Topic 14: Pigeonhole Principle

Topic 14: Pigeonhole PrincipleDr J Frost ([email protected])

Last modified: 18th July 2013

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Key to question types:

SMC Senior Maths Challenge

BMO British Maths Olympiad

The level, 1 being the easiest, 5 the hardest, will be indicated.

Those with high scores in the SMC qualify for the BMO Round 1. The top hundred students from this go through to BMO Round 2.Questions in these slides will have their round indicated.

Frost A Frosty SpecialQuestions from the deep dark recesses of my head.

Uni University InterviewQuestions used in university interviews (possibly Oxbridge).

Classic ClassicWell known problems in maths.

MAT Maths Aptitude TestAdmissions test for those applying for Maths and/or Computer Science at Oxford University.

STEP STEP ExamExam used as a condition for offers to universities such as Cambridge and Bath.

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? Any box with a ? can be clicked to reveal the answer (this works particularly well with interactive whiteboards!).Make sure you’re viewing the slides in slideshow mode.

A: London B: Paris C: Madrid

For multiple choice questions (e.g. SMC), click your choice to reveal the answer (try below!)

Question: The capital of Spain is:

1 2 3 4 5 6 7 8

Starter Question

Show that if I pack 5 different numbers from 1 to 8, at least two of them add up to 9.

Solution: There are four pairs of numbers which add up to 9: {1, 8}, {2,7}, {3,6}, {4,5}.For the numbers we choose, each must belong to one of the four pairs. But there’s 5 numbers, so two must belong to the same pair.

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The Pigeonhole Principle

Imagine that we have a number of slots, or ‘pigeon holes’ in which we can put letters.

A DumbledoreJ Frost

R Hagrid R Lupin F Flitwick P Sprout

A Filch S Snape

The pigeonhole principle is very simple: if we have more letters than pigeonholes, then one of the pigeonholes must have at least one letter.

* The Pigeonhole Principle is also sometimes known as Dirichlet’s Box Principle.

The Pigeonhole Principle

There’s many problems in which we can model as pigeonholes. If we take the previous problem:

{2,7}{1,8} {3,6} {4,5}

Each pigeonhole represents a number pair.

The letters represent the numbers from 1 to 8 we choose.

8 2 7 6 4

By the pigeonhole principle, as we used more numbers then pairs, two of the numbers must be in the same pair.

In general, we use the pigeonholes to model ‘possible values’, and letters to model ‘items’, which is put into a pigeonhole hole according to its value.

Another Example of Modelling

Question: Show that if we pick 5 cards from a pack, two must be of the same suit.

DiamondClub Heart Spade

Each pigeonhole represents a suit.

The letters represent the cards we choose. We put it in the correct pigeonhole according to its suit.

Since we have more cards than suits, one pigeonhole must have at least two cards in it, and thus at least two cards of the same suit.

Dijkstra’s Principle

Given there are 8 pigeonholes and 18 letters, what can we say about the maximum number of letters in a pigeonhole?

A DumbledoreJ Frost


A Filch S Snape

When we spread out the letters as much as possible, we found that the maximum number of letters was 3.Dijkstra’s Principle (just a variant of the ‘Pigeonhole Principle’) states that all the maximum number of letters must be at least the average number of letters.


A DumbledoreJ Frost


A Filch S Snape

There average number of letters in each pigeonhole is 18/8 = 2.25. But since the maximum number of letters must be at least this, the minimum this maximum can be is the nearest whole number above, i.e. 3.

In general, if there are n letters and k pigeonholes, there maximum letters in a pigeonhole must be at least:

These special brackets are known as the ‘ceiling’ function, which rounds the number up.


Question: There are 1000 people in the room. What is the minimum number of people that share the same birthday?

Solution: 3. By the pigeonhole principle, the maximum number of birthdays is at least 1000/365 = 2.74. Therefore at least 3 people share the same birthday.

A simple example:

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Restricted Allocation

There are n people at a party, where each person can shake hands with others at the party (or if they’re antisocial, no one at all), although they can only shake a given person’s hand at most once. Show that at least 2 people must have had the same number of handshakes.

Sometimes how we put letters in some pigeonholes restricts how we put them in others.

