Download presentation

Presentation is loading. Please wait.

Published byBradley Farro Modified over 2 years ago

1
P vs. NP, AKS, RSA: The Acronyms of Mathematics Awareness Month Emily List Wittenberg University

2
April 2006: Mathematics Awareness Month Mathematics and Internet Security

3
Definitions P: yes or no decision problems that can be solved by an algorithm that runs in polynomial time. Polynomial time: the number of steps needed to solve a problem can be expressed as a function. Where x is the size of the input and n is a constant.

4
Whats so great about polynomial time? Running time of algorithm t(n) Maximum size solvable in 1 second nN 0 =100 million 100 N N 0 100nN 1 =1 million100 N N 1 n2n2 N 2 =10,00010 N N 2 n3n3 N 3 = N 3 10 N 3 2n2n N 4 =26N N Current computer 100 times faster 1000 times faster Ramachandran, Vijaya. P versus NP

5
NP: a problem that can be verified using an algorithm that runs in polynomial time IMPORTANT: This does not mean not polynomial time Definitions Continued

6
What would a solution to P = NP? look like? or P NP P

7
Why is P vs NP important? Clay Mathematics Institute: $1,000,000 prize Internet security implications Public Key Encryption Whitfield Diffie and Martin Hellman, 1976 RSA public-key cryptosystem Ronald Rivest, Adi Shamir, and Leonard Aldeman, 1977

8
RSA Encryption Uses a function that is NP but not known to be P to encrypt information. Fermats Little Theorem: Let a and p be integers such that p is prime and gcd(a, p) =1, then

9
Theorem: If m is an integer, n = pq, p and q are primes, and ef 1 mod ((p-1)(q-1)), then (m e ) f (mod n) m. Proof.

10
Theorem: If m is an integer, n = pq, p and q are primes, and ef 1 mod ((p-1)(q-1)), then (m e ) f (mod n) m. Proof. ef = (p-1)(q-1)k + 1 By substitution, (m e ) f = m (p-1)(q-1)k+1 = m (p-1)(q-1)k m.

11
Theorem: If m is an integer, n = pq, p and q are primes, and ef 1 mod ((p-1)(q-1)), then (m e ) f (mod n) m. Proof. ef = (p-1)(q-1)k + 1 By substitution, (m e ) f = m (p-1)(q-1)k+1 = m (p-1)(q-1)k m. Then by Fermats little theorem: (m (p-1) ) (q-1)k 1 (m e ) f m (p-1)(q-1)k m m (mod p)

12
Theorem: If m is an integer, n = pq, p and q are primes, and ef 1 mod ((p-1)(q-1)), then (m e ) f (mod n) m. Proof. ef = (p-1)(q-1)k + 1 By substitution, (m e ) f = m (p-1)(q-1)k+1 = m (p-1)(q-1)k m. Then by Fermats little theorem: (m (p-1) ) (q-1)k 1 (m e ) f m (p-1)(q-1)k m m (mod p) Similarly, (m e ) f m (p-1)(q-1)k m m (mod q).

13
Theorem: If m is an integer, n = pq, p and q are primes, and ef 1 mod ((p-1)(q-1)), then (m e ) f (mod n) m. Proof. ef = (p-1)(q-1)k + 1 By substitution, (m e ) f = m (p-1)(q-1)k+1 = m (p-1)(q-1)k m. Then by Fermats little theorem: (m (p-1) ) (q-1)k 1 (m e ) f m (p-1)(q-1)k m m (mod p) Similarly, (m e ) f m (p-1)(q-1)k m m (mod q). Therefore, by the Chinese Remainder Theorem we have (m e ) f (mod n) m.

14
RSA Example Necessary InformationWhere is comes fromWhat it is for this example p,qprimep=67 q=89 npq5963 Φ(n)Number of integers less than n that are relatively prime to n. (p-1)(q-1) 5808 e,fe,f >1 such thate = 37 f = 157 We want to encrypt the number 17: x e (mod n) (mod 5963) 5064 To decrypt: 5064 f (mod 5963)

15
Why is RSA secure? Its nearly impossible to find f without the factors of n. Since we do not have an algorithm that runs in polynomial time to find factorizations, finding the factors n is nearly impossible.

16
Is this number prime, if so what are its factors?

17
Sieve of Eratosthenes

18
Sieve of Eratosthenes

19
Sieve of Eratosthenes

20
Sieve of Eratosthenes

21
Does the Sieve of Eratosthenes run in polynomial time? NO. Why not? For a number with N digits, the number of steps the sieve needs is [10 N ] 1/2 which is exponential.

22
Primes is in P In 2002, Manindra Agrawal, Neeraj Kayal and Nitin Saxena came up with an algorithm that runs in polynomial and give the primality of a number. This algorithm is beautiful Carl Pomerance The proof is simple, elegant and beautiful R. Balasubramanian

23
AKS Algorithm From PRIMES is in P

24
Explanation of AKS Lemma 2.1 Let a be an integer, n is a natural number, n > 2 and gcd(a,n)=1. Then n is prime iff (X+ a) n X n +a(mod n). Proof. By the binomial theorem: the coefficient of x i in ((X+a) n –(X n +a) is a n-i.

25
Explanation of AKS Lemma 2.1 Let a be an integer, n is a natural number, n > 2 and gcd(a,n)=1. Then n is prime iff (X+ a) n X n +a(mod n). Proof. By the binomial theorem: the coefficient of x i in ((X+a) n –(X n +a) is a n-i. Suppose n is prime. Then 0 (mod n) and hence all of the coefficients are zero.

26
Explanation of AKS Lemma 2.1 Let a be an integer, n is a natural number, n > 2 and gcd(a,n)=1. Then n is prime iff (X+ a) n X n +a(mod n). Proof. By the binomial theorem: the coefficient of x i in ((X+a) n –(X n +a) is a n-i. Suppose n is prime. Then 0 (mod n) and hence all of the coefficients are zero. Suppose n is composite. Consider a prime q that is a factor of n and let q k divide n, but q k+1 does not. Then q k does not divide and gcd( a n-q, q k ) =1 Hence, the coefficient of X q is not zero (mod n). Therefore (X+a) n X n +a (mod n).

27
Does AKS ruin RSA? NO!! Why not? AKS does not factor a number, it only tells us if it is prime or not. RSA is secure as long as we dont have an algorithm that can factor in polynomial time.

28
Acknowledgements Manindra Agrawal, Neeraj Kayal, and Nitin Saxena. PRIMES is in P. (http://www.cse.iitk.ac.in/news/primality_v3.ps), Februaruy 2003.http://www.cse.iitk.ac.in/news/primality_v3.ps P vs NP Problem. Clay Mathematics Institute, (http://www.claymath.org/millennium/P_vs_NP/) Ramachandran, Vijaya. P versus NP. University of Texas Lectures on the Millennium Prize Problems, May (http://www.claymath.org/video/) Stewart, Ian. Ian Stewart on Minesweeper. Clay Mathematics Institute, (http://www.claymath.org/Popular_Lectures/Minesweeper) Kaliski, Burt. The Mathematics of the RSA Public-Key Cryptosystem. RSA Laboratories. Polynomial time. Wikipedia, (http://en.wikipedia.org/wiki/Polynomial _time)

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google