Simple question that bugs this beginner - hitting the head

DrEntropy · #1

Ok, this is my first beginners question. It is probably obvious, but it has been bugging me! In this situation:

Click Here To Show Diagram Code: [go]$$c After hitting the head, white cross cuts! $$ ......... $$ ......... $$ ......... $$ ..XX2b... $$ .cOO1a... $$ ....o.... $$ ......... $$ ......... $$ .........[/go]

Here, black hit the head of the two white stones, and then instead of continuing at 'o' (which i understand is normal) white cross cuts with 2! This seems on the face of it bad for white, but I am not sure what best response for black is. Does black play at a and threaten a future ladder against 2? (Which white could respond to by extending) Does he play at b and attack 2 directly (white responds by extending)? Does he play at c and threaten the now weakened pair? The answer depends on what is going on in the neighborhood here I am quite sure, but is there a standard response to 2?

Kirby · #2

I don't know if I would call this a beginner's question... :-)

When white cuts with :w2:

, the stones on the board have been split into four groups. A fight will likely erupt in that area, and there are many possibilities of what could happen.

In general, if you want a heuristic, here's a process you can use:
1.) Can I read out a way to capture one of the groups? For example, if you were close enough to the wall to verify that you could capture :w2:

, then you can do it! If you can read out a way, go ahead and do it. Then you'll have an advantage.

2.) If I can't capture a group, which one of my groups is weakest? This depends highly on the other stones on the board in the nearby vicinity, but taking this example alone, probably :b1:

is the weakest.

3.) Now that I've identified the weakest of my groups, can I strengthen my weakest group without being captured? If so, how do I make it the strongest?

It comes down to reading what variation will occur. But in general, you can use a thought process kind of like this.

In this scenario, with no other stones nearby, and if the stones are in the middle of the board, I would suggest either 'a' or 'o' in your diagram (see step 3, above). The atari at 'b' should be used if you know that you can capture :w2:

. Otherwise, the result of the atari will be that white's :w2:

is strengthened, and :b1:

is not really strengthened that much (if you atari at 'b', after white responds, there will be a weakness at 'a', for example).

Joaz Banbeck · #3

It is really hard to judge without knowing anything about the edge of the board or nearby stones, but if there is no orientation here, 'c' is as much the head of two stones as :b1:

. So I'd play it.

Kirby wrote:

I don't know if I would call this a beginner's question... :-)

...

He gets down to fundamental issues faster than most beginners. :clap:

Chew Terr · #4

You guessed it, it's highly dependent on lots of local factors. I think the biggest thing to consider, though, is the health of both sides' groups. If your stones are near friendly stones and you have all possible ladder breakers and all that, play as aggressively as you want. However, if you're trying to gain some stability for stones closer to an opponent's sphere of influence, you will want to find a move that buys you more liberties or eyespace. You probably wouldn't have hane'd to hit the head in the first place if you were SUPER-weak, but it's hard to say. As you guessed, hitting the head AND foot of stones is very severe, if you can afford to do it. Just have to make sure you can't get punished for it.

edit: Seems like Kirby answered more fully. Read his message twice. =D

Kirby · #5

I suppose that I'd also say that 'c' might work in this situation, too, if there are no other stones or walls nearby. This doesn't really go along with the heuristic I outlined above, but it does have the advantage of removing the liberties of the two white stones...

It's a tough call for me to make.

I suppose that's what makes go interesting. :-)

I do still think, though, that if no other stones are nearby, you should probably not atari at 'a' unless you know you can capture the stone, as it makes your opponent stronger without helping yourself that much...

Joaz Banbeck · #6

I tend to disagree with Kirby's reasoning. :b1:

is not all that weak, for it has miai to run at 'a' and 'o'. I still like 'c' better.

Tryphon · #7

Joaz's famous "you did hane his head ? Now hane his ass !"

I agree with him too (depends on the context of course) but since I'm weak, it's not a good support

Joaz Banbeck · #8

Kirby wrote:

...
I do still think, though, that if no other stones are nearby, you should probably not atari at 'a' unless you know you can capture the stone, as it makes your opponent stronger without helping yourself that much...

Typo? Perhaps you mean 'b'?

Kirby · #9

Joaz Banbeck wrote:

I tend to disagree with Kirby's reasoning. :b1:

is not all that weak, for it has miai to run at 'a' and 'o'. I still like 'c' better.

