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- # MuSig2: Simple Two-Round Schnorr Multi-Signatures
- # https://eprint.iacr.org/2020/1261.pdf
- # This scheme is n-of-n, not threshold.
- from hashlib import sha256
- # Nonces
- v = 3
- # Participants
- n = 5
- # Pallas
- p = 0x40000000000000000000000000000000224698fc094cf91b992d30ed00000001
- q = 0x40000000000000000000000000000000224698fc0994a8dd8c46eb2100000001
- Fp = GF(p)
- Fq = GF(q)
- Ep = EllipticCurve(Fp, (0, 5))
- Ep.set_order(q)
- # NullifierK Generator: g
- nfk_x = 0x25e7aa169ca8198d2e375571faf4c9cf5e7eb192ccb5db9bd36f6aa7e447ca75
- nfk_y = 0x155c1f851b1a3384880473442008ff755fe0a49ec1c1b4332db8dce21ae001cc
- g = Ep([nfk_x, nfk_y])
- def hash_domain(domain, *args):
- concat = domain.encode() + b"".join(str(arg).encode() for arg in args)
- return Fq(int(sha256(concat).hexdigest(), 16))
- # Domain separator for H_agg
- def H_AGG(*args):
- return hash_domain("musig2_H_agg", *args)
- # Domain separator for H_non
- def H_NON(*args):
- return hash_domain("musig2_H_non", *args)
- # Domain separator for H_sig
- def H_SIG(*args):
- return hash_domain("musig2_H_sig", *args)
- # =================
- # 1. Key generation
- # =================
- x = [Fq.random_element() for _ in range(n)]
- X = [x_i * g for x_i in x]
- # ==================
- # 2. Key aggregation
- # ==================
- L = b"".join(str(X_i).encode() for X_i in X)
- X_tilde = Fq(0) * g
- for i in range(n):
- a_i = H_AGG(L, X[i])
- X_tilde += X[i] * a_i
- # ======================
- # 3. First signing round
- # ======================
- R_i = [] # Each participant's public nonces
- r_i = [] # Each participant's secret nonces
- for _ in range(n):
- r_j = [Fq.random_element() for _ in range(v)]
- R_j = [r * g for r in r_j]
- r_i.append(r_j)
- R_i.append(R_j)
- # Sum up the nonces for all participants for each j
- R = [sum(R_ij[j] for R_ij in R_i) for j in range(v)]
- assert len(R) == v
- # =======================
- # 4. Second signing round
- # =======================
- message = "Hello MuSig2"
- s_i = []
- b = H_NON(X_tilde, *R, message)
- R_total = sum(R[j] * b * (j+1) for j in range(v))
- c = H_SIG(X_tilde, R_total, message) # Compute the challenge based on R_total
- for i in range(n):
- a_i = H_AGG(L, X[i])
- s_partial = c * a_i * x[i] + sum(r_i[i][j] * (b * (j+1)) for j in range(v))
- s_i.append(s_partial)
- s = sum(s_i)
- # ===============
- # 5. Verification
- # ===============
- c = H_SIG(X_tilde, R_total, message)
- assert g * s == R_total + X_tilde * c
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