demetria: I'm not sure there's a way to solve this without sacrifice (if there is, I hope someone puts it up soon). I think black can make life by forcing white's hand with and .
Then, I think black can live. If white A then black B, and black can capture A to make a solid territory. I think black might even come out a point ahead in this exchange.
David: Your solution is mostly correct.
Black is alive after in your first diagram. ( is not a sacrifice, because it captures three white stones.)
a and b are now miai for life (that is, if White plays a, Black must play b, and if White plays b, Black must play a).
demetria: Oops! I don't know why I got it backwards - dyslexic I guess.
unkx80: There is something wrong with this failure diagram too.
demetria: Ah, now I remember the sacrifice bit. Black has to have that vital point so he prevents white from taking it at and now black has a snapback if white captures. Better?
David: No, at makes a bulky five, killing Black. If Black captures the five stones, White will play again at (the vital point of the bulky five eye shape), making it impossible for Black to make two eyes.
threatens to make an eye, requiring (see below).
Neither player wants to play at a or b. If Black plays one of them, White makes a bulky five. If White plays a, Black captures and makes two eyes. If White plays b, Black must capture. White may not recapture if the rules forbid repeating positions, but even if the rules allow it, it's a loss of one point in gote.
I think this position is an example of case S1 at Almost Almost Fill, meaning that the inside shape is a killing shape, and neither of its extensions makes a killing shape. According to that page, the groups in such a position are alive in seki regardless of whose turn it is and the number of external liberties.
Here's another failure diagram, showing what happens if Black starts from the other side. This makes an S0, L1 position, in Almost Almost Fill terms. If White doesn't play , Black will play there to create a ko for life.
demetria: I think you may have misread the progression from my solution (well, I guess I put the order wrong, I should have shown the capture better)
After white captures, A and B are miai. If white tries to connect at A then black captures three to make an eye. If white tries to connect out then black captures .
demetria: Not really. Black gets there first. If I'm reading the suicide rule correctly, this is okay because black is removing his opponent's liberty, not his own. (Isn't that why one eye, especially a false one, is not enough to live?)
After the plays of the previous diagram, white plays as indicated (not at the circled point which is what was previously assumed). Even if black now captures the five stones, white responds by playing again at , and black has only one eye. See bulky five and killable eye shapes. The capture of the five white stones actually does not hurt white's score, because after the dead black group is removed at the end of the game, the five points on the board where white had the stones are now white's territory.
The solution resulting in seki (as shown on the solution page) is correct; it accepts that black cannot prevent white from making a bulky five, and handles the situation by making sure there are two liberties inside the black eyespace at that point.
In this case a play by white at the circled point is irrelevant. Suppose white fills the outside liberties. Then white still needs a play at a or b to try to capture the black stones. But then black just captures, and either capture of six stones (which are not a rabbitty six) results in black living with eyes. So white will not try to capture. If black tries to capture, he has to play a or b first before he can play the circled point; and then white can connect and reduce the situation to the previous one - so this is suicidal for black. Thus the result is seki - mutual life.