Minerval

← 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 →

Székely's crossing-lemma 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

Draw the unit circle around each of the n points, discard circles carrying fewer than three points, and form the multigraph whose vertices are the points and whose edges are the arcs between consecutive points on each circle; this graph has about u edges, where u is the number of unit distances, at most two parallel copies of any edge, and at most 2·(n choose 2) = n(n − 1) crossings, since two unit circles meet in at most two points. Because a graph with n vertices and m ≥ 4n edges has crossing number Ω(m^3/n^2), and the bounded edge multiplicity costs only a constant factor, u^3/n^2 = O(n^2), which gives u = O(n^(4/3)).

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 goes through. The only premise beyond elementary geometry is the crossing number inequality, a textbook theorem with a short proof from Euler's formula, and the remaining steps (two unit circles meet in at most twice, so the arc graph has at most n(n − 1) crossings; edge multiplicity is at most two) are checked in a line each. The argument also shows the bound for every strictly convex norm, since only the two-intersection property of the unit circles is used.

Székely's crossing-lemma proof · Minerval