Super-supercube challenge!!

Hmmm. That is weird. Well here is my notation. On the 5x5x5 we have layers R r M l L going from left to right. From top to bottom we have U u E d D. From front to back we have F f S b B. M turns same way as L. E same way as D and S same way as F. With this notation my complete and improved solution is:

1. l' u' (l m) d' (l'M') u' (l M) d (l' M') u2 l (11 turns block turn metric)

2. f l (f' S')d (f S) l (f' S') d' (f S) l2 f' (11 turns)

A full explanation will follow shortly. Part 1. and 2. are 2 block 3-cycles.

Per
 
Sometimes easily solutions are buried in bad notation. I had the same 20 turn solution all the time. But i had written my block cycles with solely outer turns (on 3x3x3). Changing to proper commutator turns (with block turns in this case) was the way to go(to avoid side effects).

Coincidentally there is at least 18 different 20 turn solutions. 3 ways to cycle first 3 blocks. Then 3 ways for the last 3. In addition there are 2 ways to split it into 6 blocks ...

Per
 
What i meant is that my normal cube in cube is a 20 turn solution, involving 2 block 3-cycles. I realised i should "normalise" that one right after i posted my 22-turn solution. Oh well, you beat me to it cause of my "stupidity":tu

Check this thread!!

Per
 
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