Archive for May, 2013

3.8.24

May 27th, 2013

A regular 24-gon, octagon, and triangle together form one of the 17 ways (called vertex figures) to surround a vertex with regular polygons. This vertex figure can’t do anything nice like tile the plane. You can surround the 24-gon with octagons and triangles, but then you have gaps that can’t be filled with regular polygons, and you have to stop.

24-8-3-rose

Or, instead of stopping, you could take that 24-gon surrounded by octagons and triangles, and surround it with 12 more 24-gons surrounded by octagons and triangles, overlapping in an elegant sort of way:

24-8-3-doodle

And then, instead of stopping there, you could surround that whole thing with more copies of that whole thing…

But sometimes art is where you stop.

Hat tip to John Baez, who brought up the 3.7.42 vertex figure in a Google+ post.

A Polyformist’s Toolkit: Practical Topology

May 23rd, 2013

In polyomino puzzles, we would frequently like to tile the simplest shape possible, and a rectangle usually seems to fit the bill. But sometimes a rectangle isn’t possible. For example, we can never make a 4×5 rectangle with the five tetrominoes. One way to prove this is with a checkerboard parity argument. Four of the 5 tetrominoes must always occupy even numbers of both black and white squares if they are placed on a checkerboard. The T tetromino must occupy odd numbers of each color. Therefore a rectangle must have odd numbers of each color, but any rectangle of size 20 will have colors evenly divided, 10 and 10. A similar argument can be made to show that the 35 hexominoes cannot tile a rectangle.

The tetrominoes, and a 5×4 rectangle.

This will never work…

Rather than give up and accept that we’ll need to find a less elegant shape to tile, we have another option. If we wrap the edges of a 5×4 rectangle around to form a cylinder, (so that the cylinder is 4 squares tall and 5 squares in circumference) tiling is once again possible. To see why this might be so, imagine that you are coloring the squares as in a checkerboard. Once you got back around to where you began, you would find that in order to continue the pattern, you would need to use the opposite colors from those you already used. Note that this would not work if you wrapped the rectangle in the other direction; because the other side has even length, the checkering colors remain consistent.

The tetrominoes tiling a 5×4 cylinder a cylinder

…until we wrap the rectangle into a cylinder.

There is a video by Edo Timmermans showing how a tetromino cylinder can be made with toy magnets. He claims that there are seven distinct tilings of a cylinder with the tetrominoes, and poses an interesting puzzle involving them. A commercially produced cylindrical polyomino puzzle is Logiq Tower, designed by Marko Pavlović, which uses wooden pentomino-based pieces that form a cylinder together with some other pieces. Because these pieces are inflexible, they lack some of the allowable symmetry actions of free pentominoes.

A cylinder isn’t our only option. We could give the rectangle a half-twist before connecting the ends; this gives us a Möbius strip. We could also connect both pairs of sides instead of one; this gives a structure that is topologically equivalent to a torus or doughnut. And then we could add twists to that— well, at this point it would be nice to be systematic so we can be sure that we’ve found all of the possibilities. One thing to note is that adding more twists doesn’t actually give us more possibilities. A strip with two twists will have exactly the same tilings as a strip with no twists, and in general, a strip with an even number of half-twists will have the same tilings as the no-twist strip, and a strip with an odd number of half-twists will have the same tilings as the Möbius strip. So for each dimension, we have three options: no connection, connection without a twist, and connection with a half-twist. This gives us the following matrix of possibilities:
Topologies for polyomino tilings
Only six possibilities here, not nine, because the ones in the lower left are equivalent to the ones across the main diagonal from them. Note that the Möbius strip, Klein bottle, and projective plane are nonorientable surfaces, which means that they effectively have only one side.

An important consideration when working with these is that one-sided polyominoes don’t exist on nonorientable surfaces. With one-sided polyominoes, translation is allowed, but reflection isn’t. However, on a non-orientable surface, translating far enough leaves an object in a reflected state.

