The Ulam distance of two permutations on $[n]$ is $n$ minus the length of their longest common subsequence. In this paper, we show that for every $\varepsilon>0$, there exists some $\alpha>0$, and an infinite set $\Gamma\subseteq \mathbb{N}$, such that for all $n\in\Gamma$, there is an explicit set $C_n$ of $(n!)^{\alpha}$ many permutations on $[n]$, such that every pair of permutations in $C_n$ has pairwise Ulam distance at least $(1-\varepsilon)\cdot n$. Moreover, we can compute the $i^{\text{th}}$ permutation in $C_n$ in poly$(n)$ time and can also decode in poly$(n)$ time, a permutation $\pi$ on $[n]$ to its closest permutation $\pi^*$ in $C_n$, if the Ulam distance of $\pi$ and $\pi^*$ is less than $ \frac{(1-\varepsilon)\cdot n}{4} $. Previously, it was implicitly known by combining works of Goldreich and Wigderson [Israel Journal of Mathematics'23] and Farnoud, Skachek, and Milenkovic [IEEE Transactions on Information Theory'13] in a black-box manner, that it is possible to explicitly construct $(n!)^{\Omega(1)}$ many permutations on $[n]$, such that every pair of them have pairwise Ulam distance at least $\frac{n}{6}\cdot (1-\varepsilon)$, for any $\varepsilon>0$, and the bound on the distance can be improved to $\frac{n}{4}\cdot (1-\varepsilon)$ if the construction of Goldreich and Wigderson is directly analyzed in the Ulam metric.

翻译：两个在$[n]$上的排列的Ulam距离是指$n$减去它们最长公共子序列的长度。本文证明，对任意$\varepsilon>0$，存在某个$\alpha>0$以及无限集$\Gamma\subseteq \mathbb{N}$，使得对所有$n\in\Gamma$，存在一个显式集合$C_n$，包含$(n!)^{\alpha}$个$[n]$上的排列，且$C_n$中任意两个排列的Ulam距离至少为$(1-\varepsilon)\cdot n$。此外，我们可在poly$(n)$时间内计算$C_n$中的第$i$个排列，并在poly$(n)$时间内对任意$[n]$上的排列$\pi$解码为$C_n$中最近的排列$\pi^*$（当$\pi$与$\pi^*$的Ulam距离小于$ \frac{(1-\varepsilon)\cdot n}{4} $时）。此前，通过将Goldreich和Wigderson [Israel Journal of Mathematics'23] 与Farnoud、Skachek和Milenkovic [IEEE Transactions on Information Theory'13] 的工作以黑盒方式结合，已隐含可知：对任意$\varepsilon>0$，可显式构造$(n!)^{\Omega(1)}$个$[n]$上的排列，使得任意两两排列的Ulam距离至少为$\frac{n}{6}\cdot (1-\varepsilon)$；且若直接对Goldreich和Wigderson的构造在Ulam度量下进行分析，该距离界可改进至$\frac{n}{4}\cdot (1-\varepsilon)$。