# DIGITAL SIGNATURES 1/1

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1 DIGITAL SIGNATURES 1/1

2 Signing by hand COSMO ALICE ALICE Pay Bob \$100 Cosmo Alice Alice Bank =? no Don t yes pay Bob 2/1

3 Signing electronically Bank Internet SIGFILE } {{ } ALICE Pay Bob \$100 scan Alice 3/1

4 Signing electronically Bank Internet SIGFILE } {{ } ALICE Pay Bob \$100 scan Alice Problem: signature is easily copied Inference: signature must be a function of the message that only Alice can compute 3/1

5 What about a MAC? Let Bank and Alice share a key K Bank Internet ALICE Pay Bob \$100 MAC K T A digital signature will have additional attributes: Even the bank cannot forge Verifier does not need to share a key with signer or, indeed, have any secrets 4/1

6 Digital signatures A digital signature scheme DS = (K,S,V) is a triple of algorithms where pk K sk V σ M A σ S M 0/1 Correctness: V(pk,M,S(sk,M)) = 1 with probability one for all M. 5/1

7 Usage Step 1: key generation Alice lets (pk,sk) \$ K and stores sk (securely). Step 2: pk dissemination Alice enables any potential verifier to get pk. Step 3: sign Alice can generate a signature σ of a document M using sk. Step 4: verify Anyone holding pk can verify that σ is Alice s signature on M. 6/1

8 Dissemination of public keys The public key does not have to be kept secret but a verifier needs to know it is authentic, meaning really Alice s public key and not someone else s. Could put (Alice,pk) on a trusted, public server (cryptographic DNS.) Common method of dissemination is via certificates as discussed later. 7/1

9 Signatures versus MA schemes In a MA scheme: Verifier needs to share a secret with sender Verifier can impersonate sender! In a digital signature scheme: Verifier needs no secret Verifier cannot impersonate sender 8/1

10 Security of a DS scheme Possible adversary goals find sk Forge Possible adversary abilities can get pk known message attack chosen message attack 9/1

11 uf-cma adversaries M 1 pk σ 1. sk A M q σ q M σ S V pk d A wins if d = 1 M / {M 1,...M q } 10/1

12 Security of a DS scheme Interpretation: adversary cannot get a verifier to accept σ as Alice s signature of M unless Alice has really previously signed M, even if adversary can obtain Alice s signatures on messages of the adversary s choice. As with MA schemes, the definition does not require security against replay. That is handled on top, via counters or time stamps. 11/1

13 Formalization: UF-CMA Let DS = (K,S,V) be a signature scheme and A an adversary. Game UF-CMA DS procedure Initialize (pk,sk) \$ K; S return pk procedure Finalize(M, σ) d V(pk,M,σ) return (d = 1 M / S) procedure Sign(M): σ \$ S(sk,M) S S {M} return σ The uf-cma advantage of A is Adv uf-cma DS (A) = Pr [ UF-CMA A DS true] 12/1

14 A difference with MACs The UF-CMA game for MA schemes gave the adversary a verification oracle which is not given in the DS case. Why? 13/1

15 A difference with MACs The UF-CMA game for MA schemes gave the adversary a verification oracle which is not given in the DS case. Why? Verification in a MA scheme relies on the secret key but in a DS scheme, the adversary can verify on its own anyway with the public key, so the oracle would not provide an extra capability. 13/1

16 Strong unforgeability Adversary can t get receiver to accept σ as Alice s signatre on M unless UF: Alice previously signed M SUF: Alice previously signed M and produced signature σ Adversary wins if it gets receiver to accept σ as Alice s signature on M and UF: Alice did not previously sign M SUF: Alice may have previously signed M but the signature(s) produced were different from σ 14/1

