[Computer-go] Combinatorics of Go

[Robert Finking]

Hi Alvaro,

I think you have perhaps misunderstood. As I read it, Arthur was 
refering to his own analytic result (1.232) as being "on the high side", 
not John's result in the paper. Arthur is implicitly assuming that 
John's number is correct (which I think we all are), and then 
rationalising what the discrepancy is between his analytic result and 
John's. Personally I found his analysis very helpful.

The way I read the reply from Jonas, he is similarly is referring to 
Arthur's calculation, not John's (specifically he is referring to 
Arthur's reasoning), other than that he also is assuming implicitly that 
John's calculation is correct.

I hope this helps smooth over ruffled feathers.

Happy New Year =)

Raffles

On 01/01/2011 14:08, Álvaro Begué wrote:
> The people that think the number is low or high have bad intuitions,
> that's all. Writing a program that generates random configurations and
> checks whether they are valid is fairly trivial. If you don't trust
> John's numbers, that's what you can do.
>
> Alvaro.
>
>
> On Saturday, January 1, 2011, Kahn Jonas<jonas.kahn at math.u-psud.fr>  wrote:
>> 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.
>>
>>
>> I think it's rather (much) on the low side. The probabilities are not
>> independent. And the correlations are positive: in particular, if a
>> single point is legal, it may be that it is next to a free intersection,
>> which immediately makes legal four (at the center of the board) other
>> points.
>>
>> Jonas
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