We study the problem of computing a longest increasing subsequence in a sequence S of n distinct elements in the presence of persistent comparison errors. In this model, Braverman and Mossel (Noisy sorting without resampling, SODA 2008, pages 268–276, 2008) every comparison between two elements can return the wrong result with some fixed (small) probability p, and comparisons cannot be repeated. Computing the longest increasing subsequence exactly is impossible in this model, therefore, the objective is to identify a subsequence that (i) is indeed increasing and (ii) has a length that approximates the length of the longest increasing subsequence. We present asymptotically tight upper and lower bounds on both the approximation factor and the running time. In particular, we present an algorithm that computes an \(O(\log n)\)-approximation in \(O(n\log n)\) time, with high probability. This approximation relies on the fact that we can approximately sort (Geissmann et al. Optimal Sorting with Persistent Comparison Errors, ArXiv e-prints 1804.07575, 2018) n elements in \(O(n\log n)\) time such that the maximum dislocation of an element is \(O(\log n)\). For the lower bounds, we prove that (i) there is a set of sequences, such that on a sequence picked randomly from this set every algorithm must return an \({\Omega }(\log n)\)-approximation with high probability, and (ii) any \(\log n\)-approximation algorithm for longest increasing subsequence requires \({\Omega }(n \log n)\) comparisons, even in the absence of errors.

中文翻译：

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持续比较误差下最长的递增子序列

我们研究了在存在持续比较误差的情况下计算n个不同元素的序列S中最长的递增子序列的问题。在此模型中，Braverman和Mossel（无重采样的噪声排序，SODA 2008，第268-276页，2008年）两个元素之间的每次比较都可能以一定的固定（小）概率p返回错误结果。，并且无法重复比较。在此模型中，准确地计算出最长的增长子序列是不可能的，因此，目标是确定一个子序列，其中（i）确实在增长，并且（ii）的长度近似于最长的增长子序列的长度。我们给出了逼近因子和运行时间的渐近上界和下界。特别地，我们提出了一种算法，该算法以高概率计算\（O（n （log n）\））时间中的\（O（ logn ）\）逼近。该近似值依赖于以下事实：我们可以对元素中的n个元素进行近似排序（Geissmann等人，带有持续比较误差的最优排序，ArXiv电子版1804.07575，2018）。\（O（n \ log n）\）时间，以使元素的最大位错为\（O（\ log n）\）。对于下限，我们证明（i）有一个序列集，这样，从该集合中随机选择的序列上，每个算法都必须返回\（{\ Omega}（\ log n）\） - （ii）任何最长延长子序列的\（\ log n \） -逼近算法都需要进行\（{\ Omega}（n \ log n）\）比较，即使没有错误也是如此。