[bitcoin-dev] Blinded 2-party Musig2

[moonsettler <moonsettler@protonmail.com>]

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@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.
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