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Re: chessboard rectangles
Posted:
Oct 11, 1998 2:13 PM
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Sarah,
I agree with your total (1296). With a little rearranging your method
will produce the general formula for the number of rectangles on an n by n
chessboard.
Cont the rectangles by rows.
In row 1 there are n 1 by 1s , n-1 1 by 2s, ... 2 1 by n-1s and 1 1 by n:
Row Total = n + (n-1) + (n-2) + ... + 3 + 2 + 1 = n(n+1)/2 rectangles.
But of course there are n rows giving n ( row sum )
Now count all rectangels of height 2 . Start in the bottom row.
There are n 2 by 1s and n-1 2 by 2s and ... and 1 2 by n for a
Row Total = n + (n-1) + (n-2) + ... + 3 + 2 + 1 = n (n+1)/2
The row total is the same, as it will be for all the rectangles of height
3, 4, ... n because they all share the same bases at the bottom of the
board. However, there are only n-1 rows of height 2 and n-2 rows of
height 3 etc. Thus,
# 1 by any totals: n [n(n+1)/2]
# 2 by any totals: (n-1)[n(n+1)/2]
# 3 by any totals: (n-2) [n(n+1)/2]
...
#n by any totals: 1 [n(n+1)/2]
So the total number of rectangles is [n(n+1)/2](n + n-1) + ... + 3+ 2+1)
or Total = [n(n+1)/2]^2
which, as you mentioned, is the sum of the cubes from 1 to n.
Alan Lipp
alipp@crocker.com
-------------------------------------------------
On Sun, 11 Oct 1998, Sarah Seastone wrote:
>
> Someone asked the number of rectangles on a chessboard. The answer is
> "Many, in fact 1296."
>
...
> Have I missed any? Or added wrong? What if we reduce or augment the
> size of the chessboard? Can we find a general formula? What is it? How
> do we use induction to show that it is indeed general?
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