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No three in line: more than 1.5n points in the n × n grid

Construct, for arbitrarily large n, more than 1.8n points of the n × n grid with no three on a line; the best asymptotic family gives 1.5n.

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No three in line: more than 1.5n points in the n × n grid

How many points of an n×nn \times n grid can be chosen with no three on a line? At most 2n2n, since each row holds at most two, and 2n2n is achievable for every nn from 2 to 74. For large nn, though, the best construction, from 1975, gives only about 1.5n1.5n, while a heuristic predicts about 1.81n1.81n; the task is to beat 1.5n1.5n by a constant factor.

Problem

Let t(n)t(n) be the largest size of a set S⊆{1,…,n}2S \subseteq \{1, \dots, n\}^2 with no three points on a common line. The pigeonhole principle gives t(n)≤2nt(n) \le 2n, and a 1975 construction gives t(n)≥(32−o(1)) nt(n) \ge (\tfrac{3}{2} - o(1))\,n [1].

Task. Give an explicit family of no-three-in-line sets with more than 1.8n1.8n points in the n×nn \times n grid, for arbitrarily large nn. Any family with at least (32+c) n(\tfrac{3}{2} + c)\,n points, for a constant c>0c > 0, would already improve the asymptotic lower bound.

Background

Special cases go back to Dudeney's puzzles more than a century ago. Hall, Jackson, Sudbery and Wild obtained (32−o(1)) n(\tfrac{3}{2} - o(1))\,n points with a construction modulo a prime [1], and nothing better is known asymptotically. Guy and Kelly gave a heuristic suggesting that t(n)∼c nt(n) \sim c\,n with c=(2π2/3)1/3≈1.87c = (2\pi^2/3)^{1/3} \approx 1.87 [2]; in 2004 Gábor Ellmann found an error in their derivation, and the corrected heuristic gives c=π/3≈1.814c = \pi/\sqrt{3} \approx 1.814 [3]. Not everyone believes the heuristic: Ben Green's list of open problems (Problem 72) records his suspicion that 32\tfrac{3}{2} is optimal.

What is known

  • t(n)=2nt(n) = 2n for every 2≤n≤742 \le n \le 74 and for n=76n = 76, according to Achim Flammenkamp's record page (updated 15 October 2026), with several of the largest cases found in 2026 [3].
  • Asymptotically, the best lower bound is (32−o(1)) n(\tfrac{3}{2} - o(1))\,n [1].
  • For the analogous no-(k+1)(k+1)-in-line problem with k≥3k \ge 3, the maximum is exactly knkn for large nn (Ghosal, Goenka, Grebennikov, Keevash, Kwan and Pham, 2026), but their method does not reach k=2k = 2 [4].
  • formal-conjectures states Green's question (whether 2n2n points become impossible for all large nn) in Lean as green_72 in GreensOpenProblems/72.lean, with the opposite question as a variant.

What would count as progress

  • A program that, given mm, returns some nn with m≤n≤2mm \le n \le 2m and more than 1.8n1.8n points of {1,…,n}2\{1, \dots, n\}^2 with no three collinear. Collinearity among NN points can be checked in O(N2log⁡N)O(N^2 \log N) time by sorting slopes from each point.
  • Epoch scores submissions with its own verifier, which calls such a function solution(min_n) for increasing lower bounds and requires strictly increasing nn.
  • As Epoch warns, a construction with 1.5n+C1.5n + C points can pass finitely many tests without improving the ratio; the real goal is a family with a provable constant-factor gain over 1.5n1.5n.

References

  1. R. R. Hall, T. H. Jackson, A. Sudbery and K. Wild, Some advances in the no-three-in-line problem, J. Combin. Theory Ser. A 18 (1975), 336–341. doi:10.1016/0097-3165(75)90043-6
  2. R. K. Guy and P. A. Kelly, The no-three-in-line problem, Canad. Math. Bull. 11 (1968), 527–531. doi:10.4153/CMB-1968-062-3
  3. A. Flammenkamp, The no-three-in-line problem (record page). wwwhomes.uni-bielefeld.de/achim/no3in/readme.html
  4. A. Ghosal, R. Goenka, A. Grebennikov, P. Keevash, M. Kwan and H. T. Pham, No-(k+1)(k+1)-in-line problem for k≥3k \ge 3, 2026. arXiv:2607.05255

Source

Imported from FrontierMath: Open Problems, no-three-in-line, curated by Epoch AI. Open according to the source as of 2026-10-20. The statement above is written in our own words; see the source for the original wording, history and discussion.