1. Chapter 3 3.1 Algorithms 3.2 The Growth of Functions
3.3 Complexity of Algorithms
3.4 The Integers and Division
3.5 Primes and Greatest Common Divisors
3.6 Integers and Algorithms
3.7 Applications of Number Theory
3.8 Matrices

2. Chapter 3 3.1 Algorithms Searching Algorithms Greedy Algorithms
The Halting Problem

3. Algorithm
Definition 1: An algorithm is a finite set of precise instructions for performing a computation or for solving a problem.
Example 1: Describe an algorithm for finding the maximum (largest) value in a finite sequence of integers.

4. We perform the following steps
Set the temporary maximum equal to the first integer in the sequence. (the temporary maximum will be the largest integer examined at any stage of the procedure.)
Compare the next integer in the sequence to the temporary maximum, and if it is larger than the temporary maximum, set the temporary maximum equal to this integer.
Repeat the previous step if there are more integers in the sequence.
Stop when there are no integers left in the sequence. The temporary maximum at this point is the largest integer in the sequence.

5. Pseudocode
Pseudocode provides an intermediate step between an English language description of an algorithm and an implementation of this algorithm in a programming language.
Algorithm 1: Finding the maximum element in a finite sequence.
procedure max(a1, a2, ,an: integers)
max := a1
for i: =2 to n
if max < ai then max := ai
{max is the largest element}

6. Property of Algorithm Input. Output.
Definiteness. The steps of an algorithm must be defined precisely.
Correctness.
Finiteness.
Effectiveness.
Generality. The procedure should be applicable for all problems of the desired form, not just for a particular set of input values.

7. Searching Algorithms Search Problem: Locating an element in an
(ordered) list.
Linear search
Binary search (ordered list)

8. The linear search Algorithm 2 : the linear search algorithm
procedure linear search (x: integer, a1, a2, …,an: distinct integers)
i :=1;
while ( i ≤n and x ≠ ai)
i := i + 1
If i ≤ n then location := i
Else location := 0
{location is the subscript of the term that equals x , or is 0 if x is not found}

9. The binary search Algorithm 3: the binary search algorithm
Procedure binary search (x: integer, a1, a2, …,an: increasing integers)
i :=1 { i is left endpoint of search interval}
j :=n { j is right endpoint of search interval}
While i < j
begin
m := (i+j)/2
if x > am then i := m+1
else j := m
end
If x = ai then location := i
else location :=0
{location is the subscript of the term equal to x, or 0 if x is not
found}
Example 3:
to search for 19 in the list

10. Sorting
Sort:
Sorting is putting elements into a list in which the elements are in increasing order.
E.g.
7,2,1,4,5,9 -> 1,2,4,5,7,9
d,h,c,a,f -> a,c,d,f,h.
Bubble sort
Insertion sort

11. Bubble Sort ALGORITHM 4: The Bubble Sort
procedure bubble sort (a1, a2, …,an: real numbers with n ≥2)
for i := 1 to n-1
for j := 1 to n- i
if aj > aj+1 then interchange aj and aj+1
{a1, a2, …,an is in increasing order}
Example 4: Use the sort to put 3, 2, 4, 1, 5 into increasing order.

12. Bubble Sort
p:\msoffice\My Projects\Rosen 6e 2007\Imagebank\JPEGs \ch03\jpeg\ jpg

13. Insertion Sort Algorithm 5: The Insertion Sort
procedure insertion sort (a1, a2, …,an: real numbers with n ≥2)
for j := 2 to n
begin
i := 1
while aj > ai
i := i + 1
m := aj
for k :=0 to j-i-1
aj-k := a j-k-1
ai := m
end {a1, a2, …,an are sorted}
Example 5:
Use the insertion sort to put the elements of the list 3, 2, 4, 1, 5 into increasing order.

14. Greedy Algorithm Optimization Problem: find the best solution.
Algorithms that make what seems to be the best choice at each step are called greedy algorithms.

15. Example 6: Consider the problem of making n cents change with quarters, dimes, nickels, and pennies, and using the least total number of coins.
Algorithm 6: Greedy Change-Marking Algorithm
procedure change (c1, c2, …, cr: values of denominations of coins, where c1 > c2 > … > cr ; n: a positive integer)
for i := 1 to r
while n ≥ ci
begin
add a coin with value ci to the change
n := n – ci
end

16. The Halting Problem
There is a problem that cannot be solved using any procedure.
That is, there are unsolvable problems.
Halting Problem
FIGURE 2 Showing that the Halting Problem is Unsolvable.

17. Chapter 3 3.1 Algorithms 3.2 The Growth of Functions
3.3 Complexity of Algorithms
3.4 The Integers and Division
3.5 Primes and Greatest Common Divisors
3.6 Integers and Algorithms
3.7 Applications of Number Theory
3.8 Matrices

18. Chapter 3 3.2 The Growth of Functions Big-O Notation
Some Important Big-O Results
The Growth of Combinations of Functions
Big-Omega and Big-Theta Nation

19. The Growth of Functions
We quantify the concept that g grows at least as fast as f.
What really matters in comparing the complexity of algorithms?
We only care about the behavior for large problems.
Even bad algorithms can be used to solve small problems.
Ignore implementation details such as loop counter incrementation, etc. we can straight-line any loop.

