def pollard_pm1(n, B1=100, B2=1000):
from _primefac._arith import ispower, gcd, ilog
from _primefac._prime import isprime, primegen
import six
if isprime(n):
return n
m = ispower(n)
if m:
return m
while True:
pg = primegen()
q = 2 # TODO: what about other initial values of q?
p = six.next(pg)
while p <= B1:
q, p = pow(q, p ** ilog(B1, p), n), six.next(pg)
g = gcd(q - 1, n)
if 1 < g < n:
return g
while p <= B2:
q, p = pow(q, p, n), six.next(pg)
g = gcd(q - 1, n)
if 1 < g < n:
return g
# These bounds failed. Increase and try again.
B1 *= 10
B2 *= 10
def pollardRho_brent(n):
from _primefac._arith import gcd
from _primefac._prime import isprime
from six.moves import xrange
from random import randrange
if isprime(n):
return n
g = n
while g == n:
y, c, m, g, r, q = randrange(1, n), randrange(1, n), randrange(1, n), 1, 1, 1
while g == 1:
x, k = y, 0
for _ in xrange(r):
y = (y ** 2 + c) % n
while k < r and g == 1:
ys = y
for _ in xrange(min(m, r - k)):
y = (y ** 2 + c) % n
q = q * abs(x - y) % n
g, k = gcd(q, n), k + m
r *= 2
if g == n:
while True:
ys = (ys ** 2 + c) % n
g = gcd(abs(x - ys), n)
if g > 1:
break
return g
def ecm(n, B1=10, B2=20):
"""
TODO: Determine the best defaults for B1 and B2 and the best way to
increment them and iters "Modern" ECM using Montgomery curves and an
algorithm analogous to the two-phase variant of Pollard's p-1 method
TODO: We currently compute the prime lists from the sieve as we need them,
but this means that we recompute them at every iteration. While it
would not be particularly efficient memory-wise, we might be able
to increase time-efficiency by computing the primes we need ahead of
time (say once at the beginning and then once each time we increase
the bounds) and saving them in lists, and then iterate the inner
while loops over those lists.
"""
from _primefac._arith import ispower, gcd, ilog
from _primefac._prime import isprime, primegen
from six.moves import xrange
from random import randrange
import six
if isprime(n):
return n
m = ispower(n)
if m:
return m
iters = 1
while True:
for _ in xrange(iters):
seed = randrange(6, n)
u, v = (seed**2 - 5) % n, 4 * seed % n
p = pow(u, 3, n)
Q, C = (pow(v - u, 3, n) * (3 * u + v) % n,
4 * p * v % n), (p, pow(v, 3, n))
pg = primegen()
p = six.next(pg)
while p <= B1:
Q, p = _ecmul(p**ilog(B1, p), Q, C, n), six.next(pg)
g = gcd(Q[1], n)
if 1 < g < n:
return g
while p <= B2:
"""
"There is a simple coding trick that can speed up the second
stage. Instead of multiplying each prime times Q, we iterate
over i from B1 + 1 to B2, adding 2Q at each step; when i is
prime, the current Q can be accumulated into the running
solution. Again, we defer the calculation of the greatest
common divisor until the end of the iteration."
TODO: Implement this trick and compare performance.
"""
Q = _ecmul(p, Q, C, n)
g *= Q[1]
g %= n
p = six.next(pg)
g = gcd(g, n)
if 1 < g < n:
return g
# This seed failed. Try again with a new one.
# These bounds failed. Increase and try again.
B1 *= 3
B2 *= 3
iters *= 2
def williams_pp1(n):
from _primefac._arith import ispower, ilog, isqrt, gcd
from _primefac._prime import isprime, primegen
from six.moves import xrange
import itertools
if isprime(n):
return n
m = ispower(n)
if m:
return m
for v in itertools.count(1):
for p in primegen():
e = ilog(isqrt(n), p)
if e == 0:
break
for _ in xrange(e):
v = mlucas(v, p, n)
g = gcd(v - 2, n)
if 1 < g < n:
return g
if g == n:
