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Prove the sharp fallback child count: 33269, not 49911, at m = 64 (no new κ) - #4237
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Mister-Meeseeks merged 1 commit intoOct 21, 2026
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FALLBACK.md charges three children per swap. Opposite-bank swaps pay one, and the head-bank elimination makes exactly r of them, so the m=64 maximum is 33269 children rather than 49911.
This was referenced Oct 20, 2026
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Independent scoped audit of #4237 sharp fallback child count at Code-disjoint checks; nested No packaging defect in SOURCES scope. All 7 What passes:
Not covered here: physical fallback execution; BIT recurrence retuning; new κ; Lean; merge/close. Finite pin/seal/census replay ≠ formal proof / all-size admission. James Reed (@rcfc-jreed), with substantial AI assistance (Claude, in Cursor). |
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Summary
FALLBACK.md in
research/unconditional-batched-crt(#4100) bounds the retained BIT fallback program at m = 64 by 16637 swaps and 3 × 16637 = 49911 reversed children. This PR gives the exact count for every header and its sharp maximum:Claim type: scoped proof with a finite certificate for one tape-cost term. No new κ, no audit of other #4100 premises, no supplier admission and no change to the selected frontier. Package:
research/fallback-child-count. Claim.Result
For a rank-r projector header (1 ≤ r < m), let s be the number of same-bank elimination swaps.
dim(H+V_k) − k, which drops by one at each head pivot and stays fixed at each tail pivot. It follows that s ≤ 2m − 1 − r.N = (2m−1)(2m+2) + r + sswaps, of whichS = 2(m²−1) + sare same-bank.paid_childrenpays8m² + 2m − 6 + r + 3schildren when f ≥ 2 and N children when f = 1. The total width isfN + 2⌊f/2⌋S.8m² + 8m − 11children when f ≥ 2, and at most(2m−1)(2m+3)swaps. The rank-one header P = 1·e₀ᵀ attains both, for every odd prime power and every 2 ≤ m ≤ 64. Over Q, every intermediate entry of its elimination stays in {−1, 0, 1}, so its itinerary is the same modulo every odd prime power.This answers PR4215's Open next question 2 and the matching open question in PR4228's package.
Evidence
verify.pyloads the unchangedprepare/paid_childrenpath after checking SHA256 pins. It checks:Exact reproduction from the repository root:
Any CPython 3.12 works. Expected markers:
PASS_REGISTERED_RESEARCH_PACKAGE research/fallback-child-count outcome=completed,PASS_RESEARCH_COMMIT_BYTESand"status": "PASS_FALLBACK_CHILD_COUNT". Locally, all four passed, and each verify command took about 25 seconds.Limitations and unrun obligations
This counts what
paid_childrencharges for the retained program. It does not execute physical reversals, price preparation or parking, decide which headers reach the fallback, or audit BIT.md's recurrence. The maximum is attained at rank one, and in the censuses only there, so an actual caller may pay less. The bound s ≤ 2(m − r) for every rank is an observation from the censuses, not a proof. Independent human and formal proof-kernel review are unclaimed.Open next questions
Related open work
research/selected-bit-bad-class-fallback) price an earlier Gaussian local-ring fallback at m = 78. They use the uniform count 6N(N−1) + 3N + 6(N−1) rank-one children with N = 2m, which charges every swap alike and has no orientation or rank dependence; at m = 64 it would be 98,682. None of them states the orientation-aware count or the rank law for the fallback program Propose unconditional κ = 0.00173097911567079247276057 via batched CRT #4100 retains.CI notes
Hosted run 39488451544 is green on head
34af2cd420ccf5339ce05e3aefe5c569aaf17f85:planpassed in 2m55s and the packagereplaypassed in 3m32s. All four reproduction commands also passed locally. No local obligation was left to CI.Produced by Nick Petalas. Apache-2.0. The fallback program and descriptors are retained from source527, commdata2338, Ross Richardson (PR2159/PR2188) and Jose Manuel; FALLBACK.md is from #4100.