ADVANCED ALGORITHMS-UNIT-3-Final.ppt

Module 3:
Number-Theoretic Algorithms
1

2
Prime and composite numbers
 The first 20 primes, in order, are
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67,
71
 Example, 39 is composite because 3 | 39

3
Common divisors and greatest common divisors
 If d is a divisor of a and d is also a divisor of b, then d is a common divisor of a
and b.
 For example:
 The divisors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30,
 The divisors of 24 are 1, 2, 3, 6, 8, 12, and 24
 and so the common divisors of 24 and 30 are 1, 2, 3, and 6.
 The greatest common divisor of two integers a and b, not both zero,
is the largest of the common divisors of a and b.
 For example: gcd(24,30)=6

4
 Greatest common divisor
 Euclid’s algorithm used to compute greatest
common divisors.

5
Example of the running of EUCLID, consider the computation
of gcd (30,21)

6
EXAMPLE
Calculate: gcd(22, 60) = gcd(60,22)
60 = 2 x 22 + 16 = Euclid(22,16)
22 = 1 x 16 + 6 = Euclid(16,6)
16 = 2 x 6 + 4 = Euclid(6,4)
6 = 1 x 4 + 2 = Euclid(4,2)
4 = 2 x 2 + 0 = Euclid(2,0)
= 2.

7
The running time of Euclid’s algorithm
 The overall running time of EUCLID is proportional to the number of recursive
calls it makes.
 The number of recursive calls in EUCLID is O(log b)

8
The extended form of Euclid’s algorithm
 The procedure EXTENDED_EUCLID takes as input a pair of nonnegative integers
and returns a triple of the form (d,x,y).

9
Figure : illustrates how EXTENDED-EUCLID computes gcd(99,78)

10
a b [a/b] d x y
99 78
Example: gcd(99,78)

11
a b [a/b] d x y
99 78 1
gcd(99,78)

12
a b [a/b] d x y
99 78 1
78 21
Remainder
Example: gcd(99,78)

13
a b [a/b] d x y
99 78 1
78 21 3
Example: gcd(99,78)

14
a b [a/b] d x y
99 78 1
78 21 3
21 15
Remainder
Example: gcd(99,78)

15
a b [a/b] d x y
99 78 1
78 21 3
21 15 1
15 6 2
6 3 2
3 0 -
Example: gcd(99,78)

16
a b [a/b] d x y
99 78 1
78 21 3
21 15 1
15 6 2
6 3 2
3 0 - 3 1 0
Example: gcd(99,78)
When b=0 , then d=b and initialize x=1
and y=0

17
a b [a/b] d x y
99 78 1 3
78 21 3 3
21 15 1 3
15 6 2 3
6 3 2 3 0 1
3 0 - 3 1 0
Example: gcd(99,78)
For every step, calculate x and y
Step 1: New x value = previous y value
x=0
New y value = previous x value – ([a/b] * (previous y value)
y= 1- (2*0) = 1

18
a b [a/b] d x y
99 78 1
78 21 3
21 15 1
15 6 2 3 1 -2
6 3 2 3 0 1
3 0 - 3 1 0
x=1
y= 0- (2*1) = -2
Example: gcd(99,78)

19
a b [a/b] d x y
99 78 1
78 21 3
21 15 1 3 -2 3
15 6 2 3 1 -2
6 3 2 3 0 1
3 0 - 3 1 0
x= -2
y= 1- (1*-2) = 3
Example: gcd(99,78)

20
a b [a/b] d x y
99 78 1 3 -11 14
78 21 3 3 3 -11
21 15 1 3 -2 3
15 6 2 3 1 -2
6 3 2 3 0 1
3 0 - 3 1 0
Example: gcd(99,78)

21
 Since the number of recursive calls made in EUCLID is equal to the number of
recursive calls made in EXTENDED-EUCLID, the running times of EUCLID
and EXTENDED-EUCLID are the same, to within a constant factor.
 That is, for a > b > 0, the number of recursive calls is O(log b).
 Example:
gcd(66, 24) = 6
gcd (899,493) =29
gcd (95,355) =5

