(Referenced by 982; 1072)

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For a long period of time, it was an unanswered question, whether White might be able to benefit from creating more shared liberties, e.g. with the solid move of here.

However, after , and , ...

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... it is Black now, who takes the vital point of .

I am afraid that much more important than the matter of shared liberties will become the problems, which White will encounter by an additional ko-shape that Black will be able to establish on the upper edge - either at , or at .

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White plays sagari here.

We will examine White's hane at , instead, later.

The relationship of liberties has changed in Black's favour now.

White's group has five liberties (2 * w, 3 * c), but Black's group has eight liberties (3 * c, 3 * b, 2 * k). Therefore, we can expect that Black will win the ko-fight.

@

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Black connects the hanezeki's tail. In our opinion, this starts the most simple, and straightforward, sequence.

Ko:

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Black threatens to give atari at , so White needs atari-ko-threats.

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It does not mind that Black's ko-threat is suicidal. White's gain is too small here, compared with the forthcoming loss of her centre group.

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White needed two atari-ko-threats to win the semeai. However - in the real problem - she will be able to realise only one, so Black will win the ko-fight, and capture White's centre group.

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If White - in the preparatory sequence - plays hane with , Black throws-in with . This exchange creates another kind of ko-shape on the upper edge, this time a shared one.

Despite the changed kind of ko-shape on the upper edge, the relationship of liberties is the same as before.

White's group has five liberties (w, 3 * c, m), Black's group has eight liberties (3 * c, 2 * b, 2 * k, m). White will need two atari-ko-threats to win the ko-fight, but which she does not have in the problem.