kernel-checked, filed Tue Sep 08 2026 06:32:53 GMT+0000 (Coordinated Universal Time) by @savcab
For four edges, avoidance requires rainbow four-cycles in all three coordinate projections; a finite counterexample shows why checking two projections is insufficient.
Scope. For every fixed e≥3, HasQuadraticVanishing 3 (e+3) e is equivalent to TripartiteRelationBound e. HasQuadraticVanishing 3 7 4 is equivalent to RainbowDensityBound, with proper coloring ensured by ExplicitDisjoint in all relation bounds. The specified four-edge relation has seven coordinate vertices, is linear, and has a repeated diagonal color only in its first projection. Neither density assertion is proved by this equivalence.
prior art, filed Tue Sep 08 2026 03:38:52 GMT+0000 (Coordinated Universal Time) by @savcab
Together with the classical three-edge upper bound and finite reductions, the full original question is equivalent to its upper-bound case for triple systems with at least four edges.
Scope. For every r,e≥3 and d<(r−2)e+3, extremal(r,d,e,n) is not o(n²); for every r≥3 the e=3 upper bound is o(n²); and the entire original all-r, all-e, both-halves statement is equivalent to: for every e≥4, extremal(3,e+3,e,n)=o(n²). The right side of the equivalence remains unproved.
open, filed Tue Aug 25 2026 06:15:56 GMT+0000 (Coordinated Universal Time) by @woshuajolk
Prove d_r(e)=(r−2)e+3.
An injective copy of a graph on Fin d allows isolated labelled vertices, exactly matching e edges spanned by at most d vertices. Avoidance is non-induced, as required by the extremal-number convention.
Scope. All natural r,e at least 3.