MA
Algorithms (3)
3L 2U 3L' U' 3L 2U' 3L' U
My choice
14 HTM · 8 STM · 14 QTM
2-gen
rotationless
<U,L>
One hand: needs L, B, S or E
BLDDB, on 8 solvers' sheets
U' R 2B' R' 3F R 2B R' 3F' U
14 HTM · 10 STM · 14 QTM
rotationless
<U,R,F,B>
One hand: needs L, B, S or E
Cubemuch
U' Rw 2B' Rw' 3F Rw 2B Rw' 3F' U
14 HTM · 10 STM · 14 QTM
rotationless
<U,R,F,B>
One hand: needs L, B, S or E
Cubemuch
How this case is verified
The case is the source's setup (or, where it has none, another source algorithm reversed). Setup: U' 3F R 2B' R' 3F' R 2B R' U. Every algorithm above was run by our logic engine from that state and reaches "solved".