COMP4690 Tutorial

Cryptography

&

Number Theory

Outline

DES Example Number Theory RSA Example Diffie-Hellman Example

DES Some remarks

DES works on bits DES works by encrypting groups of 64 bits, which is the same as

16 hexadecimal numbers DES uses keys which are also apparently 64 bits long. However,

every 8th key bit is ignored in the DES algorithm, so the effective key size is 56 bits.

If the length of the message to be encrypted is not a multiple of 64 bits, it must be padded. E.g.: The plaintext message "Your lips are smoother than vaseline" is, in

hexadecimal, "596F7572206C6970 732061726520736D 6F6F746865722074 68616E2076617365 6C696E650D0A".

We then pad this message with some 0s on the end, to get a total of 80 hexadecimal digits: "596F7572206C6970 732061726520736D 6F6F746865722074 68616E2076617365 6C696E650D0A0000".

Then apply DES.

Key generation example

Let K be the hexadecimal key K = 133457799BBCDFF1. This gives us as the binary key :

K = 0001 0011 0011 0100 0101 0111 0111 1001 1001 1011 1011 1100 1101 1111 1111 0001

16 subkeys (48-bit) will be generated from K.

Key generation example

Based on table PC-1 (Permuted Choice 1), we get the 56-bit permutation

K+ = 1111000 0110011 0010101 0101111 0101010 1011001 1001111 0001111

Next, split this key into left and right halves, C0 and D0, where each half has 28 bits.

C0 = 1111000 0110011 0010101 0101111 D0 = 0101010 1011001 1001111 0001111

Key generation example

we now create sixteen blocks Cn and Dn, 1<=n<=16. Each pair of blocks Cn and Dn is formed from the previous pair Cn-1 and Dn-1, respectively, for n = 1, 2, ..., 16, using a “schedule of left shifts".

Round number

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Bits rotated

1 1 2 2 2 2 2 2 1 2 2 2 2 2 2 1

Key generation example

C0 = 1111000011001100101010101111D0 = 0101010101100110011110001111

C1 = 1110000110011001010101011111D1 = 1010101011001100111100011110

C2 = 1100001100110010101010111111D2 = 0101010110011001111000111101

C3 = 0000110011001010101011111111D3 = 0101011001100111100011110101

……

Key generation example

We now form the subkeys Kn, for 1<=n<=16, by applying the table PC-2 (Permutation Choice Two) to each of the concatenated pairs CnDn.

For the first subkey, we have C1D1 = 1110000 1100110 0101010 1011111

1010101 0110011 0011110 0011110 After we apply the permutation PC-2: K1 = 000110 110000 001011 101111 111111

000111 000001 110010

Modular Arithmetic

Two integers a and b are said to be congruent modulo n, if :

(a mod n) = (b mod n) This is written as a≡b mod n Define Zn as the set of nonnegative integers

less than n: Zn={0,1,…,(n-1)}

Modular Arithmetic

Properties of modular arithmetic

Modular Arithmetic

Define Zp as the set of nonnegative integers less than a given prime number p: Zp={0,1,…,(p-1)}

Because p is prime, all of the nonzero integers in Zp are relatively prime to p.

There exists a multiplicative inverse for all of the nonzero integers in Zp : For each nonzero w in Zp, there exists a z in Zp

such that w x z ≡ 1 mod p. z is called the multiplicative inverse of w. Or, z = w-1.

Number Theory

Fermat’s Little Theorem: ap-1 ≡ 1 mod p

where p is prime and gcd(a,p)=1 E.g.

a = 7, p = 19 72=49≡11 mod 19 74≡121≡7 mod 19 78≡49≡11 mod 19 716≡121≡7 mod 19 ap-1=718=716x72≡7x11=77≡1 mod 19

Number Theory

An alternative form of Fermat’s Little Theorem:

ap ≡ a mod p where p is prime and a is any positive integer

E.g. p=5,a=3,35=243≡3 mod 5 p=5,a=10,105=100000≡10 mod 5≡0

Number Theory

Euler’s Totient Function ø(n) The number of positive integers less than n and

relatively prime to n For prime number p,

ø(n)= p – 1 For n = pq where p and q are two different

prime numbers ø(n)= (p – 1) (q – 1)

