Sorting is a fundamental problem in computer science. In the classical setting, it is well-known that $(1\pm o(1)) n\log_2 n$ comparisons are both necessary and sufficient to sort a list of $n$ elements. In this paper, we study the Noisy Sorting problem, where each comparison result is flipped independently with probability $p$ for some fixed $p\in (0, \frac 12)$. As our main result, we show that $$(1\pm o(1)) \left( \frac{1}{I(p)} + \frac{1}{(1-2p) \log_2 \left(\frac{1-p}p\right)} \right) n\log_2 n$$ noisy comparisons are both necessary and sufficient to sort $n$ elements with error probability $o(1)$ using noisy comparisons, where $I(p)=1 + p\log_2 p+(1-p)\log_2 (1-p)$ is capacity of BSC channel with crossover probability $p$. This simultaneously improves the previous best lower and upper bounds (Wang, Ghaddar and Wang, ISIT 2022) for this problem. For the related Noisy Binary Search problem, we show that $$ (1\pm o(1)) \left((1-\delta)\frac{\log_2(n)}{I(p)} + \frac{2 \log_2 \left(\frac 1\delta\right)}{(1-2p)\log_2\left(\frac {1-p}p\right)}\right) $$ noisy comparisons are both necessary and sufficient to find the predecessor of an element among $n$ sorted elements with error probability $\delta$. This extends the previous bounds of (Burnashev and Zigangirov, 1974), which are only tight for $\delta = 1/n^{o(1)}$.

翻译：排序是计算机科学中的一个基本问题。在经典设定中，众所周知，对 $n$ 个元素进行排序需要且仅需 $(1\pm o(1)) n\log_2 n$ 次比较。本文研究有噪排序问题，其中每次比较结果以固定概率 $p\in (0, \frac 12)$ 独立翻转。我们的主要结果表明，使用有噪比较以 $o(1)$ 的错误概率对 $n$ 个元素进行排序，需要且仅需 $$(1\pm o(1)) \left( \frac{1}{I(p)} + \frac{1}{(1-2p) \log_2 \left(\frac{1-p}p\right)} \right) n\log_2 n$$ 次有噪比较，其中 $I(p)=1 + p\log_2 p+(1-p)\log_2 (1-p)$ 是交叉概率为 $p$ 的二进制对称信道的信道容量。这同时改进了该问题之前的最优下界和上界（Wang, Ghaddar and Wang, ISIT 2022）。对于相关的有噪二分查找问题，我们证明，在 $n$ 个有序元素中寻找某个元素的前驱，若允许错误概率为 $\delta$，则需要且仅需 $$ (1\pm o(1)) \left((1-\delta)\frac{\log_2(n)}{I(p)} + \frac{2 \log_2 \left(\frac 1\delta\right)}{(1-2p)\log_2\left(\frac {1-p}p\right)}\right) $$ 次有噪比较。这扩展了（Burnashev and Zigangirov, 1974）之前的界，该界仅在 $\delta = 1/n^{o(1)}$ 时紧。