Presentation is loading. Please wait.

Presentation is loading. Please wait.

Week 3 - Friday.  What did we talk about last time?  AES  Public key cryptography.

Similar presentations


Presentation on theme: "Week 3 - Friday.  What did we talk about last time?  AES  Public key cryptography."— Presentation transcript:

1 Week 3 - Friday

2  What did we talk about last time?  AES  Public key cryptography

3

4

5

6 Matt Shank

7

8  RSA depends in large part on the difficulty of factoring large composite numbers (particularly those that are a product of only 2 primes)  Recall that an integer p is prime if  p > 1  p is not divisible by any positive integers other than 1 and itself  Trivia: The largest prime currently known is – 1  It has 17,425,170 digits in base 10

9  Any integer greater than 1 can be factored into a unique series of prime factors:  Example: 52 = 2 2 ∙ 13  Two integers a and b (greater than 1) are relatively prime or coprime if and only if a shares no prime factors with b

10  We won't get into the number theory behind this (yet)  A Rabin-Miller primality test works as follows:  Let n be the number you want to prove if it's prime or not  n must be odd, thus n – 1 is even  (n – 1) = 2 s d where s and d are positive integers and d is odd  If n is prime, then for any integer 1 < a < n, exactly one of the two is true: ▪ a d  1 (mod n) or ▪ a 2 r d  -1 (mod n), 1 ≤ r < s  Pick several a values, see if either of the two cases hold  If it ever doesn't, you know you have a composite

11  What if we want to see if 221 is prime?  n – 1 = 220 = 2 2 ∙55  s = 2  d = 55  Attempt 1: Let a = 174  a 2 0 ·d mod n = mod 221 = 47 ≠ 1, n − 1  a 2 1 ·d mod n = mod 221 = 220 = n − 1 Check!  Attempt 2: Let a = 137  a 2 0 ·d mod n = mod 221 = 188 ≠ 1, n − 1  a 2 1 ·d mod n = mod 221 = 205 ≠ n − 1 Oh no!  Every successful attempt means there is only a 25% chance that the number is composite  So, after k attempts, there is a 4 -k chance that the number is composite

12  The greatest common divisor or GCD of two numbers gives the largest factor they have in common  Example:  GCD( 12, 18 ) =  GCD( 42, 56 ) =  For small numbers, we can determine GCD by doing a complete factorization

13  For large numbers, we can use Euclid's algorithm to determine the GCD of two numbers  Algorithm GCD( a, b) 1. If b = 0 ▪ Return a 2. Else ▪ temp = a mod b ▪ a = b ▪ b = temp 3. Goto Step 1  Example: GCD( 1970, 1066)

14  We can extend Euclid's algorithm to give us the multiplicative inverse for modular arithmetic  Example: Find the inverse of 120 mod 23  Let a be the number  Let b be the modular base  Find Inverse(a, b)  x = 0  lastx = 1  y = 1  lasty = 0  while b ≠ 0 ▪ quotient = a div b ▪ temp = b ▪ b = a mod b ▪ a = temp ▪ temp = x ▪ x = lastx-quotient*x ▪ lastx = temp ▪ temp = y ▪ y = lasty-quotient*y ▪ lasty = temp  Return lastx

15

16  If p is prime and a is a positive integer not divisible by p, then: a p –1  1 (mod p)

17  Assume a is positive and less than p  Consider the sequence a, 2a, 3a, …, (p – 1)a  If these are taken mod p, we will get:  1, 2, 3, …, p – 1  This bit is the least obvious part of the proof  However (because p is prime) if you add any non-zero element repeatedly, you will eventually get back to the starting point, covering all values (except 0) once  Multiplying this sequence together gives:  a ∙ 2a ∙ 3a ∙ … ∙ (p – 1)a  1 ∙ 2 ∙ 3 ∙ … ∙ (p – 1) (mod p)  a p – 1 (p – 1)!  (p – 1)! (mod p)  a p – 1  1 (mod p)

18  Euler’s totient function  (n)   (n) = the number of positive integers less than n and relatively prime to n (including 1)  If p is prime, then  (p) = p – 1  If we have two primes p and q (which are different), then:  (pq) =  (p)∙  (q) = (p – 1)(q – 1)

19  Euler’s Theorem: For every a and n that are relatively prime, a  (n)  1 (mod n)  This generalizes Fermat’s Theorem because  (p) = p – 1 if p is prime  Proof is messier

20

21  Named for Rivest, Shamir, and Adleman  Take a plaintext M converted to an integer  Create an ciphertext C as follows: C = M e mod n  Decrypt C back into M as follows: M = C d mod n = (M e ) d mod n = M ed mod n

22 TermDetailsSource MMessage to be encryptedSender CEncrypted messageComputed by sender nModulus, n = pqKnown by everyone pPrime numberKnown by receiver qPrime numberKnown by receiver eEncryption exponentKnown by everyone dDecryption exponentComputed by receiver (n)(n) Totient of nKnown by receiver

23  To encrypt: C = M e mod n  e is often 3, but is always publically known  To decrypt: M = C d mod n = M ed mod n  We get d by finding the multiplicative inverse of e mod  (n)  So, ed  1 (mod  (n))

24  We know that ed  1 (mod  (n))  This means that ed = k  (n) + 1 for some nonnegative integer k  M ed = M k  (n) + 1  M∙(M  (n) ) k (mod n)  By Euler’s Theorem M  (n)  1 (mod n)  So, M∙(M  (n) ) k  M (mod n)

25  M = 26  p = 17, q = 11, n = 187, e = 3  C = M 3 mod 187 = 185   (n) = (p – 1)(q – 1) = 160  d = e -1 mod 160 = 107  C d = mod 187 = 26  If you can trust my modular arithmetic

26  You can’t compute the multiplicative inverse of e mod  (n) unless you know what  (n) is  If you know p and q, finding  (n) is easy  Finding  (n) is equivalent to finding p and q by factoring n  No one knows an efficient way to factor a large composite number

27

28  Key management issues  Cryptographic hash functions  Yuki Gage presents

29  Finish reading Section 2.7 and 12.3  Start reading Section 2.8  Finish Assignment 1  Due tonight by midnight


Download ppt "Week 3 - Friday.  What did we talk about last time?  AES  Public key cryptography."

Similar presentations


Ads by Google