( 960)

(Referenced by 971; 971; 972; 995)

We have seen so far that Black would win the ko-fight, if he had one external ko-threat (more than White).

Let us analyse now what happens in the case that Black had an additional external liberty. For this purpose, we will take White's -stone off the board.

!!! Please be aware that White must have made a mistake earlier in the sequence to fulfil this assumption. Therefore (as an exception from the rule), we will not adjust the number of Black stones on the board here. !!!

Now, the liberty count is five liberties for White (2 * w, 3 * c), and eight liberties for Black (3 * c, 4 * b, k).

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The most straightforward way for Black is to connect the hanezeki's tail, but to not re-capture the ko thereafter.

Ko:

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Black re-captures the ko with . Now, it is White, who must look for a valid ko-threat.

Black threatens to give atari at , so only those White moves are valid ko-threats that threaten to take a surrounding Black group of the board (i.e. are also atari). Here in the diagram, Black's -group would be a suitable target (after the preparatory exchange of - played earlier), but for one atari-ko-threat only.

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Black re-captures the ko for the second time. Please note the value of Black's larger eye. Black's potential ko-threats on the right side, and , are unused so far, so White really needed a further three atari-ko-threats. However, this aim seems to be impossible for White to achieve in the problem.

Black would win the game, if White ever lost a move unnecessarily.