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# Pigeonhole principle

A photograph of pigeons in holes. Here there are n = 10 pigeons in m = 9 holes, so by the pigeonhole principle, at least one hole has more than one pigeon: in this case, both of the top corner holes contain two pigeons. The principle says nothing about which holes are empty: for n = 10 pigeons in m = 9 holes, it simply says that at least one hole here will be over-full; in this case, the bottom-left hole is empty.

In mathematics and computer science, the pigeonhole principle states that if n items are put into m pigeonholes with n > m, then at least one pigeonhole must contain more than one item. This theorem is exemplified in real-life by truisms like "there must be at least two left gloves or two right gloves in a group of three gloves". It is an example of a counting argument, and despite seeming intuitive it can be used to demonstrate possibly unexpected results; for example, that two people in London have the same number of hairs on their heads (see below).

The first formalization of the idea is believed to have been made by Johann Dirichlet in 1834 under the name Schubfachprinzip ("drawer principle" or "shelf principle"). For this reason it is also commonly called Dirichlet's box principle, Dirichlet's drawer principle or simply "Dirichlet principle"—a name that could also refer to the minimum principle for harmonic functions. The original "drawer" name is still in use in French ("principe des tiroirs"), Italian ("principio dei cassetti") and German ("Schubfachprinzip").

Though the most straightforward application is to finite sets (such as pigeons and boxes), it is also used with infinite sets that cannot be put into one-to-one correspondence. To do so requires the formal statement of the pigeonhole principle, which is "there does not exist an injective function on finite sets whose codomain is smaller than its domain". Advanced mathematical proofs like Siegel's lemma build upon this more general concept.

## Examples

### Softball team

Imagine five people who want to play softball (n = 5 items), with a limitation of only four softball teams (m = 4 holes) to choose from. A further limitation is imposed in the form of each of the five refusing to play on a team with any of the other four players. It is impossible to divide five people among four teams without putting two of the people on the same team, and since they refuse to play on the same team, by asserting the pigeonhole principle it is easily deducible that at most four of the five possible players will be able to play.

### Sock-picking

Assuming that in a box there are 10 black socks and 12 blue socks, calculate the maximum number of socks needed to be drawn from the box before a pair of the same color can be made. Using the pigeonhole principle, to have at least one pair of the same color (m = 2 holes, one per colour) using one pigeonhole per color, you need only three socks (n = 3 items). In this example, if the first and second sock drawn are not of the same color, the very next sock drawn would complete at least one same-color pair. (m = 2)

### Hand-shaking

If there are n number of people who can shake hands with one another (where n > 1), the pigeonhole principle shows that there is always a pair of people who will shake hands with the same number of people. As the 'holes', or m, correspond to number of hands shaken, and each person can shake hands with anybody from 0 to n − 1 other people, this creates n − 1 possible holes. This is 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.

### Hair-counting

We can demonstrate there must be at least two people in London with the same number of hairs on their heads as follows. A typical human head has around 150,000 hairs; therefore it is reasonable to assume that no one has more than 1,000,000 hairs on their head (m = 1 million holes). Since there are more than 1,000,000 people in London (n is bigger than 1 million items), if we assign a pigeonhole to each number of hairs on a person's head, and assign people to pigeonholes according to the number of hairs on each person's head, there must be at least two people with the same number of hair on their heads (or, n > m). In the case with the fewest overlaps, there will be at least one person assigned to every pigeonhole, at which point the 1,000,001st person is assigned to the same pigeonhole as someone else. Of course, there may be empty pigeonholes in which case this "collision" happens before we get to the 1,000,000th person. The principle just proves the existence of an overlap; it says nothing of the number of overlaps.

## Uses and applications

The pigeonhole principle 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. A hashing algorithm, no matter how clever, cannot avoid these collisions. The principle can also be used to prove that any lossless compression algorithm, provided it makes some inputs smaller (as the name compression suggests), will also make some other inputs larger. Otherwise, the set of all input sequences up to a given length l could be mapped to the (much) smaller set of all sequences of length less than l, and do so without collisions (because the compression is lossless), which possibility the pigeonhole principle excludes.

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 |nam| < e, where e > 0 is a small positive number and a is some arbitrary irrational number. But if one takes M such that 1/M < e, by the pigeonhole principle there must be n1, n2 ∈ {1, 2, ..., M + 1} such that n1a and n2a are in the same integer subdivision of size 1/M (there are only M such subdivisions between consecutive integers). In particular, we can find n1n2 such that n1a is in (p + k/M, p + (k + 1)/M), and n2a is in (q + k/M, q + (k + 1)/M), for some pq integers and k in {0, 1, ..., M − 1}. We can then easily verify that (n2n1)a is in (qp − 1/M, qp + 1/M). This implies that [na] < 1/M < e, where n = n2n1 or n = n1n2. 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{rN : r[na] < j/M}, one obtains |[(k + 1)na] − p| < 1/M < e.

## 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 $\lceil n/m \rceil$ objects, where $\lceil x\rceil$ is the ceiling function, denoting the smallest integer larger than or equal to x. Similarly, at least one container must hold no more than $\lfloor n/m \rfloor$ objects, where $\lfloor x \rfloor$ is the floor function, denoting the largest integer smaller 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

$1 - \frac{(m)_n}{m^n}, \!$

where (m)n is the falling factorial m(m − 1)(m − 2)...(mn + 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 (nm), 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.

## 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. However in this form the principle is tautological, since the meaning of the statement that the cardinality of set A is greater than the cardinality of set B is exactly that there is no injective map from A to B. What makes the situation of finite sets particular is that adding at least one element to a set is sufficient to ensure that the cardinality increases.

## References

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