POON TENG HIN

 RSA  Shamir’s Three-Pass Protocol  Other issues

 The locks in computer network  1-1 mapping function f so that c = f(m)

 The keys in computer network so that f -1 (c) = f -1 (f(m)) = (f -1 f)(m) = m

 Function  Modulo Operation  Greatest Common Divisor  Multiplicative Inverse  Number theory  Prime number

 (x × y) mod n = 1. The integer y is called a multiplicative inverse of x, usually denoted x−1 (it is unique if it exists).

People keep finding large prime numbers for computer Security. How the prime number are used?

 RSA is an algorithm for public-key cryptography  By Ron Rivest, Adi Shamir, Leonard Adleman

 Because of security, high strength  Encryption  Digital signatures  E.g electronic transactions,  software certification.

 Encryption: C = M e mod n  Decryption: M = C d mod n

 ABCDEFGHIJKLMNOPQRSTUVWXYZ  1234…………………………………26  Public key: n = 35, e = 5 C = M e mod n  Private key: d = 5 M = C d mod n  My word:  “ ”  “ ”  Also, try to give me your words

 p, q, (n) must be kept secret.  It is believed that determine (n) given n is equivalent to factoring n.  With presently known algorithms, determining d given e and n, appears to be at least as time-consuming as the factoring problem.  So use factoring as the benchmark for security evaluation.

 /HW/ASCII.html /HW/ASCII.html  A website of ASCII code

Plaintext: M ( M = {0,1}*) Cipher text: C (C = {0,1}*) It needs two distinct primes p and q Φ(n) = (p-1)(q-1) select an integer e such that gcd(e, Φ(n) ) = 1 Where n = pq, n>M Compute the d where ed = 1 (mod Φ(n)) Public key: (e,n) Private key: d

 Randomly choose p and q And n = p X q A sample n from RSA-576:

 gcd(e, Φ(n) ) = 1 and e > 1   A table to find e and d:

 Φ(n) is the number of positive integers less than n that is relative prime to n  Example Φ(6) :  the GCD(x,6) = 1 when x = 1,5 so Φ(6) = 2

Φ(p) = p-1 for any prime number p Φ(pq) = (p-1)(q-1) for any two distinct primes p and q

Euler’s: For every integer a and n that are relatively prime, a Φ(n) mod n = 1 Fermat’s : If n = p is prime, a p-1 mod p = 1

 ed = 1 (mod Φ(n)) or d = e -1 mod n  Such that ex + Φ(n) y = 1 and d is the value of x  One of the method is Euclidean algorithm

Fo example Φ(n) =20, e =3 Firstly, gcd(20,3) = 1 if the inverse exists. We use Euclidean algorithm: 20 = 3 x = 2 x = 3 – 1X2 = 3 – 1 X (20 – 6 X 3) = -1 X X 3 (ex + ny = 1) so d = 7

 66 = 1 × gcd(35, 31)  35 = 1 × gcd(31, 4)  31 = 7 × gcd(4, 3)  4 = 1 × gcd(3, 1)  3 = 3 × gcd(1, 0)  So,  gcd(66, 35) = gcd(35, 31) = gcd(31, 4) = gcd(4, 3) = gcd(3, 1) = gcd(1, 0) = 1.

 Encryption: C = M e mod n  Decryption: M = C d mod n Needs two distinct primes p and q And Φ(n) = (p-1)(q-1) select an integer e such that gcd(e, Φ(n) ) = 1 Where n = pq, n>M Compute the d where ed = 1 (mod Φ(n)) Public key: (e,n) Private key: d

 students.stanford.edu/~tjw/jsbn/rsa2.html students.stanford.edu/~tjw/jsbn/rsa2.html  /HW/RSAWorksheet.html /HW/RSAWorksheet.html

 1.A lock the box by his lock A  2.A  B (Box with lock A)  3.B lock the box by his lock B  4.B  A (Box with lock A & B)  5.A unlock his lock A  6.A  B (Box with lock B)  7. B unlock his lock B ~  ~finish~

 This is the protocol similar to the answer of the IQ question  This is different to RSA In this protocol, we need a prime p which is a public knowledge.

 A selects a random number a with gcd(a, p-1) = 1  B selects a random number b with gcd(b,p-1) = 1 a -1 and b -1 are the inverse of a and b of mod p-1

 A computes k 1 = k a mod p and send k 1 to B  B computes k 2 = k 1 b mod p and send k 2 to A  A computes k 3 = k 2 a-1 mod p and send k 3 to B  Finally, B computes k = k 3 b-1 mod p and get k.

 Q1.Using slide 13, what is the message under:  “ ”  “ ”  “”4 24  “ ”  Q2. Find d if Φ(n) = 58, e = 27 (use Euclidean algorithm)

 Others issues I would like to share.  I suggest you may think about them.

 Computer and Communications Security COMP364  By Prof. Cunsheng Ding

 People like math will like this game.

 Encryption: c = Ek(m), where Ek is usually applied to blocks of the plaintext m.  Decryption: m = Dk(c), where Dk is usually applied to blocks or characters of the ciphertext c.

 Example: Let d = 4 and define f by  i :  f(i) :  Then f is a permutation of Z 4.  The inverse permutation f−1 is given by  i :  f -1 (i) :

 E.g

A B C 1| plan1 plan3 plan2 2| plan2 plan1 plan3 3| plan3 plan2 plan1 Conclusion: Most people think that: plan1 is better than plan2 plan2 is better than plan3 plan3 is better than plan1

 ByeBye