Intersection of N rectangles

Question

I'm looking for an algorithm to solve this problem:

Given N rectangles on the Cartesian coordinate, find out if the intersection of those rectangles is empty or not. Each rectangle can lie in any direction (not necessary to have its edges parallel to Ox and Oy)

Do you have any suggestion to solve this problem? :) I can think of testing the intersection of each rectangle pair. However, it's O(N*N) and quite slow :(

Answer 1

Observation 1: given a polygon A and a rectangle B, the intersection A ∩ B can be computed by 4 intersection with half-planes corresponding to each edge of B.

Observation 2: cutting a half plane from a convex polygon gives you a convex polygon. The first rectangle is a convex polygon. This operation increases the number of vertices at most per 1.

Observation 3: the signed distance of the vertices of a convex polygon to a straight line is a unimodal function.

Here is a sketch of the algorithm:

Maintain the current partial intersection D in a balanced binary tree in a CCW order.

When cutting a half-plane defined by a line L, find the two edges in D that intersect L. This can be done in logarithmic time through some clever binary or ternary search exploiting the unimodality of the signed distance to L. (This is the part I don't exactly remember.) Remove all the vertices on the one side of L from D, and insert the intersection points to D.

Repeat for all edges L of all rectangles.

Answer 2

Abstract

Either use a sorting algorithm according to smallest X value of the rectangle, or store your rectangles in an R-tree and search it.

Straight-forward approach (with sorting)

Let us denote low_x() - the smallest (leftmost) X value of a rectangle, and high_x() - the highest (rightmost) X value of a rectangle.

Algorithm:

Sort the rectangles according to low_x().                   # O(n log n)

For each rectangle in sorted array:                         # O(n)
    Finds its highest X point.                              # O(1)
    Compare it with all rectangles whose low_x() is smaller # O(log n)
        than this.high(x)

Complexity analysis

This should work on O(n log n) on uniformly distributed rectangles.

The worst case would be O(n^2), for example when the rectangles don't overlap but are one above another. In this case, generalize the algorithm to have low_y() and high_y() too.

Data-structure approach: R-Trees

R-trees (a spatial generalization of B-trees) are one of the best ways to store geospatial data, and can be useful in this problem. Simply store your rectangles in an R-tree, and you can spot intersections with a straightforward O(n log n) complexity. (n searches, log n time for each).

Answer 3

This seems like a good application of Klee's measure. Basically, if you read http://en.wikipedia.org/wiki/Klee%27s_measure_problem there are lower bounds on the runtime of the best algorithms that can be found for rectilinear intersections at O(n log n).

Answer 4

I think you should use something like the sweep line algorithm: finding intersections is one of its applications. Also, have a look at answers to this questions

Answer 5

Since the rectangles must not be parallel to the axis, it is easier to transform the problem to an already solved one: compute the intersections of the borders of the rectangles.

• build a set S which contains all borders, together with the rectangle they're belonging to; you get a set of tuples of the form ((x_start,y_start), (x_end,y_end), r_n), where r_n is of course the ID of the corresponding rectangle
• now use a sweep line algorithm to find the intersections of those lines

The sweep line stops at every x-coordinate in S, i.e. all start values and all end values. For every new start coordinate, put the corresponding line in a temporary set I. For each new end-coordinate, remove the corresponding line from I.

Additionally to adding new lines to I, you can check for each new line whether it intersects with one of the lines currently in I. If they do, the corresponding rectangles do, too.

You can find a detailed explanation of this algorithm here.

The runtime is O(n*log(n) + c*log(n)), where c is the number of intersection points of the lines in I.

Answer 6

Pick the smallest rectangle from the set (or any rectangle), and go over each point within it. If one of it's point also exists in all other rectangles, the intersection is not empty. If all points are free from ALL other rectangles, the intersection is empty.