RTG Problem 53 / Solution

Sub-page of RTGProblem53


If White plays first, White can save some, but not all, of his stones.
If Black plays first, all White stones will be dead.

White first

White saves some stones  

If White first, White saves some of his stones with W1 and W3.

Mistake: all White stones dead  

Although W1 and W3 can capture five Black stones, all White stones are dead after B4 captures the leftmost chain, because capturing the five-stone Black chain at a gives White only one eye.

If W3 at B4, then the result is essentially a reversion to the previous diagram.

Black first

Black kills all White stones  

If Black plays first, the self-atari of B1 is the only move to kill all White stones. W2 has no other choice than to capture the three Black stones.

Black kills all White stones  

Next, B3 makes a placement, threatening to capture the leftmost White chain via a connect-and-die. We discuss W4 at a to e.

W4 at a  

If W4 proceeds to capture the five Black stones, then B5 captures the leftmost White chain. After W7, White has only one eye.

W4 at b  

W4 saves the leftmost White chain, but B5 is an atari, allowing B7 to capture the three White stones in a ladder and saving the five-stones Black chain.

W4 at c  

Seeing the failures of W4 at a or b, now W4 tries to increase the number of liberties of the three-stone chain by pressing here. However, after B5, B7 and B9 reduces White to only one liberty.

W4 at d  

W4 does not work either; after B7 White cannot cut off the two Black stones due to shortage of liberties.

W4 at e  

Against the wedge of W4, B5 is an adequate answer.

RTG Problem 53 / Solution last edited by Dieter on April 16, 2009 - 13:14
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