Author: KarinsDad
Date: 15:11:31 08/17/99
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On August 17, 1999 at 15:40:06, Michel Langeveld wrote: >>2) Storing the location of each piece. This is the typical 64 square for king >>#1, 63 squares for king #2, 63 squares for piece #3 (including off the board), >>62 for piece #4, etc. > >What about for example a promoted pawn to a knight which is on E4? If the two original knights for that side exist, then it gets stored as a pawn with a knight promotion state at location e4. Normally, with this type of algorithm, all of the main pieces (i.e. not pawns) are handled first. This allows a promoted knight to be replaced via an original knight if an original knight is off the board (there is no difference between a position with an original knight and an off board pawn; and a position with a promoted knight and an off board original knight). Hence, the reason the pawns need state information. > >As far as I can tell, there is no easy way to compress >>this type of data nor is there an easy way to derive it. The problem I see with >>this type of schema is that is allows the storage of a lot of illegal positions. >>Hence, if it does achieve a maximum number of positions greater than 2^160, it >>would be difficult (not impossible) to re-use those illegal positions by >>changing the algorithm. For example, if the opposing queen is checking our king >>and it is the opposing side to move, those positions would be dropped from the >>number of positions possible for the opposing queen. The same with if two >>knights are checking a king simultaneously or if the kings are right next to >>each other, etc. But pulling out the illegal positions would be tough since you >>have to know which are legal and which illegal and that information may not be >>available due to interposing pieces that have not yet been "placed" on the >>board. Also, one would have to come up with a way to represent the pawns in one >>of 5 states. > >White's pawn on A2 can be on (white-out promotion): >A3, A4, A5, A6, A7 = 5 >B3, B4, B5, B6, B7 = 5 > C4, C5, C6, C7 = 4 > D5, D6, D7 = 3 > E6, E6 = 2 > F7 = 1 + > --- > 20 states Yes, but the algorithm already takes into account the location of the pawn. In the case of the a pawn, it is 21 maximum locations (not states) in the not promoted state. The 5 states are: not promoted, promoted to rook, bishop, knight, or queen. And, of course, other state information like castling and en passant must also be added. KarinsDad :) > >This algorithm probably works. But, proving it may be extremely >>difficult (especially if the illegal positions bump the count above 2^160). > >I think you 're right with this. >>
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