MA
Algorithms (3)
2L 2U 2L' U' 2L 2U' 2L' U
My choice
14 HTM · 8 STM · 14 QTM
2-gen
rotationless
<U,L>
One hand: needs L, B, S or E
BLDDB, on 19 solvers' sheets
Rw 2B 2L' B' 2L 2B' 2L' B 2L Rw'
16 HTM · 10 STM · 16 QTM
3-gen
rotationless
<R,L,B>
One hand: needs L, B, S or E
Cubemuch
U' 3Rw 2B' 3Rw' 2F 3Rw 2B 3Rw' 2F' 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: Rw 2L' B' 2L 2B 2L' B 2L 2B' Rw'. Every algorithm above was run by our logic engine from that state and reaches "solved".