The Leader-Russell-Walters conjecture: is every Euclidean Ramsey set subtransitive?
A finite set in Euclidean space is Ramsey if for every number of colors there is a dimension such that every -coloring of contains a monochromatic congruent copy of it. A finite set is transitive if its isometry group acts transitively on its points, and subtransitive if it embeds isometrically in a finite transitive set. Leader, Russell and Walters (2012, Conjecture A) proposed that a finite set is Ramsey exactly when it is subtransitive. In an earlier paper they proved that the cyclic kites with transcendental are not subtransitive and conjectured that they are not Ramsey (Conjecture 3). Is every Ramsey set subtransitive, and in particular are these kites non-Ramsey?
- Result
- Disproved(see note)
- Status
- Candidate (review pending)
- AI contribution
- AI-discovered
- Method
- Argument
- Field
- Euclidean Ramsey theory
- Posed by
- Imre Leader, Paul A. Russell and Mark Walters: Transitive sets and cyclic quadrilaterals, J. Comb. 2 (2011), Conjecture 3; Transitive sets in Euclidean Ramsey theory, JCTA 119 (2012), Conjecture A
- Year posed
- 2011
- Years open
- 15y
- Solved
- 2026-09-23
- Model
- Unreleased internal OpenAI model
- Vendor
- OpenAI
- Collaborators
- —
- Verification
- Unreviewed
- Publication
- Announced
- Collection
- OpenAI math release (October 2026), version adc7f12
- Significance
- 28 / 100
- Disclosed cost
- —
- Wikipedia
- No dedicated article
What was actually shown
The manuscript proves that every nonempty set of at most five points on a circle is Ramsey, so every cyclic quadrilateral is Ramsey; in particular the kites with transcendental are Ramsey. With Leader, Russell and Walters' theorem that these kites are not subtransitive, this disproves the necessity direction of Conjecture A and disproves Conjecture 3. The sufficiency direction (subtransitive implies Ramsey) is proved in the same paper, so the conjecture fails in one direction only. The argument depends on the classification theorem of the same manuscript and on the cited non-subtransitivity result, which is not reproved.
What the AI did
The OpenAI math release (github.com/openai/math, commit adc7f12) states that its results were produced by an unreleased internal OpenAI model under one fixed procedure, averaging about three hours of ChatGPT Pro thinking compute per result, across roughly 4,000 posed problems; outputs were then grouped into families and filtered for significance. This result is not among the README's stated exceptions (the Riemann zeta zero-free region work and the Hodge conjecture for CM abelian varieties). The manuscript is credited to OpenAI alone and names no human author. The disproof is a corollary of the classification theorem in the same manuscript (separate entry).
Verification
No independent mathematician has checked this yet. Checked here: Theorem 1.1 and the consequences section (the five-circle-point corollary and the kite corollary) of the TeX source; the proof was not refereed. Lean: the challenge EuclideanRamseyCircle (OAI.EuclideanRamsey.CircleConsequence.at_most_five_circle_points_ramsey, solution module OAI/Combinatorics/EuclideanRamsey/Circle.lean) is not in the release's formalization catalogue; its JSON and solution file exist at the pinned commit. Statement read here: every injective family of one to five points of equidistant from some center is Ramsey. That is the Ramsey half of the disproof only. The non-subtransitivity of the kites is imported from Leader, Russell and Walters (2011, Corollary 2) and is not formalized, so no Lean statement asserts the disproof itself. Not rebuilt here. Listed as Unreviewed rather than Lean-checked because its formal statement covers only the Ramsey half of the disproof.