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Fun little go/math problem

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Fun little go/math problem

Posted: Sat Jun 29, 2013 11:24 pm
by shapenaji
EDIT: FIXED, accidentally cut off the white group, leading to a clear black victory, better now.

In the following sgf, white to play, komi is 7.5, assume positional superko and that the marked stone was the last stone played (you can never repeat a position)





answer the following:

A) How many moves would it take to resolve this situation and finish the game?
B) Who wins? (And by how much?)

Re: Fun little go/math problem

Posted: Sun Jun 30, 2013 12:09 am
by moyoaji
23 kos, each one will be in one of two states: taken by white or taken by black. This effectively means 2^23 - 8,388,608 - right? Well, no. White will never let black get more than two of these kos. So if we assume black having the ko is 1, then there can only be 1 or 2 1's at any given point. This number is then much smaller. However, there are only 23 states of black having 1 ko and this is the key. After 23 moves by white he will not be able to reduce black to 1 ko any longer. Black will take a second ko and white will be unable to take it back. After this, no matter what white does on his next move, black will take a third ko and capture the group on the bottom. This would mean there would be 23 white takes + 23 black takes + 1 white filling a ko or passing + 1 black ko take + 1 white ko take + black ataring the bottom + white taking another ko + black taking the group. 32 moves total.

However, black only has 42 points from his captures and the bottom. White will have 42.5 with his center and komi. So white wins by .5 anyway.

Is this correct?

Re: Fun little go/math problem

Posted: Sun Jun 30, 2013 12:22 am
by shapenaji
Right! But you should hide your answer using the hide tags, so that others can try!

The reason I came up with this problem:
I found the asymmetry in the number of black states and white states in the ko really interesting, you have Choose(23,2) states where black can approach, but only 23 states where white can hold off the approach, hence, white gets stuck long before black, even though white takes the ko first.
Uh-oh, interesting follow-up on the final score:
I think black actually wins this one, the same principle should allow black to keep winning the ko situation until the states balance. So black actually should win by around 10, of course that would take so many moves as to be counterproductive

Re: Fun little go/math problem

Posted: Sun Jun 30, 2013 8:55 am
by Bill Spight
With White taking a ko first, Black can play a triple ko, forcing White to stop taking a ko at move 8. Black can atari at move 9 and take at move 11. Territory result: B +5.5.

Re: Fun little go/math problem

Posted: Sun Jun 30, 2013 11:09 am
by shapenaji
Ah... Alright, guess there's no way to force the issue...

Good point Bill

Re: Fun little go/math problem

Posted: Sun Jun 30, 2013 1:05 pm
by Solomon
Image

Re: Fun little go/math problem

Posted: Sun Jun 30, 2013 6:52 pm
by SmoothOper
moyoaji wrote:
23 kos, each one will be in one of two states: taken by white or taken by black. This effectively means 2^23 - 8,388,608 - right? Well, no. White will never let black get more than two of these kos. So if we assume black having the ko is 1, then there can only be 1 or 2 1's at any given point. This number is then much smaller. However, there are only 23 states of black having 1 ko and this is the key. After 23 moves by white he will not be able to reduce black to 1 ko any longer. Black will take a second ko and white will be unable to take it back. After this, no matter what white does on his next move, black will take a third ko and capture the group on the bottom. This would mean there would be 23 white takes + 23 black takes + 1 white filling a ko or passing + 1 black ko take + 1 white ko take + black ataring the bottom + white taking another ko + black taking the group. 32 moves total.

However, black only has 42 points from his captures and the bottom. White will have 42.5 with his center and komi. So white wins by .5 anyway.

Is this correct?
Why wouldn't black take and fill one or two before proceeding to capture the white group?

Re: Fun little go/math problem

Posted: Sun Jun 30, 2013 7:15 pm
by Bill Spight
moyoaji wrote:
23 kos, each one will be in one of two states: taken by white or taken by black. This effectively means 2^23 - 8,388,608 - right? Well, no. White will never let black get more than two of these kos. So if we assume black having the ko is 1, then there can only be 1 or 2 1's at any given point. This number is then much smaller. However, there are only 23 states of black having 1 ko and this is the key. After 23 moves by white he will not be able to reduce black to 1 ko any longer. Black will take a second ko and white will be unable to take it back. After this, no matter what white does on his next move, black will take a third ko and capture the group on the bottom. This would mean there would be 23 white takes + 23 black takes + 1 white filling a ko or passing + 1 black ko take + 1 white ko take + black ataring the bottom + white taking another ko + black taking the group. 32 moves total.

However, black only has 42 points from his captures and the bottom. White will have 42.5 with his center and komi. So white wins by .5 anyway.
Is this correct?
SmoothOper wrote:Why wouldn't black take and fill one or two before proceeding to capture the white group?
Well, Black needs to avoid leaving Black ko stones en prise.

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