Download presentation

Presentation is loading. Please wait.

Published byKerry Patterson Modified over 2 years ago

1
Public Key Cryptosystems - RSA Receiver Sender Eavesdroppe r 175753411947 p q p q p q p and q prime

2
Public Key Cryptosystems - RSA Receiver Sender Eavesdroppe r 175753411947 p q p q p q p and q prime Compute numbers n = p*q 2337 323 2491

3
Public Key Cryptosystems - RSA Receiver Sender Eavesdroppe r 175753411947 p q p q p q p and q prime Choose e prime relative to (p-1)(q-1) 2337 323 2491 43 23 41

4
Public Key Cryptosystems - RSA Receiver Sender Eavesdroppe r 175753411947 p q p q p q p and q prime Publish pair as the public key 2337 323 2491 43 23 41

5
Public Key Cryptosystems - RSA Receiver Sender Eavesdroppe r 175753411947 p q p q p q p and q prime Find d such that (e*d -1) is divisible by (p-1)(q-1) 263 323 2491 43 23 41 16672337 2217

6
Public Key Cryptosystems - RSA Receiver Sender Eavesdroppe r 175753411947 p q p q p q p and q prime Keep as the private key 263 323 2491 43 23 41 16672337 2217

7
Public Key Cryptosystems - RSA Receiver Sender Eavesdroppe r Toss p and q 263 323 2491 43 23 41 16672337 2217

8
Public Key Cryptosystems - RSA Receiver Sender Eavesdroppe r 263 323 2491 43 23 41 16672337 2217 m mod 2337 43

9
Public Key Cryptosystems - RSA Receiver Sender Eavesdroppe r 263 323 2491 43 23 41 16672337 2217 m mod 2337 43 ( ) 1667 mod 2337

10
Public Key Cryptosystems - RSA, signing Receiver Sender Eavesdroppe r 263 323 2491 43 23 41 16672337 2217 m mod 2337 263

11
Public Key Cryptosystems - RSA, signing Receiver Sender Eavesdroppe r 263 323 2491 43 23 41 16672337 2217 m mod 2337 263 ( ) 23 mod 2337

12
Public Key Cryptosystems - RSA, exponentiating 25663 55637 mod 78837

13
Public Key Cryptosystems - RSA, exponentiating 25663 55637 mod 78837 Yikes!

14
Public Key Cryptosystems - RSA, exponentiating 25663 55637 mod 78837 25663 2 2 mod 78837... That is, do the modular reduction after each multiplication.

15
Public Key Cryptosystems - RSA, find primes 2 Probability that a random number n is prime: 1/ln(n) For 100 digit number this is 1/230.

16
Public Key Cryptosystems - RSA, find primes 2 Probability that a random number n is prime: 1/ln(n) For 100 digit number this is 1/230. But how to test for being prime?

17
Public Key Cryptosystems - RSA, find primes 2 Probability that a random number n is prime: 1/ln(n) For 100 digit number this is 1/230. But how to test for being prime? If p is prime and 0 < a < p, then a p-1 = 1 mod p Ex: 3 (5-1) = 81 = 1 mod 5 36 (29-1) = 37711171281396032013366321198900157303750656 = 1 mod 29

18
Public Key Cryptosystems - RSA, find primes 2 Probability that a random number n is prime: 1/ln(n) For 100 digit number this is 1/230. But how to test for being prime? If p is prime and 0 < a < p, then a p-1 = 1 mod p Ex: 3 (5-1) = 81 = 1 mod 5 36 (29-1) = 37711171281396032013366321198900157303750656 = 1 mod 29 Pr(p isn't prime but a p-1 = 1 mod p) = 1/1000000000000000

19
Public Key Cryptosystems - RSA, find primes Can always express a number n-1 as 2 b c for some odd number c. ex: 48 = 2 4 3 Then can compute a n-1 mod n by computing a c mod n and squaring the result b times. If the result is not 1 then n is not prime.

20
Public Key Cryptosystems - RSA, find primes Choose a random odd integer p to test. Calculate b = # times 2 divides p-1. Calculate m such that p = 1 + 2 m. Choose a random integer a such that 0 < a < p. If a ≡ 1 mod p || a ≡ -1 mod p, for some 0 ≤ j≤b-1, then p passes the test. A prime will pass the test for all a. b m 2 m j

21
Public Key Cryptosystems - RSA, find primes Choose a random odd integer p to test. Calculate b = # times 2 divides p-1. Calculate m such that p = 1 + 2 m. Choose a random integer a such that 0 < a < p. If a ≡ 1 mod p || a ≡ -1 mod p, for some 0 ≤ j≤b-1, then p passes the test. A prime will pass the test for all a. A non prime number passes the test for at most 1/4 of all possible a. So, repeat N times and probability of error is (1/4). b m 2 m j N

22
Public Key Cryptosystems - RSA, picking d and e Choose e first, then find p and q so (p-1) and (q-1) are relatively prime to e RSA is no less secure if e is always the same and small Popular values for e are 3 and 65537 For e = 3, though, must pad message or else ciphertext = plaintext Choose p ≡ 2 mod 3 so p-1 = 1 mod 3 So, choose random odd number, multiply by 3 and add 2, then test for primality Or, start with any random number, multiply by 6 and add 5

Similar presentations

OK

ECE-8843 Prof. John A. Copeland 404 894-5177 fax 404 894-0035 Office: GCATT Bldg.

ECE-8843 Prof. John A. Copeland 404 894-5177 fax 404 894-0035 Office: GCATT Bldg.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on financial services in india Ppt on peak load pricing lecture Ppt on latest technology in computer science Ppt on 2004 tsunami in india Free ppt on loch ness monster Ppt on drugs and medicines Ppt on fibonacci numbers formula Animated ppt on magnetism Ppt on right to education in india Doc convert to ppt online ticket