4x4 Orientation parity

For those interested in this thread, check out my thread in puzzle theory:
Methods for Forming 2-Cycle Odd Parity Algorithms for Big Cubes

There I show that common brief OLL parity algorithms can be understood in full...and yes they involve a commutator. In fact all brief symmetrical 2-cycle parity algorithms of the conventional move set use the same commutator (some use it more directly than others), as far as I can see. I haven't come across one which I couldn't derive from the same commutator.;)

I still have a lot to add to it yet, but making videos and writing out derivations takes time. I will add to it very soon.
 
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Thanks for the link, I can't read it at the moment (thesis takes priority over my curiosity) but it looks excellent. I'll read it over later!
 
For those interested in this thread, check out my thread in puzzle theory:
Methods for Forming 2-Cycle Odd Parity Algorithms for Big Cubes

There I show that common brief OLL parity algorithms can be understood in full...and yes they involve a commutator. In fact all brief symmetrical 2-cycle parity algorithms of the conventional move set use the same one commutator (some use it more directly than others), as far as I can see. I haven't come across one which I couldn't derive from the same commutator.;)

I still have a lot to add to it yet, but making videos and writing out derivations takes time. I will add to it very soon.

By the way I would like to know how you derived this one.
 
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A more intuitive way to fix the orientation parity is to simply perform an odd number of quarter turns on the inner layers. You will most likely have the centers already solved at this point though, so I suggest to use the following cyclic way to break up the centers and fix them again.

(r' U2)*4 r' or written out that's r' U2 r' U2 r' U2 r' U2 r'

This breaks apart and restores the centers in the middle M slice, and does so in 5 quarter turns of the inner r layer, thus "fixing", or really changing, the parity of the wings from odd to even.

After this you will need to restore the 4 edge groups in the M slice that got destroyed. This can be done simply with commutators or with your regular edge pairing method.

This algorithm is far longer than the parity algorithm you found, but it is much more intuitive.

Chris



YES! That is exactly what I was looking for. its so similar to the PLL parity alg (at least in my mind) so it makes it easy to memorize. Thanks!!!

:fp

please don't bump a really old thread just to say thanks to someone for something they said 2 and a half years ago..

At least he was using the search function. :o
 
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