17 Formalization: SUF-CMA Let DS = (K,S,V) be a signature scheme and A an adversary. Game SUF-CMA DS procedure Initialize: (pk,sk) \$ K; S return pk procedure Finalize(M, σ): return V(pk,M,σ) = 1 and (M,σ) / S procedure Sign(M): σ \$ S(sk,M) S S {(M,σ)} return σ The suf-cma advantage of A is Adv suf-cma DS (A) = Pr [ SUF-CMA A DS true] 15/1

18 RSA signatures Fix an RSA generator K rsa and let the key generation algorithm be Alg K (N,p,q,e,d) \$ K rsa pk (N,e); sk (N,d) return pk, sk We will use these keys in all our RSA-based schemes and only describe signing and verifying. 16/1

19 Plain RSA signatures: Idea Signer pk = (N,e) and sk = (N,d) Let f,f 1 : Z N Z N be the RSA function (encryption) and inverse (decryption) defined by f(x) = x e mod N and f 1 (y) = y d mod N. Sign by decrypting the message y: x = S N,d (y) = f 1 (y) = y d mod N Verify by encrypting signature x: V N,e (x) = 1 iff f(x) = y iff x e y mod N. 17/1

20 Plain RSA signature scheme Signer pk = (N,e) and sk = (N,d) Alg S N,d (y): x y d mod N return x Alg V N,e (y,x): if x e y (mod N) then return 1 return 0 Here y Z N is the message and x Z N is the signature. 18/1

21 Security of plain RSA signatures To forge signature of a message y, the adversary, given N,e but not d, must compute y d mod N, meaning invert the RSA function f at y. But RSA is 1-way so this task should be hard and the scheme should be secure. Correct? 19/1

22 Security of plain RSA signatures To forge signature of a message y, the adversary, given N,e but not d, must compute y d mod N, meaning invert the RSA function f at y. But RSA is 1-way so this task should be hard and the scheme should be secure. Correct? Of course not... 19/1

23 Attacks on plain RSA Existential forgery under no-message attack: Given pk = (N, e) adversary outputs message y = 1 and signature x = 1 message y = x e mod N and signature x for any x Z N of its choice Adversary wins because in both cases we have x e y (mod N) 20/1

24 Homomorphic properties of RSA Let pk = (N,e) and sk = (N,d) be RSA keys. Then x 1,x 2 Z N and y 1,y 2 Z N (x 1 x 2 ) e x e 1 xe 2 mod N (y 1 y 2 ) d y d 1 yd 2 mod N That is f(x 1 x 2 ) f(x 1 ) f(x 2 ) mod N f 1 (y 1 y 2 ) f 1 (y 1 ) f 1 (y 2 ) mod N where f(x) = x e mod N and f 1 (y) = y d mod N are the RSA function and its inverse respectively. 21/1

25 Another attack on plain RSA For all messages y 1,y 2 Z N we have S N,d (y 1 y 2 ) = S N,d (y 1 ) S N,d (y 2 ) } {{ } } {{ } x 1 So given x 1,x 2 one can forge signature of message y 1 y 2 mod N Adversary A(N, e): Pick some distinct y 1,y 2 Z N {1} x 1 Sign(y 1 ); x 2 Sign(y 2 ) return (y 1 y 2 mod N,x 1 x 2 mod N) x 2 22/1

26 DH signatures When Diffie and Hellman introduced public-key cryptography they suggested the DS scheme S(sk,M) = D(sk,M) V(pk,M,σ) = 1 iff E(pk,σ) = M where (E,D) is a public-key encryption scheme. But This views public-key encryption as deterministic; they really mean trapdoor permutations in our language Plain RSA is an example It doesn t work! Nonetheless, many textbooks still view digital signatures this way. 23/1

27 Other issues In plain RSA, the message is an element of Z N. We really want to be able to sign strings of arbitrary length. 24/1

