# Signature Schemes. CSG 252 Fall Riccardo Pucella

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1 Signature Schemes CSG 252 Fall 2006 Riccardo Pucella

2 Signatures Signatures in real life have a number of properties They specify the person responsible for a document E.g. that it has been produced by the person, or that the person agrees with the document Attached to a particular document Easily verifiable by third parties We want a similar mechanism for digital documents Some difficulties: Need to bind signature to document Need to ensure verifiability (and avoid forgeries)

3 Formal Definition A signature scheme is a tuple (P,A,K,S,V) where: P is a finite set of possible messages A is a finite set of possible signatures K (the keyspace) is a finite set of possible keys For all k, there is a signature algorithm sig k in Sand a verification algorithm ver k in V such that sig k : P A ver k : P A {true,false} ver k (x,y) = true iff y=sig k (x) A pair (x,y) P A is called a signed message

4 Example: RSA Signatures The RSA cryptosystem (in fact, most public key cryptosystems) can be used as a signature scheme Take: sig k (x) = d k (x) ver k (x,y) = (x =? e k (y)) Only user can sign (because decryption is private) Anyone can verify (because encryption is public)

5 Signing and Encrypting Suppose you want to sign and encrypt a piece of data Where encryption is public key (why is this important?) Public key cryptography does not say anything about the sender Two possibilities: First encrypt, then sign: x (e ke (x), sig ks (e ke (x))) But adversary could replace by sig ke (e ke (x)) making it seem the message came from someone else First sign, then encrypt: x (e ke (x), sig ks (x)) Better make sure signature does not leak info!

6 Possible Attacks (Alice is the signer, Oscar the attacker) Key-only attack Oscar possesses Alice s public verification algorithm Known message attack Oscar possesses a list of signed messages (x i, y i ) Chosen message attack Oscar queries Alice for the signatures of a list of messages x i

7 Possible Adversarial Goals Total break Oscar can derive Alice s private signing algorithm Selective forgery Oscar can create a valid signature on a message chosen by someone else, with some nonnegligible probability Existential forgery Oscar can create a valid signature for at least one message

8 Some Comments Cannot have unconditional security, only computational or provable security Attacks above are similar to those against MACs We mostly concentrated on existential forgeries against chosen message attacks What Graham was asking was for total break against (some) attacks Existential forgeries against chosen message attacks: Least damage against worst attacker The minimum you should ask for

9 Security of RSA Signatures Existential forgery using a key-only attack: Choose a random y Compute x = e k (y) We have y = sig k (x), a valid signature of x Existential forgery using a known-message attack: Suppose y = sig k (x) and y = sig k (x ) Can check e k (y y mod n) = x x mod n So y y mod n = sig k (x x mod n) Existential forgery using a chosen message attack: To get a signature for x, find x 1 x 2 = x mod n Query for signatures of x 1 and x 2 Apply previous attack

10 Signatures and Hashing The easiest way to get around the above problems is to use a cryptographic hash function Given message x Produce digest h(x) Sign digest h(x) to create (x,sig k (h(x))) To verify: Get (x,y) Compute h(x) Check ver k (h(x),y)

11 Use of Hashing for Signatures Existential forgery using a chosen messge attack Oscar finds x,x s.t. h(x)=h(x ) He gives x to Alice and gets her to sign h(x) Then (x,sig k (h(x))) is a valid signed message Prevented by having h collision resistant Existential forgery using a known message attack Oscar starts with (x,y), where y = sig k (h(x)) He computes h(x) and tries to find x s.t. h(x ) = h(x) Prevented by having h second preimage resistant Existential forgery using a key-only attack (If signature scheme has existential forgery using a key-only attack) Oscar chooses message digest and finds a forgery z for it Then tries to find x s.t. h(x)=z Prevented by having h preimage resistant

12 Example: ElGamal Signature Scheme Let p be a prime s.t. discrete log in Z p is hard Let a be a primitive element in Z p P = Z p, A = Z p Z p-1 K = {(p,α,a,β) β= α a (mod p)} For k = (p,α,a,β) and t Z p-1 γ = α t mod p sig k (x,t) = (γ, (x-aγ)t -1 (mod p-1)) ver k (x,(γ,δ)) = ( β γ γ δ =? α x (mod p) )

13 Security of ElGamal Scheme Forging a signature (γ,δ) without knowing a Choosing γ and finding corresponding δ amounts to finding discrete log Choosing δ and finding corresponding γ amounts to solving β γ γ δ = α x (mod p) No one knows the difficulty of this problem (believed to be hard) Choosing γ and δ and solving for the message amounts to finding discrete log Existential forgery with a key-only attack: Sign a random message by choosing γ, δ and message simultaneously (p.289)

