Does anybody know an efficient mechanism to prove the possession of a digital signature (e.g. RSA) on a certain attribute (message) in zero-knowledge? That is, without revealing the actual signature (against tracking, for privacy) prove that you have one? Thanks a lot in advance!

  • $\begingroup$ It would be trivial if you could generate the signature in question. Is this your scenario? $\endgroup$
    – rath
    Commented May 5, 2014 at 13:05
  • $\begingroup$ My scenario is that the attribute is signed by another party. That is, I can not generate the signature. Instead, I'd like to prove I have this signature without revealing it. This should happen in such a way that two different proof session are unlinkable to each other. $\endgroup$
    – OnTarget
    Commented May 5, 2014 at 13:22
  • $\begingroup$ If both of you possess the signature you could ask for a nonce and reply with the hash of the nonce and the signature. I take it only one party has the signature? $\endgroup$
    – rath
    Commented May 5, 2014 at 13:29
  • 2
    $\begingroup$ Are you familiar with the concept of multi-show (unlinkable) anonymous credentials? They are built upon re-randomizable signature schemes which have the properties you require and are used in exactly the way you want it. Well known candidates are CL credentials Strong RSA or Pairing based or built upon the pairing based BBS signature scheme. $\endgroup$
    – DrLecter
    Commented May 5, 2014 at 13:36
  • 1
    $\begingroup$ I meant, the requirement of user anonymity (with respect to user-side secrets) against the issuer during the issuance of the credential. Verifier and Issuer are parts of one system in my case, so they do communicate with each other. With respect to "one-show" I really meant "multi-show and unlinkable". Since both CL and BBS are fairly complex, I would like to know if there is any high-level description of such systems (e.g. a lecture, etc.). For example, both of them essentially use a public reference parameter with certain trapdoor built in (to issuer credentials)..... And so on. Thanks! $\endgroup$
    – OnTarget
    Commented May 5, 2014 at 17:31

3 Answers 3


Guillou and Quisquater (link) present a zero-knowledge proof of an RSA signature. Basically, the scheme is as follows:

Public knowledge: RSA modulus $n$, public RSA exponent $v$, preimage $X$.

Secret knowledge for prover: $A$, such that $A^v = X \mod n$.

$$ \begin{matrix} \mathcal{P} & & \mathcal{V} \\ r \xleftarrow{\$} \mathbb{Z}_n^* \phantom{\mod n} & & \\ T \leftarrow r^v \mod n & & \\ & \xrightarrow{\quad{}T\quad{}} & \\ & & d \xleftarrow{\$} \{0,1,\ldots,v-1\} \\ & \xleftarrow{d} & \\ t \leftarrow A^dr \mod n & & \\ & \xrightarrow{\quad{}t\quad} & \\ & & t^v \stackrel{?}{=} X^{d}T \mod n \end{matrix} $$

In this diagram, $\leftarrow$ denotes assignment of a value to a variable and $\xleftarrow{\$}$ denotes uniformly random selection from a finite set.

  • 1
    $\begingroup$ Nice, but here you require that signer and verifier are distinct entities and do not communicate, because any verifier must not know the factorization of $n$ (and in a previous comment the OP wrote "Verifier and Issuer are parts of one system in my case, so they do communicate with each other."). But this could be solved by requiring the issuer issue a blind signature (anyways, one would then still not achieve unlinkability of showings when conducting several proofs - the OP wants multi-show unlinkability - under the assumption that issuer and verifier collude). $\endgroup$
    – DrLecter
    Commented May 6, 2014 at 11:33
  • $\begingroup$ Thanks, @Alan, for an interesting link. The only problem in this case is that even if verifiers and the issuer do not collude, the verification sessions are traceable, since for each verification the prover has to deliver his "identity" which is used to check the proof (in my previous terms, to "bootstrap" verification). I guess that was one of the motivations to resort to more complex CL-like signatures. Any ideas on how to make Guillou and Quisquater method "multi-show" in a privacy-preserving sense (let's assume at first that verifiers and the issuer do no collude)? $\endgroup$
    – OnTarget
    Commented May 6, 2014 at 13:49
  • 2
    $\begingroup$ You can use Ferguson's randomized blind signature protocol (oai.cwi.nl/oai/asset/5290/05290D.pdf) for issuing the credentials. (You want the protocol from figure 1, in which the bank signs a jointly generated and perfectly blinded random value.) $\endgroup$
    – Alan
    Commented May 6, 2014 at 14:17
  • $\begingroup$ How about a "small patch": simply randomizing the reference parameter $X$ on each session? Then, on each verification, additionally generate $s \in_{R} Z_{n}^{*}$ and deliver $X_{r}=X^{s}$ (instead of $X$) along with T as well as $K=s^{v}$. Then compute $t$ as $t \leftarrow A^{ds}rs$. To verify, compare $t^{v} \overset{?}{=} X_{r}^{d} TK$. What do you think, guys? $\endgroup$
    – OnTarget
    Commented May 6, 2014 at 14:25
  • 1
    $\begingroup$ Generating the credential: $z \xleftarrow{\$} \{0,1,\ldots,2^{|n|}-1\}; y \leftarrow zv; X \leftarrow g^y \mod n; A \leftarrow g^z \mod n$. Since $y$ is necessarily kept secret, the verifier cannot tell it is a multiple of $v$. $\endgroup$
    – Alan
    Commented May 8, 2014 at 11:15

