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 Post subject: I don't get this very easy problem
Post #1 Posted: Thu Mar 13, 2014 1:06 pm 
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Hello,

So this is a basic problem for super beginners..

Image

The solution says to play the next black move exactly where the black stone just got captured, so black captures 4 white stones, which makes sense.

but when white played 1, what made it that it was the black stone that was captured, and not the 4 white stones?

Do you understand what I mean? :s

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 Post subject: Re: I don't get this very easy problem
Post #2 Posted: Thu Mar 13, 2014 1:32 pm 
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That situation is called a "snap-back." The rule in go is that you can play a move that would be suicide if it would capture an enemy's stone(s). If that is the case then the move is not suicide, instead the enemy's stone(s) are captured.

You can think of it like the standard ko situation:

Click Here To Show Diagram Code
[go]$$ White at 'a' is not suicide
$$ . . . . . .
$$ . . X O . .
$$ . X a X O .
$$ . . X O . .
$$ . . . . . .
$$[/go]


When white plays it looks like the white stone has no liberties. However, because it was white who played the stone, first the black stone is captured. Then you count the liberties that the white stone has.
Click Here To Show Diagram Code
[go]$$ The black stone is removed
$$ . . . . . .
$$ . . X O . .
$$ . X W X O .
$$ . . X O . .
$$ . . . . . .
$$[/go]

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"You have to walk before you can run. Black 1 was a walking move.
I blushed inwardly to recall the ignorant thoughts that had gone through
my mind before, when I had not realized the true worth of Black 1."

-Kageyama Toshiro on proper moves


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 Post subject: Re: I don't get this very easy problem
Post #3 Posted: Thu Mar 13, 2014 1:42 pm 
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moyoaji wrote:
That situation is called a "snap-back." The rule in go is that you can play a move that would be suicide if it would capture an enemy's stone(s). If that is the case then the move is not suicide, instead the enemy's stone(s) are captured.

You can think of it like the standard ko situation:

Click Here To Show Diagram Code
[go]$$ White at 'a' is not suicide
$$ . . . . . .
$$ . . X O . .
$$ . X a X O .
$$ . . X O . .
$$ . . . . . .
$$[/go]


When white plays it looks like the white stone has no liberties. However, because it was white who played the stone, first the black stone is captured. Then you count the liberties that the white stone has.
Click Here To Show Diagram Code
[go]$$ The black stone is removed
$$ . . . . . .
$$ . . X O . .
$$ . X W X O .
$$ . . X O . .
$$ . . . . . .
$$[/go]



Well that makes perfect sense, thanks a lot for the explaination with diagrams :clap:

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 Post subject: Re: I don't get this very easy problem
Post #4 Posted: Thu Mar 13, 2014 3:52 pm 
Oza

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The simplest way to think about it is that a move is not complete until any surrounded stones are captured. After the move is complete white is no longer surrounded and his stones stay on the board. Black, on the other hand, had his stones removed as part of the move.

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 Post subject: Re: I don't get this very easy problem
Post #5 Posted: Thu Mar 13, 2014 4:46 pm 
Judan
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With apologies to the other posters, the discussion of 'snapback' is a bit off topic.

The issue at hand is how a rule set should handle a move that simultaneously takes the last liberty of TWO groups.
There are 4 possibilities:
1) The group that filled the last liberty should die
2) The other group should die.
3) Both groups should die.
4) Neither group should die.

There is nothing inherently better about any of these. The use of any of the four will create a rule set that defines a playable game. ( It is sort of like the parallel postulate in geometry. There are several possible definitions, and each implies a different universe. Those other universes, although seemingly bizarre to us, are nonetheless internally consistent. )

We chose #2. And it was arbitrary.

The best that we can say about it is the subjective opinion that the rule set which incorporates #2 is more fun.

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 Post subject: Re: I don't get this very easy problem
Post #6 Posted: Thu Mar 13, 2014 5:15 pm 
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Joaz Banbeck wrote:
We chose #2. And it was arbitrary.

The best that we can say about it is the subjective opinion that the rule set which incorporates #2 is more fun.


I think it's deeper than that. I think #2 leads to more varied play and that, if someone could devise an objective measure of varied play, it could be proven rigorously. Of course, this leads to a more enjoyable game.

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 Post subject: Re: I don't get this very easy problem
Post #7 Posted: Thu Mar 13, 2014 6:02 pm 
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Having no rules would lead to the most variation but that wouldn't be nearly as fun. Imho it's about finding the right balance of freedom and structure, which is completely subjective.

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