George Danezis Microsoft Research, Cambridge, UK

Transcription

1 George Danezis Microsoft Research, Cambridge, UK

2 Identity as a proxy to check credentials Username decides access in Access Control Matrix Sometime it leaks too much information Real world examples Tickets allow you to use cinema / train Bars require customers to be older than 18 But do you want the barman to know your address?

3 Usual way: Identity provider certifies attributes of a subject. Identity consumer checks those attributes Match credential with live person (biometric) Examples: E-passport: signed attributes, with lightweight access control. Attributes: nationality, names, number, pictures,... Identity Cards: signatures over attributes Attributes: names, date of birth, picture, address,...

4 The players: Issuer (I) = Identity provider Prover (P) = subject Verifier (V) = identity consumer Properties: The prover convinces the verifier that he holds a credential with attributes that satisfy some boolean formula: Simple example age=18 AND city=cambridge Prover cannot lie Verifier cannot infer anything else aside the formula Anonymity maintained despite collusion of V & I

5 1. Issuing protocol: Prover gets a certified credential. Issuer Passport Issuing Authority Name=Peggy, age=25, address=cambridge, Status=single Cannot learn anything beyond age Prover Peggy 2. Showing Protocol: Prover makes assertions about some attributes age=25 Verifier Victor (Bar staff Checking age)

6 Single-show credential (Brands & Chaum) Blind the issuing protocol Show the credential in clear Multiple shows are linkable BAD Protocols are simpler GOOD We will Focus on these Multi-show (Camenisch & Lysyanskaya) Random oracle free signatures for issuing (CL) Blinded showing Prover shows that they know a signature over a particular ciphertext. Cannot link multiple shows of the credential More complex no implementations

7 Cryptographic preliminaries The discrete logarithm problem Schnorr s Identification protocol Unforgeability, simulator, Fiat-Shamir Heuristic Generalization to representation Showing protocol Linear relations of attributes AND-connective Issuing protocol Blinded issuing

8 Assume p a large prime (>1024 bits 2048 bits) Detail: p = qr+1 where q also large prime Denote the field of integers modulo p as Z p Example with p=5 Addition works fine: 1+2 = 3, 3+3 = 1,... Multiplication too: 2*2 = 4, 2*3 = 1,... Exponentiation is as expected: 2 2 = Choose g in the multiplicative group of Z p Such that g is a generator Example: g=

9 Exponentiation is computationally easy: Given g and x, easy to compute g x But logarithm is computationally hard: Given g and g x, difficult to find x = log g g x If p is large it is practically impossible Related DH problem Given (g, g x, g y ) difficult to find g xy Stronger assumption than DL problem

10 Efficient to find inverses Given c easy to calculate g -c mod p (p-1) c mod p-1 Efficient to find roots Given c easy to find g 1/c mod p c (1/c) = 1 mod (p-1) Note the case N=pq (RSA security) No need to be scared of this field.

11 Exemplary of the zero-knowledge protocols credentials are based on. Players Public g a generator of Z p Prover knows x (secret key) Verifier knows y = g x (public key) Aim: the prover convinces the verifier that she knows an x such that g x = y Zero-knowledge verifier does not learn x! Why identification? Given a certificate containing y

12 Knows: x Public: g, p Knows: y=g x P->V: g w = a (witness) Peggy (Prover) Random: w V->P: c P->V: cx+w = r (challenge) (response) Victor (Verifier) Check: g r = y c a g cx+w = (g x ) c g w

13 Assume that Peggy (Prover) does not know x? If, for the same witness, Peggy forges two valid responses to two of Victor s challenges r 1 = c 1 x + w r 2 = c 2 x + w Then Peggy must know x 2 equations, 2 unknowns (x,w) can find x

14 The verifier learns nothing new about x. How do we go about proving this? Verifier can simulate protocol executions On his own! Without any help from Peggy (Prover) This means that the transcript gives no information about x How does Victor simulate a transcript? (Witness, challenge, response)

15 Need to fake a transcript (g w, c, r ) Simulator: Trick: do not follow the protocol order! First pick the challenge c Then pick a random response r Then note that the response must satisfy: g r = (g x ) c g w -> g w = g r / (g x ) c Solve for g w Proof technique for ZK but also important in constructions (OR)

