Английская Википедия:Ham sandwich theorem

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In mathematical measure theory, for every positive integer Шаблон:Mvar the ham sandwich theorem states that given Шаблон:Mvar measurable "objects" in Шаблон:Mvar-dimensional Euclidean space, it is possible to divide each one of them in half (with respect to their measure, e.g. volume) with a single Шаблон:Math-dimensional hyperplane. This is even possible if the objects overlap.

It was proposed by Hugo Steinhaus and proved by Stefan Banach (explicitly in dimension 3, without taking the trouble to state the theorem in the Шаблон:Mvar-dimensional case), and also years later called the Stone–Tukey theorem after Arthur H. Stone and John Tukey.

Naming

Файл:Ham sandwich.jpg
A ham sandwich

The ham sandwich theorem takes its name from the case when Шаблон:Math and the three objects to be bisected are the ingredients of a ham sandwich. Sources differ on whether these three ingredients are two slices of bread and a piece of ham Шаблон:Harv, bread and cheese and ham Шаблон:Harv, or bread and butter and ham Шаблон:Harv. In two dimensions, the theorem is known as the pancake theorem to refer to the flat nature of the two objects to be bisected by a line Шаблон:Harv.

History

According to Шаблон:Harvtxt, the earliest known paper about the ham sandwich theorem, specifically the Шаблон:Math case of bisecting three solids with a plane, is a 1938 note in a Polish mathematics journal Шаблон:Harv. Beyer and Zardecki's paper includes a translation of this note, which attributes the posing of the problem to Hugo Steinhaus, and credits Stefan Banach as the first to solve the problem, by a reduction to the Borsuk–Ulam theorem. The note poses the problem in two ways: first, formally, as "Is it always possible to bisect three solids, arbitrarily located, with the aid of an appropriate plane?" and second, informally, as "Can we place a piece of ham under a meat cutter so that meat, bone, and fat are cut in halves?" The note then offers a proof of the theorem.

A more modern reference is Шаблон:Harvtxt, which is the basis of the name "Stone–Tukey theorem". This paper proves the Шаблон:Mvar-dimensional version of the theorem in a more general setting involving measures. The paper attributes the Шаблон:Math case to Stanislaw Ulam, based on information from a referee; but Шаблон:Harvtxt claim that this is incorrect, given the note mentioned above, although "Ulam did make a fundamental contribution in proposing" the Borsuk–Ulam theorem.

Two-dimensional variant: proof using a rotating-knife

Файл:Ham sandwich theorem 2d.png
A two-dimensional ham sandwich theorem example with noncontiguous regions: lines at 5° increments bisect the similarly coloured region (pink ham and green vegetable) into two equal areas, the black line denoting the common bisector of both regions

The two-dimensional variant of the theorem (also known as the pancake theorem) can be proved by an argument which appears in the fair cake-cutting literature (see e.g. Robertson–Webb rotating-knife procedure).

For each angle <math>\alpha\in[0,180^\circ]</math>, a straight line ("knife") of angle <math>\alpha</math> can bisect pancake #1. To see this, translate [move parallelly] a straight line of angle <math>\alpha</math> from <math>-\infty</math> to <math>\infty</math>; the fraction of pancake #1 covered by the line changes continuously from 0 to 1, so by the intermediate value theorem it must be equal to 1/2 somewhere along the way. It is possible that an entire range of translations of our line yield a fraction of 1/2; in this case, it is a canonical choice to pick the middle one of all such translations.

When the knife is at angle 0, it also cuts pancake #2, but the pieces are probably unequal (if we are lucky and the pieces are equal, we are done). Define the 'positive' side of the knife as the side in which the fraction of pancake #2 is larger. We now turn the knife, and translate it as described above. When the angle is <math>\alpha</math>, define <math>p(\alpha)</math> as the fraction of pancake #2 at the positive side of the knife. Initially <math>p(0) > 1/2</math>. The function <math>p</math> is continuous, since small changes in the angle lead to small changes in the position of the knife.

