An analysis of a bounded resource search puzzle

TL;DR

This paper analyzes the optimal strategies for a classic bounded resource search puzzle involving glass balls and floors, providing new bounds and algorithms for the general case with multiple balls.

Contribution

It introduces a new analysis and bounds for the general case of the puzzle, extending previous work limited to specific cases like two balls.

Findings

For k ≥ m, the minimum attempts m = log(n+1).

For k < m, bounds are given by combinatorial sums.

Provides an alternative algorithm with improved efficiency.

Abstract

Consider the commonly known puzzle, given $k$ glass balls, find an optimal algorithm to determine the lowest floor of a building of $n$ floors from which a thrown glass ball will break. This puzzle was originally posed in its original form in \cite{focs1980}and was later cited in the book \cite{algthc}. There are several internet sites that presents this puzzle and its solution to the special case of $k=2$ balls. This is the first such analysis of the puzzle in its general form. Several variations of this puzzle have been studied with applications in Network Loading \cite{cgstctl} which analyzes a case similar to a scenario where an adversary is changing the lowest floor with time. Although the algorithm specified in \cite{algthc} solves the problem, it is not an efficient algorithm. In this paper another algorithm for the same problem is analyzed. It is shown that if $m$ is the minimum number of attempts required then for $k \geq m$ we have $m = \log (n+1)$ and for $k < m$ we have, $1 + \sum_{i=1}^{k}{{m-1}\choose{i}} < n \leq \sum_{i=1}^{k}{{m}\choose{i}}$