For many signature schemes based on zero knowledge proofs of discrete log knowledge, (EG:Schnorr signature, ECDSA) it is possible to perform an interactive proof that a signature exists without revealing the signature. This proof is interactive only, and a dishonest verifier cannot turn it into a non-interactive proof.

A Schnorr signature for a public key Y=xG (x is secret key) is a tuple (e,s) such that for an R R=eY+sG, e=H(R,m). Where H is a cryptographic hash.

It is verified by computing R=eY+sG from the (e,s) signature tuple, then checking that e=H(R,m) holds.

Suppose Alice has a signature on a string m and wishes to prove that a signature (e,s) exists to Bob. Alice first sends everything but the s scalar.

  • A-->B:(m,R)
  • Bob: compute e=H(R,m),S=R-e*Y

Bob can, with R and m compute the point S in the equation for the valid signature but not the discrete log s of that point. If Bob could do that he could sign things.

Alice does know the scalar s and can prove this by using it as a key during a Diffie-Hellman key exchange. DH acting as an interactive only proof.

  • Bob: generate ephemeral public key Y_b=b*G
  • B-->A: Y_b
  • Alice: compute K=Y_b*s,
  • A-->B: H(K,context)
  • Bob: compute K=b*S, verify H(K,context)=(received_value)

Alice has proven knowledge of the scalar s in the signature (e,s) tuple without revealing it to Bob. DH key exchange shouldn't reveal anything about s especially with that hash in there to destroy any algebraic structure.

Note that there are a lot of extra back and forth steps in the protocol above. If Bob has a known public key, Alice can just send Bob (e,H(K,context),context,m) or (R,H(K,context),context,m), with Bob computing S and verifying H(K,context). Bob can provide an ephemeral challenge key as a first step if there's no such widely known key.

Sending K directly may be fine, although it makes Alice into a "multiply by s oracle". The context helps if using Bob's public key since Eve the eavesdropper who listens in on the exchange doesn't learn K and so can't compute the hash value for a different (proverID,verifierID,time) context, preventing replay attacks.

The following python code uses the NACL library to generate signatures. Additionally the ECPy library is used to implement EC calculations concisely. You can use NACL._sodium.(whatever) but that's a lot more verbose. AFAIK this should just work securely assuming you have an ED25519 signature. Vanilla Schnorr signatures require slight changes to a few sections.

The following graphical overview of the ED25519 signature system is useful for understanding the code below.

how ED25519 signatures work

from nacl.signing import SigningKey
import json


    {"name":"Alice Liddell",

from ecpy.curves import Curve
cv = Curve.get_curve('Ed25519')
import hashlib
def decodeint(s):
    return sum(256**i * b for i,b in enumerate(bytearray(s)))
def encodeint(y,n):
    return bytes(bytearray((y>>i*8)&255 for i in range(n)))
def sha512(m):return hashlib.sha512(m).digest()
def h_int(m):return decodeint(sha512(m))

def verify(sig,pk,m):
    #verify an ed25519 signature
    assert len(sig)==64
    R = cv.decode_point(sig[0:32])
    A = cv.decode_point(pk)
    s = decodeint(sig[32:])
    h = h_int(cv.encode_point(R) + pk + m)
    return cv.generator*s == R+A*h

