[Computer-go] Komi and the value of the first move
Jean-loup Gailly
jloup at gailly.net
Mon Apr 5 15:25:25 PDT 2010
> the value of the first move has to be twice the value of the Komi
Here is a proof that I already posted in this list (slightly reworded thanks
to comments
from Robert Jasiek).
Assume a 19x19 no-handicap game played by perfect players with
New-Zealand rules with komi k. The fair komi k is defined as the integer
value which guarantees a draw (this is more fair than a non-integer
value hich guarantees a win by one of the players). k exists because
with New Zealand rules the game is finite. It is not possible to win
by an infinite amount of points, so k is bounded therefore k exists.
Now give the choice to black to either play the first move, or pass
and receive x extra points, x constrained to be an integer. Let m be
the smallest value of x that black will accept to pass instead of
playing, and still be sure of getting a draw. m exists because the
game is finite, and black being perfect can determine m exactly.
m is my definition of the value of the first move. I am not attempting
to define the value of subsequent moves, and not assuming that these
values decrease constantly. I am only interested in the relation
between m and k.
If black passes, black has m points, white has k points, white to
play. White is now in exactly the same position that black was at
initially, so white playing perfectly for maximum score will also
get a draw. The player next to move on the empty board, who can
get a draw but not win, has a cash deficit of k points in the
first case (black to play) and m-k points in the other (black
passed and white to play). So k = m-k or m = 2*k.
Jean-loup
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