20. Big-O Notation
Definition 1: let f and g functions from the set of integers or the set of real numbers to the set of real number. We say that f(x) is O(g(x)) if there are constants C and k such that |f(x)| ≤ C |g(x)| whenever x > k.
This is read as “ f(x) is big-oh of g(x) ”.
The constants C and k in the definition of big-O notation are called witnesses to the relationship f(x) is O(g(x)).
Note:
Choose k
Choose C ; it may depend on your choice of k
Once you choose k and C, you must prove the truth of the implication (often by induction).
Example 1: show that f(x)= x2+ 2x + 1 is O(x2)

21. Big-O Notation
FIGURE 1 The Function x2 + 2x + 1 is O(x2).

22. FIGURE 2 The Function f(x) is O(g(x)).
Big-O Notation
FIGURE 2 The Function f(x) is O(g(x)).

23. Big-O Notation Example 2: show that 7x2 is O( x3 ).
Example 4: Is it also true that x3 is O(7x2)?
Example 3: show that n2 is not O(n).

24. Little-O Notation Definition: if then f is o(g), called little-o of g.
An alternative for those with a calculus background:
Definition: if then f is o(g),
called little-o of g.

25. Theorem: if f is o(g) then f is O(g).
Proof: by definition of limit as n goes to infinity, f(n)/g(n) gets arbitrarily small.
That is for any ε >0 , there must be n integer N such
that when n > N, | f(n)/g(n) | < ε.
Hence, choose C = ε and k= N . Q.E.D.
It is usually easier to prove f is o(g)
Using the theory of limits
Using L’Hospital’s rule
Using the properties of logarithms
etc

26. Example : 3n + 5 is O(n2).
Proof: it’s easy to show using the theory of limits.
Hence, 3n+5 is o(n2) and so it is O(n2).
Q.E.D.

27. Some Important Big-O Results
Theorem 1: let
where a0, a1, . . .,an-1 , an are real numbers
then f(x) is O(xn) .
Example 5: how can big-O notation be used to estimate the sum of the first n positive integers?

28. Some Important Big-O Results
Example 6: give big-O estimates for the factorial function and the logarithm of the factorial function, where the factorial function f(n) =n! is defined by
n! = 1* 2 * 3 * . . .*n
Whenever n is a positive integer, and 0!=1.

29. Some Important Big-O Results
Example 7: In Section 4.1 ,we will show that n <2n whenever n is a positive integer.
Show that this inequality implies that n is O(2n) , and use this inequality to show that log n is O(n).

30. The Growth of Combinations of Functions1
logn n
n log n n2 2n n!
FIGURE 3 A Display of the Growth of Functions Commonly Used in Big-O Estimates.

31. Important Complexity Classes
Where j > 2 and c> 1.
Example :Find the complexity class of the function
Solution: this means to simplify the expression.
Throw out stuff which you know doesn’t grow as fast.
We are using the property that if f is O(g) then f + g is O(g).

32. Important Complexity Classes
if a flop takes a nanosecond, how big can a problem be solved (the value of n ) in a minute? a day? a year? For the complexity class O(n n! nn)

33. Important Complexity Classes
a minute= 60*109= 6*1010 flops a day= 24*60*60= 8.65*1013 flops a year= 365*24*60*60*109= *1016 flops We want to find the maximal integer so that n*n!*nn < 6*1010 n*n!*nn < 8.65*1013 n*n!*nn < *1016

34. Important Complexity Classes
Maple Program: for k from 1 to 10 do (k,k*factorial(k)*kk)end do; 1, 1 2, 16 3, 486 4, , , , , , , So, n=7,8,9 for a minute, a day, and a year.

35. The Growth of Combinations of Functions
Theorem 2: suppose that f1(x) is O(g1(x)) and f2(x) is O(g2(x)). Then (f1 + f2)(x) is O(max( |g1(x)| , |g2(x)| )).
Corollary 1: suppose that f1(x) and f2(x) are both O(g(x)). Then (f1 + f2)(x) is O(g(x)).

