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Kanazawa Solution 55
White forces with 1 and 3 (the order of which is important) then with 5 threatens to make a second eye: Black can't falsen it at the marked point, because White will push through at A and take two stones. So he plays 6 upon which White connects at 7. Now at first sight, the shape is one of a dead group: a three point nakade, and a false eye at the top.
Surprisingly enough, White captures with 1 and 3, making miai of replaying at the marked (black) spots !! For reference ...
When Black throws in, White makes two eyes at 2, sacrificing three stones at the top. There are MANY opportunities for White to go wrong. Let's analyze some: (to be ctd - please contribute if you want)
At move 9, if White plays 1 instead of A, Black answers at 2 to keep White at one eye. If next White captures two stones at A, he throws in at the marked stone. White should capture simultaneously to make miai of Black's answers. The order of A and 1 (9 and 11) is important.
If White plays 7 at 1 here (instead of connecting at 2) Black captures at 2 and the miai is destroyed.
White has now one big eye AlmostFilled with three stones. There is no threat in A.
If White changes the order of 3 and 5, Black can answer at 6 and make it ko with 8.
In the main line, White hasn't yet exchanged b for 5, so when she ataris at 3, and White captures at 4, White 5 makes miai of A and B. If the order of 1 and 3 wouldn't matter, then either one of the answers would be different.
Suppose 3 is not answered at A, then 5 makes life.
But Black can answer 1 at 2, then 3 at 4 as in this diagram. Next, White comes one move too late: after 8, A and B are miai. Finally, what if White ataris from the other side ?
I thought, well, now Black can destroy the eye with 6, since the aji now works against White: Black 8 is atari. But ...
... the capture makes miai of A and B. So ? This is a copy of the living page "Kanazawa Solution 55" at Sensei's Library. (C) the Authors, published under the OpenContent License V1.0. |