# Pigeonhole principle

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In mathematics, the **pigeonhole principle**, also known as **Dirichlet's box** (or **drawer'**) **principle**, is exemplified by such things as the fact that in a family of three children there must be at least two of the same gender. This principle states that, given two natural numbers *n* and *m* with *n* > *m*, if *n* items are put into *m* pigeonholes, then at least one pigeonhole must contain more than one item. Another way of stating this would be that *m* holes can hold at most *m* objects with one object to a hole; adding another object will force one to reuse one of the holes, provided that *m* is finite. More formally, the theorem states that there does not exist an injective function on finite sets whose codomain is smaller than its domain.

The pigeonhole principle is an example of a counting argument which can be applied to many formal problems, including ones involving infinite sets that cannot be put into one-to-one correspondence. In Diophantine approximation the quantitative application of the principle to the existence of integer solutions of a system of linear equations goes under the name of *Siegel's lemma*.

The first statement of the principle is believed to have been made by Dirichlet in 1834 under the name *Schubfachprinzip* ("drawer principle" or "shelf principle"). In Italian too, the original name "principio dei cassetti" is still in use; in some other languages (for example, Russian) this principle is called the *Dirichlet principle* (not to be confused with the minimum principle for harmonic functions of the same name).

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## [edit] Examples

- An easy example of the pigeonhole principle involves the situation when there are five people who want to play softball (
*n*= 5 objects), but there are only four softball teams (*m*= 4 holes). This would not be a problem except that each of the five refuses to play on a team with any of the other four. To prove that there is no way for all five people to play softball, the pigeonhole principle says that it is impossible to divide five people among four teams without putting two of the people on the same team. Since they refuse to play on the same team, at most four of the people will be able to play. *Assume that in a box there are 10 black socks and 12 blue socks and you need to get one pair of socks of the same colour. Supposing you can take socks out of the box only once and only without looking, what is the minimum number of socks you'd have to pull out at the same time in order to guarantee a pair of the same color?*The correct answer is three. To have at least one pair of the same colour (*m*= 2 holes, one per colour), using one pigeonhole per colour, you need only three socks (*n*= 3 objects). In this example, if the first and second sock drawn are not of the same colour, the very next sock drawn would complete at least one same-colour pair. (*m*= 2)*If there are n persons (where n >*1*) who can shake hands with one another, there is always a pair of persons who shake hands with the same number of people.*Here, the 'holes' correspond to number of hands shaken. Each person can shake hands with anybody from 0 to*n*− 1 other people. This creates*n*− 1 possible holes because either the '0' or the '*n*− 1' hole must be empty (if one person shakes hands with everybody, it's not possible to have another person who shakes hands with nobody; likewise, if one person shakes hands with no one there cannot be a person who shakes hands with everybody). This leaves*n*people to be placed in at most*n*− 1 non-empty holes, guaranteeing duplication.- Although the pigeonhole principle may seem to be intuitive, it can be used to demonstrate possibly unexpected results. For example,
*there must be at least two people in London with the same number of hairs on their heads*. A typical head of hair has around 150,000 hairs; it is reasonable to assume that no one has more than 1,000,000 hairs on his head (*m*= 1 million holes). There are more than 1,000,000 people in London (*n*is bigger than 1 million objects). If we assign a pigeonhole for each number of hairs on a head, and assign people to the pigeonhole with their number of hairs on it, there must be at least two people with the same number of hairs on their heads.

## [edit] Uses and applications

The pigeonhole principle often arises in computer science. For example, collisions are inevitable in a hash table because the number of possible keys exceeds the number of indices in the array. No hashing algorithm, no matter how clever, can avoid these collisions. This principle also proves that any general-purpose lossless compression algorithm that makes at least one input file smaller will make some other input file larger. (Otherwise, two files would be compressed to the same smaller file and restoring them would be ambiguous.)

