IC
Algorithms (3)
2R 2U2 2R' U 2R 2U2 2R' U'
My choice
14 HTM · 8 STM · 18 QTM
2-gen
rotationless
oh
<U,R>
One hand: R and U only
BLDDB, on 20 solvers' sheets
U' Rw U 2R' U' 2L' U 2R U' 2L Rw' U
16 HTM · 12 STM · 16 QTM
3-gen
rotationless
<U,R,L>
One hand: needs L, B, S or E
BLDDB, on 1 solver's sheet
U' 2B 2U 2B' U 2B 2U' 2B'
14 HTM · 8 STM · 14 QTM
2-gen
rotationless
<U,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: U' Rw 2L' U 2R' U' 2L U 2R U' Rw' U. Every algorithm above was run by our logic engine from that state and reaches "solved".