36. Theorem: If f1 is O(g1) and f2 is O(g2) then
f1 f2 is O(g1g2)
f1+f2 is O(max {g1 ,g2})

37. The Growth of Combinations of Functions
Theorem 3 :suppose that f1(x) is O(g1(x)) and f2(x) is O(g2(x)).
Then (f1f2)(x) is O(g1(x) g2(x)).
Example 8: give a big-O estimate for
f(n)=3n log(n!) + (n2 +3) log n
where n is a positive integer.
Example 9: give a big-O estimate for
f(x)=(x+1)log(x2+1) + 3x2

38. Properties of Big-O f is O(g) iff If f is O(g) and g is O(f) then
The set O(g) is closed under addition:
if f is O(g) and h is O(g) then f+h is O(g)
The set O(g) is closed under multiplication by a scalar a (real number):if f is O(g) then af is O(g)
That is ,O(g) is a vector space. (The proof is in the book.)
Also, as you would expect,
If f is O(g) and g is O(h), then f is O(h) .
In particular

39. Algorithm 1 has complexity n2 – n +1
Note : we often want to compare algorithms in the same complexity class
Example:
Suppose
Algorithm 1 has complexity n2 – n +1
Algorithm 2 has complexity n2/2 + 3n + 2
Then both are O(n2) but Algorithm 2 has a smaller leading coefficient and will be faster for large problems.
Hence we write
Algorithm 1 has complexity n2 +O(n)
Algorithm 2 has complexity n2/2 + O(n)

40. Big-Omega and Big-Theta Nation
Definition 2: Let f and g be functions from the set of integers or the set of real numbers to the set of real numbers.
We say that f(x) is Ω(g(x)) if there are positive constants C and k such that |f(x)|≥ C|g(x)|
Whenever x > k. ( this is read as “f(x) is big-Omega of g(x)” .)
Example 10 :The function f(x) =8x3+ 5x2 +7 is Ω(g(x)) , where g(x) is the function g(x) =x3.
This is easy to see because f(x) =8x3+ 5x2 +7 ≥ x3 for all positive real numbers x. this is equivalent to saying that
g(x) = x3 is O(8x3+ 5x2 +7 ) ,which can be established directly by turning the inequality around.

41. Example 12: show that 3x2 + 8x(logx) is Θ(x2).
Definition 3: Let f and g be functions from the set of integers or the set of real numbers to the set of real numbers.
We say that f(x) is Θ(g(x)) if f(x) is O(g(x)) and f(x) is Ω(g(x)).
When f(x) is Θ(g(x)) , we say that” f is big-Theta of g(x)” and we also say that f(x) is of order g(x).
Example 11: we showed (in example 5) that the sum of the first n positive integers is O(n2). Is this sum of order n2?
Example 12: show that 3x2 + 8x(logx) is Θ(x2).

42. Example 13: the ploynomials
Theorem 4: let
, where a0, a1, . . .,an-1 , an are real numbers with an≠0 . Then f(x) is of order xn .
Example 13: the ploynomials
3x8+10x7+221x2+1444
x19-18x
-x x x
are of orders x8, x19 and x99 ,respectively.

43. Chapter 3 3.3 Complexity of Algorithms Time Complexity
Understanding the complexity of Algorithms

44. Complexity of Algorithm
Computational Complexity (of the Algorithm)
Time Complexity: Analysis of the time required.
Space Complexity: Analysis of the memory required.

45. Time Complexity
Example 1: Describe the time complexity of Algorithm 1 of section 3.1 for finding the maximum element in a set (in terms of number of comparisons).
Algorithm 1: Finding the maximum element in a finite sequence.
procedure max(a1, a2, ,an: integers)
max := a1
for i: =2 to n
if max < ai then max := ai
{max is the largest element}

46. procedure linear search (x: integer, a1, a2, …,an: distinct integers)
Example 2: Describe the time complexity of the linear search algorithm.
Algorithm 2 : the linear search algorithm
procedure linear search (x: integer, a1, a2, …,an: distinct integers)
i :=1;
while ( i ≤n and x ≠ ai)
i := i + 1
If i ≤ n then location := i
Else location := 0
{location is the subscript of the term that equals x , or is 0 if x is
not found}

47. Example 3: Describe the time complexity of the binary search algorithm in terms of the number of comparisons used .
(and ignoring the time required to compute m= in each iteration of the loop in the algorithm)
Algorithm 3: the binary search algorithm
Procedure binary search (x: integer, a1, a2, …,an: increasing integers)
i :=1 { i is left endpoint of search interval}
j :=n { j is right endpoint of search interval}
While i < j
begin
m :=
if x > am then i := m+1
else j := m
end
If x = ai then location := I
else location :=0
{location is the subscript of the term equal to x, or 0 if x is not found}

48. Example 4: Describe the average-case performance of the linear search algorithm, assuming that the element x is in the list.
Example 5: What is the worst-case complexity of the bubble sort in terms of the number of comparisons made?
ALGORITHM 4: The Bubble Sort
procedure bubble sort (a1, a2, …,an: real numbers with n ≥2)
for i := 1 to n-1
for j := 1 to n- i
if aj > aj+1 then interchange aj and aj+1
{a1, a2, …,an is in increasing order}

49. Example 6: What is the worst-case complexity of the insertion sort in terms of the number of comparisons made?
Algorithm 5: The Insertion Sort
procedure insertion sort (a1, a2, …,an: real numbers with n ≥2)
for j := 2 to n
begin
i := 1
while aj > ai
i := i + 1
m := aj
for k :=0 to j-i-1
aj-k := a j-k-1
ai := m
end {a1, a2, …,an are sorted}

50. Understanding the complexity of Algorithms

51. Solvable (in polynomial time, or in exponential time)
Tractable: A problem that is solvable using an algorithm with polynomial worst-case complexity.
Intractable: The situation is much worse for problems that cannot be solved using an algorithm with worst-case polynomial time complexity. The problems are called intractable.
NP problem.
NP-complete problem.
Unsolvable problem: no algorithm to solve them.

