Learning with Errors

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1 Learning with Errors Chethan Kamath IST Austria April 22, 2015

2 Table of contents Background PAC Model Noisy-PAC Learning Parity with Noise The Parity Function Learning Parity with Noise BKW Algorithm Cryptography from LPN Background/LWE Bit-Encryption from LWE Security

3 BACKGROUND

4 Notation X : input set; Y : binary label-set {0, 1} D: distribution on the input set χ, η: distribution of the noise C: concept class, c: target concept R(h): generalisation error for a hypothesis h R(h) := P (h(x) c(x)) x D

5 PAC Model request D, c C L x D, b = c(x) h S

6 PAC Model Definition 1 A concept class C is called PAC-learnable if there exists an algorithm L and a function q 0 = q 0 (ɛ, δ) s.t. for any 1. ɛ > 0 (accuracy: approximately correct) 2. δ > 0 (confidence: probably) 3. distribution D on X 4. target concept c C outputs a hypothesis h S C s.t. for any sample size q q 0 : P S D q(r(h S) ɛ) (1 δ) If L runs in poly(1/ɛ, 1/δ)-time, C is efficiently PAC-learnable Distribution-free 1 Valiant, 1984

7 Noisy-PAC Model request D, c, η L x D, b η c(x) h S

8 Noisy-PAC Model Definition 2 A concept class C is efficiently learnable in presence of random classification noise if there exists an algorithm L and a function q 0 = q 0 (ɛ, δ) s.t. for any 1. ɛ > 0 (accuracy: approximately correct) 2. δ > 0 (confidence: probably) 3. distribution D on X 4. target concept c C and fixed noise-rate η < 1/2 outputs a hypothesis h S C s.t. for any sample size q q 0 : P S D q(r(h S) ɛ) (1 δ) and L runs in poly(1/ɛ, 1/δ)-time 2 Angluin and Laird, 1998

9 LEARNING PARITY WITH NOISE

10 The Parity Function: Definition Denoted by f s, where s Z n 2 determines it The value of the function is given by the rule f s (x) := s, x (mod 2) C := {f s : s Z n 2 } and C = 2n Restricted parity function: f s depends on only the first k bits if all non-zero components of s lies in the first k bits

11 Learning the Parity Function s Z n 2 request L s x Z n 2, b = s, x (mod 2) Find s, given s, x 1 = b 1 (mod 2). s, x q = b q (mod 2) where s Z n 2, x i Z n 2 (D=uniform), b i Z 2 and q poly(n) It is possible to learn s using O(n) samples and poly(n) time: Gaussian elimination Learning for arbitrary D possible 3 3 Helmbold et al., 1992

12 Learning Parity with Noise s Z n 2 request L s x Z n 2, b η s, x (mod 2) Find s, given s, x 1 η b 1 (mod 2) s, x 2 η b 2 (mod 2). s, x q η b q (mod 2) where s Z n 2, x i Z n 2, b i Z 2, q poly(n) and η < 1/2 Let A s,χ denote this distribution

13 Hardness of LPN: Intuition Consider applying Gaussian elimination to the noisy samples to find the first bit Find S [q] s.t. i S x i = (1, 0,..., 0) But the noise is amplified: solution correct only with probability 1/2 + 2 Θ(n) Therefore, the procedure needs to be repeated 2 Θ(n) times Alternative: maximum likelihood estimation of s using O(n) samples and 2 O(n) time

14 Hardness of LPN Statistical Query 4 Model: the learning algorithm has access to statistical queries, that is instead of the label, it get the probability of a property holding for the particular example C is learnable in SQ-model imples it is learnable in the Noisy-PAC model LPN: Hard to learn efficiently in the SQ-model 4 Kearns, 1998

15 BKW ALGORITHM

16 Overview Best known algorithm for LPN Solves LPN in time O(2 n/ log n ) Block-wise Gaussian elimination Works by iterative zeroising Focus: LPN on uniform distribution; algorithm works for arbitrary distributions

17 Setting Two parameters: a and b s.t. n ab Each sample is partitioned into a blocks of size b. That is, a sample, x = x 1,..., x n Z n 2 is split as x 1,..., x }{{ b... x } b(i 1)+1,..., x b(i 1)+b... x k b,..., x n }{{}}{{} block 1 block i block a Definition: V i, i-sample V i : the subspace of Z ab 2 consisting of those vectors whose last i blocks have all bits equal to zero i-sample of size s: a set of s vectors independently and uniformly distributed over V i. Example: 1-sample x 1,..., x }{{ b... x } b(i 1)+1,..., x b(i 1)+b... 0, 0,..., 0 }{{}}{{} block 1 block i block a

18 Main Theorem Theorem 5 LPN can be solved with a sample-size and total computation time poly(( 1 1 2η )2a, 2 b ). Corollary LPN for constant noise-rate η < 1/2 can be solved with sample-size and total computation time 2 O(n/ log n). Proof: Plug in a = (log n)/2 and b = 2n/ log n 5 Blum et al., 2003

19 Zeroising Input: i-samples x 1,..., x s Output: (i + 1)-samples u 1,..., u s Zeroise i (x 1,..., x s ). 1. Partition x 1,..., x s based on the values in block a i 2. For each partition p pick a vector x jp at random 3. Zeroise by x jp to each of the other vectors in the partition 4. Return the resulting vectors u 1,..., u s Lemma 1. u 1,..., u s are (i + 1)-samples with s s 2 b 2. Each vector in u 1,..., u s is written as the sum of two vectors in x 1,..., x s 3. The run-time O(s)

