Ch02 algorithm 127836129812y12y823873893.ppt

Fall 2002 CMSC 203 - Discrete Structures 1
Enough Mathematical Appetizers!
Enough Mathematical Appetizers!
Let us look at something more interesting:
Let us look at something more interesting:
Algorithms
Algorithms

Algorithms
Algorithms
What is an algorithm?
What is an algorithm?
An algorithm is a finite set of precise instructions
An algorithm is a finite set of precise instructions
for performing a computation or for solving a
for performing a computation or for solving a
problem.
problem.
This is a rather vague definition. You will get to
This is a rather vague definition. You will get to
know a more precise and mathematically useful
know a more precise and mathematically useful
definition when you attend CMSC441.
definition when you attend CMSC441.
But this one is good enough for now…
But this one is good enough for now…

Algorithms
Algorithms
Properties of algorithms:
Properties of algorithms:
• Input
Input from a specified set,
from a specified set,
• Output
Output from a specified set (solution),
from a specified set (solution),
• Definiteness
Definiteness of every step in the computation,
of every step in the computation,
• Correctness
Correctness of output for every possible input,
of output for every possible input,
• Finiteness
Finiteness of the number of calculation steps,
of the number of calculation steps,
• Effectiveness
Effectiveness of each calculation step and
of each calculation step and
• Generality
Generality for a class of problems.
for a class of problems.

Algorithm Examples
Algorithm Examples
We will use a pseudocode to specify algorithms,
We will use a pseudocode to specify algorithms,
which slightly reminds us of Basic and Pascal.
which slightly reminds us of Basic and Pascal.
Example:
Example: an algorithm that finds the maximum
an algorithm that finds the maximum
element in a finite sequence
element in a finite sequence
procedure
procedure max(a
max(a1
1, a
, a2
2, …, a
, …, an
n: integers)
: integers)
max := a
max := a1
1
for
for i := 2
i := 2 to
to n
n
if
if max < a
max < ai
i then
then max := a
max := ai
i
{max is the largest element}
{max is the largest element}

Algorithm Examples
Algorithm Examples
Another example:
Another example: a linear search algorithm, that is,
a linear search algorithm, that is,
an algorithm that linearly searches a sequence for a
an algorithm that linearly searches a sequence for a
particular element.
particular element.
procedure
procedure linear_search(x: integer; a
linear_search(x: integer; a1
1, a
, a2
2, …, a
, …, an
n:
:
integers)
integers)
i := 1
i := 1
while
while (i
(i 
 n and x
n and x 
 a
ai
i)
)
i := i + 1
i := i + 1
if
if i
i 
 n
n then
then location := i
location := i
else
else location := 0
location := 0
{location is the subscript of the term that equals x,
{location is the subscript of the term that equals x,
or is zero if x is not found}
or is zero if x is not found}

Algorithm Examples
Algorithm Examples
If the terms in a sequence are ordered, a binary
If the terms in a sequence are ordered, a binary
search algorithm is more efficient than linear
search algorithm is more efficient than linear
search.
search.
The binary search algorithm iteratively restricts
The binary search algorithm iteratively restricts
the relevant search interval until it closes in on
the relevant search interval until it closes in on
the position of the element to be located.
the position of the element to be located.

Algorithm Examples
Algorithm Examples
a c d f g h j l m o p r s u v x z
binary search for the letter ‘j’
center element
center element
search interval
search interval

Algorithm Examples
Algorithm Examples
a c d f g h j l m
o p r s u v x z
center element
center element
search interval
search interval

Algorithm Examples
Algorithm Examples
a c d f g
a c d f g h j l m
h j l m o p r s u v x z
o p r s u v x z
center element
center element
search interval
search interval

Algorithm Examples
Algorithm Examples
a c d f g
a c d f g h j
h j l m
l m o p r s u v x z
o p r s u v x z
center element
center element
search interval
search interval

Algorithm Examples
Algorithm Examples
a c d f g
a c d f g h
h j
j l m
l m o p r s u v x z
o p r s u v x z
center element
center element
search interval
search interval
found !
found !

Algorithm Examples
Algorithm Examples
procedure
procedure binary_search(x: integer; a
binary_search(x: integer; a1
1, a
, a2
2, …, a
, …, an
n:
:
integers)
integers)
i := 1
i := 1 {i is left endpoint of search interval}
{i is left endpoint of search interval}
j := n
j := n {j is right endpoint of search interval}
{j is right endpoint of search interval}
while
while (i < j)
(i < j)
begin
begin
m :=
m := 
(i + j)/2
(i + j)/2

if
if x > a
x > am
m then
then i := m + 1
i := m + 1
else
else j := m
j := m
end
end
if
if x = a
x = ai
i then
then location := i
location := i
else
else location := 0
location := 0
{location is the subscript of the term that equals
{location is the subscript of the term that equals
x, or is zero if x is not found}
x, or is zero if x is not found}

Complexity
Complexity
In general, we are not so much interested in the
In general, we are not so much interested in the
time and space complexity for small inputs.
time and space complexity for small inputs.
For example, while the difference in time
For example, while the difference in time
complexity between linear and binary search is
complexity between linear and binary search is
meaningless for a sequence with n = 10, it is
meaningless for a sequence with n = 10, it is
gigantic for n = 2
gigantic for n = 230
30
.
.

