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← the claim it bears on

A line of reasoning from premises to a conclusion. Like a claim, it belongs to no one who makes it.constitution →

Spencer–Szemerédi–Trotter incidence proof

This argument, if it holds, bears in favour of the claim.constitution →The maximum number of unit distances among n planar points is O(n^(4/3)).

The argument

Centre a unit circle at each of the n points; a pair of points at unit distance is then a pair of point–circle incidences, so the number of unit distances is half the number of incidences between the n points and these n unit circles. Because the number of incidences between n points and m unit circles is O(n^(2/3) m^(2/3) + n + m), setting m = n gives O(n^(4/3)) incidences and hence O(n^(4/3)) unit distances.

Does it work?

Not yet evaluated.

Premises

Where it is made

Passages where sources make, criticize or mention this argument. They record where it appears, not how strong it is, and none has precedence over another.

No source is recorded as making it.

Earlier ratings

  • Sep 16, 2026

    The inference is immediate once the premise is granted: the argument stands or falls with the incidence bound for points and unit circles, which Spencer, Szemerédi and Trotter proved by cell decomposition in 1984 and which Clarkson, Edelsbrunner, Guibas, Sharir and Welzl reproved by cuttings in 1990; it is also the two-degrees-of-freedom case of the Pach–Sharir theorem. The reduction from unit distances to incidences with m = n circles is exact, each unit-distance pair contributing two incidences.

Spencer–Szemerédi–Trotter incidence proof · Minerval