Lattice-Based Cryptography: From Practice to Theory to Practice Vadim Lyubashevsky

Lattice-Based Cryptography: From Practice to Theory to Practice VadimLyubashevsky INRIA / CNRS / ENS Paris (September 12, 2011)

La CryptographieReposantsurles Réseaux: de la Pratique à la Théorie à la Pratique VadimLyubashevsky INRIA / CNRS / ENS Paris (Septembre12, 2011)

Lattice-Based Encryption Schemes • NTRU [Hoffstein, Pipher, Silverman ‘98] • LWE-Based [Regev ‘05] • Ring-LWE Based [L, Peikert, Regev ’10] • “NTRU-like” with a proof of security [Stehle, Steinfeld ‘11]

CryptosystèmesReposantsur les Réseaux • NTRU [Hoffstein, Pipher, Silverman ‘98] • Reposantsur LWE [Regev ‘05] • ReposantsurAnneau-LWE [L, Peikert, Regev ’10] • “NTRU” avec unepreuve de la sécurité [Stehle, Steinfeld ‘11]

Subset Sum Problem • Subset-Sum Based [L, Palacio, Segev ‘10] • LWE-Based [Regev ‘05] • Ring-LWE Based [L, Peikert, Regev ’10] • “NTRU-like” with a proof of security [Stehle, Steinfeld ‘11] • NTRU [Hoffstein, Pipher, Silverman ‘98]

Part 0.The Subset Sum Problem

ai ,T in ZM ai are chosen randomly T is a sum of a random subset of the ai a1 a2 a3 … anT Find a subset of ai's that sums to T (mod M) Subset Sum Problem

ai ,T in Z49 ai are chosen randomly T is a sum of a random subset of the ai 15 31 24 3 1411 15 + 31 + 14=11(mod 49) Subset Sum Problem

ai ,T in ZM a1 a2 a3 … anT Find a subset of ai's that sums to T (mod M) Hardness Depends on: Size of n and M Relationship between n and M How Hard is Subset Sum?

Complexity of Solving Subset Sum M 2log²(n) 2n 2n log(n) 2n² poly(n) 2Ω(n) poly(n) run-time “generalized birthday attacks” [FlaPrz05,Lyu06,Sha08] “lattice reduction attacks” [LagOdl85,Fri86]

Subset Sum Crypto • Why? • Simple operations • Exponential hardness • Seems very different from number theoretic assumptions • Seems to resist quantum attacks

Subset Sum is “Pseudorandom” • [Impagliazzo-Naor 1989]: • For random a1,...,an in ZM and random x1,...,xn in {0,1} • distinguishing the distribution • (a1,...,an, a1x1+...+anxn mod M) • from the uniform distribution U(ZMn+1) is as hard as finding x1,...,xn

Part 1.Cryptosystem Based on Subset Sum[L, Palacio, Segev 2010]

Public Key Encryption Allows for secure communication between parties who have previously never met

Public Key Encryption public key: p secret key: s Encrypt, Decrypt Decrypt(s,Encrypt(p,M))=M c=Encrypt(p,M) M=Decrypt(s,c)

Public Key Encryption What does “secure” mean? Intuitive answer: The adversary should not be able to read the message. c=Encrypt(p,M)

Public Key Encryption What does “secure” mean? Intuitive answer: The adversary should not be able to read the message. But what about other information about the message? c=Encrypt(p,M), c=Encrypt(p,M) E.g. If adversary can figure out that the same message was sent twice, is the scheme “secure”?

Public Key Encryption What does “secure” mean? Semantic Security: For every 2 messages M, M', it's impossible to distinguish Encrypt(p,M) from Encrypt(p,M') in polynomial time The Encrypt algorithm must be randomized!

Subset Sum Cryptosystem • Semantically secure based on Subset Sum for M ≈ nn • Main tools Subset sum is pseudo-random Addition in (Zq)n is “kind of like” addition in ZM where M=qn • The proof is very simple

Want to add 4679+3907+8465+1343 mod 104 4 6 7 9 3 9 0 7 8 4 6 5 1 3 4 3 Facts About Addition 2 1 2 4 6 7 9 3 9 0 7 8 4 6 5 1 3 4 3 6 2 7 4 8 3 9 4 Adding n numbers (written in base q) modulo qm → carries < n If q>>n, then Adding with carries ≈ Adding without carries (i.e. in ZM) (i.e. in (Zq)n )

So... 4 6 7 9 3 9 0 7 8 4 6 5 1 6 4 3 4 6 7 9 3 9 0 7 8 4 6 5 1 6 4 3 1 1 0 1 1 1 0 1 2 1 1 0 + = 8 1 1 9 0 2 2 9 = NOT Pseudorandom! Pseudorandom based on Subset Sum!