Complexity
Complexity
For example, let us assume two algorithms A and
For example, let us assume two algorithms A and
B that solve the same class of problems.
B that solve the same class of problems.
The time complexity of A is 5,000n, the one for
The time complexity of A is 5,000n, the one for
B is
B is 
1.1
1.1n
n

 for an input with n elements.
for an input with n elements.
For n = 10, A requires 50,000 steps, but B only 3,
For n = 10, A requires 50,000 steps, but B only 3,
so B seems to be superior to A.
so B seems to be superior to A.
For n = 1000, however, A requires 5,000,000
For n = 1000, however, A requires 5,000,000
steps, while B requires 2.5
steps, while B requires 2.5
10
1041
41
steps.
steps.

Complexity
Complexity
This means that algorithm B cannot be used for
This means that algorithm B cannot be used for
large inputs, while algorithm A is still feasible.
large inputs, while algorithm A is still feasible.
So what is important is the
So what is important is the growth
growth of the
of the
complexity functions.
complexity functions.
The growth of time and space complexity with
The growth of time and space complexity with
increasing input size n is a suitable measure for
increasing input size n is a suitable measure for
the comparison of algorithms.
the comparison of algorithms.

Complexity
Complexity
Comparison:
Comparison: time complexity of algorithms A and B
time complexity of algorithms A and B
Algorithm A
Algorithm A Algorithm B
Algorithm B
Input Size
Input Size
n
n
10
10
100
100
1,000
1,000
1,000,000
1,000,000
5,000n
5,000n
50,000
50,000
500,000
500,000
5,000,000
5,000,000
5
5
10
109
9

1.1
1.1n
n


3
3
2.5
2.5
10
1041
41
13,781
13,781
4.8
4.8
10
1041392
41392

The Growth of Functions
The growth of functions is usually described
The growth of functions is usually described
using the
using the big-O notation
big-O notation.
.
Definition:
Definition: Let f and g be functions from the
Let f and g be functions from the
integers or the real numbers to the real numbers.
integers or the real numbers to the real numbers.
We say that f(x) is O(g(x)) if there are constants
We say that f(x) is O(g(x)) if there are constants
C and k such that
C and k such that
|f(x)|
|f(x)| 
 C|g(x)|
C|g(x)|
whenever x > k.
whenever x > k.

When we analyze the growth of
When we analyze the growth of complexity
complexity
functions
functions, f(x) and g(x) are always positive.
, f(x) and g(x) are always positive.
Therefore, we can simplify the big-O requirement
Therefore, we can simplify the big-O requirement
to
to
f(x)
f(x) 
 C
C
g(x) whenever x > k.
g(x) whenever x > k.
If we want to show that f(x) is O(g(x)), we only
If we want to show that f(x) is O(g(x)), we only
need to find
need to find one
one pair (C, k) (which is never unique).
pair (C, k) (which is never unique).

The idea behind the big-O notation is to establish
The idea behind the big-O notation is to establish
an
an upper boundary
upper boundary for the growth of a function
for the growth of a function
f(x) for
f(x) for large
large x.
x.
This boundary is specified by a function g(x) that
This boundary is specified by a function g(x) that
is usually much
is usually much simpler
simpler than f(x).
than f(x).
We accept the constant C in the requirement
We accept the constant C in the requirement
f(x)
f(x) 
 C
C
g(x) whenever x > k,
g(x) whenever x > k,
because
because C does not grow with x.
C does not grow with x.
We are only interested in large x, so it is OK if
We are only interested in large x, so it is OK if
f(x) > C
f(x) > C
g(x) for x
g(x) for x 
 k.
k.

Example:
Example:
Show that f(x) = x
Show that f(x) = x2
2
+ 2x + 1 is O(x
+ 2x + 1 is O(x2
2
).
).
For x > 1 we have:
For x > 1 we have:
x
x2
2
+ 2x + 1
+ 2x + 1 
 x
x2
2
+ 2x
+ 2x2
2
+ x
+ x2
2

 x
x2
2
+ 2x + 1
+ 2x + 1 
 4x
4x2
2
Therefore, for C = 4 and k = 1:
Therefore, for C = 4 and k = 1:
f(x)
f(x) 
 Cx
Cx2
2
whenever x > k.
whenever x > k.

 f(x) is O(x
f(x) is O(x2
2
).
).

