KX
Algorithms (3)
U2 2L2 U' 2R U 2L2 U' 2R' U'
My choice
13 HTM · 9 STM · 18 QTM
3-gen
rotationless
<U,R,L>
One hand: needs L, B, S or E
BLDDB, on 6 solvers' sheets
x 2L' U 2L 2D2 2L' U' 2L 2D2 x'
14 HTM · 8 STM · 18 QTM
3-gen
rotation
<U,D,L>
One hand: needs L, B, S or E
BLDDB, on 1 solver's sheet
2B Rw 2F Rw' 2B2 Rw 2F' Rw' 2B
14 HTM · 9 STM · 16 QTM
3-gen
rotationless
<R,F,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: x 2D2 2L' U 2L 2D2 2L' U' 2L x'. Every algorithm above was run by our logic engine from that state and reaches "solved".