# BQM 401

Keywords: Question

Here's the final position from a game between Chen Zude (White) and Yang Yilun (Black). It's Black to move, and Black resigned. I've been trying to count the liberty battle on the upper side, starting with a Black play at a. I don't see how White can win the race...

Thanks! golearner

xela: It looks to me like ko: (edit: Bill has a simpler way, where white wins without ko; scroll down!)

( at a, at b doesn't seem to change the liberty count. Also, at c seems to give the same result, if I've read correctly.)

continuation: at ; takes ko

White will ignore any black ko threat and play at a. I don't think black has any threats big enough.

Black loses

Bill: I guess we agree that if , wins the capturing race. Black has only 4 liberties, while White has 5 (not counting the shared liberty).

Eye vs. no eye

Bill: does not help, because takes away Black's eye, for an easy win, I think. (White does not have to capture Black to win, and the shared liberty belongs to White.)

golearner: I think I've got it. It seems to me that Black wins the race. But I still see it as a race - what do you mean, "White does not have to capture Black"? I mean, White's group doesn't have two eyes - or were you meaning if it ended up being seki?

Bill: In an eye vs. no eye capturing race if the side with no eye cannot capture the side with an eye, it loses. To win, the side with the eye does not have to capture the side with no eye unless the side with no eye forces it to by filling all shared liberties.

xela: To make it absolutely clear, imagine that you end up with a position something like this:

Eye vs. no eye

Here, black can't play a because it would be self-atari. Therefore there is no way for black to capture white--so white is alive. However, white can put black into atari by playing b, c, d, etc. There is no way for black to prevent this, so black is dead. Does this make sense?

BQM 401 last edited by Dieter on July 5, 2008 - 12:49