What is it?
1. RSA – Rivest, Shamir,
Adleman
2. Algorithm for public key
cryptography. Public-key
cryptography refers to a
cryptographic system
requiring two separate keys,
one to lock or encrypt the
plaintext and one to unlock
or decrypt the cypher text.
3. Wildly used in e-commerce
4. Involves 3 steps – key
generation, encryption and
decryption
RSA involves a public key and a private key. The public
key can be known to everyone and is used for
encrypting messages. Messages encrypted with the
public key can only be decrypted using the private key.
The keys for the RSA algorithm are generated the
following way:
Key Generation
0 Two large distinct prime numbers p and q, chosen at random
0 Block B of text has been encoded by some function g into an integer T such that T
is an integer and 0 < T < n (Calculate n=pq, where n is the mod of p and q)
0 Compute the Euler phi function. φ(n) = (p – 1)(q – 1) because n is the product of
2 primes.
0 Choose an integer e such that 1 < e < φ(n) and
gcd(e,φ(n)) = 1 (e and φ(n) are co-prime)
0 gcd(e,φ(n)) = 1 means that 1 is a linear combination of e and φ(n):
Use Euclidean algorithm to find unique value for d and such that 1 < d < φ(n)
d * e = 1 modφ(n) therefore, d = e -1 (mod φ(n) )
0 Public key is pair of values (e,n).
0 Private key is pair values (d,n)
Example 1
0 Select 2 prime numbers p=17, q=13
0 Calculate n = pq = 17x13 = 221
0 Calculate φ(n) = (p – 1)(q – 1) = 16x12 = 192
0 Select e such that e is relatively prime to φ(n) = 160
and less than φ(n) ; we choose e = 11
Example 1 cont’d
0 So n = 221, φ(n) = 192, e = 11
0 Determine d such that de = 1(mod 221) and d<221 using extended
Euclidean algorithm (pg 273) with 11 and 192 (gcd(192, 11)).
192 = (17 * 11) + 5
11 = (2 * 5) + 1
5 = (5 * 1) + 0
from which
1 = 11 – 2 * 5
= 11 – 2[192 – 17 * 11]
= 35 * 11 – 2 * 192
Therefore, d = 35
or
or
5 = 192 – (17 * 11)
1 = 11 – (2 * 5)
Encryption
Plaintext : T < n
Cipher Text: C = T e (mod n)
Decryption
Cipher Text : C
Plaintext : T = C d (mod n)
Example 1 Cont’d
Encrypt:
Plaintext : Let T = 8 < n
Cipher Text: C = T e (mod n) = 8 11 (mod 221)
= 83 * 83 * 83 * 82 mod 221
= 512 * 512 * 512 * 64 mod 221
=70 * 70 * 70 * 64 mod 221
= 4900 * 4480 mod 221
= 38 * 60 mod 221
= 2280 mod 221
= 70
Computational Aspects
0 Exponentiation in Modular Arithmetic:
[(a mod n) x (b mod n)]mod n = ( a x b ) mod n
X16 = X1 x X2 x X3 x … x X15
x11 = x1+2+8 = (x)(x2)(x8)
Example 1 Cont’d
Decrypt:
Cipher Text : C = 70
Plaintext : T = C d (mod n) = 70 35 (mod 221)
= (702)17 * 70 mod 221 = (4900)17 * 70 mod 221
= 3817 * 70 mod 221 = (382)8 * 38 * 70 mod 221
= (1444)8 * 38 * 70 mod 221 = (118)8 * 38 * 70 mod 221
= (1182)4 * 38 * 70 mod 221 = (13924) 4 * 38 * 70 mod 221
= 14 * 38 * 70 mod 221 = 2660 mod 221
=8
Example 2
0 In a public key system using RSA, you interpret the
ciphertext C=10 sent to a use whose public key is e =
5, n = 35. What is the plaintext M?
The Security of RSA
Four approaches to attacking the RSA Algorithm are:
0 Brute Force – trying all possible keys (small value of d
is vulnerable to this)
0 Mathematical Attacks – factoring the product of 2
primes (p and q should only differ in length by a few
digits)
0 Timing Attacks – Depend on running time of
algorithms
0 Chosen Ciphertext attacks – exploits properties of RSA
system
Home Work
Section 4.5
#20 and #21