String Sanitization Under Edit Distance

G. Bernardini, Huiping Chen, G. Loukides, N. Pisanti, S. Pissis, L. Stougie, Michelle Sweering
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引用次数: 10

Abstract

textabstractLet W be a string of length n over an alphabet Σ, k be a positive integer, and be a set of length-k substrings of W. The ETFS problem asks us to construct a string X_{ED} such that: (i) no string of occurs in X_{ED}; (ii) the order of all other length-k substrings over Σ is the same in W and in X_{ED}; and (iii) X_{ED} has minimal edit distance to W. When W represents an individual’s data and represents a set of confidential substrings, algorithms solving ETFS can be applied for utility-preserving string sanitization [Bernardini et al., ECML PKDD 2019]. Our first result here is an algorithm to solve ETFS in (kn²) time, which improves on the state of the art [Bernardini et al., arXiv 2019] by a factor of |Σ|. Our algorithm is based on a non-trivial modification of the classic dynamic programming algorithm for computing the edit distance between two strings. Notably, we also show that ETFS cannot be solved in (n^{2-δ}) time, for any δ>0, unless the strong exponential time hypothesis is false. To achieve this, we reduce the edit distance problem, which is known to admit the same conditional lower bound [Bringmann and Kunnemann, FOCS 2015], to ETFS.
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编辑距离下的字符串处理
设W是字母表Σ上长度为n的字符串,k是一个正整数,并且是W的长度为k的子字符串的集合。ETFS问题要求我们构造一个字符串X_{ED},满足:(i)在X_{ED}中不存在字符串;(ii) Σ上所有其他长度为k的子串的顺序在W和X_{ED}中是相同的;(iii) X_{ED}与W的编辑距离最小。当W代表个人数据并代表一组机密子字符串时,求解ETFS的算法可以应用于保持效用的字符串处理[Bernardini等人,ECML PKDD 2019]。我们在这里的第一个结果是在(kn²)时间内求解ETFS的算法,该算法比目前的技术水平(Bernardini et al., arXiv 2019)提高了一个因子|Σ|。我们的算法是基于经典动态规划算法的一个重要修改,用于计算两个字符串之间的编辑距离。值得注意的是,我们还表明,对于任何δ>0,除非强指数时间假设为假,否则ETFS不能在(n^{2-δ})时间内求解。为了实现这一点,我们减少了编辑距离问题,该问题承认相同的条件下界[Bringmann和Kunnemann, FOCS 2015], ETFS。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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