The Pohlig-Hellman exponentiation cipher as a bridge between classical and modern cryptography
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1 The Pohlig-Hellman exponentiation cipher as a bridge between classical and modern cryptography Joshua Holden Rose-Hulman Institute of Technology Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 1 / 17
2 The Pohlig-Hellman Exponentiation Cipher: Background Originally proposed in 1976 (about same time as Diffie-Hellman) Not published until after RSA and Diffie-Hellman Private key cipher In some ways, logical successor to classical ciphers But also uses ideas from RSA and Diffie-Hellman Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 2 / 17
3 A Classical (Block) Additive Cipher m letters 2m-digit number Example: m = 2, blocks are 4 digit numbers ca 0200 ts 1918 an 0013 dd 0303 og 1406 sx 1823 Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 3 / 17
4 Additive Encryption Pick a modulus n larger than the largest possible block e.g. n > 2525, say n = 3000 Pick a key, e, between 1 and n Add the key to the plaintext block to get the ciphertext block Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 4 / 17
5 Example Let e = 1640, then mod mod mod mod mod mod 3000 Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 5 / 17
6 Additive Decryption Subtract the key to get the plaintext block back mod 3000 ca mod 3000 ts mod 3000 an mod 3000 dd mod 3000 og mod 3000 sx Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 6 / 17
7 Multiplicative Encryption A small variation Probably not used classically, but could have been This time the modulus and the key need to be relatively prime e.g. n = 3000, e = 1801 Multiply the key by the plaintext block to get the ciphertext block mod mod mod mod mod mod 3000 Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 7 / 17
8 Multiplicative Decryption Since gcd(e, n) = 1, we can use the Euclidean Algorithm to find d e mod n In this case mod 3000 Multiply the decryption key d by the ciphertext block to get the plaintext block mod 3000 ca mod 3000 ts mod 3000 an mod 3000 dd mod 3000 og mod 3000 sx Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 8 / 17
9 Pohlig-Hellman Encryption Seems like a logical extension Pick a prime number larger than the largest possible block e.g. p > 2525, say p = 3001 Pick a key, e C P e mod p (C is ciphertext block; P is plaintext block) Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 9 / 17
10 Pohlig-Hellman Encryption, continued Let e = 7, then (0200) mod 3001 (1918) mod 3001 (0013) mod 3001 (0303) mod 3001 (1406) mod 3001 (1823) mod 3001 Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 10 / 17
11 Pohlig-Hellman Decryption Now we have ciphertext. Recipient needs to recover plaintext P. Need to be able to take e-th roots! Need an inverse of some sort again. What sort? Need a number d such that C d P mod p (P e ) d P mod p P ed P mod p P ed P 1 1 mod p P ed 1 1 mod p So it s very important to know what numbers x have P x 1 mod p Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 11 / 17
12 Theorem (Fermat s Little Theorem, 1640) If p is a prime number and a is a positive integer, p a, then a p 1 1 mod p. Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 12 / 17
13 Now what do we need to decrypt? Need P ed 1 1 mod p This works if ed 1 = k(p 1) because then P ed 1 P k(p 1) (P (p 1) ) k 1 k mod p 1 mod p mod p mod p ed 1 = k(p 1) means ed 1 mod p 1 d e mod p 1 (not modulo p!) So for decryption we figure out d e mod p 1 using gcd(e, p 1), which had better = 1 and let P C d C e mod p. Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 13 / 17
14 Back to our Example Luckily, if p = 3001, e = 7, then gcd(e, p 1) = gcd(7, 3000) = 1 so we can decrypt. d e 2143 mod 3000 (using the Euclidean algorithm) Voilà! (1640) mod 3001 ca (0213) mod 3001 ts (0608) mod 3001 an (1140) mod 3001 dd (2918) mod 3001 og (0094) mod 3001 sx Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 14 / 17
15 What s the advantage? Known-plaintext attacks Consider the additive cipher Suppose an attacker obtains a plaintext-ciphertext pair E.g. ca 0200 corresponds to mod 3000 obtains the key The multiplicative cipher works similarly Finding the key is approximately the same speed as decryption with key Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 15 / 17
16 Pohlig-Hellman resists known-plaintext attacks Suppose the attacker obtains C = P e and P Needs to find the discrete logarithm e = log P C mod p Best known algorithm is substantially slower than decryption with key Why isn t Pohlig-Hellman used in practice? Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 16 / 17
17 Pohlig-Hellman resists known-plaintext attacks Suppose the attacker obtains C = P e and P Needs to find the discrete logarithm e = log P C mod p Best known algorithm is substantially slower than decryption with key Why isn t Pohlig-Hellman used in practice? It s slower than other private key ciphers which also resist known-plaintext attacks Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 16 / 17
18 So why should we care? RSA also uses modular exponentiation Composite modulus Find decryption key using Euler s Theorem (extension of Fermat s Little Theorem to composite modulus) But RSA also introduces idea of public and private keys Encryption key is public Decryption key is private Factorization of modulus is secret; need it to find decryption key Pohlig-Hellman provides a gentle introduction Other related topics Fast exponentiation Fast primality testing Always good to know another cipher! Joshua Holden (RHIT) The Pohlig-Hellman exponentiation cipher 17 / 17
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