Discrete Structures
Li Tak Sing (李德成)
Lectures 9-10
1
Pigeonhole principle
If m pigeons fly into n pigeonholes where
m>n, then one pigeonhole will have two or
more pigeons.
2
Examples in pigeonhole principle
1. How many people are needed in a group to be
sure that three were born on the same day of
the week?
2. How many people are needed in a group to be
sured that four were born in the same month?
3. Why does any set of 10 nonempty strings over
{a,b,c} contain two different strings whose
starting agree and whose ending letters agree?
4. Find the size needed for a set of nonempty
strings over {a,b,c,d} so that it must contain at
least three strings whose starting letters agree
and whose ending letters agree.
3
1. 27+1
2. 3 12+1
3. maximum number of strings with different
starting and ending=3 3=9. Therefore if
there are 10 strings, at least two of them
must have the same starting and ending.
4. 4 4 2+1
4
The Mod function and inverses
Let n>1 and let f:NnNn be defined as
follows, where a and b are integers.
f(x)=(ax+b)mod n.
Then f is a bijection iff gcd(a,n)=1. When
this is the case, the inverse function f-1 is
defined by
f-1(x)=(kx+c) mod n
where c is an integer such that f(c)=0, and
k is an integer such that 1=ak+nm for
some integer m.
5
Proof

