The problem lives at the intersection of graph theory, combinatorics, and measure theory. Here is what is known and why each piece is hard.
The upper bound: 7 colors always suffice. Tile the plane with regular hexagons of diameter slightly less than 1. Color the hexagons with 7 colors in a repeating pattern so that no two same-colored hexagons share a boundary or are within distance 1 of each other. Any two points in the same hexagon are less than 1 apart; any two points exactly 1 apart land in differently-colored hexagons. Done — 7 colors work.
The lower bound: 4 colors are not enough. For decades the best lower bound was 4, proved in the 1950s using the Moser spindle, a unit-distance graph on just 7 vertices that requires 4 colors. But 4 was stuck for nearly 70 years.
The 2018 breakthrough: 5 colors are not enough. In April 2018, Aubrey de Grey posted a paper on arXiv showing a unit-distance graph with 1581 vertices that cannot be properly colored with 4 colors — so the plane needs at least 5. The proof was computer-assisted: he generated and filtered candidate graphs until finding one where a SAT-like search confirmed no 4-coloring exists. Smaller examples (around 500–600 vertices) were quickly found by others, and a graph with just 510 vertices suffices today. This is the first improvement to the lower bound in 65 years.
Why 5, 6, or 7 seems so hard to close. No one has found a unit-distance graph requiring 5 colors (which would push the lower bound to 6), and no one has proved 5 colors are enough (which would close the problem). The difficulty is that unit-distance graphs are an extraordinarily rich family — there are infinitely many, with no simple structure to exploit. Measure-theoretic approaches (assigning colors randomly and using ergodic arguments) also hit walls. The problem connects to graph coloring in deep ways, but the infinite nature of the plane makes finite techniques hard to transfer.
Comments
Loading comments...