Absolute beginner's question regarding a certain problem in the Learn To Play Go Book 1
Page 25 Problem 2
Where can Black play to prevent White from cutting off a stone?

The answer found on page 26 confuses me because the white stone on the right will be cut.
Is this an english error in the way the question was asked?
Or is there something so obvious I am not getting?

Where can Black play to prevent White from cutting off a stone?
The problem looks like this.. (smily face are blanks)






the answer in the book is this.







but what about the black on the right side? won't that get cut if white does this?






Does the problem really just ask that only one stone is prevented from being cut so letting the black stone on the right get cut is ok with regards to the question being asked?

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . O X B . . . .
$$ | . . O O X O X . .
$$ | . . . . . . . . .
$$-------------------[/go]

Maybe some better diagrams will make it easier for all of us. I'm assuming that this position is near the edge of the board.

If this is indeed the position that is intended:

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . O X . . . . .
$$ | . . O O X . X . .
$$ | . . . . . . . . .
$$-------------------[/go]

...then this is the proper answer:

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . O X 1 . . . .
$$ | . . O O X . X . .
$$ | . . . . . . . . .
$$-------------------[/go]

...for the cut like this:

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . O X 1 . . . .
$$ | . . O O X 2 X . .
$$ | . . . . . . . . .
$$-------------------[/go]

...just gets killed:

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . O X 1 3 . . .
$$ | . . O O X 2 X . .
$$ | . . . . . . . . .
$$-------------------[/go]

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Actually, on a large scale it may be better to leave 'a' and 'b' as miai and play like this:

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . 3 . . . . . .
$$ | . . O X 1 b . . .
$$ | . . O O X 2 X . .
$$ | . . . . a . . . .
$$-------------------[/go]

But, regardless of , the move of is so solid that is no good for white.

EDIT: I'm getting off subject. If the goal of the problem is to not get cut, then 3 is best at 'b'.

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Sorry for my beginner's question, but wouldn't 4 cut the black stone under it?

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . O X 1 3 4 . .
$$ | . . O O X 2 X . .
$$ | . . . . . . . . .
$$-------------------[/go]

I guess I am just obsessing because of the question itself
"Where can black play to prevent white from cutting off a stone?"

I was looking for a way to connect all the black stones in this beginners problem.
Is it even possible? Without all the eating thing?
I can't believe I'm having such a hard time with a beginner's problem
Hmmm...

Please don't apologize for asking a question. Answering beginner's questions is one of the major goals of this forum.
BTW, welcome to the forums.

Click Here To Show Diagram Code

[go]$$Bc Black is connected with 5
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . O X 1 3 4 . .
$$ | . . O O X . X . .
$$ | . . . . . 5 . . .
$$-------------------[/go]

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Still confused.. I know problems are just problems and not everything applies to the real game as much, but this problem really bugs me.

So does this mean that if I put a stone on the eaten white space they will be connected? Is it necessary to put a black stone inside the eaten space to connect them?

Also, if that's the case, wouldn't

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . . 3 . . . .
$$ | . . O X 2 1 . .
$$ | . . O O X 4 X . .
$$ | . . . . . 5 . . .
$$-------------------[/go]

or

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . . 5 . . . .
$$ | . . O X 4 1 . .
$$ | . . O O X 2 X . .
$$ | . . . . . 3 . . .
$$-------------------[/go]

wouldn't the any of the 2 above ways done the trick?

Or is there something about this that i just don't see?

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . O X . 1 . .
$$ | . . O O X . X . .
$$ | . . . . . . . . .
$$-------------------[/go]

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . O X . 1 . .
$$ | . . O O X . X . .
$$ | . . . . . . . . .
$$-------------------[/go]

I think this is a solution also. The other diagram is slightly preferable in my opinion, because white can atari in sente this way:

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . 2 . . . . .
$$ | . . O X 3 1 . .
$$ | . . O O X . X . .
$$ | . . . . . . . . .
$$-------------------[/go]

And white 2 might be useful. But that's not really the focus of the problem, so I actually think either solution is acceptable.

Woo... oh boy... how I go about explaining diagram choice to you...


Short answer: your answer is also correct. The stones are connected. However, in a real game, the correct answer is the better choice.


Long answer:

The book you are reading is trying to get you to internalize good shapes, so they try not to show you bad shapes, even in example problems where shapes are irrelevant.

In a real game, your shape would be taken advantage of.

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . 2 . . . . .
$$ | . . O X . 1 . .
$$ | . . O O X . X . .
$$ | . . . . . . . . .
$$-------------------[/go]

White's position becomes strong, and black's becomes weaker.

