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Most of the other math blogs in the blog tour for the book release are about early childhood math education, so I thought I’d write something in a slightly more advanced vein, exploring a bit of the underlying mathematics of making tessellations. My hope is that you’ll learn some things and also come away with ideas of new kinds of tessellations to explore. There is way more than I could ever fit in a single blog post (if you want to explore more, John Golden has a great list of resources on Math Hombre), but let’s see how far we get!

Let’s start with using *regular* polygons (that is, polygons whose sides and angles are all equal) to tile the plane. Most everyone is familiar with the idea that we can do this with regular (equilateral) triangles, regular quadrilaterals (*i.e.* squares), and regular hexagons:

- Every vertex of an equilateral triangle has an angle of , so six triangles can meet around every vertex to make a total of .
- Four squares meet around a vertex to make a total of .
- Three hexagons meet to make a total of .

In addition, the triangle and hexagon tilings are closely related, since we can get one from the other by subdividing the hexagons:

It’s easy to see that these are the only regular polygons that will work: regular pentagons have angles of , which does not evenly divide . And anything with more than six sides will have angles bigger than , so more than two of them will not be able to fit around a vertex.

Now, as explained in the back of Tessalation!, and as reproduced in this blog post on Kids Math Teacher, we can take a square tessellation and modify the squares to produce more intricate tessellations which still follow the same underlying pattern. In particular, if you add some shape to one side, you have to remove it from the opposite side, and vice versa. For example, beginning with a square, we might change the right side like this:

But if we do that we need to change the left side in a symmetric way:

Now the altered squares will still line up in a row:

Likewise, we can make symmetric modifications to the top and bottom, like so:

The resulting thingy can still tile the plane:

So far so good. But if we take a step back to think about what’s really going on here, a whole world of possibilities opens up.

What we’ve really done with the square is match up certain edges, so that matching edges always meet in the tessellation.

Here I’ve marked the top and bottom edge both with a single arrow, and the left and right edges with a double arrow. (I’ve also put a letter “P” in the middle; I’ll explain why later.) In the tessellation, corresponding markings always have to match up. Like this:

Now, instead of matching up the edges of a bunch of copies of the same square, we can think about taking *one* square and gluing matching edges together. First, we glue the top and bottom edges together, resulting in a cylinder; then bend the ends of the cylinder around to match up the left and right edges, resulting in a torus (a donut shape).

Now imagine a very tiny ant who lives by itself on the surface of the torus. The ant is so small that it can’t tell that the surface it lives on is curved. To the ant, it just looks flat. (You may know some tiny creatures in a similar situation who live on a sphere.) Unlike those tiny creatures on the sphere, however, the ant has nothing it can use to draw with, no objects to leave behind, *etc.*, so it has no way to tell whether it has ever been to a particular location before. The ant starts walking around, exploring its world. Occasionally there is a straight line drawn on the ground, extending off into the distance. Sometimes it finds places where two lines cross at right angles. Sometimes it finds places where the ground is black, and after making some maps the ant realizes that these places are shaped like a giant letter “P”. After exploring for quite a while, the ant thinks its world looks something like this:

Or perhaps it lives on a torus? (Or an infinitely long cylinder?) The point is that there is *no way for the ant to tell the difference*. The ant cannot tell whether there are infinitely many copies of the letter “P”, or if there is only one letter “P” that it keeps coming back around to. So **a square tessellation is “what a torus looks like to an ant”**, that is, what we get if we cut open a torus and glue infinitely many copies together so that each copy picks up exactly where the previous copy left off.

But there are lots of ways to cut a torus open so it lays flat! And *all* of them will produce some shape which tiles the plane just like a square. This is another way to think about what we are doing when we modify matching edges of a square—we are really just cutting the torus along different lines.

This blog post has gotten long enough so I think I will stop there! But I plan to write another followup post or three, because we have only just scratched the surface. In the meantime, I will leave you with some things to think about. First, what if we match up the edges of a square in a different way?