Question: If f(x) is O(x
Question: If f(x) is O(x2
2
), is it also O(x
), is it also O(x3
3
)?
)?
Yes.
Yes. x
x3
3
grows faster than x
grows faster than x2
2
, so x
, so x3
3
grows also
grows also
faster than f(x).
faster than f(x).
Therefore, we always have to find the
Therefore, we always have to find the smallest
smallest
simple function g(x) for which f(x) is O(g(x)).
simple function g(x) for which f(x) is O(g(x)).

“
“Popular” functions g(n) are
Popular” functions g(n) are
n log n, 1, 2
n log n, 1, 2n
n
, n
, n2
2
, n!, n, n
, n!, n, n3
3
, log n
, log n
Listed from slowest to fastest growth:
Listed from slowest to fastest growth:
• 1
1
• log n
log n
• n
n
• n log n
n log n
• n
n2
2
• n
n3
3
• 2
2n
n
• n!
n!

A problem that can be solved with polynomial
A problem that can be solved with polynomial
worst-case complexity is called
worst-case complexity is called tractable
tractable.
.
Problems of higher complexity are called
Problems of higher complexity are called
intractable.
intractable.
Problems that no algorithm can solve are called
Problems that no algorithm can solve are called
unsolvable
unsolvable.
.
You will find out more about this in CMSC441.
You will find out more about this in CMSC441.

Useful Rules for Big-O
Useful Rules for Big-O
For any
For any polynomial
polynomial f(x) = a
f(x) = an
nx
xn
n
+ a
+ an-1
n-1x
xn-1
n-1
+ … + a
+ … + a0
0, where
, where
a
a0
0, a
, a1
1, …, a
, …, an
n are real numbers,
are real numbers,
f(x) is O(x
f(x) is O(xn
n
).
).
If f
If f1
1(x) is O(g
(x) is O(g1
1(x)) and f
(x)) and f2
2(x) is O(g
(x) is O(g2
2(x)), then
(x)), then
(f
(f1
1 + f
+ f2
2)(x) is O(max(g
)(x) is O(max(g1
1(x), g
(x), g2
2(x)))
(x)))
If f
If f1
1(x) is O(g(x)) and f
(x) is O(g(x)) and f2
2(x) is O(g(x)), then
(x) is O(g(x)), then
(f
(f1
1 + f
+ f2
2)(x) is O(g(x)).
)(x) is O(g(x)).
If f
If f1
1(x) is O(g
(x) is O(g1
1(x)) and f
(x)) and f2
2(x) is O(g
(x) is O(g2
2(x)), then
(x)), then
(f
(f1
1f
f2
2)(x) is O(g
)(x) is O(g1
1(x) g
(x) g2
2(x)).
(x)).

Complexity Examples
Complexity Examples
What does the following algorithm compute?
What does the following algorithm compute?
procedure
procedure who_knows(a
who_knows(a1
1, a
, a2
2, …, a
, …, an
n: integers)
: integers)
m := 0
m := 0
for
for i := 1 to n-1
i := 1 to n-1
for
for j := i + 1 to n
j := i + 1 to n
if
if |a
|ai
i – a
– aj
j| > m
| > m then
then m := |a
m := |ai
i – a
– aj
j|
|
{m is the maximum difference between any two
{m is the maximum difference between any two
numbers in the input sequence}
numbers in the input sequence}
Comparisons: n-1 + n-2 + n-3 + … + 1
Comparisons: n-1 + n-2 + n-3 + … + 1
= (n – 1)n/2 = 0.5n
= (n – 1)n/2 = 0.5n2
2
– 0.5n
– 0.5n
Time complexity is O(n
Time complexity is O(n2
2
).
).

Complexity Examples
Complexity Examples
Another algorithm solving the same problem:
Another algorithm solving the same problem:
procedure
procedure max_diff(a
max_diff(a1
1, a
, a2
2, …, a
, …, an
n: integers)
: integers)
min := a1
min := a1
max := a1
max := a1
for
for i := 2 to n
i := 2 to n
if
if a
ai
i < min
< min then
then min := a
min := ai
i
else
else if a
if ai
i > max
> max then
then max := a
max := ai
i
m := max - min
m := max - min
Comparisons: 2n - 2
Comparisons: 2n - 2
Time complexity is O(n).
Time complexity is O(n).

Let us get into…
Let us get into…
Number Theory
Number Theory

Introduction to Number Theory
Introduction to Number Theory
Number theory is about
Number theory is about integers
integers and their
and their
properties.
properties.
We will start with the basic principles of
We will start with the basic principles of
• divisibility,
divisibility,
• greatest common divisors,
greatest common divisors,
• least common multiples, and
least common multiples, and
• modular arithmetic
modular arithmetic
and look at some relevant algorithms.
and look at some relevant algorithms.

