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[go]$$B
$$ -----------------------
$$ . . . . . C C C . . . . |
$$ . . . O . C C Y Y O O . |
$$ . . O . Y Y Y O O X X . |
$$ O . O Y . . . O X O . . |
$$ . . X O O . O O X . O . |
$$ . O . X O X O X . X . . |
$$ . . X X O O X X . . . . |
$$ . . . . X X . . . . . . |
$$ . . . . . . . . . . . . |[/go]
Black's marked stones at the top cannot live on their own.
Eye-space is too small and maximum eye-space would consist of a dead Nakade-shape.
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[go]$$B
$$ -----------------------
$$ . . . . . . . . . . . . |
$$ . . . Q . . . X X Q Q . |
$$ . . Q . X X X O O X X . |
$$ Q . Q X . . . O # O . . |
$$ . . X O O . O O # . O . |
$$ . O . # O X O # . # . . |
$$ . . # # O O # # . . . . |
$$ . . . . # # . . . . . . |
$$ . . . . . . . . . . . . |[/go]
Black cannot kill any of White's group at the top, so he will have to connect to his stones below.
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[go]$$B
$$ -----------------------
$$ . . . . . . . . . . . . |
$$ . . . O . . . X X O O . |
$$ . . O . X X X O O X X . |
$$ O . O X . . . O X O . . |
$$ . . X P P . O O X . O . |
$$ . O . X P X O X . X . . |
$$ . . X X P P X X . . . . |
$$ . . . . X X . . . . . . |
$$ . . . . . . . . . . . . |[/go]
To connect, he has to take either White's left-hand group ...
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[go]$$B
$$ -----------------------
$$ . . . . . . . . . . . . |
$$ . . . O . . . X X O O . |
$$ . . O . X X X P P X X . |
$$ O . O X . . . P X O . . |
$$ . . X O O . P P X . O . |
$$ . O . X O X P X . X . . |
$$ . . X X O O X X . . . . |
$$ . . . . X X . . . . . . |
$$ . . . . . . . . . . . . |[/go]
... or White's right-hand group off the board.
What may be overlooked:
It is not necessary to take both of White's groups off the board !
These two groups seem to be connected, but actually they are NOT.
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[go]$$B
$$ -----------------------
$$ . . . . . . . . . . . . |
$$ . . . O . . . X X O O . |
$$ . . O . X X X O O Y Y . |
$$ O . O X . . . O X O . . |
$$ . . Y O O . O O X . O . |
$$ . O . X O X O X . X . . |
$$ . . X X O O X X . . . . |
$$ . . . . X X . . . . . . |
$$ . . . . . . . . . . . . |[/go]
What makes it difficult for Black: Both of his groups on the right and on the left have only two liberties.
But:
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[go]$$W
$$ -----------------------
$$ . . . . . . . . . . . . |
$$ . . . O . . . X X O O . |
$$ . . O . X X X O O X X . |
$$ O . O X . . . O X O . . |
$$ . 3 X O O . O O X . O . |
$$ . O 1 X O X O X . X . . |
$$ . . X X O O X X . . . . |
$$ . . . . X X . . . . . . |
$$ . . . . . . . . . . . . |[/go]
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[go]$$W
$$ -----------------------
$$ . . . . . . . . . . . . |
$$ . . . O . . . X X O O . |
$$ . . O . X X X O O X X . |
$$ O . O X . . . O X O . . |
$$ . O 5 O O . O O X . O . |
$$ . O O X O X O X . X . . |
$$ . . X X O O X X . . . . |
$$ . . . . X X . . . . . . |
$$ . . . . . . . . . . . . |[/go]
What may be overlooked:
In will take White three moves to connect one of her encircled groups securely to the open.
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[go]$$B
$$ -----------------------
$$ . . . . . . . . . . . . |
$$ . . . O . . . X X O O . |
$$ . . O . X X X O O Y Y . |
$$ O . O X . . . O X O . . |
$$ . . Y O O . O O X . O . |
$$ . O . X O X O X . X . . |
$$ . . X X O O X X . . . . |
$$ . . . . X X . . . . . . |
$$ . . . . . . . . . . . . |[/go]
What may be overlooked:
Black's group on the right and the left are indeed separated. But they have to be though about as somewhat connected, because they have to do with White's encircled stones.
