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I have a set of $n$, $d$-dimensional hyperrectangles which may be overlapping in arbitrary ways. I would like to partition the area covered by this set into a set of non-overlapping hyperrectangles. I am not sure of the hardness of this problem, so first and foremost I am interested in any correct algorithm; any notions of optimality would be a bonus.

This problem is related to, may be a special case of or identical to the problem of finding a rectangular covering of a rectilinear polygon, as discussed in the answer to another stack overflow question. One difference is that I am interested in the $n$, not 2 dimensions, and I am unsure of how their results generalise. Secondly, I am starting not with an arbitrary orthogonal polyhedron, but with a set of orthotopes of known dimensions; hopefully this will make it easier.

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Dear @zenna: I removed the deprecated tag 'geometry' from this post. Feel free to add some other more specific and descriptive tags. – Ricardo Andrade Aug 6 '13 at 7:20

Perhaps explore using a Binary Space Partition (BSP). Here are two papers on the topic, from which you can reach more. (Google scholar lists 26 papers that cite the 2001 paper below.)

(1) Viet Hai Nguyen, Peter Widmayer. "Binary space partitions for sets of hyperrectangles." Algorithms, Concurrency and Knowledge. Lecture Notes in Computer Science Volume 1023, 1995, pp 59-72. (Springer link)

(2) Adrian Dumitrescu , Joseph S. B. Mitchell , Micha Sharir. "Binary Space Partitions for Axis-Parallel Segments, Rectangles, and Hyperrectangles" Proc. Sympos. Comput. Geom. 2001. (Siteseer link)

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In response to the zenna's comment, here is another reference that specifically explores cutting overlapping hyperrectangles by "balanced cuts," cuts which can then be employed by a BSP algorithm.

(3) F. d'Amore, V. H. Nguyen, T. Roos, P. Widmayer. "On Optimal Cuts of Hyperrectangles." Computing. 1995, Volume 55, Issue 3, pp 191-206.

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Thanks, but both of these references require that no two rectangles intersect. – zenna Aug 6 '13 at 0:29
I believe that the complexity results require nonoverlapping boxes, but the algorithms function the same for overlapping boxes. You just couldn't be guaranteed the bounds they quote. I'll give another reference... – Joseph O'Rourke Aug 6 '13 at 1:06

I just happen to have written code to do exactly this yesterday. Think in terms of subtracting sets, and think recursively.

For sets $A_1 ... A_n$, compute $A_1' = A_1$, then $A_2' = A_2 \backslash A_1'$, then $A_3' = (A_3 \backslash A_1') \backslash A_2'$, and so on. By a simple inductive argument, $A_1'...A_n'$ are disjoint. If it helps, here's a Haskell function that does this:

disjoint [] = []
disjoint (a:as) = a : disjoint (concatMap (\\ a) as)
where "\\" is a set subtraction operator that returns a list of disjoint sets (to represent a union), and (by Haskell's sectioning rules) "(\\ a)" is a function that subtracts "a" from its input. This function preserves the head of the list of sets, subtracts the head from the rest of them, flattens the resulting list of lists, and applies itself to the result.

For $d$-dimensional sets, think recursively again. A $d$-dimensional rectangle is a cartesian product $I_d \times \prod_{j=1}^{d-1} I_j$, where $I_d$ is, say, a one-dimensional thing like an interval. Thus, you need functions for subtracting intervals and for subtracting two-dimensional products of any kind of set, such as products of two intervals, products of an interval and a product, etc.

Subtracting intervals isn't too hard, but it's a bit tedious to work out all the cases if you track whether endpoints are open or closed.

You can turn the following formula into a function for subtracting products:

$$(I_1 \times I_2) \backslash (J_1 \times J_2) = ((I_1 \backslash J_1) \times J_2) \uplus ((I_1 \cap J_1) \times (I_2 \backslash J_2))$$

Because subtraction can yield a list (i.e. union) of intervals or products, it's a little trickier to implement than just typing it in, but it's not too hard. Haskell code again, in case there are native speakers about:

  PairSet a1 a2 \\ PairSet b1 b2 =
    let cs = do c1 <- a1 \\ b1
                return (PairSet c1 a2)
        ds = do d2 <- a2 \\ b2
                return (PairSet (a1 /\ b1) d2)
      in cs ++ ds
where "/\" computes intersections of arbitrary sets. So if you have subtraction and intersection (which can be defined in terms of subtraction) for intervals and products, something like the above code will recursively subtract rectangular sets of any data shape you have, with $d$-dimensional rectangles (i.e. lists of intervals) as a special case.

Now, "disjoint" is obviously $O(n^2)$, which isn't bad. But each interval subtraction can generate two intervals, so you're looking at exponential time (in $d$) in the worst case. I haven't discovered any worst cases yet, but I haven't tried very hard yet.

The code above distributes lists (i.e. unions) of intervals over products. In some past code, I kept this from happening by representing sets of reals as sorted, disjoint intervals instead of just as intervals, but I've never measured how much it helps. If it does, you could probably do even better by representing sets of pairs as a union of products in a BSP tree, as Joseph suggests. It would basically be a method to compress disjoint unions of rectangles, in a way that doesn't complicate common operations (i.e. product, union, intersect, subtract, measure volume).

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