Episode 172, October 2026

Euclidean Ramsey: classified?

A preprint claims an exact algebraic test for which finite point patterns every finite colouring of high-dimensional space must contain, at their original size

Covers result 172 in OpenAI’s catalog: its one paper. See its connections to other results and the work it cites.

Play Episode 172: Euclidean Ramsey: classified? A kite inscribed in a circle, over rings of colour: a pattern the paper shows is Ramsey though it is part of no symmetric set (our drawing).

6:23. English captions available. The video loads from YouTube when you press play.

Chapters

The claim, in plain words

“A finite nonempty set A in a Euclidean space is Ramsey if, for every integer r ≥ 2, there is an integer D ≥ 1 such that every map c: ℝ^D → {1,…,r} is constant on a congruent copy of A. Congruence means preservation of every pairwise distance, at the original scale.” Classically, two points at distance 1 are Ramsey, and three points on a line, one apart, are not: colour each point by the whole-number part of its squared distance from the origin, modulo 4, and no such triple is ever one colour. As the paper recounts, Erdős, Graham, Montgomery, Rothschild, Spencer and Straus (1973) proved that every Ramsey set lies on a sphere, and Graham conjectured the converse. Frankl and Rödl proved the property for triangles and all simplices, Kříž for sets with a soluble transitive symmetry group and for cyclic trapezoids, and Cantwell for regular polytopes. Leader, Russell and Walters proposed that a set is Ramsey exactly when it is part of a finite transitive set, one that looks the same from each of its points. On September 20, 2026, Pálvölgyi posted a seven-point circle pattern that is not Ramsey, “disproving the spherical conjecture”; the paper notes that he attributes the writing, proofs and ideas of that manuscript to ChatGPT.

Result 172 is a single OpenAI preprint, A classification of finite Euclidean Ramsey configurations (26 pages, dated September 23, 2026). Theorem 1.1: for a set of points a_i spanning ℝ^d, with p_i = (1, a_i), coordinate field F and B = F ⊗_ℚ F, the set is Ramsey if and only if some matrix P with entries in B satisfies (p_i ⊗ 1)^T P (1 ⊗ p_i) = 0 for every point and, after multiplying the two sides, has the identity as its spatial block. Multiplying turns the certificate into an ordinary sphere equation; “the tensor condition requires the evaluation identities to hold before multiplication. This is the distinction between sphericity and the Ramsey property that the classification detects.” Consequences: “every nonempty subset of a finite transitive Euclidean set is Ramsey” (Pálvölgyi had announced this, with the write-up still being prepared); “every nonempty set of at most five distinct points on a circle is Ramsey”; for transcendental a the kite K_a is Ramsey, while Leader, Russell and Walters had proved it is not part of any finite transitive set, so “the necessity direction of the Leader–Russell–Walters subtransitive characterization fails”; and nine generic circle points, and twelve points made of three squares turned by a transcendental angle (twice the arctangent of Liouville’s constant), are not Ramsey.

What it doesn’t claim

The proof gives no bound on the dimension needed. The test “uses exact relations in the coordinate field. It is not a procedure for recovering those relations from numerical approximations to arbitrary real coordinates.” The colourings that avoid a non-Ramsey pattern come from linear functionals over the rationals (“these are purely algebraic functionals; no continuity or measurability is asserted or needed”), with at most (2s+1)^s colours for s points in every dimension. The paper credits Pálvölgyi with the first refutation of Graham’s conjecture, and notes that his generic seven-point claim already implies its nine-point example. It says nothing about colouring the plane itself.

Where it stands

Posted as a preprint: OpenAI, 26 pages, dated September 23, 2026
Peer reviewed: Not yet
Checked in Lean: The classification and six consequences, formalized by OpenAI; not run by us

Last checked October 19, 2026. The preprint has not been revised since it was added to the repository in its initial commit on October 6, 2026. OpenAI published a Lean formalization with seven comparator statements: the classification, cosphericity, subsets of transitive sets, a quadratic-independence criterion, at most five circle points, and the nine- and twelve-point examples (the last stated with an explicit 50-colour avoiding colouring in every dimension, which the paper does not state). By our count the proofs take 65 files and 9,589 lines excluding Mathlib (the main development), with no unproved steps by our search; we have not rebuilt them. The kite corollary also relies on Leader, Russell and Walters’s 2011 theorem that the kites are not subtransitive, which is not formalized. Several works it cites are 2026 preprints, though its proofs do not depend on them. The paper is not among those OSVerify reviewed, and we have seen no expert review.

