Here I present a solution, which I attempt to make as simple as possible, as well as involving the smallest area on the go board as possible. There are other possible variations, please see /Discussion. --unkx80
There is death in the hane. Black reduces White's eye space twice, and then makes the nakade (placement) inside, reducing White to one eye. If White captures at and
, Black can answer at
and
and White has no way out.
Note: can be played at a instead. Likewise,
can be played at b instead.
If White descends at ,
reduces the eye space further and
makes a placement, reducing White to just one eye.
captures two stones...
makes a throw-in and turns it into a false eye. If White tries to escape with
, Black makes a net with
. It is not possible for White to get out.
is tricky.
is one way, but there are other possible answers. When
captures, Black can actually defend at
as White has not enough eye space.
In the Solution diagram, when captures a stone, Black should refrain from playing the atari at
. It gives
a possibility to open up a ko. If Black tries to avoid the ko with
onwards, eventually he is caught in a net by
and still has to fight the ko.
fractic: No this doesn't work. White can make an eye with
and then Black
can't prevent her from capturing
to make a second eye.
See also