Division
Division
If a and b are integers with a
If a and b are integers with a 
 0, we say that
0, we say that
a
a divides
divides b if there is an integer c so that b = ac.
b if there is an integer c so that b = ac.
When a divides b we say that a is a
When a divides b we say that a is a factor
factor of b
of b
and that b is a
and that b is a multiple
multiple of a.
of a.
The notation
The notation a | b
a | b means that a divides b.
means that a divides b.
We write
We write a
a χ
χ b
b when a does not divide b
when a does not divide b
(see book for correct symbol).
(see book for correct symbol).

Divisibility Theorems
Divisibility Theorems
For integers a, b, and c it is true that
For integers a, b, and c it is true that
• if a | b and a | c, then a | (b + c)
if a | b and a | c, then a | (b + c)
Example:
Example: 3 | 6 and 3 | 9, so 3 | 15.
3 | 6 and 3 | 9, so 3 | 15.
• if a | b, then a | bc for all integers c
if a | b, then a | bc for all integers c
Example:
Example: 5 | 10, so 5 | 20, 5 | 30, 5 | 40, …
5 | 10, so 5 | 20, 5 | 30, 5 | 40, …
• if a | b and b | c, then a | c
if a | b and b | c, then a | c
Example:
Example: 4 | 8 and 8 | 24, so 4 | 24.
4 | 8 and 8 | 24, so 4 | 24.

Primes
Primes
A positive integer p greater than 1 is called prime
A positive integer p greater than 1 is called prime
if the only positive factors of p are 1 and p.
if the only positive factors of p are 1 and p.
Note: 1 is not a prime
Note: 1 is not a prime
A positive integer that is greater than 1 and is not
A positive integer that is greater than 1 and is not
prime is called composite.
prime is called composite.
The fundamental theorem of arithmetic:
The fundamental theorem of arithmetic:
Every positive integer can be written
Every positive integer can be written uniquely
uniquely as
as
the
the product of primes
product of primes, where the prime factors
, where the prime factors
are written in order of increasing size.
are written in order of increasing size.

Primes
Primes
Examples:
Examples:
3·5
3·5
48 =
48 =
17 =
17 =
100 =
100 =
512 =
512 =
515 =
515 =
28 =
28 =
15 =
15 =
2·2·2·2·3 = 2
2·2·2·2·3 = 24
4
·3
·3
17
17
2·2·5·5 = 2
2·2·5·5 = 22
2
·5
·52
2
2·2·2·2·2·2·2·2·2 = 2
2·2·2·2·2·2·2·2·2 = 29
9
5·103
5·103
2·2·7
2·2·7

Primes
Primes
If n is a composite integer, then n has a prime
If n is a composite integer, then n has a prime
divisor less than or equal .
divisor less than or equal .
This is easy to see: if n is a composite integer, it
This is easy to see: if n is a composite integer, it
must have at least two prime divisors. Let the
must have at least two prime divisors. Let the
largest two be p
largest two be p1
1 and p
and p2
2. Then p
. Then p1
1
p
p2
2 <= n.
<= n.
p
p1
1 and p
and p2
2 cannot both be greater than
cannot both be greater than
, because then p
, because then p1
1
p
p2
2 > n.
> n.
n
n

The Division Algorithm
Let
Let a
a be an integer and
be an integer and d
d a positive integer.
a positive integer.
Then there are unique integers
Then there are unique integers q
q and
and r
r, with
, with
0
0 
 r < d
r < d, such that
, such that a = dq + r
a = dq + r.
.
In the above equation,
In the above equation,
• d
d is called the divisor,
is called the divisor,
• a
a is called the dividend,
is called the dividend,
• q
q is called the quotient, and
is called the quotient, and
• r
r is called the remainder.
is called the remainder.

Example:
Example:
When we divide 17 by 5, we have
When we divide 17 by 5, we have
17 = 5
17 = 5
3 + 2.
3 + 2.
• 17 is the dividend,
17 is the dividend,
• 5 is the divisor,
5 is the divisor,
• 3 is called the quotient, and
3 is called the quotient, and
• 2 is called the remainder.
2 is called the remainder.

Another example:
Another example:
What happens when we divide -11 by 3 ?
What happens when we divide -11 by 3 ?
Note that the remainder cannot be negative.
Note that the remainder cannot be negative.
-11 = 3
-11 = 3
(-4) + 1.
(-4) + 1.
• -11 is the dividend,
-11 is the dividend,
• 3 is the divisor,
3 is the divisor,
• -4 is called the quotient, and
-4 is called the quotient, and
• 1 is called the remainder.
1 is called the remainder.