Source: Uni + Classic

Solution: Each pigeonhole represents the quantity of handshakes a person had.Consider the ‘edge cases’: if someone shook hands with no one, the no one can have shook hands with everyone, and vice versa. Thus we only have n-1 available pigeonholes, but n items, so one pigeonhole must have at least 2 items.

? Click for solution ?

10 n-1 n

BobJim Tim....

Minimal Arrangements

20 cities are visited by up to 20 people each, such that no two cities are visited by the same number of people.

a) Minimum total number of visitors to the cities: 190

b) Maximum total number of visitors to the cities: 210

Sometimes we’re required to do some calculation (e.g. the sum) involving letters in the minimal valid pigeonhole arrangement.

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The Lowest Maximum

Question: A cinema sells 31 tickets in total one evening, and each movie can be watched by up to 3 people (but won’t screen if no one shows up). Show that the minimum number of movies seen by the same number of people is 6.

When we want to consider the lowest maximum, we spread out the letters as evenly as possible over the pigeon holes.

Source: Frosty Special

What are the pidgeonholes? The number of people attending a screening.

Let’s deconstruct the problem:

What are the letters? Each movie/screening.

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The Lowest Maximum



Source: Frosty Special

21 3

Batman

Spiderman

Meerkatman

The arrangement that will minimise the maximum number of letters in a pidgeonhole (i.e. minimising the number of movies same by the same number of people) involves putting 1 letter in each pidgeonhole each time.Each time we do so, this represents 1 + 2 + 3 = 6 tickets.

This means that Meerkatman was seen by 3 people. This letter thus represents 3 tickets.

The Lowest Maximum



1

Fishman

Given that putting one letter in each pigeonhole represents 6 total tickets, we can put 5 letters in each ticket to get us to 30 tickets.

Catman

Ratman

Dogman

Batman

2

Limaman

Foxman

Voleman

Tigerman

Spiderman

3

Hamman

Oxman

Antman

Liceman

Rabbitman

We have 1 ticket left. This must have been for a movie watched by 1 person. And thus, six movies were watched by the same number of people. (Although note that there’s other solutions in which 6 movies were watched by the same number of people, e.g. Once we got to 4 letters in each hole, we could have put 2 letters in the ‘3 hole’ and one in the ‘1 hole’ – the point is that we’ve proved that 6 is the minimum)

Toucanman

BMO ProblemQuestion: Let S be a subset of the set of numbers {1, 2, 3, … , 2008} which consists of 756 distinct numbers. Show that there are two distinct elements, a, b of S such that a + b is divisible by 8.

Round 2

Round 1

BMO

Things that might be going through your head when you see this problem:

“What are my pigeonholes and what are my letters?”

“It sounds like from the question that we can have 755 numbers without any two of them adding to give a multiple of 8. The problem is therefore equivalent to ‘show that 755 is the maximum numbers of numbers we can choose without this happening’”.

“How might I be restricted in putting letters into pigeonholes?”


10

4 5 6 7

2 3

Whether the sum of a and b is divisible by 8 is equivalent to using the remainder of each of the numbers when we divide each by 8. e.g. If 15 + 9 is divisible by 8, then so is 7 + 1 (we’ll encounter modulo arithmetic in the ‘Number Theory’ topic).So make the pigeonholes 0-7, and put a number in this hole if it’s remainder is that number when dividing by 8.

1026

80


10

4 5 6 7

2 3

1026

80

We’re restricted in how we can put letters in holes.If there’s any in the 3 hole, we can’t put any the 5 hole because then we’d have two numbers which add to 8.

Similarly, we could only have one number in the 4 hole; if there were two they’d again add up to 8. The same applies for the 0 hole.


10

4 5 6 7

2 3

There’s 2008 numbers, so 2008/8 = 251 letters belong to each pigeonhole (e.g. all the multiples of 8, of which there are 251, would go in the ‘0’ pigeonhole).

1026

80

We would put 251 letters in either hole 1 or 7. Similarly, we could put 251 in hole 2 or 6. And another 251 in hole 3 or 5. And we could put 1 in holes 0 and 4. That gives us 251 + 251 + 251 + 1 + 1 = 755. This therefore is the maximum number before two of the numbers add to a multiple of 8. With 756, we’d therefore violate this property.

