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### ADVANCED ALGORITHMS

9 26 14 31 49 36 9

Number-Theoretic Algorithms (UNIT-4)

Elementary Number-theoretic Notions :

a) Divisibility and Divisors :

The notation d | a (d divides a) means :

a = kd for some integer k.

Here, ‘a’ is multiple of ‘d’.

Here, if d 0, then d is ‘divisor’ of a.

The ‘trivial divisors’ of a are : 1, a

The nontrivial divisors of a are called factors of a

Ex-1 : Find the divisors and trivial divisors of 24.

The trivial divisors of 24 : 1 24

The divisors of 24 : 1,2,3,4,6,8,12, and 24

The factors of 24 : 2,3,4,6,8,12

b) Prime and Composite Numbers :

An integer a > 1, whose only divisors are trivial

divisors ‘1’ and ‘a’ is a ‘Prime Number’.

An integer a > 1, which is not a prime number,

is called ‘Composite Number’.

Ex-2 : Find all the first 5 prime numbers.

2,3,5,7,11

39 is a composite no. since it divides by 3.

1 is called unit & is neither prime nor composite.

Similarly 0 and all –ve nos. are neither prime nor composite.

TH-4.1 : Division Theorem

“ For any integer ‘a’ and any positive integer ‘n’,

there exists unique integers ‘q’ and ‘r’ such that

0 ≤ r < n and a = qn + r”.

The value q = a/n is the quotient of the division.

The value r = a mod n is the remainder of the division.

Here n | a ( n divides a), if and only if a mod n = 0.

Ex-3 : Find the quotient and remainder of 12 and 67.

The quotient : 5

The remainder : 7

c) Common Divisors & Greatest CD :

If ‘d’ is a divisor of ‘a’

and ‘d’ is also divisor of ‘b’

then ‘d’ is a common divisor of ‘a’ and ‘b’.

Note : a) ‘1’ is a common divisor of any two integers.

b) If a | b and b | a then a = b

Important Property :

Ifd | a andd | b thend | (a + b) &d | (a – b)

Ifd | a andd | b thend | (ax + by)

Ex-4: Find all the common divisors of 24 and 30.

1 2 3 6

The GCD of two integers a and b , not both Zero, is the largest of the common divisors of a and b.

GCD(24, 30) = 6

Note : GCD(a,0) = |a|

Relatively Prime Integers :

Two integers ‘a’ and ‘b’ are relatively prime

if their only common divisor is 1.

i.e., gcd(a,b) = 1

The relatively primes are : (8,15)

-do- : (10,21)

TH-4.2 : If a and b are any two integers, then

gcd(a,b) is the smallest positive element s of the set

s = (ax + by).

Ex-5: Let a =6 & b = 21, Find the values of x , y.

TH-4.3 : For any integers a, b and p, if both

gcd(a,p) = 1 and gcd(b,p) = 1, then gcd(ab,p) = 1.

TH-4.4 : For all primes p and all integers a and b,

if p | abthen p | a or p | b or both.

Unique Factorization :

There is exactly one way to write any composite integer ‘a’ as a product of the form

a = p1^ e1. p2^ e2. p3^ e3…… pr^ er

where all pi are prime, p1<p2<..<pr and ei are +ve integers.

Common Divisors & Greatest CD :

Let there are two positive integers ‘a’ and ‘b’

a = p1^ e1. p2^ e2. p3^ e3…… pr^ er

b = p1^ f1. p2^ f2. p3^ f3…… pr^ fr

Here, gcd(a,b) = p1 ^ min(e1 , f1 ). p2 ^ min(e2 , f2 ).

……. pr ^ min(er , fr ).

Ex-6 : Let a = 90 b = 150

Find the value of gcd(a,b) using above rule.

Here, a = 2 x 32 x 5

b = 2 x 3 x 52

gcd(a,b) = 2 x 3 x 5

TH-4.5 : GCD recursion theorem :

For any non-negative integer ‘a’ and

any positive integer ‘b’, we have

gcd (a,b) = gcd (b, a mod b)

Proof : case-1 :

Let d = gcd (a,b) d | a & d | b

Here, a mod b = a – q b where q = a / b

Since, a mod b is a linear combination of ‘a’ and ‘b’,

we can say that d | (a mod b).