23
1. Modular Arithmetic :
a) Group : A group (S,) is a set S together with
binary operation  defined on S for which
the following properties hold :
i) Closure : For all a, b  S, a  b  S.
ii) Identity : There exists an element e  S,
called the identity of the group, 
a  e = e  a = a for all a  S.
iii) Associativity : For all a, b, c  S, we have
(a  b)  c = a  (b  c)
iv) Inverse : For each a  S, there exists a
unique element b  S, called the
inverse of ‘a’, such that
(a  b) = (b  a) = e

24
The familiar group (Z,+) of the integers Z under the
operation of addition: 0 is the identity, and the inverse of a
is -a.
Abelian Group : A group (S,) is said to be ‘Abelian
Group’, if it satisfies the commutative property.
(a  b) = (b  a)

25
Finite Group : A group (S,) is said to be ‘Finite
Group’, if it satisfies the property.
|S| < 
Sub-Group : If (S,) is a group, and S’  S and
(S’,) is also a group, then (S’,) is a sub
group of (S’,)

26
The groups defined by modular addition and multiplication
For any positive integer n, let S be the complete set of
residues {0, 1, 2,…, n−1}.
Addition modulo n on S is defined as follows.
For a and b in S, take the usual sum of a and b as
integers, and let r be the element of S to which the result
is congruent (modulo n); the sum a+b (mod n) is equal
to r.
(A + B) mod C = (A mod C + B mod C) mod C

28
For example, addition modulo 5 is given by the following
table: Finite group (Z6,+6)
Additive group modulo n

29
Multiplication modulo n is defined by taking ab (mod n)
to be equal to s, where s is the element of S to which the
usual product of a and b is congruent (modulo n).
(A * B) mod C = (A mod C * B mod C) mod C

30
For example, Multiplication modulo 15 is given by the
following table: Finite group (Z15,*15)
Multiplicative group modulo n

31
Solving modular linear equations

32
a x ≡ b (mod n)
where a > 0 and n > 0.
We assume that a, b, and n are given, and we wish to find all values of x,
modulo n.

34
Example : a= 14 b= 30 and n=100
Step1: gcd(14 , 100) = (a, n)
a b [a/b] d x y
14 100
100 14 7 2 1 -7
14 2 7 2 0 1
2 0 - 2 1 0
Solution of Extended Euclid Algorithm
d= 2
x= 1
y= -7

35
Step 2:
If d |b
if 2 | 30 , yes it divides
Then
x0 = -7 ( 30/2) mod 100 = 95

36
Step 3:
i=0
95 + 0 (100/2) mod 100 = 95
i=1
95 + 1 (100/2) mod 100 = 45
The loop on lines 4-5 prints the two solutions 95
and 45.

37
Assignment:
1. 35 x ≡ 10 (mod 50)
Sol: x1 = 16, x2 = 26, x3 = 36, x4 = 46 x5=6
2. 56x ≡ 1 mod 93
Sol: x=5
3. 5x ≡ 12 mod 19
Sol: x=48 x=12
4. 15x ≡ 12 mod 57
Sol: x ≡ 16, 35 or 54 mod 57.

38
The Chinese Remainder Theorem

39
 Around A.D. 100, the Chinese mathematician Sun-Ts˘u solved the
problem of finding those integers x that leave remainders 2, 3, and 2
when divided by 3, 5, and 7 respectively.
 One such solution is x =23
 The “Chinese remainder theorem” provides a correspondence between a
system of equations modulo a set of pairwise relatively prime moduli (for
example, 3, 5, and 7) and an equation modulo their product (for example,
105).

40
1. x ≡ 2 (mod 5) x ≡ 3 (mod 13)
Standard Equation : x ≡ a (mod n)
Solution:
a1= 2 a1 , a2 be integers
a2= 3
n1= 5 n= n1,n2 …… nk, where the ni are pairwise relatively prime
n2= 13
n = n1* n2 = 13*5 = 65
m1= n / n1 =13 moduli (m)
m2= n / n2 = 5
Calculate : ci = mi (mi
-1 mod ni)

