Erdős-Szekeres Maker-Breaker Games

TL;DR

The paper investigates Erdős-Szekeres-inspired Maker-Breaker games in the plane, focusing on empty convex $k$-holes. It introduces three main strategies: (i) constructing $k$-strips that force $k$-holes in the monochromatic setting, (ii) a Breaker tactic using perturbed polygons to block large holes in the bichromatic setting, and (iii) a Density Hales–Jewett–based approach for a one-round bichromatic game, enabling Maker to win near the critical bias of $2$. The results establish that Maker wins the monochromatic game for all $k$ (even under any fixed constant Breaker bias), wins the bichromatic game when Breaker’s bias is below $2$, and Breaker can win in certain large-bias regimes (e.g., $k\ge 8$ with bias $1:12$ or higher). The paper also provides detailed analyses of round counts under various biases and offers open questions about optimal bias thresholds and the efficiency of Maker’s strategies. Overall, the work advances our understanding of geometric Ramsey-type games and bridges combinatorial geometry with positional game strategies, with implications for extremal configurations in the plane and related computational questions.

Abstract

We present new results on Maker-Breaker games arising from the Erdős-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, Erdős 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.

Figures (12)

• Figure 1: Example of a 4-cap, a 5-cup and a 5-strip with respect to $\vec{d}=\overrightarrow{(0,-1)}$.
• Figure 2: A 4-cap $S_i=\{s_{i,1},s_{i,2},s_{i,3},s_{i,4}\}$ and 4-strip $\mathrm{conv}(S_i)+\vec{d}$. Cones $C_i^-$ and $C_i^+$ are colored in pink. The half-lines $L(s_{i,1})$ and $L(s_{i,k-1})$, resp., are rotated clockwise and counterclockwise until they pass through a point in the point set or until they hit $L_i^-$ or $L_i^+$.
• Figure 3: Placing a point $p_j$ on the halfline $\vec{\ell}_j$, Maker creates two candidates for possible $k$-strips.
• Figure 4: Example of a group of $4$-strips for $s=1$. The point $p_j$ creates two possible $k$-strips.
• Figure 5: If Breaker has a bias $s\geq 2$ in the bichromatic game, Maker's strategy to create $k$-strips no longer works. Breaker can place one point left and right of $p_j$.

Theorems & Definitions (20)

• Definition 1
• Definition 2
• Theorem 3
• proof
• Theorem 4
• proof
• Theorem 5
• proof : Proof of \ref{['thm:s2']}
• Proposition 7
• proof