Previously, I looked at problems concerning colorings of individual cells of polyominoes. These were not map coloring problems, (i. e., problems of giving a set of shapes a limited number of colors so no two adjacent shapes share the same color.) Map coloring the cells of a square grid isn’t very interesting, beyond noting that the grid is 2-colorable, with a checker pattern being the 2-coloring.

But suppose our polyform components are more complicated than individual cells. For example, the components could themselves be polyominoes. Now component-wise map coloring can be a source of interesting problems.

Since 4-coloring is always possible, 3-coloring is the usual place to go to when we want a challenge. Given three colors, the di-dominoes can be component colored in 15 ways. (There are 4 di-dominoes, and because the L-tetromino is asymmetrical, there are two ways to color it for each color pair.) Here is a tiling with 3-colored components:

Moving up to the tri-dominoes, there are 26, which can tile a 12 × 13 rectangle. Problem #58: Find a 3-coloring of the dominoes in such a tiling where each tri-domino contains all three colors. Edit: As Bryce Herdt pointed out in a comment, this is impossible, because there are tri-dominoes where all three dominoes surround a square that could not then take any of the three colors.

Four-coloring can be a worthwhile problem, provided that we can find a good additional restriction on color usage. With the 11 heterogeneous di-trominoes, we can restrict ourselves to two colors each for the I and L trominoes. Then we can find a component-wise 4-coloring of the set using those colors:

Notice that this is a “non-strict” coloring, since two red L’s meet at a vertex. Problem #59: find a strict 4-coloring of the components of the heterogeneous di-trominoes in a 6 × 11 rectangle.

There are undoubtedly other fruitful directions to take component coloring. Perhaps there is something to do with poly-polyiamonds, or poly-polyhexes. I would be delighted to see what you can find!