41
1. x ≡ 2 (mod 5) x ≡ 3 (mod 13)
Standard Equation : x ≡ a (mod n)
Solution:
a1=2 a2= 3 n1= 5 n2= 13 n = 65 m1=13 m2= 5
Calculate : ci = mi (mi
-1 mod ni)
c1 = m1 (m1
-1 mod n1)
= 13 (13-1 mod 5) = 13 (2 mod 5) = 26
c2 = m2 (m2
-1 mod n2) = 5 (5-1 mod 13) = 5 (8 mod 13) = 40
x= (c1 *a1 + c2 *a2 ) mod n
= (26* 2 + 40 * 3 ) mod 65
= (52 +120) mod 65
= 42 (mod 65)
= 42

42
1. 13-1 (mod 5) = 13 * 0 ≡ 1 (mod 5) WRONG
13 * 1 ≡ 1 (mod 5) WRONG
13 * 2 ≡ 1 (mod 5) RIGHT
The modular inverse of 13 (mod 5) is 13-1 is 2
A x ≡ 1 (mod C)
The modular inverse of A (mod C) is A-1
2. 5 -1 (mod 13) = 5 * 0 ≡ 1 (mod 13) WRONG
5 * 1 ≡ 1 (mod 13) WRONG
5 * 2 ≡ 1 (mod 13) WRONG
5 * 3 ≡ 1 (mod 13) WRONG
5 * 4 ≡ 1 (mod 13) WRONG
5 * 5 ≡ 1 (mod 13) WRONG
5 * 6≡ 1 (mod 13) WRONG
5 * 7 ≡ 1 (mod 13) WRONG
5 * 8 ≡ 1 (mod 13) RIGHT

43
Find all solutions to the equations
1. x ≡ 2 (mod 5) x ≡ 3 (mod 13)
2. x ≡ 4 (mod 5) x ≡ 5 (mod 11)
3. x ≡ 6 (mod 11) x ≡ 13 (mod 16)
Solu:51669
4. x ≡ 9 (mod 21) x ≡ 19 (mod 25)
Solu:51669
5. x ≡ 1 (mod 5) x ≡ 2 (mod 7)
Solu:1731
6. x ≡ 3 (mod 9) x ≡ 4 (mod 11)
Solu: 1731

45
Consider the sequence of powers of ‘a’, modulo n
where a ∊ Zn
*. For example,
i 0 1 2 3 4 5 6 7 8 9
3i mod 7 1 3 2 6 4 5 1 3 2 6
i 0 1 2 3 4 5 6 7 8 9
2i mod 7 1 2 4 1 2 4 1 2 4 1
Now, < 2 > = {1, 2, 4} in Z7
*
< 3 > = {1, 3, 2, 6, 4, 5} in Z7
*
Here, ord7 (2) = 3 & ord7 (3) = 6
Powers of an Element :

46
Compute : ab (mod n)
Example: 7560 (mod 561)
Is it possible using Calculator ?
If yes, How much time?
So, Use MODULAR-EXPONENTIATION algorithm

47
MODULAR-EXPONENTIATION (a, b, n)
1. c = 0
2. d = 1
3. Let (bk , bk-1 , …..,b1 , b0 )
4. for i = k downto 0
5. c = 2c
6. d = (d.d) mod n
7. if bi = = 1
8. c = c + 1
9. d = (d.a) mod n
10. return d

48
Example: 7560 (mod 561)
a=7 b= 560 n=561
Step 1: The binary representation of b =560 is 1000110000
k=10 (count of 1’s and 0’s)
Step 2: c=0 d=1
for ( i = 10 downto 0 )
i=9
c= 2.c = 2.0 =0
d= (d*d) mod n = 1*1 mod 561
= 1
if b9 is 1
update c and d
c= 0+1 =1
d= 1* 7 mod 561 = 7 Result of this step: c=1 d=7

49
i=8
c= 2.c = 2.1 =2
d= (d*d) mod n = 7*7 mod 561
= 49
if b8 is not 1
So don’t update c and d
Result of this step: c=2 d=49
560 = 1000110000

50
i=7
c= 2.c = 2.2 =4
d= (d*d) mod n = 49*49 mod 561
= 157
if b7 is not 1
So don’t update c and d
Result of this step: c=4 d=157
560 = 1000110000

51
Final result is 7560 (mod 561) = 1

52
The RSA (Rivest–Shamir–Adleman)
public-key cryptosystem

53
In RSA Cryptosystem, the public and private keys are generated
as follows :
a) Select at random two large prime numbers
p and q such that p ≠ q.
b) Compute n = p*q
n is used as the modulus for both the public and private keys. Its length, usually
expressed in bits, is the key length.

