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P vs. NP, AKS, RSA: The Acronyms of Mathematics Awareness Month Emily List Wittenberg University [email protected] April 2006: Mathematics Awareness Month “Mathematics and Internet Security”. Definitions.

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P vs. NP, AKS, RSA: The Acronyms of Mathematics Awareness MonthEmily ListWittenberg [email protected]


April 2006 mathematics awareness month mathematics and internet security
April 2006: Mathematics Awareness Month“Mathematics and Internet Security”


Definitions
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.


What s so great about polynomial time
What’s so great about polynomial time?

Current

computer

100 times

faster

1000 times

faster

Ramachandran, Vijaya. P versus NP


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”



Why is p vs np important
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


Rsa encryption
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


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
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.


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 m1
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.


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 m2
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)


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 m3
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).


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 m4
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.


Rsa example
RSA Example

We want to encrypt the number 17:

xe(mod n) 1716(mod 5963) 5064

To decrypt:

5064f (mod 5963) 5064157 17


Why is rsa secure
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.


Is this number prime if so what are it s factors
Is this number prime, if so what are it’s factors?

203956878356401977405765866929034577280193993314348263094772646453283062722701277632936616063144088173312372882677123879538709400158306567338328279154499698366071906766440037074217117805690872792848149112022286332144876183376326512083574821647933992961249917319836219304274280243803104015000563790123


Sieve of eratosthenes
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


Sieve of eratosthenes1
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


Sieve of eratosthenes2
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


Sieve of eratosthenes3
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


Does the sieve of eratosthenes run in polynomial time
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.


Primes is in p
“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


Aks algorithm
AKS Algorithm

From “PRIMES is in P”


Explanation of aks
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 .


Explanation of aks1
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.


Explanation of aks2
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).


Does aks ruin rsa
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.


Acknowledgements
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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