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Prove the narrow-row half-gap budget at 1/(16p^2). - #4648
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Mister-Meeseeks merged 1 commit intoOct 22, 2026
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Alias removal and band truncation preserve the ideal certificate 1/(8p^2). A rounding defect of at most 1/(16p^2) leaves row gap 1/(16p^2) and row norm below 2. No new kappa. Co-authored-by: unpaidmoon <unpaidmoon@users.noreply.github.com>
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Change and mathematical scope
Scoped audit of one #4100 precision sentence. No new κ. The selected conditional value stays
60403/20000000 = 0.00302015.vendor/FINE-PRIME.mdsays that deleting diagonal aliases preserves a polynomial gap, and that exponentially accurate band truncation and rounding preserve half of it.GAUSSIAN.mdconcludes that the rounded reference has diagonal 1, row gap at least1/(16p^2)and row norm below 2. This package makes that budget explicit.Under
a ≥ 4 log(16p)andθ ≥ 1/(4p^2), withlogthe natural logarithm,Alias removal (reset the nonnegative diagonal to 1) and band truncation (zero any off-diagonal entries) spend none of this certificate. A rounding perturbation whose off-diagonal absolute row sum is at most
τ ≤ 1/(16p^2)leaves row gap at least1/(16p^2)and absolute row sum strictly below 2. The factor one half is sharp for that certificate: a row that meets it and a defect of1/(8p^2)leave gap 0. A sufficient special case is at most2poff-diagonal entries, each moved by at most2^{-5p}. The same count at2^{-4p}fails atp = 1. This does not identify #4100's unnamed precision constant with 5.This is not a supplier admission, a Gaussian-circuit replay, or a complete multiplication theorem.
Task and overlap
uniform-transfer-audit, scope refined to this one sentence. The card asks for one quantified implication at its caller.fbc9ea0249746acca4f504d4fda2b84e0e36cf0d, priority revision 7. Propose unconditional κ = 0.00173097911567079247276057 via batched CRT #4100 remains the family-record head337d04939571a637aeab54a0f58f54fde811663a.2^{-4p}sentence and left the row gap open.1/(10p^2)only after that larger gap is already known.Validation
Python 3.12, standard library only, from the repository root:
All three passed.
verify.pypins the source sentences and checks the series comparisons, the uniform integer certificate, the three operations forp = 1..200, theC = 5induction, the polynomial thresholds, and a singular control. It does not build a Gaussian matrix or replay #4100. LOG-A §3's elimination identity is cited, not re-proved.Credit
The ideal bound is LOG-A §1 from PR1673. PR1786 and #4100's
vendor/FINE-PRIME.mdstate the one-sentence claim. The1/(8p^2)certificate is the mergedpr1786-fine-prime-transfer-audit. #4217 asked the question. #4438 (James Reed) separated the column gauge from the row.Prepared by unpaidmoon with substantial Cursor Grok assistance. Apache-2.0.
Open next questions
2poff-diagonal entries per row, and does the note's rounding meet absolute error2^{-5p}? The total budget is proved either way. Blocked untilWand the rounding mode are pinned.p = 19entry above 1. This row budget does not choose the origin.