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . 3 . . . . . .
$$ | . . . b c d . . .
$$ | . . a 2 . . . . .
$$ | . . O X 1 . . . .
$$ | . . O O X . X . .
$$ | . . . . . . . . .
$$-------------------[/go]

Now white may not want to play 2 because he leaves a bit of a weakness(aji) at a which can be targeted by moves like 3, or b, c, or d.

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . 2 . . . . . .
$$ | . . . . . . . . .
$$ | . . O X 1 . . . .
$$ | . . O O X . X . .
$$ | . . . . . . . . .
$$-------------------[/go]

White may opt to extend, but in this case, black has come out with a decent position, compared to the variation with your shape.



Blargh... I'm sure that alot of what I said is going to cause you more confusion, and I don't really know how else to explain it... If you do get confused, simply look at the short answer and forget everything I just told you and come back and read it in the future when you are stronger.

You are indeed obsessing about the wrong things, but it is understandable because some of this stuff is not obvious. Don't worry about it, and don't be afraid to keep asking.



Your problem (as others have explained in various ways) is misunderstanding what 'connect' means in go. It doesn't mean that there must be an unbroken chain of black stones linking the two in the problem. Instead, it means that there must be no sequence of moves that white can use to cut those stones apart.

This is what Joaz showed; white can play a move such that there is not an unbroken chain of stones, but we still call black's stones connected because there is no sequence of moves that white can use to permanently disconnect them. In this case, as has been shown, black can capture white's stone and there is nothing white can do about it (you may like to convince yourself of this). Because of this, we still call the black stones 'connected', as white cannot do anything to prevent this even if there is not a currently existing chain of black stones.



Without 'the eating thing', the game is no longer go.



Beginners problems wouldn't exist if their content was too easy to be worth showing. Don't worry about it.



Have you thought about the possibilities for learning go online? The popular go server KGS (www.gokgs.com) has a very good teaching community, and would certainly have people happy to answer your questions. The advantage there would be that you could ask questions (and modify the board) in real time, so that it's much clearer what is going on. I would strongly recommend this if you are able, though don't feel you have to.

Wow, thanks! Some of the stuff is still beyond my understanding, but I'm sure most of it will be clearer to me in the future, though I'm sure I will experience tons more of beyond my understanding, but somehow they feel nice!

Thanks to all!! Really helpful indeed to a newbie such as myself.

Great forum with great people!

Not sure if the teaching community in KGS being mentioned is the beginner's room. If it is, I will check it when I can.

Thanks again to everyone!!!!

Inspires me to become try harder to become better so I can someday be of some help to newbies such as myself.

Yeah, I agree with this. Violence's answer is way more info that you are ready for. Ignore it for now and look at it in a year or so.


Maybe I can get to the core of the confusion. Connecting can include killing your opponent's stones to complete the connection. Indeed, this is often a good way to do it because you make the connection and you get a point for the stone that you killed.

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For more details about diagrams, look at viewtopic.php?f=5&t=226

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I don't know... I thought Violence's answer made pretty good sense (but maybe only a true beginner will be able to judge that)

Let me add something about language. Go has been around for a long time, and a lot of go terms are like words in regular language. They have more than one meaning, which are not always precise. Here there are two distinct meanings of "connected".

The first meaning is the basic one, where stones of the same color may be connected along the lines of the board.

Click Here To Show Diagram Code

[go]$$Bc
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . O B B . . .. .
$$ | . . O O B . X . .
$$ | . . . . . . . . .
$$-------------------[/go]

The marked Black stones are connected in that sense. I think that that is clear to you.

There is another sense of connected that is a little hard to explain, and different people explain it differently. In that sense the other Black stone is connected to the marked stones.

Suppose that we are playing a go like game where Black wins if she can connect all four Black stones in the first, basic sense, even if White plays first. If Black can win that game, then the Black stones are connected in the second sense.

As you already know, the game might go this way:

Click Here To Show Diagram Code

[go]$$Wc Black is connected
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . . . . . . . .
$$ | . . O X X 2 3 . .
$$ | . . O O X 1 X 5 .
$$ | . . . . . 4 . . .
$$-------------------[/go]

at 1.

With Black connects in the first sense, and wins the game, so Black is connected in the second sense.

Now, we may also consider the Black stones to be connected after , even without , because for White to play at 1 would be illegal. That is a third sense of connected.

Most people do not think about these things in so formal a manner, they just picked up the idea as they learned the game.

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