Marlowe: Please fix this problem, it can only be possible for the answer to be either black is dead or it is unsettled depending on liberties. The answer "white is dead" is literally impossible because if white goes first, he takes at L3 and he cannot lose after that, any attempt to capture anything after that is a self-atari for black.
semelis: I thought that this problem was not randomized because I always get the same problem, with 2 black stones at L1-L2, a mostly symmetric position, and a black lump of stones with 5 liberties.
hoshi911: [current position]
solution says:
Semeai Type 3: Eye vs no eye.
No Seki possible.
All inside liberties count for the group with the eye.
Black has 6 liberties.
White has 5 liberties.
White is dead. (1 move behind).
I think this is incorrect. It should be unsettled. But let me explain. In a way this has to do with the "L3 argument" of semelis which has something. In case White starts with L3, the black top group is killed in one move! In particular, after White L3, the dotted square-shaped group of White centered at L13 lives immediately. Ok, Black could now have the idea to threaten the White group in the top
right corner, e.g. with Black R16. But as far as I can see, White's group in the top right corner
should win the capturing race against Black's top group. So Black should be dead.
In a way, we here a have a similar situation as in Problems 199-200 and 200-200:
"Semeai Type 3: Eye vs no eye, but with other groups involved. Standard counting principles do not apply."
semelis: In the hovering image it seems to be a black stone at L1, while in the problem there's another one at L2. If white plays L3 black can't do anything.
If black captures at L3, can follow up with L6, L9, L11, and L13, liberating the upper clump that has 5 liberties in 5 moves.
d4rkm4tter: You know that the positions are randomized. Of course we don't randomize the thumbnails.
I changed the position.