04 Oct

[Note this post is a bit outdated, and not maintained; the most up-to-date version of this information can be found in my pages on the OEIS wiki.]

I’ve recently become interested in a family of interrelated sequences that can be found in the Online Encyclopedia of Integer Sequences (OEIS). These sequences all have to do with one approach to measuring the “complexity” of an integer. This approach assumes that you have some basic “building blocks” for constructing integers: starting numbers you’re allowed to use, and operations that you may apply to them. The idea, then, is that the number of these building blocks that you need to arrive at a particular integer is one way of characterizing how “complex” that integer is.

This approach to constructing integers may seem familiar from the “Four fours” puzzle: Given four instances of the number 4 to start with, and a variety of operations that you might use, how many different whole numbers can you construct? For some examples, $1 = \frac44 \quad 2 = \frac{4+4}4 \quad 3 = \frac{4+4+4}4 \quad 5 = 4+\frac44 \quad \ldots \quad 8 = 4 + 4 \quad \ldots$ Here it seems quite reasonable that in terms of “expressing with fours,” the number eight is “simpler” than the number three.

The lowest-numbered sequence (that I’ve found) in the OEIS that has to do with integer complexity is A000792. In the case of this sequence, the only number you’re allowed to start with is one, and the only operations you’re allowed to use are plus and times. The $n$th entry of sequence A000792 then tells you the largest number you can produce using at most $n$ ones.

Formally, let $A$ be an “alphabet” of numbers and operators; for simplicity we’ll assume all operators are unary or binary. One can then form finite sequences of elements from $A$; such a sequence is a well-formed RPN expression if no initial subsequence has as many binary operators as numbers. Each well-formed RPN expression unambiguously determines its value: imagine you have an initially-empty stack of numbers, and go through the sequence from left to right. Any time you encounter a number, put it on top of the stack; when you encounter a unary operator, modify the top number on the stack by applying that operator; and when you encounter any binary operator $\ast$, remove $a$ and $b$ (respectively) from the second-to-top and top positions from the stack, and then put $a\ast b$ on top of the stack (reducing the height of the stack by one). The value of the RPN expression is the top number on the stack when the entire expression has been scanned. The well-formed condition guarantees the stack never empties.

We can now define the $A$-operand complexity of a number $n$, which we’ll write as $c_A(n)$, as the minimum number of numbers in any well-formed RPN expression with value $n$, the $A$-operator complexity $d_A(n)$ as the minimum number of operators in any well-formed RPN expression with value $n$, and similarly the $A$-expression complexity $e_A(n)$ as the minimum length of a well-formed RPN expression with value $n$. There is no general relationship between $c_A$, $d_A$, and $e_A$, but if $A$ contains only binary operators, then $e_A(n)$ $= 2c_A(n)-1$ $= 2d_A(n)+1$. (Briefly, count the number of operators needed to reduce the operands to a single number; clearly the minimal expression in any case ends with a stack of just one number.)

Since it’s the most common notion of complexity used in the OEIS, we take “complexity” without an adjective to refer to $c_A$, the $A$-operand complexity.

I’ve also assembled a table that organizes sequences in the OEIS related to these notions of integer complexity, according to the alphabet $A$ and the particular measure or property of the complexity.

30 Aug

In both the Woven SSD and the Woven GSD, you calculate the inset (from the points where the rods meet near their ends) for putting marks on the rods by multiplying their lengths by 2/(3+ā5). Where does that strange-looking number come from?

The key is that both figures consist of (regular) pentagrams, just interlocked in different ways. In other words, in both structures, looking only at any 5 rods in the same plane, we see:

What we’re interested in the fraction of the length of one rod (that is, the length between the points of the star where the rods cross; we’ll ignore the small overhangs beyond that from here on) represented by the distance from a point to the first internal intersection. In other words, we want the ratio of the red segment to the green segment in this diagram:

Now imagine that these segments are roads, and you start driving in a car from the red-blue point, and you follow all five of the roads until you end up back where you started, pointing in the same direction. You make five turns through the angle $\alpha$ indicated in the diagram below, but end up overall making two full turns all the way around (i.e., overall your car turns through an angle of $720^{\circ} = 4\pi = 2\tau.$) Therefore, $5\alpha = 2\tau,$ so the angle $\alpha = 2\tau/5,$ or 2/5 of a turn. We can then conclude that the point angle $\beta$ in the following diagram must be $\tau/2 – 2\tau/5 = \tau/10,$ or one-tenth of a turn (36 degrees).

