# Rubik's Cube Conjecture

One day when I was holding a solved Rubik's cube, I wondered that if I kept on moving the cube U and R (Singmaster notation), would it ever return to its solved position? After some time, the cube actually did become solved again! Here is my conjecture:

After being repeatedly applied a sequence of moves, a Rubik's cube will always return to its original position.

In other words, if a Rubik's cube is in position P, and there is a sequence of moves S (such as {R, U, F}), the cube will be in position P after it is moved according to S a certain number of times. For example, take a cube in the solved position. Apply the move U. Apply U again. After two more Us, the cube will return to its original solved position. To deal with more complex sequences, such as {R, U, B, D, D, D, R, F, L}, I wrote the Java applet above.

In the applet, type in a sequence of moves and press the "Start" button. The cube will start moving according to the sequence. The "Reset" button repositions the cube and clears the sequence. Note that R' is the same as RRR, U' = UUU, F' = FFF, etc.

I'm hoping to prove (or disprove) this conjecture.

Here's a beautiful proof by Allen Yuan (submitted via feedback):

It's actually not too bad to prove the Rubik's cube conjecture. Note the following facts:
1) The Rubiks Cube has a finite number of possible configurations.
2) Thus, while you're moving it in a certain way, you'll eventually have two points in time in which it is at the same configurations (if not, then there would be an infinite number of configurations!).
3) Let these times be after x repetitions and then after y repetitions. However, take both of these and go back x repetitions. That means that the first of the two becomes 0 moves, or your original thing, and the next becomes y-x repetitions. But since they were the same after x and y repetitions, they must be the same after 0 and y-x repetitions. Thus, we've found a place where it's repeated!

## Source Code

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