White forces with and
(the order of which is important) then with
threatens to make a second eye: Black can't make it false at the marked point, because White will push through at a and take two stones. So he plays
, upon which White connects at
.
Now at first sight, the shape is one of a dead group: a three point big eye, and a false eye at the top.
Surprisingly enough, White captures with and
, making miai of replaying under the
stones!! For reference ...
When Black throws in, White makes two eyes at , sacrificing three stones at the top.
There are many opportunities for White to go wrong. Let's analyze some:
(To be continued - please contribute if you want.)
At , if White plays
instead of a, Black answers at
to keep White down to one eye. If next White captures two stones at a, he throws in at
. White should capture simultaneously to make miai of Black's answers. The order of a and
(
and White 11) is important.
If White plays at
here (instead of connecting at
) Black captures at
and the miai are destroyed.
White has now one big eye almost filled with three stones. There is no threat in a.
If White changes the order of and
, Black can answer at
and make it ko with
.
In the main line, White hasn't yet exchanged b for , so when she plays atari at
, and
captures,
makes miai of a and b.
If the order of and
didn't matter, then either one of the answers would be different.
Suppose w3 is not answered at a; then makes life.
But Black can answer at
, then
at
as in this diagram. Next, White comes one move too late: after
, a and b are miai.
Finally, what if White ataris from the other side ?
I thought, well, now Black can destroy the eye with , since the aji now works against White:
is atari. But ...
... the capture makes miai of a and b. So ?