Concept in computational geometry
In computational geometry, a maximum disjoint set (MDS) is a largest set of non-overlapping geometric shapes selected from a given set of candidate shapes.
Every set of non-overlapping shapes is an independent set in the intersection graph of the shapes. Therefore, the MDS problem is a special case of the maximum independent set (MIS) problem. Both problems are NP complete, but finding a MDS may be easier than finding a MIS in two respects:
Finding an MDS is important in applications such as automatic label placement, VLSI circuit design, and cellular frequency division multiplexing.
The MDS problem can be generalized by assigning a different weight to each shape and searching for a disjoint set with a maximum total weight.
In the following text, MDS("C") denotes the maximum disjoint set in a set "C".
Greedy algorithms.
Given a set "C" of shapes, an approximation to MDS("C") can be found by the following greedy algorithm:
For every shape "x" that we add to "S", we lose the shapes in "N(x)", because they are intersected by "x" and thus cannot be added to "S" later on. However, some of these shapes themselves intersect each other, and thus in any case it is not possible that they all be in the optimal solution "MDS(S)". The largest subset of shapes that "can" all be in the optimal solution is "MDS(N(x))". Therefore, selecting an "x" that minimizes "|MDS(N(x))|" minimizes the loss from adding "x" to "S".
In particular, if we can guarantee that there is an "x" for which "|MDS(N(x))|" is bounded by a constant (say, "M"), then this greedy algorithm yields a constant "M"-factor approximation, as we can guarantee that:
formula_1
Such an upper bound "M" exists for several interesting cases:
1-dimensional intervals: exact polynomial algorithm.
When "C" is a set of intervals on a line, "M"=1, and thus the greedy algorithm finds the exact MDS. To see this, assume w.l.o.g. that the intervals are vertical, and let "x" be the interval with the "highest bottom endpoint". All other intervals intersected by "x" must cross its bottom endpoint. Therefore, all intervals in "N(x)" intersect each other, and "MDS(N(x))" has a size of at most 1 (see figure).
Therefore, in the 1-dimensional case, the MDS can be found exactly in time "O"("n" log "n"): 
This algorithm is analogous to the earliest deadline first scheduling solution to the interval scheduling problem.
In contrast to the 1-dimensional case, in 2 or more dimensions the MDS problem becomes NP-complete, and thus has either exact super-polynomial algorithms or approximate polynomial algorithms.
Fat shapes: constant-factor approximations.
When "C" is a set of unit disks, "M"=3, because the leftmost disk (the disk whose center has the smallest "x" coordinate) intersects at most 3 other disjoint disks (see figure). Therefore the greedy algorithm yields a 3-approximation, i.e., it finds a disjoint set with a size of at least "MDS(C)"/3.
Similarly, when "C" is a set of axis-parallel unit squares, "M"=2.
When "C" is a set of arbitrary-size disks, "M"=5, because the disk with the smallest radius intersects at most 5 other disjoint disks (see figure).
Similarly, when "C" is a set of arbitrary-size axis-parallel squares, "M"=4.
Other constants can be calculated for other regular polygons.
Divide-and-conquer algorithms.
The most common approach to finding a MDS is divide-and-conquer. A typical algorithm in this approach looks like the following:
The main challenge with this approach is to find a geometric way to divide the set into subsets. This may require to discard a small number of shapes that do not fit into any one of the subsets, as explained in the following subsections.
Axis-parallel rectangles with the same height: 2-approximation.
Let "C" be a set of "n" axis-parallel rectangles in the plane, all with the same height "H" but with varying lengths. The following algorithm finds a disjoint set with a size of at least |MDS("C")|/2 in time "O"("n" log "n"):
Axis-parallel rectangles with the same height: PTAS.
Let "C" be a set of "n" axis-parallel rectangles in the plane, all with the same height but with varying lengths. There is an algorithm that finds a disjoint set with a size of at least |MDS("C")|/(1 + 1/"k") in time "O"("n"2"k"−1), for every constant "k"  ; 1.
The algorithm is an improvement of the above-mentioned 2-approximation, by combining dynamic programming with the shifting technique of Hochbaum and Maass.