break
def _isprime(n, tb=(3, 5, 7, 11), eb=(2, ), mrb=()): # TODO: more streamlining
from _primefac import _arith
# tb: trial division basis
# eb: Euler's test basis
# mrb: Miller-Rabin basis
# This test suite's first false positve is unknown but has been shown to
# be greater than 2**64.
# Infinitely many are thought to exist.
if n % 2 == 0 or n < 13 or n == _arith.isqrt(n)**2:
# Remove evens, squares, and numbers less than 13
return n in (2, 3, 5, 7, 11)
if any(n % p == 0 for p in tb):
return n in tb # Trial division
for b in eb: # Euler's test
if b >= n:
continue
if not pow(b, n - 1, n) == 1:
return False
r = n - 1
while r % 2 == 0:
r //= 2
c = pow(b, r, n)
if c == 1:
continue
while c != 1 and c != n - 1:
c = pow(c, 2, n)
if c == 1:
return False
s, d = pfactor(n)
if not sprp(n, 2, s, d):
return False
if n < 2047:
return True
# BPSW has two phases: SPRP with base 2 and SLPRP.
# We just did the SPRP; now we do the SLPRP:
if n >= 3825123056546413051:
d = 5
while True:
if _arith.gcd(d, n) > 1:
p, q = 0, 0
break
if _arith.jacobi(d, n) == -1:
p, q = 1, (1 - d) // 4
break
d = -d - 2 * d // abs(d)
if p == 0:
return n == d
s, t = pfactor(n + 2)
u, v, u2, v2, m = 1, p, 1, p, t // 2
k = q
while m > 0:
u2, v2, q = (u2 * v2) % n, (v2 * v2 - 2 * q) % n, (q * q) % n
if m % 2 == 1:
u, v = u2 * v + u * v2, v2 * v + u2 * u * d
if u % 2 == 1:
u += n
if v % 2 == 1:
v += n
u, v, k = (u // 2) % n, (v // 2) % n, (q * k) % n
m //= 2
if (u == 0) or (v == 0):
return True
for _ in xrange(1, s):
v, k = (v * v - 2 * k) % n, (k * k) % n
if v == 0:
return True
return False
if not mrb:
if n < 1373653:
mrb = [3]
elif n < 25326001:
mrb = [3, 5]
elif n < 3215031751:
mrb = [3, 5, 7]
elif n < 2152302898747:
mrb = [3, 5, 7, 11]
elif n < 3474749660383:
mrb = [3, 5, 6, 11, 13]
elif n < 341550071728321:
# This number is also a false positive for primes(19+1).
mrb = [3, 5, 7, 11, 13, 17]
elif n < 3825123056546413051:
# Also a false positive for primes(31+1).
mrb = [3, 5, 7, 11, 13, 17, 19, 23]
# Miller-Rabin
return all(sprp(n, b, s, d) for b in mrb)
def primefac(n,
trial_limit=1000,
rho_rounds=42000,
verbose=False,
methods=(_primefac.pollardRho_brent, _primefac.pollard_pm1,
_primefac.williams_pp1, _primefac.ecm, _primefac.mpqs,
_primefac.fermat, _primefac.factordb),
timeout=60):
from _primefac import isprime, isqrt, primegen
from random import randrange
timeout = timeout - 1
if n < 2:
return
if isprime(n):
yield n
return
factors, nroot = [], isqrt(n)
# Note that we remove factors of 2 whether the user wants to or not.
for p in primegen():
if n % p == 0:
while n % p == 0:
yield p
n //= p
nroot = isqrt(n)
if isprime(n):
yield n
return
if p > nroot:
if n != 1:
yield n
return
if p >= trial_limit:
break
if isprime(n):
yield n
return
if rho_rounds == "inf":
factors = [n]
while len(factors) != 0:
n = min(factors)
factors.remove(n)
f = _primefac.pollardRho_brent(n)
if isprime(f):
yield f
else:
factors.append(f)
n //= f
if isprime(n):
yield n
else:
factors.append(n)
return
factors, difficult = [n], []
while len(factors) != 0:
rhocount = 0
n = factors.pop()
try:
g = n
while g == n:
x, c, g = randrange(1, n), randrange(1, n), 1
y = x
while g == 1:
if rhocount >= rho_rounds:
raise Exception
rhocount += 1
x = (x**2 + c) % n
y = (y**2 + c) % n
y = (y**2 + c) % n
g = gcd(x - y, n)
