Author: Sune Fischer
Date: 03:55:11 01/19/02
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On January 19, 2002 at 06:26:17, Uri Blass wrote: >>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 believe that a better upper bound can be achieved by giving every pawn 48 >squares. That isn't possible with 32 pieces, and with 31 I assumed a promotion and 64 squares. My plan was to get a descending sequence, but the promotion factor does blow things up beyond my wildest dreams. >The advantage is that the order of pawns is not important and when you assume >only 24 squares for the d or the e pawns the order of pawns is important. > >I remember that the upper bound of my program(the last number that was posted) >was better than the upper bound by your calculation. What is your lowest upper bound? I get 10^46 when running your code. >> >>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;) > >It is possible to get 12 promotions even with 28 pieces(one capture can lead to >3 promotions) You are correct Sir. But the number can be cut down, if e.g. you choose 9 queens, there is a 9!^2 pulling the other way. The largest weight is achieved with equal number of Q,R,N and B. The number of ways that can be chosen is way lower than 4^16, but not sure by how much. -S.
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