28 Throwing in a hash function Let H: {0,1} Z N be a public hash function and let pk = (N,e) and sk = (N,d) be the signer s keys. The hash-then-decrypt scheme is Alg S N,d (M): y H(M) x y d mod N return x Alg V N,e (M,x): y H(M) if x e y (mod N) then return 1 return 0 Succinctly, S N,d (M) = H(M) d mod N Different choices of H give rise to different schemes. 25/1

29 What we need from H Suppose an adversary can find a collision for H, meaning distinct M 1,M 2 with H(M 1 ) = H(M 2 ). Then H(M 1 ) d H(M 2 ) d (mod N) meaning M 1,M 2 have the same signature. So forgery is easy: Obtain from signing oracle the signature x 1 = H(M 1 ) d mod N of M 1 Output M 2 and its signature x 1 Conclusion: H needs to be collision-resistant 26/1

30 Preventing previous attacks For plain RSA 1 is a signature of 1 S N,d (y 1 y 2 ) = S N,d (y 1 ) S N,d (y 2 ) But with hash-then-decrypt RSA H(1) d 1 so 1 is not a signature of 1 S N,d (M 1 M 2 ) = H(M 1 M 2 ) d H(M 1 ) d H(M 2 ) d (mod N) A good choice of H prevents known attacks. 27/1

31 RSA PKCS#1 signatures Signer has pk = (N,e) and sk = (N,d) where N = Let h: {0,1} {0,1} 160 be a hash function (like SHA-1) and let n = N 8 = 1024/8 = 128. Then And Then H PKCS (M) = FF... FF } {{ } n 22 S N,d (M) = H PKCS (M) d mod N h(m) }{{} 20 H PKCS is CR as long as h is CR H PKCS (1) 1 (mod N) H PKCS (y 1 y 2 ) H PKCS (y 1 ) H PKCS (y 2 ) (mod N) etc 28/1

32 Does 1-wayness prevent forgery? Forger s goal Inverter s goal A N,e M A y d mod N N,e y A y d mod N H y y here need not be random y here is random Problem: 1-wayness of RSA does not imply hardness of computing y d mod N if y is not random 29/1

33 H PKCS revisited Recall H PKCS (M) = FF... FF h(m) But first n 20 = 108 bytes out of n are fixed so H PKCS (M) does not look random even if h is a RO or perfect. We cannot hope to show RSA PKCS#1 signatures are secure assuming (only) that RSA is 1-wayno matter what we assume about h and even if h is a random oracle. 30/1

34 Goal We will validate the hash-then-decrypt paradigm S N,d (M) = H(M) d mod N by showing the signature scheme is provably UF-CMA assuming RSA is 1-way as long as H is a RO. This says the paradigm has no structural weaknesses and we should be able to get security with good choices of H. 31/1

35 Choice of H A good choice of H might be something like H(M) = first n bytes of SHA1(1 M) SHA1(2 M) SHA1(11 M) 32/1

36 Full-Domain-Hash (FDH) [BR96] Signer pk = (N,e) and sk = (N,d) algorithm S H N,d (M) return H(M) d mod N algorithm V H N,e (M,x) if x e H(M) (mod N) then return 1 else return 0 Here H: {0,1} Z N is a random oracle. 33/1

37 UF-CMA in RO model Let DS = (K,S,V) be a signature scheme and A an adversary. Game UF-CMA DS procedure Initialize: (pk,sk) \$ K; S return pk procedure Finalize(M, σ): return V H (pk,m,σ) = 1 and M / S procedure Sign(M): σ \$ S H (sk,m) S S {M} return σ procedure H(M): \$ if H[M] = then H[M] R return H[M] Here R is the range of H. The uf-cma advantage of A is Adv uf-cma DS (A) = Pr [ UF-CMA A SD true] 34/1

38 Security of FDH in RO model Theorem: [BR96] Let K rsa be a RSA generator and DS = (K,S,V) the associated FDH RO-model signature scheme. Let A be a uf-cma adversary making q s signing queries and q H queries to the RO H and having running time at most t. Then there is an inverter I such that Adv uf-cma DS (A) (q s +q H +1) Adv owf K rsa (I). Furthermore the running time of I is that of A plus the time for O(q s +q H +1) computations of the RSA function. 35/1

