Re: computer-go: How many bits are needed to encode an N x N Go position?

[Ray Tayek <rtayek@xxxxxxxxx>]

At 11:21 AM 10/16/02 +0000, you wrote:
> ...
> What is the minimum number of bits needed to encode a 5 x 5 Go position?
> And
> generalizing, how many bits are needed to encode an N x N position?
seems like each position can be represented as a base 3 integer, so you 
would have 3^(n*n) positions. this should require 
ceil(log2(3^(n*n)))=ceil(n*n*log2(3)). for n=5, we have 
ceil(25*log2(3))=ceil(25*ln(3)/ln(2))=ceil(25*1.09861/.69314)=40. checking 
3^25=847288609443 and 2^40=1099511627776

thanks

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