The same argument applied to a regular pentagon shows that its interior angles are $3\tau/10$ (or 108 degrees). In other words, in the following diagram, the edges of the pentagram trisect the vertex angles of the circumscribed pentagon:

This angle relationship in turn means that the orange-pink-green triangle is similar to the brown-purple-green triangle. But it’s also clear that the long green is the sum of the brown and the purple. This similarity tells us that the sum of the purple and brown is to the orange as the purple is to the brown. But the orange is clearly the same as the purple, so the sum of the purple and brown is to the purple as the purple is to the brown. And this is exactly the definition that the purple cuts the long green in the golden ratio $\phi,$ i.e. that the purple is $1/\phi$ as long as the long green.

On the other hand, the brown-purple-green triangle is also similar to the acute pink-blue-red triangle. Applying the exact same argument as before shows that the red segment is $1/\phi$ as long as the purple segment.

Combining these two facts tells us that the red segment is $1/\phi^2$ as long as the long green segment. By solving the equation $(\phi+1)/\phi = \phi/1$, you can find the value $\phi=(1+\sqrt5)/2$. Using this value and some algebraic manipulations lets us calculate that the red segment is $2/(3+\sqrt5)$ as long as the green one, as used in the posts linked above.

23 Mar

This post was for announcing a week-long summer workshop on Illustrating Mathematics at the Park City Mathematics Institute, this past 2021 July 19-23. It was an exciting week with lots of interesting programming including several different hands-on, how-to tutorials, keynotes by Vernelle Noel, Ingrid Daubechies, and Daniel Piker, and mathematical “show-and-ask” sessions in which a wide array of mathematicians displayed some of the intriguing and beautiful images and objects they’ve created, as well as highlighting the questions these projects have raised. (I led one minicourse on using CAD/CAM software like LibreCAD or FreeCAD in creating physical mathematical models.)

23 Mar

This is a sequel to a (pre-pandemic) post about weaving a stellated polyhedron. This time, I’d like to show how similar techniques can also be used to create a “great stellated dodecahedron” (“GSD” for short; illustration below).

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12 Mar

When I showed this recent post to my friend and colleague Laura Taalman, aka mathgrrl, she suggested that another approach to creating a model of the underlying structure would be to construct the icosahedra themselves (rather than the negative space), except use wireframes of the icosahedra rather than solid ones to avoid obscuring all of the internal structure. Her encouragement motivated me to create a new OpenSCAD file for this.

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09 Mar

If you look again at the diagram of an icosahedron in a cube (at the right), you’ll see that because of its symmetry, all three of the icosahedron vertices nearest the top front cube vertex are the same distance from that cube vertex. That equidistance means that the body diagonal of the cube passes through the center of the icosahedron and is perpendicular to the nearest face of the icosahedron.

All of these properties hold at every vertex of the cube in relation to one of the eight faces of the icosahedron that do not have an edge lying in a face of the cube. Because the eight faces of the inscribed dual octahedron of the cube have these same properties, we’ve demonstrated that an icosahedron may also be inscribed in an octahedron like so:

It also means that if you extend the plane of one these particular eight faces of the icosahedron, it slices off an isosceles right triangular pyramid from the corner of the cube:

In a lattice of such cubes, eight of these pyramids fuse together at each vertex to create the octahedra seen inside the Anticos.

09 Mar

As mentioned and illustrated in the post on the Anticos, it’s possible to inscribe an icosahedron in a cube. (In this case, that technically means that given a cube, you can choose two points on each face of the cube such that the convex hull of the resulting set of twelve points is a regular icosahedron.)

But why should this be so? To see this, it’s easiest to start with a regular dodecahedron, say with unit edge length. Notice the interesting pattern of the blue face diagonals in this diagram:

Notice that during this transformation, each of the edges remains in the plane perpendicular to that axis. Therefore, twelve of the vertices of an icosahedron lie on the surface of a cube, two on each face. Since an icosahedron only has twelve vertices, it is inscribed in a cube.

09 Mar

Judging from at least one of the previous projects, Studio Infinity is intrigued with connecting polyhedra edge-to-edge. (Of course, connecting them face-to-face is interesting, too, but that’s pretty familiar from Legos and such; and vertex-to-vertex is the same as connecting dual polyhedra face-to-face.)

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07 Dec

Here’s a large-scale model I designed of the Weaire-Phelan space packing, built by the participants of the Fall 2019 semester on Illustrating Mathematics at ICERM in Providence.

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