Musical chairs

TL;DR

This paper analyzes the 'Musical Chairs' game, establishing tight bounds on the number of chairs needed for the team to guarantee termination and proposing strategies based on topological methods.

Contribution

It proves that for at least 2n-1 chairs, the team has a winning strategy, and for fewer, the scheduler can prevent termination indefinitely, with tight bounds demonstrated.

Findings

Team wins if m ≥ 2n-1 chairs

Scheduler wins if m ≤ 2n-2 chairs

Topological methods used to prove bounds

Abstract

In the {\em Musical Chairs} game $MC(n,m)$ a team of $n$ players plays against an adversarial {\em scheduler}. The scheduler wins if the game proceeds indefinitely, while termination after a finite number of rounds is declared a win of the team. At each round of the game each player {\em occupies} one of the $m$ available {\em chairs}. Termination (and a win of the team) is declared as soon as each player occupies a unique chair. Two players that simultaneously occupy the same chair are said to be {\em in conflict}. In other words, termination (and a win for the team) is reached as soon as there are no conflicts. The only means of communication throughout the game is this: At every round of the game, the scheduler selects an arbitrary nonempty set of players who are currently in conflict, and notifies each of them separately that it must move. A player who is thus notified changes its chair according to its deterministic program. As we show, for $m\ge 2n-1$ chairs the team has a winning strategy. Moreover, using topological arguments we show that this bound is tight. For $m\leq 2n-2$ the scheduler has a strategy that is guaranteed to make the game continue indefinitely and thus win. We also have some results on additional interesting questions. For example, if $m \ge 2n-1$ (so that the team can win), how quickly can they achieve victory?