Author: Michael Neish
Date: 23:46:30 01/13/00
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Hi,
About the number of Pawn positions, I think you can get an
exact upper bound in this way (excuse me if someone already
said this and I missed it):
We ignore illegal positions, and so the number we will get
is an upper limit. We also ignore the first and last ranks
(is it correct to do so?). Also I'm not taking into account
mirror images or whatever. So this is an absolute upper
bound (for what it's worth). I have no idea how important
illegal positions are, and such like. Maybe someone can
enlighten me. Pardon the "naive" analysis.
There are 48 squares that a Pawn can occupy.
Starting with all 16 Pawns on the board, the number of
combinations are
48!
----------- = 2.902e16
32! 8! 8!
i.e., the factorial of the number of possible squares
divided by the factorial of the number of empty squares,
and the factorial of the number of White and Black Pawns
separately.
Now we consider 15 Pawns, 8 White, 7 Black (and 7 White,
8 Black, which is mathematically the same). The number
of combinations is:
48!
----------- x 2 = 7.035e15
33! 8! 7!
For fourteen Pawns there are more possibilities, 8W6B,
7W7B, 6W8B
So for this we get
48! 48!
----------- x 2 + ----------- = 4.552e15
34! 8! 6! 34! 7! 7!
Thirteen Pawns (8W5B, 7W6B, 6W7B, 5W8B):
48! 48!
----------- x 2 + ----------- x 2 = 8.277e14
35! 8! 5! 35! 7! 6!
Twelve Pawns (8W4B, 7W5B, 6W6B, 5W7B, 4W8B)
48! 48! 48!
----------- x 2 + ----------- x 2 + -----------= 2.437e14
36! 8! 4! 36! 7! 5! 36! 6! 6!
and so on.
Eleven Pawns (8W3B, 7W4B, 6W5B, 5W6B, 4W7B, 3W8B)
= 4.324e13
Ten Pawns (8W2B, 7W3B, 6W4B, 5W5B, 4W6B, 3W7B, 2W8B)
= 6.554e12
Nine Pawns (8W1B, 7W2B, 6W3B, 5W4B, 4W5B, 3W6B, 2W7B,
1W8B)
= 8.553e11
Eight Pawns (8W0B, 7W1B, 6W2B, 5W3B, 4W4B, 3W5B,
2W6B, 1W7B, 0W8B)
= 9.660e10
We could probably stop calculating here, because the
numbers start getting small, but anyway:
Seven Pawns (7W0B, 6W1B, 5W2B, 4W3B, 3W4B, 2W5B,
1W6B, 0W7B)
= 9.425e9
Six Pawns (6W0B, 5W1B, 4W2B, 3W3B, 2W4B, 1W5B,
0W7B)
= 7.854e8
Five Pawns (5W0B, 4W1B, 3W2B, 2W3B, 1W4B, 0W5B)
= 53,081,424
Four Pawns (4W0B, 3W1B, 2W2B, 1W3B, 0W4B)
= 3,113,280
Three Pawns (3W0B, 2W1B, 1W2B, 0W3B)
= 138,368
Two Pawns (2W0B, 1W1B, 0W2B)
= 4512
One Pawn (1W0B, 0W1B)
= 48
Which gives a total of about 4,173e16. No need to
Calculate more than just the first couple of cases!
If we assume White/Black mirror images, then it's
about half this value, though not exactly I think.
I did this by hand so I may have made a typo on my
calculator.
I'd like to hear some comment about how many of these
combinations are expected to be illegal.
Thanks,
Mike.
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