在线性时间内将一串分解成平方

Yoshiaki Matsuoka, Shunsuke Inenaga, H. Bannai, M. Takeda, F. Manea
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引用次数: 5

摘要

字符串w的平方分解是每个因子都是平方的w的因数分解。Dumitran等人[SPIRE 2015, pp. 54-66]展示了如何在O(n log n)时间内找到给定长度为n的字符串的平方分解,他们提出了一个问题,即是否可以在O(n)时间内完成。在本文中,我们积极地回答了他们的问题,在机器字长omega = omega (log n)的标准字RAM模型中,我们展示了一个O(n)时间的平方分解算法。我们还展示了一个O(n + (n log^2 n) / omega)时间(分别,O(n log n)时间)的算法,以找到包含最大(分别,最小)平方数的平方分解。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Factorizing a String into Squares in Linear Time
A square factorization of a string w is a factorization of w in which each factor is a square. Dumitran et al. [SPIRE 2015, pp. 54-66] showed how to find a square factorization of a given string of length n in O(n log n) time, and they posed a question whether it can be done in O(n) time. In this paper, we answer their question positively, showing an O(n)-time algorithm for square factorization in the standard word RAM model with machine word size omega = Omega(log n). We also show an O(n + (n log^2 n) / omega)-time (respectively, O(n log n)-time) algorithm to find a square factorization which contains the maximum (respectively, minimum) number of squares.
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