Does something seem off in this problem?
Posted: Mon Aug 26, 2013 3:39 pm
by illluck
I came across another interesting post on Baidu Tieba (http://tieba.baidu.com/p/2554736555).
The below is a problem taken from a collection, with my translation of the book's comments:
The original poster asked whether there is something strange about the solution.
It turns out I was wrong, though:
The below is a problem taken from a collection, with my translation of the book's comments:
$$Bc Black to play: Black is separated into three groups, the only way to connect is to capture the three white stones in the middle. What is the correct way to utilize white's cutting points?
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- Click Here To Show Diagram Code
[go]$$Bc Black to play: Black is separated into three groups, the only way to connect is to capture the three white stones in the middle. What is the correct way to utilize white's cutting points?
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$$ | . . . . . . . . . . . .[/go]
$$Bc Solution: The cut at 1 is the correct way to make use of white's cutting points. After the exchange of 2 to 3, the atari at 4 is white's strongest resistance, but in the sequence up to 9, white is caught in a connect-and-die.
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$$ | . . 9 O 5 O 1 8 . . . .
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- Click Here To Show Diagram Code
[go]$$Bc Solution: The cut at 1 is the correct way to make use of white's cutting points. After the exchange of 2 to 3, the atari at 4 is white's strongest resistance, but in the sequence up to 9, white is caught in a connect-and-die.
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$$ | . X X O 6 7 4 O . . . .
$$ | . . 9 O 5 O 1 8 . . . .
$$ | . X X O 2 X O . . . . .
$$ | . . O X X 3 . . . . . .
$$ | . . O . . . . . . . . .
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$$ | . . . . . . . . . . . .[/go]
$$Bc Failure: The wedge at 1 is an easy move to make, when white plays atari at 2 black squeezes at 3, but white's capture at 4 is a fatal blow to black.
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- Click Here To Show Diagram Code
[go]$$Bc Failure: The wedge at 1 is an easy move to make, when white plays atari at 2 black squeezes at 3, but white's capture at 4 is a fatal blow to black.
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$$ | . X X O 2 3 . O . . . .
$$ | . . . O 1 O 6 . . . . .
$$ | . X X O 4 X O . . . . .
$$ | . . O X X 5 . . . . . .
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$$ | . . . . . . . . . . . .[/go]
$$Bc Failure variation: When black wedges at 1, white atari at 2 is very bad. Black connects at 3 and reverts back to the solution diagram.
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$$ | . X X O 4 7 6 O . . . .
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- Click Here To Show Diagram Code
[go]$$Bc Failure variation: When black wedges at 1, white atari at 2 is very bad. Black connects at 3 and reverts back to the solution diagram.
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$$ | . . . O . . . . . . . .
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$$ | . X X O 4 7 6 O . . . .
$$ | . . . O 1 O 5 . . . . .
$$ | . X X O 2 X O . . . . .
$$ | . . O X X 3 . . . . . .
$$ | . . O . . . . . . . . .
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$$ | . . . . . . . . . . . .[/go]
The original poster asked whether there is something strange about the solution.
I took a look and realized that the "Failure" diagram works for black as well as black could play as below, which reverts back to the solution:
So, satisfied that I understood the question, I responded saying that indeed something is off with the problem - it is off-the-mark as it has 2 solutions (generally frowned upon). It's a bit amusing how certain the comments were about the failure, but mistakes such as this are understandable in any large collection of problems.
$$Bc Not a failure: Black captures the three stones.
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- Click Here To Show Diagram Code
[go]$$Bc Not a failure: Black captures the three stones.
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$$ | . X X O 2 . . O . . . .
$$ | . . . O 1 O 3 . . . . .
$$ | . X X O . X O . . . . .
$$ | . . O X X . . . . . . .
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$$ | . . . . . . . . . . . .[/go]
So, satisfied that I understood the question, I responded saying that indeed something is off with the problem - it is off-the-mark as it has 2 solutions (generally frowned upon). It's a bit amusing how certain the comments were about the failure, but mistakes such as this are understandable in any large collection of problems.
It turns out I was wrong, though:
Another poster pointed out that in fact the "solution" isn't correct either! Black cannot capture white with the cut.
It turns out that the question was right all along - it was just that the "answer" turned out to be spectacularly more wrong than what I thought originally.
$$Bc Not a solution: "a" and "b" are miai.
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$$ | . . . O 2 O 1 . . . . .
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- Click Here To Show Diagram Code
[go]$$Bc Not a solution: "a" and "b" are miai.
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$$ | . . . O 2 O 1 . . . . .
$$ | . X X O . X O . . . . .
$$ | . . O X X b . . . . . .
$$ | . . O . . . . . . . . .
$$ | . . . O . . . . . . . .
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$$ | . . . . . . . . . . . .[/go]
It turns out that the question was right all along - it was just that the "answer" turned out to be spectacularly more wrong than what I thought originally.