def pm1_pollard(n,B,nbRM=20,verbose=False):
"""Pollard's p-1 factoring algorithm.
Args:
- *n (int)*: an integer
- *B (int)*: smooth boundary
Optional Args:
- *nbRM (int)*: number of repeats of Rabin-Miller primality test.
- *verbose (bool)*: set to True if you want a display.
Returns:
- *(int)*: a subfactor p of n such that p-1 is B-smooth if it exists.
0 if attack failed.
"""
lB = itools.ilog(B,2)
primes = itools.get_primes(1,B,nbRM)
a = random.randint(1,n)
for q in primes:
e = lB//itools.ilog(q,2)
a = itools.exp_mod(a,q**e,n) #q**e ~ B
g = itools.gcd(a-1,n)
if g>1 and g<n:
if(verbose): print("pm1_pollard :\n",n,"=",g,"x",n//g)
return g
else:
if(verbose): print("pm1_pollard failed with", B, ":(")
return 0
def pm1_pollard_auto(n,Bmax,verbose=False):
"""Pollard's p-1 factoring algorithm with automatic Boundary adjustment.
Args:
- *n (int)*: an integer
- *Bmax (int)*: Maximum smooth boundary
Optional Args:
- *verbose (bool)*: set to True if you want a display.
Returns:
- *(int)*: a subfactor p of n such that p-1 is B-smooth if it exists.
0 if attack failed.
"""
#optimized parameters
B = 16000
nbRM = 10
alternate = False
g = 1
while(g==1 and B<=Bmax):
lB = itools.ilog(B,2)
primes = itools.get_primes(1,B,nbRM)
g = n
while(g==n):
a = random.randint(1,n)
for q in primes:
e = lB//itools.ilog(q,2)
a = itools.exp_mod(a,q**e,n) #q**e ~ B
g = itools.gcd(a-1,n)
if(g==n and verbose):
print("N reached, resuming...")
if(g==1):
c = a
primes = itools.get_primes(B,2*B,nbRM)
d = primes[0]
a = itools.exp_mod(c,d,n)
for i in range(1,len(primes)):
g = itools.gcd(a-1,n)
if(g!=1):
alternate=True
break
d = primes[i]-primes[i-1]
a *= itools.exp_mod(c,d,n) #peut être amélioré en mémorisant les c**(2k) dans une table
if(g==1):
B *= 4
if(g!=1 and g!=n):
if(verbose):
if(alternate): print("pm1_pollard_auto ( B' =",2*B,") :\n",n,"=",g,"x",n//g)
else: print("pm1_pollard_auto ( B =",B,") :\n",n,"=",g,"x",n//g)
return g
else:
if(verbose): print("pm1_pollard_auto failed with", Bmax, ":(")
return 0
def pp1_williams(n,B,nbRM=20,verbose=False):
"""Williams' p+1 factoring algorithm.
Args:
- *n (int)*: an integer
- *B (int)*: smooth boundary
Optional Args:
- *nbRM (int)*: number of repeats of Rabin-Miller primality test.
- *verbose (bool)*: set to True if you want a display.
Returns:
- *(int)*: a subfactor p of n such that p+1 is B-smooth if it exists.
0 if attack failed.
"""
lB = itools.ilog(B,2)
primes = itools.get_primes(3,B,nbRM)
a = random.randint(1,n)
v = a
i=0
for q in primes:
e = lB//itools.ilog(q,2)
#plus efficace que : for i in range(e): v = lucal_mul(v,q,n)
#évite de calculer e fois la décomposition binaire de q et d'appeler e fonctions, au prix d'une exponentiation entière
v = lucas_mul(v,q**e,n) #q**e ~ B
g = itools.gcd(v-2,n)
if(g!=1 and g!=n): break
if(g>1 and g<n):
if(verbose): print("pp1_williams :\n",n,"=",g,"x",n//g)
return g
else:
if(verbose): print("pp1_williams failed with", B, ":(")
return 0
def pp1_williams_auto(n,Bmax,verbose=False):
"""Williams' p+1 factoring algorithm with automatic Boundary adjustment.
Args:
- *n (int)*: an integer
- *Bmax (int)*: Maximum smooth boundary
Optional Args:
- *verbose (bool)*: set to True if you want a display.
Returns:
- *(int)*: a subfactor p of n such that p+1 is B-smooth if it exists.
0 if attack failed.
"""
B = 1000
nbRM = 10
alternate = False
g = 1
while(g==1 and B<=Bmax):
lB = itools.ilog(B,2)
B2 = (10 ** (itools.ilog(B,10)//5))*B #tiré de resultats expérimentaux : http://www.loria.fr/~zimmerma/records/Pplus1.html
primes = itools.get_primes(3,B,nbRM)
g = n
#Step 1
while(g==n):
a = random.randint(1,n)
v = a
for q in primes:
e = lB//itools.ilog(q,2)
#plus efficace que : for i in range(e): v = lucal_mul(v,q,n)
#évite de calculer e fois la décomposition binaire de q et d'appeler e fonctions, au prix d'une exponentiation entière
v = lucas_mul(v,q**e,n) #q**e ~ B
#les valeurs atteintes par g sont croissantes et stagent à n lorsqu'atteints
#=> pas besoin de les évaluer à chaque multiplication de l'indice
#si g == n -> on a dépassé les facteurs, il faut recommencer avec la même borne (voire plus petite ?)
#le tirage de a sera déterminant
#si g==1 la borne n'est pas suffisante, on passe au Step 2 puis on augmente
g = itools.gcd(v-2,n)
if(g==n and verbose):
print("N reached, resuming...")
#Step 2 TODO : optimal? L'idée des multiples de 6 peut-elle être reprise pour p-1 Pollard ?
if(g==1):
c = (B//6)*6
v6 = lucas_mul(v,6,n)
Vl = []
Vl.append(lucas_mul(v,c,n))
Vl.append(lucas_mul(v,c+6,n))
i = 2
c = c+12
while(c<B2):
Vl.append( (Vl[i-1]*v6 - Vl[i-2]) % n )
c += 6
i += 1
produit = 1
for vi in Vl:
produit = ( produit * (vi-v) ) % n
g = itools.gcd(produit,n)
if(g==1):
B *= 10
else:
alternate = True
if(g!=1 and g!=n):
if(verbose):
if(alternate): print("pp1_williams_auto ( B' =",B2,") :\n",n,"=",g,"x",n//g)
else: print("pp1_williams_auto ( B =",B,") :\n",n,"=",g,"x",n//g)
return g
else:
if(verbose): print("pp1_williams_auto failed with", Bmax, ":(")
return 0