# We now have a nontrivial factor g of n. If we took too long to get here, we're actually at the except statement.
if isprime(g):
yield g
else:
factors.append(g)
n //= g
if isprime(n):
yield n
else:
factors.append(n)
except Exception:
difficult.append(
n
) # Factoring n took too long. We'll have multifactor chug on it.
factors = difficult
while len(factors) != 0:
n = min(factors)
factors.remove(n)
f = multifactor(n, methods=methods, verbose=verbose, timeout=timeout)
if isprime(f):
yield f
else:
factors.append(f)
n //= f
if isprime(n):
yield n
else:
factors.append(n)
def mpqs(n):
"""
When the bound proves insufficiently large, we throw out all our work and
start over.
TODO: When this happens, get more data, but don't trash what we already
have.
TODO: Rewrite to get a few more relations before proceeding to the
linear algebra.
TODO: When we need to increase the bound, what is the optimal increment?
"""
from _primefac._arith import ispower, isqrt, ilog, gcd, mod_sqrt, legendre
from _primefac._arith import modinv
from _primefac._prime import isprime, nextprime
from _primefac._util import listprod, mpz
from six.moves import xrange
from math import log
# Special cases: this function poorly handles primes and perfect powers:
m = ispower(n)
if m:
return m
if isprime(n):
return n
root_2n = isqrt(2 * n)
bound = ilog(n**6, 10)**2 # formula chosen by experiment
while True:
try:
prime, mod_root, log_p, num_prime = [], [], [], 0
# find a number of small primes for which n is a quadratic residue
p = 2
while p < bound or num_prime < 3:
leg = legendre(n % p, p)
if leg == 1:
prime += [p]
# the rhs was [int(mod_sqrt(n, p))].
# If we get errors, put it back.
mod_root += [mod_sqrt(n, p)]
log_p += [log(p, 10)]
num_prime += 1
elif leg == 0:
return p
p = nextprime(p)
x_max = len(prime) * 60 # size of the sieve
# maximum value on the sieved range
m_val = (x_max * root_2n) >> 1
"""
fudging the threshold down a bit makes it easier to find powers of
primes as factors as well as partial-partial relationships, but it
also makes the smoothness check slower. there's a happy medium
somewhere, depending on how efficient the smoothness check is
"""
thresh = log(m_val, 10) * 0.735
# skip small primes. they contribute very little to the log sum
# and add a lot of unnecessary entries to the table instead, fudge
# the threshold down a bit, assuming ~1/4 of them pass
min_prime = mpz(thresh * 3)
fudge = sum(log_p[i] for i, p in enumerate(prime) if p < min_prime)
fudge = fudge // 4
thresh -= fudge
smooth, used_prime, partial = [], set(), {}
num_smooth, num_used_prime, num_partial = 0, 0, 0
num_poly, root_A = 0, isqrt(root_2n // x_max)
while num_smooth <= num_used_prime:
# find an integer value A such that:
# A is =~ sqrt(2*n) // x_max
# A is a perfect square
# sqrt(A) is prime, and n is a quadratic residue mod sqrt(A)
while True:
root_A = nextprime(root_A)
leg = legendre(n, root_A)
if leg == 1:
break
elif leg == 0:
return root_A
A = root_A**2
# solve for an adequate B. B*B is a quadratic residue mod n,
# such that B*B-A*C = n. this is unsolvable if n is not a
# quadratic residue mod sqrt(A)
b = mod_sqrt(n, root_A)
B = (b + (n - b * b) * modinv(b + b, root_A)) % A
C = (B * B - n) // A # B*B-A*C = n <=> C = (B*B-n)//A
num_poly += 1
