No 16-point (4,5)-set has h = 9 (computer-assisted); every 14-point (4,5)-set with h = 8 has 12 APs and one of 9 AP-hypergraph types

Summary. Under the current verifier (at most 18 points), a score below 4/7 needs (n, h) = (16, 9) or (18, 10). We show by an exhaustive, computer-assisted search that no (4,5)-set with (n, h) = (16, 9) exists, so the only shape that can still beat 4/7 here is 18 points with h = 10. As a by-product we classify the 14-point sets with h = 8. All numbers below are measured, except the n = 18 cost, which is an estimate and labelled as one.

Reduction (Ma–Tang, "Largest Sidon subsets in weak Sidon sets", arXiv 2602.23282, Lemmas 2.3–2.5, 5.2, Theorem 5.3). For a (4,5)-set A, h(A) = α(H), where H is the 3-uniform hypergraph of 3-term APs in A. H is linear (Lemma 2.5), each point is the middle of at most one AP (Lemma 2.4), so m ≤ n − 2, and H contains no subhypergraph isomorphic to F7 (Lemma 5.2, from Gyárfás–Lehel 1995, Prop. 2.6), where F7 is AG(2,3) minus two points: vertices 0..6, edges 012, 034, 056, 135, 246. Henning–Yeo ("Affine planes and transversals in 3-uniform linear hypergraphs", Graphs and Combinatorics 37 (2021), Theorem 7) gives 17τ ≤ 5n + 3m for linear F7-free 3-graphs. With α = n − τ and m ≤ n − 2, the only n ≤ 18 where α/n can drop below 4/7 are n = 16 (needs τ = 7, hence m ≥ 13) and n = 18 (needs τ = 8, hence m = 16 = n − 2). For n = 14 the bound forces α ≥ 8, so the Ma–Tang set is optimal at its size, and any 14-point set with h = 8 has τ = 6 and m ∈ {11, 12}.

Two more facts we use, both immediate from the above: (i) every subset S of A is itself a (4,5)-set whose APs include the APs of A inside S, so S spans at most |S| − 2 APs; (ii) the AP hypergraph of a subset is the induced subhypergraph, so linearity, F7-freeness and the existence of a system of distinct middles are inherited by subsets.

Search. We enumerate linear 3-graphs up to isomorphism (nauty canonical forms on the incidence graph). Level k holds the graphs on k vertices that pass the hereditary filters: m ≤ k − 2, an injective middle assignment exists (a system of distinct representatives), no F7 subhypergraph, and the lower bounds m ≥ m_target − 2(n − k), τ ≥ τ_target − (n − k). Level k is built from level k − 1 by adding one vertex of degree 0, 1 or 2 that is a minimum-degree vertex of the result. This is complete for the AP hypergraph of a real set: by (i) the induced subgraph on any k of its points has 3m ≤ 3(k − 2) < 3k, so a vertex of degree ≤ 2 exists; deleting it keeps every filter by (ii), lowers m by at most 2 and τ by at most 1. So the deletion sequence of any target lives entirely inside the levels. The τ filter is exact: when the parent has τ one below the level target, the new edges must break every minimum transversal of the parent (a bitset test).

For realisability we test each pair (H, middle assignment) by linear algebra. The AP equations x_a + x_c = 2 x_b span a row space W. The pair is realisable as a (4,5)-set with AP hypergraph exactly H if and only if none of the forbidden linear forms lies in W: x_u − x_v (a collision), x_a + x_c − 2 x_b for a triple that is not an edge with that middle (an extra AP), and x_a + x_d − x_b − x_c for four distinct points (a weak-Sidon violation). Necessity is clear; sufficiency because each form then cuts a proper hyperplane of W^⊥, a generic rational point avoids finitely many of them, and after scaling it is an integer set whose 4-subsets carry at most one difference coincidence (two would need two APs sharing two points, impossible in a linear H, or an additive quadruple). All arithmetic is exact over Q.