You can take up an argument with Yilun Yang, then. :-)

The general idea for the heuristic is actually his (I believe it was a discussion on crosscuts from one of his workshop lecture books).

I don't have the book in front of me, but I do know that components of his algorithm included:
1.) Can I capture a cutting stone?
2.) Which of my groups is weakest?
3.) Can I help my weakest group?

That said, I don't typically like treating "algorithms" as gospel, so that's why I don't feel that 'c' is necessarily a bad move. That's just based on my feeling, though.

Joaz Banbeck wrote:

Kirby wrote:

...
I do still think, though, that if no other stones are nearby, you should probably not atari at 'a' unless you know you can capture the stone, as it makes your opponent stronger without helping yourself that much...

Typo? Perhaps you mean 'b'?

Yes, I meant 'b'.

Joaz Banbeck · **#10**

If we do extend from :b1:

, I'd think that 'o' is better, for it threatens his 2-stone group, wheras 'a' threatens one.

topazg · **#11**

I would play, "b", "c", or "o" depending on context. "o" is a bit tricky:

Click Here To Show Diagram Code: [go]$$c After hitting the head, white cross cuts! $$ ......... $$ ......... $$ ......... $$ ..XX2.... $$ .4OO1.... $$ ....3.... $$ ......... $$ ......... $$ .........[/go]

It's very hard now for Black to strongly attack :w2:

, and White's main group is fairly strong. If Black's thick up and right, this makes a lot of sense though.

Click Here To Show Diagram Code: [go]$$c After hitting the head, white cross cuts! $$ ......... $$ ......... $$ ....4.... $$ .aXX2.... $$ .5OO13... $$ ......... $$ ......... $$ ......... $$ .........[/go]

If the ladder works to the top right and the bottom right, I'll probably play here, as after :b5:

the two stones can be laddered in both directions. Which means White can't cut. If the ladder to the lower right doesn't work, "a" seems a reasonable :b5:

to me.

Click Here To Show Diagram Code: [go]$$c After hitting the head, white cross cuts! $$ ......... $$ ......... $$ ......... $$ ..XX2.... $$ .3OO1.... $$ ......... $$ ......... $$ ......... $$ .........[/go]

This would probably be the first move I'd want to play though. Hane at the head and foot of two is almost always beneficial with respect to fighting, but the continuations are insanely complex.

Kirby · **#12**

Looks like daal also read the same thing:
http://www.lifein19x19.com/forum/viewto ... 918#p64918

daal wrote:

For example, in a crosscut fight you should first read whether an opponent's stone can be captured. If not, you should read whether you can save both your stones, and if so, you should help the weaker (often inside) one first by increasing its liberties. If you can't you should save your most important stone.

This was Yilun Yang's algorithm.

Now that I read this, I think the algorithm in the book was pretty much like what daal said:
1.) See if you can capture opponent's stone.
2.) If not, see if you can save both stones. If you can, help weaker stone by increasing its liberties (not atari).
3.) If you cannot save both stones, then save the most important one.

The last step was one that I didn't remember, I think, but on an empty board, I'm pretty sure you can save both stones.

Kirby · **#13**

Joaz Banbeck wrote:

If we do extend from :b1:

, I'd think that 'o' is better, for it threatens his 2-stone group, wheras 'a' threatens one.

I think I agree.

Joaz Banbeck · **#14**

topazg wrote:

I would play, "b", "c", or "o" depending on context...

I think the typo is contagious. You probably meant 'a', not 'b'.

I think that 'b' is the one move that nobody would recommend unless the surrounding stones were absolutely requiring it.

Click Here To Show Diagram Code: [go]$$c 'B' is almost always a bad move $$ ......... $$ ......... $$ ......... $$ ..XX23... $$ ..OO1.... $$ ......... $$ ......... $$ ......... $$ .........[/go]

topazg · **#15**

Joaz Banbeck wrote:

topazg wrote:

I would play, "b", "c", or "o" depending on context...

I think the typo is contagious. You probably meant 'a', not 'b'.

I think that 'b' is the one move that nobody would recommend unless the surrounding stones were absolutely requiring it.