BMO ProblemQuestion: A booking office at a railway station sells tickets to 200 destinations. One day, tickets were issued to 3800 passengers. Show that (i) there are (at least) 6 destinations at which the passenger arrival numbers are the same; (ii) the statement in (i) becomes false if ‘6’ is replaced by ‘7’.

Round 1

Round 2

BMO

Yes, this is a Round 2

question! Don’t

panic!

Things that might be going through your head when you see this problem:

“What are my pigeonholes and what are my letters?”

“A station is limited in the number of people that can buy a ticket for there. For example, if 3800 all bought a ticket for the same station, there wouldn’t be 200 destinations! So we must have to restrict the allocation of letters to pigeonholes in some way.”

“What’s the average number of people at each station?”


10 18

Bath ....Kings CrossParis

19 20...Oslo

Hull

My solution: Model the pigeonholes as the number of tickets for a station. Let the letters be the stations (so Bath goes in pigeonhole 1 if 1 person is going to Bath).

If there’s 3800 passengers, and 200 destinations, there’s an average of 19 people per station. Note that this relates to the average value of the pigeonhole, NOT the average number of letters in each pigeonhole.The key to this strategy is thinking how we allocate letters to pigeonholes so the average pigeonhole value is 19.


10 37

Bath ....Kings Cross Kings

Cross

38 39...Oslo Hull

We want to minimise the number of letters in each pigeonhole. The way to do this is put one letter in each of the pigeonholes 0 to 38. As highlighted earlier, this ensures the average people per destination is still 19 whilst minimising the number of stations with the same number of passengers.


10 37

Bath ....Kings Cross Kings

Cross

38 39...Oslo Hull

We keep adding in this way for each 39 stations. Each time we do, this represents 0 + 1 + 2 + 3 + ... + 38 tickets. Using the standard summation formula, this gives 741 tickets. 741 goes into 3800 five whole times, so we have 5 letters in each of the pigeonholes 0 to 38 and have 741x5 = 3705 tickets.

How do we deal with the remaining 95 tickets? The average per station needs to be 19, which means we’re spreading these over 95/19 = 5 stations. So using the 17, 18, 19, 20 and 21 pigeonholes will do to keep the quantities of tickets for each station spread out..

Thus in the minimal case, 5 quantities of tickets will have 6 stations with that quantity. This proves (ii) as well, because we’ve proved this minimum can be 6.

Paris Madrid Berne Hell

The mathematician Paul Erdős once asked a student what he wanted to when he graduated. The student said he wanted to be a mathematical researcher. Erdős said in response:

Let . Prove that if you take a set of numbers from then you can find

distinct such that .

One final problem…

( means that divides )

i.e. Show that if you pick numbers from to , one must divide another.Big Hint: Suppose that each pigeonhole contains the integers of the form , for each odd number , i.e. the first contains numbers of the form , the second of the form and so on.

Suppose n is 5. Then we have 5 odd numbers, 1, 3, 5, 7, 9, and the numbers in each pigeonhole (using the hint above) will be:

We have 5 pigeonholes (as there are 5 odd numbers), but 5+1 = 6 numbers to pick. So by the pigeonhole principle, we’ll have chosen at least 2 numbers from one set. And clearly the numbers in each set divide each other, since when .

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Summary

For some problems, we use can use pigeonholes to represent different values, and letters to represent items can have one of these values.Be careful how you model these.

There are two forms of the pigeonhole principle:1. If there are more items than pigeonholes, at least one of the

pigeonholes has more than one item in it. (Also known as ‘Dirichlet’s Box Principle’)

2. If there are n items and k pigeonholes, then the maximum number of items in a pigeonhole is at least the average, n/k. (This is Dijkstra’s take on the pigeonhole principle, although this variant is also know as the ‘Fubini Principle’)

Often, putting items in one pigeonhole prevents items from being added to another.

When considering problems in which we wish the minimise the maximum number of letters in a hole, spread the letters out as much as possible, without violating any given/implied constraints.

Recommended Resource

http://www.cut-the-knot.org/do_you_know/pigeon.shtml has a gigantic collection of problems from past international Olympiads (and elsewhere) concerning the pigeonhole principle.

http://www.cut-the-knot.org/do_you_know/pigeon.shtml

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Topic 14: Pigeonhole Principle