Cubing Conundrum

Here's a sketch of the argument:

Let the solved state of the cube be $\mathcal{C}_0$ and, for simplicity, take this to be the initial state.

There are a finite number of arrangement of the cube, $\Gamma$ . Assume we apply some move sequence, $\mathcal{M}$ to the cube $n$ times so that the state of the cube becomes $\mathcal{M}^n\mathcal{C}_0$ .

This move can only be applied so many times without repeating a state of the cube. The maximum is $\Gamma$ , the size of the state space for the cube.

By definition, the process would have to repeat after this point as it would have already visited every state once. Therefore, there are no algorithms that never repeat a state of the cube.

tl;dr : the state space of the Rubik's cube is finite and can only be partitioned into finite cycles of states. These must repeat.

Josh, but that sketch of an argument needs a small refinement. Let's say we start with the solved state. With each repeated move sequence, it goes to another state. Let's say that it fails to return to the original solved state, but instead hits a state other than the solved one (it has to eventually, because there's only a finite number of them). Then it repeats. If this were possible, I could repaint the cube in any state in that loop chain so that it is a "solved state". Hence, it will come back to the solved state. How the cube cycles through its states is actually color-blind, it's only relative to some initial state.

This might be a stretch to say this, but this is analogous to how gauge symmetries work in particle physics.

That was easy!!!!! There isnt any scientific or mathematically explainable reason.(or maybe i dont know that) but it can be explained as an algorithm is like any point on a circle. Moving by its circumference we will reach again on our initial point. Like the alg. Ri F Ri B2 R Fi Ri B2 R2 will move 3 corners clockwise and by reapplying twice it will orient permute corners back to their original position