This is almost like the square from before, but notice that the arrow on the top edge is flipped. This means that we can’t just stack two copies of this square on top of each other, because the edges wouldn’t match:

But we can stack them if we flip one of the squares over, like this:

Finally you can see why I included the letter “P”—it lets us keep track of how the square has been flipped and/or rotated.

Can you complete the above to a tiling of the whole plane? What do such tessellations look like? Is it still possible to modify the edges to make other shapes that tile the plane in the same pattern?

How about this square?

Or this one?

Or this one?

And what about triangles and hexagons? What are different ways you can match up their edges to make tessellations? (Related challenge question: when we glue opposite pairs of sides on a square, we got a torus. If you glue opposite pairs of sides on a hexagon, what shape do you get?)

Happy tessellating!

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(Click for a larger version.) I didn’t make it, and I have no idea where I got it from (do you know?). But in any case, wherever it comes from, I think it’s a really great puzzle. I did find the number that can make it through the diagram, but I never did completely finish proving that the solution is unique.

Can you solve it? Let’s see if we can prove it together. Please **don’t** post the **number** in the comments. But please **do** post proofs that certain combinations of nodes are impossible. For example, you might post a proof that no triangular number can be one more than a prime; that would mean the leftmost path is impossible.

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Note how is one more than ; is more than ; is more than ; and so on.

But that’s not quite how the Recamán sequence is defined. In the Recamán sequence, “wants” to be *less* than : it will be so if is nonnegative and *has not already appeared in the sequence*. Otherwise, it “settles” for being *more* than (as with the triangular numbers).

So . Then:

- has to differ from by —that is, it must be or . But it can’t be negative, so it is.
- How about ? It must be away from , so either or —but again, it can’t be negative, so it is .
- Now, must be away from , so it must be or . is nonnegative, but it has already appeared in the sequence as , so we choose .
- So far, this is looking a lot like the triangular numbers! But let’s see what happens with . It must be , that is, either or . And here something different happens: is positive and has not appeared yet in the sequence, so .

Continuing this pattern, we get

From the definition, you might initially think that no number will ever be repeated: we explicitly avoid picking numbers that have already occurred in the sequence, right? Well, we don’t pick if it has already occurred, but in that case we definitely pick whether it has already occurred or not—and in fact sometimes it has. You don’t have to continue the sequence very much farther before you find, for example, .

Here’s a scatterplot of the first terms, with the -axis scaled by 1/2 to make it easier to see:

From the graph we can see that the numbers tend to form long parallel alternating runs where the top numbers are increasing by and the bottom decreasing. For example, we can see the first of these starting at :

This makes sense: we are alternately adding , then subtracting , then adding , then subtracting , and so on. The run will be broken when we hit a number that has already occurred: in this case, the number after is not , because that has already occurred, so instead it jumps up again to . After , it jumps up one more time (since was already in the sequence) and we get a very short parallel sequence before it falls back down and starts another alternating sequence .

So far, this looks regular-ish—one might hope to discover some regular patterns that could, for example, lead to a closed formula. But our hopes are dashed when we look further out in the sequence:

It doesn’t look very regular-ish anymore! And this is what I find fascinating about the Recamán sequence: the *self-reference* in its definition (we choose only when it has not already appeared) throws a giant monkey wrench of chaos into the works, so that it is very difficult to find any patterns or prove anything definite about it, even though it is still highly structured. To see what I mean, here’s a graph of the first 5000 terms:

There is most definitely structure—for example, all the terms seem to fall along these curving “bands” radiating out from the origin. I imagine one might even be able to say something approximate about the shape of those curves. But there is still obviously lots of chaos—it’s hard to discern any regular patterns.

