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Ok, I am looking for an OLL super cube 4x4 parity algorithm. I cannot find it on the speedsolving wiki. I remember seeing on Youtube but cannot find the video again. It was such an easy alg to do too.... It was something like r' F R U R2 F' r l F R2 U' R' F' l... but that is not it. Any idea what I might be thinking of? It definitely started with an internal r move r' then some other moves ending in r followed by l with the same moves reverse and then l'.

Edit: Ok, I think I am actually thinking of a PLL alg.... same issue though.

Edit 2: found it
r' (U' R U F R2 F') r l (F R2 F' U' R' U) l'
 
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U M E' M' E U' E' M E M' ? holding center that needs to be twisted on U face

EDIT: 333th post

That turns two faces 90 degrees. Ok. So I've tried solving my new Megamorphox about half a dozen times and overtime I get this (photo). I have looked at CBC's as well as many other tutorials, but nothing seems to touch upon this case, or at least I am not understanding it.
Screenshot 2015-11-09 23.57.57.jpg
attachment.php
 
That turns two faces 90 degrees. Ok. So I've tried solving my new Megamorphox about half a dozen times and overtime I get this (photo). I have looked at CBC's as well as many other tutorials, but nothing seems to touch upon this case, or at least I am not understanding it.
View attachment 5623
https://www.speedsolving.com/forum/attachment.php?attachmentid=5623&d=1447131494

The problem is that ONLY rotating a center 90 degrees is 3 swaps, which violates parity. If you swap two identical centers somewhere else on the puzzle I think you should be good.
 
swap two identical centers?

There is no algorithm out there to just rotate the four x-center pieces?

No, that's not possible.

It's difficult for me to work out a solution since I don't actually have one of these cubes, but I'll try:

Hold the four unsolved x-center pieces on the front with blue-red-yellow corner in the picture at UFL. Then do (small letters are slice moves):
r U l' U' r' U l U' - this cycles FDR to UFR to FDL, and since FDR and UFR are identical blue pieces it appears to be a 2-cycle (it's really a 3-cycle)
x - reposition so the unsolved group is on top
r' u r U r' u' r U r' u r U2 r' u' r - cycles UBR to UBL to UFL
 
No, that's not possible.

It's difficult for me to work out a solution since I don't actually have one of these cubes, but I'll try:

Hold the four unsolved x-center pieces on the front with blue-red-yellow corner in the picture at UFL. Then do (small letters are slice moves):
r U l' U' r' U l U' - this cycles FDR to UFR to FDL, and since FDR and UFR are identical blue pieces it appears to be a 2-cycle (it's really a 3-cycle)
x - reposition so the unsolved group is on top
r' u r U r' u' r U r' u r U2 r' u' r - cycles UBR to UBL to UFL

Perfect. Thank you
 
I am looking for a 2-gen 2x2 alg that solves the case that can be set up with:
R U' R F2 R' U R' x2 y'
Just curious and maybe I could use it in solves as well?
 
I am looking for a 2-gen 2x2 alg that solves the case that can be set up with:
R U' R F2 R' U R' x2 y'
Just curious and maybe I could use it in solves as well?

This is the algs I use. It's basically transcribed from the square-1 cp alg for that case.
R2 U R2 U' R2 U R2 U' R2
 
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