Christopher Mowla
Premium Member
Thanks for sharing your method.
The algorithm you created to flip the two edges at the end is a commutator indeed:
B E B2 E2 B U2 B' E2 B2 E' B' U2 = [B E B2 E2 B, U2].
I enjoyed how you explained your thought process regarding the trial and error you did to slowly develop each stage of your method as well.
This mental process reminds me of a "derivation" I came up with of the Nikolas commutator, which I show in one of my solving guides that variations of which can twist corners and flip edges in addition to cycling them.
The algorithm you created to flip the two edges at the end is a commutator indeed:
B E B2 E2 B U2 B' E2 B2 E' B' U2 = [B E B2 E2 B, U2].
I enjoyed how you explained your thought process regarding the trial and error you did to slowly develop each stage of your method as well.
This mental process reminds me of a "derivation" I came up with of the Nikolas commutator, which I show in one of my solving guides that variations of which can twist corners and flip edges in addition to cycling them.