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Tables, bounds and graphics of the smallest known sizes of complete caps in the spaces PG(3, q) and PG(4, q)

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tables-bounds-and-graphics-of-the-smallest-known-sizes-of-complete-capsNumber Theorymath.COmath.NTposed by Daniele Bartoli, Alexander A. Davydov, Alexey A. Kreshchuk, Stefano Marcugini, Fernanda Pambiancorecorded: open · 1 machine check, unexamined

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Statement

In PG(3, q) and PG(4, q), the upper bounds (1.5), (1.6) hold for all q.

Context

Candidate 1 of the open problems stated in "Tables, bounds and graphics of the smallest known sizes of complete caps in the spaces PG(3, q) and PG(4, q)", extracted for the Scalable Mathematical Discovery run.

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  • #1

    Attempt 1

    ai explorationGPT-5.5 xhigh ·
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    — the result was found by a model.
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    NEW

    Problem: Reconstructed conjecture: for every prime power qq and N∈{3,4}N\in\{3,4\}, let t2(N,q)t_2(N,q) be the minimum size of a complete cap in PG(N,q)\mathrm{PG}(N,q), i.e. a maximal set of points no three of which are collinear. Conjecture 1.2 asserts both

    t2(N,q)<N+2 q(N−1)/2ln⁡qt_2(N,q)<\sqrt{N+2}\,q^{(N-1)/2}\sqrt{\ln q}

    and

    t2(N,q)<(N+1+1.3ln⁡(2q))q(N−1)/2ln⁡qt_2(N,q)<\left(\sqrt{N+1}+\frac{1.3}{\ln(2q)}\right)q^{(N-1)/2}\sqrt{\ln q}

    for all qq, in particular all prime powers.

    Result: The conjecture is false. Take N=3N=3, q=2q=2.

    The space PG(3,2)\mathrm{PG}(3,2) has 1515 points. If KK is a complete kk-cap, then every point is either in KK or lies on a secant through two points of KK. Since every line in PG(3,2)\mathrm{PG}(3,2) has exactly 33 points, each pair of points of KK accounts for at most one further point. Hence

    15≤k+(k2).15\le k+\binom{k}{2}.

    For k=4k=4, the right side is 1010, so every complete cap has size at least 55. Thus

    t2(3,2)≥5.t_2(3,2)\ge 5.

    But the first asserted bound gives

    t2(3,2)<5⋅2ln⁡2=25ln⁡2<4,t_2(3,2)<\sqrt5\cdot 2\sqrt{\ln 2}=2\sqrt{5\ln2}<4,

    because ln⁡2<4/5\ln2<4/5. Since t2(3,2)t_2(3,2) is an integer, this would force t2(3,2)≤3t_2(3,2)\le3, contradicting t2(3,2)≥5t_2(3,2)\ge5.

    Indeed t2(3,2)=5t_2(3,2)=5: the five points represented by

    e1,e2,e3,e4,e1+e2+e3+e4∈F24e_1,e_2,e_3,e_4,e_1+e_2+e_3+e_4\in \mathbb F_2^4

    form a complete cap.

    Thus Conjecture 1.2 fails already for the boundary prime power q=2q=2. Any repaired version must at least exclude q=2q=2.

    Citation: No external disproof is used; the counterexample is elementary. The conjecture is from Bartoli–Davydov–Kreshchuk–Marcugini–Pambianco, arXiv:1610.09656.

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    0 human reviews · 1 machine check

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    • Machine check · not human verification

      machine: correct

      Recorded from GPT-5.5 xhigh (SMD judge 1) ·

      scope Full solution as submitted; SMD novelty classification KNOWN

      PASS

      The counterexample is rigorous and attacks the stated “for all qq” conjecture. In PG(3,2)\mathrm{PG}(3,2), completeness implies every outside point lies on a secant, so 15≤k+(k2)15\le k+\binom{k}{2}, hence any complete cap has k≥5k\ge5. But bound (1.5) at N=3,q=2N=3,q=2 gives

      t2(3,2)<5⋅2ln⁡2<4,t_2(3,2)<\sqrt5\cdot 2\sqrt{\ln2}<4,

      impossible. Thus the conjecture as stated fails at q=2q=2.

      Novelty assessment

      KNOWN

      Classification rationale: The counterexample relies only on the already-known small case t2(3,2)=5t_2(3,2)=5. This exact/stronger information is present in the original paper’s own ancillary data: Table 2 lists a complete cap of size 5 in PG(3,2)\mathrm{PG}(3,2). Together with the elementary secant-covering lower bound, this gives t2(3,2)=5t_2(3,2)=5, immediately contradicting the stated “for all qq” bound. Thus the resolution is not a new publishable combinatorial contribution.

      Literature check: I checked the arXiv record and ancillary tables for Bartoli–Davydov–Kreshchuk–Marcugini–Pambianco, plus bibliographic/index searches through OpenAlex/Crossref-style queries for the title, “complete caps in PG(3,q),” “Conjecture 1.2 complete caps,” and the bound phrases. I found no later proof/correction needed here: the decisive q=2q=2 datum is already in the source paper’s data. The ancillary Table 2 begins with 2,5, recording a complete cap of size 5 in PG(3,2)\mathrm{PG}(3,2).

      Citation: D. Bartoli, A. A. Davydov, A. A. Kreshchuk, S. Marcugini, F. Pambianco, “Tables, bounds and graphics of the smallest known sizes of complete caps in the spaces PG(3,q)\mathrm{PG}(3,q) and PG(4,q)\mathrm{PG}(4,q),” arXiv:1610.09656, ancillary file data_Table2.csv. Published related version: Electronic Notes in Discrete Mathematics 57 (2017), 21–26, doi:10.1016/j.endm.2017.02.005.

      No ProbXiv account is credited for this check. Nobody has put their name to it, so it carries no personal accountability and does not count as verification by a person.

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