Skip to content

research: exclude 15-call five-b UT4 splits across dirty workspaces - #2828

Merged
Mister-Meeseeks merged 1 commit into
CrocSwap:mainfrom
steph4n-gh:research/ut4-fifteen-call-five-b-seed09aa8c
Oct 16, 2026
Merged

Mister-Meeseeks merged 1 commit into
CrocSwap:mainfrom
steph4n-gh:research/ut4-fifteen-call-five-b-seed09aa8c

Conversation

@steph4n-gh

Copy link
Copy Markdown
Contributor

Summary

Resolves the explicit 15-call five-$b$ weighted-tariff handoff formulated by Gareth Davidson (@bitplane) in PR #2314 (research/ut4-mixed-weighted-boundary) and PR #2343 (research/register-pairing-compression, Discussion #2019 comment) for $\operatorname{diag}([z], [uz], I_{\text{rest}})$ in $\mathbb{F}_2[\mathrm{UT}_4(\mathbb{F}_2)]$ across arbitrary finite restored dirty workspaces ($r \ge 2$):

  1. Nilpotent Same-Label Dirty Compression for (6,5,4) and (4,5,6) to Two Roles (Lemma 1): Because $n_b = 5 \implies e_b^x = e_b^y = 0$ and $n_c = 4 \implies e_c^x = e_c^y = 0$ (or $n_a = 4 \implies e_a^x = e_a^y = 0$), only one outer generator label can have nonzero dirty row or column components. Any dirty interior contraction $v_{i_m}(u_{i_{m+1}}) \ne 0$ in a product subsequence therefore occurs between two adjacent calls of the same outer label, which vanishes identically in $R = \mathbb{F}_2[\mathrm{UT}_4(\mathbb{F}_2)]$ by $([a]+1)^2 = ([c]+1)^2 = 0$. Active-corner projection preserves the $2\times 2$ active block over $R$, and the predecessor exclusions at length $\le 14$ (PR Research: prove the mixed UT4 z/uz endpoint needs fourteen calls #2175, PR Prove a sharp weighted region for the mixed UT4 target #2314) prevent any projected coupling from vanishing.
  2. Three-Role Active-Plus-Dirty Quotient Decomposition for (5,5,5) (Lemma 2): All common-orientation excess profiles compress to two roles. For the remaining two opposite-orientation three-role modes (Case I: $(e_a^x, e_a^y, e_c^x, e_c^y) = (1, 0, 0, 1)$, Case II: $(0, 1, 1, 0)$), in $M_3(Q)$:
    • the 14 basis monomials without adjacent ${A, C}$ decouple completely from the dirty role on $W_{HH}$, restricting the active projections to the exact 5,664 compatible (5,5,5) profile pairs;
    • the off-diagonal blocks $W_{KH} = 0$ and $W_{HK} = 0$ receive zero dirty interior contractions (by $ACA = CAC = 0$ and $\deg \ge 4$ vanishing in $Q$), yielding exact homogeneous linear equations on the dirty prefix profiles;
    • $W_{HH}$ on AC, ACB, BAC receives only the rank-one cross-role contraction.
  3. Complete 5-Gap Rank-Sum Profile Characterization and Exact Censuses (Lemma 3):
    • A 5-gap prefix profile $P \in M_2(\mathbb{F}2)^5$ with gap increments $S_g = P{g+1} \oplus P_g$ is realizable by $n \in {4, 5, 6}$ nonzero rank-one couplings iff $\sum_{g=0}^5 \operatorname{rank}(S_g) \le n$ (9,721, 47,521, and 174,670 profiles; 55,566, 907,200, and 15,471,918 realizations).
    • Two-role censuses: Reproduces PR Research: prove the mixed UT4 z/uz endpoint needs fourteen calls #2175 (4,5,4) (5,536 words) and PR Prove a sharp weighted region for the mixed UT4 target #2314 (5,5,4) / (4,5,5) (211,968 / 209,920 words) byte-for-byte, and exhausts (6,5,4) (3,376 profile pairs, 6,049,824 words), (5,5,5) (5,664 profile pairs, 8,655,168 words), and (4,5,6) (3,280 profile pairs, 5,874,848 words) — 20,579,840 new 15-call words, 0 survivors.
    • Three-role (5,5,5) censuses: Across all 5,664 active profile pairs, both Case I and Case II leave 17,864 dirty profile pairs passing $W_{KH} = W_{HK} = 0$ and 42,447,616 expanded 3-role words per mode (84,895,232 total), 0 survivors.
  4. Main Theorem & Weighted Price Cone:
    • Every 15-call realizing word has $n_b \ge 6$, and every realizing word with $n_b = 5$ requires $n_a + n_c \ge 11$ ($L \ge 16$) on any finite dirty workspace $r \ge 2$.
    • For all $(\alpha, \beta, \gamma) \in \mathbb{R}_{\ge 0}^3$, every realizing word costs at least $\min(4\alpha + 6\beta + 4\gamma,; 4\alpha + 5\beta + 4\gamma + 3\min(\alpha, \gamma))$.
    • The direct 14-call (4,6,4) word is globally optimal on the expanded cone $\beta \le 3\min(\alpha, \gamma)$; in particular, at $(\alpha, \beta, \gamma) = (1, 3, 1)$, cost $25$ is impossible and $26$ is the exact minimum.

Scope & Attribution

Verification

python3.12 -I -B scripts/research_package.py --package research/ut4-fifteen-call-five-b
python3 -B -m unittest discover -s tests -p 'test_research_*.py'

Both verify.py (PASS_UT4_FIFTEEN_CALL_FIVE_B, receipt 9b7ad621553bb1517c52128298e5ea66f3841f51cebed67dc9839d0b6ffb65bf) and controls.py (PASS_UT4_FIFTEEN_CALL_CONTROLS) complete in ~8 seconds total.

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