2x10^43 (two times ten to the 43rd power) according to the book The Complete Chess Addict.
Mathematics
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I am not sure how do you come to the answer 2*10^43?
Is this the possible settings legaly allowed in chess? Or, it includes all choices?
This is the very first link when I google search for "number of chess positions"
http://en.wikipedia.org/wiki/Shannon_number
Sounds like a difficult homework assignment.
I don`t think its known but its been estimated.
I don`t know how reasonable the estimate is even though it may be so.
I wonder if Grandmaster John Nunn has ever made an estimate since I think he is a Mathematician as well as a GM.
I think this question is not answerable exactly because the problem is too complex. First try to solve the much "simpler" problem: How many different (legal) positions are there with 2 kings and 8 white queens? (all 8 white pawns have been promoted), then solve the other constellations with other pieces. A legal position in the case of 8 queens, e. g., is a position where either the black king is not in check or is in check given by exactly one queen (provided black is to move)... - That looks quite solvable, you might say, but be aware that, for example, a position with a white queen on a1 and a2 and the black king on a8 (with black to move) cannot result from a legal move sequence...
gwhuebner wrote:
I think this question is not answerable exactly because the problem is too complex. First try to solve the much "simpler" problem: How many different (legal) positions are there with 2 kings and 8 white queens? (all 8 white pawns have been promoted), then solve the other constellations with other pieces. A legal position in the case of 8 queens, e. g., is a position where either the black king is not in check or is in check given by exactly one queen (provided black is to move)... - That looks quite solvable, you might say, but be aware that, for example, a position with a white queen on a1 and a2 and the black king on a8 (with black to move) cannot result from a legal move sequence...
I think it is not easy but there is a way and might take time to do.
How would one improve, free of charge, his chess skills?
and by improvement I mean serious improvment.
You need to open a separate thread to get some attention to this question. "Mathematics" just don't cut it.
What is a thread?
Jul 13, 2010
0
#12
Thread is another term for topic.
There is a greater variety of pizzas than there are chess positions. (I have discovered a wonderful proof for this, but the margin is not big enough for me to write it.)
Jul 13, 2010
0
#13
artfizz wrote:
Thread is another term for topic.
There is a greater variety of pizzas than there are chess positions. (I have discovered a wonderful proof for this, but the margin is not big enough for me to write it.)
Jul 13, 2010
0
#14
These previous discussions may shed some light on it ...
how-many-different-chess-positions-are-there
what-chess-position-has-the-most-number-of-possible-moves
possible-positions-after-40-moves
Jul 13, 2010
0
#15
I read randomly one of those threads I think it will no shed any light
The number of positions is in the neighbourhood of the number of atoms in the universe. How can you store all that information? Is it even physically possible?
Not all of the positions are very meaningfull or dificult.
30th December 2009, 06:53pm
#3
Member Since: Sep 2008
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Maybe in the year 3000-4000 something, super computers will be invented that can store that much information. Chess will eventually be solved but not in this century.
Yes but how do you store that much information? Even if you save one position to as small as an atom...
30th December 2009, 06:56pm
#5
Member Since: Feb 2008
Member Points: 3635
Member Points: 3635
duplicate thread...the basic answer is no.
30th December 2009, 06:58pm
gwhuebner wrote:
I think this question is not answerable exactly because the problem is too complex. First try to solve the much "simpler" problem: How many different (legal) positions are there with 2 kings and 8 white queens? (all 8 white pawns have been promoted), then solve the other constellations with other pieces. A legal position in the case of 8 queens, e. g., is a position where either the black king is not in check or is in check given by exactly one queen (provided black is to move)... - That looks quite solvable, you might say, but be aware that, for example, a position with a white queen on a1 and a2 and the black king on a8 (with black to move) cannot result from a legal move sequence...
Well my king is on a7 not a8 but the idea is the same.
Jul 13, 2010
0
#19
Crazychessplaya wrote:
2x10^43 (two times ten to the 43rd power) according to the book The Complete Chess Addict.
This looks like an over-estimate.
Here is one way to break the problem apart in terms of getting a count of how to arrange 32 pieces randomly on the board (note: this does not describe the number of legal positions! Only the ways of randomly selecting positions for all 32 pieces).
First, consider picking 32 squares at random to be considered for occupation by a piece. There are 64-choose-32 ways of doing this (64!/(32! * 32!))
Now, label each of the pieces to be distinct -- so that you can identify each and every piece by a number between 1 and 32. These 32 pieces can be placed into their 32 selected squares in one of 32! permutations.
So now, we have 32! * 64!/(32! * 32!) = 64!/32!
Now, we have to deal with the fact that not all arrangements are unique. For a given unique arrangement, there are 8! ways of arranging the black pawns. Similarly, 8! ways of arranging the white pawns, and 2 ways of arranging each of the white rook, knight, or bishop. Altogether, we get:
64!/(32! * 8! * 8! * 2 * 2 * 2 * 2 * 2 * 2)
=
64!/(32! * 64 * 8!^2)
This figure will tell us precisely how many ways to arrange all 32 pieces on the board randomly. The answer according to Google Search:
(64 !) / ((32 !) * 64 * ((8 !)^2)) = 4.6347267 × 1042
This number looks surprisingly close to the estimate provided in the provided reference. (which gave 2.0 * 10^43, which is only slightly larger).
I'm not sure, but the answer seems still a bit fishy. In any case, it would be highly unlikely that the answer was derived at by means other than the kind of calculation I just performed, but adding more details and constraints.
Jul 13, 2010
0
#20
billprovince wrote:
Crazychessplaya wrote:
2x10^43 (two times ten to the 43rd power) according to the book The Complete Chess Addict.
This looks like an over-estimate.
Here is one way to break the problem apart in terms of getting a count of how to arrange 32 pieces randomly on the board (note: this does not describe the number of legal positions! Only the ways of randomly selecting positions for all 32 pieces).
First, consider picking 32 squares at random to be considered for occupation by a piece. There are 64-choose-32 ways of doing this (64!/(32! * 32!))
Now, label each of the pieces to be distinct -- so that you can identify each and every piece by a number between 1 and 32. These 32 pieces can be placed into their 32 selected squares in one of 32! permutations.
So now, we have 32! * 64!/(32! * 32!) = 64!/32!
Now, we have to deal with the fact that not all arrangements are unique. For a given unique arrangement, there are 8! ways of arranging the black pawns. Similarly, 8! ways of arranging the white pawns, and 2 ways of arranging each of the white rook, knight, or bishop. Altogether, we get:
64!/(32! * 8! * 8! * 2 * 2 * 2 * 2 * 2 * 2)
=
64!/(32! * 64 * 8!^2)
This figure will tell us precisely how many ways to arrange all 32 pieces on the board randomly. The answer according to Google Search:
(64 !) / ((32 !) * 64 * ((8 !)^2)) = 4.6347267 × 1042
This number looks surprisingly close to the estimate provided in the provided reference. (which gave 2.0 * 10^43, which is only slightly larger).
I'm not sure, but the answer seems still a bit fishy. In any case, it would be highly unlikely that the answer was derived at by means other than the kind of calculation I just performed, but adding more details and constraints.
Ah. I see the problem. This was Shannon's computation. The result is exactly the same, except that using Google Calendar loses precision for calculating n!. Shannon apparently gave this calculation as a back-of-the envelope calculation for the total number of positions.
Somebody here already posted the Wikipedia link to Shannon's number, but here it is again for reference. :) http://en.wikipedia.org/wiki/Shannon_number
How many possible settings of chess pieces are there?
That is # of different settings legaly possible from 32 pieces to 2 pieces (only kings)?