Presentation is loading. Please wait.

Presentation is loading. Please wait.

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

Similar presentations


Presentation on theme: "P vs. NP, AKS, RSA: The Acronyms of Mathematics Awareness Month Emily List Wittenberg University"— Presentation transcript:

1 P vs. NP, AKS, RSA: The Acronyms of Mathematics Awareness Month Emily List Wittenberg University s07.elist@wittenberg.edu

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 0 1000 N 0 100nN 1 =1 million100 N 1 1000 N 1 n2n2 N 2 =10,00010 N 2 31.6 N 2 n3n3 N 3 =4644.64 N 3 10 N 3 2n2n N 4 =26N 4 +6.64N 4 +9.97 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) 17 16 (mod 5963) 5064 To decrypt: 5064 f (mod 5963) 5064 157 17

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? 2039568783564019774057658669290345772 8019399331434826309477264645328306272 2701277632936616063144088173312372882 6771238795387094001583065673383282791 5449969836607190676644003707421711780 5690872792848149112022286332144876183 3763265120835748216479339929612499173 1983621930427428024380310401500056379 0123

17 1 2 3 4 5 6 7 8 910 11121314151617181920 21222324252627282930 31323334353637383940 41424344454647484950 51525354555657585960 61626364656667686970 71727374757677787980 81828384858687888990 919293949596979899 100 Sieve of Eratosthenes

18 1 2 3 4 5 6 7 8 910 11121314151617181920 21222324252627282930 31323334353637383940 41424344454647484950 51525354555657585960 61626364656667686970 71727374757677787980 81828384858687888990 919293949596979899 100 Sieve of Eratosthenes

19 1 2 3 4 5 6 7 8 910 11121314151617181920 21222324252627282930 31323334353637383940 41424344454647484950 51525354555657585960 61626364656667686970 71727374757677787980 81828384858687888990 919293949596979899 100 Sieve of Eratosthenes

20 1 2 3 4 5 6 7 8 910 11121314151617181920 21222324252627282930 31323334353637383940 41424344454647484950 51525354555657585960 61626364656667686970 71727374757677787980 81828384858687888990 919293949596979899 100 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 2001. (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)


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

Similar presentations


Ads by Google