How the proof goes

The test on a kite (our calculation). Write the kite’s points with a parameter, one copy T on the left and one copy U on the right. The plain circle equation leaves −2(T − U)²/((1 + T²)(1 + U²)) when left and right are kept apart; reweighting its coefficients by ratios of left and right numbers makes it vanish at all four points exactly, and multiplying the two sides gives the circle again.
Necessity. If no certificate exists, linear algebra over the rationals gives functions H_i whose sum over every congruent copy is −1 but whose sum at each single point is 0. Colouring each point by which of 2s + 1 intervals each H_i mod 2 falls in avoids the pattern with at most (2s+1)^s colours, in every dimension (Proposition 2.4). For the twelve points, with weights 1, 1 and −2 on the three squares, a derivation turns any certificate into 0 = 8 (Proposition 7.7).
Two scales. From a certificate, the paper builds two copies of the pattern, one at full size and one as small as needed, with each point of the small copy a coordinate shuffle of its partner (Lemma 4.1).
Paths and Hales–Jewett. Chains of copies form paths whose weight is the sum of squared sizes. A one-colour Hales–Jewett line chooses how many positions t move, so paths of weight 1, 1/2, …, 1/n are prepared before colouring; the t moving positions give t · (1/t) = 1, a copy at exactly the original size. Ultrafilters turn words into points, and compactness gives a finite dimension (Proposition 5.1).
Equal endpoints. Averaging over coordinate permutations puts the discrepancies between the two copies’ paths into a subgroup where cheap corrections exist, giving equal endpoints with arbitrarily small weight ratios (Section 6).

The numbers in the video

Our test of the classical ring colouring: of 200,000 random collinear triples with spacing 1 in ℝ³, none was one colour with 4 colours; with 3 colours, 16,670 were.
Our kite certificate: P = diag(−λ, λ, μ) with λ = b_R²/b_L² and μ = b_R(1 − a_La_R)/b_L³, checked with sympy.
The twelve-point example: t = 0.110001…, rotations of about ±12.55°, weights 1, 1, −2; our symbolic check of the three moment identities.
The Lean count: 65 files and 9,589 lines, our count of the union of the import closures.

Read the source

A classification of finite Euclidean Ramsey configurations (PDF), 26 pages.
The Lean scope note.
The collection’s catalog, where this is result 172.

The paper

OpenAI groups related papers into one result; this one has a single paper.

A classification of finite Euclidean Ramsey configurations

Transcript

Read the full script

0:00The question

Colour every point of space, in as many dimensions as you like, using finitely many colours. Which finite patterns of points must then appear in a single colour, at exactly their original size? Two points at distance one always do: in r dimensions there are r + 1 points all at distance one from each other, and with r colours two of them share a colour. Three points on a line, one apart, do not. Colour each point by the whole-number part of its squared distance from the origin, counted modulo 4. For every such triple, the two outer squared distances minus twice the middle one is exactly two, wherever the triple sits, so its three colours can never agree. In our test, none of 200,000 random triples came out in one colour; with 3 colours instead of 4, about 1 in 12 did. Patterns that always appear, once the dimension is high enough for the number of colours, are called Ramsey. In 1973, Erdős, Graham, Montgomery, Rothschild, Spencer and Straus started this subject and proved that every Ramsey pattern lies on a sphere. Graham conjectured the converse. Frankl and Rödl proved the Ramsey property for triangles and then for all simplices; Kříž for patterns with a soluble transitive symmetry group, and for cyclic trapezoids; Cantwell for the vertices of regular polytopes. Leader, Russell and Walters proposed a different answer: a pattern is Ramsey exactly when it is part of a finite transitive set, one that looks the same from each of its points. Then, on September 20, 2026, Pálvölgyi posted a seven-point pattern on a circle that is not Ramsey, disproving Graham’s conjecture, in a manuscript whose writing, proofs and ideas he attributes to ChatGPT, with appendix claims he says he has not checked.

