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

[Jean-Pierre Vesinet <jpvesinet@xxxxxxxxxx>]

It is difficult to further diminish in practice ceil(log2(3^(N*N))) bits (=
40 for N=5).
- Eliminating illegal positions gets you another factor 2 for N=5 (only 49%
of the 3^25 positions are legal).
- Taking into account symmetry / rotations / color reversal to avoid
computing several times game values that you already have does divide almost
by 16 the useful state space with komi 0 (some positions are already
symmetric).

But there is no simple function to map between this reduced number and the
actual board state.

We also need 1 bit for the next player and ceil(log2(N*N)) (=5 for N=5) for
a ko-forbidden point (basic ko rule).

Regards,

Jean-Pierre Vesinet