Yes.
How many rectangles on a chessboard?
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Hold on, I'm still drawing them...
FancyKnight wrote:
As a sequel to the "How many squares on a chessboard" thread, a much more interesting question is the number of possible rectangles on a chessboard, or rather, the method to arrive at this number. This includes squares as well.
The best way I can think of is to seperate these rectangles into three groups:
a) 1x1 squares
b) 1xN rectangles, 1<N<=8
c) Nx1 rectangles, this should be the exact same number as in b)
d) MxN rectangles, 1<N,M<=8
The reason for this split is that i thought of a nice way to get group d):
I believe that every d) rectangle is uniquely defined by a pair of opposite corner squares.
The amount of possible such pairs is 64*49/2=1568,
with the reasoning of picking the first square, and then having 49 squares left to pick which are not on the same rank/file.
Lastly, each rectangle in d) has two possible diagonal pairs which uniquely define it.
Therefore, we have 1568/2=784 rectangles in group d).
That leaves the b)/c) symmetrical groups.
What do you think?
Because you already counted the nxn (squares) don't you also have to not count diagonal. e.g. not on the same rank, file, or diagonal, so it would be 42 instead of 49 if I'm following you... Although I don't understand part d "opposite corner squares" thing.
I did something similar. I counted all squares first. Then like you noticed 1x8 will have the same number as 8x1.
Then basically I just clumsily added 8x1, 8x2, 8x3... 7x1, 7x2, 7x3... etc.
I've forgotten how to simplify n+(n+1)+(n+2)... type series :p If you know how and you note the pattern I'm sure you can simplify it. In fact if you google you can find a general equation for the rectangles of an nxn board.
I was thinking that each pair of squares not on the same rank or file correspond to one unique rectangle in d) by using those squares as opposite corner squares of the rectangle. On the other hand every unique rectangle in d) has two possible pairs of opposite corner squares.
Oh, I can see what you mean by opposite corners now. Don't know why that didn't make sense to me before :p
Gentlemen, the problem is simpler than you think and I shall solve it for d dimensions and n "squares" for each Dimension.
The length of one of the edges of the considered rectangle is independent of the length in the other Dimensions. Therefore, we can solve the one Dimension problem and generalize to d Dimensions.
Solving for 1 Dimension: (I shall use LaTeX notation, just copy/paste into http://www.codecogs.com/latex/eqneditor.php)
No. rectangles = \sum_{i=1}^{n} i = \frac{1}{2}n(n+1)
For d Dimensions we only need to power the number to the number of Dimensions
No. rectangles(d-Dimensions) = \left(\sum_{i=1}^{n} i = \frac{1}{2}n(n+1)\right)^d
For d=2 and n=8, which is our problem, we have
No. rectangles = \frac{1}{2}8(8+1)\right)^2 = 36^2 = 1296
(correct me if I am wrong, but I think it is that simple)
Stay cool
Easiest method.
Whenever two vertical lines and two horizontal lines are taken, a rectangle is formed. There are 9 vertical and 9 horizontal lines in a chess board. Hence 9C2 * 9C2 = 1296
Jan 6, 2016
0
#9
koshyaria wrote:
Easiest method.
Whenever two vertical lines and two horizontal lines are taken, a rectangle is formed. There are 9 vertical and 9 horizontal lines in a chess board. Hence 9C2 * 9C2 = 1296
That is the best solution of this I have ever seen :)
Very nice.
the answer would be 1296 or if you want to be picky and exclude the amount of squares that would then go down to 1232 - have a look here
https://chesslovin.com/how-many-squares-on-the-perimeter-of-a-chess-board/#How-many-Rectangles-are-on-a-chess-board - theres even a table with a list of how many rectangles of each size will fit on 8 x 8 board
As a sequel to the "How many squares on a chessboard" thread, a much more interesting question is the number of possible rectangles on a chessboard, or rather, the method to arrive at this number. This includes squares as well.
The best way I can think of is to seperate these rectangles into three groups:
a) 1x1 squares
b) 1xN rectangles, 1<N<=8
c) Nx1 rectangles, this should be the exact same number as in b)
d) MxN rectangles, 1<N,M<=8
The reason for this split is that i thought of a nice way to get group d):
I believe that every d) rectangle is uniquely defined by a pair of opposite corner squares.
The amount of possible such pairs is 64*49/2=1568,
with the reasoning of picking the first square, and then having 49 squares left to pick which are not on the same rank/file.
Lastly, each rectangle in d) has two possible diagonal pairs which uniquely define it.
Therefore, we have 1568/2=784 rectangles in group d).
That leaves the b)/c) symmetrical groups.
What do you think?