The tiling world is a bit aflutter recently with the drop of Smith, Meyers, Kaplan, and Goodman-Strauss's paper showing an einstein – a simply-connected polygon – that must aperiodically tile the plane. Their einstein is a concave 13-gon that they have called the "hat" monotile.
The image above is from Kaplan's tweet announcing the paper. While playing with his discovery Smith supposed that it was necessarily aperiodic; Meyers, Kaplan, and Goodman-Strauss worked with Smith to give two separate proofs of aperiodicity.
Earlier I think in "Mathematical Games" in a Scientific American article from the '70s, Martin Gardner relayed a question considered by John Conway about whether a (rhombic) Penrose tiling of the plane could admit a three-coloring; this was algorithmically answered in the affirmative by Sibley and Wagon. I dug up some plastic Penrose rhombi that I had bought about 25 years ago, and indeed they came in only three colors. It's a straightforward but fun puzzle to tile with only these three colors.
The same question can be asked about the SMKG einstein – can we three-color any tiling with the above 13-gon monotile? If so, can we give an efficient algorithm?
The construction is gloriously simple and could be taught to bright young children. How many colors would be needed to 3D-print a bunch of these?
Added later
As we know, at least for a finite part of the plane four colors suffice, but some questions may remain about how easy it is to four-color the monotiling. Jesse Clark suggested that it could be easily four-colored along radial stripes here.
Best Answer
No, you cannot three-color that tiling.
Here's a finite part of the tiling from page 10 top left of the article, the tile numbered 2 here is the one darkened on that figure. This part cannot be three-colored.
To prove this, start from the three adjacent tiles marked 0, 1, 2, and call their colors yellow, blue and red respectively. Tile 0 is surrounded by a sequence of adjacent tiles 1, 2, 3, 4, 5, 6, so these must be colored blue and red alternatingly, 5 being blue, 6 being red. Tile 7 is adjacent to tile 5 and 6 and so must be yellow; while tile 8 is adjacent to tile 1 and 6 and so also must be yellow. Now tile 9 is adjacent to tiles 6 and 8 and so must be blue, but tile 10 is adjacent to 6 and 7 and so also must be blue, but 9 and 10 are adjacent so this coloring is invalid.