Rubik's Cube in 1 move

By: Gijs Bellaard

You might have heard of the "urban myth" that a scrambled Rubik's cube can always be solved by repeating a certain special move (that is, a sequence of twists) sufficiently enough times. But this is trivially true; if we let our special move be a (absolutely gigantically long) move that goes through all possible arrangements we can be certain that if we apply this to our scrambled cube we at some point arrive at the unscrambled cube during our twists. In this case we even only need to do the move once! However, this is an extremely unsatisfying answer, and that is why I will create the extra constraint that the move must be performed an exact number of times. So, if during our special move we come across the unscrambled cube it doesn't count; we must have the solved cube after doing the special move a integer amount of times. It turns out that, after we add this constraint, such a special move can not exist! In this article I will show why this is the case using elementary arguments (I hope) everyone can follow.

We start by assuming that such a special move does exist. Take a solved Rubik's cube and apply the special move \(1\) time, then we have a scrambled cube. This means that we are able to perform the special move sufficiently enough times and end up with the solved cube again. We can conclude that there exists a \(k\) such that repeating our special move \(k\) times does nothing; it returns the cube to the position it was in.

Take the solved cube again and carry out any move to make a scrambled cube. Let \(l\) be the smallest amount of times the special move needs to be repeated such that it becomes solved again. Notice that \(l\) must be smaller than \(k\); if it wasn't \(l-k\) would also solve our scrambled cube but we assumed \(l\) to be the smallest. To repeat: applying the special move \(l\) times on our scrambled cube solves it, executing it \(k\) times does nothing. Thus performing the special move \(k-l\) times is equivalent to the original move. This means any move can be encoded by repeating the special move enough times.

Again, take a solved cube and do the following move: twist the right face and afterwards the top face one quarter turn clockwise. This will give a different cube than if we did it the other way around: that is first twisting the top face and afterwards the right face. Lets return to what we said earlier: any move can be encoded by the special move. So lets encode the top twist and say it is the same as repeating the special move \(t\) times, analogously we encode the right twist with \(r\) special moves. This means our move is equal to doing the special move \(r\) and then \(t\) times, and our other move \(t\) and then \(r\) times. But wait... \(r+t = t+r\) so the two different moves should be the same however we know that they aren't. All the reasoning we did was sound so our assumptions must have been wrong; we have to conclude that such a special move can not exist!

In group theory the argument is as follows: If such a special move exists the Rubik's group would be generated by a single element, which means it is cyclic, which means it is abelian which it obviously isn't.