39 RO proofs for encryption There is a crucial hash query Q such that If A does not query Q it has 0 advantage If A queries Q an overlying algorithm can see it and solve some presumed hard computational problem Example: In the RO EG KEM, Q = g xy where pk = g x and g y is in challenge ciphertext. 36/1

40 Programming the RO For signatures we exploit the RO model in a new way by replying to RO queries with carefully constructed objects. In particular the inverter I that on input y aims to compute y d mod N might reply to a RO query M made by A via y or some function thereof x e mod N for x Z \$ N chosen by I Thus I is programming H(M) to equal values of I s choice. 37/1

41 The case q s = 0 and q H = 1 Assume A Makes no Sign queries Makes exactly one H-query M Then outputs a forgery (M,σ) Let us see how to build I so that Adv uf-cma DS (A) = Adv owf K rsa (I). 38/1

42 The case q s = 0 and q H = 1 N,e A M w H (M,σ) Adv uf-cma DS (A) = Pr[σ e w (mod N)] 39/1

43 The inverter for the case q s = 0 and q H = 1 Inverter I N,e A M w HSim w? y (M,σ)? Q: How should I choose w and what should it output? 40/1

44 The inverter for the case q s = 0 and q H = 1 Inverter I N,e A M w HSim w? y (M,σ)? Q: How should I choose w and what should it output? A: Let w = y and output σ! 40/1

45 The inverter for the case q s = 0 and q H = 1 Inverter I N,e A M w HSim w y y M σ σ Adv uf-cma DS (A) = Pr[σ e w (mod N)] = Pr[σ e y (mod N)] = Adv owf K rsa (I) 41/1

46 The inverter for the case q s = 0 and q H = 1 Inverter I(N,e,y): (M,σ) A HSim (N,e) return σ subroutine HSim(M): return y Then Adv uf-cma DS (A) = Adv owf K rsa (I). 42/1

47 The case q s = 0 and q H > 1 Assume A Makes no Sign queries Makes H-queries M 1,...,M qh Then outputs a forgery (M,σ) such that M {M 1,...,M qh } Let us see how to build I so that Adv uf-cma DS (A) = q H Adv owf K rsa (I). 43/1

48 The case q s = 0 and q H > 1 (N,e) A M 1 y 1. H M qh y qh (M,σ) Let i be such that M = M i. Adv uf-cma DS (A) = Pr[σ e y i (mod N)] 44/1

49 Inverter for the case q s = 0 and q H > 1 As before, return y in response to a H-query: Inverter I(N,e,y): (M,σ) \$ A HSim (N,e) return σ subroutine HSim(M): return y 45/1

50 Inverter for the case q s = 0 and q H > 1 As before, return y in response to a H-query: Inverter I(N,e,y): (M,σ) \$ A HSim (N,e) return σ subroutine HSim(M): return y Say A s queries are M 1,...,M q and M = M i. Then if σ e H(M i ) (mod N) we have σ e y (mod N) so I wins so Adv uf-cma DS (A) = Pr[σ e y i (mod N)] = Adv owf K rsa (I). 45/1

51 Inverter for the case q s = 0 and q H > 1 As before, return y in response to a H-query: Inverter I(N,e,y): (M,σ) \$ A HSim (N,e) return σ subroutine HSim(M) return y This is wrong because the answers to A s queries are not independent, meaning HSim does not look like a real RO. What if A made queries M 1 M 2 and on getting back y 1,y 2 aborted if y 1 = y 2? A s advantage in the simulation could be 0. 46/1

