MJ
Algorithms (3)
2U 2R U' 3L' U 2R' U' 3L U 2U'
My choice
16 HTM · 10 STM · 16 QTM
3-gen
rotationless
<U,R,L>
One hand: needs L, B, S or E
BLDDB, on 2 solvers' sheets
F' 2U 3L' U' 3L 2U' 3L' U 3L F
16 HTM · 10 STM · 16 QTM
3-gen
rotationless
<U,F,L>
One hand: needs L, B, S or E
BLDDB, on 1 solver's sheet
3D R 3D' 2R' 3D R' 3D' 2R
14 HTM · 8 STM · 14 QTM
2-gen
rotationless
<R,D>
One hand: R, U and M with F or D
Cubemuch
How this case is verified
The case is the source's setup (or, where it has none, another source algorithm reversed). Setup: F' 3L' U' 3L 2U 3L' U 3L 2U' F. Every algorithm above was run by our logic engine from that state and reaches "solved".