Karl Knechtel: The eye at a is already secured. White needs to make one in the space around b.
First, White chases Black around. If is played elsewhere, then White will play at to seal off the second eye. Similar reasoning is applied for .
threatens capture of more than enough eye space, so Black must capture.
threatens to capture again so Black fills in, but it's a connect-and-die - finishes the job since Black can't escape.
dent: I don't understand why Black is compelled to play at here and die. Why not as Black just connect at ? Then recapture the White capturing stone at and connect after?
Dieter : yes that would be better but then white lives by capture three to make An eye. This variation shows what happens if B tries to prevent that.
This seems like a much better outcome for Black than the final diagram in the solution. Connecting at a might not even be the best move on the board for Black. What am I missing?
Uberdude dent: you are missing that if black is able to read that he can't kill, it would be better to not lose points trying but simply play as shown left, as this is better for black than the result in "Dent's Question 2" (more black points in corner, fewer points for white). However, as a more advanced point, black shouldn't play this too early as a move at a is sente to kill with the attachment posed in the problem, and there is a chance that move could be useful on the outside later.
splice: Small technicality. Karl's comment: " threatens capture of more than enough eye space ...". As far as I can read, anything less will leave White with one eye. To wit:
So playing until the black stone is hit is not more than enough eye space, it's actually just enough. At least, the way I see it :)
unkx80: In the diagram entitled "Not enough", is wrong, but is also wrong. lives.
aigo: Following the same line of logic, is another mistake, it should go where is to make an eye instead of capturing the black stone.
I'm just a beginner, but I've played through a bunch of variations here. If black attempts to atari white can play double atari at a. If black attempts to connect in any way at a, b, or c, white can capture at least the three black stones on the first line to make another eye. Am I missing anything?
-- Brent Amersbach
Shaydwyrm: This basically works the same way as the throw-in Karl showed above, but allows black an extra move on the outside. This may not matter, if black is strong on the outside, but it may matter an awful lot if white can mount an attack later on black's outside group. In other words, exchanging for gives black better shape on the outside than just exchanging for , while gaining no real benefit for white. Just as a question of life and death, however, does seem to work fine.
I think this variation gives way too much to black to be considered. He follows with a and controls the corner, which was not the case with the original solution. Once white finally threatens the corner with b, black safely fills in . If white plays b earlier by playing at b, black plays at . Then if white plays next at or , black still responds with a and controls the corner, and actually makes white's group dead. If white plays at , then at b instead, black follows by playing at (or a), white plays at , and black secures the corner with a (or ) again. The b position really does need to be played at move .
Moosky: for the "I think this works..." diagram, here is why it doesn't. Black connects at and now just count the liberties for the black group (circles) and (squares). Black would capture before White can eat the Black group, and so the entire White group dies.
Uberdude: Actually Moosky, black dies with all the corner stones with your black 6 if white then extends to 7. So 5 can live, but it's better to directly throw-in at 6 as Shaydwyrm correctly explained above.