有效地找到所有最大$α$缺口重复序列
Efficiently Finding All Maximal $α$-gapped Repeats
摘要：
对于$α\geq 1$，单词$w$中的$α$ -缺口重复序列是$w$的因子$uvu$，使得$|uv|\leq α|u|$ ;在这样的重复序列中的两个因子$u$被称为臂，而因子$v$被称为缺口.这样的重复被称为最大的，如果它的武器不能同时延长相同的符号，以权利或，分别向左。在本文中，我们表明，可能出现在一个字的最大$α$缺口重复的数量是上界的$18αn$。这允许我们构建一种算法，找到$O(αn)$中一个单词的所有最大$α$ -缺口重复;在最坏的情况下，这是最优的，因为存在具有$Θ(αn)$最大$α$ -缺口重复的单词.我们的技术可以被扩展以在$α$ -缺口回文的情况下得到可比较的结果，即，因子$uvu^\mathrm{T}$与$|uv|\leq α|u|$。
英文摘要：
For $α\geq 1$, an $α$-gapped repeat in a word $w$ is a factor $uvu$ of $w$ such that $|uv|\leq α|u|$; the two factors $u$ in such a repeat are called arms, while the factor $v$ is called gap. Such a repeat is called maximal if its arms cannot be extended simultaneously with the same symbol to the right or, respectively, to the left. In this paper we show that the number of maximal $α$-gapped repeats that may occur in a word is upper bounded by $18αn$. This allows us to construct an algorithm finding all the maximal $α$-gapped repeats of a word in $O(αn)$; this is optimal, in the worst case, as there are words that have $Θ(αn)$ maximal $α$-gapped repeats. Our techniques can be extended to get comparable results in the case of $α$-gapped palindromes, i.e., factors $uvu^\mathrm{T}$ with $|uv|\leq α|u|$.
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。
