PA
Algorithms (3)
U 2R 2D 2R' U' 2R 2D' 2R'
My choice
14 HTM · 8 STM · 14 QTM
3-gen
rotationless
<U,R,D>
One hand: R, U and M with F or D
BLDDB, on 16 solvers' sheets
3Uw 2R 2D 2R' 3Uw' 2R 2D' 2R'
14 HTM · 8 STM · 14 QTM
3-gen
rotationless
<U,R,D>
One hand: R, U and M with F or D
Cubemuch
Uw 2R 2D 2R' Uw' 2R 2D' 2R'
14 HTM · 8 STM · 14 QTM
3-gen
rotationless
<U,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: 2R 2D 2R' 3Uw 2R 2D' 2R' 3Uw'. Every algorithm above was run by our logic engine from that state and reaches "solved".