Author: Sune Fischer
Date: 02:21:43 01/19/02
Go up one level in this thread
On January 19, 2002 at 03:02:41, Uri Blass wrote: >>>>>>>Does this take into account all the possible promotions? It is possible to have >>>>>>>9 queens on one side or 9 rooks or 9 bishops or 9 knights, or any combination >>>>>>>thereof. Each possible promotion will make the number larger. 31 pieces on the >>>>>>>board has more possible positions than 32 because you have to take into account >>>>>>>the promotions. If 31 pieces are remaining there is upto 3 possible promotions. >>>>>>>Makes the calculation a little more difficult. >>>>>>> >>>>>>>Chris >>>>>>> >>>>>> >>>>>>Oops, make that 9 queens or 10 rooks, or 10 bishops or 10 knights. Some days my >>>>>>fingers work faster than my brain. >>>>>> >>>>>>Chris >>>>> >>>>>You do have a point Chris. >>>>>I tried to refine the number a little bit: >>>>> >>>>>All 32 pieces: >>>>>(64*(64 - 4)*23^4*21^4*18^4*14^4*62!)/((62 - 14)!*(8!*2!*2!*2*2)^2) >>>>>= 5.13*10^36 >>>>> >>>>>I put in the factor: 23^4*21^4*18^4*14^4 >>>>>because the d and e pawns has 23 squares, the c and f pawns has 21 squares and >>>>>the b and g has 18, a and h has 14 (unless I miscounted;) >>>> >>>>I do not understand it >>> >>>Actually there are 24 squares for a white d pawn: >>8/pppppppp/PPPPPPPP/PPPPPPP1/1PPPPP2/2PPP3/3P4/8 w - - 0 1 > >I understand now your idea but with 32 pieces there are only 5 options for white >d pawns because the white d pawn can go to another file only by capture. > >d2 d3 d4 d5 d6 d7 are the options. > >My 15 options are for white and black pawn on the d file and the same for every >file. > >d2 d7 >d2 d6 >d2 d5 >d2 d4 >d2 d3 >d3 d7 >d3 d6 >d3 d5 >d3 d4 >d4 d5 >d4 d6 >d4 d7 >d5 d6 >d5 d7 >d6 d7 > >are the possible squares for d pawns when there are no captures. > >It is exactly 15 and the same for every file. > >My program did not use the 15^8 limit for the pawns squares and assumed that >every pawn can be chosen from 48 squares. > >It is not close to be truth when you have 32 pieces but it becomes closer to the >truth when there are less pieces. > >It is clear that if both sides have exactly 2 pawns you can choose them from the >48 squares in the board in every possible way. > >It is not truth for 3 pawns and h2 h3 g2 is an example for squares that cannot >be squares of white pawns but it is still truth for almost all the cases of 3 >pawns. > >Uri Oh, so that's what you mean, well I'm not concerned with counting *too many* squares since is was supposed to be an upper bound anyway. That it was easier when the capturings began. I did think of one interesting example; 24 pieces with 12 pawn promotions (I think that is possible): 2*4^12*(64*(64 - 4)*62!)/((62 - 22)!*(2!*3!*3!*2*2)^2) =5.99*10^43 It is the highest number I can produce;) Depending on what the question is, we can reduce a bit by using the 8 symmetries, or increase by side-to-move and en-passent rights etc. Althought the latter is what is needed for hashing, then when I think about it does seems a bit silly to count a position like this 192 times: [D]r3k2r/8/8/ppp5/5PPP/8/8/R3K2R w KQkq - 0 1 -S.
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