a one-move solution? (not chess moves...)

on a rubick's cube, is there a single "move" (where a move is a combination of simple moves, like left "up" twice, rotate front face 90 deg then left "down" twice, etc) that will cycles through every possible position which the cube can be in? thus, no matter where we start, applying this "move" will solve the cube every time.

i really surprised myself with this one. hints can be given if needed...although i'm unsure about being able to give a clearer explanation of what a "move" is...

yea i think that there is such a thing

Originally posted by O Artem O
yea i think that there is such a thing

...proofs should be given...😛

Rubiks cube has a parity of 2.

Take one apart and put it back together and there is only a 50% chance you can solve it.

Originally posted by genius
on a rubick's cube, is there a single "move" (where a move is a combination of simple moves, like left "up" twice, rotate front face 90 deg then left "down" twice, etc) that will cycles through every possible position which the cube can be in? thus, no matter where we start, applying this "move" will solve the cube every time.

i really surprised myself wit ...[text shortened]... ugh i'm unsure about being able to give a clearer explanation of what a "move" is...

pXsXs

??

Originally posted by Freddie2006
pXsXs

??

say what?

Originally posted by genius
say what?

That's google for you. 😕

Originally posted by Freddie2006
That's google for you. 😕

effective equals product times stakeholders times setting?

...i can give a clue?...

Originally posted by genius
on a rubick's cube, is there a single "move" (where a move is a combination of simple moves, like left "up" twice, rotate front face 90 deg then left "down" twice, etc) that will cycles through every possible position which the cube can be in? thus, no matter where we start, applying this "move" will solve the cube every time.

i really surprised myself wit ...[text shortened]... ugh i'm unsure about being able to give a clearer explanation of what a "move" is...

Such a move does not exist. If it existed, the group of permutations of the cube would be cyclic and thus Abelian. However, it's easy to find pairs of moves which don't commute. QED

Originally posted by David113
Such a move does not exist. If it existed, the group of permutations of the cube would be cyclic and thus Abelian. However, it's easy to find pairs of moves which don't commute. QED

yup-that's it. i knew my group theory would come in useful one day... can you say how many different moves we would need to cycle through all possible positions?

Any position to solved can be done in 7 moves.

This either implies that any position to any other position is 7 moves, or any position to any other position is 14 moves.

Also, there are 4.3*10^19 positions, so if you intend solving a cube by cycling it through every possible position, you have a lot of time on your hands.

Originally posted by doodinthemood
Any position to solved can be done in 7 moves.

Defining "move" as in the original post, any position to be solved can be done in one move.

I was defining move as one side turning, as soon as you switch and do another side, you're onto the second move.

Also, it's 17 and 34. Not 7 and 14. 7 moves would be incredible.

Originally posted by doodinthemood
I was defining move as one side turning, as soon as you switch and do another side, you're onto the second move.

Also, it's 17 and 34. Not 7 and 14. 7 moves would be incredible.

You are wrong. There is a position that is 20 moves away from the monochromatic position.
http://en.wikipedia.org/wiki/Optimal_solutions_for_Rubik%27s_Cube
See the paragraph just above the "contents" box.

What you probably meant is that 17 is a theoretical lower bound, since you have 18 possible first moves (any face can be turned 90, 180 or 270 degrees) and after that for each move you have 15 possibilities (you don't want to turn the same face again).

So in 0 moves you can reach at most 1 position; in 1 move, 18 positions; in 2 moves, 18*15 positions; in 3 moves, 18*15*15 positions, and so on - in n moves, 18*15^(n-1) positions. This number is bigger than ~4.3*10^19 (number of possible positions) only if n>=17, which proves there are positions which need 17 moves. But this does not prove that all positions can actually be solved in 17 moves.