Author: Dann Corbit
Date: 09:55:13 05/21/99
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On May 21, 1999 at 12:28:15, KarinsDad wrote: [snip] >>Yes, what a fascinating rejoinder! In this case, if 10^52 is correct, then 173 >>bits should be the minimum, since 2^173 = 1.197e52 >>If we can encode in less, then the number of board positions is less than we >>thought (or we have an error in our thinking and the scheme won't work). > >Actually, I think that this estimate is based on the number of combinations of >piece placement on a board (regardless of whether any set of legal moves can get >you to that position), not the legal number of positions that can be achieved >via legal moves. An example of this would be black having a pawn on a2, white >having pawns on a3, a4, a5, a6, and a7 without having at least 5 captures >somewhere along the way (and at least 4 of those captures had to be white >captures of black pieces). You could have a lot of positions with just this >example alone where the position is not legal, but it has pieces all on legal >squares for those particular pieces (e.g. if you had this example with all 32 >original pieces still on the board). > >A simpler example is having your king in check when it is your opponent's turn >to move. This cannot occur via legal moves, but may not be excluded at all times >in the calculations. In any case, it still puts a lid on the maximum number possible. If, for instance, I can encode all chess board positions in 100 bits, then there are only about 10^30 possible board positions. The legal positions will be a subset of that. Hence, finding a minimal coding for board positions also has a fascinating mathematical result: It puts a cap on the maximum number of possible board positions.
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