Is there a way to evenly distribute face turns for 12 flip?
Submitted by cubex on Mon, 11/14/2016 - 22:10.
Back in Jan 1995 Mike Reid found this process for the 12 flip:
R3 U2 B1 L3 F1 U3 B1 D1 F1 U1 D3 L1 D2 F3 R1 B3 D1 F3 U3 B3 U1 D3 24q
This process has 24 q turns, so I'm wondering could there be a 24 q turn process that evenly distributes the turns so that each side turns 4 q? The idea just seemed elegant to me, 6 faces each turning 4 q turns.
Mark
R3 U2 B1 L3 F1 U3 B1 D1 F1 U1 D3 L1 D2 F3 R1 B3 D1 F3 U3 B3 U1 D3 24q
This process has 24 q turns, so I'm wondering could there be a 24 q turn process that evenly distributes the turns so that each side turns 4 q? The idea just seemed elegant to me, 6 faces each turning 4 q turns.
Mark