52 Inverter for the case q s = 0 and q H > 1 We could return y in response to a random query and random values in response to the rest: Inverter I(N,e,y): g \$ {1,...,q H }; j 0 (M,σ) \$ A HSim (N,e) return σ subroutine HSim(M) j j +1 if j = g then y i y \$ else y j Z N return y j Say A s queries are M 1,...,M qh and M = M i. Then if σ e y i (mod N) and i = g we have σ e y (mod N), so Adv uf-cma DS (A) = q H Adv owf K rsa (I). 47/1

53 The case q s > 0 (N,e) M 1 A y 1. H M qh y qh M 1 σ 1. M q s Sign (N,d) σ qs (M,σ) 48/1

54 Replying to Sign queries How can the Inverter I (not knowing d) return the signature H(M) d mod N in response to Sign query M? Trick: When M i is queried to H, Inverter will \$ pick x i Z N and let y i xi e mod N Return H(M i ) = y i Then if there is a Sign(M i ) query it can return x i as the signature. 49/1

55 Simplification Assume that if A Makes Sign query M, it has previously made H-query M Outputs (M,σ) then it has previously made H-query M and not made Sign query M Can easily modify A to have these properties at the cost of increasing the number of H-queries to q = q s +q H +1. Also assume A never repeats a H-query. 50/1

56 Inverter for case q s > 0 Inverter I(N,e,y): g \$ {1,...,q H }; j 0 (M,σ) \$ A HSim,SignSim (N,e): return σ subroutine SignSim(M) j Ind(M) return x j subroutine HSim(M) j j +1; M j M; Ind(M) j if j = g then H[M] y; x j \$ else x j Z N ; H[M] xe j mod N return H[M] 51/1

57 Analysis intuition Let i be such that A outputs (M,σ) with M = M i. Then if i = g σ e H(M i ) y (mod N) so inverter finds σ = y d mod N All A s queries are correctly answered Since i = g with probability 1/q we have Adv owf K rsa (I) 1 q Advuf-cma DS (A). 52/1

58 Fundamental Lemma variant Lemma [BR06] Let G i,g j be identical-until-bad games and A an adversary. Then for any y [ Pr Gi A y Gi A doesn t set bad ] [ = Pr Gj A y Gj A doesn t set bad ] 53/1

59 Analysis Games G 0, G 1 procedure Initialize (N,p,q,e,d) \$ K rsa g \$ {1,...,q H }; j 0; y \$ Z N return (N, e) procedure Sign(M) j Ind(M) if j = g then x j y d mod N return x j procedure H(M) j j +1; M j M; Ind(M) j if j = g then H[M] y; x j \$ else x j Z N ; H[M] xe j mod N return H[M] procedure Finalize(M, σ) j Ind(M) if j g then bad true return (σ e H[M] (mod N)) 54/1

60 Analysis Let Bad i be the event that G i sets bad. Then [ ] Adv owf K rsa (I) Pr G0 A true Bad 0 [ ] = Pr G1 A true Bad 1 where last line is due to Fundamental Lemma variant. But the events G1 A true and Bad 1 are independent so ] = Pr [G1 A true Pr [ ] Bad 1 = Adv uf-cma DS (A) 1 q 55/1

61 Security of FDH in RO model Theorem: [BR96] Let K rsa be a RSA generator and DS = (K,S,V) the associated FDH RO-model signature scheme. Let A be a uf-cma adversary making q s signing queries and q H queries to the RO H and having running time at most t. Then there is an inverter I such that Adv uf-cma DS (A) (q s +q H +1) Adv owf K rsa (I). Furthermore the running time of I is that of A plus the time for O(q s +q H +1) computations of the RSA function. 56/1

62 Choosing a modulus size Say we want 80-bits of security, meaning a time t attacker should have advantage at most t 2 80 : For inverting RSA, this is provided by a 1024 bit modulus assuming NFS is the best attack. But according to the BR96-reduction, FDH could be less secure than RSA by a factor of q s +q H +1, so that a bigger modulus would be needed for 80-bit security. 57/1

