posetcay:
I suggest replacing L1 with a white stone to avoid confusion.

luki:
I agree with semelis. If white plays L3 black can't do anything.

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:
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.

semelis:
Oh ! Hover-on image seems to be different problem. The right one ?

semelis:
Whoever captures first at L3 wins or what am I missing ?