This algorithm can be generalized to "d" dimensions. If the labels have the same size in all dimensions except one, it is possible to find a similar approximation by applying dynamic programming along one of the dimensions. This also reduces the time to n^O(1/e).
Axis-parallel rectangles: Logarithmic-factor approximation.
Let "C" be a set of "n" axis-parallel rectangles in the plane. The following algorithm finds a disjoint set with a size of at least formula_10 in time 
formula_11:
It is provable by induction that, at the last step, either formula_25 or formula_27 have a cardinality of at least formula_10.
Chalermsookk and Chuzoy have improved the factor to formula_35.
Chalermsook and Walczak have presented an formula_35-approximation algorithm to the more general setting, in which each rectangle has a "weight", and the goal is to find an independent set of maximum total weight.
Axis-parallel rectangles: constant-factor approximation.
For a long time, it was not known whether a constant-factor approximation exists for axis-parallel rectangles of different lengths and heights. It was conjectured that such an approximation could be found using "guillotine cuts". Particularly, if there exists a guillotine separation of axes-parallel rectangles in which formula_37 rectangles are separated, then it can be used in a dynamic programming approach to find a constant-factor approximation to the MDS.
To date, it is not known whether such a guillotine separation exists. However, there are constant-factor approximation algorithms using non-guillotine cuts:
Fat objects with identical sizes: PTAS.
Let "C" be a set of "n" squares or circles of identical size. Hochbaum and Maass presented a polynomial-time approximation scheme for finding an MDS using a simple shifted-grid strategy. It finds a solution within (1 − ε) of the maximum in time "n""O"(1/"ε"2) time and linear space. The strategy generalizes to any collection of fat objects of roughly the same size (i.e., when the maximum-to-minimum size ratio is bounded by a constant).
Fat objects with arbitrary sizes: PTAS.
Let "C" be a set of "n" fat objects, such as squares or circles, of arbitrary sizes. There is a PTAS for finding an MDS based on multi-level grid alignment. It has been discovered by two groups in approximately the same time, and described in two different ways.
Level partitioning.
An algorithm of Erlebach, Jansen and Seidel finds a disjoint set with a size of at least (1 − 1/"k")2 ⋅ |MDS("C")| in time "n""O"("k"2), for every constant "k"  ; 1. It works in the following way.
Scale the disks so that the smallest disk has diameter 1. Partition the disks to levels, based on the logarithm of their size. I.e., the "j"-th level contains all disks with diameter between ("k" + 1)"j" and ("k" + 1)"j"+1, for "j" ≤ 0 (the smallest disk is in level 0).
For each level "j", impose a grid on the plane that consists of lines that are ("k" + 1)"j"+1 apart from each other. By construction, every disk can intersect at most one horizontal line and one vertical line from its level.
For every "r", "s" between 0 and "k", define "D"("r","s") as the subset of disks that are not intersected by any horizontal line whose index modulo "k" is "r", nor by any vertical line whose index modulu "k" is "s". By the pigeonhole principle, there is at least one pair "(r,s)" such that formula_38, i.e., we can find the MDS only in "D"("r","s") and miss only a small fraction of the disks in the optimal solution:
Shifted quadtrees.
An algorithm of Chan finds a disjoint set with a size of at least (1 − 2/"k")⋅|MDS("C")| in time "n""O"("k"), for every constant "k"  ; 1.
The algorithm uses shifted quadtrees. The key concept of the algorithm is "alignment" to the quadtree grid. An object of size "r" is called "k-aligned" (where "k" ≥ 1 is a constant) if it is inside a quadtree cell of size at most "kr" ("R" ≤ "kr").
By definition, a "k"-aligned object that intersects the boundary of a quatree cell of size "R" must have a size of at least "R"/"k" ("r" ; "R"/"k"). The boundary of a cell of size "R" can be covered by 4"k" squares of size "R"/"k"; hence the number of disjoint fat objects intersecting the boundary of that cell is at most 4"kc", where "c" is a constant measuring the fatness of the objects.
