Short Non-interactive Zero-Knowledge Proofs

Jens Groth

University College London

Non-interactive zero-knowledge proof

Prover VerifierSoundness:Statement is true

Zero-knowledge:Nothing but truth revealed

CRS: 0100…11010 Statement: xL

Proof:

(x,w)RL

Non-interactive zero-knowledge proofs

• Statement C is satisfiable circuit• Perfect completeness• Statistical soundness• Computational zero-knowledge

• Uniformly random common reference string• Efficient prover – probabilistic polynomial time• Deterministic polynomial time verifier

Adaptive soundness:Adversary sees CRS before attempting to cheat with false (C,)

Our results

• Security level: 2-k

• Trapdoor perm size: kT = poly(k)

• Circuit size: |C| = poly(k)• Witness size: |w| |C|

CRS in bits Proof in bits Assumption

Kilian-Petrank |C|∙kT∙k∙(log k) |C|∙kT∙k∙(log k) Trapdoor perms

This work |C|∙kT∙polylog(k) |C|∙kT∙polylog(k) Trapdoor perms

CRS in bits Proof in bits Assumption

Gentry poly(k) |w|∙poly(k) Lattice-based

G-Ostrovsky-Sahai k3/polylog(k) |C|∙k3/polylog(k) Pairing-based

This work |C|∙polylog(k) |C|∙polylog(k) Naccache-Stern

Hidden random string - soundness

Statement: xL

(x,w)RL0

1

0

1

Hidden random string – zero-knowledge

Statement: xL

0

1

Two new techniques

• More efficient use of hidden random bits– Kilian-Petrank: |C|∙k∙(log(k)) hidden random bits– This work: |C|∙polylog(k) hidden random bits

• More efficient implementation of hidden bits– Trapdoor permutations:

kT = poly(k) bits per hidden random bit

– Naccache-Stern encryption:O(log k) bits per hidden random bit

Implementing the hidden random bits model

Statement: xL

(x,w)RL01...0

11…1

00…1

10…0K(1k) (pk,sk)

c1

c2

c3

c4

Epk(0;r1)

Epk(1;r2)

Epk(0;r3)

Epk(1;r4)

c1

1 ; r2

c3

0 ; r4

Naccache-Stern encryption

• pk = (M,P,g) sk = (M)– M is an RSA modulus

– P = p1p2…pd where p1,…,pd are O(log k) bit primes

– P | ord(g) = (M)/4 and |P| = O(|M|)

• Epk(m;r) = gmrP mod M

• Dsk(c): For each pi compute m mod pi

c(M)/pi = (g(M)/pi)m

Chinese remainder gives m mod P

Naccache-Stern implementation of hidden bits

Statement: xL

(x,w)RL01...0

11…1

00…1

10…0K(1k) (pk,sk)

c1

c2

c3

c4

Epk(010;r1)

Epk(101;r2)

Epk(011;r3)

Epk(110;r4)

?1? ; 1

10? ; 2

??1 ; 3

??? ; 4

0 if m mod pi even1 if m mod pi odd if m mod pi is -1

Revealing part of Naccache-Stern plaintext

• Ciphertext c = gmrP

• How to prove that m = x mod pi?

• Prover reveals such that P = (cg-x)P/pi

• Shows (M) = (gm-xrP)(M)/pi = (g(M)/pi)m-x

• Can compute the proof as = (cg-x)(P-1 mod (M)/P)P/pi

• Can randomize proof by multiplying with s(M)/P

• Generalizes to reveal m mod iSpi with a proof consisting of one group element

Zero-knowledge

• Simulator sets up pk = (M,P,g) such that ord(g) = (M)/4P and g = hP mod M

• Simulator also sets up the CRS such that it only contains ciphertexts of the form gt mod M

• For any m ZP we can compute r = ht-m mod M such that

gt = gm(gt-m) = gmrP mod M• This means the simulator can open each

ciphertext to arbitrary hidden bits

Efficient use of the hidden random bits

Statement: xL

(x,w)RL0

1

0

1

Kilian-Petrank

• Random bits not useful; need bits with structure• Use statistical sampling to get “good” blocks

10

11

00

01

Probably hidden pairs

are 00 and 11

Kilian-Petrank continued

• Reveal blocks of bits so remaining “good” blocks of bits have a particular structure (statistically)

• Reduce C to a 3SAT formula • Assign remaining “good” blocks to variables in • For each clause reveal some bits in the blocks

assigned to the literals of the clause• An unsatisfied clause has some probability of the

revealed bits not satisfying certain criterion• Repeat many times to make the probability of

cheating negligible for each clause

Probabilistically checkable proofs

• Polynomial time algorithms f, fw:

f: C belongs to gap-3SAT5fw: w x if C(w)=1 then (x)=1

is a gap-3SAT5 formula – All variables appear in exactly 5 clauses – thrice as

positive literal and twice as negative– Either all clauses are simultaneously satisfiable or a constant fraction are unsatisfiable

Strategy

• Compute = f(C) and prove that it is satisfiable• With the most efficient probabilistically checkable

proofs (Dinur 07 combined with BenSasson-Sudan 08) we have || = |C| polylog(k)

• Seems counterintuitive to make statement larger• However, since allows for a constant fraction of

“errors” less repetition is needed to make the overall soundness error negligible

• It is ok if the prover cheats on some clauses as long as cannot cheat on a constant fraction

Summary

• Technique 1: Reduce soundness error with probabilistically checkable proofs

• Technique 2: Implement hidden random bit string with Naccache-Stern encryption

Hidden bits Proof in bits Assumption

Kilian-Petrank |C|∙kT∙k∙(log k) |C|∙kT∙k∙(log k) Trapdoor perms

This work |C|∙kT∙polylog(k) |C|∙kT∙polylog(k) Trapdoor perms

CRS in bits Proof in bits Assumption

Gentry poly(k) |w|∙poly(k) Lattice-based

G-Ostrovsky-Sahai k3/polylog(k) |C|∙k3/polylog(k) Pairing-based

This work |C|∙polylog(k) |C|∙polylog(k) Nacache-Stern