So we know that the Recamán sequence is not 1-1: some numbers can appear multiple times. But is it *onto*? That is, does every number appear *at least once*, somewhere in the sequence? It is *conjectured* that this is true, but *no one knows*—it is an open question! According to the OEIS entry, after looking at the first terms, the smallest number that is still missing is ; every number smaller than that has appeared somewhere in the first terms of the sequence. Crazily though, is *still* missing after looking at the first terms!!! (That’s a staggeringly large number of terms; it is *far* more than the estimated number of atoms in the universe.) So perhaps the Recamán sequence is not onto after all—is there something special about so that it *never* apears? Or does it eventually appear *very very far* into the sequence? No one knows, and to be honest, I think the latter actually seems more likely. But it just underscores how difficult it would be to prove this.

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(Remember that the *signed bend* of a circle is like the *curvature* , except that if one circle contains the other circles inside it, we give the outer circle a *negative* bend.)

As I explained previously, if we have the bends of three mutually tangent circles, we can compute the possible bends of a fourth circle using the formula , where is the sum of the other bends and

Moreover, the two solutions for have the property that

so if we have one of them we can easily “flip” to the other one. Doing these repeated “flips” results in an infinite tree of kissing sets, and drawing all of them (until they get too tiny to draw) results in an Apollonian gasket.

But the *bend* of a circle only tells us *how big* it is—how do we decide *where* to draw the circles?

In their paper “Beyond the Descartes Circle Theorem”, Jeffrey Lagarias, Colin Mallows, and Allan Wilks prove the following remarkable theorem. Let denote the signed bend of circle as before. Now let denote the *center* of circle *expressed as a complex number*. For example, if circle has its center at the point , then . Now here is the theorem:

That is, the *products* of the signed bend and complex center of each circle *also* satisfy Descartes’ Theorem! So for each circle we just keep track of the signed bend as well as the product of the signed bend and the center. For each operation we compute the same formula twice, once for the bends and once for the bend-center products. To recover the center of a circle, just divide the bend-center product by the bend. And voila!

As a result of this beautiful theorem, the code to generate the above picture is very short—we just have to implement a few formulas and apply them to both bends and bend-center products, and then recursively generate the tree of kissing sets I explained in a previous post.

I really wish I could explain the proof of this theorem, but alas, I do not understand it. Lagarias *et al.* go on to generalize the theorem four or five more times—the final theorem involves a matrix version of the equation for kissing sets of arbitrary -dimensional hyperspheres in spherical space, or something like that—and then prove the most general version using some sophisticated, abstract machinery. The generalized theorem implies the others, so it is technically a proof of the version of Descartes’ Theorem for bend-center products, but it doesn’t give much intuition into why it is true!

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- The Nuclear Pennies Game
- Distributing cookies: solutions
- The hyperbinary sequence and the Calkin-Wilf tree
- m-bracelets
- Triangular number equations via pictures: solutions
- Computing with decadic numbers
- Mystery curve, animated
- Post without words #5, explained
- An amazingly symmetric icosahedron edge coloring

Probably my most controversial post (and the one that generated the most traffic!) was this one, where I (somewhat snarkily) debunked a claimed proof of the Collatz conjecture.

Also, a lot of my most popular posts have been part series of posts exploring topics or explaining proofs in depth, which are all collected here.

It’s been a ton of fun and I’ve learned a lot. Thanks for reading—here’s to the next ten years of beautiful math!

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Back in January I mentioned that I had been enjoying Siobhan Roberts’s new biography, Genius at Play: The Curious Mind of John Horton Conway. I finally finished it, and can now *officially* say that I really enjoyed it.

This is not a conventional biography. It is all jumbled together, verging on stream-of-consciousness—anything but chronological. In places, Roberts simply quotes Conway at length rather than trying to explain. But then again, John Conway is not a conventional human being. The biographical style fits him.

Sometimes, reading a biography can be inspiring, as you read about a person who has achieved extraordinary things. Even if you could never achieve the same things, you still come away with effective habits to emulate, insights to ponder, or perhaps even the new realization that you *can* achieve the same things if you just work hard enough.

This is not that sort of biography. Through Roberts’s portrait it becomes clear that there are no conscious habits to emulate, no insights to abide by—the only way to be like John Conway is to *be* John Conway. And there is only one John Conway.

