CoCalc Public Files18.783 Lecture 11 Montgomery ECM.ipynb
Author: Andrew Sutherland
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Compute Environment: Ubuntu 20.04 (Default)

Montgomery curve group operations

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def madd(P,Q,R):
"""Montgomery addition: returns (x_{m+n},z_{m+n}) given P=(x_m,z_m), Q=(x_n,z_n), and R=(x_{m-n},z_{m-n})"""
a = P[0]-P[1]; b=P[0]+P[1]; c=Q[0]-Q[1]; d=Q[0]+Q[1]
e=a*d; f=b*c;
return (R[1]*(e+f)^2,R[0]*(e-f)^2)

def mdbl(P,C):
"""Montgomery doubling: returns (x_{2n},z_{2n}) given P=(x_n,z_n) and C=(A+2)/4 where By^2=x^3+Ax^2+x is the Montgomery curve equation"""
a=P[0]+P[1]; b=P[0]-P[1]
c=a^2; d=b^2; e=c-d
return (c*d,e*(d+C*e))

def mmul(P,n,C):
"""Montgomery multiplication: returns (x_n,z_n) given P=(x_1,z_1) and C=(A+2)/4 where By^2=x^3+Ax^2+x is the Montgomery curve equation"""
Q = [P,mdbl(P,C)]
b=n.digits(2)
for i in range(len(b)-2,-1,-1):
Q[b[i]] = mdbl(Q[b[i]],C)
return Q[0]

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F=GF(random_prime(2^256,2^255))
A=F.random_element(); B=F.random_element()
C=(A+2)/4
while true:
x = F.random_element()
if ((x^3+A*x^2+x)/B).is_square(): break;
P=(x,1);
t=cputime()
for p in primes(0,10000): P=mmul(P,p,C)
print("Montgomery time: %s" % (cputime()-t))
a4=1/B^2-A^2/(3*B^2); a6=-A^3/(27*B^3)-a4*A/(3*B)
E=EllipticCurve([a4,a6])
P=E.random_element()
t=cputime()
for p in primes(0,10000): P=p*P
print("Weierstrass time: %s" % (cputime()-t))


Single stage ECM using Montgomery curves

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def L(a,c,N):
z = ceil(exp(c*log(N)^a*log(log(N))^(1-a)))
return ceil(z)

def ECM_1curve(N,p,M,B1):
R=Integers(N)
# generate Montgomery curve using parameterization that guarantees a point of order 3
c=R(randint(1,10^10))
a=6*c/(c^2+6)
b=R(randint(1,10^10))
A=(-3*a^4-6*a^2+1)/(4*a^3)
B=(a^2-1)^2/(4*a*b^2)
C=(A+2)/4
P=(3*a/4,1)
a4=1/B^2-A^2/(3*B^2); a6=-A^3/(27*B^3)-a4*A/(3*B)
F=GF(p)
E=EllipticCurve([F(a4.lift()),F(a6.lift())])
print(factor(E.cardinality()))
m=ceil(M+2*sqrt(M)+1)
for p in primes(B1):
q = p; r=p*q
while r <= m: q=r; r=p*q
P=mmul(P,q,C)
d=gcd(N,P[1].lift())
if d == N: print("N fail"); return 0
if d > 1: return d
return 0

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k=50; n=200
q=random_prime(2^(n-k),2^(n-k-1))
p=random_prime(2^k,2^(k-1))
B=2.5*L(1/2,1/sqrt(2),2^k) # includes an empirical fudge factor
print("B = %d" % B)
print("Expected number of iterations is about %s " % L(1/2,1/(2*sqrt(2)),2^k))
t=cputime()
i=1
while true:
print(i),
d = ECM_1curve(p*q,p,2^50,B)
if d: break
i += 1
print(d)
print("ECM time: %s" % (cputime()-t))
print(p)

B = 6340 Expected number of iterations is about 51 1 2^2 * 3^3 * 5^2 * 17 * 89 * 1499 * 37243
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