Pi (B)
OLL parity (1 flip)
0 Corners
Count the flipped dedges (one here), then read the corner shape as for OLL.
Your status
Algorithms (5)
(U') first, then:
R U 2R U R' U' Rw2 R U2 Rw2 U Rw U2 Rw' U2 Rw
U' Rw2 U2 Rw'
My choice
21 HTM · 20 STM · 28 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 Rw' U2 Rw U2 Rw2 U' Rw' U' Rw U' Rw' U2 Rw' U2
Rw2 U2 Rw' U' Rw' U Rw U Rw
25 HTM · 25 STM · 33 QTM
2-gen
rotationless
oh
<U,R>
One hand: R and U only
Cubemuch
U' R' U' R U' R' U2 R Rw U2 x Rw U2 Rw U2 Rw'
U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw'
25 HTM · 25 STM · 34 QTM
3-gen
rotation
<U,R,L>
One hand: needs L, B, S or E
Cubemuch
Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 Rw' U2 Rw U2 Rw'
U2 Rw' U' 3Rw U R' U' 3Rw' F R F'
26 HTM · 26 STM · 34 QTM
rotation
<U,R,F,L>
One hand: needs L, B, S or E
Cubemuch
R U' L' U R' 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 U
L
42 HTM · 33 STM · 50 QTM
rotationless
<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: Rw' U' Rw' U' Rw U Rw U2 Rw2 U2 Rw U2 Rw U Rw' U Rw U Rw2 U2 Rw' U2 Rw U2 Rw'. Every algorithm above was run by our logic engine from that state and reaches "oll-solved", keeping first-layers.