Re: computer-go: perfect players

[Don Dailey <drd@xxxxxxxxxxxxxxxxxxxxx>]

But your  translation was that  the game was  always a draw.  This was
not an accurate paraphrase.   No one is trying  to refute  the author,
just your original  interpretation.   With your  clarification  it now
makes more sense.

Don



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   Date: Fri, 4 May 2001 08:30:24 -0400 (EDT)
   From: Azathoth <shoffman@xxxxxxxxxxxxxxxxx>
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   I have read many of the responses to this thread, and some of the
   suggestions seem to ignore the word "perfect".
   The book I am referencing is full of proofs written in hard to read formal
   logical notation, so, even though they might be hard to understand, they
   are still validly constructed proofs.

   A better paraphrase from the book (book talks about chess, not go):

   Similar to chess, go is a strictly competitive game with finitely many
   histories (the ko rule prevents a game from returning to a previous
   state), and thus, as proved by Zermelo in 1913, has subgame perfect
   equilibrium and Nash equilibrium. Despite this fact, the equilibrium has
   yet to be calculated, and, even if were, it is doubtful that a human
   player would be able to implement the resultant strategy.


   And so it goes,
   -seth

   On Fri, 4 May 2001, Mika Kojo wrote:

   > 
   > Seth, 
   > 
   > perhaps you should consider your translation again. There surely are
   > games where even a perfect player would lose if not given the *choice*
   > of color.
   > 
   > -- Mika
   > 
   > Azathoth writes:
   > > from the book: A Course in Game Theory(1994, Martin J. Osborne & Ariel
   > > Rubinstein)
   > > "Every finite extensive game with perfect information has a subgame
   > > perfect equilibrium"(p.99)
   > > 
   > > In english: assuming perfect ability to analyze a game, he can develop a
   > > strategy that will ensure him of at least an tie, no matter what strategy
   > > the opponent is using. 
   > > 
   > > And so it goes,
   > > -seth
   > > 
   > > On Fri, 4 May 2001, Allan Crossman wrote:
   > > 
   > > > 
   > > > >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
   > > > ------------------------------------------------------
   > > > PGP Keys: 0x497F13C8 (New) and 0xCEC9FAE1 (Compatible)
   > > > 
   > > > 
   > > 
   >