16 Schnorr s protocol Requires interaction between Peggy and Victor Victor cannot transfer proof to convince Charlie (In fact we saw he can completely fake a transcript) Fiat-Shamir Heuristic H[ ] is a cryptographic hash function Peggy sets c = H[g w ] Note that the simulator cannot work any more g w has to be set first to derive c Signature scheme Peggy sets c = H[g w, M]

17 Traditional Schnorr For fixed g, p and public key h = g x Peggy proves she knows x such that h = g x General problem Fix prime p, generators g 1,..., g l Public key h =g 1 x 1g2 x 2... gl x l Peggy proves she knows x 1,..., x l such that h =g 1 x 1 g 2 x 2... g l x l

18 Knows: x 1,..., x l Public: g, p Knows: h = g 1 X1 g 2 X2... g l Xl Peggy (Prover) l random: w i P->V: 0<i<l g w i = a (witness) V->P: c (challenge) Victor (Verifier) r i = cx i +w i P->V: r 1,..., r l Check: ( 0<i<l g i r i) = hc a (response) Let s convince ourselves: ( 0<i<l g i r i ) = ( 0<i<l g i x i ) c ( 0<i<l g w i) = h c a

19 Knows: x 1,..., x l Public: g, p Knows: h = g 1 X1 g 2 X2... g l Xl Peggy (Prover) l random: w i P->V: 0<i<l g w i = a (witness) V->P: c (challenge) Victor (Verifier) r i = cx i +w i P->V: r 1,..., r l Check: ( 0<i<l g i r i) = hc a (response) Lets convince ourselves: ( 0<i<l g i r i ) = ( 0<i<l g i x i ) c ( 0<i<l g w i) = h c a

20 Relation to DL representation Credential representation: Attributes x i Credential h =g 1 X1 g 2 X2... g l Xl, Sig Issuer (h) Credential showing protocol Peggy gives the credential to Victor Peggy proves a statement on values x i X age = 28 AND x city = H[Cambridge] Merely DL rep. proves she knows x i

21 Remember: Attributes x i, i = 1,...,4 Credential h =g 1 x 1g2 x 2 g3 x 3 g 4 x 4, Sig Issuer (h) Example relation of attributes: (x 1 + 2x 2 10x 3 = 13) AND (x 2 4x 3 = 5) Implies: (x 1 = 2x 3 +3) AND (x 2 = 4x 3 +5) Substitute into h h = g 1 2x 3 +3 g 2 4x 3 +5 g 3 x 3 g4 x 4= (g13 g 25 )(g 12 g 24 g 3 ) x 3 g4 x 4 Implies: h / (g 13 g 25 ) = (g 12 g 24 g 3 ) x 3 g4 x 4

22 Example (continued) (x 1 + 2x 2 10x 3 = 13) AND (x 2 4x 3 = 5) Implies: h / (g 13 g 25 ) = (g 12 g 24 g 3 ) x 3 g4 x 4 How do we prove that in ZK? DL representation proof! h = h / (g 13 g 25 ) g 1 = g 12 g 24 g 3 g 2 = g 4 Prove that you know x 3 and x 4 such that h = (g 1 ) x 3 (g2 ) x 4

23 Knows: x 1, x 2, x 3, x 4 Public: g, p Knows: h = g 1 X1 g 2 X2 g 3 X3 g 4 X4 random: w 1, w 2 Peggy (Prover) P->V: g 1 w 1 g2 w 2 = a V->P: c (witness) (challenge) Victor (Verifier) r 1 = cx 3 +w 1 r 2 = cx 4 +w 2 P->V: r 1, r 2 Check: (g 1 ) r 1 (g2 ) r 2 = (h )c a (response)

