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Bandersnatch/banderwagon endomorphism acceleration #380

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@mratsim mratsim commented May 8, 2024

This implements:

  • endomorphism acceleration support for Twisted Edwards curves
  • endomorphism acceleration for Bandersnatch and Banderwagon
  • benchmarks for scalar multiplication and MSM

Departures from the generic derivation

λ

The endomorphism is equivalent to a scalar multiplication by a solution of λ² = -2.
There are 2 such solutions and for a generic implementation we should try
to ensure which of λ₀ = √-2 and λ₁ = -√-2 corresponds to the computation at

func computeEndoBander[F](r {.noalias.}: var ECP_TwEdwards_Prj[F], P: ECP_TwEdwards_Prj[F]) =
static: doAssert F.C in {Bandersnatch, Banderwagon}
var xy {.noInit.}, yy {.noInit.}, zz {.noInit.}: F
xy.prod(P.x, P.y)
yy.square(P.y)
zz.square(P.z)
const b = F.fromHex("0x52c9f28b828426a561f00d3a63511a882ea712770d9af4d6ee0f014d172510b4")
const c = F.fromHex("0x6cc624cf865457c3a97c6efd6c17d1078456abcfff36f4e9515c806cdf650b3d")
r.x.diff(zz, yy)
r.x *= c
zz *= b
r.y.sum(yy, zz)
r.y *= b
r.z.diff(yy, zz)
r.x *= r.z
r.y *= xy
r.z *= xy

This is what is done for cubic root endomorphisms:

try:
check_cubic_root_endo(G1, Fp, r, cofactor, int(lambda1), phi1)
except:
print('Failure with:')
print(' 𝜑 (mod p): 0x' + Integer(phi1).hex())
print(' λᵩ (mod r): 0x' + Integer(lambda1).hex())
phi1, phi2 = phi2, phi1
check_cubic_root_endo(G1, Fp, r, cofactor, int(lambda1), phi1)
finally:
print('Success with:')
print(' 𝜑 (mod p): 0x' + Integer(phi1).hex())
print(' λᵩ (mod r): 0x' + Integer(lambda1).hex())

def check_cubic_root_endo(G1, Fp, r, cofactor, lambdaR, phiP):
## Check the Endomorphism for p mod 3 == 1
## Endomorphism can be field multiplication by one of the non-trivial cube root of unity 𝜑
## Rationale:
## curve equation is y² = x³ + b, and y² = (x𝜑)³ + b <=> y² = x³ + b (with 𝜑³ == 1) so we are still on the curve
## this means that multiplying by 𝜑 the x-coordinate is equivalent to a scalar multiplication by some λᵩ
## with λᵩ² + λᵩ + 1 ≡ 0 (mod r) and 𝜑² + 𝜑 + 1 ≡ 0 (mod p), see below.
## Hence we have a 2 dimensional decomposition of the scalar multiplication
## i.e. For any [s]P, we can find a corresponding [k1]P + [k2][λᵩ]P with [λᵩ]P being a simple field multiplication by 𝜑
## Finding cube roots:
## x³−1=0 <=> (x−1)(x²+x+1) = 0, if x != 1, x solves (x²+x+1) = 0 <=> x = (-1±√3)/2
assert phiP^3 == Fp(1)
assert lambdaR^3 % r == 1
Prand = G1.random_point()
P = Prand * cofactor
assert P != G1([0, 1, 0])
(Px, Py, Pz) = P
Qendo = G1([Px*phiP, Py, Pz])
Qlambda = lambdaR * P
assert P != Qendo
assert P != Qlambda
assert Qendo == Qlambda
print('Endomorphism OK')

However, SageMath has no builtin support for Edwards curves so we would would need to reimplement scalar multiplication to do these checks. Hence we hardcode the λ from the paper.

Lattice decomposition

The lattice decomposition also follows the reference implementation, i.e. instead of

def derive_lattice(r, lambdaR, m):
lat = Matrix(matrix.identity(m))
lat[0, 0] = r
for i in range(1, m):
lat[i, 0] = -lambdaR^i
return lat.LLL()

we do https://github.com/asanso/Bandersnatch/blob/e280929/python-ref-impl/bandersnatch.py#L20-L21

        M = Matrix([[-L,1], [r,0]])
        self.N = M.LLL()

This only moves signs around.

@mratsim mratsim marked this pull request as ready for review June 7, 2024 15:43
@mratsim mratsim closed this Jun 27, 2024
@mratsim mratsim deleted the bander_endo branch June 27, 2024 20:07
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mratsim commented Jun 27, 2024

Rebased and reopened in #405

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Bandersnatch / Banderwagon endomorphism acceleration
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