AC
Algorithms (3)
2L 2U' 2L' U2 2L 2U 2L' U 2L 2U' 2L' U 2L 2U 2L'
My choice
27 HTM · 15 STM · 28 QTM
2-gen
rotationless
<U,L>
One hand: needs L, B, S or E
BLDDB, on 1 solver's sheet
Lw2 U' 2L2 U 2F2 U' 2L2 U 2F2 Lw2
14 HTM · 10 STM · 24 QTM
3-gen
rotationless
<U,F,L>
One hand: needs L, B, S or E
Cubemuch
Lw2 Uw' 2L2 Uw 2F2 Uw' 2L2 Uw 2F2 Lw2
14 HTM · 10 STM · 24 QTM
3-gen
rotationless
<U,F,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: Lw2 2F2 U' 2L2 U 2F2 U' 2L2 U Lw2. Every algorithm above was run by our logic engine from that state and reaches "solved".