Black sacrifices one more stone and plays atari at .
and capture four white stones (including ). Interestingly the eye space of the White group at the top is not enough to make two eyes, c.f., 212SS notcher. For example, if White a then Black b (bulky five), if White c then Black b (pyramid four).
Jasonred- ah, I stopped thinking once I saw Black's entire structure there die... I was also wondering about the whole title for this page! (why it's about eye spaces) but ... I was wondering about this following sequence? After , are a and b miai for life?
Dieter: The point is, White is dead. In order to prove/disprove that, we have to verify what happens if White goes first, not Black.
For instance: then makes miai of a and b: White needs both a and b to live, so Black can always play one of them to kill. Verify there is no eye in the corner.
Jasonred- verify this for me, but, ummm... modifying the above, I think I got a nakade for a 17-space eye... is this correct?
jvt: This eye space cannot possibly result from a capture. If such eye spaces are allowed, it is easy to add cutting points and stones in atari.
Dieter: The eyespace and ensuing nakade you propose, would have to come from this capture. In your diagram, however there is a stone at a but then Black gets a double capture and is out of trouble already.