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

[schraudo@xxxxxxxxxxx]

> So we have an upper bound of, for a 5 x 5 board:
> 
> ((log(3) / log (2) ) * 25) - 3 + 2 + log2(25) = (rounding up) 44 bits

Two corrections:

1) Subtract 3, not 2, bits for the 8-fold symmetry of the square

2) "no active ko" can be encoded as "active ko off the board",
   so instead of log2(25) + 1 you have log2(26)

The answer, then, is 42.  ;-)

Best,

- Nici.

-- 
    Dr. Nicol N. Schraudolph               http://www.inf.ethz.ch/~schraudo/
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