20 Main Algorithm Input: s labelled examples (x 1, b 1 ),..., (x s, b s ) Output: set S [s] s.t. i S x i = (1, 0,..., 0) Solve(x 1,..., x s ): 1. For i = 1,..., a 1, iteratively call Zeroise i ( ) 2. Let u 1,..., u s be the resulting (a 1)-samples 3. If (1, 0,..., 0) {u 1,..., u s } output the index of the 2 a 1 vectors subset of x 1,..., x s that resulted in (1, 0,..., 0) The first bit of s is: i S b i (mod 2) Analysis If s = a2 b, then s 2 b Probability of output is (1 1/e) Probability that output is correct is 1/2 + 1/2(1 2η) 2a 1 1 Repeat poly(( 1 2η )2a, b) times to reduce the error probability

21 Main Algorithm The rest of the bits of s can be found using Solve( ) on cycling shifting all the examples. Thus the effective computation time is poly(( 1 1 2η )2a, 2 b ) Recall: Restricted parity function depends only on k bits of s If k = O(log n) then we can learn the parity in O(n) Leads to separation between SQ-Model (where restricted-lpn is hard) and the noisy-pac model

22 CRYPTOGRAPHY FROM LPN

23 In some sense, cryptography is the opposite of learning. Shalev-Schwartz and Ben-David

24 Cryptography 101 How to build protocols? 1. Assume a hard problem π (e.g., factorisation, discrete-log) 2. Build a protocol Π on π 3. Aim: η is hard = Π is not breakable Π is breakable = π is not hard Reductions: π Π 1. Assume an adversary A against Π and use it to break π C π π Π B Π A 2. Since η is assumed to be hard, this leads to a contradiction.

25 Recall: LPN Find s, given s, x 1 η b 1 (mod 2) s, x 2 η b 2 (mod 2). s, x q η b q (mod 2) where s Z n 2, x i Z n 2, b i Z 2, q poly(n) and η < 1/2

26 Learning with Errors: LPN for higher moduli Find s, given s, x 1 χ b 1 (mod p) s, x 2 χ b 2 (mod p). s, x q χ b q (mod p) where s Z n p, x i Z n p, b i Z p, q poly and χ is a probability distribution on Z p LPN=LWE if p = 2 and χ(0) = 1 η, χ(1) = η

27 Hardness of LWE Conjectured to be hard to break Lattice problems reduce 6 to LWE for appropriate choice of p and χ Example: p = O(n 2 ), α = O( n log n) and χ = Ψ α, discrete Gaussian on Z p with s.d. αp For the above parameters SVP, SIVP LWE SVP: shortest-vector problem SIVP: shortest independent vectors problem The above parameters used for the encryption scheme 6 Regev, 2005

28 REGEV S ENCRYPTION SCHEME

29 Encryption Scheme: Definitions Consists of three algorithms Π = {K, E, D} Key Generation. K : N K (pk, sk) $ K(1 n ) Encryption. E : M C c $ E(m, pk) Decryption. D : C M { } m D(c, sk) Requirements: 1. Correctness: for all (pk, sk) $ K(1 n ), m $ M D(E(pk, m), sk) = m 2. Security: ciphertext c should not leak any information about the plaintext m

30 Bit-Encryption from LWE Bit-Encryption: M = {0, 1} Parameters: 1. n N: the security parameter 2. p: prime modulus of the underlying group (p = O(n 2 )) 3. l: length of the public key (l = 5n) 4. χ = Ψ α

31 Key Generation, K(1 n ): Bit-Encryption from LWE 1. Secret key: sk := s $ Z n p 2. Public key: pk := {x i, b i } l i=1, where $ x 1,..., x l Z n $ p, e 1,..., e l χ and b i := x i, s + e i Encryption, E(m, pk): 1. Choose { random S [l] ( i S 2. c := x i, i S b i) if m = 0 ( i S x i, p/2 + i S b i) if m = 1 Decryption, D(c, sk): Note that c = (x, b) { 1. m 0 if b x, s is closer to 0 than p/2 (modulo p) := 1 otherwise

32 Correctness Intuition: since the noise is sampled from appropriate discrete Gaussian, it does not drown the message Argument Decryption: e := { i S e b x, s if m = 0 i = b x, s p/2 if m = 1 m = 0 m = 1 p/4 0 p/4 p/2 3p/4 Error in decryption only if e < p/4 Let s χ denote the distribution of e Claim: for χ = Ψ α P e χ (e < p/4) > 1 δ for some δ > 0

33 Security Distributions involved: 1. A s,η : LWE sampling 2. C m : ciphertext corresponding to encryption of bit m 3. U : uniform distribution on Z n p Z p D X Y : denotes that D distinguishes X from Y Argument 1. Assume that the ciphertexts are distinguishable 2. A s.t. C 0 C 1 = A A 3. A s.t. C 0 U [shifting + averaging] = 4. A s.t. A s,η A U [Leftover Hash Lemma]

34 More LWE Post-Quantum Cryptosystems Fully-Homomorphic Encryption 7 7 Brakerski and Vaikuntanathan, 2011

35 Sources Mohri et al. Foundations of Machine Learning Shalev-Schwartz and Ben-David Understanding Machine Learning Regev On Lattices, Learning with Errors, Random Linear Codes, and Cryptography Blum et al. Noise-Tolerant Learning, the Parity Problem and the SQ Model

36 THANK YOU!

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