Pi (L)
OLL parity (1 flip)
0 Corners
Count the flipped dedges (one here), then read the corner shape as for OLL.
Your status
Algorithms (4)
(U') first, then:
Rw' U2 Rw' U2 Rw R U Rw2 R' U2 Rw' R' U R U Rw'
U2 Rw2 U' Rw2
My choice
20 HTM · 20 STM · 27 QTM
2-gen
rotationless
oh
<U,R>
One hand: R and U only
Do U' first (AUF) on the case as drawn. · Cubemuch
Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 Rw' U2 Rw U2 Rw'
U2 Rw' R' U2 R U R' U R
24 HTM · 24 STM · 33 QTM
3-gen
rotation
<U,R,L>
One hand: needs L, B, S or E
Cubemuch
U' R' U2 R U R' U R U' Rw U2 x Rw U2 Rw U2
Rw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw'
26 HTM · 26 STM · 35 QTM
3-gen
rotation
<U,R,L>
One hand: needs L, B, S or E
Cubemuch
y' L U R 2R R 2R B2 2R' R' U2 2R' R' U2 B2 2R'
R' B2 R 2R B2 2R' R' B2 R 2R R 2R B2 R' U L'
U' R
42 HTM · 33 STM · 50 QTM
rotation
<U,R,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: R' U' R U' R' U2 R Rw U2 Rw U2 Rw' U2 Rw U2 Lw' U2 Rw U2 Rw' U2 Rw' x' U2 Rw'. Every algorithm above was run by our logic engine from that state and reaches "oll-solved", keeping first-layers.