July 25th, 2009

5x5x5 Solved!



Okay, after getting this as a gift (a pay stub would be better, but compared to nothing...), I managed to figure it out.
I was concerned about the parity cases, since those plagued my 4x4x4 cube.

Good news. If you reduce the 5x5x5 to a distorted 3x3x3, you can solve with no parity errors to worry about.
HOWEVER, when you make the edge pieces (1x3), there are two parity cases possible. One is when the middle of the three is flipped, and the other is when you are almost finished making the edges, but the end piece of two separate rows of three are 'switched'.

Annoying, but not quite as bad as the 4x4x4. I fear the 6x6x6, but not so much the 7x7x7