Hadwiger–Nelson problem

Put three dots at the corners of a triangle whose sides are all one unit long. Every pair must have different colors, so two colors cannot work. Now color every point of the ordinary flat plane under that same rule: points exactly one unit apart must differ. What is the fewest colors that works everywhere?[1] [2]

Points exactly one unit apart must receive different colors, even when the task includes every point of the plane. How many colors are necessary and sufficient? The triangle with corners $(0,0),(1,0),(1/2,\sqrt3/2)$ has all three pairwise distances equal to one, so it forces three colors.

Let $G$ have all points of $\mathbb R^2$ as vertices and an edge exactly when Euclidean distance is one. This is the plane's unit-distance graph. Its chromatic number $\chi(G)$ is the smallest number of color classes avoiding a same-colored edge. No measurability assumption or extra boundary rule is imposed on those classes.[1] [2]

The triangle only forces three colors. Larger finite configurations force at least five, while a coloring of the whole plane works with seven. The exact answer is still among five, six and seven. A successful example using seven does not prove that six is impossible.[1] [2]

The established bounds in the reviewed sources are $5\leq\chi(G)\leq7$. De Grey's finite unit-distance construction rules out four colors. The seven-color upper bound uses a repeating hexagonal tiling, with tiles small enough to avoid unit distances within one tile and repeated colors sufficiently separated. Covering the tile boundaries is part of coloring every point.

The September 2026 typical-norm preprint changes the distance rule and explicitly distinguishes the Euclidean case. Its results do not supply a five- or six-color Euclidean construction, or an obstruction forcing six or seven.[1] [2]

A rule about one pair of points can constrain an unlimited region. The interest is in how small arrangements create unavoidable conflicts, and how a repeating geometric construction can avoid all such conflicts across the plane.[1] [2]

The same pairwise rule has to survive both dense local arrangements and an infinite extension. Finite configurations can certify impossibility with a finite witness; a plane-wide coloring certifies possibility by a rule for every point. Determining the minimum connects these two kinds of evidence. It would also clarify which coloring restrictions come from ordinary Euclidean distance, rather than from features shared by other distance rules.[1] [2]