14 Variant 1: Schnorr Signature Scheme ElGamal requires a large modulus p to be secure A 1024 bit modulus leads to a 2048 bit signature Too large for some uses of signatures (smartcards) Idea: use a subgroup of Z p of size q (q << p) Let p be a prime s.t. discrete log is hard in Z p Let q be a prime that divides p-1 Let α in Z p be a q-th root of 1 mod p Let h : {0,1} Z q be a secure hash function P = {0,1}, A = Z q Z q K = {(p,q,α,a,β) β = α a (mod p)} For k=(p,q,α,a,β) and 1 t q-1: γ = h(x α t mod p) sig k (x,t) = (γ, t+aγ mod q) ver k (x,(γ,δ)) = (h(x α γ β -γ mod p) =? γ

15 Variant 2: DSA Let p be a prime s.t. discrete log is hard in Z p bitlength of p = 0 (mod 64), 512 bitlength 1024 Let q be a 160 bit prime that divides p-1 Let α in Z p be a q-th root of 1 mod p Let h : {0,1} Z q be a secure hash function P = {0,1}, A = Z q Z q K = {(p,q,α,a,β) β = α a (mod p)} For k=(p,q,α,a,β) and 1 t q-1: γ = (α t mod p) mod q sig k (x,t) = (γ, (SHA1(x)+aγ)t -1 mod q) ver k (x,(γ,δ)) = (α e1 β e2 mod p) mod q =? γ e1 = SHA1(x)δ -1 mod q e2 = γδ -1 mod q

16 Variant 3: Elliptic Curve DSA Modification of the DSA to use elliptic curves Instead of choosing α, β, use A and B two points on an elliptic curve over Z p Roughly speaking, instead of: (α t mod p) mod q use the x coordinate of the point ta, mod q The rest of the computation is as before

17 Provably Secure Signature Schemes The previous examples were (to the best of our knowledge) computationally secure signature scheme Here is a provably secure signature scheme As long as only one message is signed Let m be a positive integer Let f : Y Z be a one-way function P = {0,1} m, A = Y m Choose y i,j in Y at random for 1 i m, j=0,1 Let z i,j = f(y i,j ) A key = 2m y s and 2m z s (y s private, z s public) sig k (x 1,...,x m ) = (y 1,x1,...,y m,xm ) ver k ((x 1,...,x m ),(a 1,...,a m )) = (f (a i ) =? z i,xi ) for all i

18 Argument for Security Argument for provable security: Existential forgeries using a key-only attack Assume that f is a one-way function Show that if there is an existential forgery using a key-only attack, then there is an algorithm that finds preimage of random elements in the image of f with probability at least 1/2 We need the restriction to one signature only If the attacker gets two messages signed with the same key, then can easily construct signatures for other messages (0,1,1) and (1,0,1) can give signatures for (0,0,1), (1,1,1)

19 Undeniable Signature Schemes Introduced by Chaum and van Antwerpen in 1989 A signature cannot be verified without the signer Prevent signer from disavowing signature Let p,q primes, p = 2q+1, and discrete log hard in Z p Let α in Z p be an element of order q G = multiplicative subgroup of Z p of order q P = A = G K = {(p,α,a,β} β = α a mod p} For key k=(p,α,a,β) and x in G: sig k (x) = x a mod p To verify (x,y): pick e 1,e 2 at random in Z q Compute c = y e1 β e2 Signer computes d = c inv(a) mod q mod p (where inv(a) = a -1 ) y is a valid signature iff d = x e1 α e2 mod p

20 Disavowal Protocol Can prove that Alice cannot fool Bob into accepting a fraudulent signature (except with very small probability = 1/q) What if Bob wants to make sure that a claimed forgery is one? 1. Bob chooses e 1,e 2 at random in Z q 2.Bob computes c = y e1 β e2 mod p; sends it to Alice 3.Alice computes d = c inv(a) mod q mod p; sends it to Bob 4.Bob verifies d x e1 α e2 mod p 5.Bob chooses f 1,f 2 at random, in Z q 6.Bob computes C = y f1 β f2 mod p; sends it to Alice 7.Alice computes D = C inv(a) mod q mod p; sends it to Bob 8.Bob verifies D x f1 α f2 mod p 9.Bob concludes y is a forgery iff (dα -e2 ) f1 = (Dα -f2 ) e1 mod p

21 Why Does This Work? Alice can convince Bob that an invalid signature is a forgery If y x a mod p and Alice and Bob follow the protocol, then the check in last step succeeds Alice cannot make Bob believe that a valid signature is a forgery except with a very small probability Intuition: since she cannot recover e 1,e 2,f 1,f 2, she will have difficulty coming up with d and D that fail steps 4 and 8, but still pass step 9 See Stinson for details

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