重新审视k-映射性问题
The k-mappability problem revisited
摘要：
$k$可映射性问题具有两个整数参数$m$和$k$。对于文本$S$中大小为$m$的每个子字，我们希望报告$S$中出现最多$k$不匹配单词的索引数。这个问题最近由Alzamel等人解决。对于具有恒定字母表$Σ$和$k \in O(1)$的文本，他们提出了一种具有线性空间和$O(n\log^{k+1}n)$时间的算法。对于$k = 1$和恒定大小字母表的情况，Alzamel等人在2020年的一篇论文中提出了一种具有线性空间和$O(n\log(n)\log\log(n))$时间的更快算法。的2020年论文，以获得用于$k = 1$的具有线性空间和$O(n \log(n))$时间的算法。我们的算法消除了约束的字母是恒定的大小。我们还提出了线性算法的情况下$k=1$，$|Σ|\in O(1)$和$m=Ω(\sqrt{n})$。
英文摘要：
The $k$-mappability problem has two integers parameters $m$ and $k$. For every subword of size $m$ in a text $S$, we wish to report the number of indices in $S$ in which the word occurs with at most $k$ mismatches. The problem was lately tackled by Alzamel et al. For a text with constant alphabet $Σ$ and $k \in O(1)$, they present an algorithm with linear space and $O(n\log^{k+1}n)$ time. For the case in which $k = 1$ and a constant size alphabet, a faster algorithm with linear space and $O(n\log(n)\log\log(n))$ time was presented in a 2020 paper by Alzamel et al. In this work, we enhance the techniques of Alzamel et al.'s 2020 paper to obtain an algorithm with linear space and $O(n \log(n))$ time for $k = 1$. Our algorithm removes the constraint of the alphabet being of constant size. We also present linear algorithms for the case of $k=1$, $|Σ|\in O(1)$ and $m=Ω(\sqrt{n})$.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
