1. POON TENG HIN

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

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

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

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

7.  (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).

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

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

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

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

12.  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:  “17 21 14 33 8”  “ 6 30 11”  Also, try to give me your words

13.  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.

14.  http://www.cs.drexel.edu/~jpopyack/IntroCS /HW/ASCII.html http://www.cs.drexel.edu/~jpopyack/IntroCS /HW/ASCII.html  A website of ASCII code

15. 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

16.  Randomly choose p and q And n = p X q A sample n from http://www.rsa.com/rsalabs/node.asp?id=2093 RSA-576: 18819881292060796383869723946165043980 71635633794173827007633564229888597152 34665485319060606504743045317388011303 396716199692321205734031879550656996 221305168759307650257059

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

18.  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

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

20. 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

21.  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 http://www.di-mgt.com.au/euclidean.html

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

23.  66 = 1 × 35 + 31 gcd(35, 31)  35 = 1 × 31 + 4 gcd(31, 4)  31 = 7 × 4 + 3 gcd(4, 3)  4 = 1 × 3 + 1 gcd(3, 1)  3 = 3 × 1 + 0 gcd(1, 0)  So,  gcd(66, 35) = gcd(35, 31) = gcd(31, 4) = gcd(4, 3) = gcd(3, 1) = gcd(1, 0) = 1.

24.  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

25.  http://www-cs- students.stanford.edu/~tjw/jsbn/rsa2.html http://www-cs- students.stanford.edu/~tjw/jsbn/rsa2.html  http://www.cs.drexel.edu/~jpopyack/IntroCS /HW/RSAWorksheet.html http://www.cs.drexel.edu/~jpopyack/IntroCS /HW/RSAWorksheet.html

29.  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~

30.  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.

31.  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

32.  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.

33.  Q1.Using slide 13, what is the message under:  “12 21 10 24 20 4 15 14”  “15 14 10”  “”4 24  “6 4 14 4 24 8 10 9”  Q2. Find d if Φ(n) = 58, e = 27 (use Euclidean algorithm)

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

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

36.  People like math will like this game.

37.  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.

38.  Example: Let d = 4 and define f by  i : 0 1 2 3  f(i) : 2 0 3 1  Then f is a permutation of Z 4.  The inverse permutation f−1 is given by  i : 0 1 2 3  f -1 (i) : 1 3 0 2

39.  E.g

42. 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

43.  ByeBye