Author: Les Fernandez
Date: 19:26:34 07/11/99
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On July 11, 1999 at 13:09:56, blass uri wrote:
Hello Blass,
Not to long ago several people out here including Dann Corbit began a similar
study, you might have missed the post. In either case Blass I wiould be curious
to know if the positions you generated were not only legal positions but also
unique. In the past post here Dann found what appears to be a rather
tantalizing difference comparing legal positions to unique positions. When the
two are compared there is a tremendous reduction in position count if you go
with the unique positions. I dont remember the num but I am sure you can find
it in past posts here. I seem to remember it being a magnitude difference. If
you Blass have looked at this or hear of anyone who has I would be very greatful
if you could bring it to my attention. Well good work and I am sure more
significant finds will be made. BTW I think Dann and several otheres had taken
it to I beleive ply 10.
Les
>I did a short program in C to calculate an upper bound to the number of legal
>position in chess and I need exact representation of numbers like 64! to have an
>exact bound.
>
>This program run for some minutes and give the number 2.88456157249518055*10^47
>as an upper bound to the number of legal positions in chess if I was not wrong
>in copying the number.
>
>I did not consider side to move, castling, en passant rule and the 50 move rule
>so the real upper bound is bigger but on the other side it is possible to
>improve the bound because there are some inequalities that I did not use.
>
>
>This program calculates an upper bound for the number of positions for every
>material configuration and add the numbers
>
>some explanation about this program:
>
>m[k]=k!=1*2*3...*k
>
>x1 is the number of white pawns
>x2 is the number of white knights
>x3 is the nuber of white bidhops
>x4 is the number of white rooks
>x5 is the number of white queens
>y1 is the the number of black pawns
>y2 is the number of black knights
>y3 is the nuber of black bidhops
>y4 is the number of black rooks
>y5 is the number of black queens.
>
>m[48]/(m[x1]*m[y1]*m[48-x1-y1] is an upper bound for the number of pawns
>configuration in the board(asuuming x1 white pawns and y1 black pawns).
>
>I can improve this program
>
>There are some inequalities that I did not use(for example if x2>2 then
>y1+y2+y3+y4+y5<15)
>
>Here is the program
>
>#include <stdio.h>
>#include <conio.h>
>main()
>{
>clrscr();
>char x1,x2,x3,x4,x5,y1,y2,y3,y4,y5,i;
>long double k,n=0;
>long double m[65];
>m[0]=1;
>for (i=1; i<65; i++)
> m[i]=m[i-1]*i;
>for (x1=0; x1<9; x1++)
> for (x2=0;x2<11-x1;x2++)
> for (x3=0;(x3<11-x1&&x1+x2+x3<13);x3++)
> for (x4=0;(x4<11-x1&&x1+x2+x4<13&&x1+x3+x4<13&&x1+x2+x3+x4<15);x4++)
> for (x5=0;(x5<10-x1&&x1+x2+x5<12&&x1+x3+x5<12&&x1+x4+x5<12&&\
> x1+x2+x3+x5<14&&x1+x2+x4+x5<14&&x1+x3+x4+x5<14&&x1+x2+x3+x4+x5<16)\
> ;x5++)
>{
>k=m[x2]*m[x3]*m[x4]*m[x5];
>for (y1=0; y1<9; y1++)
> for (y2=0;y2<11-y1;y2++)
> for (y3=0;(y3<11-y1&&y1+y2+y3<13);y3++)
> for (y4=0;(y4<11-y1&&y1+y2+y4<13&&y1+y3+y4<13&&y1+y2+y3+y4<15);y4++)
> for (y5=0;(y5<10-y1&&y1+y2+y5<12&&y1+y3+y5<12&&y1+y4+y5<12&&\
> y1+y2+y3+y5<14&&y1+y2+y4+y5<14&&y1+y3+y4+y5<14&&y1+y2+y3+y4+y5<16)\
> ;y5++)
>
> n=n+(m[48]/(m[x1]*m[y1]*m[48-x1-y1]))*(m[64-x1-y1]/\
> (k*m[y2]*m[y3]*m[y4]*m[y5]*m[62-x2-x3-x4-x5-y2-y3-y4-y5]));
>}
>printf("%Lf\n",n);
>}
>
>
>
>
>Uri
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