A notable problem in mathematical analysis is, for a fixed irrational number *a*, to show that the set {[*na*]: *n* is an integer} of fractional parts is dense in [0, 1]. After a moment's thought, one finds that it is not easy to explicitly find integers *n*, *m* such that |*na* − *m*| < *e*, where *e* > 0 is a small positive number and *a* is some arbitrary irrational number. But if you take *M* such that 1/*M* < *e*, by the pigeonhole principle there must be *n*_{1}, *n*_{2} ∈ {1, 2, ..., *M* + 1} such that *n*_{1}*a* and *n*_{2}*a* are in the same integer subdivision of size 1/*M* (there are only *M* such subdivisions between consecutive integers). In particular, we can find *n*_{1}, *n*_{2} such that *n*_{1}*a* is in (*p* + *k*/*M*, *p* + (*k* + 1)/*M*), and *n*_{2}*a* is in (*q* + *k*/*M*, *q* + (*k* + 1)/*M*), for some *p*, *q* integers and *k* in {0, 1, ..., *M* − 1}. We can then easily verify that (*n*_{2} − *n*_{1})*a* is in (*q* − *p* − 1/*M*, *q* − *p* + 1/*M*). This implies that [*na*] < 1/*M* < *e*, where *n* = *n*_{2} − *n*_{1} or *n* = *n*_{1} − *n*_{2}. This shows that 0 is a limit point of {[*na*]}. We can then use this fact to prove the case for *p* in (0, 1]: find *n* such that [*na*] < 1/*M* < *e*; then if *p* ∈ (0, 1/*M*], we are done. Otherwise *p* in (*j*/*M*, (*j* + 1)/*M*], and by setting *k* = sup{*r* ∈ *N* : *r*[*na*] < *j*/*M*}, you get |[(*k* + 1)*na*] − *p*| < 1/*M* < *e*.

## [edit] Generalizations of the pigeonhole principle

A generalized version of this principle states that, if *n* discrete objects are to be allocated to *m* containers, then at least one container must hold no fewer than objects, where is the ceiling function, denoting the smallest integer larger than or equal to *x*.

A probabilistic generalization of the pigeonhole principle states that if *n* pigeons are randomly put into *m* pigeonholes with uniform probability 1/*m*, then at least one pigeonhole will hold more than one pigeon with probability

where (*m*)_{n} is the falling factorial *m*(*m* − 1)(*m* − 2)...(*m* − *n* + 1). For *n* = 0 and for *n* = 1 (and *m* > 0), that probability is zero; in other words, if there is just one pigeon, there cannot be a conflict. For *n* > *m* (more pigeons than pigeonholes) it is one, in which case it coincides with the ordinary pigeonhole principle. But even if the number of pigeons does not exceed the number of pigeonholes (*n* ≤ *m*), due to the random nature of the assignment of pigeons to pigeonholes there is often a substantial chance that clashes will occur. For example, if 2 pigeons are randomly assigned to 4 pigeonholes, there is a 25% chance that at least one pigeonhole will hold more than one pigeon; for 5 pigeons and 10 holes, that probability is 69.76%; and for 10 pigeons and 20 holes it is about 93.45%. If the number of holes stays fixed, there is always a greater probability of a pair when you add more pigeons. This problem is treated at much greater length at birthday paradox.

A further probabilistic generalisation is that when a real-valued random variable *X* has a finite mean *E*(*X*), then the probability is nonzero that *X* is greater than or equal to *E*(*X*), and similarly the probability is nonzero that *X* is less than or equal to *E*(*X*). To see that this implies the standard pigeonhole principle, take any fixed arrangement of *n* pigeons into *m* holes and let *X* be the number of pigeons in a hole chosen uniformly at random. The mean of *X* is *n*/*m*, so if there are more pigeons than holes the mean is greater than one. Therefore, *X* is sometimes at least 2.

## [edit] Infinite sets

The pigeonhole principle can be extended to infinite sets by phrasing it in terms of cardinal numbers: if the cardinality of set A is greater than the cardinality of set B, then there is no injection from A to B.

## [edit] See also

- Combinatorial principles
- Cardinal number
- Dedekind-infinite set
- Hilbert's paradox of the Grand Hotel
- Multinomial theorem
- Ramsey's theorem
- Combinatorial proof

## [edit] References

- Grimaldi, Ralph P.
*Discrete and Combinatorial Mathematics: An Applied Introduction*. 4th edn. 1998. ISBN 0-201-19912-2. pp. 244–248. - Jeff Miller, Peter Flor, Gunnar Berg, and Julio González Cabillón. "Pigeonhole principle". In Jeff Miller (ed.)
*Earliest Known Uses of Some of the Words of Mathematics*. Electronic document, retrieved November 11, 2006.

## [edit] External links

- "The strange case of The Pigeon-hole Principle"; Edsger Dijkstra investigates interpretations and reformulations of the principle