I learned a bit of new math from the book; got inspired to re-learn the Doomsday Algorithm (my current average is about 12 seconds per date); and gained a new appreciation for the various ways that Conway’s interests weave together and for the scope of his career. Mostly, though, I got a fascinating glimpse at a strange and wondrous life. I certainly didn’t come away from the book inspired to be like Conway. But I am thankful and amazed that he exists.

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With one week to go in the campaign, they are still about $3,000 short of their funding goal—go support them if you haven’t already! If you don’t support this campaign it doesn’t get funded my son doesn’t get a deck of cards my son is sad. Don’t be that jerk who makes a 4-year-old sad. Support the campaign today!

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So far, however, I have only claimed that various things *exist*. But to actually *draw* an Apollonian gasket, we have to start dealing concretely with centers, radii, and so on—how do we actually *compute* the positions and sizes of the circles involved?

This is where something called *Descartes’ Theorem* begins to play a starring role. Given a circle with radius , define the *curvature* of , denoted , by

that is, the curvature is the reciprocal of the radius. Circles with a *small* radius have a *large* curvature —intuitively, they have to curve a lot to fit into a small space; whereas circles with a large radius have a small curvature, since they can curve at a more leisurely pace. Curvature is like how much you have to turn the steering wheel of a car (or the handlebars of a bike) to turn in a circle: you have to turn the steering wheel a lot (large curvature) to turn in a small circle, and only a little (small curvature) to turn in a big circle.

OK, so now for the theorem: if we look at the curvatures of four mutually tangent circles (that is, a kissing set), then the sum of their squares is equal to half the square of the sum. That is,

For example, in this configuration:

the curvatures of the four circles are—in order from the biggest circle to the smallest—2, 4, 7, and (approximately) 27.1421356… (you’ll have to take my word on these, I guess). We can verify that , and also , half of which is , as expected.

This is an amazing theorem, which (at the moment) I have no idea how to prove. (Perhaps later this summer I will track down the proof and write about it here.) But it gets much, *much* cooler. It turns out that this theorem can be generalized in all sorts of ways.

First, as stated, the theorem only applies to kissing sets where the circles are all *externally* tangent, as in the example above. With one small change, however, it applies to other configurations as well. If a circle is *internally* tangent to the other circles—that is, it has other circles inside it—then we will think of it as having a “negative” radius, and therefore a “negative” curvature as well. This new notion of “positive or negative curvature” we will call the (*signed*) *bend* of a circle, denoted by the letter *b*. It turns out that using bends, the theorem continues to hold even when some of the circles are inside another (in fact, it even holds when one or two of the “circles” are straight lines, in which case we say they have a bend of zero):

For example:

In the above configuration, the circles have bends of 2, 4, 7, and -1.1421356…, and one can verify that

The theorem can also be generalized to higher dimensions: in three dimensions, *five* spheres can be mutually tangent, and the sum of the squares of their bends is equal to one-*third* the square of the sum; generally, in dimensions, hyperspheres can be mutually tangent, and the sum of the squares of their bends is equal to times the square of the sum. (As a fun historical note, these generalizing results were first published in *poem form*!)

Using Descartes’ Theorem, given the signed bends of three mutually tangent circles, we can solve for a fourth. We end up with a quadratic equation, which yields two different solutions: in particular (as you might like to verify yourself), , where is the sum of the other bends, and

Even better, once we have four bends which satisfy Descartes’ Theorem, if we remove one of them (say, ) and want to compute its “alternate”, we don’t have to solve the whole quadratic equation again, since the two solutions to the quadratic satisfy

So if we already have one value for , we can just subtract it from double the sum of the other three bends to find the second value of .

Of course, finding the *bends* of circles is not quite enough—how do we actually compute the positions of the circle *centers*? It turns out that comes from yet another generalization of Descartes’ Theorem, which I’ll explain in my next post!

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