General question

In 1979, Helge Tverberg^[1] raised the following question. For two positive integers k, l, what is the smallest number n(k,l) such that, for any family of pairwise-disjoint convex objects in the plane, there exists a straight line that has at least k objects on one side and at least l on the other side?

The following results are known.

• Obviously, n(1,1)=1.
• Hope and Katchalski^[2] proved that n(k,1) ≤ 12(k-1) for all k ≥ 2.
• Villanger^[1] proved that n(2,2) = ∞: he showed an infinite family of pairwise-disjoint segments such that no straight line has two segments in each side. Pach and Tardos^[3] showed a simpler construction using only unit segments, and another construction using only discs (or squares).

Separators for axes-parallel rectangles

Given a set of N=4k disjoint axis-parallel rectangles in the plane, there is a line, either horizontal or vertical, such that at least N/4 rectangles lie entirely to each side of it (thus at most N/2 rectangles are intersected by the separator line).

Optimality

The number of intersected shapes, guaranteed by the above theorem, is O(N). This upper bound is asymptotically tight even when the shapes are squares, as illustrated in the figure to the right. This is in sharp contrast to the upper bound of O(√N) intersected shapes, which is guaranteed when the separator is a closed shape (see previous section).

Moreover, when the shapes are arbitrary rectangles, there are cases in which no line that separates more than a single rectangle can cross less than N/4 rectangles, as illustrated in the figure to the right.^[4]

Proof

Define a 2-fat rectangle as an axis-parallel rectangle with an aspect ratio of at most 2.

Let R₀ be a minimal-area 2-fat rectangle that contains the centers of at least n/3 squares. Thus every 2-fat rectangle smaller than R₀ contains fewer than n/3 squares.

For every t in [0,1), let R_t be a 2-fat rectangle with the same center as R₀, inflated by 1 + t.

• R_t contains R₀, so it contains the centers of at least n/3 squares.
• R_t is less than twice as large as R₀, so it can be covered by two 2-fat rectangles that are smaller than R₀. Each of these 2-fat rectangles contains the centers of less than n/3 squares. Therefore R_t contains the centers of less than 2n/3 squares.

Now it remains to show that there is a t for which R_t intersects at most O(sqrt(n)) squares.

First, consider all the "large squares" – the squares whose side-length is at least ${\displaystyle \operatorname {width} (R_{0})/2{\sqrt {n}}}$. For every t, the perimeter of R_t is at most 2·perimeter(R₀) which is at most 6·width(R₀), so it can intersect at most ${\displaystyle 12{\sqrt {n}}}$ large squares.

Next, consider all the "small squares" – the squares whose side-length is less than ${\displaystyle \operatorname {width} (R_{0})/2{\sqrt {n}}}$.

For every t, define: intersect(t) as the set of small squares intersected by the boundary of R_t. For every t₁ and t₂, if ${\displaystyle |t_{1}-t_{2}|\geq 1/{\sqrt {n}}}$, then ${\displaystyle |\operatorname {width} (R_{t_{1}})-\operatorname {width} (R_{t_{2}})|\geq \operatorname {width} (R_{0})/{\sqrt {n}}}$. Therefore, there is a gap of at least ${\displaystyle \operatorname {width} (R_{0})/2{\sqrt {n}}}$ between the boundary of R_t1 and the boundary of R_t2. Therefore, intersect(t₁) and intersect(t₂) are disjoint. Therefore:

${\displaystyle \sum _{j=0}^{{\sqrt {n}}-1}{|\operatorname {intersect} (j/{\sqrt {n}})|}\leq n}$

Therefore, by the pigeonhole principle there is a certain j₀ for which:

${\displaystyle |\operatorname {intersect} (j_{0}/{\sqrt {n}})|\leq {\sqrt {n}}}$

The separator we look for is the rectangle R_t, where ${\displaystyle t=j_{0}/{\sqrt {n}}}$.^[7]

Application example

Using this separator theorem, we can solve certain problems in computational geometry in the following way:

• Separate the input set of squares to two disjoint subsets;
• Solve the problem on each subset separately;
• Combine the solutions to the two sub-problems and get an approximate solution to the original problem.

Generalizations

The above theorem can be generalized in many different ways, with possibly different constants. For example:

• Instead of squares, the input collection can contain arbitrary fat objects, such as: circles, rectangles with a bounded aspect ratio, etc.
• Instead of two-dimensional shapes in a plane, the input collection can contain objects of any dimension, and they can be situated in a d-dimensional torus.
• Instead of requiring that the shapes in the input collection be disjoint, we can put a weaker requirement, that the collection is:^[5]
  • k-thick, i.e., each point is covered by at most k different shapes.
  • l-k-thick, i.e., each point is covered by at most k different shapes with a size ratio (size of largest shape divided by size of smallest shape) at most l.
  • k-overloaded, i.e., for any subcollection of shapes, the sum of their individual measures is at most k times the measure of their union.
• Instead of a rectangle separator, the separator can be any shape that can be covered by smaller copies of itself.
• Instead of bounding the number of shapes in each side of the separator, it is possible to bound any measure which satisfies certain axioms.^[6]

Optimality

The ratio of 1:2, in the square separator theorem above, is the best that can be guaranteed: there are collections of shapes that cannot be separated in a better ratio using a separator that crosses only O(sqrt(n)) shapes. Here is one such collection (from theorem 34 of ^[5]):

Consider an equilateral triangle. At each of its 3 vertices, put N/3 shapes arranged in an exponential spiral, such that the diameter increases by a constant factor every turn of the spiral, and each shape touches its neighbours in the spiral ordering. For example, start with a 1-by-Φ rectangle, where Φ is the golden ratio. Add an adjacent Φ-by-Φ square and get another golden rectangle. Add an adjacent (1+Φ)-by-(1+Φ) square and get a larger golden rectangle, and so on.

Now, in order to separate more than 1/3 of the shapes, the separator must separate O(N) shapes from two different vertices. But to do this, the separator must intersect O(N) shapes.

Proof sketch

Define the centerpoint of Q as a point o such that every line through it has at most 2n/3 points of Q in each side of it. The existence of a centerpoint can be proved using Helly's theorem.

For a given point p and constant a>0, define Pr(a,p,o) as the probability that a random line through o lies at a distance of less than a from p. The idea is to bound this probability and thus bound the expected number of points at a distance less than a from a random line through o. Then, by the pigeonhole principle, at least one line through o is the desired separator.

Applications

Bounded-width separators can be used for approximately solving the protein folding problem.^[9] It can also be used for an exact sub-exponential algorithm to find a maximum independent set, as well as several related covering problems, in geometric graphs.^[8]