dorsal/arxiv
View SchemaToken positional games
| Authors | Guillaume Bagan, Quentin Deschamps, Florian Galliot, Mirjana Mikalački, Nacim Oijid |
|---|---|
| Categories | |
| ArXiv ID | 2601.08967vv1 |
| URL | https://arxiv.org/abs/2601.08967 |
| License | http://arxiv.org/licenses/nonexclusive-distrib/1.0/ |
Abstract
The classical Maker-Breaker positional game is played on a board which is a hypergraph $\mathcal{H}$, with two players, Maker and Breaker, alternately claiming vertices of $\mathcal{H}$ until all the vertices are claimed. When the game ends, Maker wins if she has claimed all the vertices of some edge of $\mathcal{H}$; otherwise, Breaker wins. Playing this game in real life can be done by placing tokens on the vertices of the board. In this paper, we study the unfortunate case in which one or both players do not have enough tokens to cover all the vertices and, as such, will have to move their tokens around at some point instead of placing new ones. There may be a bias, in that Maker and Breaker do not necessarily have the same amount of tokens. The present paper initiates the study of this generalization of positional games, called token positional games. A particularly interesting case is when Maker has a winning strategy in the classical game: what is the lowest number of tokens with which she still wins against Breaker's unlimited stock? We notably show that, for $k$-uniform hypergraphs on an arbitrarily large number $n$ of vertices, this number equals $k$ if $k \in\{2,3\}$ but can vary from $k$ to $\Omega(n)$ if $k \geq 4$. From an algorithmic point of view, PSPACE-hardness in general is inherited from classical positional games, but we get a polynomial-time algorithm to solve the case where Breaker only has one token. We also establish EXPTIME-completeness for a "token sliding" variation of the game.
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"abstract": "The classical Maker-Breaker positional game is played on a board which is a hypergraph $\\mathcal{H}$, with two players, Maker and Breaker, alternately claiming vertices of $\\mathcal{H}$ until all the vertices are claimed. When the game ends, Maker wins if she has claimed all the vertices of some edge of $\\mathcal{H}$; otherwise, Breaker wins. Playing this game in real life can be done by placing tokens on the vertices of the board. In this paper, we study the unfortunate case in which one or both players do not have enough tokens to cover all the vertices and, as such, will have to move their tokens around at some point instead of placing new ones. There may be a bias, in that Maker and Breaker do not necessarily have the same amount of tokens. The present paper initiates the study of this generalization of positional games, called token positional games. A particularly interesting case is when Maker has a winning strategy in the classical game: what is the lowest number of tokens with which she still wins against Breaker\u0027s unlimited stock? We notably show that, for $k$-uniform hypergraphs on an arbitrarily large number $n$ of vertices, this number equals $k$ if $k \\in\\{2,3\\}$ but can vary from $k$ to $\\Omega(n)$ if $k \\geq 4$. From an algorithmic point of view, PSPACE-hardness in general is inherited from classical positional games, but we get a polynomial-time algorithm to solve the case where Breaker only has one token. We also establish EXPTIME-completeness for a \"token sliding\" variation of the game.",
"arxiv_id": "2601.08967",
"authors": [
"Guillaume Bagan",
"Quentin Deschamps",
"Florian Galliot",
"Mirjana Mikala\u010dki",
"Nacim Oijid"
],
"categories": [
"cs.DM",
"math.CO"
],
"license": "http://arxiv.org/licenses/nonexclusive-distrib/1.0/",
"title": "Token positional games",
"url": "https://arxiv.org/abs/2601.08967",
"version": "v1"
},
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"variant": "snapshot-2026-01-17",
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