Number Theory

Example: ø(21) From 1 to 21, totally 21 numbers 21 = 3x7, 3 and 7 are prime 3’s multiples:

3, 6, 9, 12, 15, 18, 21 7’s multiples:

7, 14, 21 Other numbers are all relatively prime to 21

21-7-3+1 = (3-1)x(7-1)

Number Theory

Euler’s Theorem aø(n) ≡ 1 mod n

where gcd(a,n)=1 E.g.

a=3;n=10; ø(10)=4; hence 34 = 81 ≡ 1 mod 10 a=2;n=11; ø(11)=10; hence 210 = 1024 ≡ 1 mod 11

Number Theory The powers of an integer a, modulo n

a, a2, a3, … (mod n) If a and n are relatively prime, based on Euler’s theorem, we

have aø(n) ≡ 1 mod n a, a2, a3, … will have a repeated pattern E.g., ø(5)=4, 3ø(5)=81≡1 mod 5

3, 4, 2, 1, 3, 4, 2, 1, … There may exist lots of m such that am ≡ 1 mod n The least positive exponent m such that am ≡ 1

mod n is referred to as the order of a (mod n) the exponent to which a belongs (mod n) the length of the period generated by a

Number Theory

Number Theory

Primitive root If a number’s order (mod n) is ø(n), this number is called a

primitive root of n Property of primitive root

If a is a primitive root of n, then its powers a,a2, a3,…, aø(n) are distinct (mod n), and are all relatively prime to n.

In particular, for a prime number p, if a is a primitive root of p, then a,a2, a3,…, ap-1 are distinct (mod p).

From the previous table, we can see that prime number 19’s primitive roots are 2, 3, 10, 13, 14, and 15.

RSA Example

1. Select primes: p=17 & q=11

2. Compute n = pq =17×11=187

3. Compute ø(n)=(p–1)(q-1)=16×10=160

4. Select e: gcd(e,160)=1; choose e=7

5. Determine d: de=1 mod 160 and d < 160 • d=23 since 23×7=161= 10×160+1

6. Publish public key KU={7,187}

7. Keep secret private key KR={23,17,11}

RSA Example

given message M = 88 encryption:C = 887 mod 187 = 11

decryption:M = 1123 mod 187 = 88

RSA Example Fast Modular Exponentiation

To calculate 887 mod 187 881 mod 187 = 88 882 mod 187 = 7744 mod 187 = 77 884 mod 187 = 772 mod 187 = 132 887 mod 187 = 884+2+1 mod 187 = 132x77x88 mod 187 = 894,432 mod

187 = 11 To calculate 1123 mod 187

111 mod 187 = 11 112 mod 187 = 121 114 mod 187 = 14,641 mod 187 = 55 118 mod 187 = 552 mod 187 = 33 1116 mod 187 = 332 mod 187 = 154 1123 mod 187 = 1116+4+2+1 mod187 = 154x55x121x11 mod 187 =

11,273,570 mod 187 = 88

Diffie-Hellman Key Exchange

Diffie-Hellman Key Exchange

users Alice & Bob who wish to swap keys: agree on prime q=7 and α=5 select random secret keys:

A chooses xA=3, B chooses xB=2

compute public keys: yA=5

3 mod 7 = 6 (Alice)

yB=52 mod 7 = 4 (Bob)

compute shared session key as:

Alice: KAB= yB

xA mod 7 = 43 mod 7 = 1

Bob: KAB= yA

xB mod 7 = 62 mod 7 = 1

Diffie-Hellman Key Exchange

users Alice & Bob who wish to swap keys: agree on prime q=353 and α=3 select random secret keys:

A chooses xA=97, B chooses xB=233

compute public keys: yA=3

97 mod 353 = 40 (Alice)

yB=3233 mod 353 = 248 (Bob)

compute shared session key as:

Alice: KAB= yB

xA mod 353 = 24897 mod 353 = 160

Bob: KAB= yA

xB mod 353 = 40233 mod 353 = 160