How many rectangles do you have on a cylindrical chessboard?

Question

Say I roll up a square chessboard into a cylinder. How many rectangles, if finite, will it have? Provided that no lateral sides of the rectangle must overlap (so you can't basically go round and round and over and over the same rectangle by completing rotations). i.e. you may start and end a rectangle anywhere, but not overlap it over itself. (Top and bottom of a rectangle may, however, overlap). Can you generalise this for any rectangular grid m×n rolled up into a cylinder?

Answer 1

I am assuming that rectangles that occupy the same squares are different here.

You can handle the height and length of the rectangles independently.

You can pick any 2 rows (non-wrapped dimension) and have these be the top and bottom of the rectangle. If the top row is the first row of the board, then there are $8$ choices for the bottom row. If the top row is the second, then there are $7$ choices for the bottom. Therefore we get $8 + 7 +...+1 = 36$ choices for those pairs.

For the other dimension, we are not so limited. The start and end of the rectangle can be at any column so there is just $8 \times 8 = 64$ possibilities.

$36 \times 64 = 2304$ rectangles

Answer 2

Let a chessboard of size $M \times N$.

Suppose we roll it as a cylinder so it has a width of $N$. You have $N$ cases from one boundary to the other and $M$ for a whole circle around it.

Consider all the rectangles with width $n$. We count all the possibilities starting from one side of the cylinder. We can chose $M$ unit height rectangle, $m=1$, but $M$ rectangles also for $m \in\{2,\dots,M\}$ (just take one and rotate it around till your reach the first one again). This can be done till we reach the $(N+1-n)$-th line.

Then the number of rectangles with height $n$ is then $$(N+1-n)M^2=(N+1-n)M^2.$$

Then the total number of rectangle is :

$$\sum_{n=1}^N (N+1-n)M^2 = M(M-1)\sum_{n=1}^N (N+1-n)=M^2\frac{(N+1)N}{2}$$

The last formula is due to the fact that $\sum_{n=1}^N (N+1-n)$ is the sum of $n$ terms of an arithmetic sequence.

So the number of rectangles is : $\frac 1 2 M^2(N+1)N$.