NT
Algorithms (3)
U' 2R2 3D' 2R U 2R' 3D 2R U' 2R U
My choice
18 HTM · 11 STM · 20 QTM
3-gen
rotationless
<U,R,D>
One hand: R, U and M with F or D
BLDDB, on 2 solvers' sheets
y R2 2R U' 2R' 3D' 2R U 2R' 3D R2 y'
16 HTM · 10 STM · 18 QTM
3-gen
rotation
<U,R,D>
One hand: R, U and M with F or D
BLDDB, on 1 solver's sheet
2R' 3B' R 3B 2R 3B' R' 3B
14 HTM · 8 STM · 14 QTM
2-gen
rotationless
<R,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: y R2 3D' 2R U' 2R' 3D 2R U 2R' R2 y'. Every algorithm above was run by our logic engine from that state and reaches "solved".