Author: Uri Blass
Date: 17:24:52 01/18/02
Go up one level in this thread
On January 18, 2002 at 19:49:25, Sune Fischer wrote: >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;) I do not understand it if there are 32 pieces you can say that for every file you have exactly 15 options to put the pawns so you get 15^8 options to put the pawns. I do not understand how do you get 23 squares for the d and e pawns. It is clear that you can reduce the number of possibilities by a big factor with 32 pieces. Things are less clear with less pieces. I guess that 10^40 is close to the right number of legal positions and I am not sure if the real number is lower or bigger. It is going to be hard to prove 10^40 as a bound for the number of legal position and I believe that it may be possible to get an estimate by the monta karlo method of generating random positions and finding the number of legal positions from them. I have ideas to improve my program for the upper bound and it can help in generating random positions but I did not find it as important to waste more time about it. Uri
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