Author: Sune Fischer
Date: 16:49:25 01/18/02
Go up one level in this thread
On January 18, 2002 at 19:26:08, Chris Hull wrote: >On January 18, 2002 at 19:06:26, Chris Hull wrote: > >>On January 18, 2002 at 17:13:29, Sune Fischer wrote: >> >>> >>>Hehe, this was the old: >>>(64*(64 - 4)*62!)/((62 - 30)!*(8!*2!*2!*2*2)^2)= 1.1*10^42 >>>capture a piece and turn a pawn into a queen: >>>(64*(64 - 4)*62!)/((62 - 29)!*7!*8!*2!(2!*2!*2*2)^2)= 1.33*10^41 >>> >>>OK, so its about a factor 10 or maybe about same order of magnitude, they drop >>>fairly quickly though. >>> >>>>Here is the table of numbers Uri's program dumps as it goes (which is the list >>>>by category): >>> >>>I would like to know his approach ;) >>> >>> >>>I'm quite sure it is, unless someone can find a flaw in the product (which there >>>could be of cause;) >>>A number like 64 squares for the king is very optimistic, most of the squares >>>will be attacked, you can't be in check if it isn't your turn, and you can't be >>>in check by more than two pieces (right?) and so on, many rules that will >>>diminish the final product. In particular the pawn movement, pawns have a rather >>>limited number of squares they can go to, if just one pawn has half of 64 >>>squares, then that is a factor 2 smaller yet. >>> >>>-S. >>> >> >>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;) The number fell quite a bit. Now if there are only 31 pieces, the cheapest to remove is an h or a pawn, so lets do that: 31 pieces: 8*(64*(64 - 4)*23^4*21^4*18^4*14^2*62!)/((62 - 13)!*2!*(7!*2!*2!*2*2)^2) =1.36*10^35 The new '8' in front represents: 2 possible colors * 4 possible pieces there is a new factor 2! because there is now two queens (again an upper limit, since 2!*2!<3! if we choose a knight or rook) Same procedure with 30 pieces, I don't see this going above 10^40, but it is starting to get a little tricky to do ;) -S.
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