Elements of Coding and Encryption

1. Elements of Coding and Encryption Continuation

2. Greatest Common Divisor • Given integers m and n, not both zero, we define the greatest common divisor of m and n to be the largest integer that divides both m and n: GCD(m, n). • Note that if m≠ 0, then GCD(m, 0) = |m|.

3. Greatest Common Divisor: Euclidian Algorithm • Determine the greatest common divisor (GCD) for two numbers. • Euclidean algorithm: GCD (a,b) can be recursively found from the formula • Theorem. Let a, b, c, and q be integers with b>0. If a=qb+c, then GCD(a, b) = GCD (b, c).

4. Euclidean Algorithm.Example GCD(22,77):

5. Euclidean Algorithm.Example GCD(22,77):

6. Euclidean Algorithm.Example GCD(22,77):

7. Euclidean Algorithm.Example GCD(22,77):

8. Simple Encryption • Variations on the following have been used to encrypt messages for thousands of years. • Convert a message to capitals. • Encode each letter by a number between 1 and 26. • Apply an invertible modular function to each number. • Convert back to letters (0 becomes 26).

9. Letter  Number Conversion Table

10. Encryption example • Let the encryption function (“secret key”) be f (a) = (3a + 9) mod26 • Encrypt “Stop Thief” • STOP THIEF (capitals) • 19,20,15,16 20,8,9,5,6 • 14,17,2,5 17,7,10,24,1 • Scipher obtained: NQBE QGJXA

11. Decryption example • Decryption works the same, except that we apply the inverse key function. • Find the inverse of f (a) = (3a + 9) mod26 • If we didn’t have to deal with mod 26, inverse would be g (b) = 3-1 (b- 9) • We’ll see that since GCD(3,26) = 1, the inverse of 3 (mod 26) is the number 9. This gives: g (b)= 9(b- 9) mod 26 = (9b – 3) mod 26

12. Open key Encryption • Encryption methods, which are based on the “secret key” function have one, but significant disadvantage: it is necessary to distribute a key among all users and all of them must take care of keeping the key unavailable to others. • The alternative is an “open key” encryption, which became very popular during last 15-20 years. In this method, which is based on the modular arithmetic, two different keys are used for encryption and decryption. The encrypting key is a “public key”, it is open, while the decrypting key is hidden.

13. RSA Encryption • In “open key” encryption methods, it is extremely hard (actually, it is impossible within some reasonable long time interval) to reverse a key, which is used for the encryption. • One of the most popular, efficient (and simple!) modern “open key” encryption methods is RSA named for its inventors (Rivest, Shamir, Adleman).

14. Modular Exponentiation • Modular exponentiation is a key operation in the RSA encryption. It is raising of a number to a power e in Zn: • In the RSA method, both e and n are very large numbers. In the real world applications, n is an integer of about 400 decimal digits.

15. Modular Exponentiation • To perform modular exponentiation, congruence properties are used. • To reduce computations, exponent e should be presented as a sum of powers of 2 (any integer number should be easily presented in such a way: ) • Then the following technique should be applied:

16. Modular Exponentiation • For example, let • Then

17. RSA Encryption • In the RSA encryption method, a message is converted into numbers, as in the modular encryption that we saw before. • The letters are then put together into number blocks with each block less than n. • Then each number block is exponentiated by the exponent e modulo n , which completes the encryption procedure.

18. RSA Encryption • In the RSAencryption method, n is chosen to be the product of prime numbers. • If n=pq, then e is chosen so that GCD (e, r) = 1, where r= (p-1)(q-1). There are many such es (almost all positive integers satisfy the given condition). • In the RSAdecryption, the exponent d is chosen as the smallest positive solution x to the congruence ex1 (mod r): (be(modn))d(modn)= b

19. RSA Encryption  b e(mod n) n = 4559, e = 13. Smiley Transmits: “Last name Smiley”

20. RSA Encryption  b e(mod n) n = 4559, e = 13. Smiley Transmits: “Last name Smiley” • L A S T N A M E S M I L E Y

21. RSA Encryption  b e(mod n) n = 4559, e = 13. Smiley Transmits: “Last name Smiley” • L A S T N A M E S M I L E Y • 1201 1920 0014 0113 0500 1913 0912 0525

22. RSA Encryption  b e(mod n) n = 4559, e = 13. Smiley Transmits: “Last name Smiley” • L A S T N A M E S M I L E Y • 1201 1920 0014 0113 0500 1913 0912 0525 • 120113mod 4559, 192013mod 4559, …

23. RSA Encryption  b e(mod n) n = 4559, e = 13. Smiley Transmits: “Last name Smiley” • L A S T N A M E S M I L E Y • 1201 1920 0014 0113 0500 1913 0912 0525 • 120113mod 4559, 192013mod 4559, … • 2853 0116 1478 2150 3906 4256 1445 2462

24. RSA Encryption  b e(mod n) n = 4559, e = 13. Smiley Transmits: “Last name Smiley” • L A S T N A M E S M I L E Y • 1201 1920 0014 0113 0500 1913 0912 0525 • 120113mod 4559, 192013mod 4559, … • 2853 0116 1478 2150 3906 4256 1445 2462

25. RSA Encryption • a private decryption exponent d ,which when applied to n , recovers the original blocks b : (b emod n )dmod n = b • For n = 4559, e = 13 the decryptor d= 3397

26. Homework • Read Sections 3.2 – 3.3 • Problems (Exercises 3.3) 1, 2.