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## Chess and a Half

Posted by Karl Sharman on May 28, 2011 – 6:33 am

If you were to construct a 8 × 12 chequered square (i.e., an 8 × 12 chess board), how many rectangles would there be in total?

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This post is under “Tom” and has 16 respond so far.

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12 * 8 i.e 96

2. 2. jaymie Said：

97 rectangles includingg the board it self

3. 3. Acleron Said：

My best guesses are
1) if rectangles only and joining squares together not allowed=1
2) all rectangles and squares =2448
3) rectangles only =2100
Can’t find an elegant method, but will be fascinated to see one

4. 4. xoe Said：

0 (zero) rectangles as it is made up of squares.

5. 5. kelly_rocks Said：

0 duh

6. 6. Chris Said：

Let the board be W by D (12 by 8 in this case). For a rectangle of size w by d, then for a given d, w can range from W to 1. For a given w there are W -w +1 locations, altogether that gives 1+2+3+…+W = W(W+1)/2 variations. The same for D. So the total number of rectangles is W(W+1)D(D+1)/4. Using W=12, D=8 => 2808 rectangles (including squares).

7. 7. Chris Said：

Calculating the squares only is a little harder. Orient the board so that W ≥ D. Then for each d we have (D-d+1) positions vertically and (W-d+1) horizontally. Altogether that’s Σd=1D (D-d+1)(W-d+1).
But W = D+(W-D), so Σd=1D (D-d+1)(D-d+1 + (W-D))
= Σd=1D ((D-d+1)² +(D-W)(D-d+1)) = (1+2²+3²+…+D²)+(W-D)(1+2+3+…+D)

Standard formula: 1+2+3+…+n = n(n+1)/2, I’d actually used that in my last post.
Standard formula: 1+2²+3²+…+n² = n(n+1)(2n+1)/6.

So the total number of squares is D(D+1)(2D+1)/6 +(W-D)D(D+1)/2
= D(D+1)(3W-D+1)/6

In the problem W=12,D=8 => 348 squares
Using the previous 2808 for mixed => 2460 non-square rectangles

8. 8. Chris Said：

Here’s a nicer way to calculate the number of rectangles (I nearly did it this way originally). For a W by D board there are W+1 and D+1 lines defining the squares. We need to choose 2 at a time (for each direction). So we get C(W+1,2)*C(D+1,2) where C(m,n) = m!/((m-n)! n!). This also gives 2808 rectangles.

In fact we get ((W+1)!/((W-1)! 2!)) ((D+1)!/((D-1)! 2!) = W(W+1)D(D+1)/4

9. 9. Alejandra Said：

I am guessing its 172 rectangles.

10. 10. bobbie Said：

96

11. 11. Karl Sharman Said：

I’m not going to post the answer yet, but I shall offer a little assistance for those that haven’t worked this one out yet – the definition of a rectangle:-

A rectangle is a four-sided polygon (a flat shape with straight sides) where every angle is a right angle (90°). Also opposite sides are parallel and of equal length. A square is a special type of rectangle.

Chris – orienting the board for counting squares in post 7 makes the maths easier doesn’t it! Kinda ruined the follow up question, mind!

12. 12. Saurabh Said：

The answer is 36*78 = 2808.

36 = sum of numbers upto 8.
78 = sum of numbers upto 12.

So, for a board of W X D, rects will be Summation(W) * Summation (D).

Where Summation(D) = 1 + 2 + 3…. D.

so, answer comes to as chris pointed in post 8 as W(W+1)D(D+1)/4.

Thanks & Regards,
Saurabh

13. 13. Kayla Said：

i just printed out a chess board (since that is the one example u used) and counted my self with no big equations just counting and adding and tracing an came up with 200 posible rectangles/ squares! id just like someone to tell me if i am right or wrong on this one. i am the only one to guess this number…

14. 14. Chris Said：

Hi Kayla. Just the 1 by 1 squares gets you to 96. 1 by 2 plus 2 by 1 adds another 172 to give 268 so far. There’s still many to go.

For a regular 8×8 board you get 1296 rectangles of which 204 are squares. That’s 1092 non-square rectangles.

15. 15. Karl Sharman Said：

Firstly, a square is a type of rectangle….

Chris scores in post 6 with the first correct answer.

All the rectangles on the board can be identified by connecting 2 points of each of the top corners of the 9 in the top edge (to form the length of the rectangle) and 2 points of the 13 in the left edge (to form the breadth of the rectangle).

The following table shows the number of possibilities for different lengths of the rectangles on a 8 × 12 board:
Length of rectangle Number of Possibilities
8 units 1
7 units 2
6 units 3
… …
1 unit 8

Height of rectangle Number of Possibilities
12 units 1
11 units 2
10 units 3
… …
1 unit 12

So, number of possibilities for different lengths of rectangles = 1 + 2 + 3 + … + 8 = 36.
Similarly, number of possibilities for different breadths of rectangles = 1 + 2 + 3 + … + 12 = 78.
Hence, number of rectangles = 36 × 78 = 2808.

16. 16. cazayoux Said：

n=8; m=12
Sum [i = 1 to n; j = 1 to m] (n – (i-1))*(m – (j-1))
= 2808

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