63 Choosing a modulus size Say we want 80-bits of security, meaning a time t attacker should have advantage at most t The following shows modulus size k and cost c of one exponentiation, with q H = 2 60 and q s = 2 45 in the FDH case: Task k c Inverting RSA Breaking FDH as per [BR96] reduction This (for simplicity) neglects the running time difference between A, I. This motivates getting tighter reductions for FDH, or alternative schemes with tighter reductions. 58/1

64 Better analysis of FDH in RO model Theorem: [Co00] Let K rsa be a RSA generator and DS = (K,S,V) the associated FDH RO-model signature scheme. Let A be a uf-cma adversary making q s signing queries and q H queries to the RO H and having running time at most t. Then there is an inverter I such that Adv uf-cma DS (A) O(q s ) Adv owf K rsa (I). Furthermore the running time of I is that of A plus the time for O(q s +q H +1) computations of the RSA function. 59/1

65 Choosing a modulus size Say we want 80-bits of security, meaning a time t attacker should have advantage at most t The following shows modulus size k and cost c of one exponentiation, with q H = 2 60 and q s = 2 45 in the FDH case: Task k c Inverting RSA Breaking FDH as per [BR96] reduction Breaking FDH as per [Co00] reduction /1

66 PSS [BR96] Signer pk = (N,e) and sk = (N,d) algorithm S h,g 1,g 2 N,d (M) \$ r {0,1} 160 w h(m r) r g 1 (w) r y 0 w r g 2 (w) return y d mod N algorithm V h,g 1,g 2 N,e (M, x) y x e mod N b w r P y r r g 1 (w) if (g 2 (w) P) then return 0 if (b = 1) then return 0 if (h(m r) w) then return 0 return 1 Here h,g 1 : {0,1} {0,1} 160 and g 2 : {0,1} {0,1} k 321 are random oracles where k = N. 61/1

67 Choosing a modulus size Say we want 80-bits of security, meaning a time t attacker should have advantage at most t The following shows modulus size k and cost c of one exponentiation, with q H = 2 60 and q s = 2 45 in the FDH and PSS cases: Task k c Inverting RSA Breaking FDH as per [BR96] reduction Breaking FDH as per [Co00] reduction Breaking PSS as per [BR96] reduction /1

68 Choosing a modulus size There are no attacks showing that FDH is less secure than RSA, meaning there are no attacks indicating FDH with a 1024 bit modulus has less than 80 bits of security. But to get the provable guarantees we must use larger modulii as shown, or use PSS. 63/1

69 PSS usage RSA PKCS#1 v2.1. IEEE P1363a ANSI X9.31 RFC 3447 ISO/IEC CRYPTREC NESSIE 64/1

70 ElGamal Signatures Let G = Z p = g where p is prime. Signer keys: pk = X = g x Z p and sk = x \$ Z p 1 Algorithm S x (m) k \$ Z p 1 r g k mod p s (m xr) k 1 mod (p 1) return (r,s) nnnnn Algorithm V X (m,(r,s)) if (r / G or s / Z p 1 ) then return 0 if (X r r s g m mod p) then return 1 else return 0 Correctness check: If (r,s) \$ S x (m) then X r r s = g xr g ks = g xr+ks = g xr+k(m xr)k 1 mod (p 1) = g xr+m xr = g m so V X (m,(r,s)) = 1. 65/1

71 Security of ElGamal Signatures Signer keys: pk = X = g x Z p and sk = x \$ Z p 1 Algorithm S x (m) k \$ Z p 1 r g k mod p s (m xr) k 1 mod (p 1) return (r,s) Algorithm V X (m,(r,s)) if (r / G or s / Z p 1 ) then return 0 if (X r r s g m mod p) then return 1 else return 0 Suppose given X = g x and m the adversary wants to compute r,s so that X r r s g m mod p. It could: Pick r and try to solve for s = DLog Z p,r(g m X r ) Pick s and try to solve for r...? 66/1

