Erd\H{o}s-Szekeres Maker-Breaker Games

TL;DR

This paper investigates Maker-Breaker games inspired by the Erd ext{o}s-Szekeres problem, establishing winning strategies for Maker in various biased and monochromatic/bichromatic settings, and resolving open questions about the game's dynamics.

Contribution

It provides new results on Maker-Breaker games related to convex polygons, proving Maker's winning strategies under different biases and settings, and settles an open problem from prior research.

Findings

Maker always wins in the monochromatic game.

Maker wins in the bichromatic game if Breaker places fewer than 2 points per round.

Breaker can win when she plays 12 or more points per round with k ≥ 8.

Abstract

We present new results on Maker-Breaker games arising from the Erd\H{o}s-Szekeres problem in planar geometry. This classical problem asks how large a set in general position has to be to ensure the existence of $n$ points that are the vertices of a convex $n$-gon. Moreover, Erd\H{o}s further extended this problem by asking what happens if we also require that this $n$-gon has an empty interior. In a 2-player Maker-Breaker setting, this problem inspires two main games. In both games, Maker tries to obtain an empty convex $k$-gon, while Breaker tries to prevent her from doing so. The games differ only in which points can comprise the winning $k$-gons: in the monochromatic version the points of both players can make up a $k$-gon, while in the bichromatic version only Maker's points contribute to such a polygon. Both settings are studied in this paper. We show that in the monochromatic game, Maker always wins. Even in a biased game where Breaker is allowed to place $s$ points per round, for any constant $s \geq 1$, Maker has a winning strategy. In the bichromatic setting, Maker still wins whenever Breaker is allowed to place $s$ points per round for any constant $s<2$. This settles an open problem posed by Aichholzer et al. (2019). Furthermore, we show that there are games that are not a lost cause for Breaker. Whenever $k\ge 8$ and Breaker is allowed to play 12 or more points per round, she has a winning strategy. We also consider the one-round bichromatic game (a.k.a.\ the offline version). In this setting, we show that Breaker wins if she can place twice as many points as Maker but if the bias is less than $2$, then Maker wins for large enough set of points.