Author: Ratko V Tomic
Date: 21:22:07 11/03/99
Go up one level in this thread
> Maybe you mean that for each new triplet of excess pieces an additional > piece or pawn is taken out of the maximum pawns+pieces. > > an additional pawn is not correct because of the fact that white can promote > more than 3 pawns only by taking black pieces. You're right here about excess pieces. The limit of 12 excess pieces that several folks mentioned (and which seemed plausible at the time) isn't actually the correct limit (while it may be useful for a particular type of encoding discussed at the time, which is different than my proposal). Namely, white can take with his pawn a black piece to get around the black pawn and promote. So white can create, say, 8 excess Queens by capturing 6 black pieces (2r, 2b, 2n) and 2 black pawns. Then black, which now has 1 Queen and 6 pawns can promote his remaining 6 pawns (without a need to capture white pieces since white has no pawns) and produce 6 excess Queens, for total of 14 excess pieces (i.e. pieces which could come only from promotions). The maximum excess outcome is white with 9 Q, 2R, 2B, 2N (white can distribute excess pieces in any other way across his Q,R,B,N) and black with 7Q for total of 22 pieces on board (plus 2 Kings, as always). Instead of 8+6 excess pieces, in similar way one can obtain 6+8 or 7+7 excess pieces. Now, for white to get a passed pawn either white or black has to do a capture with a pawn. To maximize the excess pieces, we could have white capture 4 black pieces and black capture 4 white pieces, giving white and black 8 passed pawns each, hence producing 8+8 excess pieces. These 8+8 promotions would have to go into the remaining 3 white and 3 black pieces, which weren't subject to capture (so that we avoid cases of capturing piece and later promoting into that same piece, since that doesn't create excess pieces but is equivalent to a simple pawn loss). As with the 8+6 or 7+7 excess cases above, the final number of pieces is 22 (instead of 26). By combining one or more pawn captures with piece captures we could have total number of pieces above 22 (and below 26), with excess still above 12 (i.e. up to 16). By doing pawn captures only, the two sides could produce at most 12 excess pieces, but if no other pieces were captured the total of pieces on the board is 12+2*(1+2+2+2)=26 (and 0 pawns). That still seems to be the maximum for non-pawn pieces on board. Thanks for bringing up this issue for reconsideration, now we know that the constraint on the number of excess pieces is much weaker than I thought. It only limits somewhat more the distribution of the excess pieces compared to the pawn only capture method). But in creating the material content (MC) tables this kind of highly irregular and sparse residual constraint may not be worthwile imposing, i.e. when the MC table for one color is produced, the pairing for the full material content tables might as well match all 1 color MC elements with all 1 color MC elements to generate all 2 color MC elements without any pair type elimination. Any excess piece would only drop the pawn maximum for that side by 1 and one would allow for as many as 8+8 excess pieces. That simplifies the algorithm great deal and probably won't generate too many impossible MC entires. After all, other impossible positions (especially with checks) are counted by the algorithm as Ok, so this wouldn't exactly ruin the purity of the produced positions. (Since the total number of positions involved is 2^150 at least, individual one by one checking is out of question, so some impurity will always remain, i.e. the final numbers are only upper bounds.)
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