- 91 Views
- Uploaded on
- Presentation posted in: General

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

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

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.

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”

P

NP

or

NP

P

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

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

Proof.

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.

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)

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).

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.

We want to encrypt the number 17:

xe(mod n) 1716(mod 5963) 5064

To decrypt:

5064f (mod 5963) 5064157 17

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.

203956878356401977405765866929034577280193993314348263094772646453283062722701277632936616063144088173312372882677123879538709400158306567338328279154499698366071906766440037074217117805690872792848149112022286332144876183376326512083574821647933992961249917319836219304274280243803104015000563790123

1 2 3 4 5 6 7 8 910

11121314151617181920

21222324252627282930

31323334353637383940

41424344454647484950

51525354555657585960

61626364656667686970

71727374757677787980

81828384858687888990

919293949596979899 100

1 2 3 4 5 6 7 8 910

11121314151617181920

21222324252627282930

31323334353637383940

41424344454647484950

51525354555657585960

61626364656667686970

71727374757677787980

81828384858687888990

919293949596979899 100

1 2 3 4 5 6 7 8 910

11121314151617181920

21222324252627282930

31323334353637383940

41424344454647484950

51525354555657585960

61626364656667686970

71727374757677787980

81828384858687888990

919293949596979899 100

1 2 3 4 5 6 7 8 910

11121314151617181920

21222324252627282930

31323334353637383940

41424344454647484950

51525354555657585960

61626364656667686970

71727374757677787980

81828384858687888990

919293949596979899 100

NO.

Why not?

For a number with N digits, the number of steps the sieve needs is [10N]1/2 which is exponential.

- 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

From “PRIMES is in P”

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 .

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.

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).

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.

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)