hadwiger — chromatic number of the plane

max colors allowed

click a vertex to cycle

The Moser spindle: 7 vertices, 11 edges. Every edge connects two points exactly 1 unit apart in the plane. Try 3 colors first — find a coloring with no edge between same-coloured vertices. You won't be able to. That impossibility is the proof that the chromatic number of the plane is at least 4.

max colors allowed

click a vertex to cycle

Six Moser spindles rotated 60° around a shared centre. All edges below are genuine unit-distance pairs — some inside a single spindle, others crossing between adjacent ones (the 60° rotation forces extra unit-distance constraints). De Grey's actual graph is 1,581 vertices of the same shape: spindles glued through a carefully chosen scheme until even 4 colours can't satisfy every edge. Use reveal coloring to backtrack to a proper colouring at the chosen palette size.

difficulty

click a vertex to cycle

A procedurally generated unit-distance graph. Each one is anchored by either a triangle (χ=3) or a Moser spindle (χ=4), then grown by equilateral-triangle completions on random existing edges. A backtracking solver verifies the chromatic number before the puzzle ships, so the colour target is always honest. Try to finish using exactly the target palette — every vertex coloured, no edge with matching endpoints.

hex side length · s

0.42

Isbell's 1950 upper bound: the plane tiles with regular hexagons of side s, 7-coloured in a repeating pattern. If s < ½, no two points in one hex are at distance 1; if same-coloured hexes sit ≥ √7 · s√3 ≈ 4.58s apart, no unit-distance pair shares a colour. Drag a point to draw a unit circle and see for yourself — its colours never collide.

The problem

Colour every point in the Euclidean plane. How few colours suffice if you forbid any two points at distance exactly 1 from sharing a colour? This number is the chromatic number of the plane, χ(ℝ²). Hugo Hadwiger and Edward Nelson asked the question independently around 1950; it's been called the chromatic number of the plane problem ever since.

By 1950 the answer was bracketed: 4 ≤ χ(ℝ²) ≤ 7. Then nothing moved for 68 years.

The lower bound — 4 (Moser brothers, 1961)

The spindle tab shows the Moser spindle: 7 points in the plane, 11 unit-distance pairs, requiring 4 colours. The proof that 3 don't suffice is purely combinatorial — you try every assignment and fail. The proof that the configuration is geometrically realisable comes from a single calculation: two unit rhombi share a vertex, rotated by θ = arccos(5/6) so their far vertices land exactly 1 apart. χ(ℝ²) ≥ 4 follows.

The upper bound — 7 (Isbell, 1950)

Tile the plane with regular hexagons of side s < ½, and 7-colour them in a repeating pattern (visible in the tiling tab) such that same-coloured hexes sit further than 1 unit apart. Then any unit-distance pair either lives in one hex (impossible, since the hex's diameter is < 1) or crosses into a hex of a different colour. χ(ℝ²) ≤ 7.

The 68-year gap

From 1950 to 2018, the bounds 4 ≤ χ(ℝ²) ≤ 7 didn't budge. Some people pushed for χ = 7, some for χ = 4, but nobody could improve either inequality. The problem became a fixture in popular-math books — Martin Gardner wrote about it; it appears in Open Problems in Combinatorial Geometry; it was the kind of problem you'd mention at a party.

Why bigger constructions are needed

The de grey tab shows a Moser flower — six Moser spindles rotated 60° around a shared centre, giving 37 vertices and 126 unit-distance edges (the spindles share more edges than you'd think, because the rotation puts vertices at unit distance across spindle boundaries). It's richer than the basic spindle, but a backtracking solver finds a proper 4-colouring. So the flower itself doesn't push χ(ℝ²) past 4.

That's the lesson: to force the chromatic number above 4, you need many more vertices, and the identifications between copies have to be just right. De Grey's 2018 graph has 1,581 vertices for exactly this reason.

April 2018 — Aubrey de Grey publishes

Aubrey de Grey is a biogerontologist. He runs a foundation researching longevity. He is not a research mathematician. In April 2018, after years of amateur work, he posted "The chromatic number of the plane is at least 5" to the arXiv with a construction: a unit-distance graph on 1,581 vertices that requires 5 colours.

The graph was constructed by combining many copies of the Moser spindle in a carefully chosen pattern, building up enough simultaneous constraints to force any 4-colouring to fail somewhere. Verification was done by SAT solver — the graph is far too large to 4-colour by hand.

"I had a sneaking suspicion that 4 wasn't enough… I'd been thinking about this on and off for years, but never quite got there. Then one day the construction clicked."

Within weeks of de Grey's announcement, Marijn Heule (Carnegie Mellon, SAT-solver specialist) and the open collaborative Polymath16 project shrank the graph dramatically — first to 826 vertices, then 553, eventually 510. The fastest reductions came from running SAT solvers to find smaller graphs with the same chromatic-number property.

The 4 in the lower bound became a 5, and it held.

Still open

5 ≤ χ(ℝ²) ≤ 7. The remaining question — is it 5, 6, or 7? — is what we don't know.

A few measurable facts:

Lower-bound improvements would require even larger unit-distance graphs that need 6 colours. None has been found.
The upper bound of 7 has stood since 1950. Improvement would require a more clever colouring than Isbell's hexagonal tiling — perhaps using shapes other than hexagons.
Variants of the problem are settled or partially settled: for example, if colour classes are restricted to be measurable, the answer is at least 5 (Falconer 1981); for ℝ³, similar bounds exist but at different magnitudes.

What's in this site

spindle — the Moser spindle (7 vertices, 11 edges). Click a vertex to cycle its colour. Cap the palette at 3 and try to find a valid colouring — you can't. Cap at 4 and many work. The conflict count and verdict update live.
de grey — a Moser flower (six rotated spindles around a shared centre), with the same colour-cycling interaction at a richer scale. A backtracking solver finds a proper colouring on demand.
play — procedurally generated unit-distance graph puzzles. Pick a difficulty (3 or 4 colour target); the page generates a fresh graph, anchors it with a structure that forces the target chromatic number, grows it by equilateral-triangle completions, then verifies χ(G) with the solver before letting you near it. The colour target on each puzzle is always honest — the solver has confirmed both that the graph needs the target and that it doesn't need more.
tiling — Isbell's 7-colour hexagonal tiling. The hex-side slider shows the constraint: hexes too small leave gaps in the unit-distance argument; too large and same-colour hexes get within 1 of each other. The unit-circle overlay drags around and demonstrates no colour collision.
docs — this page.

Why this story

The Hadwiger–Nelson problem is part of the same genre as erdős and guthkatz — extremal questions about point configurations in the plane, with a small integer answer that's been stuck for decades. The shape of the breakthrough is different from both:

erdős: an AI built a configuration that beats the conjectured optimum.
guthkatz: a polynomial-method proof matched the conjectured optimum from below.
hadwiger: an amateur outside the mathematical establishment found a small-but-large hand-designed graph that improved a 68-year-old bound by one.

One construction; one proof; one piece of patient amateur work. The genre keeps surprising in different directions.

sources

de Grey, A. D. N. J. (2018). The chromatic number of the plane is at least 5.
Heule, M. (2018). Computing small unit-distance graphs with chromatic number 5.
Polymath16 project page
Quanta — Decades-Old Graph Problem Yields to Amateur Mathematician
Soifer, A. (2009). The Mathematical Coloring Book. Springer — the standard reference for χ(ℝ²) history.
Isbell, J. R. — the 1950 upper-bound proof appeared in problem sessions; first written record is in Hadwiger's papers.