Therefore, if all objects are fat and "k"-aligned, it is possible to find the exact maximum disjoint set in time "n""O"("kc") using a divide-and-conquer algorithm. Start with a quadtree cell that contains all objects. Then recursively divide it to smaller quadtree cells, find the maximum in each smaller cell, and combine the results to get the maximum in the larger cell. Since the number of disjoint fat objects intersecting the boundary of every quadtree cell is bounded by 4"kc", we can simply "guess" which objects intersect the boundary in the optimal solution, and then apply divide-and-conquer to the objects inside.
If "almost" all objects are "k"-aligned, we can just discard the objects that are not "k"-aligned, and find a maximum disjoint set of the remaining objects in time "n""O"("k"). This results in a (1 − "e") approximation, where e is the fraction of objects that are not "k"-aligned.
If most objects are not "k"-aligned, we can try to make them "k"-aligned by "shifting" the grid in multiples of (1/"k",1/"k"). First, scale the objects such that they are all contained in the unit square. Then, consider "k" shifts of the grid: (0,0), (1/"k",1/"k"), (2/"k",2/"k"), ..., (("k" − 1)/"k",("k" − 1)/"k"). I.e., for each "j" in {0...,"k" − 1}, consider a shift of the grid in (j/k,j/k). It is possible to prove that every label will be 2"k"-aligned for at least "k" − 2 values of "j". Now, for every "j", discard the objects that are not "k"-aligned in the ("j"/"k","j"/"k") shift, and find a maximum disjoint set of the remaining objects. Call that set "A"("j"). Call the real maximum disjoint set is "A"*. Then:
formula_39
Therefore, the largest "A"("j") has a size of at least: (1 − 2/"k")|"A"*|. The return value of the algorithm is the largest "A"("j"); the approximation factor is (1 − 2/"k"), and the run time is "n""O"("k"). We can make the approximation factor as small as we want, so this is a PTAS.
Both versions can be generalized to "d" dimensions (with different approximation ratios) and to the weighted case.
Geometric separator algorithms.
Several divide-and-conquer algorithms are based on a certain geometric separator theorem. A geometric separator is a line or shape that separates a given set of shapes to two smaller subsets, such that the number of shapes lost during the division is relatively small. This allows both PTASs and sub-exponential exact algorithms, as explained below.
Fat objects with arbitrary sizes: PTAS using geometric separators.
Let "C" be a set of "n" fat objects, such as squares or circles, of arbitrary sizes. Chan described an algorithm finds a disjoint set with a size of at least (1 − "O"(√"b"))⋅|MDS("C")| in time "n""O"("b"), for every constant "b"  ; 1.
The algorithm is based on the following geometric separator theorem, which can be proved similarly to the proof of the existence of geometric separator for disjoint squares:
For every set "C" of fat objects, there is a rectangle that partitions "C" into three subsets of objects – "Cinside", "Coutside" and "C"boundary, such that:
* |MDS("C"inside)| ≤ "a"|MDS("C")|
* |MDS("C"outside)| ≤ a|MDS("C")|
* |MDS("C"boundary)| "c"√
where "a" and "c" are constants. If we could calculate MDS("C") exactly, we could make the constant "a" as low as 2/3 by a proper selection of the separator rectangle. But since we can only approximate MDS("C") by a constant factor, the constant "a" must be larger. Fortunately, "a" remains a constant independent of |"C"|.
This separator theorem allows to build the following PTAS:
Select a constant "b". Check all possible combinations of up to "b" + 1 labels.
Let "E"("m") be the error of the above algorithm when the optimal MDS size is MDS("C") = "m". When "m" ≤ "b", the error is 0 because the maximum disjoint set is calculated exactly; when "m"  ; "b", the error increases by at most "c"√"m" the number of labels intersected by the separator. The worst case for the algorithm is when the split in each step is in the maximum possible ratio which is "a":(1 − "a"). Therefore the error function satisfies the following recurrence relation:
 formula_40
 formula_41
The solution to this recurrence is:
 formula_42
i.e., formula_43. We can make the approximation factor as small as we want by a proper selection of "b".
This PTAS is more space-efficient than the PTAS based on quadtrees, and can handle a generalization where the objects may slide, but