[unpaidmoon]

Change and mathematical scope

Scoped obstruction, not a construction and not a κ change. In the unchanged unsigned dirty connector, no length-42 floor word on six sources in characteristic 7 makes every pair (s, 45-s) for s = 4..22 a weight-3 mask. That is the equality case of the weight-3 count: C(6,3) = 20 masks and 20 candidate slots on s + t ∈ {3, 45}. The case is empty.

The selected conditional κ is unchanged. Whether a six-source depth-two block still exists at c=7 by using a pair off this diagonal stays open.

Task and overlap

Base: CrocSwap/integer-mult-bounds 7c49fe32c55de02c2c873655c804b16e4c3fef0a. This draft uses cursor/upstream-main-pin-b580 as its base because that branch is exactly that commit. The fork's main is an older snapshot, so it is not the review base.

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 Nick Petalas's released handoff on open question 1 of PR #6141, the tight diagonal model at c=7. 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-c7-diagonal-obstruction.

Overlap was checked with scripts/find_overlap.py on 2026-11-09 (883 open PRs, complete pagination; 6930 discussion records, complete pagination; local note inventory complete at the base commit, 8917 head records). A query for the diagonal equality case hits discussion comment 19384981 and this package. Neighbouring work is a different question:

• PR No six-source depth-two connector block in characteristics five and six (local SAT; c = 7 open; no new κ) #6141 (merged) leaves c=7 open and excludes c=5 and c=6 by SAT.
• PR Certify PR6141 characteristic-five connector exclusion (independent verification; no κ) #6216 (open) is a c=5 certificate.
• PR Unsigned connectors reach their call floor on 3c−1 sources for every c≥8 #6057 and PR Depth-two connector blocks on six, seven and eight sources: N(c) ≥ 6⌊c/2⌋ for c ≥ 8, up to 8⌊c/2⌋ for c ≥ 27 (no new κ) #6061 give words for every c≥8, which have spare diagonal slots.
• PR Prove price-null fibres need irreducible cycle exchanges (no κ) #6670 answers a symmetric-ladders price-null question and does not treat this model.

Validation

Python 3.12, standard library only, from the repository root. Author-checked on 2026-11-09:

python3.12 -I -W error -B research/six-source-c7-diagonal-obstruction/verify.py
python3.12 -I -O -W error -B research/six-source-c7-diagonal-obstruction/verify.py
python3.12 -B scripts/research_package.py --package research/six-source-c7-diagonal-obstruction --check-only

Marker: PASS_SIX_SOURCE_C7_DIAGONAL_OBSTRUCTION centres=300 type01=180 type31=120 frozen_coordinates=480 moves=+[0, 7]/-[1, 8] controls=4

The written argument is in the package README. The checker exhausts the centre classification and the one-step move table. It does not search words and does not rerun the SAT encoding. Both command modes and the registered package check passed. No Lean, Docker, or historical suite was run.

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 leaves c=7 open and records the slot count. Nick Petalas's handoff on 2026-11-03 (discussion comment 19384981) isolates the diagonal equality case. zxv's PR #6057 and JosieElliston's PR #6061 supply the c≥8 words. queelius's open PR #6216 is the c=5 certificate.

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

Open next questions

1. Does a six-source depth-two block exist at c=7 if a weight-3 mask may be realized off s + t = 45? This is now the whole of PR No six-source depth-two connector block in characteristics five and six (local SAT; c = 7 open; no new κ) #6141 question 1, and it decides whether 7 or 8 is the smallest such characteristic. First step: let exactly one index in 4..22 fail the weight-3 test and ask whether one off-diagonal pair can replace that mask. Blocked because the induction uses every index, and no replacement inequality is proved here.
2. Can the frozen-14 step become a human proof of the c=5 and c=6 exclusions? Those exclusions are still solver verdicts. First step: classify the centre of s + t = kc + 3 for c=5 and c=6. Blocked because those characteristics have fewer than 20 middle slots, so a proof must also bound off-diagonal pairs.

Workflow feedback (optional)

The handoff's centre description (non-centre counts in {0,4,7}, centre count in {0,3}) was enough to stop the word search and case-split. A forward search of the Johnson path was not needed.