Greatest Common Divisors
Let a and b be integers, not both zero.
Let a and b be integers, not both zero.
The largest integer d such that d | a and d | b is
The largest integer d such that d | a and d | b is
called the
called the greatest common divisor
greatest common divisor of a and b.
of a and b.
The greatest common divisor of a and b is denoted
The greatest common divisor of a and b is denoted
by gcd(a, b).
by gcd(a, b).
Example 1:
Example 1: What is gcd(48, 72) ?
What is gcd(48, 72) ?
The positive common divisors of 48 and 72 are
The positive common divisors of 48 and 72 are
1, 2, 3, 4, 6, 8, 12, 16, and 24, so gcd(48, 72) = 24.
1, 2, 3, 4, 6, 8, 12, 16, and 24, so gcd(48, 72) = 24.
Example 2:
Example 2: What is gcd(19, 72) ?
What is gcd(19, 72) ?
The only positive common divisor of 19 and 72 is
The only positive common divisor of 19 and 72 is
1, so gcd(19, 72) = 1.
1, so gcd(19, 72) = 1.

Using prime factorizations:
a = p
a = p1
1
a
a
1
1 p
p2
2
a
a
2
2 … p
… pn
n
a
a
n
n , b = p
, b = p1
1
b
b
1
1 p
p2
2
b
b
2
2 … p
… pn
n
b
b
n
n ,
,
where p
where p1
1 < p
< p2
2 < … < p
< … < pn
n and a
and ai
i, b
, bi
i 
 N
N for 1
for 1 
 i
i 
 n
n
gcd(a, b) = p
gcd(a, b) = p1
1
min(a
min(a
1
1, b
, b
1
1 )
)
p
p2
2
min(a
min(a
2
2, b
, b
2
2 )
)
… p
… pn
n
min(a
min(a
n
n, b
, b
n
n )
)
Example:
Example:
a = 60 =
a = 60 = 2
22
2
3
31
1
5
51
1
b = 54 =
b = 54 = 2
21
1
3
33
3
5
50
0
gcd(a, b) =
gcd(a, b) = 2
21
1
3
31
1
5
50
0
= 6
= 6

Relatively Prime Integers
Definition:
Definition:
Two integers a and b are
Two integers a and b are relatively prime
relatively prime if
if
gcd(a, b) = 1.
gcd(a, b) = 1.
Examples:
Examples:
Are 15 and 28 relatively prime?
Yes, gcd(15, 28) = 1.
Yes, gcd(15, 28) = 1.
Yes, gcd(55, 28) = 1.
Yes, gcd(55, 28) = 1.
No, gcd(35, 28) = 7.
No, gcd(35, 28) = 7.

Definition:
Definition:
The integers a
The integers a1
1, a
, a2
2, …, a
, …, an
n are
are pairwise relatively
pairwise relatively
prime
prime if gcd(a
if gcd(ai
i, a
, aj
j) = 1 whenever 1
) = 1 whenever 1 
 i < j
i < j 
 n.
n.
Examples:
Examples:
Are 15, 17, and 27 pairwise relatively prime?
No, because gcd(15, 27) = 3.
No, because gcd(15, 27) = 3.
Yes, because gcd(15, 17) = 1, gcd(15, 28) = 1 and
Yes, because gcd(15, 17) = 1, gcd(15, 28) = 1 and
gcd(17, 28) = 1.
gcd(17, 28) = 1.

Least Common Multiples
Definition:
Definition:
The
The least common multiple
least common multiple of the positive
of the positive
integers a and b is the smallest positive integer
integers a and b is the smallest positive integer
that is divisible by both a and b.
that is divisible by both a and b.
We denote the least common multiple of a and b
We denote the least common multiple of a and b
by lcm(a, b).
by lcm(a, b).
Examples:
Examples:
lcm(3, 7) =
lcm(3, 7) = 21
21
lcm(4, 6) =
lcm(4, 6) = 12
12
lcm(5, 10) =
lcm(5, 10) = 10
10

a = p
a = p1
1
a
a
1
1 p
p2
2
a
a
2
2 … p
… pn
n
a
a
n
n , b = p
, b = p1
1
b
b
1
1 p
p2
2
b
b
2
2 … p
… pn
n
b
b
n
n ,
,
where p
where p1
1 < p
< p2
2 < … < p
< … < pn
n and a
and ai
i, b
, bi
i 
 N
N for 1
for 1 
 i
i 
 n
n
lcm(a, b) = p
lcm(a, b) = p1
1
max(a
max(a
1
1, b
, b
1
1 )
)
p
p2
2
max(a
max(a
2
2, b
, b
2
2 )
)
… p
… pn
n
max(a
max(a
n
n, b
, b
n
n )
)
Example:
Example:
a = 60 =
a = 60 = 2
22
2
3
31
1
5
51
1
b = 54 =
b = 54 = 2
21
1
3
33
3
5
50
0
lcm(a, b) =
lcm(a, b) = 2
22
2
3
33
3
5
51
1
= 4
= 4
27
27
5 = 540
5 = 540

