It has to exist: it's a mindbogglingly large but finite number of possible permutations, and to go from any one permutation to any other takes a finite number of moves. Therefore, if you can enumerate all possible permutations in some way, any arbitrary way will do, you have a Devil's algorithm by going through them in that order. The question is not whether a Devil's algorithm exists but what the shortest one is.
Worth noting that the Devil's algorithm is not a sequence of turns you repeat over and over, but one that takes you through every possible cube state. You "abort" it partway through when you reach your desired state.
Somewhat related: I was playing with cube recently, and, starting with solved state, I started doing sequence of two moves that came to my mind: rotate right side down, then the bottom, clockwise. After maybe a 100-150 repetitions (I did not count, just did the moves mindlessly for couple minutes) I went back to solved state. I wonder if there are more such sequences, and how long it takes to go back to original state. The obvious difference from Devil's algorithm was that the 2x2 sub-cube remained untouched the whole time, only the edges were messed up.
Every sequence behaves this way (For all sequences S, there exists n_S such that S ^ n_S equals S^0, the starting state.) Proving this is an introductory problem in Rubik's theory. Try it!
hvdijk|3 years ago
owalt|3 years ago
988747|3 years ago
lupire|3 years ago
zeroonetwothree|3 years ago
Also check out Lagrange’s theorem in group theory