Calvin Clark wrote:
The first thing I notice is that black has only three liberties. Next I see that white has a lot of cutting points, and black has a double atari at 'a'.
- Click Here To Show Diagram Code
[go]$$B Black to play
$$ ------------
$$ | . X . . . .
$$ | . O X X . .
$$ | . a O X . .
$$ | . . O X . .
$$ | . O X O O .
$$ | . O X S . .
$$ | O X X S O .
$$ | . O X S . .
$$ | . O O O . .
$$ | . . . , . .
$$ | . . . . . .[/go]
It also helps to count the dame of the White stones, even though, as you know, dame are not everything.
Quote:
Exploratory reading reveals the double atari isn't promising immediately.
- Click Here To Show Diagram Code
[go]$$B Black to play
$$ ------------
$$ | . X . . . .
$$ | 3 O X X . .
$$ | . 1 W X . .
$$ | . 2 W X . .
$$ | . O X O O .
$$ | . O X 4 . .
$$ | O X X . O .
$$ | . O X . . .
$$ | . O O O . .
$$ | . . . , . .
$$ | . . . . . .[/go]
The
stones had two dame, but after
they have three. Not a promising start, as you say.
However,
is a mistake.
- Click Here To Show Diagram Code
[go]$$Bc Throw-in and ko
$$ ------------
$$ | . X . . . .
$$ | 3 O X X . .
$$ | . 1 O X . .
$$ | 7 2 O X . .
$$ | 6 O X O O .
$$ | 5 O X 4 . .
$$ | O X X . O .
$$ | . O X . . .
$$ | . O O O . .
$$ | . . . , . .
$$ | . . . . . .[/go]
in the next diagram is correct.
- Click Here To Show Diagram Code
[go]$$Bc No ko
$$ ------------
$$ | . X . . . .
$$ | 3 O X X . .
$$ | . 1 O X . .
$$ | . 2 O X . .
$$ | . O X O O .
$$ | 4 O X . . .
$$ | O X X . O .
$$ | . O X . . .
$$ | . O O O . .
$$ | . . . , . .
$$ | . . . . . .[/go]
Quote:
2nd exploratory reading shows white can't play at either 'a' or 'b', so at
first this looks promising.
- Click Here To Show Diagram Code
[go]$$Bc Black to play
$$ ------------
$$ | . X . . . .
$$ | . O X X . .
$$ | a 2 O X . .
$$ | 3 1 O X . .
$$ | b O X O O .
$$ | . O X . . .
$$ | O X X . O .
$$ | . O X . . .
$$ | . O O O . .
$$ | . . . , . .
$$ | . . . . . .[/go]
But looking at liberties, it seems black has to fill three liberties to put
white in atari and white has to fill three to put black in atari. Since it's
white's move, this doesn't work for black. (Reading a sequence of alternating
triangles and squares while holding this starting position in my head takes
effort, but in this case I don't have to do it that way. I can just count.)
Discouraged, I take a nap.
- Click Here To Show Diagram Code
[go]$$Wc
$$ ------------
$$ | T X T . . .
$$ | T W X X . .
$$ | . W W X . .
$$ | B B W X . .
$$ | S O X O O .
$$ | S O X . . .
$$ | O X X . O .
$$ | S O X . . .
$$ | . O O O . .
$$ | . . . , . .
$$ | . . . . . .[/go]
The
stones have only two dame. Which, it turns out, are its only liberties. There are two ways that a unit of connected stones with only two dame, without help from nearby stones, can have more than two liberties. One is if it includes a stone on a 1-2 point, and the other is if one of its dame is in an eye. The
stones have only two dame and have no eye, but have four liberties because of help from nearby stones.
If the
stones descend to the 1-2 point, they will have two dame but three liberties. (Black can connect at C-19 and reduce the liberties back to two.)
Quote:
On my first reading, I don't see that if white plays
,
is atari.
- Click Here To Show Diagram Code
[go]$$W
$$ ------------
$$ | . X . . . .
$$ | a W X X . .
$$ | 2 W W X . .
$$ | B B W X . .
$$ | . O X O O .
$$ | 1 O X . . .
$$ | O X X . O .
$$ | . O X . . .
$$ | . O O O . .
$$ | . . . , . .
$$ | . . . . . .[/go]
Since the
stones have only two liberties and the
stones have three liberties,
is unnecessary now. Also, "a" is a slightly better way to atari the
stones, in case Black ignores a ko threat.
Quote:
I also didn't notice this:
- Click Here To Show Diagram Code
[go]$$Wc
$$ ------------
$$ | . X . . . .
$$ | . O X X . .
$$ | . O O X . .
$$ | X X O X . .
$$ | 2 O X O O .
$$ | . O X 1 . .
$$ | O X X . O .
$$ | . O X . . .
$$ | . O O O . .
$$ | . . . , . .
$$ | . . . . . .[/go]
in the next diagram is correct.
- Click Here To Show Diagram Code
[go]$$Wc
$$ ------------
$$ | . X . . . .
$$ | . O X X . .
$$ | . O O X . .
$$ | X X O X . .
$$ | . O X O O .
$$ | 2 O X 1 . .
$$ | O X X . O .
$$ | a O X . . .
$$ | . O O O . .
$$ | . . . , . .
$$ | . . . . . .[/go]
Note that connecting at "a" is not sente.
Quote:
Is it looking like this cut at
is promising.
is forcing and white has no other response but
.
At last, verify that white has no way to stop both 'a' and 'b'.
- Click Here To Show Diagram Code
[go]$$W
$$ ------------
$$ | . X . . . .
$$ | . O X X . .
$$ | a O O X . .
$$ | X X O X . .
$$ | b O X O O .
$$ | . O X . . .
$$ | O X X . O .
$$ | . O X . . .
$$ | . O O O . .
$$ | . . . , . .
$$ | . . . . . .[/go]
This kind of position is called
Golden Cock Stands on One Leg, a poetic name that suggests how it looks.
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Calvin Clark
is the natural move even without the hint. Softer moves like 'b' or 'c' would let white out more easily.
, as long as the ataris at 'a' and 'b' aren't helping white, this seems fine for black.
, because 'a' and 'b' are miai.
stone with 1 move, (a).
or (more positively) that you are not alone.
. (In fact, it does not work, which is left as an exercise for readers.)
, thinking the ladder doesn't work for black.
, white 'a' is atari. So the ladder problem is solved, but would I have seen the correct continuation?
?
connects )
white is alive:
, white 'a' and 'b' are self-atari: