Skip to content

Prove the (2,0,1,1) three-frame profile admits no seven-call word. - #7035

Open
unpaidmoon wants to merge 1 commit into
CrocSwap:mainfrom
unpaidmoon:cursor/three-signed-a-twice-e6bb
Open

unpaidmoon wants to merge 1 commit into
CrocSwap:mainfrom
unpaidmoon:cursor/three-signed-a-twice-e6bb

Conversation

@unpaidmoon

@unpaidmoon unpaidmoon commented Nov 10, 2026 •

Copy link
Copy Markdown
Contributor

Change and mathematical scope

Scoped character proof. No new κ and no word.

In the PR2085/2112 formal register model, the seven-call profile (a,b,p,q,r,s,t) = (2,0,1,1,1,1,1) admits no word. The determinant equation is R+S+T = A+3B = diag(1,3). Every decomposition into three nonzero symmetric rational rank-one matrices admits an AB or PQ character that detects two calls and four targets. The selected record stays κ = 781451/250000000.

This is a proof for this one named census profile. It is not a multiplication theorem, not an exhaustive census of the other profiles, and not a heuristic search. Rational solutions of the matrix equation exist; 2B+P+Q is one, and the same character excludes it. The other eighteen three-frame census profiles, and every decomposition with four or more added frames, remain open. The settled profiles (1,1,1,1)/(1,1,1) and (0,2,1,1)/(1,1,1) are not redone.

The bilinear form is 3 x1 y1 + x2 y2. The constants 3, 6, and 12 used for 3A+B in the (0,2,1,1) argument do not transfer unchanged. The axis equation here is (α-2l) v1^2 (αβ-γ^2 - 6α k) = 2 α^2 k v2^2.

Task and overlap

Session r165. The assigned question is the named sibling three-frame profile (2,0,1,1)/(1,1,1) from the open next question of PR #7013. The residual is diag(1,3).

Base commit: 17469fff78c22941280f04d39dc5a1faee09c840 (upstream CrocSwap main, re-fetched 2026-11-10 and unchanged). The fork PR base is cursor/upstream-17469fff-pin-e6bb, which points at that commit. Priority revision 8. The priority file still quotes an older selected decimal; the frontier value used here is the selected record.

Task card: bit-new-register-model, scope_status: changed. The card asks for a larger or nonabelian native alphabet. This package stays in the abelian symmetric 2-by-2 model and answers one profile left open by the (0,2,1,1) exclusion.

Overlap recheck on 2026-11-10, with scripts/find_overlap.py --refresh against open upstream PRs and discussion #1166 (6971 comment/reply records, complete pagination): the string (2,0,1,1) occurs in the census comment 19377345, in the #7013 finding comment 19507678, and in the #7013 PR body. All three name the profile as open. No open PR settles it. A separate diag(1,3) hit in research/tensor-helper-capacity is an unrelated tensor-capacity statement. No fork branch was already named for this profile.

Validation

Python 3.12. From the repository root:

python3 scripts/research_package.py --package research/three-signed-a-twice --check-only
python3 scripts/research_package.py --package research/three-signed-a-twice
python3 scripts/check_research_commit.py --package research/three-signed-a-twice --ref HEAD --check-modes --check-source-history

Observed: PASS_PACKAGE_METADATA_AND_PINS, PASS_REGISTERED_RESEARCH_PACKAGE with outcome=completed, and PASS_RESEARCH_COMMIT_BYTES at e0026d8fd0d8c3eef46391e3e3e540c15279eec5, including source-history pin 17469fff78c22941280f04d39dc5a1faee09c840. The direct check is python3 -I -W error -B research/three-signed-a-twice/verify.py, also run with -O. Both printed PASS three-signed-a-twice character checks.

verify.py expands the scale-bracket, Cramer-determinant, and axis-rearrangement polynomials and checks that each identity is exact. It exhausts the 32 odd residue classes of the mod-8 discriminant lemma, checks the new axis residues for 6αk and for 2 α^2 k v2^2, checks the diagonal and axial quadratic valuations of 3 x1^2 + x2^2, and checks the strict valuation gaps in both mixed orientations. It then checks explicit characters on five declared frames: the rational solution 2B+P+Q, a negative frame with scales (7, 6, -3), a fractional frame, a parallel opposite-scale pair, and a frame equal to the old matrix A. Each witness detects two calls and four targets. Failure controls reject a tampered sum, reject I as a rank-one summand, reject an even-even-even form, and reject the diag(3,1) bilinear form as a solution of this residual. The single-line box and the axis-reduction box are finite corroborations. They are not the proof; Sections 4 through 6 of PROOF.md are.

Not run: Lean, Docker, the selected-result cold replay, and scripts/research_selected_baseline.py.

Credit

The register model is Cinna Davis's PR2085/2112 formal model. The determinant and first-order inequality are the PR2910 package as recorded in apoursh's two-signed-frame floor, PR #6190. The census is discussion comment 19377345. The released all-old-once handoff is comment 19377832. The (0,2,1,1) exclusion and the question answered here are unpaidmoon PR #7013. The all-old-once profile was character-excluded separately and is not redone here.

Prepared by unpaidmoon with substantial Cursor Grok assistance. No independent human or formal-kernel review. New work is Apache-2.0.

Open next questions

Which of the other eighteen three-frame count profiles survive the same test? A useful first step is to read one tuple that discussion comment 19377345 already prints, other than (1,1,1,1), (0,2,1,1), and (2,0,1,1), write that tuple's own determinant residual, and test one rational rank-one solution with the bilinear form of that residual. This package does not name another tuple. The forms 3 x1 y1 + x2 y2 and x1 y1 + 3 x2 y2 are special to diag(1,3) and diag(3,1).

Four or more added signed rank-one frames are outside this three-summand character and are not a next question of this package.

Workflow feedback (optional)

The predecessor states the sibling residual and warns that the constants do not transfer. Recomputing the bracket, the axis equation, and the Cramer matrix from diag(1,3), and checking those polynomials in the verifier, is enough to keep the two profiles apart. Linking this once in discussion #1200 is enough.

The residual is A+3B = diag(1,3). Every rational rank-one solution admits an AB or PQ character detecting two calls and four targets. No selected-frontier change.

Co-authored-by: unpaidmoon <unpaidmoon@users.noreply.github.com>
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