*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

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.