problems / three-in-line

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Attack log — No-three-in-line at n=71

2026-08-16

2026-08-16 — q1

2026-08-16 — q2

2026-08-23 — q3: certificate shapes, ranked by cheapest first check

  1. Flammenkamp database certificate (outside this folder). The finished object is the database-native rct4 code for Heule's 17 August 2026 n=71n=71 entry, a pinned decoder, the decoded 142-line coordinate file, and an exact replay log. It settles the case because a checked 142-set meets the row upper bound D(71)≤142D(71)\le142. Cheapest check: fetch the single coded entry, decode its two column positions in each of 71 rows, and run the existing determinant checker. A database announcement alone is only a lead.
  2. Verifier gate for every claimed 142-set. The finished object is a run_all.sh that feeds the same plain coordinate file to the committed Python triple-determinant checker and to a second checker in another language using primitive line normalization, then compares their point, row, and column counts. Agreement settles the computational part of any claimed construction; disagreement kills the claim. Cheapest check: run every already committed verifier before writing D(71)=142D(71)=142 anywhere.
  3. Honest rot4 SAT — killed immediately. The hypothetical finished object would be a 142-point quarter-turn-invariant coordinate file and the same verifier replay. It would settle n=71n=71, but on an odd board every noncentral orbit has size four and the centre has size one, so its size is 00 or 11 modulo 44, never 142≡2(mod4)142\equiv2\pmod4. The one-page orbit-count check kills this shape without a solver.
  4. Flammenkamp rct4 SAT (the symmetry used at 65, 67, and 69). The finished object is an audited DIMACS or direct CP-SAT model with the anti-diagonal empty, one selected half-turn pair on the main diagonal, 35 selected four-orbits, and a decoded 142-set. It settles the case after the verifier gate. Cheapest check: regenerate the model, replay it first at n=69n=69 against a published witness, and audit its orbit weights and DIMACS header before spending solver time at 71.
  5. Lift and perturb 69, 70, or 72. The finished object is a 142-line coordinate file obtained by deleting or inserting boundary rows and columns in a published checked configuration, followed by a bounded local repair trace and the verifier replay. Any final checked file settles 71; the provenance does not. Cheapest check: replay the source witness, apply each monotone row/column map once, and count collinear triples and conflicted points. Kill a lift if its first image has no small repair support.
  6. rot2 SAT, a named weaker symmetry. The finished object is a half-turn-invariant 142-set, its exact encoding under 2-cycles, and the verifier replay. It would settle 71 while retaining more candidates than rct4. Cheapest check: construct the orbit/cardinality equations and solve small odd controls through n=25n=25; proceed only if the encoding reproduces checked witnesses and its n=71n=71 clause count fits memory.
  7. No imposed board symmetry: paired-row local search followed by SAT. The finished certificate is again the plain 142-point file, with exactly two points in every row and column; a row-pair permutation supplies a compact search state, and SAT completes only the unresolved cells. It settles 71 only after the verifier gate. Cheapest check: seed from the best lifted 69/70/72 image and measure the exact number of violated maximal lines after one minute; abandon it if it does not beat the unperturbed lift.

2026-08-23 — q3 result

2026-08-23 — q4: n=75 certificate shapes, ranked

The finished object in every surviving shape is a plain 150-point file on {0,…,74}2\{0,\ldots,74\}^2 that passes two exact, algorithmically independent verifiers. Working backwards from that object gives this order of attack:

  1. Current database certificate — checked and killed as a source. A raw Flammenkamp code plus the existing decoder would be the cheapest route to a certificate. Both an unrestricted lookup and an rct4 lookup for size 75 returned “no configurations are known” at the database cut of 19 August 2026. This establishes only that this source has no witness, not that none exists.
  2. Canonical rct4 SAT. The certificate has the anti-diagonal empty, one selected diagonal half-turn pair, and 37 selected four-orbits. The existing geometry generator, exact row and column equations, and weighted line constraints already express this shape. First replay the model and verifier at 71, then audit the generated n=75n=75 instance and run several named Kissat seeds. SAT followed by the verifier gate settles D(75)=150D(75)=150; UNKNOWN or rct4 UNSAT does not.
  3. rct4 SAT phased from the checked n=73 configuration. Embed the 146 source points into a 75-grid in symmetry-preserving ways, translate the surviving orbit choices into initial SAT phases, and let the complete rct4 instance repair them. The cheapest check counts violated maximal lines and retained source orbits for each embedding; only the best phases deserve solver time.
  4. Named rot2 SAT. A 150-set can be exactly 75 half-turn pairs, so rot2 has no orbit-cardinality obstruction and is weaker than rct4. The finished object is a half-turn-invariant point file decoded from an audited orbit model. The cheap gate is a control replay on a checked odd witness and a clause/memory count at 75.
  5. Boundary lift and bounded repair from n=73. Place the checked 146-set into the 75-grid, add two new row pairs and two new column pairs, and expose only conflicted source points plus boundary candidates to SAT. A small repair support would yield a 150-set without imposing more symmetry than the source; a large initial conflict core kills this version of the shape.
  6. Two-per-row/column permutation SAT without board symmetry. Represent the certificate as two permutations of the 75 columns, seed them from the checked 74- and 76-grid records, and use lazy line cuts or local search before a final exact SAT completion. This is the broadest credible shape, but it is ranked late because its model is much larger and a timeout leaves a less informative residue.
  7. Honest rot4 — killed by orbit cardinality. On an odd grid a quarter-turn-invariant set has size 00 or 11 modulo 4 depending on the centre, never 150≡2(mod4)150\equiv2\pmod4. No solver run is warranted.

2026-08-23 — q4 partial result: smallest n=73 repairs exhausted

2026-08-23 — q4 result: precise rct4 wall