72 Forgery of ElGamal Signatures Adversary has better luck if it picks m itself: Adversary A(X) r gx mod p; s ( r) mod (p 1); m s return (m,(r,s)) Then: X r r s = X gx (gx) gx = X gx g gx X gx = g gx =g r = g m so (r,s) is a valid forgery on m. 67/1

73 ElGamal with hashing Let G = Z p = g where p is a prime. Signer keys: pk = X = g x Z p and sk = x \$ Z p 1 H : {0,1} Z p 1 a hash function. Algorithm S x (M) m H(M) k \$ Z p 1 r g k mod p s (m xr) k 1 mod (p 1) return (r,s) Algorithm V X (M,(r,s)) m H(M) if (r / G or s / Z p 1 ) then return 0 if (X r r s g m mod p) then return 1 else return 0 68/1

74 ElGamal with hashing Let G = Z p = g where p is a prime. Signer keys: pk = X = g x Z p and sk = x \$ Z p 1 H : {0,1} Z p 1 a hash function. Algorithm S x (M) m H(M) k \$ Z p 1 r g k mod p s (m xr) k 1 mod (p 1) return (r,s) Algorithm V X (M,(r,s)) m H(M) if (r / G or s / Z p 1 ) then return 0 if (X r r s g m mod p) then return 1 else return 0 Requirements on H: Collision-resistant One-way to prevent previous attack 68/1

75 DSA Let p be a 1024-bit prime. For DSA, let q be a 160-bit prime dividing p 1. Scheme signing cost verification cost signature size ElGamal bit exp bit exp 2048 bits DSA bit exp bit exp 320 bits By a e-bit exp we mean an operation a,n a n mod p where a Z p and n is an e-bit integer. A 1024-bit exponentiation is more costly than a 160-bit exponentiation by a factor of 1024/ DSA is in FIPS /1

76 DSA Fix primes p,q such that q divides p 1 Let G = Z p = h and g = h (p 1)/q so that g G has order q H: {0,1} Z q a hash function Signer keys: pk = X = g x Z p and sk = x Z \$ q Algorithm S x (M) m H(M) k \$ Z q r (g k mod p) mod q s (m+xr) k 1 mod q return (r,s) Algorithm V X (M,(r,s)) m H(M) w s 1 mod q u 1 mw mod q u 2 rw mod q v (g u 1 X u 2 mod p) mod q if (v = r) then return 1 else return 0 Details: Signature is regenerated if s = 0. 70/1

77 Discussion DSA as shown works only over the group of integers modulo a prime, but there is also a version ECDSA of it for elliptic curve groups. In ElGamal and DSA/ECDSA, the expensive part of signing, namely the exponentiation, can be done off-line. No proof that ElGamal or DSA is UF-CMA under a standard assumption (DL, CDH,...) is known, even if H is a RO. Proofs are known for variants. 71/1

78 Schnorr Signatures The Schnorr scheme works in an arbitrary (prime-order) group. When implemented in a 160-bit elliptic curve group, it is as efficient as ECDSA. It can be proven UF-CMA in the random oracle model under the discrete log assumption [PS,AABN]. The security reduction, however, is quite loose. 72/1

79 Schnorr Signatures Let G = g be a cyclic group of prime order p H: {0,1} Z p a hash function Signer keys: pk = X = g x G and sk = x \$ Z p Algorithm S x (M) \$ r Z p R g r c H(R M) a xc +r mod p return (R, a) Algorithm V X (M,(R,a)) if R G then return 0 c H(R M) if g a = RX c then return 1 else return 0 73/1

80 Randomization in signatures We have seen many randomized signature schemes: PSS, ElGamal, DSA/ECDSA, Schnorr,... Re-using coins across different signatures is not secure, but there are (other) ways to make these schemes deterministic without loss of security. 74/1

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