Erdős–Szekeres convex polygon conjecture

Scatter dots in the plane, with no three on one straight line. How many dots force us to be able to choose a specified number as the corners of a convex polygon, one with no inward dents? Four dots can fail to give four corners: three may form a triangle with the fourth inside it. Five dots do force a convex four-corner polygon. The general conjecture predicts the exact required count for every larger number of corners; unused dots may lie inside the chosen polygon.[1]

How many points in the real plane force a convex polygon with a specified number of vertices, regardless of the arrangement? We exclude three points on one line. The chosen points must all be outer corners; other, unchosen points may lie inside their polygon.

A triangle with an interior point shows why four points need not supply a convex quadrilateral. For a checked example, use $(0,0)$, $(4,0)$, $(0,4)$ and $(1,1)$: the last point is inside the triangle, and no three are collinear. The known four-corner threshold is five.

Let $\mathrm{ES}(n)$ be the smallest count guaranteeing n vertices in convex position. The conjecture, for $n\ge3$, is

$$\mathrm{ES}(n)=2^{n-2}+1.$$

There are arrangements with one fewer point and no such n-gon. The remaining issue is whether the proposed count always suffices. Knowing an approximate exponential growth rate does not give this exact guarantee for each n.[1]

The predicted thresholds are proved for polygons with up to six corners. A proof that the proposed rule works for every number of corners is still missing.[1] [2]

The asymptotic $2^{n+o(n)}$ result and restricted computational cases do not prove this exact all-n equality. Here $o(n)$ means an exponent correction whose ratio to n tends to zero; it can still change the integer threshold substantially.[1] [2]

It is a geometric example of a wider idea: a large enough collection cannot avoid every organized pattern. The question asks how the restrictions of a real point arrangement sharpen that principle, even when the dots were placed without any regular plan.[1]

The original existence proof connects the problem to Ramsey theory; a second proof uses the geometry more directly. This makes it a way to compare arbitrary combinatorial patterns with those a real point configuration can actually realize. Sharp thresholds reveal how strongly geometric consistency constrains pattern avoidance, beyond the fact that some finite threshold exists.[1]