24 Reminder h = g 1 X1 g 2 X2 g 3 X3 g 4 X4 h = h / (g 13 g 25 ) g 1 = g 12 g 24 g 3 g 2 = g 4 a = g 1 w 1 g2 w 2 r 1 = cx 3 +w 1 r 2 = cx 4 +w 1 Check: (g 1 ) r 1 (g2 ) r 2 = (h )c a => (g 1 ) (cx 3+w 1) (g 2 ) (cx 4+w 1) = (h / (g 13 g 25 )) c g 1 w 1 g 2 w 2 => (g 1 2x 3 +3 g 2 4x 3 +5 g 3 x 3g4 x 4) = h x 1 x 2

25 Showing any relation implies knowing all attributes. Can make non-interactive (message m) c = H[h, m, a ] Other proofs: (OR) connector (simple concept) (x age =18 AND x city =H[Cambridge]) OR (x age =15) (NOT) connector Inequality (x age > 18) (Yao s millionaire protocol)

26 Standard tools Schnorr ZK proof of knowledge of discrete log. DL rep. ZK proof of knowledge of representation. Credential showing representation + certificate ZK proof of linear relations on attributes (AND) More reading: (OR), (NOT), Inequality

27 1. Issuing protocol: Prover gets a certified credential. Issuer Cannot learn anything Prover Credential h =g 1 X1 g 2 X2... g l Xl Sig Issuer (h) 2. Showing Protocol: Prover makes assertions about some attributes Verifier

28 Prover cannot falsify a credential Unlinkability Issuer cannot link a showing transcript to an instance of issuing h, Sig issuer (h) have to be unlinkable to issuing Achieving unlinkability Issuer s view: h = g 1 X1 g 2 X2...g l Xl Prover uses: h = g 1 X1 g 2 X2...g l Xl g 0 a 1

29 Knows: x 1, x 2,..., x l Public: g, p Knows: x 1, x 2,..., x l Private: x 0, (y 1,..., y l ) Public: h 0 = g x 0, g i = g y i Rand: w 0 I->P: g w 0 = a0 (witness) Prover Issuer P->I: c 0 (challenge) Rand: a 1, a 2, a 3 h = h g a 1 c 0 = H[h, g a 2(h 0 h) a 3a 0 ] c 0 = c 0 + a 3 r 0 = c 0 (x 0 + i x i y i ) +w 0 I->P: r 0 (response) Credential: h = g a 1 g i x i Signature: (c 0, r 0 ) Check: c 0 = H[h, g r 0(h 0 h ) -c 0] Check: g r o = (h 0 h) c 0a 0 r 0 = r 0 + a 2 + c 0 a 1

30 Knows: x 1, x 2,..., x l Public: g, p Knows: x 1, x 2,..., x l Private: x 0, (y 1,..., y l ) Public: h 0 = g x 0, g i = g y i Rand: w 0 I->P: g w 0 = a0 (witness) Prover Issuer Non interactive signature: c 0 = H[h, a 0 ] P->I: c 0 (challenge) Rand: a 1, a 2, a 3 h = h g a 1 c 0 = H[h, g a 2(h 0 h) a 3a 0 ] c 0 = c 0 + a 3 r 0 = c 0 (x 0 + i x i y i ) +w 0 I->P: r 0 (response) ZK knowledge proof of the representation of h 0 h = g x 0 gi x i Check: g r o = (h 0 h) c 0a 0 r 0 = r 0 + a 2 + c 0 a 1 = g (x 0 + i x i y i ) : just Schnorr!

31 Public: g, p, h 0 = g x 0, g i = g y i Knows: x 1, x 2,..., x l I->P: g w 0 = a0 (witness) P->I: c 0 (challenge) I->P: r 0 (response) Prover Rand: a 1, a 2, a 3 h = h g a 1 c 0 = H[h, g a 2(h 0 h) a 3a 0 ] c 0 = c 0 + a 3 Check: g r o = (h 0 h) c 0a 0 r 0 = r 0 + a 2 + c 0 a 1 Schnorr Verification: Issuer knows the representation of (h 0 h)! Credential: h = g a 1 g i x i Signature: (c 0, r 0 ) Check: c 0 = H[h, g r 0(h 0 h ) -c 0]

