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Elements of Coding and Encryption Continuation 1 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|. 2 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 if b=0 a GCD a, b b if a =0 GCD b, a mod b otherwise • Theorem. Let a, b, c, and q be integers with b>0. If a=qb+c, then GCD(a, b) = GCD (b, c). 3 Euclidean Algorithm. Example if b =0 a GCD a, b b if a =0 GCD b, a mod b otherwise GCD(22,77): Step r = a mod b a b 0 - 22 77 4 Euclidean Algorithm. Example if b =0 a GCD a, b b if a =0 GCD b, a mod b otherwise GCD(22,77): Step r = a mod b a b 0 - 22 77 1 22 mod 77 = 22 77 22 5 Euclidean Algorithm. Example if b =0 a GCD a, b b if a =0 GCD b, a mod b otherwise GCD(22,77): Step r = a mod b a b 0 - 22 77 77 22 22 11 1 2 22 mod 77 = 22 77 mod 22 = 11 6 Euclidean Algorithm. Example if b =0 a GCD a, b b if a =0 GCD b, a mod b otherwise GCD(22,77): Step r = a mod b a b 0 - 22 77 77 22 22 11 11 0 1 2 3 22 mod 77 = 22 77 mod 22 = 11 22 mod 11 =0 7 Simple Encryption • 1. 2. 3. 4. 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). 8 Letter Number Conversion Table A 1 B 2 C 3 D 4 E 5 F 6 G 7 H 8 I J K L M 9 10 11 12 13 N O P Q R S T U V W X Y Z 14 15 16 17 18 19 20 21 22 23 24 25 26 9 Encryption example • • 1. 2. 3. 4. Let the encryption function (“secret key”) be f (a) = (3a + 9) mod 26 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 10 Decryption example • • Decryption works the same, except that we apply the inverse key function. Find the inverse of f (a) = (3a + 9) mod 26 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 11 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. 12 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). 13 Modular Exponentiation • Modular exponentiation is a key operation in the RSA encryption. It is raising of a number to a power e in Zn: a b mod n 0 a n e • 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. 14 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:e e1 e2 ... ek ; ei 2s ) • Then the following technique should be applied: be be e ...e be be ...be 1 2 k 1 2 k be1 (mod n) be2 (mod n) ... bek (mod n) 15 Modular Exponentiation • For example, let b 90; e 101; n 1189 • Then e 101 20 22 25 26 1 4 32 64 be 90101 901 432 64 90 904 9032 9064 (mod1189) 90(mod1189) (902 ) 2 (mod1189) (90 2 ) 5 (mod1189) (90 2 )6 (mod1189) 90 980 1125 529 88200 1125 529 240750 529 240750 529(mod1189) 240750(mod1189) 529(mod1189) 572 529(mod1189) 302588(mod1189) 582 16 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. 17 RSA Encryption • In the RSA encryption 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 RSA decryption, the exponent d is chosen as the smallest positive solution x to the congruence ex 1 (mod r): (b e (mod n ))d (mod n )= b 18 RSA Encryption b e (mod n) n = 4559, e = 13. Smiley Transmits: “Last name Smiley” 19 RSA Encryption b e (mod n) n = 4559, e = 13. Smiley Transmits: “Last name Smiley” L A S T N AM E S M I L E Y 20 RSA Encryption b e (mod n) n = 4559, e = 13. Smiley Transmits: “Last name Smiley” L A S T N AM E S M I L E Y 1201 1920 0014 0113 0500 1913 0912 0525 21 RSA Encryption b e (mod n) n = 4559, e = 13. Smiley Transmits: “Last name Smiley” L A S T N AM E S M I L E Y 1201 1920 0014 0113 0500 1913 0912 0525 120113 mod 4559, 192013 mod 4559, … 22 RSA Encryption b e (mod n) n = 4559, e = 13. Smiley Transmits: “Last name Smiley” L A S T N AM E S M I L E Y 1201 1920 0014 0113 0500 1913 0912 0525 120113 mod 4559, 192013 mod 4559, … 2853 0116 1478 2150 3906 4256 1445 2462 23 RSA Encryption b e (mod n) n = 4559, e = 13. Smiley Transmits: “Last name Smiley” L A S T N AM E SM I L E Y 1201 1920 0014 0113 0500 1913 0912 0525 120113 mod 4559, 192013 mod 4559, … 2853 0116 1478 2150 3906 4256 1445 2462 24 RSA Encryption • • a private decryption exponent d ,which when applied to n , recovers the original blocks b : (b e mod n )d mod n = b For n = 4559, e = 13 the decryptor d = 3397 25 Homework • Read Sections 3.2 – 3.3 • Problems (Exercises 3.3) 1, 2. 26