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Basically, it will not be a problem for Black to connect the ko, if only he can afford to do so. It will result in a seki between the two marked groups. However, in Igo Hatsuyōron 120 White will have more territory than Black, so she will win the game.

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If White had captured the ko after her last threat (this is the exchange of , ), and ...

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... Black has no ko-threat of his own any more, White can then calmly connect the ko with . Again, there is the seki between the two marked groups.

N. B.: We will display ko-threats (and their response) in the lower left part of the board; moves which capture the ko, in the lower right. Moves that are played in the top left corner, like here, mean that there are no valid ko-threats left.

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So, Black is dependent on winning the ko, which turns out to be very unbalanced. Black's only effective ko-threats are those, which threaten to get life for his group - e.g. by capturing some surrounding White stones (), as here, with .

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White, however, does not need to fly into heavy artillery. It is sufficient - e.g. with here - to occupy one of the outside liberties of Black's group. Since it makes little sense for Black to connect the ko, it is best for him (with ) to take one liberty of White's group in the semeai, in return.

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However, in the problem, it is Black's turn after White has created the ko-shape in the top right. Black wins with the sequence shown, if he has one ko-threat () more than White does.

From here on, we will distinguish several cases by the difference "EX" of White's liberties ( + ), and Black's liberties ( + ), which is the amount of White's exceeding liberties: EX = ( + ) - ( + ).

This position has EX = 1; Black needed one more ko-threat than White to make up for White's exceeding liberty. We will use the letter "t" to reference this amount of exceeding Black ko-threats; here, t = 1 = EX.

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But, in the position shown, it would be a mistake by White to capture the ko "early", because Black saves a ko-threat.

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Black will not play a ko-threat now, to recapture the ko, but will continue taking White's liberties.

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White has no time to connect the ko.

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Now, it is White who must find the first ko-threat.

Please note a very important feature that is characteristic for this case EX = 1. White's group has only two liberties left ( would be an atari), so - in the problem - White's ko-threat must be an atari, too, on any of Black's surrounding groups.

In general, White's ko-threats must reduce a surrounding Black group to one liberty less than her group has at the moment. Otherwise, Black would simply take further liberties of her group, and win.

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Black will win now, if he has as many ko-threats as White: t = 0.

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If White connects the ko, Black wins unconditionally.

Taking the liberty in the ko has the same effect as taking any other liberty of Black's group, but the disadvantage of changing the parity (whose turn it is to take the ko) of the ko. So, White should keep this move until the very end.

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EX = 0
As a matter of course, there will be no ko if both sides have the same number of liberties, because it is Black's turn; this is the case ( + ) = ( + ), or EX = 0.

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Black wins the semeai unconditionally.

N. B.: We will not display this case "EX = 0" with the positions following below, because nothing extraordinary is supposed to happen herewith.

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EX = 2
If we give White's group an additional external liberty, it results in six liberties for White vs. four liberties for Black, so EX = 2.

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If Black has much more ko-threats than White, it is advisable for White to start the ko as soon as possible, to exhaust Black's ko-threats. Black wins with the sequence shown, if he has five ko-threats more than White does: t = 5 = ( + ) - 1 = ( + ) + 1.

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With EX = 2, Black's option of further occupying White's liberties (instead of continuing the ko) has gone. White simply connects the ko, and wins the semeai, i.e. reaches a seki. Black is unable to play at .

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If Black lacks ko-threats, it may look more elegant for White to wait until the latest possible moment to start the ko-fight. Otherwise, it will be a mistake by White to capture the ko "late" with EX = 2. This "late" capture (if suitable) is equivalent to the procedure with EX = 1, explained above, so we have here also t = EX.

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EX = 3
Further liberties for White do no longer affect the matter.

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Again, White captures the ko "early".

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And indeed, Black needed six ko-threats to win:

t = 6 = ( + ) - 1 = ( + ) + 1.

N.B.: In the following, we will restrict ourselves to displaying the case "EX = 2" (as an example for "EX >= 2"), because all further cases "EX = 3, 4, ..." follow the same formula.

Conclusion:

EX = 0. Black will win unconditionally. In the problem, this case can only happen after two White mistakes.

EX = 1. Black needs one more ko-threat than White. In the problem, Black has at least two ko-threats on the right side of the board, and all of White's ko-threats must be atari. In principle, Black will win this case, which can only happen after a White mistake, losing a move.

EX >= 2. Black needs one ko-threat less than White has ko-fight liberties, counted at the moment, White starts the ko. Smaller EX are better for Black.