6
Example
Fina a,bZ so that:
1.
2.
3.
4.
5.
10a+7b=1
5a-2b=1
-3a+4b=1
4a-5b=1
6a+7b=1
7
Solution
1.
2.
3.
4.
5.
a=-2, b=3
a=1,b=2
a=1,b=1
a=-1,b=-1
a=-1,b=1
8
Simple ciphers
N26={0,1,..25} represents letters {a,...,z}
We can use a function to transform a letter
to another. For example
f: N26N26, f(x)=(x+5) mod 26
The inverse of f, denoted as f-1, is:
f-1(x)=(x-5) mod 26
f is known as a cipher which would
transform a letter into another letter to hide
secret.
9
Multiplicative cipher
The previous cipher is called an additive
cipher because addition is use.
A multiplicative cipher is one that
multiplication is used.
An example is:
g(x)=3x mod 26
the inverse is g-1(x)=9x mod 26
10
Example
Determine where the following functions
define a cipher. If a function is a cipher,
then find the inverse function
1.
2.
3.
4.
5.
f(x)=2x mod 6
f(x)=2x mod 5
f(x)=5x mod 6
f(x)=(3x+2) mod 6
f(x)=(2x+3) mod 7
11
Solution
1. It is not a cipher because 2 and 6 are not
relatively prime.
2. It is a cipher. f-1(x)=-2x mod 5
3. It is a cipher. f-1(x)=-x mod 6
4. No, it is not a cipher.
5. It is a cipher. f-1(x)=-3x +2mod 7
12
Hash function
A hash function is a function that maps a
set S of keys to a finite set of table indexes,
which we'll assume are 0,1,...,n-1. A table
whose information is found by a hash
function is called a hash table.
13
Hash example
For example, let S be the set of threeletter abbreviations for the months of the
year. We might define a hash function
f:S{0,1,..,11} in the following way.
f(XYZ)=(ord(X)+ord(Y)+ord(Z))mod 12.
where ord(X) represents the ASCII code
for X.
14
One use of hash functions
 Assume that you have a table of m entry and
you want to store some items in the table. Each
item has a key which can be used to uniquely
identify it. Now, we can use a hash function so
that it will generate an index for the item to be
stored in the corresponding entry.
 This method has an advantage that when we
know of the key, we can find the position where
the item is stored.
15
Collisions
When two objects have the same key, a
collision is said to occur.
One technique to solve the problem of
collision is called linear probing. With this
technique the program searches the
remaining locations in a linear manner.
16
Hash examples
Let S={Monday, Tuesday, Wednesday,
Thursday, Friday, Saturday, Sunday} and
suppose that f:SN7 is the hash function
defined by f(x)=lenght(x) mod 7 where
length(x) is the number of letters in x. Use
h to place each day of S into a hash table
indexed by {0,1,..,6} starting with Monday,
then Tuesday, and so on until Sunday.
Resolve collisions by linear probing with a
gap of 2.
17
Hash examples
 Let S={one,two,three, four, five, six, seven, eight,
nine} and let f:SN9 be defined by f(x)=(3|x|)
mod 9, where |x| means the number of letters in
x. For each of the following gaps, construct a
hash table that contains the strings of S by
choosing a string for entry in the table by the
order that it is listed in S. Resolve collisions by
linear probing with the given gap and observe
whether all strings can be placed in the table.
Gap=1
Gap=2
Gap=3
18
Countability
Let A and B be sets. If there is a bijection
between A and B, we'll denote the fact by
writing
|A|=|B|.
In this case, we'll say that A and B have
the same size or have the same cardinality.
19
Finite sets examples
Find the cardinality of the following
function.
1. {2x+5|xZ and 12x+5100}.
2. {x2|x Z and 0x2 500}
3. {2,5,8,11,...,44,47}
20
Solution
1. 50
2. 23
3. 16
21
Cardinality of an infinite set
 Let Odd denote the set of odd natural numbers.
Then the function f:NOdd defined by f(x)=2x+1
is a bijection. Therefore |N|=|Odd|.
 If there is an injection from A to B, we then
|A||B|.
 If there is an surjection from B to A, we have
|A||B|.
 If there is an injection from A to B, but no
bijection between them, then
|A|<|B|.
22
Sets that are countable
A set is countable if it is finite or if there is
a bijection between it and N. In the latter
case, the set is said to be countably
infinite.
S is countable if |S|=n where n is a natural
number or |S|=|N|
23
Countable Properties
1. Every subset of N is countable.
2. S is countable iff |S||N|.
3. Any subset of a countable set is
countable.
4. Any image of a countable set is
countable.
24
Proof of rule 1
If the subset is finite, then it is countable.
If it is not finite, we can arrange the
elements in ascending order and denote
the elements as xo, x1, ...... So there is a
bijection between {xo, x1,...} and N.
25
Proof of rule 2
If SN is countable, then either |S|=n or
|S|=|N|. So |S||N|.
If |S||N|, then, there is an injection f from
S to N. Now, |S|=|f(S)|. Now, f(S) is a
subset of N, by rule 1, f(S) is countable.
Therefore f(S) is either finite or |f(S)|=|N|.
This imply either S is finite or |S|=|N|. Both
imply that S is countable.
26
Proof of rule 3
If A is a countable set, then there is an
injection from A to N. Assume that f:AN
be one of the injective functions. Now, for
any subject B of A, we define the function
g: B N such that g(x)=f(x). It is obvious
that g is also an injection. Therefore B is
also countable.
27
Proof of rule 4
 If A is a countable set, then there is an injection
from A to N. Let f:A N be an injection. Now,
for a function g: A B for every yg(A), there is
an x A such that g(x)=y. Now, we define a
function h: g(A) A such that h(y)=x so that
g(x)=y and f(x) is maximum. Now, h must be an
injection because an element in A cannot map
with two different elements in B under g.
Therefore, fh is an injection as both f and h are
injections. fh is a function from g(A) to N. So
g(A) is countable.
28
NxN is countable
(0,0) has 1 element
(0,1), (1,0) have 2 elements
(0,2),(1,1),(2,0) have 3 elements
.....
(0,n), (1,n-1),.....(0,n) have n+1 elements
29
(0,0) maps to 0,
(0,1), (1,0) map to 1,1+1
(0,2),(1,1),(2,0) map to
(1+2),(1+2)+1,(1+2)+2
.....
(0,n), (1,n-1),.....(0,n) maps to (1+2+..n),
(1+2+..n)+1,.....(1+2+...n)+n
30
(0,0) maps to 0,
(0,1), (1,0) map to 1,2
(0,2),(1,1),(2,0) map to 3,4,5
.....
(0,n), (1,n-1),.....(0,n) maps to (n2+n)/2,
(n2+n)/2,....., (n2+n)/2+n
31
Counting unions of countable sets
If So, S1,....Sn,.... is a sequence of
countable sets, then the union
SoS1 ....  Sn ....
is a countable set.
32
Proof
 List the elements of Sn as xn0, xn1,.....
 If Sn is finite, then repeat one of the elements indefinitely
so that we have an infinite sequence of elements.
 If So, S1,....Sn,.... is a finite sequence, then we repeat
one of the set indefinitely so that we have an infinite
sequence of sets.
 (m,n) of NN is associated with xmn
 The mapping is not bijective because some elements of
NN is mapped to the same xmn.
 However, the mapping is surjective because every
element of xmn is associated with (m,n). So we have
|N|=|NN|| SoS1 ....  Sn .... | and therefore SoS1
....  Sn .... is countable.
33
Countability of the rationals
Q+ ={m/n | m,nN and m,n0 }
We can associate each m/n with
(m,n)NN. Therefore, |Q+||NN|=|N|.
Therefore Q+ is countable.
Similary Q- ={-m/n | m,nN and m,n0 }
and is also countable.
Therefore Q= Q+ {0}  Q- should be
countable.
34
Examples
 Show that each of the following sets is countable
by establishing a bijection between the set and
N.
1.
2.
3.
4.
5.
6.
7.
The set of even natural numbers.
The set of negative integers.
The set of strings over {a}
The set of strings over {a} that have even length.
The set Z of integers.
The set of odd integers.
The set of even integers.
35
Solution
1. f(n)=2n
2. f(z)=-1-z
 if x  0

f
(x
)


3.
af(x - 1) otherwise
4.
5.
 if x  0

f (x)  
aaf(x - 1) otherwise
 x
 2 when x is even
f ( x)  
 x 1

when x is odd
 2
6.
7.
 x when x is odd
f ( x)  
 x  1 when x is even
 x when x is even
f ( x)  
 x  1 when x is odd
36