52. Big-O estimate on the time complexity of an algorithm provides an upper, but not a lower, bound on the worst-case time required for the algorithm as a function of the input size.
Table 2 displays the time needed to solve problems of various sizes with an algorithm using the indicated number of bit operations. Every bit operation takes nanosecond. Times of more than years are indicated with an asterisk.

53. Chapter 3 3.4 The Integers and Division Division
The Division Algorithm
Modular Arithmetic
Applications of Congruences
Cryptology

54. Division
Definition 1: if a and b are integers with a≠0, we say that a divides b if there is an integer c such that b=ac. When a divides b we say that a is a factor of b and that b is a multiple of a. the notation a|b denotes that a divides b. we write a | b when a does not divide b.
Example 1: Determine whether 3|7 and whether 3|12.
Example: Determine whether 3|0. /

55. Theorem 1: let a, b, and c be integers. Then
If a|b and a|c, then a|(b+c)
If a|b and a|bc for all integer c
If a|b and b|c, then a|c
Corollary 1: If a, b, c are integers such that a|b and a|c , then
a| mb + nc
whenever m and n are integers.

56. The Division Algorithm
Theorem 2 the division algorithm :let a be an integer and d a positive integer. Then there are unique integers q and r, with 0 ≤ r < d, such that
a= dq+r
Definition 2: In the equality give in the division algorithm, d is called the divisor, a is called the dividend, q is called the quotient, and r is called the remainder. This notation is used to express the quotient and remainder.
q = a div d, r = a mod d.
Example 4: What are the quotient and remainder when -11 is divided by 3?

57. Modular Arithmetic
Definition 3: if a and b are integers and m is a positive integer, then a is congruent to b modulo m if m divides a - b.
we use the notation a≡b (mod m) to indicate that a is congruent to b modulo m.
if a and b are not congruent modulo m, we write
a ≡b (mod m) . /

58. Modular Arithmetic
Theorem 3: let a and b be integers, and let m be a positive integer. Then a≡b (mod m) if and only if a mod m = b mod m .
Example 5: determine whether 17 is congruent to 5 modulo 6 and whether 24 and 14 are congruent modulo 6.

59. a+c≡b+d (mod m) , ac ≡ bd (mod m)
Modular Arithmetic
Theorem 4 : let m be positive integer. The integers a and b are congruent modulo m if and only if there is an integer k such that a = b + km .
Theorem 5: let m be a positive integer. If a≡b(mod m ) and c ≡d (mod m), then
a+c≡b+d (mod m) , ac ≡ bd (mod m)
Example 6: because 7≡2 (mod 5) and 11≡1 (mod 5) , it follows from theorem 5 that
18=7+11 ≡2+1=3(mod 5) , and that
77=7*11 ≡2*1=2 (mod 5)

60. Corollary 2: let m be a positive integer and let a and b be integers
Corollary 2: let m be a positive integer and let a and b be integers. Then
(a+b) mod m = ((a mod m)+(b mod m)) mod m
And
ab mod m =((a mod m)(b mod m)) mod m.

61. Applications of Congruences
Hashing Functions
Pseudorandom Numbers
Cryptology

62. Hashing Functions
How can memory locations be assigned so that customer records can be retrieved quickly?
Hashing function and key
h(k) = k mod m; m is the number of available memory locations.
Collision: one way to re solve a collision is to assign the first free location.

63. Pseudorandom Numbers
The numbers generated by systematic method are not truly random, they are called pseudorandom numbers.
Linear Congruential Method(m, a, c, x0 :integers):
Modulus m
Multiplier a, 2  a < m
Increment c, 0  c < m
Seed x0 , 0  x0 < m
xn+1= (axn+c) mod m
For example: m=9, a=7, c=4, x0 =3, then
(x1, x2, x3, x4, x5, x6, x7, x8, x9)=(7, 8, 6, 1, 2, 0, 4, 5, 3)
x10=x1

64. Cryptology Important Application of Congruences
Earliest known uses by Julius Caesar.
Shifting each letter three letters forward in the alphabet.
To express the process mathematically:
Let U={0,.., 25}, V={A, .., Z} and g: V -> U is a bijection function defined as the table below.
Define function f : U -> U, where f(p)=(p+3) mod 26.
The Encryption function h:V->V, where h(x)=g-1( f(g(x) ) )
The decryption function f-1(p)=(p-3) mod 26. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

65. Applications of Congruences
Example 9:
What is the secret message produced from the message “MEET YOU IN THE PARK” using the Caesar cipher.
HW: Example 10, p208

66. Chapter 3 3.5 Primes and Greatest Common Divisors Primes
Greatest common divisors and least common multiples

67. Primes
Definition 1: A positive integer p greater than 1 is called prime if the only positive factors of p are 1 and p. A positive integer that is greater than 1 and is not prime is called composite.
Remark: The integer n is composite if and only if there exists an integer a such that a|n and 1< a < n.
Example 1: The integer 7 is prime because its only positive factors are 1 and 7, whereas the integer 9 is composite because it is divisible by 3.