# sieve for prime factors
sums, i = [0.0] * (2 * x_max), 0
for p in prime:
if p < min_prime:
i += 1
continue
logp = log_p[i]
g = gcd(A, p)
inv_A = modinv(A // g, p // g) * g
# modular root of the quadratic
a, b, k = (mpz(((mod_root[i] - B) * inv_A) % p),
mpz(((p - mod_root[i] - B) * inv_A) % p), 0)
while k < x_max:
if k + a < x_max:
sums[k + a] += logp
if k + b < x_max:
sums[k + b] += logp
if k:
sums[k - a + x_max] += logp
sums[k - b + x_max] += logp
k += p
i += 1
# check for smooths
i = 0
for v in sums:
if v > thresh:
x, vec, sqr = x_max - i if i > x_max else i, set(), []
# because B*B-n = A*C
# (A*x+B)^2 - n = A*A*x*x+2*A*B*x + B*B - n
# = A*(A*x*x+2*B*x+C)
# gives the congruency
# (A*x+B)^2 = A*(A*x*x+2*B*x+C) (mod n)
# because A is chosen to be square, it doesn't
# need to be sieved
sieve_val = (A * x + 2 * B) * x + C
if sieve_val < 0:
vec, sieve_val = {-1}, -sieve_val
for p in prime:
while sieve_val % p == 0:
if p in vec:
"""
track perfect sqr facs to avoid sqrting
something huge at the end
"""
sqr += [p]
vec ^= {p}
sieve_val = mpz(sieve_val // p)
if sieve_val == 1: # smooth
smooth += [(vec, (sqr, (A * x + B), root_A))]
used_prime |= vec
elif sieve_val in partial:
"""
combine two partials to make a (xor) smooth that
is, every prime factor with an odd power is in our
factor base
"""
pair_vec, pair_vals = partial[sieve_val]
sqr += list(vec & pair_vec) + [sieve_val]
vec ^= pair_vec
smooth += [(vec, (sqr + pair_vals[0],
(A * x + B) * pair_vals[1],
root_A * pair_vals[2]))]
used_prime |= vec
num_partial += 1
else:
# save partial for later pairing
partial[sieve_val] = (vec, (sqr, A * x + B,
root_A))
i += 1
num_smooth, num_used_prime = len(smooth), len(used_prime)
used_prime = sorted(list(used_prime))
# set up bit fields for gaussian elimination
masks, mask, bitfields = [], 1, [0] * num_used_prime
for vec, _ in smooth:
masks += [mask]
i = 0
for p in used_prime:
if p in vec:
bitfields[i] |= mask
i += 1
mask <<= 1
# row echelon form
offset = 0
null_cols = []
for col in xrange(num_smooth):
pivot = bitfields[col - offset] & masks[
col] == 0 # This occasionally throws IndexErrors.
# TODO: figure out why it throws errors and fix it.
for row in xrange(col + 1 - offset, num_used_prime):
if bitfields[row] & masks[col]:
if pivot:
bitfields[col - offset], bitfields[
row], pivot = bitfields[row], bitfields[
col - offset], False
else:
bitfields[row] ^= bitfields[col - offset]
if pivot:
null_cols += [col]
offset += 1
# reduced row echelon form
for row in xrange(num_used_prime):
mask = bitfields[row] & -bitfields[row] # lowest set bit
for up_row in xrange(row):
if bitfields[up_row] & mask:
bitfields[up_row] ^= bitfields[row]
# check for non-trivial congruencies
# TODO: if none exist, check combinations of null space columns...
# if _still_ none exist, sieve more values
for col in null_cols:
all_vec, (lh, rh, rA) = smooth[col]
lhs = lh # sieved values (left hand side)
rhs = [rh] # sieved values - n (right hand side)
rAs = [rA] # root_As (cofactor of lhs)
i = 0
for field in bitfields:
if field & masks[col]:
vec, (lh, rh, rA) = smooth[i]
lhs += list(all_vec & vec) + lh
all_vec ^= vec
rhs += [rh]
rAs += [rA]
i += 1
factor = gcd(listprod(rAs) * listprod(lhs) - listprod(rhs), n)
if 1 < factor < n:
return factor
except IndexError:
pass
bound *= 1.2
