The chromatic number of the plane

How many colours are needed to paint the plane so that no two points at distance exactly $1$ get the same colour? Seven colours suffice, using a hexagonal tiling, and for decades only four were known to be necessary. After a burst of progress since 2018, the answer is now known to be $6$ or $7$, and deciding between them is the problem.

Problem

Let $\chi$ be the smallest number of colours with which the points of $\mathbb{R}^2$ can be coloured so that any two points at Euclidean distance $1$ receive different colours; equivalently, $\chi$ is the chromatic number of the unit-distance graph of the plane. No regularity (such as measurability) is assumed of the colour classes. Determine $\chi$.

Background

The problem dates to Edward Nelson in 1950, who showed $\chi\ge4$; John Isbell found the upper bound $7$, and Hadwiger published a formulation in terms of coverings, which is why it is called the Hadwiger–Nelson problem. Erdős included it in several of his problem papers. By the de Bruijn–Erdős compactness theorem, $\chi\ge k$ holds exactly when some finite unit-distance graph needs $k$ colours, so lower bounds can be proved by exhibiting finite graphs. The seven-vertex Moser spindle gives a small proof that $\chi\ge4$.

What is known

• Upper bound. $\chi\le7$, by colouring a tiling of the plane by regular hexagons of diameter slightly less than $1$.
• At least five. De Grey (2018) found a unit-distance graph with $1581$ vertices that is not 4-colourable. Later work found smaller examples (Heule, 553 vertices; Parts, 509 vertices) and alternative proofs (Exoo and Ismailescu, 2020).
• At least six. A September 2026 preprint posted by OpenAI proves that every 5-colouring of the plane has a monochromatic unit-distance pair, so $\chi\ge6$; the source page links a Lean formalisation of this lower bound and now gives $6\le\chi\le7$. The proof transfers an arbitrary colouring to a measurable one and does not exhibit a finite graph. By the de Bruijn–Erdős theorem a finite unit-distance graph that is not 5-colourable must exist, but none has been published.
• Fractional version. Matolcsi, Ruzsa, Varga and Zsámboki (2023) proved that the fractional chromatic number of the plane is at least $4$; Croft (1967) gave the upper bound $4.359\ldots$.

What would count as progress

• A finite unit-distance graph in the plane that is not 6-colourable, which would prove $\chi=7$; non-colourability can be certified by a SAT solver.
• A 6-colouring of the plane without monochromatic unit-distance pairs, which would prove $\chi=6$; a periodic colouring can be checked by computer.
• An explicit finite unit-distance graph that is not 5-colourable, with a SAT certificate; or smaller, more transparent witnesses for $\chi\ge5$.
• A formal proof against the Lean statement in formal-conjectures.

References

1. A. D. N. J. de Grey, The chromatic number of the plane is at least 5, Geombinatorics 28 (2018), 18–31. arXiv:1804.02385
2. G. Exoo and D. Ismailescu, The chromatic number of the plane is at least 5: a new proof, Discrete Comput. Geom. 64 (2020), 216–226. doi:10.1007/s00454-019-00058-1
3. M. Matolcsi, I. Z. Ruzsa, D. Varga and P. Zsámboki, The fractional chromatic number of the plane is at least 4 (2023). arXiv:2311.10069
4. OpenAI, The Euclidean plane is not five-colorable, preprint (2026). PDF
5. A. Soifer, The Mathematical Coloring Book, Springer (2009). doi:10.1007/978-0-387-74642-5

Source

Imported from Erdős Problems 508, curated by Thomas Bloom. Open according to the source as of 2026-10-20. The statement above is written in our own words; see the source for the original wording, history and discussion.