[unpaidmoon]

Change and mathematical scope

Scoped classification, not a construction and not a κ change. In the unchanged unsigned dirty connector, the centre pair (19, 20) on s + t = 39 never realizes a weight-3 mask. Every residue-legal centre count is a multiple of 3, and the centre length 19 is not. The frozen-14 step needs a weight-3 centre carrying the value 2c. At characteristic 6 that centre is empty, so the step does not yield a human proof of the six-source depth-two exclusion.

The exclusion itself remains the solver verdict in merged PR #6141 (CaDiCaL 1.9.5 and MapleChrono UNSAT). That UNSAT is solver evidence. This package does not rerun it, does not call it a human proof, and does not emit a DRAT certificate.

The selected conditional κ stays 781451/250000000.

Task and overlap

Base: CrocSwap/integer-mult-bounds 7c49fe32c55de02c2c873655c804b16e4c3fef0a, re-fetched at session start and again before submission; both reads were that commit. This draft uses cursor/upstream-7c49fe32-pin-f87a as its base because that branch is exactly that commit. The fork's main is an older snapshot.

Task card: connector-alphabet-change (priority revision 8). Scope status: changed. The card asks for one new connector operation. This package keeps the alphabet fixed and answers the c=6 half of the question left by the frozen-14 argument: classify the centre of s + t = kc + 3 for c = 6. A new operation is not the missing step. This is a side result on the recursive-interface schedule. It does not change a supplier, a consumer, or κ.

Package: research/six-source-c6-centre-classification.

Overlap was checked with scripts/find_overlap.py --refresh on 2026-11-10 (885 open PRs, complete pagination; 6934 discussion records, complete pagination; local note inventory complete at 7c49fe32, 8917 head records). The report is a lead, not a novelty certificate. Neighbouring work is a different question:

• Merged PR No six-source depth-two connector block in characteristics five and six (local SAT; c = 7 open; no new κ) #6141 leaves the c=6 exclusion as a solver verdict and records no human proof.
• Open PR Certify PR6141 characteristic-five connector exclusion (independent verification; no κ) #6216 certifies characteristic five only and names a characteristic-six DRAT certificate as a separate budget. This package does not start that certificate.
• Discussion comment 19384981 records the c=7 search failure, the slot count 17 at c=6, and the centre shape that frozen-14 uses at c=7.
• Discussion comment 19493159 / fork PR [Closed: Withdrawn]  #156 proves the c=7 diagonal equality case and leaves the c=7 off-diagonal question open. That off-diagonal replacement, and the c=5 centre, are not classified here.

Validation

Python 3.12, standard library only, from the repository root. Author-checked on 2026-11-10:

python3.12 -I -W error -B research/six-source-c6-centre-classification/verify.py
python3.12 -I -O -W error -B research/six-source-c6-centre-classification/verify.py
python3.12 -B scripts/research_package.py --package research/six-source-c6-centre-classification
python3.12 -B scripts/check_research_commit.py --package research/six-source-c6-centre-classification --check-source-history --check-modes

Marker: PASS_SIX_SOURCE_C6_CENTRE_CLASSIFICATION sum19=9156 examined=46956 residue_legal=0 weight3=0 mod3_shapes=0 widened_legal=0 moves=+[0, 6]/-[1, 7] frozen_value=12 controls=8

The written argument is in the package README. The checker exhausts every length-19 count vector in {0,...,6}^6 and every legal centre letter. It also checks the seven-value type table, the empty mod-3 shape list, a widened domain that admits the unreachable value 13, and the one-step move table. It does not search words and does not rerun the SAT encoding. Both command modes, the registered package replay, and the commit-byte check passed. No Lean, Docker, or historical suite was run.

Hypotheses, stated exactly: k = 6, c = 6, floor words of length 36, depth two in the PR4248 sense, and a weight-3 mask meaning every coordinate of q_s + q_t lies in {0,1,6,7,12} with exactly three coordinates in {1,7}. The census is the finite set of centre counts of length 19. It is not a census of words.

Two explicit floor words realize a weight-3 mask off this centre, one at (18, 21) and one at (0, 9). They are witnesses that the centre obstruction is local. They are not depth-two words.

Under the further hypothesis that weight-3 masks may use only s + t ∈ {3, 39}, at most 16 slots remain after the dead centre and the dead pair (1, 2), against C(6,3) = 20. That pigeonhole is a human proof of this restricted model. Discussion comment 19384981 already had the same model impossible at 17 slots. It is not the unrestricted exclusion.

Credit

The depth-two interface and the weight identity are Nick Petalas's (PR #4231, PR #4248, PR #6107), with the Ross Richardson PR #934 / PR #943 lineage. JosieElliston's PR #6141 records the c=6 solver verdict. Nick Petalas's handoff on 2026-11-03 (discussion comment 19384981) isolates the diagonal and the frozen-14 mechanism. The c=7 equality-case proof is fork PR #156, noted upstream in discussion comment 19493159. queelius's open PR #6216 is the characteristic-five certificate. zxv's PR #6057 and JosieElliston's PR #6061 supply the c ≥ 8 words.

Prepared by unpaidmoon with substantial Cursor Grok assistance. Apache-2.0.

Open next questions

1. Which human step, other than the frozen-14 centre induction, proves the six-source depth-two exclusion at c=6? The solver verdict is still not a human proof, and this centre gives that induction nothing to freeze. First step: classify the separation-3 pairs on s + t = 39, starting from q_18 = (0,0,0,6,6,6) with three coordinates equal to 12, and price the explicit mask at s + t = 9 against those slots. Blocked because the centre index is empty, so a chain through every diagonal index is unavailable, and the s + t = 9 witness lies off the diagonal the freeze follows. The c=6 DRAT certificate named in PR Certify PR6141 characteristic-five connector exclusion (independent verification; no κ) #6216 is a separate budget.

Workflow feedback (optional)

The c=7 centre type table was enough to name the c=6 calculation: non-centre coordinates of an adjacent pair are even, and 19 is not divisible by 3. A forward search of words was not needed.