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Census five-call two-relation BIT words: no three rank-two targets at one or two dirty auxiliaries, corroborating #5481 (no new κ) - #6055
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Mister-Meeseeks merged 2 commits intoOct 28, 2026
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…ies (no new kappa) Independent exhaustive corroboration of PR5481 (research/bit-fivecall-dirty-block) for W<=5 registers and every relation lattice of rank <= 2, via a cube-lattice reduction and a two-relation derivative condition over F2. Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com>
Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com>
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Independent finite corroboration of #5481; no new κ. Package
research/bit-span4-five-call-two-relation.JoyboyBrian's merged #5481 (
research/bit-fivecall-dirty-block) proves in writing that three invertible 2×2 targets need six rank-one calls in the formal abelian BIT model with any number of dirty auxiliaries. That closes open question 1 ofresearch/bit-span4-auxiliary-structureand the eight-call span-four case. Its README says no five-call word enumeration was run, and the inherited low-span, one-relation and three-register lemmas were cited, not re-proved.This PR runs that enumeration. Three data registers with one auxiliary (W = 4) are in the default replay, and two auxiliaries (W = 5) are in
--long. W = 2, 3 are also covered. The census includes every relation lattice of rank at most 2, so every five-call word of frame span at least 3. It uses neither the conic argument nor the one-relation or three-register theorems.Result. No diagonal five-call word at W ≤ 5 has three registers whose endpoint class is not
0or±A_jmodulo the relations. The maximum is two, and it is attained. An invertible 2×2 target is never such a class, so this agrees with #5481 for one and two auxiliaries.How the 120^5 search becomes small
1_S - 1_Tis inL ∩ {-1,0,1}^5. So only the saturation of that cube part matters. That leaves 4,060 rank-2 lattices with nonzero frames, plus 117 of rank ≤ 1.offdiag N_j = (b1)_j M1 + (b2)_j M2 mod 2for a Z-basisb1, b2ofL. Each call's pattern is therefore fixed by its column class. Non-zero patterns have at most 2 lifts, and the zero pattern has W.Evidence (CPython 3.12, stdlib only)
--long): 71.4M representatives (3.04G words), 53,463,710 diagonal, 0/1/2 = 28,198,385 / 21,187,785 / 4,077,540; rank ≤ 1: 785,591 diagonal.--long. Each finds identical diagonal sets and histograms, and no diagonal word violates step 2.D_N(B)D_N(-B)accepted, saturated bases.python3 -I -B research/bit-span4-five-call-two-relation/verify.py(about 3 min) →PASS_FIVE_CALL_TWO_RELATION_CENSUS ...;--longadds W = 5. Registered replay:PASS_REGISTERED_RESEARCH_PACKAGE ... outcome=negative-result(python3.12, 2m44s); the--longrun (CPython 3.14, macOS) wrote the W = 5 and (2,5) rows.Scope. This is a finite census in a formal model. It is not a proof for three or more auxiliaries and does not re-check the conic argument. Frame span ≤ 2 is cited (Lemma 2 of
bit-dirty-auxiliary-registers). The high test is a necessary condition only. No physical supplier, pricing, compiler or recursive-bill obligation is addressed. The selected record is unchanged.Open next questions. (1) Prove "at most two high registers" for every W, which would give a conic-free proof of #5481. (2) Use the derivative condition to prune #5481's remaining eight-call span-5/6 cases.
Overlap. I chose the span-four open question first. The overlap check then found #5481, so I reframed the work as corroboration. The span-four README's open question 1 is now stale. That package is frozen, so I did not edit it.
Credit. unpaidmoon (span-four reduction), JoyboyBrian #5481 (the theorem corroborated here), waynenilsen (dirty-auxiliary lemmas), Ali Romman #468, eumemic #288, huxint #203, Hanson Sun (seed04 derivative step), and the seed52 and woklerx pattern-plane censuses. Card
bit-new-register-model, seeddf1ea4ba466d07a5fa28dab974fcdd78, base721c0382(priority revision 8).Prepared for Josie Elliston with substantial assistance from Claude (Anthropic). No independent human review is claimed. Apache-2.0.
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