When the knife is at angle 180, the knife is upside-down, so <math>p(180) < 1/2</math>. By the intermediate value theorem, there must be an angle in which <math>p(\alpha)=1/2</math>. Cutting at that angle bisects both pancakes simultaneously.

n-dimensional variant: proof using the Borsuk–Ulam theorem

The ham sandwich theorem can be proved as follows using the Borsuk–Ulam theorem. This proof follows the one described by Steinhaus and others (1938), attributed there to Stefan Banach, for the Шаблон:Math case. In the field of Equivariant topology, this proof would fall under the configuration-space/tests-map paradigm.

Let Шаблон:Math denote the Шаблон:Mvar objects that we wish to simultaneously bisect. Let Шаблон:Mvar be the unit [[n-sphere|Шаблон:Math-sphere]] embedded in Шаблон:Mvar-dimensional Euclidean space <math>\mathbb{R}^n</math>, centered at the origin. For each point Шаблон:Mvar on the surface of the sphere Шаблон:Mvar, we can define a continuum of oriented affine hyperplanes (not necessarily centred at 0) perpendicular to the (normal) vector from the origin to Шаблон:Mvar, with the "positive side" of each hyperplane defined as the side pointed to by that vector (i.e. it is a choice of orientation). By the intermediate value theorem, every family of such hyperplanes contains at least one hyperplane that bisects the bounded object Шаблон:Math: at one extreme translation, no volume of Шаблон:Math is on the positive side, and at the other extreme translation, all of Шаблон:Math's volume is on the positive side, so in between there must be a translation that has half of Шаблон:Math's volume on the positive side. If there is more than one such hyperplane in the family, we can pick one canonically by choosing the midpoint of the interval of translations for which Шаблон:Math is bisected. Thus we obtain, for each point Шаблон:Mvar on the sphere Шаблон:Mvar, a hyperplane Шаблон:Math that is perpendicular to the vector from the origin to Шаблон:Mvar and that bisects Шаблон:Math.

Now we define a function Шаблон:Mvar from the Шаблон:Math-sphere Шаблон:Mvar to Шаблон:Math-dimensional Euclidean space <math>\mathbb{R}^{n-1}</math> as follows:

Шаблон:Mathvol of Шаблон:Math on the positive side of Шаблон:Math, vol of Шаблон:Math on the positive side of Шаблон:Math, ..., vol of Шаблон:Math on the positive side of Шаблон:Math.

This function Шаблон:Mvar is continuous (which, in a formal proof, would need some justification). By the Borsuk–Ulam theorem, there are antipodal points Шаблон:Mvar and Шаблон:Mvar on the sphere Шаблон:Mvar such that Шаблон:Math. Antipodal points Шаблон:Mvar and Шаблон:Mvar correspond to hyperplanes Шаблон:Math and Шаблон:Math that are equal except that they have opposite positive sides. Thus, Шаблон:Math means that the volume of Шаблон:Math is the same on the positive and negative side of Шаблон:Math (or Шаблон:Math), for Шаблон:Math. Thus, Шаблон:Math (or Шаблон:Math) is the desired ham sandwich cut that simultaneously bisects the volumes of Шаблон:Math.

Measure theoretic versions

In measure theory, Шаблон:Harvtxt proved two more general forms of the ham sandwich theorem. Both versions concern the bisection of Шаблон:Mvar subsets Шаблон:Math of a common set Шаблон:Mvar, where Шаблон:Mvar has a Carathéodory outer measure and each Шаблон:Math has finite outer measure.

Their first general formulation is as follows: for any continuous real function <math>f \colon S^n \times X \to \mathbb{R}</math>, there is a point Шаблон:Mvar of the Шаблон:Mvar-sphere Шаблон:Math and a real number s0 such that the surface Шаблон:Math divides Шаблон:Mvar into Шаблон:Math and Шаблон:Math of equal measure and simultaneously bisects the outer measure of Шаблон:Math. The proof is again a reduction to the Borsuk-Ulam theorem. This theorem generalizes the standard ham sandwich theorem by letting Шаблон:Math.

