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Puzzle #27: Number of Squares on a Chessboard

27. How many squares are there on a chessboard or checkerboard?? (the answer is not 64)

Can you extend your technique to calculate the number of rectangles on a chessboard?

This another puzzle that was also e-mailed to me through this website. My instinct was that the answer was just a lot, but I thought about it and the solution is actually fairly simple...

Before reading the answer can I interest you in a clue?

The first thing is why the answer is not just 64...

examples of valid squares on a chessboard

All the red squares in the above picture would count as valid squares, so we are asking how many squares of any dimension from 1x1 to 8x8 there are on a chess board.

The key is to think how many positions there are that each size of square can be located...

A 2x2 square, for example, can be located in 7 loactions horizontally and 7 locations vertically. ie in 49 different positions. Consider the table below...

size horizontal positions vertical positions positons
1x1 8 8 64
2x2 7 7 49
3x3 6 6 36
4x4 5 5 25
5x5 4 4 16
6x6 3 3 9
7x7 2 2 4
8x8 1 1 1
    total 204


In total there are 204 positions. this is the sum of the number of possible positions for all the different sized squares.

Formula For n x n Chessboard?

It's clear from the analysis above that the solution in the case of n x n is the sum of the squares from n2 to 12 that is to say n2 + (n-1)2 + (n-2)2 ... ... 22 + 12

Mathematically that is written as follows: sum of the squares of the first n natural numbers sigma notation

The proof of the explicit solution is beyond the scope of this site, but if you want to look it up a mathematician would refer to it as 'the sum of the squares of the first n natural numbers.' The final answer is given by

n3/3 + n2/2 + n/6

Can you extend your technique to calculate the number of rectangles on a chessboard?

Below are some examples of possible rectangles...

examples of valid rectangles on a chessboard All of the above examples would be vailid rectanges...

There is more than one way of solving this. But it makes sense to extend our technique from the squares problem first. The key to this is to think of each rectangle individually and consider the number of positions it can be located. For example a 3x7 rectangle can be located in 6 positions horizontally and 2 vertically. From this we can build a matrix of all the possible rectangles and sum.

Dimensions   1 2 3 4 5 6 7 8  
  Positions 8 7 6 5 4 3 2 1  
1 8 64 56 48 40 32 24 16 8  
2 7 56 49 42 35 28 21 14 7  
3 6 48 42 36 30 24 18 12 6  
4 5 40 35 30 25 20 15 10 5  
5 4 32 28 24 20 16 12 8 4  
6 3 24 21 18 15 12 9 6 3  
7 2 16 14 12 10 8 6 4 2  
8 1 8 7 6 5 4 3 2 1  
                    1296


In total then there are 1296 possible rectangles.

Elegant approach to rectangles, consider the vertices and diagonals.

demonstration of vertices on a chess board I've been sent an innovative solution to the problem of the number of rectangles on a chessboard by Kalpit Dixit. This solution tackles the issue from a different approach. Rather than looking at specific sizes of rectangles and working out where they can be located we start at the other end and look at locations first.

The vertices are the intersections. For our chessboard there are 81 (9 x 9). A diagonal starting at one vertex and ending at another will uniquely describe a rectangle. In order to be a diagonal and not a vertical or horizontal line we may start anywhere but the end point must not have the same vertical or horizontal coordinate. As such there are 64 (8 x 8) possible end points.

There are therefore 81 x 64 = 5184 acceptable diagonals.

However, whilst each diagonal describes a unique rectangle, each rectangle does not describe a unique diagonal.

diagonals for rectagles, there are four diagonals that describe the same rectangle
We see trivially that each rectangle can be represented by 4 diagonals.

So our number of rectangles is given by 81 x 64 /4 = 1296

n x n or n x m?

The n x n or n x m problems can now be calculated. The number of vertices being given by (n + 1)2 and (n + 1).(m + 1) respectively. Hence the final solutions are as follows.

n x n: (n + 1)2 x n2 / 4
n x m: (n + 1) x (m + 1) x (n x m) / 4

Which can obviously be arranged into something more complicated.

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