Entry Number:2007 mcs Lecture Date: 18/2/08. Scribe: Nimrita Koul. Professor: Prof. Kavitha.

Transcription

1 Scribe: Nimrita Koul Entry Number:2007 mcs 3136 Professor: Prof. Kavitha. Lecture Date: 18/2/08 1

2 A primality test is a test to determine whether or not a given number is prime without actually decomposing the number into its constituent prime factors. 2

3 The problem of primality testing Fermat's primality test Monte Carlo algorithm for primality testing Analysis of Algorithm..Claims 1,2,3 Modified Algorithm 3

4 Given an odd integer n+>=3, To determine if n is prime or composite Goal: Design an algorithm for this problem whose running time is polynomial in the input length, if n is value of input its length is ceiling log n. 4

5 Fermat's little theorem states that if p is prime and 1<=a<p, then a p-1 =1(mod p). If we want to test if p is prime, then we can pick random a's in the interval and see if the equality holds. If the equality does not hold for a value of a, then p is composite. If the equality does hold for many values of a, then we can say that p is probably prime, or a pseudoprime. It may be in our tests that we do not pick any value for a such that the equality fails. Any a such that a n-1 =1(mod n) when n is composite is known as a Fermat liar. If we do pick an a such that a n-1 1(mod n) then a is known as a Fermat witness for the compositeness of n. 5

6 1.Choose a random number a uniformly at random from the set {1,2,3,.,n-1}. To represent such a number in binary, log n bits will be required, to get these log n bits we toss a coin log n times, it may give a number greater than n and in that case we discard it and toss again. This step requires log n time. 6

7 2.If gdc(a,n) 1 then return composite Else a is relatively prime to n. Suppose n=p 2 then from 1,2,---, p 2-1 there are p such numbers which have same gcd as n, therefore the probability to pick such a no. is p/p 2 =1/p. 7

8 Claim 1:If n is prime then for all a, such that a does not divide n, gcd (a, n)=1 then a n-1 =1(mod n) For any x, y and n, If x=y(mod n),then the remainder when x is divided by n is same as remainder when y is divided by nor (x-y) is a multiple of n i.e. x=k 1 n+r y=k 2 n+r 8

9 If we partition the set of numbers with respect to the remainders we obtain different classes of integers where each class includes numbers with same remainder when divided by n. E.g. Z 0 ={0,n,2n,,-n,-2n} Z 1 ={1,n+1,2n+1,-n+1,-2n+1 }.. Z n ={0,1,2,,n-1} 9

10 Suppose Claim 1 s converse if true i.e., Converse of claim 1:if n is composite then its not the case that for all a in Z n *, a n-1 =1 mod n. i.e., there exists at least one a in Zn* such that a n-1 1 mod n. In other words, if a is composite then there exists an a in Z n * such that a n-1 1 mod n. The converse of claim one is not true, there do exist carmichael nos. with the property that for all a in Zn*,we have a n-1 = 1 mod n. 10

11 A Group is a pair consisting of a set with an operation defined on it such that the group is closed under the operation, the operation is associative, the group has an identity element with respect to the operation and for each element in the set there exists an inverse, such that the operation on an element and its inverse yields the identity element. E.g., Set of integers under binary addition forms a group(z,+), i.e., For any a,b in Z, sum (a+b) is in Z (a+b)+c=a+(b+c), i.e., the sum is associative The group has an identity element 0 for + operation. For each element a in Z we have an element a such that a+(- a)=(-a)+a=0. 11

12 (Z n, + n ) is a group, where + n is sum modulo n. If gcd( a, b)=g, where a, b,& g are integers, then there exist integers say x & y such that xa + yb=g If Z * n is a set of numbers between 1 & n-1 which are relatively prime to n, then (Z * n,*n ) forms a group. 12

13 If G is a finite group & H is a subgroup of G then size of G is divisible by size of H. If(G, o) is a group, H is a subset of G, such that (H,o ) forms a group then H is a subgroup of G. 13

14 If H=G, the proof is trivial. Let H G, & H={h1,h2,h3,.hk} Let g є G-H, then g1h={g1h1,g2h2,.g1hk} g2h={g2h1,g2h2,.g2hk}.. gnh={gnh1,gnh2, gnhk} Every time we create k new elements(cosets). 14

15 Choose a no. a uniformly at random from {1,2,..n-1}. If gcd(a,n) 1 then return composite Else compute b=a n-1 mod n. If b 1 then return composite Else { a є Z n *} compute b=a n-1 mod n If b 1 then return composite else return prime 15

16 If there exists an a in Z n * such that a n-1 1 mod n then we claim that at least half of the elements of Z n * satisfy the condition that a n-1 1 mod n. Let S be the no. of elements in Z n * such that a є Z n *, a n-1 1 mod n. Then H=Zn*-S={a є Z n *, a n-1 1 mod n} H forms a subgroup of Z n *. 16

17 Consider set H={a є Zn* :a n-1 =1 mod n} 1) aєh &b є H=>abєH (ab) n-1 = a n-1 b n-1 =1 mod n 2) aєh => a n-1 = 1mod n =(a -1 ) n-1 = 1mod n Therefore H is a subgroup of Zn* H <= Z /2 because H is a strict subgroup of Zn* i.e. atleast one element in Zn* doesn t fall in H.So max size of H is half of Zn*. 17