Their second formulation is as follows: for any Шаблон:Math measurable functions Шаблон:Math over Шаблон:Mvar that are linearly independent over any subset of Шаблон:Mvar of positive measure, there is a linear combination Шаблон:Math such that the surface Шаблон:Math, dividing Шаблон:Mvar into Шаблон:Math and Шаблон:Math, simultaneously bisects the outer measure of Шаблон:Math. This theorem generalizes the standard ham sandwich theorem by letting Шаблон:Math and letting Шаблон:Math, for Шаблон:Math, be the Шаблон:Mvar-th coordinate of Шаблон:Mvar.

Discrete and computational geometry versions

Файл:Discrete ham sandwich cut.svg
A ham-sandwich cut of eight red points and seven blue points in the plane.

In discrete geometry and computational geometry, the ham sandwich theorem usually refers to the special case in which each of the sets being divided is a finite set of points. Here the relevant measure is the counting measure, which simply counts the number of points on either side of the hyperplane. In two dimensions, the theorem can be stated as follows:

For a finite set of points in the plane, each colored "red" or "blue", there is a line that simultaneously bisects the red points and bisects the blue points, that is, the number of red points on either side of the line is equal and the number of blue points on either side of the line is equal.

There is an exceptional case when points lie on the line. In this situation, we count each of these points as either being on one side, on the other, or on neither side of the line (possibly depending on the point), i.e. "bisecting" in fact means that each side contains less than half of the total number of points. This exceptional case is actually required for the theorem to hold, of course when the number of red points or the number of blue is odd, but also in specific configurations with even numbers of points, for instance when all the points lie on the same line and the two colors are separated from each other (i.e. colors don't alternate along the line). A situation where the numbers of points on each side cannot match each other is provided by adding an extra point out of the line in the previous configuration.

In computational geometry, this ham sandwich theorem leads to a computational problem, the ham sandwich problem. In two dimensions, the problem is this: given a finite set of Шаблон:Mvar points in the plane, each colored "red" or "blue", find a ham sandwich cut for them. First, Шаблон:Harvtxt described an algorithm for the special, separated case. Here all red points are on one side of some line and all blue points are on the other side, a situation where there is a unique ham sandwich cut, which Megiddo could find in linear time. Later, Шаблон:Harvtxt gave an algorithm for the general two-dimensional case; the running time of their algorithm is Шаблон:Math, where the symbol Шаблон:Mvar indicates the use of Big O notation. Finally, Шаблон:Harvtxt found an optimal Шаблон:Math-time algorithm. This algorithm was extended to higher dimensions by Шаблон:Harvtxt where the running time is <math>o(n^{d-1})</math>. Given Шаблон:Mvar sets of points in general position in Шаблон:Mvar-dimensional space, the algorithm computes a Шаблон:Math-dimensional hyperplane that has an equal number of points of each of the sets in both of its half-spaces, i.e., a ham-sandwich cut for the given points. If Шаблон:Mvar is a part of the input, then no polynomial time algorithm is expected to exist, as if the points are on a moment curve, the problem becomes equivalent to necklace splitting, which is PPA-complete.

A linear-time algorithm that area-bisects two disjoint convex polygons is described by Шаблон:Harvtxt.

Generalizations

The original theorem works for at most Шаблон:Mvar collections, where Шаблон:Mvar is the number of dimensions. To bisect a larger number of collections without going to higher dimensions, one can use, instead of a hyperplane, an algebraic surface of degree Шаблон:Mvar, i.e., an (Шаблон:Math)–dimensional surface defined by a polynomial function of degree Шаблон:Mvar:

Given <math>\binom{k+n}{n}-1</math> measures in an Шаблон:Mvar–dimensional space, there exists an algebraic surface of degree Шаблон:Mvar which bisects them all. (Шаблон:Harvtxt).

This generalization is proved by mapping the Шаблон:Mvar–dimensional plane into a <math>\binom{k+n}{n}-1</math> dimensional plane, and then applying the original theorem. For example, for Шаблон:Math and Шаблон:Math, the 2–dimensional plane is mapped to a 5–dimensional plane via:

Шаблон:Math.

See also

References

External links