L07 msr

Functions
18 Oct 2011
A B
f( ) =

This Lecture
We will define what is a function formally, and then
in the next lecture we will use this concept in counting.
We will also study the pigeonhole principle and its applications.
• Examples and definitions (injection, surjection, bijection)
• Pigeonhole principle and applications

Functions
function, f, from set A to set B
associates an element , with an element
:f A B→
( )f a B∈ .a A∈
The domain of f is A.
The codomain of f is B.
Definition: For every input there is exactly one output.

Functions
domain = R
codomain = R+
-{0}
domain = R+
-{0}
codomain = R
domain = R
codomain = [0,1]
domain = R+
codomain = R+

Functions
f(S) = |S|
f(string) = length(string)
f(student-name) = student-ID
f(x) = is-prime(x)
domain = the set of all sets
codomain = non-negative integers
domain = the set of all strings
codomain = non-negative integers
domain = positive integers
codomain = {T,F}
not a function,
since one input could have
more than one output
[two students may have same name]

≤ 1 arrow in
A B
f( ) =
Injections (One-to-One)
is an injection iff no two inputs have the same output.:f A B→
, .
( ( ) ( ')) ( ')
′∀ ∈
= → =
a a A
f a f a a a |A| ≤ |B|

Surjections (Onto)
A B
1 arrow in
is a surjection iff every output is possible.:f A B→
f( ) =
. ( )b B a A f a b∀ ∈ ∃ ∈ = |A| ≥ |B|

Bijections
A B
is a bijection iff it is surjection and injection.:f A B→
f( ) =
exactly one arrow in
|A| = |B|

Function Domain Codomain Injective? Surjective
?
Bijective?
f(x)=sin(x) Real Real
f(x)=2x
Real Positive
real
f(x)=x2
Real Non-
negative
real
Reverse
string
Bit strings
of length n
Bit strings
of length n
Exercises
Whether a function is injective, surjective, bijective
depends on its domain and the codomain.

Inverse Sets
A B
Given an element y in B, the inverse set of y := f-1
(y) = {x in A | f(x) = y}.

Inverse Function
A B
f( ) =
exactly one arrow in
Informally, an inverse function f-1
is to “undo” the operation of function f.
There is an inverse function f-1
for f if and only if f is a bijection.

Composition of Functions
Two functions f:X->Y’, g:Y->Z so that Y’ is a subset of Y,
then the composition of f and g is the function g 。 f: X->Z, where
g 。 f(x) = g(f(x)).
X
Y
Z
Y’f g

Function f Function g g 。 f
injective?
g 。 f
surjective?
g 。 f
bijective?
f:X->Y
f surjective
g:Y->Z
g injective
f:X->Y
f surjective
g:Y->Z
g surjective
f:X->Y
f injective
g:Y->Z
g surjective
f:X->Y
f bijective
g:Y->Z
g bijective
f:X->Y f-1
:Y->X
Exercises

This Lecture
• Examples and definitions (injection, surjection, bijection)
• Pigeonhole principle and applications

If more pigeons
than pigeonholes,
Pigeonhole Principle

Pigeonhole Principle
then some hole must have at least two pigeons!
Pigeonhole principle
A function from a larger set to a smaller set cannot be injective.
(There must be at least two elements in the domain that have
the same image in the codomain.)

Example 1
Question: Let A = {1,2,3,4,5,6,7,8}
If five integers are selected from A,
must a pair of integers have a sum of 9?
Consider the pairs {1,8}, {2,7}, {3,6}, {4,5}.
The sum of each pair is equal to 9.
If we choose 5 numbers from this set,
then by the pigeonhole principle,
both elements of some pair will be chosen,
and their sum is equal to 9.

Example 2
Question: In a party of n people, is it always true that there are
two people shaking hands with the same number of people?
Everyone can shake hand with 0 to n-1 people, and there are n people,
and so it does not seem that it must be the case, but think about it carefully:
Case 1: if there is a person who does not shake hand with others,
then any person can shake hands with at most n-2 people,
and so everyone shakes hand with 0 to n-2 people,
0 to n-2 => n-1 possible values (i.e., cardinality of codomain = n-1)
There are n people (i.e., cardinality of domain = n)
so
the answer is “yes” by the pigeonhole principle.

Example 2
Question: In a party of n people, is it always true that there are
two people shaking hands with the same number of people?
Everyone can shake hand with 0 to n-1 people, and there are n people,
and so it does not seem that it must be the case, but think about it carefully:
Case 2: if everyone shakes hand with at least one person, then
any person shakes hand with 1 to n-1 people,
1 to n-1 => n-1 possible values (i.e., cardinality of codomain = n-1)
There are n people (i.e., cardinality of domain = n)
so
the answer is “yes” by the pigeonhole principle.

Birthday Paradox
In a group of 367 people, there must be two people having the same birthday.
Suppose n <= 365, what is the probability that in a random set of n people,
some pair of them will have the same birthday?
We can think of it as picking n random numbers from 1 to 365 without repetition.
There are 365n
ways of picking n numbers from 1 to 365.
[You have 365 choices to pick the first one, same for the second and so on…]
There are 365·364·363·…·(365-n+1) ways of picking n numbers from 1 to 365
without repetition.
[You have 365 choices to pick the first one, 364 for the second and so on…]
So the probability that no pairs have the same birthday is
equal to 365·364·363·…·(365-n+1) / 365n
This is smaller than 50% for 23 people, smaller than 1% for 57 people.

Generalized Pigeonhole Principle
If n pigeons and h holes,
then some hole has at least
n
h
 
  
pigeons.
Generalized Pigeonhole Principle
Cannot have < 3 cards in every hole.
♠ ♥ ♣ ♦

Subset Sum
Two different subsets of the 90 25-digit numbers shown above have the same sum.

Subset Sum
90 numbers, each with at most 25 digits.
So the total sum is at most 90x1025
Let A be the set of all subsets of the 90 numbers.
Let B be the set of integers from 0 to 90x1025
.
(pigeons)
(pigeonholes)
By pigeonhole principle, there are two different subsets with the same sum.

Club vs Strangers
Theorem: Every collection of 6 people includes a club of 3 people,
or a group of 3 strangers.
Let’s agree that given any two people, either they have met or not.
If every people in a group has met, then we’ll call the group a club.
If every people in a group has not met, then we’ll call a group of strangers.
Let x be one of the six people.
By the (generalized) pigeonhole principle, we have the following claim.
Claim: Among the remaining 5 people, either 3 of them have met x,
or 3 of them have not met x.

Club vs Strangers
Case 1: “3 people have met x”
Case 1.1: No pair among those people met each other.
Then there is a group of 3 strangers.
OK!
Case 1.2: Some pair among those people have met each other.
Then that pair, together with x, form a club of 3 people.
OK!

Club vs Strangers
Case 2: “3 people have not met x”
Case 2.1: Every pair among those people met each other.
Then there is a club of 3 people.
OK!
Case 2.2: Some pair among those people have not met each other.
Then that pair, together with x, form a group of 3 strangers.
OK!

Quick Summary
Make sure you understand basic definitions of functions.
These will be used in the next lecture for counting.
The pigeonhole principle is very simple,
but there are many clever uses of it to prove non-trivial results.

L07 msr

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