capturing race problem

macelee · Post by **macelee** » Fri May 23, 2014 1:27 am

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This is the end position of a game I just newly added to Go4Go database (http://www.go4go.net/go/games/sgfview/42035). There is a capturing race at the bottom right corner having two groups with bulky five big eyes. I think this is a very good exercise for beginners.

oca · Post by **oca** » Fri May 23, 2014 2:39 am

Thanks for posting this example, moves like white 32 kills me so many time...
I went to a situation like this just yesterday... which ended like this...

Aidoneus · Post by **Aidoneus** » Fri May 23, 2014 4:37 am

Doesn't White capture easily? Black's bulky five gives 8 - 4 = 4 liberties, plus 1 outside liberty, for a total of 5 liberties. Black's bulky five gives 8 - 2 = 6 liberties, plus 4 outside liberties (an extra as White must first capture a lone stone to avoid a self atari, yes?), for a total of 10 liberties. Of course, I am a beginner and may be missing something.

HermanHiddema · Post by **HermanHiddema** » Fri May 23, 2014 4:45 am

Aidoneus wrote:Doesn't White capture easily? Black's bulky five gives 8 - 4 = 4 liberties, plus 1 outside liberty, for a total of 5 liberties. Black's bulky five gives 8 - 2 = 6 liberties, plus 4 outside liberties (an extra as White must first capture a lone stone to avoid a self atari, yes?), for a total of 10 liberties. Of course, I am a beginner and may be missing something.

Black's shape is not a bulky five (yet). If white plays one stone inside and black captures, only then is it a bulky five (with all its liberties). Black's initial big eye is six spaces, which has 12 approach liberties.

Aidoneus · Post by **Aidoneus** » Fri May 23, 2014 5:03 am

HermanHiddema wrote:
Aidoneus wrote:Doesn't White capture easily? Black's bulky five gives 8 - 4 = 4 liberties, plus 1 outside liberty, for a total of 5 liberties. Black's bulky five gives 8 - 2 = 6 liberties, plus 4 outside liberties (an extra as White must first capture a lone stone to avoid a self atari, yes?), for a total of 10 liberties. Of course, I am a beginner and may be missing something.
Black's shape is not a bulky five (yet). If white plays one stone inside and black captures, only then is it a bulky five (with all its liberties). Black's initial big eye is six spaces, which has 12 approach liberties.

Of course, he tricked me! Still 12 - 4 = 8, plus 1 outside for 9 liberties. So White wins, yes?

Edit: I am not awake yet. I meant to type Black must capture the lone White stone before approaching.

DrStraw · Post by **DrStraw** » Fri May 23, 2014 5:11 am

So how many ko threats does each side have?

HermanHiddema · Post by **HermanHiddema** » Fri May 23, 2014 5:23 am

Aidoneus wrote: Of course, he tricked me! Still 12 - 4 = 8, plus 1 outside for 9 liberties. So White wins, yes?

Yes

Aidoneus · Post by **Aidoneus** » Fri May 23, 2014 5:24 am

DrStraw wrote:So how many ko threats does each side have?

Now I know the problem is over my head!

oca · Post by **oca** » Fri May 23, 2014 5:57 am

I even don't understand where the 12 number comes from

Aidoneus · Post by **Aidoneus** » Fri May 23, 2014 6:03 am

oca wrote:I even don't understand where the 12 number comes from

See http://senseis.xmp.net/?BigEyeLiberties

Aidoneus · Post by **Aidoneus** » Sat May 24, 2014 3:31 pm

Aidoneus wrote:
oca wrote:I even don't understand where the 12 number comes from

See http://senseis.xmp.net/?BigEyeLiberties

Another point worth mentioning that I read somewhere was that in races between two groups that each have a big eye, all of the shared liberties belong to the larger big eye. At least I find this sort of heuristic quite useful, and I wish I knew many more!

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