unkx80: This is the solution. can be played at a instead. Likewise, can be played at b instead.
Here is another solution:
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 and 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 should be answered at . But 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.
kevinwm: There's no need to make this complicated. If w tries to escape with , b can just continue to the corner with , , etc.
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. is tesuji, I think.
kevinwm: is tesuji IF white has stones close by. There are no such stones in this case. Let's keep this simple and focused on the beginner problem - killing 7 stones on the 2nd line.
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:
kevinwm: Again, 3 is not necessary. Play 3 at 5.
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:
at a is insufficient; at b will destroy the second eye and White has no more hope of escaping. ( at a will fail for the same reason; at or elsewhere lets Black capture at and it's 'game over' for White again). So 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.
at (yet another throw-in); at . 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 or , White might try to run further with another hane at :
Now at , at 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.
xela: Correct. White can tenuki and it's OK because black can't kill.
xela: In other words, playing at a to avoid the seki is worth 4 points in reverse sente, therefore tenuki is correct if white can make more than 8 points somewhere else, roughly speaking. (Actually, I think it's more like 6 points, not 4.) I don't want to get into a long discussion of endgame theory on this page; you can follow the endgame path to learn more about these things.
Didn't see this right away, until I started diagraming it as a possible solution. Ends with 2 eyes for white. Coyotebd
David: Actually, is unnecessary as White at would only create a false eye. would instead be played at , thus preventing a second eye from being formed and killing the group. However, if at instead, we end up with the following:
With a and b miai for life, as explained by JoshuaRodman? above.
AdamMarquis: Points a and b are not miai for life, as even if black gets both of them white can take the four stones and have an eye.
Florent? : As a beginning beginner i am also fond of symmetry and still do not see the flaw in this sequence :
playing 3 in the inside forces white to w4 (death) ...
... or to the option mentioned above by Unkx80, which I show in the diagramme below
any help ? or link to a concept i missed please ?
Bill: leads to seki, as the last diagram indicates.
Bill: Can White escape and make an eye in the corner? It looks pretty complicated to me. Anyway, why take the chance?