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April 2006: Mathematics Awareness Month “Mathematics and Internet Security”

P vs. NP, AKS, RSA: The Acronyms of Mathematics Awareness Month Emily List Wittenberg University s07.elist@wittenberg.edu. April 2006: Mathematics Awareness Month “Mathematics and Internet Security”. Definitions.

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April 2006: Mathematics Awareness Month “Mathematics and Internet Security”

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  1. P vs. NP, AKS, RSA: The Acronyms of Mathematics Awareness MonthEmily ListWittenberg Universitys07.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. What’s so great about polynomial time? Current computer 100 times faster 1000 times faster Ramachandran, Vijaya. P versus NP

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

  6. What would a solution to P = NP? look like? P NP or 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. Fermat’s 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 (me)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 (me)f (mod n) m. Proof. ef = (p-1)(q-1)k + 1 By substitution, (me)f = m(p-1)(q-1)k+1 = m(p-1)(q-1)km.

  11. Theorem: If m is an integer, n = pq, p and q are primes, and ef 1 mod ((p-1)(q-1)), then (me)f (mod n) m. Proof. ef = (p-1)(q-1)k + 1 By substitution, (me)f = m(p-1)(q-1)k+1 = m(p-1)(q-1)km. Then by Fermat’s little theorem: (m(p-1))(q-1)k 1 (me)fm(p-1)(q-1)kmm (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 (me)f (mod n) m. Proof. ef = (p-1)(q-1)k + 1 By substitution, (me)f = m(p-1)(q-1)k+1 = m(p-1)(q-1)km. Then by Fermat’s little theorem: (m(p-1))(q-1)k 1 (me)fm(p-1)(q-1)kmm (mod p) Similarly, (me)fm(p-1)(q-1)kmm (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 (me)f (mod n) m. Proof. ef = (p-1)(q-1)k + 1 By substitution, (me)f = m(p-1)(q-1)k+1 = m(p-1)(q-1)km. Then by Fermat’s little theorem: (m(p-1))(q-1)k 1 (me)fm(p-1)(q-1)kmm (mod p) Similarly, (me)fm(p-1)(q-1)kmm (mod q). Therefore, by the Chinese Remainder Theorem we have (me)f (mod n) m.

  14. RSA Example We want to encrypt the number 17: xe(mod n) 1716(mod 5963) 5064 To decrypt: 5064f (mod 5963) 5064157 17

  15. Why is RSA secure? It’s 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 it’s factors? 203956878356401977405765866929034577280193993314348263094772646453283062722701277632936616063144088173312372882677123879538709400158306567338328279154499698366071906766440037074217117805690872792848149112022286332144876183376326512083574821647933992961249917319836219304274280243803104015000563790123

  17. Sieve of Eratosthenes 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

  18. Sieve of Eratosthenes 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

  19. Sieve of Eratosthenes 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

  20. Sieve of Eratosthenes 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

  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 [10N]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)nXn +a(mod n). Proof. By the binomial theorem: the coefficient of xi in ((X+a)n –(Xn +a) is an-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)nXn +a(mod n). Proof. By the binomial theorem: the coefficient of xi in ((X+a)n –(Xn +a) is an-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)nXn +a(mod n). Proof. By the binomial theorem: the coefficient of xi in ((X+a)n –(Xn +a) is an-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 qk divide n, but qk+1 does not. Then qk does not divide and gcd( an-q, qk) =1 Hence, the coefficient of Xq is not zero (mod n). Therefore (X+a)nXn +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 don’t have an algorithm that can factor in polynomial time.

  28. Acknowledgements Manindra Agrawal, Neeraj Kayal, and Nitin Saxena. PRIMES is inP. (http://www.cse.iitk.ac.in/news/primality_v3.ps), Februaruy 2003. 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)

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