Math Conjectures

Hadwiger–Nelson Problem

OPENLandmarkExact constant problemProposed 1950 · Standard version

Canonical statement

Determine
χ(R2)=min⁡{k:∃c:R2→{1,…,k}, ∥x−y∥2=1⇒c(x)≠c(y)}. \chi(\mathbb R^2)= \min\bigl\{k:\exists c:\mathbb R^2\to\{1,\ldots,k\}, \ \|x-y\|_2=1\Rightarrow c(x)\ne c(y)\bigr\}.
View source LaTeX
Determine
\[
  \chi(\mathbb R^2)=
  \min\bigl\{k:\exists c:\mathbb R^2\to\{1,\ldots,k\},
    \ \|x-y\|_2=1\Rightarrow c(x)\ne c(y)\bigr\}.
\]
DG-001:[Hadwiger1950Nelson][DeGrey2018Plane][ExooIsmailescu2020Plane]
Discrete geometryGraph theory

Notes

The Hadwiger–Nelson problem asks for the chromatic number of the plane: the smallest number of colors kk with which every point of R2\mathbb R^2 can be colored so that no two points at Euclidean distance exactly 11 share a color. The question took shape around 1950 and is traditionally attributed to Nelson and to Hadwiger, whose work on coverings of Euclidean space by congruent sets contains a closely related formulation [Hadwiger1950Nelson].

Both known bounds come from explicit objects. A periodic coloring based on a hexagonal tiling of suitable mesh uses seven colors, giving χ(R2)≤7\chi(\mathbb R^2)\le 7. In the other direction, finite unit-distance graphs force lower bounds: for decades the record was 44, until de Grey exhibited a finite unit-distance graph that is not 44-colorable, proving χ(R2)≥5\chi(\mathbb R^2)\ge 5 [DeGrey2018Plane]; Exoo and Ismailescu subsequently gave a new proof [ExooIsmailescu2020Plane].

Each of the values 55, 66, and 77 remains possible: no finite unit-distance graph forcing six colors is known, nor is any coloring of the plane with fewer than seven. The problem remains open.

References (3)

The boxed statement is the canonical open formulation — not a stronger variant or a related research program. The status reflects the catalog's last review; do your own literature search before investing serious effort.