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1 Symmetric Crypto Pierre-Alain Fouque Birthday Paradox In a set of N elements, by picking at random N elements, we have with high probability a collision two elements are equal N=365, about 23 people are required Let two sets N and M of random elements in a large set D, the number of expected collisions is N M / D (Birthday paradox with boys and girls) Avoiding frequency attacks Main idea: large blocksize avoid frequency attack on small block, statistics are non-random Block cipher Cipher (E,D) «eff. algs» such that D(k,E(k,m))=c Main drawback of stream cipher: lacks of theory to construct secure PRG Iterate many times a «small» round function F Master Key k K1 K2... Kr Round Keys m F F F c

2 Data Encryption Standard DES (IBM 1973) and NBS standard in 1977 Key Length: 56 bits Block Length: 64 bits 16 rounds with 48-bit round keys K K1 K2 K3 32 bits 32 bits R0 L0 F R1 L1 F R2 L2 F R3 L3 FKi(Li,Ri)=(Ri,Li fki(ri))=(li+1,ri+1) Feistel scheme Designed by Horst Feistel at IBM Transform random function to random permutation L R K f f function Round input (32 bits) Expansion (32 to 48 bits function) Subkey (48 bits) SBox (6 to 4 bits functions) Permutation over 32 bits Round output (32 bits) Attacks against DES Before 1990: attacks against round reduced version (less than 16 rounds) : Differential cryptanalysis : Linear cryptanalysis other attacks: Davies-Murphy, side-channel In practice, the most efficient attack is the exhaustive search (EFF, copacabana)

3 Main drawback of DES Exhaustive key search in 256 (3DES) Block size (collision for 232 blocks) Differential / Linear Cryptanalysis DES: well-designed and withstands successfully 30 years of cryptanalysis 2DES 3DES Advanced Encryption Standard Substitution / Permutation Network Key Length: 128 / 192 / 256 bits Block Length: 128 bits Designed by Daemen and Rijmen Standardized by NIST in 2000 AES M ki S SubBytes ShiftRows MixColumns xi+1 xi

4 Security game Block cipher must be indistinguishable from a random permutation for all k, E(k,x) is a permutation which looks random provided the key is not known E(k,.) P Dist. x f(x) Chal. b {0,1} x f=e(k, ) or P() f(x) according to b... b Adv. Adv(A)= Pr[b=b ]-1/2 Feistel security Could you distinguish one-round Feistel? Could you distinguish two-round Feistel? Could you distinguish three-round Feistel? Modes of operation How to encipher larger messages? ECB, CBC, CTR, OFB, CFB Drawbacks: - deterministic Advantages: - parallelisable Ciphertext Block Chaining (CBC) Encrypting: C 0=IV,..., Ci=E(k,Ci-1 Mi) Decrypting: M i=d(k,ci) Ci-1 Drawbacks: - sequential Advantages: - randomized - propagation of error in decryption

5 Ciphertext FeedBack (CFB) How to use a block cipher as a stream cipher? Output FeedBack (OFB) How to use a block cipher as a stream cipher? Counter Mode (CTR) Better solution Security Confidentiality is ensure by the mode of operation Integrity: first block of CBC? Main idea: the ciphertext must be indistinguishable from random for polynomial-time adversaries Security Game: Example on CBC:

6 Def: Hash Function message M M {0,1}* H hash H(M) H(M) {0,1} n A hash function H compute a hash value, a.k.a. fingerprint of n bits for a given arbitrary long message M H : {0,1}* {0,1} n Usage: integrity, password storage, signature,... Eg: SHA-1 (160 bits), MD5 (128 bits), SHA-2,... Use cases: File integrity Idea : we want to detect if a file has been modified by recomputing its fingerprint // Fichier code.c #include <stdio.h> #include <stdlib.h> int main(int argc, char** argv) { if (argc <2) { } } SHA-1 Hash Length of 160 bits : SHA-1 (code.c) = A51F 07BB 62EC 44A3 F118 Use cases: Passwords Instead of storing a password on a machine, we store its hash h = H(password) To authenticate, the user must send h On the web, the server sends a random value N and the user must answer with H(N Password) Compression Function f a compression function f:{0,1} m+n {0,1} n Fixed-Length hashing function data For SHA-1 : n = 160 et m = 512 Chaining Variable m bits n bits f n bits output 23