GCD and LCM
GCD and LCM
a = 60 =
a = 60 = 2
22
2
3
31
1
5
51
1
b = 54 =
b = 54 = 2
21
1
3
33
3
5
50
0
lcm(a, b) =
lcm(a, b) = 2
22
2
3
33
3
5
51
1
= 540
= 540
gcd(a, b) =
gcd(a, b) = 2
21
1
3
31
1
5
50
0
= 6
= 6
Theorem: a
Theorem: a
b =
b = gcd(a,b)
gcd(a,b)
lcm(a,b)
lcm(a,b)

Modular Arithmetic
Modular Arithmetic
Let a be an integer and m be a positive integer.
Let a be an integer and m be a positive integer.
We denote by
We denote by a mod m
a mod m the remainder when a is
the remainder when a is
divided by m.
divided by m.
Examples:
Examples:
9 mod 4 =
9 mod 4 = 1
1
9 mod 3 =
9 mod 3 = 0
0
9 mod 10 =
9 mod 10 = 9
9
-13 mod 4 =
-13 mod 4 = 3
3

Congruences
Congruences
Let a and b be integers and m be a positive integer.
Let a and b be integers and m be a positive integer.
We say that
We say that a is congruent to b modulo m
a is congruent to b modulo m if
if
m divides a – b.
m divides a – b.
We use the notation
We use the notation a
a 
 b (mod m)
b (mod m) to indicate
to indicate
that a is congruent to b modulo m.
that a is congruent to b modulo m.
In other words:
In other words:
a
a 
 b (mod m) if and only if
b (mod m) if and only if a mod m = b mod m
a mod m = b mod m.
.

Congruences
Congruences
Examples:
Examples:
Is it true that 46
Is it true that 46 
 68 (mod 11) ?
68 (mod 11) ?
Yes, because 11 | (46 – 68).
Yes, because 11 | (46 – 68).
Is it true that 46
Is it true that 46 
 68 (mod 22)?
68 (mod 22)?
Yes, because 22 | (46 – 68).
Yes, because 22 | (46 – 68).
For which integers z is it true that z
For which integers z is it true that z 
 12 (mod 10)?
12 (mod 10)?
It is true for any z
It is true for any z
{…,-28, -18, -8, 2, 12, 22, 32, …}
{…,-28, -18, -8, 2, 12, 22, 32, …}
Theorem:
Theorem: Let m be a positive integer. The integers
Let m be a positive integer. The integers
a and b are congruent modulo m if and only if there
a and b are congruent modulo m if and only if there
is an integer k such that a = b + km.
is an integer k such that a = b + km.

Congruences
Congruences
Theorem:
Theorem: Let m be a positive integer.
Let m be a positive integer.
If a
If a 
 b (mod m) and c
b (mod m) and c 
 d (mod m), then
d (mod m), then
a + c
a + c 
 b + d (mod m) and ac
b + d (mod m) and ac 
 bd (mod m).
bd (mod m).
Proof:
Proof:
We know that a
We know that a 
 b (mod m) and c
b (mod m) and c 
 d (mod m)
d (mod m)
implies that there are integers s and t with
implies that there are integers s and t with
b = a + sm and d = c + tm.
b = a + sm and d = c + tm.
Therefore,
Therefore,
b + d = (a + sm) + (c + tm) = (a + c) + m(s + t) and
b + d = (a + sm) + (c + tm) = (a + c) + m(s + t) and
bd = (a + sm)(c + tm) = ac + m(at + cs + stm).
bd = (a + sm)(c + tm) = ac + m(at + cs + stm).
Hence, a + c
Hence, a + c 
 b + d (mod m) and ac
b + d (mod m) and ac 
 bd (mod m).
bd (mod m).

Congruences
Congruences
Theorem:
Theorem: Let m be a positive integer. a
Let m be a positive integer. a 
 b (mod m)
b (mod m)
iff a mod m = b mod m.
iff a mod m = b mod m.
Proof:
Proof:
Let a = mq1 + r1, and b = mq2 + r2.
Let a = mq1 + r1, and b = mq2 + r2.
Only if part:
Only if part: a mod m = b mod m
a mod m = b mod m 
 r1 = r2, therefore
r1 = r2, therefore
a – b = m(q1 – q2), and a
a – b = m(q1 – q2), and a 
 b (mod m).
b (mod m).
If part:
If part: a
a 
 b (mod m) implies
b (mod m) implies
a – b = mq
a – b = mq
mq1 + r1 – (mq2 + r2) = mq
mq1 + r1 – (mq2 + r2) = mq
r1 – r2 = m(q – q1 + q2).
r1 – r2 = m(q – q1 + q2).
Since 0
Since 0 
 r1, r2
r1, r2 
 m, 0
m, 0 
 |r1 - r2|
|r1 - r2| 
 m. The only
m. The only
multiple in that range is 0.
multiple in that range is 0.
Therefore r1 = r2, and a mod m = b mod m.
Therefore r1 = r2, and a mod m = b mod m.

