For each integer n≥ 3, consider the unbiased Maker–Breaker game on the edge set of K_n: in each round Maker claims one previously unclaimed edge, and then Breaker claims q previously unclaimed edges. Maker wins if the edges he has claimed contain a triangle K_3, while Breaker wins otherwise. Determine the threshold bias q^ast(n)=minq∈N:Breaker has a winning strategy in this game exactly.

Maker–Breaker Triangle Game

The standard Maker–Breaker triangle game asks for the threshold at which Maker can force a triangle on $K_n$ when Breaker claims $q$ edges per turn. Its exact leading constant remains unknown.

Known results

Chvátal and Erdős: Maker wins for $q<\sqrt{2n+2}-5/2$ and Breaker wins for $q\ge 2\sqrt n$ .
Balogh and Samotij improved Breaker’s upper-bound constant to approximately $1.935\sqrt n$ .
Glazik and Srivastav, 2018: Breaker wins for sufficiently large $n$ when $q\ge\sqrt{(8/3+o(1))n}$ , approximately $1.633\sqrt n$ .

September 2026 budget-variant result

The preprint Maker Breaker Games on a Budget by Sebastian Lüderssen, Fabien Nießen, and Silas Rathke reports an exact threshold for a natural budget variant and a sharp asymptotic threshold for a restricted strategy. This is progress on a variant, not a solution of the standard game, and the claim is unverified.

Current status (as of September 2026): the standard Maker–Breaker triangle game remains open with only constant-factor threshold bounds, while an exact budget-variant result is claimed in an unrefereed preprint.

Maker–Breaker Triangle Game

Primary source

Additional references

Progress summary

Solutions 0