AC
Algorithms (3)
3Lw' 2R' 2D2 2R U 2R' 2D2 2R U' 3Lw
My choice
16 HTM · 10 STM · 20 QTM
rotationless
<U,R,D,L>
One hand: needs L, B, S or E
BLDDB, on 1 solver's sheet
2L 2U' 2L' U2 2L 2U 2L' U 2L 2U' 2L' U 2L 2U 2L'
27 HTM · 15 STM · 28 QTM
2-gen
rotationless
<U,L>
One hand: needs L, B, S or E
BLDDB, on 1 solver's sheet
3Fw U' 2B 2D2 2B' U 2B 2D2 2B' 3Fw'
16 HTM · 10 STM · 20 QTM
rotationless
<U,F,D,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: 2L 2U' 2L' U' 2L 2U 2L' U' 2L 2U' 2L' U2 2L 2U 2L'. Every algorithm above was run by our logic engine from that state and reaches "solved".