Skip to content

Census five-call two-relation BIT words: no three rank-two targets at one or two dirty auxiliaries, corroborating #5481 (no new κ) - #6055

Merged
Mister-Meeseeks merged 2 commits into
CrocSwap:mainfrom
JosieElliston:research/s37-fivecall-two-relation-census
Oct 28, 2026
Merged

Mister-Meeseeks merged 2 commits into
CrocSwap:mainfrom
JosieElliston:research/s37-fivecall-two-relation-census

Conversation

@JosieElliston

Copy link
Copy Markdown
Contributor

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 of research/bit-span4-auxiliary-structure and 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 0 or ±A_j modulo 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. Two call subsets share an address iff 1_S - 1_T is in L ∩ {-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.
  2. Derivative condition over F2 (seed04's step 1, with one more relation): offdiag N_j = (b1)_j M1 + (b2)_j M2 mod 2 for a Z-basis b1, b2 of L. 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.
  3. One representative per register-permutation orbit, weighted by orbit size. The weighted counts equal the unreduced run at W = 2.

Evidence (CPython 3.12, stdlib only)

  • W = 4, rank-2 lattices: 18.8M orbit representatives (222M words), 9,388,016 diagonal, high histogram 0/1/2 = 2,843,284 / 4,763,716 / 1,781,016. Rank ≤ 1: 183,432 diagonal. W = 5 (--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.
  • Unpruned cross-checks enumerate every word for every lattice at (W,k) = (2,4) and (3,3), plus (2,5) with --long. Each finds identical diagonal sets and histograms, and no diagonal word violates step 2.
  • Five controls: the detector fires on six coordinate calls, a cancelling pair is low, non-diagonal rejection, 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 ...; --long adds W = 5. Registered replay: PASS_REGISTERED_RESEARCH_PACKAGE ... outcome=negative-result (python3.12, 2m44s); the --long run (CPython 3.14, macOS) wrote the W = 5 and (2,5) rows.
  • OpenAI Codex read-only adversarial review: no gap in steps 1–2, the enumeration, orbit weights, diagonal acceptance or the high test. Its must-fix was the then-pending W = 5 receipt (now recorded). Its should-fix on control 5's description was applied. This is not independent human review.

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, seed df1ea4ba466d07a5fa28dab974fcdd78, base 721c0382 (priority revision 8).

Prepared for Josie Elliston with substantial assistance from Claude (Anthropic). No independent human review is claimed. Apache-2.0.

🤖 Generated with Claude Code

JosieElliston and others added 2 commits October 27, 2026 23:22
…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>
@Mister-Meeseeks
Mister-Meeseeks merged commit c62bc64 into CrocSwap:main Oct 28, 2026
1 of 2 checks passed
Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment

Labels

None yet

Projects

None yet

Development

Successfully merging this pull request may close these issues.

2 participants