54
c) Select a small odd integer ‘e’ that is relatively prime to
ø(n)= (p-1 * q-1) (public exponent)
d) Compute the integer ‘d’ (private exponent) from e, p and q
such that d * e ≡ 1 mod ø(n)

55
e) Publish P = (e,n) RSA Public Key
Keep Secret S = (d,n) RSA Secret
Key
Here, e = ENCRYPT(m) = me mod n
d = DECRYPT(c) = cd mod n

56
Apply RSA algorithm for the following.
p = 11 q = 29 e = 3
Encryption: Here n = pq = 319
(n) = p-1 * q-1 = 10 * 28 = 280
and d * e ≡ 1 mod 280
So, d = 187
Example:
Consider an RSA key set with p = 11, q = 29, n = 319, and e = 3.
What value of d should be used in the secret key?
What is the encryption of the message M =100?

57
Hence, (e, n) = (3, 319) is the public key and (d, n) = (187, 319) is the
private key.
For M = 100,
the cipher text is
1003 mod 319 = 254.
Here, e = ENCRYPT(m) = me mod n
d = DECRYPT(c) = cd mod n
Decryption:
the Plain text is
254187 mod 319 = 100.

58
Problems:
Sl No. Message P Q E Cipher Plain
Text
1. 7 11 3 ? ? ?
2. 88 17 11 ? ? ?
3. 513 137 131 ? ? ?
4. 123 61 53 ? ? ?
5. 148 23 37 ? ? ?
6. 42 13 19 ? ? ?

59
Problem with solution:
Sl No. Message P Q E Cipher Plain
Text
1. 7 11 3 3 13 7
2. 88 17 11 7 11 88
3. 513 137 131 3 8363 513
4. 123 61 53 17 855 123
5. 148 23 37 5 111 148
6. 42 13 19 7 120 42

61
Primality Testing :
Methods for primality testing
1. Pseudoprimality testing
2. The Miller-Rabin randomized primality test
The problem of finding large primes.

62
Pseudoprimality testing
This procedure can make errors, but only of one type. That is, if it says
that n is composite, then it is always correct.
If it says that n is prime, however, then it makes an error only on
specific cases.
The first four such values are 341, 561, 645, and 1105. <10000

64
The Miller-Rabin randomized primality test

65
a) Carmichael number
A Carmichael number is a composite positive integer
which satisfies the following formula.
bn-1 ≡ 1 ( mod n)
for all integers ‘b’ which are relatively
prime to ‘n’.
The first Carmichael Number is : 561

66
MILLER-RABIN (n,s)
for j = 1 to s
a = RANDOM(1, n-1)
if WITNESS (a,n)
return COMPOSITE
return PRIME
 The Procedure MILLER-RABIN is a probabilistic search for a
proof that n is composite.
 In the following procedure, ‘s’ is the number of times the value
of ‘a’ is to be chosen at random.

67
WITNESS(a,n)
1. Let t and u be such that t ≥ 1.
u is odd, such that n-1 = 2t u
2. x0 = MODULAR-EXPONENTIATION(a,u,n)
3. for i = 1 to t
4. xi = x2
i-1 mod n
5. if ( xi = = 1) and ( xi-1 ≠ 1) and ( xi-1 ≠ n-1)
6. return TRUE
7. if xt ≠ 1
8. return TRUE
9. return FALSE

68
Example: Let ‘n’ be a carmichael number. n = 561
So, here n – 1 = 560
If n – 1 is written in the form of n-1 = 2t u, then
t = 4 and u = 35
Let the value of ‘a’ is chosen from the
algorithm as : 7
From the WITNESS algorithm,
find the value of x0.
Here, call the
MODULAR_EXPONENTIATION(a,u,n)
where a = 7 & u = 35 & n = 561

