JP
Algorithms (3)
F2 2L' U2 2L 3D' 2L' U2 2L 3D F2
My choice
16 HTM · 10 STM · 20 QTM
rotationless
<U,F,D,L>
One hand: needs L, B, S or E
BLDDB, on 2 solvers' sheets
U' 2R U 2R 3D 2R' U' 2R 3D' 2R2 U
18 HTM · 11 STM · 20 QTM
3-gen
rotationless
<U,R,D>
One hand: R, U and M with F or D
BLDDB, on 1 solver's sheet
Rw' 3D R' 3D' 2R 3D R 3D' R
14 HTM · 9 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: U' 2R2 3D 2R' U 2R 3D' 2R' U' 2R' U. Every algorithm above was run by our logic engine from that state and reaches "solved".