# Tag Info

19

In tsumegos where you should kill a group, the best (and thus correct) answer is always the solution where you kill unconditionally (if possible). Only if there is no unconditional kill the ko would be the best (correct) solution. The basic idea is that a tsumego is a local fighting position which you should solve without knowledge of the rest of the board. ...

10

The key is that killing the black stones doesn't give White an eye. After White captures the two stones, Black will throw in at the same place he played in #3. White can capture that stone too, but later in the game the stone above the capture will be placed in atari, and White will have to fill in the "eye" on the right. Because that can happen, it is a ...

7

This has been discussed a lot in Sensei's Library (http://senseis.xmp.net/?BiggestCorner and http://senseis.xmp.net/?10x10CornerGame1). The consensus seems to be that white dies with 8 free spaces, but can live with 9 free spaces using 3-3 point and some clever tesujis. I copied some variations from Sensei's to EidoGo for easier studying: http://eidogo.com/#...

6

TimK's answer is right on point on how to kill white in the corner, but your question belies a more general concern: Thinking "I need to kill all the things" is a very common beginner's mistake. Take a look at your final outcome; yeah, white lives. He lives small, blocked into the corner. You've got pretty much the entire lower left quadrant on lock, and ...

5

Actually, White is dead after you fill that liberty. He can never attack the Black stones, but Black can capture the stone in the corner. After that there is a ko that will kill White. There's no way for White to win the ko - filling it in leaves the white stones in atari. At the end of the game when White has no more ko threats, Black can take the ko, ...

5

Black lives: \$\$B \$\$ ------------------- \$\$ | . 1 . . X O . . . \$\$ | O X . X X O . . . \$\$ | . X X O O O . . . \$\$ | O O O . . . . . . \$\$ | . . . . . . . . . \$\$ | . . . . . . . . . White kills unconditionally: \$\$W \$\$ ------------------- \$\$ | . 2 1 . X O . . . \$\$ | O X . X X O . . . \$\$ | 3 X X O O O . . . \$\$ | O O O . . . . . . \$\$ | . . . . . . . . . \$\$ | . ...

4

Seems Black has to take the 1-2 vital point to avoid White making 2 eyes. If White begin to deprive Black's liberty from outside, then Black get connected by taking on 1-3 and White deprives outside again. After that, both Black and White have same 3 liberties, but 2 of them are in common used. This becomes seki and it should be the best result for White.

3

I think black should play 2-1 then 3-1 and 1-3 are miai for b life, no need for ko

3

Yes, that's the best you can do.

3

This seems like a mistake to me. I don't see a way to win the race. I wonder if the point was that White now has a stable group, so this is a failure for black. White can hane underneath the stone on the right side, and should probably be happy with that - Black was stronger here to begin with.

3

This is a really broad question, and it depends on your level. The general idea is to reduce the opponent's eye space. D8-C8-B6 would probably be a good way to start: \$\$Bc \$\$ --------------------- \$\$ - . . . . . . . . . - \$\$ - . . 2 1 . . . . . - \$\$ - . . W . X . O . . - \$\$ - . 3 . . . . . . . - \$\$ - . . . . X . . . . - \$\$ - . . . . . . . . . - \$\$ - . . X ....

2

To answer the original question: White's corner is unconditionally dead, it cannot be rescued even with White playing first. The points P1 and R1 are miai, when White takes one of them Black takes the other. In the actual game White uses the aji left in the corner for some Ko threats, that's all White can get out of the local situation.

2

I'm not familiar with this as folklore, but the setup is similar to the setup a person might use to work on corner problems. I believe that this could be answered by looking a many examples of corner joseki. Assuming white's goal here is simply to live, the border formed by black is too far removed from the corner to have a significant effect of white's ...

2

The Black move 1 reduces the "three in a row" from two eyes to one. White cannot protect his "second" eye at 3 until he kills the marked stone (in red) with 2, because it would be self atari. Black then quashes the potential second eye with 3, and White is dead (for sure). If Black plays 1 at 3, White will play in the middle of "three in a row" to create a ...

1

Oh sorry now I see: Pretty tricky sequence :-)

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