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Subject: Re: Approximate # of Unique chess positions = (current estimated #)/4

Author: Chris Hull

Date: 16:26:08 01/18/02

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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



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