RE: computer-go: perfect players

["Fant, Chris" <chris.fant@xxxxxxxxxx>]

Yes, that's a good point.  If a perfect player has a choice of two moves,
each of which will lead to a loss against an opposing perfect player with
the same final score and in the same number of moves, how does he choose?
So my definition of a perfect player was not completely unambiguous.  Let me
add one more attribute: He must make the move that leads to the fewest
number of tied-for-best-move choices for his opponent over the remaining
course of the game.  This is a worthwhile thing to do in case he is only
playing against a near-perfect player.  But, obviously, we are still not
quite there.  If I were coding a perfect player (I just haven't had the
time) and wanted him to be completely deterministic, I would just take the
first move in my move list.  But since that varies with architecture, and we
(or maybe just I) want an architecture independent definition of the perfect
player, I think the only thing left to do is to make the move nearest the
beginning of the board (in a row-major sense).  Yes, its completely
inelegant, but how else can you define the perfect move for a situation like
that one?


-----Original Message-----
From: Allan Crossman [mailto:a.crossman@xxxxxxxxxxxxxxxxx]
Sent: Thursday, May 03, 2001 7:35 PM
To: computer-go@xxxxxxxxxxxxxxxxx
Subject: RE: computer-go: perfect players



>If you define perfect the way I define it (no mind reading or learning
about
>your opponent), you would only need to play two games, one with player A as
>Black, one with player A as white.  Both games would be exactly the same
>because both players are exactly the same (perfect).

This is only true if at no point does either player have two or more moves
that lead to the same result. In fact I suspect that for Go and many other
games there are probably millions of possible "perfect games".




                    Allan Crossman
               http://www.faldara.co.uk
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