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Hardness Amplification via Group Theory

Tejas Nareddy, Abhishek Mishra

TL;DR

This work shows that for almost all choices of the $\delta 2^{n \choose 2}$ corrupt answers within the average-case solver, there is 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 $\delta<1 / 2$.

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 $δ< 1 / 2$, 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 $δ$. In this work, we show that for almost all choices of the $δ2^{n \choose 2}$ 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 $δ$ in the average-case solver for any constant $δ< 1 / 2$. 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 $1-2^{-Ω\left( n^2 \right)}$, 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 $p = Θ\left( 2^n \right)$, 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 $Ω\left( 1 / 2^n \right)$-fraction of instances.

Hardness Amplification via Group Theory

TL;DR

This work shows that for almost all choices of the corrupt answers within the average-case solver, there is a reduction taking -time and tolerating an error probability of in the average-case solver for any constant .

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 , there is an -time randomized reduction from computing the number of -cliques modulo with a success probability of greater than to computing the number of -cliques modulo with an error probability of at most . In this work, we show that for almost all choices of the corrupt answers within the average-case solver, we have a reduction taking -time and tolerating an error probability of in the average-case solver for any constant . By "almost all", we mean that if we choose, with equal probability, any subset with , then with a probability of , we can use an average-case solver corrupt on 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 , the problem of counting the number of unique Hamiltonian cycles modulo on -vertex directed multigraphs and the problem of counting the number of unique half-cliques modulo on -vertex undirected multigraphs, both require exponential time to compute correctly on even a -fraction of instances. Meanwhile, simply printing on all inputs is correct on at least a -fraction of instances.

Paper Structure

This paper contains 47 sections, 41 theorems, 122 equations, 10 algorithms.

Key Result

Theorem 1

For any $k \in \mathbb{N}$ (not necessarily a constant), given an $\epsilon = \omega \left( n^{3/2} / \sqrt{n!} \right)$, given an oracle $O$ sampled from $O^{H_n}_{1 / 2 + \epsilon}$, where $H_n: \{\, 0,1 \,\}^{n \choose 2} \to \mathbb{D}$ is any function defined over $n$-vertex undirected simple g

Theorems & Definitions (107)

  • Definition 1
  • Theorem 1
  • Theorem 2
  • Theorem 3
  • Theorem 4
  • Lemma 1
  • Lemma 2
  • Definition 2
  • Definition 3
  • Definition 4
  • ...and 97 more