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Prove a fourth dirty role keeps the S3 xyx double at six calls (no new κ) - #5947
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Mister-Meeseeks merged 1 commit intoOct 28, 2026
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Answer the fourth-role open question of PR CrocSwap#4084: in M4(F2[S3]) with paid labels {x,y} and all 120 rank-one idempotent couplings, no word of length at most five equals diag([xyx],[xyx],[e],[e]). An involution meet-in-the-middle in the faithful image M8(F2) x M4(F2[C2]) finds zero hits; the direct six-call word attains the target.
This was referenced Oct 27, 2026
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Result
Finite exact minimum (exhaustive finite certificate). No new κ. In
M4(F2[S3])with paid labelsx = (1 2),y = (2 3)and callsD_N(g) = I + ([g]+[e]) Nover all 120 rank-one idempotentsNofM4(F2), the targetdiag([xyx],[xyx],[e],[e])needs exactly six paid calls. Roles 2 and 3 are arbitrary dirty roles returned unchanged.This answers open question 1 of Wentao Li's #4084 (
research/s3-three-role-xyx), which proves six with three roles: a fourth restored dirty role does not save a call. The claim covers four roles only.Evidence
Phi : F2[S3] -> M2(F2) x F2[C2],[g] -> (rho(g), c^sign(g)), is checked to be an algebra isomorphism. In theC2block each paid factor isI + delta Nwithdelta^2 = 0. In theM2block it is a transvection ofF2^8.ball(I,3)meetsT * ball(I,2).ball(I,2)has layers1, 240, 32650. Every element ofball(I,2)and every one-call extension was tested: zero hits. TheM8(F2)image alone also has zero hits.[1, 56, 1750, 45396], matching the pinned Research: one dirty role does not lower S3 longest Coxeter double minimum (no new kappa) #4084 certificate.M4(F2[S3])produces the same 32,891-stateball(I,2)underPhi.diag([xyx],[xy],[e],[e])anddiag([x],[y],[xy],[e]).x y xon role 0, then on role 1, equals the target. Each of its six one-call omissions does not.Reproduce
From the repository root, with CPython 3.12 (here
~/git/integer-mult-bounds/.venv/bin/python3.12):Local results: both direct runs print
PASS_S3_XYX_FOUR_ROLE ... ball_i2_layers=[1, 240, 32650] ... short_hits=0 rho_quotient_hits=0 direct=6 target_minimum=6in about 11 s each. The registered replay printsPASS_REGISTERED_RESEARCH_PACKAGE(about 22 s for both commands), and the commit check printsPASS_RESEARCH_COMMIT_BYTES.Limitations
One target, four roles, and the optimistic unit-cost formal alphabet. Five or more roles, other
S3targets, free address maps, native source generation, hosting and supplier pricing are not covered. No selected-frontier, κ or multiplication-theorem change. No unrun proof obligation is claimed as discharged.Open next questions
ball(I,2)is estimated, not computed, at up to about5 x 10^5states, which is slow for this pure-Python search. A uniform argument in theM_{2m}(F2)block is the likely route.rhoquotient enough for every role count? At four roles theM8(F2)image alone excludes length five. Characterizing short products of these restricted transvections that act as the partial swap would give a cleaner certificate.Related open work
S3alphabet, for the different targetdiag([s],[t],[st]). This package uses the same representation as one block of its faithful encoding and does not compute that target.Credit
Target, alphabet, three-role minimum and the open question: Wentao Li / VictorLiwentao, #4084. Alphabet lineage: Cinna Davis's two-register
S3census and the representation bound of #3893. The four-role search, controls and note were produced by Nick Petalas.CI notes
All checks passed for head
b96ecfbda694cf4917d54b0c567d0ff5cbee8155in run https://github.com/CrocSwap/integer-mult-bounds/actions/runs/40530572035:plan(5m55s) andreplay (research/s3-xyx-four-role-minimum)(2m6s).