Column Subset Sum Addition Is Also Pseudorandom 4 6 7 9 3 9 0 7 8 4 6 5 1 6 4 3 1 1 0 1 1 1 1 0 0 9 8 0 + =

“Hybrid” Subset Sum Addition Is Also Pseudorandom 4 6 7 9 0 3 9 0 7 9 8 4 6 5 8 1 6 4 3 0 1 0 0 1 pseudorandom 1 1 1 0 0 + 6 3 2 2 0 =

Encryption Scheme (for 1 bit) A s t A t r + = {0,1}n + Zqn x n {0,1}n = u v Public Key

Encryption Scheme A s t A t r + = + = u v Is pseudo-random based on the hardness of the subset sum problem

Encryption Scheme A s t A t r + = + = u v = + v A A s s r r + + = A A s s r

Encryption Scheme A s t A t r + = + = u v + = A r s s u + ≈ = v s A r

Encryption Scheme A s t A t r + = + = u v Encryption of 0 - = s u v

Encryption Scheme A s t A t r + = + = u v + Encryption of 1 0 q/2 = - + = v’ q/2 s u v’ u

Part 2.Cryptosystem Based on Learning With Errors andWorst-Case Lattice Problems[Regev 2005]

Encryption Scheme(what we needed) A s t A t r + = + = “small” u v Pseudorandom

Picking the “Carries” • In Subset Sum: carries were deterministic • What if … we pick the “carries” at random from some distribution?

So... 4 6 7 9 3 9 0 7 8 4 6 5 1 6 4 3 4 6 7 9 3 9 0 7 8 4 6 5 1 6 4 3 1 1 0 1 2 3 0 1 2 1 1 0 + 1 3 2 1 + 0 2 2 9 = 7 2 0 3 = Pseudorandom based on Subset Sum! Pseudorandom based on LWE and worst-case lattice problems [Reg ‘05] (with a lemma from [ACPS ‘09])

Learning With Errors (LWE) Problem a1 . . . s e b a2 + = am Given aiand<ai,s>+eifind s. (eiand s are “small”) (Once there are enough ai , the s is uniquely determined) Theorem [Regev '05] : There is a polynomial-time quantum reduction from solving certain lattice problems in the worst-case to solving LWE.

Decision LWE Problem World 1 World 2 a1 a1 . . . b . . . s e b a2 a2 + = am am uniformly random in Zpm Lemma[Reg ’05]: Search-LWE < Decision-LWE

LWE vs. Subset Sum • The Subset Sum assumption has deterministic “noise” • The LWE assumption is more “versatile” LWE Problem Subset Sum Problem a1 . . . s e b s b a2 EASY !! a1 a2 … an n2 + = n2 + = am n n

LWE / Subset Sum Encryption A s t A t r + = + = u v

Part 3.Cryptosystem Based on Learning With Errors over Rings andWorst-Case Ideal Lattice Problems[L, Peikert, Regev 2010]

Source of Inefficiency of LWE Getting just one extra random-looking number requires n random numbers and a small error element. 2 8 7 3 1 2 1 + = * 0 2 1 Wishful thinking: get n random numbers and produce n pseudo-random numbers in “one shot” 2 1 8 0 + = * 7 2 3 1

Use Polynomials • f(x) is a polynomial xn+ an-1xn-1 + … + a1x + a0 • R = Zp[x]/(f(x)) is a polynomial ring with • Addition mod p • Polynomial multiplication mod p and f(x) • Each element of R consists of n elements in Zp • In R: • small+small = small • small*small = small (depending on f(x) )

Polynomial Interpretation of the LWE-based cryptosystem a s + = t r a + = u + r t = v Public Key v - u s = + - r a + s r t + - r a s + r a + s + r - s =

Security a s + = t r a + = u + r t = v Pseudorandom??

Learning With Errors over Rings a1 s e1 b1 a2 e2 b2 a3 e3 b3 + = … … … am em bm Theorem [LPR ‘10]: Finding s is as hard as solving lattice problems in all ideals of the ring Z[x]/(f(x))

Decision Learning With Errors over Rings World 1 World 2 a1 s b1 a1 b1 a2 b2 a2 b2 a3 b3 a3 b3 + = … … … … am bm am bm Theorem [LPR ‘10]: In cyclotomic rings, Search-RLWE < Decision-RLWE

Use Polynomials in Zp[x]/(f(x)) a s + = t r a + = u + r t = v

Part 4.1-Element Cryptosystem Based on Learning With Errors over Rings andWorst-Case Ideal Lattice Problems[Stehle, Steinfeld 2011]

Number of Ring Elements a s + = t r a + = u + r t = v p Encryption of m: u v + m , 2 Can you have a ciphertext with just 1 ring element?

Stehle, Steinfeld Cryptosystem “small” coefficients f a u a r + + m mod p = = 2 mod p g Uniformly random Pseudorandom based on Ring-LWE u g f r + g + g m = 2 u g mod 2 = g m u g mod 2 m = g

Part 5.NTRU Cryptosystem[Hoffstein, Pipher, Silverman 1998]

NTRU Cryptosystem f g - Very small f a u a r + + m mod p = = 2 mod p g “looks” random If a is random, then pseudorandom based on Ring-LWE u g f r + g + g m = 2 Since f, g are smaller, p can be smaller as well

Lattice-Based Cryptography: From Practice to Theory to Practice Vadim Lyubashevsky