Another consideration is that coloring is harder when we leave the plane behind. On the plane, we have a theorem stating that we never need to use more than four colors to make all of the tiles differ in color from all of their neighbors. On a torus, this may require seven colors. In 2001, Roger Phillips found 18 heptominoes that could tile a 7×7 torus, and sent these tilings to MathPuzzle.com. Here’s one:

7-colored 7-omino torus

Depending on the dimensions of the torus, it may be possible for a polyomino to wrap around and touch itself. In a strict sense, this makes any coloring impossible, since we don’t let tiles of the same color touch. However, we can follow a looser standard, and allow self-touching polyominoes in our colored tilings. Patrick Hamlyn found a 3-coloring of a tiling of the 35 hexominoes in 7 3×10 tori using this scheme in 2003:

The 35 hexominoes in 7 3×10 tori, 3-colored

This problem has no solutions if the tori are replaced with rectangles or cylinders.

Problems #31-37:
Though it seems like a pretty basic problem, if anyone has counted the number of pentomino tilings of cylinders, I am not aware of it. Wrapping the short sides of the 3×20 together should not give any solutions beyond the two obtained by wrapping the solutions on the 3×20 rectangle. That leaves the 3×20 wrapped the other way, and both ways of wrapping the 4×15, the 5×12, and the 6×10 rectangles.

Problems #38-40: Find the solution counts for the 4×15, 5×12, and 6×10 tori. I don’t know if these are all computationally tractable, but I can hope. (The 3×20 will be the same as the 3×20 cylinder with long sides wrapped together.)

Even more possibilities for tiling become available when you choose parallelograms with diagonal sides to wrap around, but this post is long enough, so that will have to be a matter for another post.

Chromatic Number of the Plane Is Still Less Than Or Equal To Seven

May 6th, 2013

Well, it was worth a shot.

Not a breakthrough in the CNP problem

The Chromatic Number of the Plane is defined as the number of colors required to paint the plane such that no two points one unit distance apart are the same color. There are very simple figures that demonstrate a lower bound of 4 and an upper bound of 7 for this number. But this is more or less where knowledge of this problem has stood for the last 50 years. (See the Wikipedia page on this problem for more info.)

I didn’t really expect that I could make a breakthrough, given that real mathematicians have given serious thought to this problem and gotten nowhere, but I thought this would be fun to play with. One potentially useful starting point is to consider how much of the plane one single color could cover. A disk of unit diameter is the biggest blob we can make without two points a unit distance apart. (We’ll allow some fudging about the points on the perimeter of the disk.) We can then pack these disks into a triangular array such that the distance between nearby disks in the array is also one unit. (It’s not clear to me whether we couldn’t get a slightly higher proportion of the plane in one color by flattening the parts of the circles that are closest to other circles. But in any case, the packing with circles should be close to the best we can do.)

It was easy enough to make an array of disks in this fashion using Inkscape. I then copied the array five times, gave each of the six copies its own color, and made them all translucent, so that the overlapping areas could be readily distinguished. The point of this exercise is this: if I could find a way to arrange the six arrays of circles so that they completely covered a region of the plane, I would be able to demonstrate the existence of a coloring that uses only six colors. In other words, I would have decreased the upper bound of the chromatic number of the plane by one, which would have been a significant mathematical result. Note that the circles could and would overlap; any areas of overlap would consist of points that could belong to either color in a valid coloring.

As you can see, I failed. The illustration above was about the best I could do. The white gaps represent areas that could not be colored with any of the six colors. Failing here really proves nothing. Even if I could prove that no coloring using disk arrays like these could work, it wouldn’t prove that there wasn’t some other way to partition the plane that would work better. In fact, Ed Pegg found a configuration where the amount of space needed by the seventh color is very tiny indeed, which gives some hope that 6 colors is possible.

If I could prove that this array of circles really represented the greatest possible proportion of the plane that could be taken up by a single color, (which I can’t) I would in fact be able to improve the lower bound of the CNP. If I didn’t make a mistake, the proportion of the plane taken up by the array of circles works out to π/(8·sqrt(3)), or about .227. Since this is less than ¼, that would mean that the total area that could be taken by 4 colors could not cover the whole plane. Part of the problem here is that even if one could show that the array of circles had the highest proportion of the plane of any “nice” arrangement, it might turn out that an actual coloring of the plane with the fewest colors would end up taking the form of interpenetrating fractal foams, where no individual piece has a measurable area.