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Subject: Re: number of pawn positions in chess...

Author: Michael Neish

Date: 01:10:33 01/14/00

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On January 13, 2000 at 02:54:31, Timothy J. Frohlick wrote:

>I don't think the "a" pawn is going to end up on the e,f,g,or h files and don't
>forget that the black pawns prevent occupation except by capture of the white
>pawns. You also have to account for en passant which usually occurs only once
>per game at most.
>
>[D] 8/pppppppp/8/8/8/8/PPPPPPPP/8
>
>So the progression should be 20x19x18....etc and be extended out for a total of
>sixteen pawns.  20x19x18x17x16x15x14x13x12x11x10x9x8x7x6x5x4= 8.109673360589e+17
>which is a pretty big number of possibilities.  Even if we limited  the combos
>to (Whitemoves x 10x9x8x7x6x5x4x3) + (Blackmoves x 10x9x8x7x6x5x4x3)=3,628,800
>possibilities you can see that we will not run out of moves soon.  If the
>average game is 75 moves or 150 ply we would have to play at least 24,192 games
>to exhaust the possibilities.  Again, since not every move is a pawn move we
>would have a variable number of pawn moves and positions.  So if only one
>quarter of our moves are pawn moves we would have to play about 100,000 games to
>come to a correct estimation of the number of different pawn structures.
>
>This is just my opinion and is probably wrong.
>
>Tim Frohlick

Very interesting comments.  I never thought of that.  The best I can
do -- for sixteen Pawns, which seems to be the one giving the most
possibilities -- is 15^8, which is 2.563e9, which is a whole lot less
than the "naive" calculation I slogged through in another post of
about 2.9e16 (for 16 Pawns).  For sixteen Pawns my assumption is
that all Pawns are on their original files, although there are
surely many more possiblities if pieces have been captured.

Okay, let's look at some possibilities.  If White has captured just
one piece, creating doubled Pawns on one file, the number of
combinations becomes 3.35e10, by treating each file individually
and multiplying the combinations for each of them.  For one doubled
Black Pawn, it's the same, so just multiply this figure by two to
get the combinations for White and Black.

For one set of doubled Pawns for each side, well, get the computer out.

For doubled White A-Pawns, Black can have Doubled Pawns on files
A to F, and in addition, for files B-G the file to the left or
right of the doubled Pawns can be empty.

For the case of doubled White A-Pawns, we get

White Doubled A, Black Doubled A = 1.025e9
White Doubled A, Black Doubled B, no B-Pawn on A-file = 2.567e9
White Doubled A, Black Doubled B, no B-Pawn on C-file = 4.101e9
White Doubled A, Black Doubled C, no B-Pawn on B-file = 2.734e9
White Doubled A, Black Doubled C, no B-Pawn on D-file = 6.561e9
White Doubled A, Black Doubled D, no B-Pawn on C-file = 6.561e9
White Doubled A, Black Doubled D, no B-Pawn on E-file = 6.561e9
White Doubled A, Black Doubled E, no B-Pawn on D-file = 6.561e9
White Doubled A, Black Doubled E, no B-Pawn on F-file = 6.561e9
White Doubled A, Black Doubled F, no B-Pawn on E-file = 6.561e9
White Doubled A, Black Doubled F, no B-Pawn on G-file = 6.561e9
White Doubled A, Black Doubled G, no B-Pawn on F-file = 6.561e9
White Doubled A, Black Doubled G, no B-Pawn on H-file = 6.561e9
White Doubled A, Black Doubled H, no B-Pawn on G-file = 6.561e9

Which comes to 7.5012e10.

And this is for a doubled White A-Pawns only.  I couldn't
be bothered to write a program to do it for all possibilities
for one set of doubled Pawns for each colour, so if someone
does it could you post the results on here?  Thanks.

By the way, why are we doing this?  Academic reasons or is
someone thinking of setting up a Pawn database?

This is a tricky business!

Cheers,

Mike.





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