So, d | b and d | (a mod b)

d | gcd (b, a mod b)

gcd (a,b) | gcd (b, a mod b) ….(1)

Case-2 :Let d = gcd (b, a mod b).

d | b & d | (a mod b)

Since, a = q b + (a mod b) where q = a / b

we have that a is a linear combination of ‘b’ and

‘a mod b’ d | a

Hence, we can say that d | a & d | b

d | gcd(a,b)

gcd(b,a mod b) | gcd(a,b) ……(2)

From (1) and (2) we can say that

gcd (a,b) = gcd (b, a mod b) //

Let a and b are non-negative integers.

EUCLID (a,b)

If (b = = 0)

2 return a

3 else return EUCLID(b, a mod b)

Ex-7 : Find the value of gcd(30,21) using Euclid algorithm.

EUCLID(30,21) = EUCLID (21,9)

= EUCLID (9,3)

= EUCLID (3,0)

= 3.

This computation calls EUCLID recursively three times.

b) Extended Euclid’s Algorithm :

In this algorithm we find additional information

like the values of ‘x’ and ‘y’, where

d = gcd (a,b) = ax + by

EXTENDED-EUCLID(a,b)

1 If b = = 0

return (a,1,0)

else (d’, x’, y’) = EXTENDED-EUCLID(b, a mod b)

(d,x,y) = (d’, y’, x’ - a / b y’)

return (d, x, y)

d = a x + b y

d’ = bx’ + (a mod b) y’

because d = d’, we have

ax + by = bx’ + (a mod b) y’

= bx’ + (a – b a / b ) y’

= a y’ + b (x’ - a / b y’)

So, x = y’ & y = (x’ - a / b y’)

Ex-8 :Find the value of gcd(99,78) and corres-

ponding x, y values using EE algorithm.

Step-1 : a = 99 b =78

a / b = 1 d = gcd(99,78) = 3

Here, a = 99 = 1. 78 + 21

78 = 3. 21 + 15

21 = 1. 15 + 6

15 = 2.6 + 3

6 = 2. 3 + 0

And 3 = 15 - 2 . 6 = 15 – 2 (21 – 1. 15)

= 3.15 - 2.21 = 3(78 – 3.21) – 2.21

= 3. 78 – 11. 21 = 3. 78 – 11(99 – 1.78)

= 3.78 - 11.99 + 11.78 = -11.99 + 14.78

3 = gcd(99,78) = -11.99 + 14. 78 …(1)

Step-2 : a = 78 b = 21

a / b = 3 d = gcd(78,21) = 3

Here, a = 78 = 3. 21 + 15

21 = 1. 15 + 6

15 = 2.6 + 3

6 = 2. 3 + 0

And 3 = 15 - 2 . 6 = 15 – 2 (21 – 1. 15)

= 3.15 - 2.21 = 3(78 – 3.21) – 2.21

= 3. 78 – 11. 21

So, x = 3 y = -11

3 = gcd(78,21) = 3.78 - 11. 21 …(2)

Step-3 : a = 21 b = 15

a / b = 1 d = gcd(21,15) = 3

Here, a = 21 = 1. 15 + 6

15 = 2.6 + 3

6 = 2. 3 + 0

And 3 = 15 - 2 . 6 = 15 – 2 (21 – 1. 15)

= 3.15 - 2.21

So, x = -2 y = 3

3 = gcd(21,15) = -2.21 + 3. 15 …(3)

Step-4 : a = 15 b = 6

a / b = 2 d = gcd(15,6) = 3

Here, a = 15 = 2.6 + 3

6 = 2. 3 + 0

And 3 = 15 - 2 . 6

So, x = 1 y = -2

3 = gcd(15,6) = 1.15 - 2. 6 …(4)

Step-5 : a = 6 b = 3

a / b = 2 d = gcd(6,3) = 3

Here, a = 6 = 2. 3 + 0

And 3 = 0.6 + 1.3

So, x = 0 y = 1

3 = gcd(6,3) = 0.6 + 1. 3 …(5)

Step-6 : a = 3 b = 0

a / b = - d = gcd(3,0) = 3

Here, a = 3 = 1. 3 + 0.0

And 3 = 1.3 + 0.0

So, x = 1 y = 0

3 = gcd(3,0) = 1.3 + 0. 0 …(6)

So, the final output of EE algorithm is as follows :

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

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

Abelian Group : A group (S,) is said to be ‘Abelian Group’, if it satisfies the commutative property.

