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arXiv 1406.7017math.CO
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词集合中最长的共同连续性

Longest common subsequences in sets of words

Boris Bukh, Jie Ma

摘要:

给定一个$t$字母表上的一组长度为$n$的$t$字,证明了对于某些依赖于$k$和$t$的$c>0$,其中两个字之间存在一个长度至少为$\frac{n}{k}+cn^{1-1/(t-k-2)}$的公共子序列.这是急剧上升到值$c$。

英文摘要:

Given a set of $t$ words of length $n$ over a $k$-letter alphabet, it is proved that there exists a common subsequence among two of them of length at least $\frac{n}{k}+cn^{1-1/(t-k-2)}$, for some $c>0$ depending on $k$ and $t$. This is sharp up to the value of $c$.

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。

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