Snack Foods, etc.

It snowed all night last night, apparently, so when I got up this morning, I was surprised to see about a foot of snow on top of my car. that was sweet. I have a couple pictures, but they are on the phone for right now. Instead enjoy these:


Valentine's Peeps. Thanks, ma! I just got these last night. These are the vanilla flavored ones, and they certainly live up to that claim. When I opened the box, I was almost overwhelmed by the smell of vanilla. They are excellent. I've also a box of strawberry flavored ones, but I like my peeps relatively fresh, so I don't open up multiple boxes at once.


Choice Biscuit. According to the box, "The New Standard Biscuit." I'm not sure what they mean by that. They're just butter cookies, but I get a kick out of the use of the word "biscuit" to mean "cookie." I mean, come on, England.

Last night I was thinking how I get annoyed when people show off that they've memorized someone's method for solving Rubik's cubes. It doesn't even seem like it would be fun. Then I thought that perhaps the rotations on such a cube could form a group. They seem rather like permutations, is why. Here's a little detail if you're not familiar.

A group is a set, G, equipped with an operation, *, such that for any a, b, c in G, we have:

1) a * b is also in G [In this case, * is said to be a binary operation on G]
2) (a * b) * c = a * (b * c) [This should be familiar as the associative property]
3) There exists an element of G, e, called the identity element that e * a = a * e = a
4) There exists an element of G, a', called the inverse of a, such that a * a' = e

So, nothing extra special, and you can probably come up with a couple examples. I should hope so, anyway. Note that one seemingly common property was not included,

5) a * b = b * a [That is, the commutative property].

If 5) also holds, then G is said to be an Abelian group, named for one Henrik Abel, the only mathematician known to have a nude statue erected in his honor.

It may be a bit weird to think about these rotations as elements, but that is what we are doing. The operation (*) that we would be using here would be composition. So, imagine the Rubik's cube, in whatever configuration you want. Suppose we denote a rotation right (it doesn't actually matter which way) about the z-axis as z, and similarly one about the x-axis as x, and so on for y. Then we could see turning about the x-axis and then the y-axis as y(x(c)), with c being whatever our cubes initial "value" was. Or in group notation, x * y. Actually, we need to specify which part of the cube we are rotating, but let's just stick to this notation for the time being.

Ok, so, now we have a set of moves on a cube. Is it a group? Let's check

1) Is G closed? [Is * a binary operation?] Well, yeah, if you rotate part of the cube, and then rotate it again, do we still have a way of moving the cube? Yeah, pretty obviously. More technically, a function of this sort (really the only sort short of breaking the cube) composed with another of this sort is still a function of this sort. Neat, but hardly worth checking.

2) Is * associative? This is the one I'm not sure of, but I'm not very good at picturing Rubik's cubes in my head, and the only store where I think I can get one here was closed today. Too bad.

3) Is there an identity rotation? Yeah, just don't rotate it. Genius.

4) Is there an inverse for any rotation? Sure, just do it backwards, and you are set. In fact, let's look at little closer. What is the inverse of x? It's -x, if you will, rotating left, if x is rotating right. But what is rotating left if not rotating three times right? Or in group notation,

(x * x * x) * x = e.

That should be pretty obvious because turning the thing around four times is the same as not doing anything to it.

5) Is G abelian? Well, no. Here's what I mean, rotate once along the x-axis and then once along the y-axis. Now undo that (inverses!) and rotate again, but this time along the y-axis first, then the x-axis. Does the cube end up the same? No, so it's not abelian, if it is indeed a group. Shucks.

Why would it matter if this is a group? Well, groups let us do neat little things like solve equations, so maybe we could come up with a general solution to the Rubik's cube. For those more Turing-inspired, once you could figure a way to solve the Rubik's algebraically, you'd be just a hop, a skip, and a jump away from a computer program with the ability to take the "value" of the cube from you (this would be tedious to input, what with 54 squares) and spit out a set of directions for your individual cube. So, hey, how about that. If you come up with a proof or disproof of the associativity, let me know.