[Computer-go] Combinatorics of Go

[Arthur Cater]

Intriguing!

A position is obviously illegal if any point is occupied by a stone  
surrounded by opposite-colour stones.
At the 4 corners, 25 out of 27 combinations will be legal. The  
proportion (25/27)^4 will survive.
At the 68 edges, 79 out of 81: (79/81)^68 will survive.
At the 289 interior points, 241 out of 243: (241/243)^289.

Multiply those, I get 0.012321913.

So presumably the number is on the high side, because this calculation  
only takes account of
single stone blocks illegally on the board.

Arthur


On Jan 1, 2011, at 10:16 AM, Ray Tayek wrote:

> At 01:09 AM 1/1/2011, you wrote:
>> ... If I understand correctly, they computed the State-space  
>> complexity of 19x19 Go to be 2.08168199382· 10^170, which is really  
>> a big number.
>
> 3^(19*19)=1.740896506590319E172 is all combinations of black, white  
> and vacant intersections on a 19 by 19 board. but some of these are  
> illegal. off the top of my head, that number seems a bit low as it  
> seems to be saying that only about 1.2 percent of the combinations  
> are legal board states.
>
> thanks
>
> ---
> co-chair http://ocjug.org/
>
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