Longest Increasing Subsequence under Persistent Comparison Errors

TL;DR

This paper investigates the challenge of approximating the longest increasing subsequence in sequences where comparison errors are persistent and cannot be repeated, providing tight bounds and an efficient approximation algorithm.

Contribution

It introduces an $O(rac{ ext{log} n}{n})$-approximation algorithm with optimal time complexity and establishes fundamental lower bounds for approximation and comparison requirements.

Findings

An $O( ext{log} n)$-approximation algorithm runs in $O(n ext{log} n)$ time.

Any $O( ext{log} n)$-approximation requires $ ext{Omega}(n ext{log} n)$ comparisons.

Approximate sorting with maximum dislocation $O( ext{log} n)$ is achievable in $O(n ext{log} n)$ time.

Abstract

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, 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 time $O(n\log n)$, with high probability. This approximation relies on the fact that that we can approximately sort $n$ elements in $O(n\log n)$ time such that the maximum dislocation of an element is at most $O(\log n)$. 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 $O(\log n)$-approximation algorithm for longest increasing subsequence requires $\Omega(n \log n)$ comparisons, even in the absence of errors.