(a b) = (b a)

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’,)

Galois Field : The set of integers (0,1,2,…,p-1), where p is a prime, is called GF(p).

Multiplicative Inverse : The factor b-1 is the ‘multiplicative inverse’ of b in GF(p).

b b-1 mod p = 1

Ex-9 :Find the multiplicative inverses of the following, where p = 7.

1 2 3 4 5 6

Answer : 1 4 5 2 3 6

Ex-10 :Find the multiplicative inverses of the 1 2 3 4 5 6 7 8 9 10 (p = 11)

Answer : 1 6 4 3 9 2 8 7 5 10

Let the moduli be p1 = 3, p2 = 5, p3 = 7

Let us consider the integers : 10, 15

Here, 10 = (10 mod 3, 10 mod 5, 10 mod 7) = (1, 0, 3)

Here, 15 = (15 mod 3, 15 mod 5, 15 mod 7) = (0, 0, 1)

Modular Addition :

10 + 15 = (25 mod 3, 25 mod 5, 25 mod 7) = (1, 0, 4)

& (1+0 mod 3, 0+0 mod 5, 3+1 mod 7) = (1, 0, 4)

Modular Subtraction :

15 – 10 = (5 mod 3, 5 mod 5, 5 mod 7) = (2, 0, 5)

& (0 – 1 mod 3, 0 – 0 mod 5, 1 – 3 mod 7) = (2, 0, 5)

10 * 15 = (150 mod 3, 150 mod 5, 150 mod 7) = (0, 0, 3)

& (1*0 mod 3, 0*0 mod 5, 3*1 mod 7) = (0, 0, 3)

Prime Divisors : The divisors, which are prime numbers are called ‘Prime Divisors.

Euler’s Phi Function : For a given integer ‘n’,

the following function is called ‘EPF’.

(n) = n. (1 – 1/p)

Ex-12 : Find the value of EPF where n = 45.

(45) = 45 (1-1/3) (1-1/5) = 24

Basis for Chinese Remainder Theorem :

Ex-13 : Find the lowest integer x such that it leaves remainders 2, 3 and 2 when divided by

3, 5 and 7.

The Answer : 23

4. Chinese Remainder Theorem :

TH : Let n = n1. n2. n3…nk, where n are pairwise

relatively prime.

Find the value of ‘a’, where

a ai mod nifor i = 1,2,3,…,k

i.e., a (a1. a2. a3…ak)

Here ai = a mod ni

Proof : Let us define mi = n / nifor i = 1,2,3,…,k

i.e, m = n1. n2.…ni-1 ni+1.…nk

Now let ci = mi(mi -1 mod ni ) for i = 1,2,…,k

Here mi, ni are relatively prime.

Finally, the value of ‘a’ is :

a (a1 c1 + a2 c2 + a3 c3 + … + ak ck) (mod n )

Ex-14 : Find the value of ‘a’ for the following equations using Chinese Remainder Theorem :

a 2 (mod 5)

a 3 (mod 13)

Here a1 = 2 n1 = 5 m2 = 5

n = 65 a2 = 3 n2 = 13 m1 = 13

Because 13 -1 2 (mod 5) and 5 8 (mod 3)

We have c1 = 13 ( 2 mod 5) = 26

c2 = 5 ( 8 mod 13) = 40

a 2. 26 + 3.40 (mod 65)

52 + 120 (mod 65) 42

Ex-15 : Find the value of ‘x’ using CRT,

x 4 (mod 5) x 5 (mod 11)

The answer is : 49

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

For any integer n > 1

a ^ (n) 1 (mod n) for all a ∊ Zn*

6 (b) Fermat’s Theorem :

If p is a prime, then ap - 1 1 (mod p)

Note that if p is a prime, then a ^(p) = p - 1

Ex-16 : Prove the Euler theorem for the following.