The Euclidean Algorithm
The
Euclidean Algorithm finds the
finds the greatest
greatest
common divisor
common divisor of two integers a and b.
of two integers a and b.
For example, if we want to find gcd(287, 91), we
For example, if we want to find gcd(287, 91), we
divide
divide 287 by 91:
287 by 91:
287 = 91
287 = 91
3 + 14
3 + 14
We know that for integers a, b and c,
We know that for integers a, b and c,
if a | b and a | c, then a | (b + c).
if a | b and a | c, then a | (b + c).
Therefore, any divisor (including their gcd) of 287
Therefore, any divisor (including their gcd) of 287
and 91 must also be a divisor of 287 - 91
and 91 must also be a divisor of 287 - 91
3 = 14.
3 = 14.
Consequently, gcd(287, 91) = gcd(14, 91).
Consequently, gcd(287, 91) = gcd(14, 91).

In the next step, we divide 91 by 14:
In the next step, we divide 91 by 14:
91 = 14
91 = 14
6 + 7
6 + 7
This means that gcd(14, 91) = gcd(14, 7).
This means that gcd(14, 91) = gcd(14, 7).
So we divide 14 by 7:
So we divide 14 by 7:
14 = 7
14 = 7
2 + 0
2 + 0
We find that 7 | 14, and thus gcd(14, 7) = 7.
We find that 7 | 14, and thus gcd(14, 7) = 7.
Therefore, gcd(287, 91) = 7.
Therefore, gcd(287, 91) = 7.

In
In pseudocode
pseudocode, the algorithm can be implemented
, the algorithm can be implemented
as follows:
as follows:
procedure
procedure gcd(a, b: positive integers)
gcd(a, b: positive integers)
x := a
x := a
y := b
y := b
while
while y
y 
 0
0
begin
begin
r := x
r := x mod
mod y
y
x := y
x := y
y := r
y := r
end
end {x is gcd(a, b)}
{x is gcd(a, b)}

Representations of Integers
Let b be a positive integer greater than 1.
Let b be a positive integer greater than 1.
Then if n is a positive integer, it can be expressed
Then if n is a positive integer, it can be expressed
uniquely
uniquely in the form:
in the form:
n = a
n = ak
kb
bk
k
+ a
+ ak-1
k-1b
bk-1
k-1
+ … + a
+ … + a1
1b + a
b + a0
0,
,
where k is a nonnegative integer,
where k is a nonnegative integer,
a
a0
0, a
, a1
1, …, a
, …, ak
k are nonnegative integers less than b,
are nonnegative integers less than b,
and a
and ak
k 
 0.
0.
Example for b=10:
Example for b=10:
859 = 8
859 = 8
10
102
2
+ 5
+ 5
10
101
1
+ 9
+ 9
10
100
0

Example for b=2 (binary expansion):
Example for b=2 (binary expansion):
(10110)
(10110)2
2 = 1
= 1
2
24
4
+ 1
+ 1
2
22
2
+ 1
+ 1
2
21
1
= (22)
= (22)10
10
Example for b=16 (hexadecimal expansion):
Example for b=16 (hexadecimal expansion):
(we use letters A to F to indicate numbers 10 to 15)
(we use letters A to F to indicate numbers 10 to 15)
(3A0F)
(3A0F)16
16 = 3
= 3
16
163
3
+ 10
+ 10
16
162
2
+ 15
+ 15
16
160
0
= (14863)
= (14863)10
10

How can we construct the base b expansion of an
How can we construct the base b expansion of an
integer n?
integer n?
First, divide n by b to obtain a quotient q
First, divide n by b to obtain a quotient q0
0 and
and
remainder a
remainder a0
0, that is,
, that is,
n = bq
n = bq0
0 + a
+ a0
0, where 0
, where 0 
 a
a0
0 < b.
< b.
The remainder a
The remainder a0
0 is the rightmost digit in the base b
is the rightmost digit in the base b
expansion of n.
expansion of n.
Next, divide q
Next, divide q0
0 by b to obtain:
by b to obtain:
q
q0
0 = bq
= bq1
1 + a
+ a1
1, where 0
, where 0 
 a
a1
1 < b.
< b.
a
a1
1 is the second digit from the right in the base b
is the second digit from the right in the base b
expansion of n. Continue this process until you obtain a
expansion of n. Continue this process until you obtain a
quotient equal to zero.
quotient equal to zero.

