Finding overlaps between two lists of axis-aligned rectangles

Question

I am trying to find an efficient solution for finding overlapping of n rectangles where rectangles are stored in two separate lists. We are looking for all rectangles in listA that overlap with rectangles in listB (and vice versa). Comparing one element from the first list to second list could take immensely large amount of time. I am looking for an efficient solution.

I have two lists of axis-aligned rectangles:

rect = Rectangle(10, 112, 56, 15)
rect2 = Rectangle(0, 0, 1, 15)
rect3 = Rectangle (10, 12, 56, 15)

listA = [rect, rect2]
listB= [rect3]

which is created from the class:

import numpy as np
import itertools as it


class  Rectangle(object):
    def __init__(self, left, right, bottom, top):
        self. left = left
        self.bottom = bottom
        self.right = right
        self.top = top

    def overlap(r1, r2):
        hoverlaps = True
        voverlaps = True
        if r1.left > r2.right or r1.right < r2.left:
            hoverlaps = False
        if r1.top < r2.bottom or r1.bottom > r2.top:
            voverlaps = False
        return hoverlaps and voverlaps

I need to compare rectangle in listA to listB the code goes like this which is highly inefficient - comparing one by one

for a in it.combinations(listB) :
    for b in it.combinations(listA):
        if a.overlap(b):

Any better efficient method to deal with the problem?

Answer 1

Dec. 10, 2018

The simple answer is to use a 2-d segment tree.

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Given that we have low value for dimension (i.e. two), an appropriate approach would be to use a 2-d segment tree. We use what is known as stabbing query or intersection query. Let list one be the list that is smaller. Store all rectangles from list one in such a tree. Query using rectangles from list two. Time to construct and query are \$O(n_1 \cdot \log^2(n_1))\$ and \$O(\log^2(n_1) \cdot k)\$, respectively, s.t. \$k\$ is number of matches to report. If we use fractional cascading and interval tree, times become \$O(n_1 \cdot \log(n_1))\$ and \$O(\log(n_1) \cdot k)\$, respectively. Assuming these better times, time for all queries is \$O(\textrm{max}(n_2, k_\textrm{overall}) \cdot \log(n_1))\$. Overall time becomes \$O((\textrm{max}(n_2, k_\textrm{overall}) + n_1) \cdot \log(n_1))\$. Given that \$k_\textrm{overall}\$ is in \$O(n_1 \cdot n_2)\$, time could be loosely \$O((n_1 \cdot n_2) \cdot \log(n_1))\$.

If rectangles are not axis-aligned, one might wish to have bounding boxes and then use a tree. In principle we then could use a segment tree again, but a natural alternative (though we usually do not have good guaranteed theoretical running time for it unless we use "look-ahead") could be an R-tree.