KJ
Algorithms (3)
F' U' 2L' U 2R U' 2L U 2R' F
My choice
14 HTM · 10 STM · 14 QTM
rotationless
<U,R,F,L>
One hand: needs L, B, S or E
Cubemuch
3Uw2 2L 3Uw' 2R 3Uw 2L' 3Uw' 2R' 3Uw'
13 HTM · 9 STM · 14 QTM
3-gen
rotationless
<U,R,L>
One hand: needs L, B, S or E
Cubemuch
Uw2 2L Uw' 2R Uw 2L' Uw' 2R' Uw'
13 HTM · 9 STM · 14 QTM
3-gen
rotationless
<U,R,L>
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: 3Uw 2R 3Uw 2L 3Uw' 2R' 3Uw 2L' 3Uw2. Every algorithm above was run by our logic engine from that state and reaches "solved".