Let n = 7 (n) = 6 & a = {1, 2, 4}

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 = pq

c) Select a small odd integer ‘e’ that is relatively prime to p-1 and q-1. (public exponent)

d) Compute the integer ‘d’ (private exponent) from e, p and q such that de ≡ 1 mod L, where L = LCM [ (p-1), (q-1) ]

e) Publish P = (e,n) RSA Public Key

Secret S = (d,n)RSA Secret Key

Here, e = ENCRYPT(m) = me mod n

d = DECRYPT(c) = cd mod n

Ex-17 : Apply RSA algorithm for the following.

p = 5 q = 11 e = 3

Here n = pq = 55 (n) = 40

and d : ed ≡ 1 mod LL = 20

So, d = 7

LetA = Message(m) B = m2 mod n

C = m3 mod n (encrypted message)

D = c2 mod n E = c3 mod n F = c6 mod n

G = c7 mod n (decrypted message)

A B C D E H G

0 0 0 0 0 0 0

1 1 1 1 1 1 1

2 4 8 9 17 14 2

3 9 27 14 48 49 3

4 16 9 26 14 31 4

5 25 15 5 20 15 5

6 36 51 16 46 26 6

7 49 13 4 52 9 7

- 8 9 17 14 18 49 8

Here, the first column is message sent.

the third column is cipher text

the last column is decrypted message.

8. Primality Testing :

a) Carmichael number

ACarmichael numberis a composite positive integer which satisfies the following formula.

bn-1 ≡ 1 ( mod n)

for all integers ‘b’ which are relatively

prime to ‘n’.

Def : A positive composite integer ‘n’ is a CN, iff ‘n’ is square-free and for all prime divisors

p of n, it is true that (p – 1) | (n – 1).

The first Carmichael Number is : 561

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.

b) MILLER-RABIN (n,s)

for j = 1 to s

a = RANDOM(1, n-1)

if WITNESS (a,n)

return COMPOSITE

return PRIME

c) WITNESS(a,n)

1. Let t and u be such that t ≥ 1.

u is odd, and n-1 = 2t u

2. x0 = MODULAR-EXPONENTIATION(a,u,n)

3. for i = 1 to t

4. xi = x2i-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

d) MODULAR-EXPONENTIATION (a, b, n)

1. c = 0

d = 1

3. Let (bk , bk-1 , …..,b1 , b0 )

for i = k downto 0

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

Ex-18 : 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

bi -- 1 0 0 0 1 1

c 0 1 2 4 8 17 35

d 1 7 49 157 526 160 241

Here, d = ac mod n (c = b = u)

From above, the value of ‘d’ returned is : 241

Here, x0 ≡ a35≡ 241 (mod 561).

Note : Further we can have

a70 ≡ 298 (mod n) a140 ≡ 166 (mod n)

a280 ≡ 67 (mod n) a560 ≡ 1 (mod n)

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

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

x1 = RANDOM(0, n-1)

y = x1

k = 2

5. While TRUE

i = i + 1

xi = (x2i-1 - 1 ) mod n

d = gcd(y – xi , n)

if ( d 1) and (d n)

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.

Ex-19 :Pollard’s Rho Heuristic

Let n = 1387

So,Initialization :

i = 1 x1 = 2 y = 2 k = 2

WHILE : STEP-1 :

i= 2

xi = (x2i-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

STEP-2 : i= 3

xi = (x2i-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 = (x2i-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

STEP-4 : i= 5

xi = (x2i-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 = (x2i-1 - 1 ) mod n x6 = 1186

d = gcd(y – xi , n) d = 1

if [ (d 1) and ( d n) ] FALSE

if ( i = = k) FALSE

STEP-6 : i = 7

xi = (x2i-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 :

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