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Math Question

Aquilarion
Nov 26, 2012, 8:40 AM 0

So, I saw another blog title that was asking if chess and math are related.  Being that I graduated with a physics degree from college, I think they are related.  Physics in a nutshell is equating physical things to a mathematical formula.  If you think about it, math is what's called an exact science.  That is, there is one answer.

 

If you have the equation x + y = 5 and you say y = 3, then x has to be equal to 2.  There is no other answer.  If you can take something physical and turn it into a mathematical equation, then you have something that represents exactly what that physical thing is.

 

Being that I come from a physics background, I want to find a way to link chess to math.  The way to do this, is to essentially see how many different positions are possible on the board.

 

Now, this is an extremely difficult task, but I propose starting at the basics:  ignore "impossible" scenarios, no piece can occupy the same space as another piece, and for simplicity sake at the moment, let's have no pawns on the board, because their movement is fairly complex.

 

So how do you go about doing that?  Well, let's start with a blank board, and think about which pieces are limited.  The first one that comes to mind for me is a bishop.  A board has 64 squares on it, but a bishop can only go along one color, so it is limited to only 32 squares, and the other bishop is limited to the other 32 squares.  So, if you were to place a black-squared bishop on the board, you have 32 possibilities, and then if you place the white-squared bishop on the board, you have 32 possibilities again.  So the total number of positions on a chess board with just two bishops of opposite color is:

 

(32) x (32) = 1024

 

So, you can easily see how quickly this number is going to grow, and how many positions can be possible when you start throwing on other pieces.  Staying with the bishops though, what if we throw the other 2 bishops on the board as well?  They can't occupy the same space as the other bishops, so instead of having 32 squares available, they only have 31 squares.  So, with 4 bishops on the board (2 black-squared and 2 white-squared), you get:

 

(32) x (32) x (31) x (31) = 984,064 positions available.

 

That's a lot of possibilities!

 

Let's throw in the other major pieces as well, starting with the kings and queens.  Since the bishops are occupying 4 of the 64 available squares, you can only place the king on 60 squares, and once that king is placed, the other king can only go on 59 squares, and you can see that the two queens can go on 58, and 57 squares.  So, now you get:

 

(32) x (32) x (31) x (31) x (60) x (59) x (58) x (57) = 11,516,737,167,360

 

Geez!  That's a lot!  And we still have the knights and rooks to worry about!  So, let's throw them in to see what happens.  The thing about rooks and knights though, is that there is nothing preventing them from being able to occupy the same space as the other like how there was with the bishops being constrained to one color.

 

Because of this, yes, there are 56 squares remaining to put a rook on the board, and then there would be 55 squares remaining for the other rook.  But let's say you put rook 1 on A1, and you put rook 2 on A2.  Now in another scenario, you put rook 1 on A2, and put rook 2 on A1.  You will have the exact same position, because there is nothing that makes the rooks unique from the other.  Because of this, you have to divide your rook positions by 2, because every position will be doubled.  And the same applies for the knights, so:

 

(32) x (32) x (31) x (31) x (60) x (59) x (58) x (57) x (56) x (55) x (54) x (53) / 4  = 2.5379894 x 10 ^ 19 positions

 

My calculator couldn't display all the digits of it because it's that large of a number!  And this is using the simplified criteria I specified earlier of no pawns, and not taking into account that not all the pieces will always stay on the board.

 

Now, even though the number is very large, it really is just a matter of sitting down and figuring out every possible piece combination, taking into account how the pieces move.  Now, that's all fine and dandy, but here's where the monumental challenge comes into play. Using the criteria above, it is entirely possible that you can come up with a board where a king is in check 6 different ways, and the kings are right next to each other.

 

 

 

As you can clearly see, this is a problem!  So the million-dollar question is, how do you account for the rules of chess?  It is entirely possible to have a position where a king is in check, or even in double check, so you can't just put the constraint in that a piece can't be placed on the board that would put the kings in check.

 

And I still haven't taken into account the fact that there are 16 pawns that weren't placed on the board, those pawns can promote to other pieces, pieces can be captured and removed from the board entirely, and there are positions that physically can not be attained using the rules and movements of the pieces.

 

So, that's where I am currently.  I'm stuck trying to reason out a way to relate the game mechanics to mathematics.  Because there are a set number of pieces and squares on the board, there definitely is a finite number of positions possible, and that's the magic number that I'm trying to find.  If anyone has any ideas, I'd love to hear what you have to say!  I will probably update this at a later time with a new blog about my progress if I can make any.

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