Originally posted by doodinthemood
Can someone provide an algorithm order 3 for rubik's cube that looks messed up after 2?
I.E. The algorithm is performed twice, and the cube looks messed up, then it is performed a third time, and this solves the cube.
Thanks if you can.
Explain... If you start from A and go to B, C, and then D, you want B & C to be messed up and D to be the solution. But then if you start from B, then after two steps you are at D, which is not messed up...
OK, so the algorithm must have a memory - what you do next can't be only a function of the current situation - you must remember where you came from.
But then,' you have a simple algorithm:
counter = 1
if counter < 3 then mess up the cube else solve the cube by some known algorithm and stop
counter = counter + 1
go to line x
Without cheating like this you have the problem I explained in the first paragraph.