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.



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.



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.

[Diagram]

Variation II

In the previous variation, we see that allowing White to play at B2 will kill the group. What is 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 unneccessary, but it clearly shows that Black is dead.



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 points or ko-threats discussion.






unkx80: Older discussion follows.

KarlKnechtel, filling in a solution.

[Diagram]

Solution.

Quickest explanation (not at all useful to those for whom this was challenging): "W1 is tesuji. B2 and W3 are miai for the kill."


How I'd explain it to someone a bit below my level: "The throw-in at W1 threatens to capture BC. Black cannot connect because he is in shortage of liberties - B2 at W3 is answered by the capture with W3 at B2. So Black captures, making one eye; but W3 destroys the other eye."

In easier terms: Black needs two eyes to live. There is already effectively one at a. White must prevent the centre space from becoming an eye. The sacrifice changes the shape of that space, achieving the goal.

This problem is small enough that we can consider all the alternatives:

[Diagram]

Reference.

Playing elsewhere is answered by Black either at a or b. (b claims an extra point of territory, at the expense of one ko threat (at a) for White[1]. So clearly White has to do something.

(Incidentally, the eye around a is an example of what I call a two-space elbow. Not that it really matters.)

[Diagram]

Failure.

One try is to threaten the existing eye; but the capture solves all of Black's problems. (On the other hand, connecting at a would be very bad for Black; White doesn't even have to capture.) This is also a throw-in, but it's the wrong one and doesn't help.

[Diagram]

Failure.

Moving to make the other eye false looks good at first, because Black is in atari to start off with. But all Black has to do is connect. Again, the two eyes are solid.

ProtoDeuteric- Don't the failure sequences prove that "a" in the first diagram below the horizontal rule is not eyespace? Shouldn't the "a" be at the point to the left?

byorgey: You're right, the point marked a in that diagram (the one titled "Reference") is not eyespace; however, that diagram doesn't claim it is. The text next to the diagram refers to a as a ko threat for white if black plays b. It is somewhat confusing having different points mared a and b in different diagrams, though...


How does W3 in the solution destroy the eye? It looks as if a White attempt to capture at W1 will only start a ko fight... -- Gregory Price

[Diagram]

Position at the end of solution sequence

unkx80: A group requires two eyes (that is, two real eyes) to live. But the position at the end of solution sequence, which is reproduced here, shows a real eye at a and a false eye (which is not an eye) at b. So this group is dead, because White can capture at b followed by capturing at a.


I'm really new to Go (just started learn last week). I'm a little confused about the terminology "played B2 at W3". What does that mean? I guess that B2 is same as the numbered black stone, but what does "at W3" mean?

Confused: It's a convenient shorthand for describing simple variations of a diagram. B2 at W3 means, the situation if Black would have played the second move (B2) at the location of the White stone at W3. Here Karl Knechtel was writing about the following case.

[Diagram]

Situation after B2 at W3.


Jasonred (I think) wrote:

tesuji or nakade?

Confused: It's a tesuji, although I'm not sure if that play counts as nakade too. If the definition on the nakade page is correct, I'd say it isn't one.

This is one of the reasons, why I avoid throwing around Japanese terms I don't understand completly or for which there is a good English translation. Some of those concepts are complicated enough on their own, that I don't need to confuse the matter more with the wrong Japanese names.

It's a throw-in


[1] A black play at a is safe from that, as I'd originally thought, but I managed to confuse myself somehow. Thanks SAS. :) Anyway - see Points Or Ko Threats Discussion for more information about that sort of tradeoff.

unkx80: I think that b is better than a in the context of this question, for two reasons: (1) It claims an extra point of territory, (2) It creates a cutting point. The group in Points Or Ko Threats Discussion is different in the sense that either of the two points there has equal value except for ko threats, so the one giving less ko threats might be better.
Coyote?: Depending on the situation, I would go with "b". That is - if I have more Ko threats then my opponent, especially if the threat of the cut can give me Sente. If it looks like my opponent is getting more Ko threats then me, I can still close it off, losing the point of territory. But not until I have to, because whether now or later, the move is Gote. - Coyote, 17kyu


This is a copy of the living page "Beginner Exercise 4 / Solution" at Sensei's Library.
(OC) 2004 the Authors, published under the OpenContent License V1.0.
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