The Prisoners and the Swap: Less than Half is Enough

TL;DR

This paper improves the classical prisoners and drawers puzzle by demonstrating that with a single swap, prisoners can find their numbers opening significantly fewer drawers than previously possible, approaching a theoretical lower bound.

Contribution

It introduces a new optimal strategy for the puzzle that minimizes the number of drawers each prisoner must open, and provides an explicit efficient method to implement this strategy.

Findings

All prisoners can find their numbers by opening approximately n log log n / log n drawers.

A lower bound shows no strategy can do better than twice this amount.

An explicit efficient strategy matching the optimal asymptotic bound is constructed.

Abstract

We improve the solution of the classical prisoners and drawers riddle, where all prisoners can find their number using the pointer-following strategy, provided that the prisoners can send a spy to inspect all drawers and swap one pair of numbers. In the traditional approach, each prisoner may need to open up to half of the drawers. We show that this strategy is sub-optimal. Remarkably, a single swap allows all $n$ prisoners to find their number by opening only $\frac{n \ln \ln n}{\ln n} (1 + o(1))$ drawers in the worst case. We show that no strategy can do better than that by a factor larger than two. Efficiently constructing such a strategy is harder, but we provide an explicit efficient strategy that requires opening only $O(\frac{n \log \log n}{\log n})$ drawers by each prisoner in the worst case.