

Simple algorithms like printing zero are correct on a significant fraction of instances.

Abstract

We employ techniques from group theory to show that, in many cases, counting problems on graphs are almost as hard to solve in a small number of instances as they are in all instances. Specifically, we show the following results. 1. Goldreich (2020) asks if, for every constant $\delta < 1 / 2$, there is an $\tilde{O} \left( n^2 \right)$-time randomized reduction from computing the number of $k$-cliques modulo $2$ with a success probability of greater than $2 / 3$ to computing the number of $k$-cliques modulo $2$ with an error probability of at most $\delta$. In this work, we show that for almost all choices of the $\delta 2^{n \choose 2}$ corrupt answers within the average-case solver, we have a reduction taking $\tilde{O} \left( n^2 \right)$-time and tolerating an error probability of $\delta$ in the average-case solver for any constant $\delta < 1 / 2$. By "almost all", we mean that if we choose, with equal probability, any subset $S \subset \{0,1\}^{n \choose 2}$ with $|S| = \delta2^{n \choose 2}$, then with a probability of $1-2^{-\Omega \left( n^2 \right)}$, we can use an average-case solver corrupt on $S$ to obtain a probabilistic algorithm. 2. Inspired by the work of Goldreich and Rothblum in FOCS 2018 to take the weighted versions of the graph counting problems, we prove that if the RETH is true, then for a prime $p = \Theta \left( 2^n \right)$, the problem of counting the number of unique Hamiltonian cycles modulo $p$ on $n$-vertex directed multigraphs and the problem of counting the number of unique half-cliques modulo $p$ on $n$-vertex undirected multigraphs, both require exponential time to compute correctly on even a $1 / 2^{n/\log n}$-fraction of instances. Meanwhile, simply printing $0$ on all inputs is correct on at least a $\Omega \left( 1 / 2^n \right)$-fraction of instances.