IE
Algorithms (3)
3L' U' 3L 2U' 3L' U 3L 2U
My choice
14 HTM · 8 STM · 14 QTM
2-gen
rotationless
<U,L>
One hand: needs L, B, S or E
BLDDB, on 4 solvers' sheets
3Lw 2F 3Lw' 3F' 3Lw 2F' 3Lw' 3F
12 HTM · 8 STM · 12 QTM
2-gen
rotationless
<F,L>
One hand: needs L, B, S or E
Cubemuch
3Lw' 2B 3Lw 3F' 3Lw' 2B' 3Lw 3F
12 HTM · 8 STM · 12 QTM
3-gen
rotationless
<F,L,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: 3F' 3Lw 2F 3Lw' 3F 3Lw 2F' 3Lw'. Every algorithm above was run by our logic engine from that state and reaches "solved".