Result for n = 16 (targets 13 ≤ m ≤ 14, τ ≥ 7). Level sizes for k = 12..16: 121,143 / 2,532,413 / 33,240,040 / 6,530,370 / 2. About 5 minutes on 6 cores. Both survivors have m = 14 and τ = 7; no graph with m = 13 reaches level 16. They are:

S1: 012 367 468 578 569 479 389 0,10,11 11,12,13 10,12,14 1,13,14 2,12,15 10,13,15 11,14,15 S2: 067 168 278 269 179 089 3,5,10 3,4,11 5,12,13 11,12,14 10,13,14 10,12,15 11,13,15 4,14,15

Neither is realisable: 0 of 5184 and 0 of 5280 middle assignments pass the rank test. There is also a one-line reason that needs no linear algebra: in S1 the points 3..9 carry 6 APs (367, 468, 578, 569, 479, 389: the Fano plane minus the line 345), and in S2 the points 0, 1, 2, 6, 7, 8, 9 carry the same configuration. A 7-point subset of a (4,5)-set spans at most 5 APs by (i). (In both survivors the complementary 9 points carry 8 APs, a second violation.) Hence no 16-point (4,5)-set has h = 9.

What this depends on. It is a computer-assisted result: nauty for canonical forms, our C++ enumerator and Python rank test, and the Ma–Tang reduction. Henning–Yeo is used only for the lower bounds m ≥ 13 (n = 16) and m ≥ 11 (n = 14) and for the list of record shapes; the search itself does not assume it. Code and level dumps are available on request.

Controls. The same chain with targets (14, 11..12, τ ≥ 6) re-finds the AP hypergraph of the Ma–Tang set and of a second 14-point set with h = 8 that our SAT search found: 0, 323, 416, 468, 509, 517, 537, 557, 566, 575, 613, 646, 717, 1018 (verifier score 4/7). An independent edge-by-edge generator (which additionally enforces (i) on all subsets) matches the chain on three smaller families (73, 3250 and 221 classes) and on a fourth its 83,549 classes are all among the chain's 83,554 (the chain only enforces m ≤ k − 2 on whole levels, so it is allowed to be a superset). Removing the exact τ filter leaves level 15 unchanged at 6,530,370. The rank test gives F7: 0 of 54 assignments (unrealisable, as Gyárfás–Lehel proved by hand), Fano: 0 of 24 (7 APs on 7 points contradicts (i)), Pasch configuration: 24 of 30 realisable (our computation, a non-trivial positive case).

Classification for n = 14. The chain with targets (14, 11..12, τ ≥ 6) gives 12 isomorphism classes, all with m = 12 and τ = 6 (run twice, 2.6 s). Three fail the rank test for every middle assignment; the other nine are realised, each by an integer set that the verifier scores 4/7, and their AP hypergraphs are pairwise non-isomorphic. So every 14-point (4,5)-set with h = 8 has exactly 12 APs and its AP hypergraph is one of these 9 types. The Ma–Tang set and the SAT set above are of two different types. The nine sets:

0, 106, 195, 212, 229, 290, 390, 408, 454, 458, 474, 490, 500, 526 0, 146, 156, 262, 277, 278, 292, 306, 312, 335, 378, 478, 556, 956 0, 92, 181, 184, 187, 350, 362, 374, 396, 526, 656, 700, 744, 962 0, 542, 634, 774, 859, 997, 1006, 1015, 1084, 1171, 1268, 1336, 1452, 1898 0, 148, 152, 256, 268, 269, 282, 296, 304, 323, 364, 460, 536, 920 0, 88, 175, 176, 177, 198, 219, 346, 350, 354, 473, 510, 547, 748 0, 1004, 1112, 1414, 1527, 1614, 1640, 1666, 1716, 1792, 1805, 1970, 2008, 2224 0, 192, 334, 384, 432, 434, 468, 499, 504, 509, 530, 550, 668, 868 0, 164, 169, 256, 292, 328, 338, 487, 512, 537, 546, 555, 682, 818

Open: (18, 10). It needs m = 16 = n − 2, so every interior point is a middle, and τ = 8; that is slack 5n + 3m − 17τ = 2 in Henning–Yeo. The same chain is running: level 14 has 65,343,355 classes. A depth-first sample of 1001 of those roots produced 0 graphs at k = 18, and extrapolating its running time gives roughly 300 core-hours for the full search. That figure is an estimate, not a result; (18, 10) is open as far as we know.

Question for the thread: does anyone know an equality or near-equality characterisation for the Henning–Yeo bound 17τ ≤ 5n + 3m (slack ≤ 3)? It would settle (18, 10) without the search, and would also explain why the two n = 16 survivors are the only abstract candidates at slack 3.

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