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Transport three-frame character exclusions across count profiles: (1,1,2,0) excluded, twelve profiles not character-excludable (no κ) - #7128
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…nt profiles Eight linear symmetries of the address space permute A, B, P, Q and the four targets, and carry the character table of a decomposition to its image. The profile (1,1,2,0)/(1,1,1) is therefore excluded as the image of (2,0,1,1)/(1,1,1), conditional on the cited exclusion theorem. The sixteen census profiles still open form three orbits, and two explicit decompositions pass every character on twelve of them.
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Result
No κ claim and no word. The selected record is unchanged.
In the PR2085/2112 register model with three added signed rank-one frames, eight linear symmetries of the address space permute the old labels
A, B, P, Qand the four targets. They carry the whole character table of a decomposition of one old-count profile to its image. Three results follow for the census in discussion comment 19377345.(1,1,2,0)/(1,1,1)admits no seven-call word. It is the image of(2,0,1,1)/(1,1,1)under the mapM -> W M W^T / 2,W = [[1,1],[1,-1]], which exchangesAwithPandBwithQ. Prove the (1,1,0,2) three-frame profile admits no seven-call word. #7051 names this profile as not settled.(2,1,1,0)/(1,1,1)and the 4-orbit of(0,1,1,1)/(1,1,2)an explicit rank-one decomposition passes every additive character. The argument of Prove the (0,2,1,1) three-frame profile admits no seven-call word. #7013, Prove the (2,0,1,1) three-frame profile admits no seven-call word. #7035 and Prove the (1,1,0,2) three-frame profile admits no seven-call word. #7051 cannot be repeated there.O1(1,1,1,1)/(1,1,1)O2(2,0,1,1),(0,2,1,1),(1,1,2,0),(1,1,0,2), all/(1,1,1)(1,1,2,0)is new hereO3(2,1,1,0)/(1,1,1)O4(0,1,1,1)and its three images,/(1,1,2)O5/(1,1,1)The surviving decompositions are
-B + (5/2)Q + (1/4)(1,3)(1,3)^Tfor(2,1,1,0)/(1,1,1)and(1/2)B + (1/2)(2,1)(2,1)^T + 2Qfor(0,1,1,1)/(1,1,2). On each, all eight characters of the label lattice detect at least as many calls as targets. Both fail target support, so they are not candidate words.Claim types
PROOF.md, Sections 2 and 3).(1,1,2,0)/(1,1,1). It uses the exclusion theorem of open PR Prove the (2,0,1,1) three-frame profile admits no seven-call word. #7035 (the theorem of open PR Prove the (1,1,0,2) three-frame profile admits no seven-call word. #7051 gives the same conclusion through the mapP <-> Q) and the determinant and character premises of PR Prove eight-call floor for two arbitrary signed rank-one frames (no κ) #6190, Section 1. None of these is reproved or replayed here. Prove the (2,0,1,1) three-frame profile admits no seven-call word. #7035 and Prove the (1,1,0,2) three-frame profile admits no seven-call word. #7051 are open and unreviewed.O3andO4.This is not a complete multiplication theorem.
Evidence and reproduction
Package
research/three-frame-profile-orbits. CPython 3.12, standard library only, from the repository root. The interpreter used locally was~/git/integer-mult-bounds/.venv/bin/python3.12.Observed locally:
Both registered commands take about four seconds each.
The verifier checks, in exact rational arithmetic:
I, preserve the determinant, permute the labels and the targets, and act unimodularly onZA + ZB + ZP;A + B + 2Qof Prove the (2,0,1,1) three-frame profile admits no seven-call word. #7035's named decomposition2B + P + Q, with its character that detects two calls and four targets;(1,1,2,0)/(1,1,1)are character-excluded, directly on the new profile;O3andO4;Limitations and unrun obligations
(1,1,2,0)/(1,1,1)is conditional on the theorem of Prove the (2,0,1,1) three-frame profile admits no seven-call word. #7035 or Prove the (1,1,0,2) three-frame profile admits no seven-call word. #7051 and on the premises of Prove eight-call floor for two arbitrary signed rank-one frames (no κ) #6190, Section 1. Those proofs were not replayed.O3,O4andO5are not settled.O3orO4. They do not produce a word.Open next questions
O5, represented by(0,1,1,1)/(1,1,1)withR+S+T = diag(3,2), character-excluded? All 1870 box triples are. A proof would close four profiles. First step: run the parity-image argument of Prove the (2,0,1,1) three-frame profile admits no seven-call word. #7035 with3 x1^2 + 2 x2^2in place of3 x1^2 + x2^2. Blocker:diag(3,2)has determinant 6, so the 2-adic case split has to be redone.(2,1,1,0)/(1,1,1)fail a character or target support? In the box none of 1936 triples passes both. First step: describe the decompositions that pass every character, starting from the latticeZ^3 + Z(1/2,1/2,1/2)of the witness, and test the subset sums forA+Qon that family.(0,1,1,1)/(1,1,2), where 385 box triples survive the characters and none passes target support.Related open work
(2,0,1,1)/(1,1,1). Its theorem is the input of the transported exclusion here. Its named decomposition2B + P + Qis transported and checked.(1,1,0,2)/(1,1,1)and poses the question answered here. It maps to(1,1,2,0)underP <-> Q, so it gives the same conclusion independently of Prove the (2,0,1,1) three-frame profile admits no seven-call word. #7035.(0,2,1,1)/(1,1,1), the fourth profile of the same orbit. By the transport lemma the three exclusions Prove the (0,2,1,1) three-frame profile admits no seven-call word. #7013, Prove the (2,0,1,1) three-frame profile admits no seven-call word. #7035 and Prove the (1,1,0,2) three-frame profile admits no seven-call word. #7051 are images of one another.Credit
The register model is Cinna Davis's PR2085/2112 formal model. The determinant, support and character premises are the PR2910 package as recorded by apoursh in #6190. The census is discussion comment 19377345. The exclusions of
(1,1,1,1),(0,2,1,1),(2,0,1,1)and(1,1,0,2)are unpaidmoon's #6942, #7013, #7035 and #7051. The symmetry lemma, the transported exclusion, the two surviving witnesses and the checker are produced by Nick Petalas.CI notes
CI did not certify this package. At head
633774e4d9442ccd07f9e5138991d3ebb226547c, run 42906716667 failed in theplanjob (job 130146489645), and the packagereplayjob was skipped becauseplanfailed. The evidence for this PR is the local runs listed above.The
planfailure is one test of 760 (24 skipped) in the "Onboarding helper tests" step:The cause is on
main, not in this PR. Ate54523517d0f03e58352e647c403a5af3e525a57,certificates/selected-result.jsonnamesmake frontier7000-verify, whiledocs/ci-verification.mdis still headed "Current selected-family replay: #6030" and does not contain that command. This PR changes only files underresearch/three-frame-profile-orbits/.The same single test is the only failure, with the same 760/24 counts and
replayskipped, in five other runs from today that I read: 42905094314 and 42903313790 on other contributors' branches, and 42905202459 (#7125), 42905330981 (#7126) and 42905806694 (#7127). No fix or retrigger was pushed, and the job was not re-run.