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Record the square i-block obstruction for mixed negative Q8 targets. - #4378
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A square i-call block forces at least twice as many j-calls, so the inherited 11t/5 bound is not attained on the rays t_i=4 t_j and t_j=4 t_i, and the (7,1) count witness (7,13) is unrealizable. The other 20-call splits stay open. No new kappa. Co-authored-by: unpaidmoon <unpaidmoon@users.noreply.github.com>
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Change and mathematical scope
Scoped obstruction in the existing full formal
F2[Q8]dirty-register model. No new κ. The selected conditional value stays0.001672660197076918.If an exact word uses only
t_icalls labelledi, it needs at least2 t_icalls labelledj. Trace parity raises that floor when2 t_iandt_jhave opposite parity. Dirty coordinates can change the unitriangular factorK_i, but not its invertibility or this active-image count. The label swap gives the symmetric statement.The inherited inequality
5n >= 11tis therefore never attained on the rayst_i = 4 t_jandt_j = 4 t_i. For4kcopies of[zi]andkcopies of[zj], length11khas a unique cubic split(4k, 7k), and that split is a square i-block with too few j-calls. Every exact word on this ray has length at least11k+2. The direct three-call word still costs15k.At seven
[zi]targets and one[zj]target, the same lemma excludes the relaxation witness(n_i, n_j) = (7, 13). The other 20-call splits(9, 11),(11, 9)and(13, 7)remain open. No 20-call word is constructed or ruled out. The direct word has length 24, so the exact minimum stays in{20, 22, 24}.This is not a new alphabet, a physical caller, a supplier admission, or a multiplication theorem.
Task and overlap
12fa8c5c844be21bfefb38a19503fbd867a11244(canonical main, priority revision 7).da82b198bd06dfc248bdcdf4ab477da50688fb07.research/q8-square-iblock-obstruction.bit-new-register-model, scope changed. The card asks whether a new alphabet escapes the three-register obstruction. This package does not change the alphabet; it removes unrealizable equality cases inside the model already on main. Reason inTASK.json.(7,13)as a minimizing count witness. It does not decide either open question. This package does not replay its enumeration and does not depend on the unmerged tree. Premises are the pinned proofsresearch/q8-mixed-boundary-rank,research/q8-homogeneous-mixed-floorandresearch/q8-count-region-rank-floor.Item 9 of the chronological #1166 queue continued from unpaidmoon PR #4161, after dropping delivered rows and keeping comment 19035513 because the census lever that #4161 stopped on was closed by #4090. The #4100 audit map is left to open PR #4311.
(7,18)/(7,20)functional census (comment 19021170, question 1).Validation
Python 3.12, standard library only.
Passed locally as
PASS_REGISTERED_RESEARCH_PACKAGE research/q8-square-iblock-obstruction outcome=partial. The verifier printsPASS_Q8_SQUARE_IBLOCK_OBSTRUCTION.Not run: an endpoint search of
(9,11),(11,9),(13,7)or of the 13-call splits at shape(4,1); Lean; Docker; the selected full cold replay. The image argument isPROOF.md. The executable checks are corroboration.Credit
Nick Petalas, PR3570 / PR3578 / PR3589 / PR3736 and open PR #4234, for the cubic ranks, the homogeneous quadratic method, the count region, and the two open questions. PR3536 and PR3558 retain the chronological and cyclic methods. The lineage in those proofs stays with PR3517, Tuafo/PR2900, ywz65/PR2587, PR3487, Aaron Wang/PR2318, Cinna Davis, steph4n-gh and the credited Q8 register sources.
Prepared by unpaidmoon with substantial Cursor Grok assistance. Apache-2.0. No independent human or formal proof-kernel review is claimed.
Open next questions
(7,1)have a 20-call word on(9,11),(11,9)or(13,7)? First step: the cubic-tight split(13,7), whererank(X_j) = rank(Y_j) = 4. Blocker: no endpoint search is included.(4,1)minimum 13 or 15? The square lemma still allows(4,9),(6,7),(8,5)and(10,3)at 13 calls. Blocker: no exhaustive search at that length.