def solve(x):
# Solve for y, given x
# There are two possible points that satisfy the curve,
# an even and an odd. We choose the odd one.
y = sqrt(x**3 + 7)
if y.n % 2 == 0: y = -y
if not curve.testPoint(x, y): raise ValueError
return Point(curve, x, y)
def fast__mul__(self, n):
if n < 0:
return -self * -n
if n == 0:
return Ideal(self.curve)
p = secp256k1_openssl.SPoint(self.x.n, self.y.n)
x, y = p.mult(n)._coords()
#slow = self._slow_mul(n)
#assert slow.x == x
#assert slow.y == y
return Point(self.curve, Fq(x), Fq(y))
# Then the curve, always of the form y^2 = x^3 + {a6}
# The curve E: y2 = x3+ax+b over Fp is defined by:
# a = 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000
# b = 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000007
curve = GeneralizedEllipticCurve(a6=Fq(7)) # Ex: y2 = x3+7
# base point, a generator of the group
# The base point G in compressed form is:
# G = 02 79BE667E F9DCBBAC 55A06295 CE870B07 029BFCDB 2DCE28D9 59F2815B 16F81798
Gx = Fq(0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798)
Gy = Fq(0x483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8)
G = Point(curve, Gx, Gy)
# Finally the order n of G and the cofactor are:
# n = FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE BAAEDCE6 AF48A03B BFD25E8C D0364141
# h = 01
# This is the order (# of elements in) the curve
p = order = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
Fp = FiniteField(p, 1)
##
# Convenience functions
##
def random_oracle_string_to_Zp(s):
return sha2_to_long(s) % p