CoCalc Public Files18.783 Lecture 11 Montgomery ECM.ipynbOpen with one click!
Author: Andrew Sutherland
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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[1-b[i]] = madd(Q[1],Q[0],P) 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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