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[go]$$W
$$ -----------------------
$$ . . . . . . . . . . . . |
$$ . . . O . . . X X O O . |
$$ . . O . X X X O O X X . |
$$ O . O X . . . O X O . . |
$$ . . X O O 1 O O X . O . |
$$ . O . X O X O X . X . . |
$$ . . X X O O X X . . . . |
$$ . . . . X X . . . . . . |
$$ . . . . . . . . . . . . |[/go]
If it would be White's turn:
Simplest measure for White is to capture the single Black stone. Now she has only one group with four liberties.
One White group means that from here on only one Black group is of really interest. Black has two liberties only and no chance left.
What may be overlooked:
If you have no idea where to start as Black, change to White's point of view and ask yourself where the best point for White could be.
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[go]$$B
$$ -----------------------
$$ . . . . . . . . . . . . |
$$ . . . O . . . X X O O . |
$$ . . O . X X X O O X X . |
$$ O . O X . 2 . O X O . . |
$$ . . X O O 1 O O X . O . |
$$ . O . X O X O X . X . . |
$$ . . X X O O X X . . . . |
$$ . . . . X X . . . . . . |
$$ . . . . . . . . . . . . |[/go]
Black must prevent White to take the point 1, which is one central point of a symmetrical position, too.
It follows that Black has to start with 1 himself. This move is Double-Atari, so White has to capture with 2.
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[go]$$B
$$ -----------------------
$$ . . . . . . . . . . . . |
$$ . . . O . . . X X O O . |
$$ . . O . X X X O O X X . |
$$ O . O X . O . O X O . . |
$$ . . X O O . O O X . O . |
$$ . O . X O . O X . X . . |
$$ . . X X O O X X . . . . |
$$ . . . . X X . . . . . . |
$$ . . . . . . . . . . . . |[/go]
I have written before that Ishi-no-shita is difficult to see. So here is the position after White 2.
What may be overlooked:
Now you can see that all of White's encircled groups have three liberties. Three is the number of moves White needs to connect one of her two larger groups to the outside !
This may become a problem, because White has to save both of her two groups to become successful.
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[go]$$B
$$ -----------------------
$$ . . . . . . . . . . . . |
$$ . . . O . . . X X O O . |
$$ . . O . X X X O O X X . |
$$ O . O X 3 O 5 O X O . . |
$$ . . X O O a O O X . O . |
$$ . O 4 X O . O X . X . . |
$$ . . X X O O X X . . . . |
$$ . . . . X X . . . . . . |
$$ . . . . . . . . . . . . |[/go]
Black continues with 3. White cannot play at a, because this would mean taking an own liberty. After Black gives Atari with 5, the position is symmetrical again.
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[go]$$B
$$ -----------------------
$$ . . . . . . . . . . . . |
$$ . . . O . . . X X O O . |
$$ . . O . X X X O O X X . |
$$ O . O X X O X O X O . . |
$$ . 6 X O O 7 O O X . O . |
$$ . O O X O . O X . X . . |
$$ . . X X O O X X . . . . |
$$ . . . . X X . . . . . . |
$$ . . . . . . . . . . . . |[/go]
If White captures on the left, Black captures in the centre.
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[go]$$B
$$ -----------------------
$$ . . . . . . . . . . . . |
$$ . . . O . . . X X O O . |
$$ . . O . X X X O O X X . |
$$ O . O X X . X O X O . . |
$$ . O b O O X O O X . O . |
$$ . O O X O a O X . X . . |
$$ . . X X O O X X . . . . |
$$ . . . . X X . . . . . . |
$$ . . . . . . . . . . . . |[/go]
Due to the hole at b, White cannot connect her stones in Atari with a.
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[go]$$B
$$ -----------------------
$$ . . . . . . . . . . . . |
$$ . . . O . . . X X O O . |
$$ . . O . X X X O O X X 6 |
$$ O . O X X O X O X O . . |
$$ . . X O O 7 O O X . O . |
$$ . O O X O . O X . X . . |
$$ . . X X O O X X . . . . |
$$ . . . . X X . . . . . . |
$$ . . . . . . . . . . . . |[/go]
If White plays symmetrical on the right, Black's capture with 7 results in Double-Atari.