Erdős distinct distances problem

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Template:Short description In discrete geometry, the Erdős distinct distances problem states that every set of points in the plane has a nearly linear number of distinct distances. It was posed by Paul Erdős in 1946.[1][2] The current best result was achieved by Larry Guth and Nets Katz in 2015.[3][4][5]

Erdős considered this problem as his "most striking contribution to geometry".[6]

The conjecture

Place n distinct points in a plane. There are 12(n2−n) distinct pairs between them. Of these pairs, some have the same length, and some have different lengths. The maximum number of different lengths achievable is 12(n2−n), using the following points: (0,0),(0,1),(0,3),(0,7),…,(0,2n−1−1).

The minimum number of different lengths achievable is more difficult to find. Let this number be Template:Math. Equivalently, it is the smallest possible cardinality of their distance set.

In his 1946 paper, Erdős proved the estimates

n−3/4−1/2≤g(n)≤cn/log⁡n

for some constant c. In big-O notation, g≤O(n/log⁡n).

The lower bound was given by an easy argument. The upper bound is given by a n×n square grid. For such a grid, there are O(n/log⁡n) numbers below n which are sums of two squares, expressed in big O notation; see Landau–Ramanujan constant.

Erdős conjectured that the upper bound O(n/log⁡n) is very close to being tight: g(n)=Ω(nc) holds for every Template:Math, using big Omega notation. More succinctly, g(n)≥Ω∗(n).

He further conjectured that the upper bound is exactly tight: g(n)=Θ(n/log⁡n), and offered a prize of $500 for either proving or disproving the conjecture.[7]

Partial results

Paul Erdős' 1946 lower bound of Template:Math was successively improved to:

Variants

Restricted subsets

Instead of allowing the n distinct points to be placed anywhere in a plane, we can additionally require the points to satisfy constraints. In general, the more stringent the constraints are, the larger g gets.

If we require the points to fall on a single line, then gline(n)=n−1 by setting the points equally spaced. If we require the points to fall on a single circle, then gcircle(n)=⌊n/2⌋ by setting the points equally spaced around the circle. More generally, if we require the points to form a convex polygon, then gconvex(n)=⌊n/2⌋. The same, if we require the points to form a strictly convex polygon.[15][16]

If we require the points to be in general position, meaning no 3 points are collinear and no 4 points are cocircular, then it is an open problem. Currently the best result is ggen(n)≥Ω(n) and ggen≤n2O(log⁡n).[17]

If we require the points to be in general ,position and no 4 points form a parallelogram, then it is an open problem. Currently the best result is gpara(n)≥Ω(n) and gpara≤O(n2/log⁡n).[18][17]

See Section 3 of [19] for more problems and results of the kind.

Higher dimensions

Erdős also considered the higher-dimensional variant of the problem: for d≥3 let gd(n) denote the minimal possible number of distinct distances among n points in d-dimensional Euclidean space. He proved that gd(n)=Ω(n1/d) and gd(n)=O(n2/d) and conjectured that the upper bound is in fact sharp, i.e., gd(n)=Θ(n2/d). József Solymosi and Van H. Vu obtained the lower bound gd(n)=Ω(n2/d−2/d(d+2)) in 2008.[20]

In the other direction, it's known currently that g3(n)≥Ω∗(n3/5), by applying the recursion relation of [21] to the result g2(n)≥Ω∗(n) of Template:Harvard citation.[19]

Generic norms

The same question can be asked for any normed space. Given a norm ‖⋅‖, define g‖⋅‖(n) accordingly. The problem is solved in the generic case. Specifically, given any integer d≥2, for almost all norms ‖⋅‖,g‖⋅‖(n)=(1−o(1))nThat is, the set of norms that violate this condition is meagre in the set of all norms of ℝd, regarded as a metric space, metrized by the Hausdorff distance between the norm-unit balls.[19][22]

See also

References

Template:Reflist

Further reading