Lets make it al a bit more practical in 7x7 go under perfect play and without komi black wins by 9 points.
if Black passes on the first move then under the same perfect play white wins with 9 points (again without komi) Does that mean that the value of the first move is 18 points? 2010/2/11 Jean-loup Gailly <jl...@gailly.net>: >> Yours is not a proof because what follows is not just a single move >> of value x but a game tree of moves of various sizes, > So let me try to be more precise. > Assume a 19x19 no-handicap game played by perfect players with > New Zealand rules with komi k. k is chosen as the largest value N+0.5 > (with N integer) which guarantees a win by black. k exists because > with New Zealand rules the game is finite. Black cannot win by an > infinite amount of points, so there is an upper bound for k, 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 winning. 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 playing perfectly for maximum score will lose by exactly > 0.5 point. If white is given one extra point, white would be exactly > in the position that black was at initially: next to play and > guaranteed to win by 0.5. The player next to move on the empty board > and with a guaranteed 0.5 win has a cash deficit of k points in the > first case (black to play) and m-(k+1) points in the other (black > passed and white to play). So k = m-(k+1) or m = 2*k+1. > Again the above analysis only considers the game theoretical value of > the empty board, not any subsequent position. I welcome any > correction if there is a flaw in my reasoning (which is quite > possible). > Jean-loup > 2010/2/11 Robert Jasiek <jas...@snafu.de> >> >> Jean-loup Gailly wrote: >>> >>> I would write the proof as follows. >>> >>> Assume x is the value of one move >> >> Yours is not a proof because what follows is not just a single move of >> value x but a game tree of moves of various sizes, which need not even >> decrease constantly. Many years ago, Barry Phease was a bit farther by >> assuming a constant decrement and forming a sum. His was not a proof either >> because sizes of moves in the paths of the game tree need not decrease >> constantly. >> >> Rather than being proofs, such arguments are plausible approximations. >> >> -- >> robert jasiek >> _______________________________________________ >> computer-go mailing list >> computer-go@computer-go.org >> http://www.computer-go.org/mailman/listinfo/computer-go/ >> > > > _______________________________________________ > computer-go mailing list > computer-go@computer-go.org > http://www.computer-go.org/mailman/listinfo/computer-go/ > _______________________________________________ computer-go mailing list computer-go@computer-go.org http://www.computer-go.org/mailman/listinfo/computer-go/