Click Here To Show Diagram Code: [go]$$c 'B' is almost always a bad move $$ ......... $$ ......... $$ ......... $$ ..XX23... $$ ..OO1.... $$ ......... $$ ......... $$ ......... $$ .........[/go]

Gah, many thanks, quite right, I meant "a"

Kirby · **#16**

FWIW, I think I've also heard a proverb, "Beginners atari".

Bill Spight · **#17**

Well, as our current discussion has shown, it's not a simple question, and I doubt if strong amateurs will reach a consensus about correct play (assuming an otherwise empty board). Strong pros might, however. It might be worth doing a database search.

Let me share my thinking.

Click Here To Show Diagram Code: [go]$$c After hitting the head, white cross cuts! $$ ......... $$ ......... $$ ......... $$ ..XX2b... $$ .cOO1a... $$ ....o.... $$ ......... $$ ......... $$ .........[/go]

restores equality, but is doubtful. It may give Black a bigger play than normal.

Black "b" is surely bad, as it strengthens White and weakens Black after White extends. The other atari would not be so good, either.

Mr. Yang's heuristic is interesting, and it recommends strengthening one's weaker group. But which group is weaker? Yes, the single stone has only two liberties, but that is not the only indicator of weakness. The two stone group has a weakness by virtue of having more stones. That makes it harder to sacrifice. (Not that we want to sacrifice either group.) Also, the two stone group has only a single play to increase its dame, while the single stone has two plays to do so. (The other plays for the two stone group make bad shape.)

topazg's extension from the single stone is interesting, because it carries an immediate threat of a ladder, and then, if White saves the threatened stone, Black can hane at the base (head?) of the White two stone group, threatening it with a ladder, too. That's very nice, and it may well be the best play.

OTOH, Black's single stone is not particularly weak, and it has two different extensions, while the hane at the base of the White two stone group is the best way to reduce its liberties. And it is the play I want to make, sooner or later, anyway. Why not now?

So the hane is my choice, but topazg's extension may be better.

Kirby · **#18**

Bill Spight wrote:

...

Mr. Yang's heuristic is interesting, and it recommends strengthening one's weaker group. But which group is weaker? Yes, the single stone has only two liberties, but that is not the only indicator of weakness. The two stone group has a weakness by virtue of having more stones. That makes it harder to sacrifice. (Not that we want to sacrifice either group.) Also, the two stone group has only a single play to increase its dame, while the single stone has two plays to do so. (The other plays for the two stone group make bad shape.)
...

It could be the case that the two stones are weaker than the single stone. If that is the case, it would add to my distaste toward such algorithms, as misunderstanding a small part can lead to a totally different decision.

I do feel that the single stone is, at least, more flexible.

Bill Spight · **#19**

Kirby wrote:

Bill Spight wrote:

...

Mr. Yang's heuristic is interesting, and it recommends strengthening one's weaker group. But which group is weaker? Yes, the single stone has only two liberties, but that is not the only indicator of weakness. The two stone group has a weakness by virtue of having more stones. That makes it harder to sacrifice. (Not that we want to sacrifice either group.) Also, the two stone group has only a single play to increase its dame, while the single stone has two plays to do so. (The other plays for the two stone group make bad shape.)
...

It could be the case that the two stones are weaker than the single stone. If that is the case, it would add to my distaste toward such algorithms, as misunderstanding a small part can lead to a totally different decision.

Go is hard.

Kirby · **#20**

Bill Spight wrote:

Kirby wrote:

Bill Spight wrote:

...

Mr. Yang's heuristic is interesting, and it recommends strengthening one's weaker group. But which group is weaker? Yes, the single stone has only two liberties, but that is not the only indicator of weakness. The two stone group has a weakness by virtue of having more stones. That makes it harder to sacrifice. (Not that we want to sacrifice either group.) Also, the two stone group has only a single play to increase its dame, while the single stone has two plays to do so. (The other plays for the two stone group make bad shape.)
...

It could be the case that the two stones are weaker than the single stone. If that is the case, it would add to my distaste toward such algorithms, as misunderstanding a small part can lead to a totally different decision.

Go is hard.

Yeah. Algorithms seem to be an attempt to simplify it too much. Though, I suppose you could argue that you at least know what to look for using such an algorithm (in this case, perhaps the "weakest group").

Simple question that bugs this beginner - hitting the head

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