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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. - PowerPoint PPT Presentation
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P vs. NP, AKS, RSA: The Acronyms of Mathematics
Awareness Month
Emily ListWittenberg University
April 2006: Mathematics Awareness
Month“Mathematics and Internet Security”
Definitions
P: yes or no decision problems that can be solved by an algorithm that runs in polynomial time.
nx
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?
Running time of algorithm t(n)
Maximum size solvable in 1 second
n N0=100 million
100 N0 1000 N0
100n N1=1 million 100 N1 1000 N1
n2 N2=10,000 10 N2 31.6 N2
n3 N3=464 4.64 N3 10 N3
2n N4=26 N4+6.64 N4+9.97
Current computer
100 times faster
1000 times faster
Ramachandran, Vijaya. P versus NP
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
What would a solution to P = NP? look like?
or
PNP
P
NP
Why is P vs NP important?
Clay Mathematics Institute: $1,000,000 prize
Internet security implicationsPublic Key Encryption
•Whitfield Diffie and Martin Hellman, 1976
RSA public-key cryptosystem• Ronald Rivest, Adi Shamir, and
Leonard Aldeman, 1977
RSA Encryption
Uses a function that is NP but not known to be P to encrypt information.
)(mod11 pa p
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 (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 (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 (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)f m(p-1)(q-1)km m (mod p)
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)f m(p-1)(q-1)km m (mod p)
Similarly, (me)f m(p-1)(q-1)km m (mod q).
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)f m(p-1)(q-1)km m (mod p)
Similarly, (me)f m(p-1)(q-1)km m (mod q).
Therefore, by the Chinese Remainder Theorem we have (me)f (mod n) m.
RSA ExampleNecessary Information Where is comes from What it is for this
example
p,q prime p=67
q=89
n pq 5963
Φ(n) Number of integers less than n that are relatively prime to n .
(p-1)(q-1)
5808
e,f e,f >1 such that e = 37
f = 157ef n 1(m o d ( ))
We want to encrypt the number 17:
xe(mod n) 1716(mod 5963) 5064To decrypt:5064f (mod 5963) 5064157 17
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?
203956878356401977405765866929034577280193993314348263094772646453283062722701277632936616063144088173312372882677123879538709400158306567338328279154499698366071906766440037074217117805690872792848149112022286332144876183376326512083574821647933992961249917319836219304274280243803104015000563790123
1 2 3 4 5 6 7 8 910
11 12 13 14 15 16 17 18 1920
21 22 23 24 25 26 27 28 2930
31 32 33 34 35 36 37 38 3940
41 42 43 44 45 46 47 48 4950
51 52 53 54 55 56 57 58 5960
61 62 63 64 65 66 67 68 6970
71 72 73 74 75 76 77 78 7980
81 82 83 84 85 86 87 88 8990
91 92 93 94 95 96 97 98 99 100
Sieve of Eratosthenes
1 2 3 4 5 6 7 8 910
11 12 13 14 15 16 17 18 1920
21 22 23 24 25 26 27 28 2930
31 32 33 34 35 36 37 38 3940
41 42 43 44 45 46 47 48 4950
51 52 53 54 55 56 57 58 5960
61 62 63 64 65 66 67 68 6970
71 72 73 74 75 76 77 78 7980
81 82 83 84 85 86 87 88 8990
91 92 93 94 95 96 97 98 99 100
Sieve of Eratosthenes
1 2 3 4 5 6 7 8 910
11 12 13 14 15 16 17 18 1920
21 22 23 24 25 26 27 28 2930
31 32 33 34 35 36 37 38 3940
41 42 43 44 45 46 47 48 4950
51 52 53 54 55 56 57 58 5960
61 62 63 64 65 66 67 68 6970
71 72 73 74 75 76 77 78 7980
81 82 83 84 85 86 87 88 8990
91 92 93 94 95 96 97 98 99 100
Sieve of Eratosthenes
1 2 3 4 5 6 7 8 910
11 12 13 14 15 16 17 18 1920
21 22 23 24 25 26 27 28 2930
31 32 33 34 35 36 37 38 3940
41 42 43 44 45 46 47 48 4950
51 52 53 54 55 56 57 58 5960
61 62 63 64 65 66 67 68 6970
71 72 73 74 75 76 77 78 7980
81 82 83 84 85 86 87 88 8990
91 92 93 94 95 96 97 98 99 100
Sieve of Eratosthenes
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
• 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
From “PRIMES is in P”
Explanation of AKS
i
n
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 Xn +a(mod n).
Proof.
By the binomial theorem: the coefficient of xi in ((X+a)n –(Xn +a) is an-i .
Explanation of AKS
i
n
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 Xn +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.
i
n
Explanation of AKS
i
n
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 Xn +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) =1Hence, the coefficient of Xq is not zero (mod n).
Therefore (X+a)n Xn +a (mod n).
q
n
i
n
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
Manindra Agrawal, Neeraj Kayal, and Nitin Saxena. PRIMES is in P. (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)
