Simple question that bugs this beginner - hitting the head

[DrEntropy]

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

$$c After hitting the head, white cross cuts!
$$ .........
$$ .........
$$ .........
$$ ..XX2b...
$$ .cOO1a...
$$ ....o....
$$ .........
$$ .........
$$ .........

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]

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

When white cuts with , 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 , 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 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 . Otherwise, the result of the atari will be that white's is strengthened, and 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]

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 . So I'd play it.

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

He gets down to fundamental issues faster than most beginners.

[Chew Terr]

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]

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]

I tend to disagree with Kirby's reasoning. is not all that weak, for it has miai to run at 'a' and 'o'. I still like 'c' better.

[Tryphon]

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]

...
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]

I tend to disagree with Kirby's reasoning. 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.

...
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]

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

[topazg]

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

$$c After hitting the head, white cross cuts!
$$ .........
$$ .........
$$ .........
$$ ..XX2....
$$ .4OO1....
$$ ....3....
$$ .........
$$ .........
$$ .........

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

$$c After hitting the head, white cross cuts!
$$ .........
$$ .........
$$ ....4....
$$ .aXX2....
$$ .5OO13...
$$ .........
$$ .........
$$ .........
$$ .........

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 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 to me.

$$c After hitting the head, white cross cuts!
$$ .........
$$ .........
$$ .........
$$ ..XX2....
$$ .3OO1....
$$ .........
$$ .........
$$ .........
$$ .........

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]

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

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]

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

I think I agree.

[Joaz Banbeck]

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.

$$c 'B' is almost always a bad move
$$ .........
$$ .........
$$ .........
$$ ..XX23...
$$ ..OO1....
$$ .........
$$ .........
$$ .........
$$ .........

Click Here To Show Diagram Code

[go]$$c 'B' is almost always a bad move
$$ .........
$$ .........
$$ .........
$$ ..XX23...
$$ ..OO1....
$$ .........
$$ .........
$$ .........
$$ .........[/go]

[topazg]

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.

$$c 'B' is almost always a bad move
$$ .........
$$ .........
$$ .........
$$ ..XX23...
$$ ..OO1....
$$ .........
$$ .........
$$ .........
$$ .........

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"