I'm working my way very slowly through a Chinese book of a thousand death/life problems. (Seeing as I don't read Chinese this is going, as you can guess, very slowly.) I have come across a problem I just can't for the life of me figure out. I'd appreciate any help I could get on explaining this.
This looked pretty straightforward to me. Black can play at either A16 or C16 and will live. Escape is just not an option so far as I can tell. I played through this a dozen or more different ways and it always comes out the same: Black lives. I couldn't even see why this was a problem. So I flip to the solutions page.
Well, colour me surprised: it shows Black playing at A16 for the win. But it also showed some other symbols in there that usually indicate some kind of commentary. So I started the process of painful translation. Now according to that commentary if Black plays at the spot marked a, White plays wherever it likes (I imagine filling in one of the escape routes Black has), Black plays at , White answers back at the point marked with a circle and Black dies. And that's where my problem is. I can't for the life of me see a scenario where Black will die without going out of the way to play stupidly.
Am I missing something obvious (or inobvious, for that matter!) here, or is the book just plain wrong?
If Black starts with , White adds one more stone at . After White throws in one stone at 2 giving:
Now the shape point at a and the throw in at b are miai for White to kill Black.
Michael Richter: Nasty. And yeah, I completely overlooked that option. I was mapping the shape into five internal liberties which I had decided long ago were always a guaranteed life. I didn't notice that the white stone at would turn things into a false eye.
So just how embarrassed should I be for not spotting that despite dozens of tries?
unkx80: There isn't the need to feel embarrased. Everyone makes mistakes anyway. =)
Andrew Grant: But just add one white stone and ...
Chris Schack: Took a while, I had to actually go through the original solution for Black, assuming White is stubborn ...
... and Black runs White's group in the corner out of liberties. Playing 2 at 3 would let White play at 2, and the second eye is false. Now, after the first three moves of the original with the marked stone added ...