Re: [bitcoin-dev] Blinded 2-party Musig2

From: Erik Aronesty <erik@q32•com>
To: AdamISZ <AdamISZ@protonmail•com>,
	 Bitcoin Protocol Discussion
	<bitcoin-dev@lists•linuxfoundation.org>
Cc: Tom Trevethan <tom@commerceblock•com>
Subject: Re: [bitcoin-dev] Blinded 2-party Musig2
Date: Tue, 25 Jul 2023 10:12:31 -0400	[thread overview]
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posk is "proof of secret key".   so you cannot use wagner to select R

On Mon, Jul 24, 2023 at 1:59 PM AdamISZ via bitcoin-dev <
bitcoin-dev@lists•linuxfoundation.org> wrote:

> @ZmnSCPxj:
>
> yes, Wagner is the attack you were thinking of.
>
> And yeah, to avoid it, you should have the 3rd round of MuSig1, i.e. the R
> commitments.
>
> @Tom:
> As per above it seems you were more considering MuSig1 here, not MuSig2.
> At least in this version. So you need the initial commitments to R.
>
> Jonas' reply clearly has covered a lot of what matters here, but I wanted
> to mention (using your notation):
>
> in s1 = c * a1 * x1 + r1, you expressed the idea that the challenge c
> could be given to the server, to construct s1, but since a1 = H(L, X1) and
> L is the serialization of all (in this case, 2) keys, that wouldn't work
> for blinding the final key, right?
> But, is it possible that this addresses the other problem?
> If the server is given c1*a1 instead as the challenge for signing (with
> their "pure" key x1), then perhaps it avoids the issue? Given what's on the
> blockchain ends up allowing calculation of 'c' and the aggregate key a1X1 +
> a2X2, is it the case that you cannot find a1 and therefore you cannot
> correlate the transaction with just the quantity 'c1*a1' which the server
> sees?
>
> But I agree with Jonas that this is just the start, i.e. the fundamental
> requirement of a blind signing scheme is there has to be some guarantee of
> no 'one more forgery' possibility, so presumably there has to be some proof
> that the signing request is 'well formed' (Jonas expresses it below as a
> ZKP of a SHA2 preimage .. it does not seem pretty but I agree that on the
> face of it, that is what's needed).
>
> @Jonas, Erik:
> 'posk' is probably meant as 'proof of secret key' which may(?) be a mixup
> with what is sometimes referred to in the literature as "KOSK" (iirc they
> used it in FROST for example). It isn't clear to me yet how that factors
> into this scenario, although ofc it is for sure a potential building block
> of these constructions.
>
> Sent with Proton Mail secure email.
>
> ------- Original Message -------
> On Monday, July 24th, 2023 at 08:12, Jonas Nick via bitcoin-dev <
> bitcoin-dev@lists•linuxfoundation.org> wrote:
>
>
> > Hi Tom,
> >
> > I'm not convinced that this works. As far as I know blind musig is still
> an open
> > research problem. What the scheme you propose appears to try to prevent
> is that
> > the server signs K times, but the client ends up with K+1 Schnorr
> signatures for
> > the aggregate of the server's and the clients key. I think it's possible
> to
> > apply a variant of the attack that makes MuSig1 insecure if the nonce
> commitment
> > round was skipped or if the message isn't determined before sending the
> nonce.
> > Here's how a malicious client would do that:
> >
> > - Obtain K R-values R1[0], ..., R1[K-1] from the server
> > - Let
> > R[i] := R1[i] + R2[i] for all i <= K-1
> > R[K] := R1[0] + ... + R1[K-1]
> > c[i] := H(X, R[i], m[i]) for all i <= K.
> > Using Wagner's algorithm, choose R2[0], ..., R2[K-1] such that
> > c[0] + ... + c[K-1] = c[K].
> > - Send c[0], ..., c[K-1] to the server to obtain s[0], ..., s[K-1].
> > - Let
> > s[K] = s[0] + ... + s[K-1].
> > Then (s[K], R[K]) is a valid signature from the server, since
> > s[K]G = R[K] + c[K]a1X1,
> > which the client can complete to a signature for public key X.
> >
> > What may work in your case is the following scheme:
> > - Client sends commitment to the public key X2, nonce R2 and message m
> to the
> > server.
> > - Server replies with nonce R1 = k1G
> > - Client sends c to the server and proves in zero knowledge that c =
> > SHA256(X1 + X2, R1 + R2, m).
> > - Server replies with s1 = k1 + c*x1
> >
> > However, this is just some quick intuition and I'm not sure if this
> actually
> > works, but maybe worth exploring.
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