puzzlewocky

Fractalized figures.

Socolar-Taylor Tiles. 

These groundbreaking tiles are the first known solution of “the einstein problem.” No, it isn’t a physics problem–just a clever German play on words (ein Stein is german for “one tile”). 

Anyways, the problem asks for a single tile capable of tessellating space aperiodically. Let’s unpack that in two separate parts. 

Simply put, a tessellation is a tiling of the plane without any gaps or overlaps. For a given set of tiles (prototiles) to tile the plane, we arrange them to cover a flat surface (or fill space in higher dimensions). If the tiling can be continued indefinitely, the prototiles are said to tile the plane. A tiling can be described mathematically by a sequence of reflections, rotations, and translations. 

Loosely, periodicity means something repeats in regular intervals or periods. An aperiodic tiling, then, does not repeat at scale. This means we cannot find arbitrarily large patches of tile with any repeating pattern. Another way to think of this is that the “pattern” is always changing.

The first image is a 2-dimensional monotile. The black markings on the tile are used as guides for matching rules, which restrict how the tiles are arranged.

The second image is an patch of 25 tiles. Notice how the markings generate an infinite hierarchy of successively larger triangles. This concept, called limit-periodicity, is one of the main structures by which aperiodic tilings are constructed. 

Observe that the patch cannot tile the plane without leaving holes. Thus, the 2D monotile does not tessellate.

However, in the third image, we see a 3D monotile (here the guide markings are in red). This is, indeed, an einstein–it aperiodically tessellates 3D space. It should be noted that this tile does admit another tiling that is periodic. Tiles having this property are called “weakly aperiodic.” 

The fourth image depicts a 3D patch. The limit-periodic structure is exhibited by the Sierpiński Triangle-looking design on its surface. 

The final image removes a tile to show the internal structure of the patch.

One last, fascinating tidbit: don’t go and buy a bunch of 3D-printed copies hoping to fill space. The aperiodic tiling can only be constructed by allowing reflections. So, unless you have access to 4D space, you are out of luck. 

Interactive 3D tile here. Rotate it! It’s like a weird lego, I guess.

The snow art of Simon Beck.

Yes, he does this by walking. To quote him, “My life has been a competition between the mind and the body, which has been won by the body.”

Koch snowflakes… literally.

A Hopf fibration. This gorgeous visualization portrays a suitably deep idea in differential topology.

The Hopf fibration describes 3-spheres (aka hyperspheres, 4-dimensional analogues of spheres), in terms of “ordinary” spheres and circles. It is an early example of a fiber bundle.

Quoting Wikipedia, “Technically, Hopf found a many-to-one continuous function (or ‘map’) from the 3-sphere onto the 2-sphere such that each distinct point of the 2-sphere comes from a distinct circle of the 3-sphere (Hopf 1931). Thus the 3-sphere is composed of fibers, where each fiber is a circle—one for each point of the 2-sphere.”

Hopf fibrations have many useful properties. I write, however, because they are beautiful.

To remind you, here’s the question:

In reply to a question about the animals on her farm, the farmer says: “I only ever keep sheep, goats, and horses. In fact, at the moment they are all sheep except for 3, all goats except for 4, and all horses except for 5.” How many does she have of each animal?

Solution:

You know from the question that there are 1 more goats than horses and 1 more sheep than goats.  So the fewest there can be are 1 horse, 2 goats and 3 sheep.  If you test it, that works as a solution.  There are 5 non-horses, 4 non-goats, and 3 non-sheep.

You can easily see that there are no other possible solutions.  If you use any higher numbers, for example 2, 3, and 4, there will be too many animals.  With 2, 3 and 4, there are 7 non-horses, 6 non-goats, and 5 non-sheep.

Solution: 1 horse, 2 goats, 3 sheep

Too Truchet

Kokichi Sugihara of Japan is the winner of the 2018 Best Illusion of the Year Contest with his mind-bending entry, “Triply Ambiguous Object.”

Why we need a special expression for triplets

The question

If you want to write down music you will face the need for a notation to express the length for which a given note is supposed to be played, it’s note value. For reasons of perceived rhythmicality, note values arise from repeatedly halving (or occasionally doubling) a pre-set standard unit that is usually dictated by the beat. Denoting the standard unit by “1“ this is the notation we get: 

I left out the signs denoting doubling of the standard unit because they are less common. I also stopped at 1/16th, even though you can keep on halving notes to get 1/32ths, 1/64ths etc. and denote them by adding more tiny flags to the notes. Let’s call these values the standard values. Relative to the standard unit they always have the length 1/(2^n), where n is an integer. 

You can even add standard values together to create new values, like so: 

So far, the system seems pretty expressive. And I am, in fact, fairly sure that the note values of a great percentage of every musical piece I ever encountered can be expressed in this system. You only need one little addition, the triplet. You produce a triplet by taking a standard value and, instead of halving it which would yield another standard value, you split it into three parts. Here is an example of a triplet derived from the standard unit 1: 

The little “3” and the brackets above the notes signify that the notes do not represent standard values but the aforementioned third of the next bigger note value. Risking confusion I will refer to both, the group of three notes each valuing a third of a standard value and the individual notes making up the group, as triplets. 

The fact that a new notation had to be invented to express triplets already suggests that they cannot be expressed by standard values. This is in fact so. But why? Many of you mathematicians might already be anticipating the answer at this point, but I certainly did not before taking up mathematics and, to be honest, none of my music teachers could explain to me why you would never be able to create a triplet no matter how small you made the standard values. So the question I intend to answer in a second is: 

How can we be sure that we will never be able to express the note value corresponding to a triplet by using (sums of) standard values?

The answer

Because of the uniqueness of prime factorisation. By the fundamental theorem of arithmetic, every positive integer has a unique prime factorisation. It’s quite obvious that adding together standard values always produces fractions over a power of two. So in order to express a triplet through (sums of) standard values, you would need natural numbers m, n (and, if you like, k =/= 0), so that 

But 2 and 3 are both primes so the denominators (2^n and 3*k) will never be equal, no matter what numbers you choose for n and k. There is no natural number which can be expressed both as a power of two and as a multiple of three. And that’s why we need the triplet notation.