This problem seems to be mistaken, black can't kill white as far as we can see.
This is not in the solution book. Malcolm : As far as I can see White lives. unkx80 also seems to think so.
maruseru : Hm, I still only get a ko, or White dies outright.
After your sequence above, Black takes at 5, White connects at 6, then Black plays hane at 7. White can't play at 'a' because of a shortage of liberties. Continuing, if White takes at 4 and Black retakes at 1...
... White plays 1, but Black connects at 2, leaving White with a dead-three-shape in the corner. Alternatively...
... if White plays 6 to enlarge his eyespace, it's still ko. Later, when Black fills an outside liberty with 7, White will eventually have to connect at 8. Black can always connect at 4, so when White loses all his outside liberties, he will have to take the four Black stones in the corner and it's still ko.
Am I missing something?
tderz: position after last diagram.
Perhaps the should be missing, making capture a a threat?
tderz: (At least) two scenarios are possible:
unkx80: lives. It does not matter if
is exchanged with
. The thing is, (1) if later Black plays at
, then White simply captures at a and we get a bent four in the corner with two external liberties, which lives, and (2) any other move by Black cannot reduce White's two external liberties, and so cannot prevent White from living by oshitsubushi at a.
maruseru: Yes, after I edited the page and before I just got back, I thought about the problem some more and your occurred to me as well. Thanks for the clarification.
After 9, a-k (j connects left of 1). This is an unfavourable ko in which if White loses the ko she gets to play three times elsewhere.
I was lucky to be able to show this to FanHui on KGS. He didn't say there were any mistakes in these variations. So I guess the conclusion is that this is another errata.