7 Merkle-Damgard f a compression function f:{0,1} m+n {0,1} n Let M = M 1 M m a message to hash (l blocks of m bits) Construction: H f (M): h1=f(iv,), h2=f(h1,),..., hn=f(hn-1,pad) Th: If we have a collision on H f, then we have a collision on f Ml f IV f f H( M ) Security notions Collision Resistance Find M 1 and M 2 such that H(M 1 ) = H(M 2 ) (2 n/2 + Pollard) Second-preimage Resistance Given M 1, find M 2 such that H(M 1 ) = H(M 2 ) (2 n ) Preimage Resistance Given x, find M such that H(M) = x (2 n ) Length extension Attacks: Could you predict the value of H(M) without having to recompute from the beginning? Message Authentication Code (MAC) Warning: Encryption does not provide integrity Eg: CTR mode ensures confidentiality if the blockcipher used is secure. However, no integrity is guaranteed. (CBC first block) C 1 Alice Bank = «$200 on Bob s account» = (ctr) M 1 Eve M 1 = «$2000 on Eve s account» C 1 = C 1 M 1 M 1

8 Definition of Message Authentication Code Key generation: randomized alg. output: key uniformly distributed Tag MAC generation: randomized or deterministic input: M {0,1} * output: tag τ {0,1}t : τ = M K ( M ) Verification: deterministic alg. input: tag τ {0,1} t and message M output: bit if the tag is valid for this message s.t. for any K and message M, if τ = M K ( M ), then V K (τ, M ) = 1 Security game Adversary s goals: 1. key recovery attacks 2. forgery: producing a valid MAC for some message M (of his choice, or any) Adversary s ressources: 1. known message attack: interception of MACs. Adv. knows pair (M, τ) of already tagged messages 2. chosen message attack: Adv. knows the tag of message of his choice (access to a MAC generation alg. adaptively or not) Security game Def: Combining an adversary s goal and some ressources SUF-CMA: strongly inforgeability against chosen message attacks Challenger M i τ i (M, τ) 1 : valid tag Adversary A Adv ( A ) = Pr ( Expérience retourne 1) Generic Security 1. For a t-bit MAC, advantage (forgery probability) is always at least 1/2 t 2. Among 2 t/2 MACs, by the birthday paradox, there is a collision between two of them: these collisions can be used to recover the keys...

9 MAC vs. Signature Signatures: used for vertifying public keys, guarantee non-repudiation, same properties than hand-written signature MACs: very good performences, secret-key shared between two users no non-repudiation, no public verification First construction Let F : {0,1} k {0,1} * {0,1} t a random function (i.e. outputs are indistinguishable from random values) MAC construction: For message M = M 1 M m, τ = F K ( M 1 ) F K ( M m ) Is this scheme secure? Second Example Let F : {0,1} k {0,1} * {0,1} t random function For message M = M 1 M m For i = 1 to m, y i = F K ( <i>, M i ) τ = y 1 y m Is this scheme secure? unencrypted CBC-MAC C i = (M i C i-1 ) MAC = C m Secure only for constant length messages C2 Mac = C m

10 Security CBC-MAC Let 2 arbitrary messages M and M M3 MAC(M) is C 3 = Mac C2 M Mac = C 1 M 2 3 MAC(M ) is C 2 = Mac C 1 Mac = C 2 unencrypted CBC-MAC Given MACs of M and M, it is possible to forge MAC of another message M3 M 1 Mac M 2 C2 C3 C 1 Mac =C 2 Recovering the secret key is in 2 k MAC computation where k is the bit length of the used key (exhaustive search) No IV in CBC-MAC The integrity of the first block is not ensured if an IV is used IV IV IV IV Mac = C 2 Mac = C 2 ( M, IV, Mac ) ( M, IV, Mac ) 20 Encrypted CBC-MAC (EMAC) C i = (M i C i-1 ) and MAC = (C m ) Secure if less than 2 n/2 MACs are computed Keys can be recovered using 2 exhaustive search in time 2 k (for k-bit keys) C2 Cm Mac = C m+1

11 N1 N1 N2 N2 Some attacks Mac = C m+1 Nm Mac = C m +1 Some attacks Mac = C m+1 Nm Mac = C m +1 collision Some attacks Mac = C m+1 N1 N2 Nm collision Mac = C m +1 Attacks R τ N1 N2 Nm R τ = =

12 Security Analysis Assume 2 n/2 MACs computed: ( M i, τ i ), 0 i 2 n/2 and M i M j Using Birthday Paradox, there exists i,j s.t. i j and τ i = τ j Ask MAC τ of M i R, where R is a random block Claim: One can forge MAC for message M j R : τ Key Recovery DES K DES K DES K TDES K,K C2 Cm Mac = C m+1 For efficiency and security reasons, one decide E = DES with key K and E = TDES, with keys K,K. What is the complexity to recover keys K and K? Hash-based MAC Consider the following MAC scheme: MAC K ( M ) = H ( K M ) Is it secure? HMAC HMAC K ( M ) = H(K opad, H( K ipad, M )) where ipad and opad are constant values:

13 Encryption and Authentication IPSEC: MAC-Then-Encrypt SSL/TLS: Encrypt-Then-MAC SSH: MAC-And-Encrypt

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