1:50The claim

Result 172 is a single OpenAI preprint, 26 pages, dated three days later. It gives an exact test. A pattern on a sphere satisfies a sphere equation: a quadratic expression in the coordinates that vanishes at every point. The theorem says the pattern is Ramsey exactly when there is such an expression whose coefficients are built from pairs of numbers, a left one and a right one, so that putting each point’s coordinates into the left and right sides separately still gives zero, before the two sides are ever multiplied. Only after multiplying must it become the sphere. The consequences: every piece of a finite transitive set is Ramsey, which Pálvölgyi had announced without a written proof, and so is every set of at most 5 points on a circle. Certain kites, which Leader, Russell and Walters had shown are not part of any transitive set, are therefore Ramsey, refuting that half of their conjecture. On the other side, 9 generic points on a circle, and 12 points made of 3 squares turned by a special transcendental angle, are not Ramsey. Not claimed: how many dimensions are needed. The proof gives no bound. The test uses the exact algebraic relations among the coordinates; it is not a procedure for numerical coordinates. The colourings that avoid a non-Ramsey pattern come from linear algebra over the rationals; the paper does not claim they are continuous, or even measurable. The paper credits Pálvölgyi with the first refutation of Graham’s conjecture. And it says nothing about colouring the plane itself.

3:29How the proof goes

Here is the test at work on a kite, in our own calculation. Write the kite’s points using one parameter, and keep a left copy and a right copy of it. The plain circle equation, with left and right kept apart, leaves a remainder proportional to the square of their difference, so it vanishes only when they agree. Reweighting its coefficients by ratios of left and right numbers makes it vanish at every point exactly, and multiplying the two sides back together gives the circle again. When no such certificate exists, the paper turns linear algebra into a colouring: functions that add up to minus one on every copy of the pattern, but cancel at every single point. Colouring each point by which of a few intervals those values fall in avoids the pattern in every dimension, with at most (2s + 1)^s colours for s points: the ring colouring, upgraded. For the 12 points from 3 squares, with weights 1, 1 and −2, a derivative identity turns any certificate into the statement 0 = 8. Our symbolic check confirms the three identities behind it. The hard direction builds single-coloured copies. From a certificate, the paper constructs two copies of the pattern in a large space, one at full size and one tiny, where each point of the tiny copy has the same coordinates as its partner, shuffled. Chains of such copies form paths, each with a weight: the sum of their squared sizes. The Hales–Jewett theorem finds a line of words in one colour, but the line chooses how many positions move, some number t. So the paper prepares, in advance, paths of weight 1/t for every possible t, and the t moving positions add up to exactly the original size. An ultrafilter construction turns words into points, averaging over coordinate shuffles makes the path endpoints match, and compactness brings the dimension down to a finite one.

5:24Where it stands

The paper is a preprint, not peer reviewed, and not revised. OpenAI’s Lean files state the classification and 6 of its consequences, with proofs of about 9,600 lines and no gaps we could find by search; we did not run the check. The kite step also relies on Leader, Russell and Walters’ own theorem from 2011, which is not formalized. Several works it cites are 2026 preprints, though its proofs do not depend on them. It is not among the papers OSVerify reviewed, and we have seen no expert review. If it holds, whether a finite pattern is Euclidean Ramsey comes down to one exact algebraic test on its coordinates. Every piece of a transitive pattern passes, and so does every set of at most 5 points on a circle; lying on a sphere is not enough, and being part of a transitive pattern is not necessary.

Spot a mistake?

This episode simplifies on purpose, but it shouldn’t be wrong. If something is, open an issue on GitHub or leave a comment on YouTube. Corrections are noted here with a date.