Example:
Example:
What is the base 8 expansion of (12345)
What is the base 8 expansion of (12345)10
10 ?
?
First, divide 12345 by 8:
First, divide 12345 by 8:
12345 = 8
12345 = 8
1543 + 1
1543 + 1
1543 = 8
1543 = 8
192 + 7
192 + 7
192 = 8
192 = 8
24 + 0
24 + 0
24 = 8
24 = 8
3 + 0
3 + 0
3 = 8
3 = 8
0 + 3
0 + 3
The result is: (12345)
The result is: (12345)10
10 = (30071)
= (30071)8
8.
.

procedure
procedure base_b_expansion(n, b: positive integers)
base_b_expansion(n, b: positive integers)
q := n
q := n
k := 0
k := 0
while
while q
q 
 0
0
begin
begin
a
ak
k := q mod b
:= q mod b
q :=
q := 
q/b
q/b

k := k + 1
k := k + 1
end
end
{the base b expansion of n is (a
{the base b expansion of n is (ak-1
k-1 … a
… a1
1a
a0
0)
)b
b }
}

Addition of Integers
How do we (humans) add two integers?
How do we (humans) add two integers?
Example:
Example: 7583
7583
+
+ 4932
4932
5
5
1
1
5
5
2
2
1
1
1
1
1
1
1
1 carry
carry
Binary expansions:
Binary expansions: (1011)
(1011)2
2
+
+ (1010)
(1010)2
2
1
1
0
0
carry
carry
1
1
1
1
0
0
1
1
1
1
(
( )
)2
2

Let a = (a
Let a = (an-1
n-1a
an-2
n-2…a
…a1
1a
a0
0)
)2
2, b = (b
, b = (bn-1
n-1b
bn-2
n-2…b
…b1
1b
b0
0)
)2.
2.
How can we
How can we algorithmically
algorithmically add these two binary
add these two binary
numbers?
numbers?
First, add their rightmost bits:
First, add their rightmost bits:
a
a0
0 + b
+ b0
0 = c
= c0
0
2 + s
2 + s0
0,
,
where s
where s0
0 is the
is the rightmost bit
rightmost bit in the binary expansion
in the binary expansion
of a + b, and c
of a + b, and c0
0 is the
is the carry
carry.
.
Then, add the next pair of bits and the carry:
Then, add the next pair of bits and the carry:
a
a1
1 + b
+ b1
1 + c
+ c0
0 = c
= c1
1
2 + s
2 + s1
1,
,
where s
where s1
1 is the
is the next bit
next bit in the binary expansion of a +
in the binary expansion of a +
b, and c
b, and c1
1 is the carry.
is the carry.

Continue this process until you obtain c
Continue this process until you obtain cn-1
n-1.
.
The leading bit of the sum is s
The leading bit of the sum is sn
n = c
= cn-1
n-1.
.
The result is:
The result is:
a + b = (s
a + b = (sn
ns
sn-1
n-1…s
…s1
1s
s0
0)
)2
2

Example:
Example:
Add a = (1110)
Add a = (1110)2
2 and b = (1011)
and b = (1011)2
2.
.
a
a0
0 + b
+ b0
0 = 0 + 1 = 0
= 0 + 1 = 0
2 + 1, so that c
2 + 1, so that c0
0 = 0 and s
= 0 and s0
0 = 1.
= 1.
a
a1
1 + b
+ b1
1 + c
+ c0
0 = 1 + 1 + 0 = 1
= 1 + 1 + 0 = 1
2 + 0, so c
2 + 0, so c1
1 = 1 and s
= 1 and s1
1 = 0.
= 0.
a
a2
2 + b
+ b2
2 + c
+ c1
1 = 1 + 0 + 1 = 1
= 1 + 0 + 1 = 1
2 + 0, so c
2 + 0, so c2
2 = 1 and s
= 1 and s2
2 = 0.
= 0.
a
a3
3 + b
+ b3
3 + c
+ c2
2 = 1 + 1 + 1 = 1
= 1 + 1 + 1 = 1
2 + 1, so c
2 + 1, so c3
3 = 1 and s
= 1 and s3
3 = 1.
= 1.
s
s4
4 = c
= c3
3 = 1.
= 1.
Therefore, s = a + b = (11001)
Therefore, s = a + b = (11001)2
2.
.

procedure
procedure add(a, b: positive integers)
add(a, b: positive integers)
c := 0
c := 0
for j := 0 to n-1
for j := 0 to n-1
begin
begin
d :=
d := 
(a
(aj
j + b
+ bj
j + c)/2
+ c)/2

s
sj
j := a
:= aj
j + b
+ bj
j + c – 2d
+ c – 2d
c := d
c := d
end
end
s
sn
n := c
:= c
{the binary expansion of the sum is (s
{the binary expansion of the sum is (sn
ns
sn-1
n-1…s
…s1
1s
s0
0)
)2
2}
}

Ch02 algorithm 127836129812y12y823873893.ppt

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