Elements of Coding and Encryption Continuation 1

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

• Theorem. Let a, b, c, and q be integers with b>0. If a=qb+c, then GCD(a, b) = GCD (b, c).

if =0

, if =0

, mod otherwise

b

GCD

GC

a

a b ab

bD ba

3

4

Euclidean Algorithm.Example

GCD(22,77):

Step r = a mod b a b

0 - 22 77

if =0

, if =0

, mod otherwise

a b

GCD a b b a

GCD b a b

5


GCD(22,77):


0 - 22 77

122 mod 77

= 2277 22

if =0

, if =0

, mod otherwise

a b

GCD a b b a

GCD b a b

6


GCD(22,77):


0 - 22 77

122 mod 77

= 2277 22

277 mod 22

= 1122 11

if =0

, if =0

, mod otherwise

a b

GCD a b b a

GCD b a b

7


GCD(22,77):


0 - 22 77

122 mod 77

= 2277 22

277 mod 22

= 1122 11

322 mod 11

= 011 0

if =0

, if =0

, mod otherwise

a b

GCD a b b a

GCD b a b

8

Simple Encryption

• Variations on the following have been used to encrypt messages for thousands of years.

1. Convert a message to capitals.2. Encode each letter by a number between 1 and 26.3. Apply an invertible modular function to each

number.4. Convert back to letters (0 becomes 26).

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Letter Number Conversion Table

A B C D E F G H I J K L M

1 2 3 4 5 6 7 8 910

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

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Encryption example

• Let the encryption function (“secret key”) bef (a) = (3a + 9) mod 26

• Encrypt “Stop Thief”1. STOP THIEF (capitals)2. 19,20,15,16 20,8,9,5,63. 14,17,2,5 17,7,10,24,14. Scipher obtained: NQBE QGJXA

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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 beg (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

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:

• 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

mod 0e na b a n


• 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:

15

1 2 ... ; 2sk ie e e e e

1 2 1 2

1 2

... ...

(mod ) (mod ) ... (mod )

k k

k

e e e ee ee

ee e

b b b b b

b n b n b n


• For example, let• Then

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90; 101; 1189b e n

0 2 5 6

101 1 4 32 64 4 32 64

2 2 2 5 2 6

101 2 2 2 2 1 4 32 64

90 90 90 90 90 90 (mod1189)

90(mod1189) (90 ) (mod1189) (90 ) (mod1189) (90 ) (mod1189)

90 980 1125 529 88200 1125 529 240750 529 240750 529(mod1189)

2

eb

e

40750(mod1189) 529(mod1189) 572 529(mod1189) 302588(mod1189) 582

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

19

RSA Encryption

n = 4559, e = 13. Smiley Transmits: “Last name Smiley”

b e (mod n)

20

RSA Encryption

n = 4559, e = 13. Smiley Transmits: “Last name Smiley” L A S T N A M E S M I L E Y

b e (mod n)

21

RSA Encryption

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

b e (mod n)

22

RSA Encryption


120113 mod 4559, 192013 mod 4559, …

b e (mod n)

23

RSA Encryption


120113 mod 4559, 192013 mod 4559, …

2853 0116 1478 2150 3906 4256 1445 2462

b e (mod n)

24

RSA Encryption


120113 mod 4559, 192013 mod 4559, …

2853 0116 1478 2150 3906 4256 1445 2462

b e (mod n)

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.

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Documents

Elements of Coding and Encryption Continuation 1