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CS G513 / SS G513 Network Security. Agenda. Public Key Encryption Basics and Attacks RSA – Correctness and Security Strength. (Asymmetric) Public Key Encryption. Key Idea: One key (say the encryption key) is public and distributed in an open (but authentic) way.
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CS G513 / SS G513 Network Security Agenda Public Key Encryption Basics and Attacks RSA – Correctness and Security Strength
(Asymmetric) Public Key Encryption • Key Idea: • One key (say the encryption key) is public and distributed in an open (but authentic) way. • The other key (say decryption key) is kept secret/private. • Secrecy – achieved by the fact that only those who own (a private) key can decrypt a message encrypted by (the corresponding) public key. • D(kv , E(ku , m)) = m • kv is the private key • Ku is the public key Sundar B.
PKE - Types of Attacks • Chosen plain-text attack is possible by a passive adversary. • Stronger attack: Chosen ciphertext attack • Choose a ciphertext and obtain (by some means from victim) the corresponding plaintext • Indifferent chosen-ciphertext attack • Adversary provided with decryptions of ciphertexts of its choice, but the choices made a-priori. • Adaptive Chosen-ciphertext attacks • Adversary may use or have access to victim’s decryption machine but not the private key even after seeing the ciphertext to be broken. Sundar B.
Public Key Encryption - Issues • Public Key Distribution • Must be done in authentic way – otherwise leads to easy impersonation • PKE schemes may assume block wise encryption: • Plaintext is broken into blocks of fixed length • Cipher Block Chaining (CBC mode) can be used against block manipulation • CFB and OFB modes so cannot be used in PKE schemes – Why? Sundar B.
RSA • Rivest, Shamir and Adleman – 1978. • Based on modular arithmetic and prime factors. • Key generation • Generate two different, large, random primes p and q (roughly the same size as each other) • Compute n=pq and = (p-1)(q-1) • Select a random int. e such that gcd(e, ) = 1. • Compute d s.t. ed = 1 (mod ) • Let ku = (n,e) and kv = (n,d) • Does d always exist ? Sundar B.
RSA • Encryption: • Represent message as an int m in the range [0..N-1] • Compute c = me (mod n) • Decryption • Compute m = cd (mod n) Sundar B.
RSA - Correctness • Lemma 1: med = m (mod p) • Lemma 2: med = m (mod q) • Theorem: cd = (me)d = m (mod n) • Proof of Theorem: Since p and q are distinct primes, med – m divides pq (by lemma 1 and 2) i.e. med = m (mod n)
RSA - Correctness • Proof of Lemma 1: • Given ed = 1 (mod ), there exists k such that ed = 1 + k . • If gcd(m,p) = 1 then mp-1 = 1 (mod p) by Fermat’s Little Theorem. • Then med = m (mod p). • But this is true if gcd(m,p) = p as well as • both sides are congruent to 0 (mod p) Sundar B.
Security of RSA • Adversary’s task: • Given ciphertext c, and public key (n,e) • find plaintext m • find private key (n,d) • Possible approach: • Factorize n • Compute and d just as the sender does. • Then (n,d) is the private key and • m = cd (mod n) is the plaintext message. Sundar B.
Security of RSA • Security of RSA depends on 2 claims: • Factoring is difficult (Claim 1) and • Obtaining d is at least hard as factoring (Claim 2). • Factorization is difficult • No known polynomial time algorithm exists. • But not proven to be NP-hard. • It has been proved that • given any multiple of = (p-1)(q-1), n can be factored. • Since ed = 1(mod ) , ed-1 = k for some k.
