Non-locality / Solution

Jigo  

The simple answer produces jigo, which isn't good enough.

4 fills ko  

White loses, obviously.

still doesn't work.  

Now if White wins the ko, Black takes and we get the same jigo result as before. If White connects, Black takes the ko and White is without a sufficient ko threat.

Deebster: If White connects the ko, then Black at a, White block at b, and Black filling at , doesn't that leave the score at White +1 (with one prisoner each)?

Evand: No, the score is jigo -- Black has two prisoners, one from the ko and one from capturing ; White has one prisoner from the ko.

at  

This looks like the winning move. Black is without a ko threat, and so White wins by 1 point. Cute problem.

-- Evand

xela: I notice that in the "jigo" diagram at the top of this page, black has the last move; whereas in the solution diagram, white has the last move (connecting the ko) and is one point better off. Is this a sort of tedomari?

Anon: Actually, if white plays the hane the score becomes white 26 black 25 because white's group has 13 points and is fighting ko for another 12 points and and black has 23 points so the difference is 2. If white plays the hane black captures and gains 1 point (23 + 1) white then threatens to make his group live for 2 point for white and minus 17 points for black. Black responds playing inside. (23 + 1 + 1 -1) white then retakes the ko (13 + 1) and black ataris the hane white 1. white fills in ko and kills the black group (13 + 1 + 12). black then captures the hane stone. (23 + 1 + 1 - 1 + 1). white then defends with atari to cap black. black fills in and ends game. 23 + 1 + 1 - 1 + 1 = 25 13 + 12 + 1 = 26 white wins