32 Public: g, p, h 0 = g x 0, g i = g y i Knows: x 1, x 2,..., x l I->P: g w 0 = a0 (witness) Prover P->I: c 0 (challenge) I->P: r 0 (response) Unlinkable Rand: a 1, a 2, a 3 h = h g a 1 c 0 = H[h, g a 2(h 0 h) a 3a 0 ] c 0 = c 0 + a 3 Check: g r o = (h 0 h) c 0a 0 r 0 = r 0 + a 2 + c 0 a 1 1) Set c 0 2) Get r 0 such that... Credential: h = g a 1 g i x i Signature: (c 0, r 0 ) Check: c 0 = H[h, g r 0(h 0 h ) -c 0] My Goal

33 Public: g, p, h 0 = g x 0, g i = g y i Knows: x 1, x 2,..., x l I->P: g w 0 = a0 (witness) Prover Rand: a 1, a 2, a 3 h = h g a 1 c 0 = H[h, g a 2(h 0 h) a 3a 0 ] c 0 = c 0 + a 3? (response) Check: g r o = (h 0 h) c 0a 0 r 0 = r 0 + a 2 + c 0 a 1 Credential: h = g a 1 g x i i Signature: (c 0, r 0 ) Check: c 0 = H[h, g r 0(h 0 h ) -c 0] P->I: c 0 (challenge) I->P: r 0

34 Goal: c 0 = H[h, g a 2 (h0 h) a 3 a0 ] = H[h, g r 0 (h0 h ) -c 0 ] So g a 2(h 0 h) a 3a 0 = g r 0(h 0 h ) -c 0 must be true Lets follow: g r 0(h 0 h ) -c 0 = g a 2(h 0 h) a 3a 0 g (r 0 + a 2 + c 0 a 1) (h 0 h) -(c 0+a 3) g -c 0a 1 = g a 2 (h 0 h) a 3 a 0 (g r 0 (h0 h) -c 0) (g a 2(h 0 h) a 3 ) = (g a 2(h 0 h) a 3 ) a 0 Substitute r 0 and c 0 TRUE

35 Issuer sees: c o, r 0, h Such that g r o = (h0 h) c 0 a0 Verifier sees: c 0, r 0, h Relation: Random: a 1, a 2, a 3 h = h g a 1 c 0 = c 0 + a 3 r 0 = r 0 + a 2 + c 0 a 1 Even if they collude they cannot link the credential issuing and showing

36 Authentication between Issuer and Peggy Need to check that Peggy has the attributes requested Issuing protocol should not be run in parallel! (simple modifications are required)

37 Putting it all together: Issuer and Peggy run the issuing protocol. Peggy gets: Credential: h = g a 1 g i x i Signature: (c 0, r 0 ) Check: c 0 = H[h, g r 0(h 0 h ) -c 0] Peggy and Victor run the showing protocol Victor checks the validity of the credential first Peggy shows some relation on the attributes (Using DL-rep proof on h )

38 Credential issuing Proof of knowledge of DL-rep & x 0 of issuer Peggy assists & blinds proof to avoid linking Further topics Transferability of credential Double spending

39 Attribute based access control Federated identity management Electronic cash (double spending) Privacy friendly e-identity Id-cards & e-passports Multi-show credentials!

40 Core: Claus P. Schnorr. Efficient signature generation by smart cards. Journal of Cryptology, 4: , Stefan Brands. Rethinking public key infrastructures and digital certificates building in privacy. MIT Press. More: Jan Camenisch and Markus Stadler. Proof systems for general statements about discrete logarithms. Technical report TR 260, Institute for Theoretical Computer Science, ETH, Zurich, March Jan Camenisch and Anna Lysianskaya. A signature scheme with efficient proofs. (CL signatures)

41 Peggy wants to prove (A OR B) Say A is true and B is false Simple modification of Schnorr Peggy sends witness Victor sends commitment c Peggy uses simulator for producing a response r B for B (That sets a particular c B ) Peggy chooses c A such that c = c A + c B Then she produces the response r A for A Key concept: simulators are useful, not just proof tools!

42 Designated verifier proof A OR knowledge of verifier s key Simulate the second part Third parties do nor know if A is true or the statement has been built by the designated verifier! Non-interactive proof not transferable!