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Beginner Exercise 11 Discussion
Difficulty: Beginner
Keywords: Problem
unkx80: This is the solution. Black 7 can be played at a instead. Likewise, Black 9 can be played at b instead.
There is death in the hane. Black reduces White's eyespace twice, and then makes the nakade (placement) inside, reducing White to one eye. It doesn't matter that Black 1 and 3 are in atari; White's resistance with a play at either point a is met by a black play at the adjacent b; and White must solidify before she can cut, giving Black time to protect at the appropriate c.
unkx80: Sorry but I did not notice the mistake in your description earlier. Your description suggested that White 1 should be answered at 2. But it gives White 3 a possibility to open up a ko. If Black tries to avoid the ko with 4 onwards, eventually he is caught in a net by White 7 and still has to fight the ko. Even running away at the first hane won't work:
Black has to be careful not to let White escape or make a second eye near the corner here. Black 3 is tesuji, I think. At this point, c is answered by a, and otherwise the pairs a and b, d and e, and f and g are all miai; Black kills. The instant that White seals off a 1x3 space (which Black will force with this line of play), Black must play at the middle of it. Even if Black leaves two stones in atari in the left side as a result, White cannot live:
Jasonred- There's no ko in this case, so wouldn't this be the time to play the circled point? With a and b, is this miai? Anyhow, if White a, Black b, if White b, Black a.
unkx80: White 4 will make 5 and 6 miai. White 6 lives with a straight four. But if Black allows White to cut at 5, the white group escapes.
unkx80: Black 7 should simply be played at a, leaving the option of the throw-in at 8 open.
unkx80: Black 5 should play at a or b instead, because playing at 5 in the diagram leaves the possibility of Black opening up a ko at a.
One more way to resist, you say, since White still has liberties and a cutting point? Now White is too close to the corner. Things get even tricker here:
White 10 at a is insufficient; Black 11 at b will destroy the second eye and White has no more hope of escaping. (White 8 at a will fail for the same reason; White 8 at 9 or elsewhere lets Black capture at 6 and it's 'game over' for White again). So White 10 will be at b, c or d. We'll look at White b; you should be able to convince yourself that White c and d either transpose to positions in this line, or let Black play at b safely.
Black 3 at the circled stone (yet another throw-in); White 4 at 1. Black 5 finishes the kill; White is easily prevented from escaping. (White cannot cut and run on the other side because of the atari at the circled point.) At either 2 or 4, White might try to run further with another hane at 5:
Now Black 5 at the circled stone, White 6 at 1 and Black can either take in snapback or protect with a or b. White is kept in atari the whole time and never has a chance of a second eye. I think unkx80 has a point about protecting the cut first. Looks like I'm a victim of concurrent editing. ^^; unkx80: Of course protecting the cut first is better. What if I add a white stone in the problem at the 2-2 point?
So I'm feeling a bit dumb here. My natural affinity for symmetry had me consider this position first. I don't see the flaw. OK, after a lot of thought, I see the flaw. Immediate reply is here:
At this point White will get two eyes directly, or a four-point long eye which lives. Black d gives White a, trading b for c or vice versa. Black a, b, or c gives White d, which collapses to a four-point space - alive. -JoshuaRodman? Dieter: If Black b, White is more likely to answer at a, because this gives life with eyes and points. White d Black a gives a seki: no points. unkx80: If Black b then White can also play tenuki. This is a copy of the living page "Beginner Exercise 11 Discussion" at Sensei's Library. ![]() |