A rectangle is divided into $mn$ smaller rectangles, whose sides are parallel with the big rectangle. Procedure is as follows: you can point to an arbitrary smaller rectangle (rectangle which contains no more rectangles in it) and find out its area. What is the smallest number of procedures I need to find the area a) of the big rectangle; b) every other smaller rectangle?
I probably have an answer to both items: $m + n - 1$: we find out area of all rectangles across to related sides of the big rectangle. It's clear that knowing area of 3 of 4 rectangles in $2 \times 2$ rectangle, we are able to find area of the last small rectangle in $2 \times 2$. Knowing that, we calculate area of all other small rectangles, which solves b), and by taking a sum of their areas we get an area of the big rectangle, which solves a). However, I am struggling to proof that $m + n - 1$ is always the smallest amount of procedures.