Question

In this problem r is a fixed positive integer. You are given N rectangles, all the same size, in the plane. The sides are either vertical or horizontal. We assume the area of the intersection of all N rectangles has non-zero area. The problem is how to find N-r of these rectangles, so as to maximize the area of the intersection. This problem arises in practical microscopy when one repeatedly images a given biological specimen, and alignment changes slightly during this process, due to physical reasons (e.g. differential expansion of parts of the microscope and camera). I have expressed the problem for dimension d=2. There is a similar problem for each d>0. For d=1, an O(N log(N)) solution is obtained by sorting the lefthand endpoints of the intervals. But let’s stick with d=2. If r=1, one can again solve the problem in time O(N log(N)) by sorting coordinates of the corners.

So, is the original problem solved by solving first the case (N,1) obtaining N-1 rectangles, then solving the case (N-1,1), getting N-2 rectangles, and so on, until we reduce to N-r rectangles? I would be interested to see an explicit counter-example to this optimistic attempted procedure. It would be even more interesting if the procedure works (proof please!), but that seems over-optimistic.

If r is fixed at some value r>1, and N is large, is this problem in one of the NP classes?

Thanks for any thoughts about this.

David

Answer 1

Since the intersection of axis-aligned rectangles is an axis-aligned rectangle, there are O(N⁴) possible intersections (O(N) lefts, O(N) rights, O(N) tops, O(N) bottoms). The obvious O(N⁵) algorithm is to try all of these, checking for each whether it’s contained in at least N – r rectangles.

An improvement to O(N³) is to try all O(N²) intervals in the X dimension and run the 1D algorithm in the Y dimension on those rectangles that contain the given X-interval. (The rectangles need to be sorted only once.)

How large is N? I expect that fancy data structures might lead to an O(N² log N) algorithm, but it wouldn’t be worth your time if a cubic algorithm suffices.