Originally posted by doodinthemoodExplain... 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...
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.
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
x:
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.
There are two sets of cubies to consider for this. Edge cubies with 2 stickers and corner cubies with 3 stickers.
For any algorithm, you'll that there are a number of cubies moving to other spots, and possibly reorienting in the process.
For example, Cubie A might move to B's spot, which is now in C's spot but flipped, which is in D's spot, which is in A's spot, flipped.
For an Algorithm of Order 3, the following would need to be true.
1) Displacement of edge cubies would need to be in sets of 3 edge cubies, in such a way that there is an even number of flips for each set. No edges flipping without moving would be possible.
2) Displacement of corner cubes would similarly need to be in sets of 3. Cubies would need to return to same orientation after 3 repetitions. It could be possible for corners to rotate in place.
I know there's at least one I use, but I'll have to check to find it.
EDIT: I thought of one very simple order 3 algorithm although it doesn't do a lot of mixing to be honest.
R2U'R'U'RURURU'R'
EDIT 2:
RL'FB'UD'RL' is also of order 3 and is used for the "6 Boxes" pattern.