Approximate Minimum Selection with Unreliable Comparisons in Optimal Expected Time

TL;DR

This paper introduces an optimal randomized algorithm for approximate minimum selection in unreliable comparison settings, achieving expected time complexity close to the fault-free case and establishing tight lower bounds.

Contribution

It presents the first asymptotically optimal randomized algorithm for approximate minimum selection with unreliable comparisons, matching lower bounds and handling a broad range of parameters.

Findings

Algorithm achieves expected time O((n/k) log(1/q))

Lower bounds show no algorithm can do better in expectation

Optimal worst-case time is achievable for certain k values

Abstract

We consider the \emph{approximate minimum selection} problem in presence of \emph{independent random comparison faults}. This problem asks to select one of the smallest $k$ elements in a linearly-ordered collection of $n$ elements by only performing \emph{unreliable} pairwise comparisons: whenever two elements are compared, there is a constant probability that the wrong answer is returned. We design a randomized algorithm that solves this problem with probability $1-q \in [ \frac{1}{2}, 1)$ and for the whole range of values of $k$ using $O( \frac{n}{k} \log \frac{1}{q} )$ expected time. Then, we prove that the expected running time of any algorithm that succeeds w.h.p. must be $\Omega(\frac{n}{k}\log \frac{1}{q})$, thus implying that our algorithm is asymptotically optimal, in expectation. These results are quite surprising in the sense that for $k$ between $\Omega(\log \frac{1}{q})$ and $c \cdot n$, for any constant $c<1$, the expected running time must still be $\Omega(\frac{n}{k}\log \frac{1}{q})$ even in absence of comparison faults. Informally speaking, we show how to deal with comparison errors without any substantial complexity penalty w.r.t.\ the fault-free case. Moreover, we prove that as soon as $k = O( \frac{n}{\log\log \frac{1}{q}})$, it is possible to achieve the optimal \emph{worst-case} running time of $\Theta(\frac{n}{k}\log \frac{1}{q})$.