Consider tiling the plane using only square tiles like this:

The 4-fold rotational symmetry of the square allows a tile to be placed in the 4 different orientations shown here:

Despite these constraints there are still a lot of different ways to tile the plane. Shown above are a few examples constructed with an algorithm using modular arithmetic. This essentially makes the tiles along different rows follow the same sequence but shifted over by some amount.

Each of the tilings shown are actually periodic and can tile the entire plane.
(e.g. 1, 2, 3, 4)

Tilings of this variety are called Truchet tilings.

Mathematica code:

T[r_, x_, y_] :=
{EdgeForm[Thickness[0]], Polygon[{{1, 0}, {0, 0}, {0, 1}}]},
r, {.5, .5}],
{x, y}]

T[Mod[a*x + b*y, m] Pi/2, x, y],
{x, 1, X, 1}, {y, 1, Y, 1}],
PlotRange -> {{1, X}, {1, Y - .1}}, ImageSize -> 500],
{{X, 29}, 1, 100, 1}, {{Y, 29}, 1, 100, 1},
{{m, 14}, 1, 100, 1},
{{a, 5}, 1, 100, 1},
{{b, 5}, 1, 100, 1}]
Planar 2-colorability

Take some region of the plane, and any number of distinct lines that pass anywhere through this region. Consider these random lines for instance:

Notice how the lines and their crossings create polygonal shapes in the region.

Using just two colors, say black and white, is it possible to color the entire region such that any two polygons that are next to each other sharing a common edge are different colors?

This is possible in the above example as this 2-coloring indicates:

There is also another possible 2-coloring that satisfies these requirements, but its really just the same as the 2-coloring above with the colors switched.

You can convince yourself that these are the only two permissible 2-colorings meeting the criterion with this particular configuration of lines.

One may wonder if its always possible to achieve such a 2-coloring, or precisely under what circumstances it is or is not possible.

It does seem to be the case that if each line passes completely through the region, then a 2-coloring will always be possible.

The following examples show this for two particular cases, where not only are there a different number of randomly chosen lines in each case, but each line is even allowed to move. Regardless, at each instance, the 2-coloring is always preserved!

Still, these examples do not prove the claim in general since there remains an infinite number of cases left unconsidered. How would one prove this?

Well, when would such a 2-coloring not be possible?

At any intersection of lines in the region, the crossings create corners for the polygons that are formed. If there happens to be an intersection with an odd number of corners, then for any assignment of 2 colors to the parts around this intersection, there would have to exist two adjacent parts that have the same color. Otherwise, for an even number of corners around an intersection, it is always possible to assign a 2-coloring so that adjacent parts have different colors.

Therefore, as long as all the intersections formed within the region have an even number of corners, there will exist a 2-coloring. This criterion will be met if we assume that the lines always pass completely through the region as in previous considerations

These conditions are special and do limit the possible configurations that are 2-colorable.

What if configurations were allowed to have intersections with an odd number of corners?

What if lines didn’t have to pass completely through the region and were allowed to end somewhere inside of it?

What if we didn’t have to use straight lines to partition the region?

If it is not possible to color the region in the above sense with 2 colors, how many would it take?

Does there exist some maximum finite number of colors that can be used to color any possible partition of a region?

The 4-color theorem, first stated in 1852, which concerns the problem under consideration, states that only 4 colors are needed to color any configuration so that adjacent regions are not colored the same.

The truth of this theorem went without correct proof until 1976 when it was proved by Kenneth Appel and Wolfgang Haken using a computer! This computer-assisted proof may be considered controversial and has interesting implications.

It is worth mentioning that the related problem of deciding whether a given configuration is 2-colorable is easy to solve since there are efficient computer algorithms that can check. However, the problem of deciding if 3 colors are needed is hard to do in general since there are currently no known computers algorithms that can efficiently solve this problem.

If you can find a fast algorithm, or if it you can prove that no efficient algorithm can exist for deciding the 3-coloring problem, then you could win $1,000,000 solving a big open problem in computer science.

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