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Subject: Re: Why squish epd anyway?

Author: Dave Gomboc

Date: 23:40:34 05/18/99

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On May 19, 1999 at 01:54:47, KarinsDad wrote:

>Dann,
>
>Here is the Shirov Kasparov game I mentioned. I just picked a random game. As
>can be seen, this game maxed out at 142 bits (if Shirov would have castled, say
>at move 13, it could have been 150 bits or slightly higher). The two numbers on
>the left are the number of bits required after each move. The entire advantage
>of this schema is that most black pieces do not require a color bit until a
>white piece or pawn is taken. Once this happens, (in this game at move 3 since
>it was a Sicilian), that condition no longer holds. However, only 2 pawns or one
>piece have to be taken in the entire game to basically ensure that 160 bits is
>the maximum, so by move 4 in this game, the chances of it going above 160 bits
>are negligible.
>
>It would be interesting if someone could find a position from a real game that
>could get to 161 bits or more, but I doubt that will happen.
>
>KarinsDad :)
>
>125 125  1.e4 c5
>126 126  2.Nf3 d6
>126 137  3.d4 cxd4
>135 136  4.Nxd4 Nf6
>137 137  5.Nc3 a6
>139 139  6.Be3 Ng4
>139 139  7.Bg5 h6
>139 139  8.Bh4 g5
>139 140  9.Bg3 Bg7
>141 141 10.Be2 h5
>137 133 11.Bxg4 hxg4
>141 142 12.0-0 Nc6
>142 138 13.Nf5 Bxc3
>134 135 14.bxc3 Qa5
>134 134 15.Qxg4 f6
>134 132 16.Rab1 Qxc3
>132 132 17.Rfd1 Ne5
>128 124 18.Bxe5 Qxe5
>124 124 19.g3 b5
>124 122 20.c4 bxc4
>122 122 21.Qf3 c3
>122 123 22.Rbc1 Bb7
>121 119 23.Qxc3 Bxe4
>117 113 24.Nxd6+ exd6
>113 109 25.Re1 Qxc3
>106 106 26.Rxc3 d5
>106 114 27.f3 Kf7
>110 110 28.fxe4 Rhe8
>110 110 29.Rc7+ Kg6
>110 108 30.Rf1 dxe4
>110 108 31.Rc6 Rf8
>108 108 32.Re6 Rae8
>108 108 33.Rfxf6+ Rxf6
> 98  98 34.Rxe8 Kf5
> 98  98 35.Kf2 Rh6
> 98  98 36.Rf8+ Ke5
> 98  98 37.Re8+ Kf5
> 98  98 38.Rf8+ Ke5
> 1/2-1/2

What a waste of bits! :)  The entire game should have been storable in far
fewer.  To know exactly how many, take the number of legal moves from each
position, multiple them all together to get a big number, P.  Find n such that P
/ 2^n < 0.  n = the number of bits you need to store the raw positions (no
1/2-1/2, player names, etc.)

I didn't do it for this game, but n will likely be in the low hundreds.  You are
using several thousand bits to represent the game.

Dave



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