TsumegoFromGames127/Solution

Sub-page of TsumegoFromGames127

This problem really has two solutions. White can live, but white can also play ko for an additional gain.

[Diagram]
White lives  

White can simply live with W1. Because now the moves at a (making a second eye) and b (capturing some stones in the center) are miai.

So in this case, black will play B2 at b (or thereabout) to connect in the center, and white will live with W3 at a.

White has another choice however.

[Diagram]
White plays in the center  

White can play W1 directly like this. At first sight, it looks like this simply captures the marked black stones outright. But black still has a trick up his sleeve...

[Diagram]
Black makes ko  

Black plays B2 and B4 to make ko. Now if white a, black b results in ko for the life of the group on the right. If white b, black a results in a ko for the status of all the circled stones, because B4 was taken a crucial liberty from the white+circle stones. So white cannot avoid ko.

Still, it is not an easy ko for black, but better than just allowing white to capture the center stones. Note that black does not need to play B2 and B4 immediately, but could first play elsewhere (for example to create ko threats).

White should play this W1 if just living with the white group on the right is not sufficient to win the game. If it is, the first diagram is better.

[Diagram]
Internal threat (B8 at black+circle, W11 at W5)  

With W5, white starts the ko. B6 is an internal ko threat, after which black retakes the ko with B8. W9 wins the semeai in the center, forcing black to play B10, which makes the ko about the life of the white group on the right. W11 retakes the ko, and black must find the first threat.

(Follow-up problem: Is B12 at a a valid internal ko threat?)

[Diagram]
B2 Not Internal Threat (W7 at W1)  

Dave: It seems to me that after White takes the ko, B2 is not an internal threat. W3 puts the six Black center stones in atari. As a result, White can play W5 after Black takes the ko with B4. If Black connects the ko at B1, White connects at W6 and wins the fight on the side. If Black and White exchange a for b, Black needs four plays to capture Black (c, d, e, and f) but has only three liberties along the side. Without the exchange of B2 for W3, the exchange of W5 for B6 would put the center stones back in play within the ko since it reduces the liberties of the White center. Also after the W5 B6 exchange, Black can play g as an additional internal threat. So Black 2 is a net loss in terms of ko threats. It seems that Black has no internal ko threats here while White has one remaining at d.

[Diagram]
Short on liberties  

Herman: I agree, but my reasoning is like this: If white connects the ko in response to B2, then after W7 black is two liberties short. The black central chain has four, while white has six (the four circle marked points, plus the approach moves marked square)


Old Discussion

herman: I will remove this later.

[Diagram]
Continuation: B8, W13 take ko  

This is how the ko continues. After W13 takes, it is a direct ko.

[Diagram]
Continuation: W11 takes ko  

Dave: Should White play W9 here instead? This wins the center fight outright. If B10, the ko continues but it is now for the right-side White stones. Is this better for White or not?

Herman: Yes, this is the other option, it is also mentioned above (where white can play W5 at a). Allowing the B6-W7 exchange before making the ko about the group on the right is slightly better, however, so white should definitely force black to make that exchange.

The ko for the right side group is smaller than the one for the center, so white should indeed play that one instead.


Path: <= Tsumego From Games =>
TsumegoFromGames127/Solution last edited by HermanHiddema on July 18, 2009 - 11:08
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