68. Primes Theorem 1: The fundamental theorem of arithmetic
Every positive integer greater than 1 can be written uniquely as a prime or as the product of two or more primes where the prime factors are written in order of nondecreasing size.
Example 2: The prime factorizations of 100, 641 , 999 and 1024 are given by
100=2*2*5*5=2252
641=641
999=3*3*3*37=33*37
1024=2*2*2*2*2*2*2*2*2*2=210

69. Primes
Theorem 2: If n is a composite integer , then n has a prime divisor less than or equal to
Example 3: Show that 101 is prime.
Example 4: Find the prime factorization of 7007.

70. Primes Theorem 3: There are infinitely many primes . Proof:
We will prove this theorem using a proof by contradiction. We assume that there are only finitely many primes, p1, p2, … , pn.
Let Q=

71. Greatest Common Divisors
Definition 2: Let a and b be integers, not both zero.
The largest integer d such that d|a d|b is called the greatest common divisor of a and b.
The greatest common divisor of a and b is denoted by gcd(a,b).
Example 10: what is the greatest common divisor of
24 and 36?

72. Greatest Common Divisors
Definition 3: The integers a and b are relatively prime if their greatest common divisor is 1.
Example 12: Prove that y the integers 17 and 22 are relatively prime.

73. Greatest Common Divisors
Definition 4: The integers a1,a2 …,an are pairwise relatively prime if gcd(ai , aj)=1 whenever 1≦i <j ≦n.
Example 13: determine whether the integers 10 , 17 and 21 are pairwise relatively prime and whether the integers 10 , 19 and 24 are pairwise relatively prime.
Example 14: Because the prime factorizations of 120 and 500 are 120=23*3*5 and 500=22*53, the greatest common divisor is
gcd(120,500)=2 min(3 , 2) 3 min(1 , 0) 5 min(1,3)=223051=20

74. Least Common Multiples
Definition 5: The least common multiple of the positive integers a and b is the smallest positive integer that is divisible by both a and b.
The least common multiple of a and b is denoted by
lcm(a , b).
Example 15: What is the gcd and lcm of and 2433?

75. Greatest Common Divisors and Least Common Multiples
Theorem 5: Let a and b be positive integers.
Then
ab = gcd(a ,b)* lcm(a , b)

76. Chapter 3 3.6 Integers and Algorithms Representations of integers
Algorithms for integer operations
Modular Exponentiation
The Euclidean Algorithm

77. Representations of integers
Theorem 1: Let b be a positive integer greater than 1. Then if n is a positive integer, it can be expressed uniquely in the form
where k is a nonnegative integer, a0, a1, …,ak are nonnegative integers less than b, and ak ≠0.

78. Example 1: What is the decimal expansion of the integer that has ( )2 as its binary expansion?
Example 2: What is the decimal expansion of the hexadecimal expansion of (2AE0B)16 ?

79. Example 3: Find the base 8, or octal, expansion of (12345)10
Example 4: Find the hexadecimal expansion of (177130)10?

80. Algorithm 1: Construction Base b Expansions
procedure base b expansion(n:positive integer)
q: = n
k: =0
while q ≠ 0
begin
ak : =q mod b
q: =
k: =k+1
end {the base b expansion of n is (ak a1 a0)b}

82. Algorithms for integer operations
Algorithm 2: Addition of Integers
Procedure add(a , b:positive integers)
{the binary expansions of a and b are (an a1 a0)2 and
(bn b1 b0)2 respectively}
c : =0
for j: =0 to n-1
Begin
d : =
sj : = aj+bj+c-2d
c : =d
end
sn:=c {the binary expansion of the sum if (sn sn s1 s0)2 }

83. Algorithms for integer operations
Example 7:
Add a=(1110)2 and b=(1011)2.

84. Algorithms for integer operations
Algorithm 3 : Multiplying Integers
procedure multiply(a, b : positive integers)
{the binary expansions of and b are(an a1 a0)2 and
(bn b1 b0)2 respectively}
for j:=0 to n-1
Begin
if bj =1 then cj=a shifted j places
else cj:=0
end
{c0 c cn-1 are the partial products}
p :=0
for j:=0 to n-1
p: = p +cj {p is the value of ab}