69
Here, d = ac mod n (c = b = u)
From above, the value of ‘d’ returned is : 241
Here, x0 ≡ 735 ≡ 241 (mod 561).
i init 5 4 3 2 1 0
bi -- 1 0 0 0 1 1
c 0 1 2 4 8 17 35
d 1 7 49 157 526 160 241

70
Here, d = ac mod n (c = b = u)
From above, the value of ‘d’ returned is : 241
1. Here, x0 ≡ 735 ≡ 241 (mod 561).
Note : Further we can have
2. 770 ≡ 298 (mod 561) = x1
3. 7140 ≡ 166 (mod 561) = x2
4. 7280 ≡ 67 (mod 561) = x3
5. 7560 ≡ 1 (mod 561) = x4

71
So, the sequence is : (241, 298, 166, 67, 1)
Thus, WITNESS discovers 1 in the last squaring
step, since a560 ≡ 1 (mod n)
Therefore, a = 7 is the witness to the compo-
siteness of ‘n’.
WITNESS(7,N) returns TRUE.
MILLER-RABIN returns COMPOSITE
Note : 561 = 3 . 11 . 17

72
Error rate of the Miller-Rabin primality test is very
less compared to Pseudoprimality testing

74
Integer Factorization :
This is the process of integer factorization into a
product of primes.
Pollard’s rho heuristic :
This heuristic here helps in finding the product
of primes for the given integer.
POLLARD-RHO(n)
1. i = 1
2. x1 = RANDOM(0, n-1)
3. y = x1
4. k = 2

75
5. While TRUE
6. i = i + 1
7. xi = (x2
i-1 - 1 ) mod n
8. d = gcd(y – xi , n)
9. if ( d  1) and (d  n)
10. print d
11. if ( i = = k)
12. y = xi
13. k = 2k
Note : The above algorithm generates a set of factors
which are primes for the given integer.

76
Example: Pollard’s Rho Heuristic
Let n = 1387
So, Initialization :
i = 1 x1 = 2 y = 2 k = 2
WHILE : STEP-1 :
i= 2
xi = (x2
i-1 - 1 ) mod n  x2 = 3
d = gcd(y – xi , n)  d = 1
if [ (d  1) and ( d  n) ] FALSE
if ( i = =k) TRUE
y = 3 k = 4

77
STEP-2 : i= 3
xi = (x2
i-1 - 1 ) mod n  x3 = 8
d = gcd(y – xi , n)  d = 1
if [ (d  1) and ( d  n) ] FALSE
if ( i = =k) FALSE
STEP-3 : i= 4
xi = (x2
i-1 - 1 ) mod n  x4 = 63
d = gcd(y – xi , n)  d = 1
if [ (d  1) and ( d  n) ] FALSE
if ( i = =k) TRUE y = 63 k = 8

78
STEP-4 : i= 5
xi = (x2
i-1 - 1 ) mod n  x5 = 1194
d = gcd(y – xi , n)  d = 1
if [ (d  1) and ( d  n) ] FALSE
if ( i = = k) FALSE
STEP-5 : i = 6
xi = (x2
i-1 - 1 ) mod n  x6 = 1186
d = gcd(y – xi , n)  d = 1
if [ (d  1) and ( d  n) ] FALSE
if ( i = = k) FALSE

79
STEP-6 : i = 7
xi = (x2
i-1 - 1 ) mod n  x7 = 177
d = gcd(y – xi , n)  d = 19
if [ (d  1) and ( d  n) ] TRUE
 Print d = 19
if ( i = = k) FALSE
If the process is continued like this, we get another
factor : 73
The relation among the Xi values are shown in the
next slide :

80
310
996
396
814
84
x7 177
x6 1186 120
x5 1194 339 529
x4 63 595 1053
x3 8
x2 3
x1 2

ADVANCED ALGORITHMS-UNIT-3-Final.ppt

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to ADVANCED ALGORITHMS-UNIT-3-Final.ppt

Similar to ADVANCED ALGORITHMS-UNIT-3-Final.ppt (20)

Recently uploaded

Recently uploaded (20)

ADVANCED ALGORITHMS-UNIT-3-Final.ppt