k-子序列通用词的排名与取消排名
Ranking and Unranking k-subsequence universal words
AI总结：
本文针对固定长度字母表上的$k$-子序列通用词，提出了计算集合大小、排名、取消排名及延迟枚举的高效算法。
摘要：
词$w$的子序列是指满足$u = w[i_1] w[i_2] , \dots w[i_{|u|}]$的词$u$，其中索引集$1 \leq i_1 < i_2 < \dots < i_k \leq |w|$。若字母表$Σ$中的每个词均作为子序列出现在$w$中，则词$w$在字母表$Σ$上是$k$-子序列通用的。本文针对字母表$Σ= \{1,2,\dots, σ\}$上固定长度为$n$的$k$-子序列通用词提供了新算法。令$\mathcal{U}(n,k,σ)$表示$Σ$上长度为$n$的$k$-子序列通用词集合，我们提供：一个$O(n k σ)$时间算法用于计算$\mathcal{U}(n,k,σ)$的大小；一个$O(n k σ)$时间算法用于对集合$\mathcal{U}(n,k,σ)$中的词进行排名；一个$O(n k σ)$时间算法用于从集合$\mathcal{U}(n,k,σ)$中取消排名词；一个在$O(n k σ)$预处理后以$O(n σ)$延迟枚举集合$\mathcal{U}(n,k,σ)$的算法。
英文摘要：
A subsequence of a word $w$ is a word $u$ such that $u = w[i_1] w[i_2] , \dots w[i_{|u|}]$, for some set of indices $1 \leq i_1 < i_2 < \dots < i_k \leq |w|$. A word $w$ is $k$-subsequence universal over an alphabet $Σ$ if every word in $Σ^k$ appears in $w$ as a subsequence. In this paper, we provide new algorithms for $k$-subsequence universal words of fixed length $n$ over the alphabet $Σ= \{1,2,\dots, σ\}$. Letting $\mathcal{U}(n,k,σ)$ denote the set of $n$-length $k$-subsequence universal words over $Σ$, we provide: * an $O(n k σ)$ time algorithm for counting the size of $\mathcal{U}(n,k,σ)$; * an $O(n k σ)$ time algorithm for ranking words in the set $\mathcal{U}(n,k,σ)$; * an $O(n k σ)$ time algorithm for unranking words from the set $\mathcal{U}(n,k,σ)$; * an algorithm for enumerating the set $\mathcal{U}(n,k,σ)$ with $O(n σ)$ delay after $O(n k σ)$ preprocessing.
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。