85. Algorithms for integer operations
Example 9: Find the product of
a= (110)2 and b=(101)2

86. Algorithms for integer operations
Algorithm 4 : Computing div and mod
procedure division algorithm(a :integers ,d: positive integer)
q: =0
r: =|a|
while r≧d
begin
r := r-d
q :=q+1
end
if a<0 then
if r=0 then q:=-q else
r := d-r
q := -(q+1)
{q = a div d is the quotient, r = a mod d is the remainder}

87. Modular Exponentiation
In cryptography it is important to be able to find bn mod m efficiently, where b, n and m and large integers. It’s impractical to first compute bn and then find its remainder when divided by m because bn will be a huge number. Instead, we can use an algorithm that employ expansion of the exponent n , say n = (ak a1 a0)2 .
Before we present this algorithm, we illustrate its basic idea. We will explain how to use the binary expansion of n to compute bn .First , note that

88. Modular Exponentiation
To compute bn , we find the values of b, b2,(b2)2=b4, (b4)2=b8, ,
We multiply the terms in this list, where aj=1 .
This gives us
For example, to compute 311 we first note that 11 = (1011)2, so that 311=
By successively squaring, we find that 32=9, 34=81, 38=6561.
Consequently,311=383231=6561*9*3= 177,147

89. Modular Exponentiation
Algorithm 5: Modular Exponentiation
procedure modular exponentiation(b:integer ,
n=(ak a1 a0)2 ,m: positive integer)
x: = 1
power := b mod m
for i=0 to k-1
begin
for ai =1 then x :=(x*power) mod m
power :=(power*power) mod m
End
{x equals bn mod m}
Example 11: Use Algorithm 5
to find 3644 mod 645.

90. The Euclidean Algorithm
Lemma 1: Let a=bq+r ,where a, b, q, and r are
integers. Then gcd(a,b)=gcd(b,r).
Algorithm 6: The Euclidean Algorithm
procedure gcd(a.b:integers)
x: = a
y: = b
while y0
begin
r := x mod y
x := y
y := r
end {gcd(a,b) is x}

91. The Euclidean Algorithm
Example 12: Find the GCD of 414 and 662 using the Euclidean Algorithm.

92. Chapter 3 3.7 Applications of Number Theory Some Useful Results
Linear Congruences
The Chinese Remainder Theorem
Computer Arithmetic with Large Integers
Pseudoprimes
Public Key Cryptography

93. Some Useful Results
Theorem 1: If a and b are positive integers, then there exist integers s and t such that gcd(a ,b) = sa+tb .
Example 1: express gcd(252 , 198) =18 as a linear combination of 252 and 198 .

94. Some Useful Results
Lemma 1: If a, b, and c are positive integers such that gcd(a , b) = 1 and a|bc, then a|c .
Lemma 2 : If p is a prime and p|a1a2. . .an, where each ai is an integer , then p|ai for some i.
Theorem 2: Let m be a positive integer and let a, b ,and c be integers. If ac≡ bc (mod m) and gcd(c, m) = 1 , then a≡b (mod m).

95. Linear Congruences
A congruence of the form ax≡b (mod m) where m is a positive integer , a and b are integers , and x is variable, is called a linear congruence.
Such congruences arise throughout number theory and its applications.
How can we solve the linear congruence ax≡b (mod m) ? That is, find the x that satisfy this congruence.
One method that we will describe uses an integer ā such that aā≡1 (mod m), if such an integer exist.
Such an integer ā is said to be an inverse of a modulo m.
Theorem 3 guarantees that an inverse of a modulo m exists whenever a and m are relatively prime.

96. Linear Congruences
Theorem 3: If a and m are relatively prime integers and m>1, then an inverse of modulo m exist. Furthermore, this inverse is unique modulo m.
(there is a unique positive integer ā less than m that is an inverse of a modulo m and every other inverse of a modulo m is congruent to ā modulo m.)
When we have an inverse of a modulo m, that is, ax≡1 (mod m) , we can easily solve the congruence ax≡b (mod m).

97. The Chinese Remainder Theorem
Example 3: Find an inverse of 3 modulo 7?
Theorem 5, section 3.4, p204.
Let m be a positive integer. If a≡b (mod m) and c≡d (mod m), then a+c≡b+d (mod m) and ac≡bd (mod m).
Example 4: What are the solutions of the liner congruence 3x ≡4 ( mod 7)?
Example 5: In the first century, the Chinese mathematician Sun-Tsu asked:
There are certain things whose number is unknown. When divider by 3, the remainder is 2; when divided by 5, the remainder is 3; and when divided by 7 , the remainder is 2. What will be the number of things?

98. The Chinese Remainder Theorem
Theorem 4: The Chinese Remainder Theorem
Let m1, m2, ,mn be pairwise relative prime positive integers and a1, a2,. . . ,an arbitrary integers. Then the system x≡a1 ( mod m1)
x≡a2 ( mod m2) …
x≡an ( mod mn)
has a unique solution modulo m= m1, m2, ,mn .
(That is , there is solution x with 0 ≦x ＜m, and all other solutions are congruent modulo m to this solution.)

