I had not yet heard of the pantheon efforts. Thanks for sharing. I have not found a better frontier signal in AI than your Substack here. I'm piping in the RSS transcripts into my local stack. I will have something for your Get Physics Done efforts in the coming weeks.
Thank you AWG, fascinating as always. BTW, I got a helpful (for me) commentary/ summary from Claude Opus 4.7 on the Erdős unit distance conjecture disproof with implications for the diagram:
"The Erdős unit distance conjecture (1946, Amer. Math. Monthly 53), in precise form: letting u(n) denote the maximum number of unordered pairs at unit distance among n points in ℝ², Erdős proved u(n) ≥ n^(1+Ω(1/log log n)) via a √n × √n integer grid, and conjectured a matching upper bound u(n) ≤ n^(1+o(1)). The best proven upper bound, O(n^(4/3)), is Spencer–Szemerédi–Trotter (1984), still very far from the conjecture.
Intuitively: how often can the same distance recur among n points in the plane? On a square grid, plenty of horizontal, vertical, and diagonal segments share a length, giving a tiny super-linear bump above n — but a slow-growing one (vanishing slower than any honest polynomial gain). Erdős's conjecture was essentially: nothing does meaningfully better than the grid.
Why it appears wrong, per the May 20, 2026 arXiv companion paper (Alon, Bloom, Gowers, Litt, Sawin, Shankar, Tsimerman, Wang, Wood, Remarks on the disproof of the unit distance conjecture, arXiv:2605.20695): an OpenAI team — Lijie Chen using an internal model, Mark Sellke and Mehtaab Sawhney verifying — produced a counterexample. The construction abandons ℤ²: it places points using rings of integers in CM number fields that are finite layers of an infinite Golod–Shafarevich class field tower, with [K:ℚ] → ∞ at bounded root discriminant. The grid is just the K = ℚ(i) case of a far more general construction.
Will Sawin's companion paper (arXiv:2605.20579, same day) makes the exponent explicit at δ = 0.014114, after sharpening five steps of the simplified argument — which itself gave only δ ≈ 6.24 × 10⁻³⁸. Crucially this is a polynomial improvement, strictly outside the conjecture's n^(1+o(1)) form. Bloom's section confirms this resolves a $500 Erdős problem (offered in print, 1995).
One important caveat the companion paper itself raises: Sawin's Proposition 15 proves the entire method ceilings at 1 + 1/4.116 ≈ 1.243. So the qualitative refutation is settled, but the order of magnitude of u(n) is not — there remains a substantial gap between this approach's ceiling (n^1.243) and the proven upper bound (n^1.333)." — Claude Opus 4.7, prompted and edited by Stephen Charles Smith
Great as usual, but the AI voice is still pronouncing Erdős wrong. Please fix this oversight. It may seem like a little thing, but I am sure Alex would hate to have his name pronounced wrong every single time when the correct pronunciation is not hard to come by.
Thanks AWG. Awesome information.
You're welcome!
Thanks Alex - The Innermost Loop is a great instant up-dater to Moonshots
You're welcome!
Pretty sure you will not be needing the robotic hair trimming service, AWG.
Steve Wozniak knows that humans need to be reassured... they are still in the inner most loop.
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I am one who would say that the pattern identified by AI was latent within the human corpus, but I'm not a skeptic. 😉
I had not yet heard of the pantheon efforts. Thanks for sharing. I have not found a better frontier signal in AI than your Substack here. I'm piping in the RSS transcripts into my local stack. I will have something for your Get Physics Done efforts in the coming weeks.
Thank you AWG, fascinating as always. BTW, I got a helpful (for me) commentary/ summary from Claude Opus 4.7 on the Erdős unit distance conjecture disproof with implications for the diagram:
"The Erdős unit distance conjecture (1946, Amer. Math. Monthly 53), in precise form: letting u(n) denote the maximum number of unordered pairs at unit distance among n points in ℝ², Erdős proved u(n) ≥ n^(1+Ω(1/log log n)) via a √n × √n integer grid, and conjectured a matching upper bound u(n) ≤ n^(1+o(1)). The best proven upper bound, O(n^(4/3)), is Spencer–Szemerédi–Trotter (1984), still very far from the conjecture.
Intuitively: how often can the same distance recur among n points in the plane? On a square grid, plenty of horizontal, vertical, and diagonal segments share a length, giving a tiny super-linear bump above n — but a slow-growing one (vanishing slower than any honest polynomial gain). Erdős's conjecture was essentially: nothing does meaningfully better than the grid.
Why it appears wrong, per the May 20, 2026 arXiv companion paper (Alon, Bloom, Gowers, Litt, Sawin, Shankar, Tsimerman, Wang, Wood, Remarks on the disproof of the unit distance conjecture, arXiv:2605.20695): an OpenAI team — Lijie Chen using an internal model, Mark Sellke and Mehtaab Sawhney verifying — produced a counterexample. The construction abandons ℤ²: it places points using rings of integers in CM number fields that are finite layers of an infinite Golod–Shafarevich class field tower, with [K:ℚ] → ∞ at bounded root discriminant. The grid is just the K = ℚ(i) case of a far more general construction.
Will Sawin's companion paper (arXiv:2605.20579, same day) makes the exponent explicit at δ = 0.014114, after sharpening five steps of the simplified argument — which itself gave only δ ≈ 6.24 × 10⁻³⁸. Crucially this is a polynomial improvement, strictly outside the conjecture's n^(1+o(1)) form. Bloom's section confirms this resolves a $500 Erdős problem (offered in print, 1995).
One important caveat the companion paper itself raises: Sawin's Proposition 15 proves the entire method ceilings at 1 + 1/4.116 ≈ 1.243. So the qualitative refutation is settled, but the order of magnitude of u(n) is not — there remains a substantial gap between this approach's ceiling (n^1.243) and the proven upper bound (n^1.333)." — Claude Opus 4.7, prompted and edited by Stephen Charles Smith
Great as usual, but the AI voice is still pronouncing Erdős wrong. Please fix this oversight. It may seem like a little thing, but I am sure Alex would hate to have his name pronounced wrong every single time when the correct pronunciation is not hard to come by.