Scartol: Should I leave these so that actual beginners can have a crack at them?
Okay, it looks like the best black can do is get a ko. White has to fight it; white connecting at A only leads to black life at B.
Would it work better if black goes at B?
This was my initial response, but I realized (after I posted it -- stupid!) that white can connect after capturing the marked stone, and black is left with a dead shape once he captures the white stones.
This is the vital point I think. Once Black moves here, White can't play at 'a' because of a Black snapback, and can't play at 'b' because it's lacking a liberty. Black can form two eyes and live by playing at 1.
Black can respond with 1. Once Black moves to 1, if White tries to move to 'a', Black can capture White's stones by playing above at 'b'. If White chooses to respond at 'b' instead and capture the marked stone, Black can snapback and capture White.
If White plays at 'b' instead, Black can respond at 1, forcing White to capture at 2, then Black can play at 3. White is missing the liberty needed to capture Black. When Black plays at 3, White will have no choice but to use the move as sente. Black doesn't need to play there unless White tries to play at 'a' or 'b', in which case Black can simply capture the group of White stones.
unkx80: After , Black is dead.
Now W can close ko, making B take a. If B takes over ko, he lives. In other case, W closes ko, B plays a. If then W takes c, B extends to the right, and if W doesn't play c, white can take it, close it and live
is the vital point. Variations lead to life for Black.
Kayaq: This appears to be wrong; see my example three boards down.
unkx80: What happens after ?
Here, the twisted four lives.
Kayaq: above seems not to work!
After , black cannot connect at a because of shortage of liberties.
seldon: To elaborate this further, black ataris and still fails. After the atari, white cuts at 4, black captures at 5 (or is captured in due time) and is knocked out by 1 in the second diagram.
After , White can play for ko or seki but cannot kill.
unkx80: captures, Black dies.
rokirovka: Note that the marked White stone is critical, because it separates the two Black stones on the left from the rest of the group. If the marked stone were Black, then saves Black by making a seki with the bulky seven shape: White can only fill five stones inside it, after which it's a seki because if White played a sixth stone there, Black could kill the six stones and have a live shape.
But since the marked stone is White, the threat to the two Black stones on the left ruins this plan for Black. is met by , after which White captures the two Black stones on the left, and then the rest of Black's group will die.
ChrisSchack: B is dead after ... no response is needed to , and cutting off the two stones at the side doesn't matter. Once the outside liberties are gone, W can play at and a, and will have two liberties to B's one. Any attempt to block that just means a killing shape.
rokirovka: Ah, I see, thank you. Even if the marked White stone is Black, does not make a seki because White will take away the outside liberties, play at and a, and finally play at b which kills the Black group. I was mistakenly counting the point of the marked Black stone (captured by ) as a liberty for the Black group because it's inside the shape formed by the Black group, but it's not a liberty for Black because no Black stone is next to it.
Could somebody find an answer to kill surely blacks ?
If W plays at a, B at b, else at a.
zinger: White plays at c. If black a in response, white connects where the marked stone was captured. Black is dead.
tonytonychopper?: how black is supposed to survive? I can't find a solution and this page only offers solutions where black dies (that i already have found by myself), but the problem asks: "how black lives?" i tried to play the position against a software too but black seems to die in all the variations....i'm getting mad...and this is exciting
unkx80: I guess there may be a problem with the way this solution is presented, as it appears that people have been missing the solution. So I rewrote the entire solution page.
tonytonychopper?: just a question: if we take in account the possibility for black of winning the ko we have to agree to the fact that IF black can win the problem becomes global and no more local, so isn't that a violation to the tsumego?
fractic: In a tsumego problem you have to find the best way to make life. In this problem the best black can get is a ko. It doesn't matter if he can win the ko or not, at least there is a chance to save the group. Even if black loses the ko he could get compensation in the form of a ko threath that white ignored. Of course if black could live withouth a ko then the ko would be a failure since black would risk his group needlessly
[Èric]: what about this?
If W fills the ko after 3 then B can make atari at "a", then W defends at "b" and B can try an escaping route playing "c". If W does'nt fill the ko and defends at "d" B can take it and live with two eyes.
I just found out that "c" in the previous diagram is a weak point... If W ignores the atari and plays there, B can't go anywhere even if he captured the white stone due to the presence of the squared stone. However, I think that 3 could fix that, if then W plays "a" B would play "b" and then keep extending towards c until he can make an eye or two in the edge or the corner. If W plays b instead B could play "a" allowing a conection with 3 to escape; if W defends at "c" B can still escape to the center playing "d" wich may lead to a ladder but since he has an extra liberty that will not work anyway.