99. 射雕英雄傳 第一千四比二十七首
瑛姑說道: 『. . . 今有物不知其數，三三數支謄二，五五數之謄三，七七數之謄二，問物幾何?』 黃蓉笑道: 『這容易得緊，以三三數之，餘數乘以七十; 五五數之，餘數乘以二十一，七七數之，餘數乘以十五。三者相加，如不大於一百零五，即為答數; 否則須減去一百零五或其倍數。』 黃蓉道: 『也不用這般硬記，我念一首詩給你聽，那就容易記了:三人同行七十稀，五樹梅花二一枝，七子團員正半月，餘百零五便得知。』

100. The Chinese Remainder Theorem
Example 6: Solve the system of congruences in Example 5 by using theorem 4.
Example 5:there are certain things whose number is unknown. When divider by 3, the remainder is 2; when divided by 5, the remainder is 3; and when divided by 7 , the remainder is 2. What will be the number of things?

101. Computer Arithmetic with Large Integers
Suppose that m1, m2, ,mn are pairwise relatively prime integers greater than or equal to 2 and let m be their product. By the Chinese Remainder Theorem, we can show that an integer a with 0≤ a < m can be uniquely represented by the n-tuple consisting of its remainders upon division by mi , i= 1, 2,. . .,n.
We can uniquely represent a by
(a mod m1, a mod m2, . . ., a mod mn)

102. Computer Arithmetic with Large Integers
Example 7: What are the pairs used to represent the nonnegative integers less than 12 when they are represented by the ordered pair where the first component is the remainder of the integer upon division by 3 and the second component is the remainder of the integer upon division by 4?

103. Pseudoprimes Theorem 5: Fermat’s Little Theorem
If p is prime and a is an integer not divisible by p, then
ap-1 ≡1 (mod p)
Furthermore, for every integer a we have
ap ≡a (mod p)
Unfortunately, there are composite integer n, such that 2n-1≡1 (mod p). Such integers are called pseudoprimes to the base 2.
Example 9: Explain why the integer 341 is a pseudoprime to the base 2.

104. Computer Arithmetic with Large Integers
Definition 1: Let b be a positive integer. If n is a composite positive integer, and bn-1 ≡1 (mod n), then n is called a pseudoprime to the base b.
Definition 2: A composite integer n that satisfies the congruence bn-1 ≡1 (mod n) for all positive integers b with gcd(b , n)=1 is called a Carmichael number.
(This numbers are named after Robert Carmichael, who studied them in the early twentieth century)
Example 10: The integer 561 is a Carmichael number.

105. Private Key Cryptography
Private key cryptosystems (Section 3.4, Example 9, p207)
c=(p+k) mod 26, where p, c represent a letter, k is an encryption key.
Everybody knowing this key can both encrypt and decrypt messages easily.Private
Two people need to securely exchange the key in advance.

106. Public Key Cryptography
In 1976, three researchers at M.I.T. – Ronald Rivest, Adi Shamir, and Leonard Adleman – introduced to the world a public key cryptosystem, known as the RSA system.
The RSA cryptosystem is based on modular exponentiation modulo the product of two large primes, which can be done rapidly using Algorithm 5 in section 3.6.
Each individual has an encryption key consisting of a modulus n=pq, where p and q are large primes, say, with 200 digits each, and an exponent e that is relatively prime to (p-1)(q-1).

107. Public Key Cryptography
To produce a usable key, two large primes must be found. This can be done quickly on a computer using probabilistic primality test. (Example 16, Section 6.2, p text book)
However, the product of these primes n=pq, with approximately 400 digits, cannot be factored in a reasonable length of time. This is an important reason why decryption cannot be done quickly without a separate decryption key.

108. RSA Encryption
In the RSA encryption method, messages are translated into sequences of integers.
These integers are grouped together to form larger integers, each representing a block of letters.
The encryption proceeds by transforming the integer M, representing the plaintext (the original message), to an integer C, representing the ciphertext (the encryption message), using the function C=Me mod n.

109. Gcd(e, (p-1)(q-1)) = gcd(13, 42 x 58)=1.
RSA Encryption
Example 11: Encrypt the message STOP using the RSA cryptosystem with p=43 and q=59, so that n=43 x 59 = 2537, and with e=13. Note that
Gcd(e, (p-1)(q-1)) = gcd(13, 42 x 58)=1.

110. RSA Decryption
The plaintext message can be quickly recovered when the decryption key d, an inverse of e modulo (p-1)(q-1), is known. Such inverse exist because gcd(e, (p-1)(q-1))=1).
de≡1 (mod (p-1)(q-1)), there exist an integer k, such that de=k(p-1)(q-1)+1.
It follows that Cd≡(Me)d=Mde=M1+k(p-1)(q-1) (mod n).
By Fermat’s Little Theorem (theorem 5)[assuming that gcd(M,p)=gcd(M,q)=1, which holds except in rare cases], it follows that Mp-1≡1 (mod p) and Mq-1 ≡1 (mod q).