#check the signature
if verify(alice_ID["sig"],gov_pk,alice_ID["info"]):
    print("Alice verified she has a valid signature")

def get_sk_scalar(sk):
    return cv.decode_scalar_25519(sha512(sk.encode())[:32])

def make_proof(sig,challenge_key,m,context=b""):
    PKc = cv.decode_point(challenge_key)
    s = decodeint(sig[32:])
    K = cv.encode_point(s * PKc)
    H = sha512(K+context)[:32]
    proof = R+H+encodeint(len(context), 2)+context+m
    return proof

def check_proof(proof,challenge_sk,pk):
    def get(n,i=[0]):
        data=proof[i[0]:i[0]+n] if n>=0 else proof[i[0]:];i[0]+=len(data)
        if len(data)<n:raise ValueError("not enough data")
        return data
    R = cv.decode_point(get(32))
    H = get(32)
    m = get(-1)#remainder
    #calculate S from signature equation
    A = cv.decode_point(pk)
    h = h_int(cv.encode_point(R) + pk + m)
    S = R+A*h
    #check the DH shared key sha512 was correctly calculated
    K=cv.encode_point( S * get_sk_scalar(challenge_sk))
    if H_target!=H:
        raise Exception("verification failed")
    return context,m

#bob has a known key or generates an ephemeral challenge key for the protocol.

print("Bob --> Alice: challenge key:",repr(bob_pk))
#Alice makes a proof
proof=make_proof(sig=alice_ID["sig"], challenge_key=bob_pk, m=alice_ID["info"],
print("Alice --> Bob: proof:",repr(proof))

#bob can verify it
result=check_proof(proof, bob_sk, gov_pk)
print("Bob knows the following was signed:",result[1])
print("with protocol context:",result[0])

This should work for signatures based on discrete log problems. Schnorr makes the transformation clean. ECDSA is messier. Alice can prove knowledge of dlog(R) in base (H(m)*G+r*Q_A) (IE:factor inverse(S) out of the right side of the R=(H(m)/s)*G+(r/s)*Q_A equation. Bob chooses an ephemeral key with (H(m)*G+r*Q_A) as the generator and they do DH as above. There are also ways to use trapdoor one way functions to do non-interactive zero knowledge proofs that can either be generated legitimately or forged by using the trapdoor (Bob's secret key) to mess with the output of the random oracle. The resulting proof will convince Bob (and only Bob) that the signature exists.


One-Way Aggregate Signatures (OWAS) (also called composite signatures) can be used to do this. They are based on BLS signatures. I will skip the notation except mention that they are based on bilinear pairings. The above links will give the details.

Let $H$ be a hash function and $x_1$ be the private key. The public key is $y_1=g^{x_1}$. For any message $m_1$, the signature is $\sigma_1={h_1}^{x_1}$, where $h_1 = H(m_1)$. To verify the signature, test that: $\hat{e}(\sigma_1, g) \stackrel{?}{=} \hat{e}(h_1, y_1)$ holds.

Let $x_2$ be another private key. The public key is $y_2=g^{x_2}$. As before, for any message $m_2$, the signature is $\sigma_2={h_2}^{x_2}$, where $h_2 = H(m_2)$. To verify the signature, test that: $\hat{e}(\sigma_2, g) \stackrel{?}{=} \hat{e}(h_2, y_2)$ holds.

We can also combine $\sigma_1, \sigma_2$ into an aggregate signature $\sigma$ as follows: $\sigma=\sigma_{1}\cdot \sigma_{2}$. To verify, we check that: $$\hat{e}(\sigma, g) \stackrel{?}{=} \hat{e}(h_1, y_1)\cdot\hat{e}(h_2, y_2)$$

The signatures satisfy the standard security as shown in the Aggregate Signatures Paper. That is, assuming that the Diffie-Hellman problem is hard, presentation of $\sigma$ proves that $y_1$ signed $m_1$ and $y_2$ signed $m_2$ (even if the original signatures are not provided).

The security of OWAS relies on the fact that given just $\sigma$, it is hard to compute either $\sigma_1$ or $\sigma_2$ (as hard as solving Diffle-Hellman problem). In fact, for two combined signatures, the resulting signature is a zero-knowledge proof of knowledge of each individual signature. This can be shown as follows:

Suppose I can control the output of the hash function (i.e., we are using the "random oracle model"), then instead of doing it the correct way as described above, given $(g, h_1, y_1)$, I can compute a fake tuple $(\sigma, h_2, y_2)$ such that the last verification equation above holds. First generate random $a, b$. The compute:

$$y_2=g^a y_1$$ $$h_2=\frac{g^b}{h_1}$$ $$\sigma=\frac{{y_1}^b\cdot g^{ab}}{{h_1}^a}$$

It can be checked that the above values satisfy the aggregate signature verification equation. Yet, I did this without knowing $\sigma_1$. Hence this is zero-knowledge.


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