Beginner Exercise 4 / Solution

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Analysis

[Diagram]
Analysis  

In life and death, a group needs two eyes (two real eyes) to live. If a group is totally surrounded and has only one eye (false eyes do not count), then it would be a dead group.

This group has effectively an eye at a, because White playing at b will be self-atari. However, what is not so sure is that whether the spaces at b and c is really an eye.



Solution

[Diagram]
Solution  

The solution is to make a throw-in at W1. This is a tesuji, or technique, for destroying the eye on the outside.

Notice that the W1 stone can be captured, but...

[Diagram]
Variation I  

Suppose B2 captures...

[Diagram]
Variation I  

... then W3 turns b into a false eye. The entire group only has an eye at a, there is no eye at b. So Black is dead.

[Diagram]
Variation I  

The proof that Black is dead is that W5 can capture...

[Diagram]
Variation I  

... followed by W7. However, there really is no need to play W5 and W7 immediately, because White will waste two moves. B6 cannot recapture the point without causing ko.

[Diagram]
Variation II  

In the previous variation, we see that allowing White to play at B2 will kill the group. What if Black plays at B2?

[Diagram]
Variation II  

This is what we call shortage of liberties. More precisely, Black suffers from a shortage of liberties. The reason is that W3 can capture the four Black stones chain.

[Diagram]
Variation II  

This diagram is probably unnecessary, but it clearly shows that Black is dead.



Failure

[Diagram]
Failure I  

If White makes an atari at W1, then B2 connects and has two real eyes. Note that now White cannot play at either a or b - it is suicide.

[Diagram]
Failure II  

W1 here is also a failure. Black B2 captures the W1 stone...

[Diagram]
Failure II  

... and the result is no different from the previous failure sequence.



Discussion

[Diagram]
Black to play  

This discussion is at a higher level than the problem and may be skipped by introductory level players on their first reading.

We now turn the problem around and ask how Black can live. There are two ways, each with their pros and cons, but usually the consensus is that one of them is better.

[Diagram]
Black lives I  

From the previous discussion, it should be easy to deduce that B1 makes two solid real eyes.

[Diagram]
Black lives II  

However, B1 here makes two eyes and lives as well.

[Diagram]
Black lives II  

Of course, W2 is a legal move, but B3 captures and lives.

[Diagram]
Black lives II  

Similarly, if W2 here, B3 makes two eyes.

[Diagram]
Black lives I  

Now we compare both methods.

For the first way to live, Black has two points of territory at a and b. Also, White has no ko-threat.

[Diagram]
Black lives II  

For the second way to live, Black has three points of territory at a, b and c. However, the disadvantage is that White has one ko-threat at either b or c. On the other hand, B1 creates a cutting-point at d, which may be used for Black's advantage in the later part of the game.

We can see that the pros of the second method significantly outweighs the cons, so the second method is better. Additionally, this is consistent with the observation that each ko-threat is probably worth 1/6 points, then at the bare minimum, we can say that the second method is 1/3 points better than the first method.

For more discussion, please refer to Ko threat exercise 3.



Beginner Exercise 4 / Solution last edited by 130.85.235.201 on December 7, 2008 - 04:48
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