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.

翻译：我们运用群论技术证明，在许多情况下，图上的计数问题在少量实例中的求解难度几乎与其在所有实例中的求解难度相当。具体而言，我们展示了以下结果。1. Goldreich (2020) 提出疑问：对于任意常数 $\delta < 1 / 2$，是否存在一个 $\tilde{O} \left( n^2 \right)$ 时间随机归约，能够将以大于 $2 / 3$ 的成功概率计算 $k$-团模 $2$ 数量的问题，归约至以至多 $\delta$ 的错误概率计算 $k$-团模 $2$ 数量的问题。在本工作中，我们证明，对于平均情况求解器中 $\delta 2^{n \choose 2}$ 个错误答案的几乎所有可能选择，我们都可以构造一个 $\tilde{O} \left( n^2 \right)$ 时间的归约，且该归约能够容忍平均情况求解器具有任意常数 $\delta < 1 / 2$ 的错误概率。所谓"几乎所有"，是指如果我们以等概率随机选择任意满足 $|S| = \delta2^{n \choose 2}$ 的子集 $S \subset \{0,1\}^{n \choose 2}$，那么以 $1-2^{-\Omega \left( n^2 \right)}$ 的概率，我们可以利用一个在 $S$ 上出错的平均情况求解器来得到一个概率算法。2. 受 Goldreich 和 Rothblum 在 FOCS 2018 中关于图计数问题加权版本工作的启发，我们证明：如果 RETH 成立，那么对于一个素数 $p = \Theta \left( 2^n \right)$，在 $n$ 顶点有向多重图上计算唯一哈密顿环模 $p$ 数量的问题，以及在 $n$ 顶点无向多重图上计算唯一半团模 $p$ 数量的问题，即使在 $1 / 2^{n/\log n}$ 比例的实例上正确计算也需要指数时间。与此同时，简单地在所有输入上输出 $0$，则至少在 $\Omega \left( 1 / 2^n \right)$ 比例的实例上是正确的。