111. RSA Decryption Consequently, Cd ≡M(Mp-1)k(q-1) ≡M (mod p)
and
Cd ≡M(Mq-1)k(p-1) ≡M (mod q)
Because gcd(p,q)=1, it follows by the Chinese Remainder Theorem that
Cd ≡M (mod pq)

112. RSA Decryption
Example 12: We receive the encrypted message What is the decrypted message if it was encrypted using the RSA cipher form example 11.

113. Chapter 3 3.8 Matrices Matrix Arithmetic
Algorithms for Matrix Multiplication
Transposes and Powers of Matrices
Zero-One Matrices

114. Matrix Arithmetic Definition 1:
A matrix is a rectangular array of numbers.
A matrix with m rows and n columns is called an
m × n matrix.
The plural of matrix is matrices. A matrix with the same number of rows as columns is called square.
Two matrices are equal if they have the same number of rows and the same number of columns and the corresponding entries in every position are equal.

115. Matrix Arithmetic Definition 2: Let The ith row of A is the
1 x n matrix [ai1,ai2,. . .,ain].
The jth column of A is the n x 1 matrix
The (i, j)th element or entry of is the element aij ,
that is , the number in the ith row and jth column of A.
A convenient shorthand notation for expressing the matrix A is to write A =[aij], which indicates that A is the matrix with its (i, j)th element equal to aij.

116. Matrix Arithmetic
Definition 3: Let A=[aij] and B=[bij] be m x n matrices. The sum of A and B, denoted by A+B,
is the m x n matrix that has aij+bij as its (i, j)th element. In other words, A+B= [aij+bij].
Example 2: we have

117. cij = ai1b1j + ai2b2j +. . . +aikbkj
Matrix Arithmetic
Definition 4:
Let A be an m x k matrix and B be k x n matrix.
The product of A and B, denoted by AB, is the m x n matrix with its (i , j )th entry equal to the sum of the products of the corresponding elements from the
ith row of A and the jth column of B.
In other words, if AB=[cij], then
cij = ai1b1j + ai2b2j aikbkj

118. Matrix Arithmetic

119. Algorithms for Matrix Multiplication
Algorithm 1 : Matrix Multiplication
procedure matrix multiplication
(A, B: matrices)
for i := 1 to m
for j := 1 to n
begin
cij :=0
for q := 1 to k
cij :=cij + aiqbqj
end
{C= [cij] is the product of A and B}
Example 6: In which order should the matrices A1, A2, and A3, where
A1 is 30x20 , A2 is 20x40 , A3 is 40x10,
all with integer entries – be multiplied to use the least number of multiplications of integers?

120. Transposes and Powers of Matrices
Definition 5:
the identity matrix of order n is the n x n matrix
In = [δij]
where δij =1 if i = j and δij = 0 if i ≠ j. Hence,

121. Transposes and Powers of Matrices
Definition 6: Let A=[aij] be an m x n matrix.
The transpose of A, denoted by At, is the n x m matrix obtained by interchanging the rows and columns of A .
In other words, if At=[bij], then bij = aji for i=1,2,. . .,n and j = 1,2,. . .,m .
Definition 7: A square matrix A is called symmetric
if A = At.
Thus A =[aij] is symmetric if aij = aji for all i and j with 1≤ i ≤ n and 1 ≤ j ≤ n .

122. Symmetric Matrix

123. Zero-One Matrices
Definition 8: Let A=[aij] and B=[bij] be m x n zero-one matrices.
Then the join of A and B is the zero-one matrix with (i , j )th entry aij v bij.
The join of A and B is denoted by A v B.
The meet of A and B is the zero-one matrix with (i , j )th entry aij Λ bij.
The meet of A and B is denoted by A Λ B.

124. Zero-One Matrices
Definition 9: Let A=[aij] be an m x k zero-one matrix and B=[bij] be a k x n zero-one matrix .
Then the boolean product of A and B,denote by A⊙B , is the m x n matrix with with (i , j)th entry cij where
Example 10: find the Boolean product of A and B,
where

125. Zero-One Matrices Algorithm 2: The Boolean Product
procedure Boolean product(A, B: zero-one matrices)
for i := 1 to m
for j := 1 to n
begin
cij :=0
for q := 1 to k
end
{C= [cij] is the Boolean product of A and B}

126. Zero-One Matrices
Definition 10: Let A be a square zero-one matrix ant let r be a positive integer.
The rth Boolean power of A is the Boolean product of r factors of A. The rth Boolean product of A is denoted by A[r]
Hence,
(this is well defined because the Boolean product of matrices is associative.)
We also define A[0] to be In

127. Zero-One Matrices Example 11: Let .
Find A[n] for all positive integers n.