Nostr Web Client

📅 Original date posted:2023-07-24

🗒️ Summary of this message: The sender is discussing with Jonas the need for a method to blind the value of c in order to prevent the server from learning the value of m.

📝 Original message:

Hi Jonas,

Seems you are right: for every tx, compute c from the on-chain data, and

the server can match the c to the m (tx). So there would need to be a

method for blinding the value of c.

On Mon, Jul 24, 2023 at 4:39 PM Jonas Nick wrote:

> > Party 1 never learns the final value of (R,s1+s2) or m.

> Actually, it seems like a blinding step is missing. Assume the server

> (party 1)

> received some c during the signature protocol. Can't the server scan the

> blockchain for signatures, compute corresponding hashes c' = H(R||X||m) as

> in

> signature verification and then check c == c'? If true, then the server

> has the

> preimage for the c received from the client, including m.

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eae21eb2... 2y ago

📅 Original date posted:2023-07-26

🗒️ Summary of this message: Blind Schnorr signatures can solve the issue of blinding, but not the problem of client-controlled forged signatures. Recent work proposes alternative approaches for blind Schnorr signatures.

📝 Original message:

While this may solve blinding, I don't see how it solves the problem that the

client can forge signatures because the client is in control of challenge e'.

This is not special to MuSig(2), but is also the reason why original blind

Schnorr signatures are insecure (as demonstrated in David Wagner's "A

Generalized Birthday Problem" paper).

For some more recent work on blind Schnorr signatures, see:

- https://eprint.iacr.org/2019/877.pdf Blind Schnorr Signatures and Signed

ElGamal Encryption in the Algebraic Group Mode

- https://eprint.iacr.org/2020/1071.pdf On Pairing-Free Blind Signature Schemes

in the Algebraic Group Model

In particular, the first paper proposes a less-efficient variant of blind

Schnorr signatures that is secure under concurrent signing if the "mROS" problem

is hard (which is imho plausible). Another potential approach is using

commitments and a ZKP as I mentioned earlier in this thread. This scheme is

"folklore", in the sense that it is being discussed from time to time but isn't

specified and does not have a security proof as far as I am aware.

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9b3cb906... 2y ago

📅 Original date posted:2023-07-26

🗒️ Summary of this message: The protocol described in the text is an interesting idea for incorporating 2FA authentication into blind signing. However, there may be vulnerabilities in the protocol that need to be addressed.

📝 Original message:

It's an interesting idea for a protocol. If I get it right, your basic idea here is to kind of "shoehorn" in a 2FA authentication, and that the blind-signing server has no other function than to check the 2FA?

This makes it different from most uses of blind signing, where *counting* the number of signatures matters (hence 'one more forgery etc). Here, you are just saying "I'll sign whatever the heck you like, as long as you're authorized with this 2FA procedure".

Going to ignore the details of practically what that means - though I'm sure that's where most of the discussion would end up - but just looking at your protocol in the gist:

It seems you're not checking K values against attacks, so for example this would allow someone to extract the server's key from one signing:

1 Alice, after receiving K2, sets K1 = K1' - K2, where the secret key of K1' is k1'.

2 Chooses b as normal, sends e' as normal.

3 Receiving s2, calculate s = s1 + s2 as normal.

So since s = k + ex = (k' + bx) + ex = k' + e'x, and you know s, k' and e', you can derive x. Then x2 = x - x1.

(Gist I'm referring to: https://gist.github.com/moonsettler/05f5948291ba8dba63a3985b786233bb)

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------- Original Message -------

On Wednesday, July 26th, 2023 at 03:44, moonsettler via bitcoin-dev wrote:

> Hi All,

> I believe it's fairly simple to solve the blinding (sorry for the bastard notation!):

> Signing:

> X = X1 + X2

> K1 = k1G

> K2 = k2G

> R = K1 + K2 + bX

> e = hash(R||X||m)

> e' = e + b

> s = (k1 + e'*x1) + (k2 + e'*x2)

> s = (k1 + k2 + b(x1 + x2)) + e(x1 + x2)

> sG = (K1 + K2 + bX) + eX

> sG = R + eX

> Verification:

> Rv = sG - eX

> ev = hash(R||X||m)

> e ?= ev

> https://gist.github.com/moonsettler/05f5948291ba8dba63a3985b786233bb

> Been trying to get a review on this for a while, please let me know if I got it wrong!

> BR,

> moonsettler

> ------- Original Message -------

> On Monday, July 24th, 2023 at 5:39 PM, Jonas Nick via bitcoin-dev bitcoin-dev at lists.linuxfoundation.org wrote:

> > > Party 1 never learns the final value of (R,s1+s2) or m.

> >

> > Actually, it seems like a blinding step is missing. Assume the server (party 1)

> > received some c during the signature protocol. Can't the server scan the

> > blockchain for signatures, compute corresponding hashes c' = H(R||X||m) as in

> > signature verification and then check c == c'? If true, then the server has the

> > preimage for the c received from the client, including m.

> > _______________________________________________

> > bitcoin-dev mailing list

> > bitcoin-dev at lists.linuxfoundation.org

> > https://lists.linuxfoundation.org/mailman/listinfo/bitcoin-dev

> _______________________________________________

> bitcoin-dev mailing list

> bitcoin-dev at lists.linuxfoundation.org

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