```
"""
Integer factorization
"""
from __future__ import print_function, division
import random
import math
from .primetest import isprime
from .generate import sieve, primerange, nextprime
from sympy.core import sympify
from sympy.core.evalf import bitcount
from sympy.core.numbers import igcd, oo, Rational
from sympy.core.power import integer_nthroot, Pow
from sympy.core.mul import Mul
from sympy.core.compatibility import as_int, SYMPY_INTS, xrange
from sympy.core.singleton import S
from sympy.core.function import Function
from sympy.core.symbol import Dummy
small_trailing = [i and max(int(not i % 2**j) and j for j in range(1, 8))
for i in range(256)]
[docs]def smoothness(n):
"""
Return the B-smooth and B-power smooth values of n.
The smoothness of n is the largest prime factor of n; the power-
smoothness is the largest divisor raised to its multiplicity.
>>> from sympy.ntheory.factor_ import smoothness
>>> smoothness(2**7*3**2)
(3, 128)
>>> smoothness(2**4*13)
(13, 16)
>>> smoothness(2)
(2, 2)
See Also
========
factorint, smoothness_p
"""
if n == 1:
return (1, 1) # not prime, but otherwise this causes headaches
facs = factorint(n)
return max(facs), max(m**facs[m] for m in facs)
[docs]def smoothness_p(n, m=-1, power=0, visual=None):
"""
Return a list of [m, (p, (M, sm(p + m), psm(p + m)))...]
where:
1. p**M is the base-p divisor of n
2. sm(p + m) is the smoothness of p + m (m = -1 by default)
3. psm(p + m) is the power smoothness of p + m
The list is sorted according to smoothness (default) or by power smoothness
if power=1.
The smoothness of the numbers to the left (m = -1) or right (m = 1) of a
factor govern the results that are obtained from the p +/- 1 type factoring
methods.
>>> from sympy.ntheory.factor_ import smoothness_p, factorint
>>> smoothness_p(10431, m=1)
(1, [(3, (2, 2, 4)), (19, (1, 5, 5)), (61, (1, 31, 31))])
>>> smoothness_p(10431)
(-1, [(3, (2, 2, 2)), (19, (1, 3, 9)), (61, (1, 5, 5))])
>>> smoothness_p(10431, power=1)
(-1, [(3, (2, 2, 2)), (61, (1, 5, 5)), (19, (1, 3, 9))])
If visual=True then an annotated string will be returned:
>>> print(smoothness_p(21477639576571, visual=1))
p**i=4410317**1 has p-1 B=1787, B-pow=1787
p**i=4869863**1 has p-1 B=2434931, B-pow=2434931
This string can also be generated directly from a factorization dictionary
and vice versa:
>>> factorint(17*9)
{3: 2, 17: 1}
>>> smoothness_p(_)
'p**i=3**2 has p-1 B=2, B-pow=2\\np**i=17**1 has p-1 B=2, B-pow=16'
>>> smoothness_p(_)
{3: 2, 17: 1}
The table of the output logic is:
====== ====== ======= =======
| Visual
------ ----------------------
Input True False other
====== ====== ======= =======
dict str tuple str
str str tuple dict
tuple str tuple str
n str tuple tuple
mul str tuple tuple
====== ====== ======= =======
See Also
========
factorint, smoothness
"""
from sympy.utilities import flatten
# visual must be True, False or other (stored as None)
if visual in (1, 0):
visual = bool(visual)
elif visual not in (True, False):
visual = None
if type(n) is str:
if visual:
return n
d = {}
for li in n.splitlines():
k, v = [int(i) for i in
li.split('has')[0].split('=')[1].split('**')]
d[k] = v
if visual is not True and visual is not False:
return d
return smoothness_p(d, visual=False)
elif type(n) is not tuple:
facs = factorint(n, visual=False)
if power:
k = -1
else:
k = 1
if type(n) is not tuple:
rv = (m, sorted([(f,
tuple([M] + list(smoothness(f + m))))
for f, M in [i for i in facs.items()]],
key=lambda x: (x[1][k], x[0])))
else:
rv = n
if visual is False or (visual is not True) and (type(n) in [int, Mul]):
return rv
lines = []
for dat in rv[1]:
dat = flatten(dat)
dat.insert(2, m)
lines.append('p**i=%i**%i has p%+i B=%i, B-pow=%i' % tuple(dat))
return '\n'.join(lines)
[docs]def trailing(n):
"""Count the number of trailing zero digits in the binary
representation of n, i.e. determine the largest power of 2
that divides n.
Examples
========
>>> from sympy import trailing
>>> trailing(128)
7
>>> trailing(63)
0
"""
n = int(n)
if not n:
return 0
low_byte = n & 0xff
if low_byte:
return small_trailing[low_byte]
# 2**m is quick for z up through 2**30
z = bitcount(n) - 1
if isinstance(z, SYMPY_INTS):
if n == 1 << z:
return z
t = 0
p = 8
while not n & 1:
while not n & ((1 << p) - 1):
n >>= p
t += p
p *= 2
p //= 2
return t
[docs]def multiplicity(p, n):
"""
Find the greatest integer m such that p**m divides n.
Examples
========
>>> from sympy.ntheory import multiplicity
>>> from sympy.core.numbers import Rational as R
>>> [multiplicity(5, n) for n in [8, 5, 25, 125, 250]]
[0, 1, 2, 3, 3]
>>> multiplicity(3, R(1, 9))
-2
"""
try:
p, n = as_int(p), as_int(n)
except ValueError:
if all(isinstance(i, (SYMPY_INTS, Rational)) for i in (p, n)):
try:
p = Rational(p)
n = Rational(n)
if p.q == 1:
if n.p == 1:
return -multiplicity(p.p, n.q)
return S.Zero
elif p.p == 1:
return multiplicity(p.q, n.q)
else:
like = min(
multiplicity(p.p, n.p),
multiplicity(p.q, n.q))
cross = min(
multiplicity(p.q, n.p),
multiplicity(p.p, n.q))
return like - cross
except AttributeError:
pass
raise ValueError('expecting ints or fractions, got %s and %s' % (p, n))
if p == 2:
return trailing(n)
if p < 2:
raise ValueError('p must be an integer, 2 or larger, but got %s' % p)
if p == n:
return 1
m = 0
n, rem = divmod(n, p)
while not rem:
m += 1
if m > 5:
# The multiplicity could be very large. Better
# to increment in powers of two
e = 2
while 1:
ppow = p**e
if ppow < n:
nnew, rem = divmod(n, ppow)
if not rem:
m += e
e *= 2
n = nnew
continue
return m + multiplicity(p, n)
n, rem = divmod(n, p)
return m
[docs]def perfect_power(n, candidates=None, big=True, factor=True):
"""
Return ``(b, e)`` such that ``n`` == ``b**e`` if ``n`` is a
perfect power; otherwise return ``False``.
By default, the base is recursively decomposed and the exponents
collected so the largest possible ``e`` is sought. If ``big=False``
then the smallest possible ``e`` (thus prime) will be chosen.
If ``candidates`` for exponents are given, they are assumed to be sorted
and the first one that is larger than the computed maximum will signal
failure for the routine.
If ``factor=True`` then simultaneous factorization of n is attempted
since finding a factor indicates the only possible root for n. This
is True by default since only a few small factors will be tested in
the course of searching for the perfect power.
Examples
========
>>> from sympy import perfect_power
>>> perfect_power(16)
(2, 4)
>>> perfect_power(16, big = False)
(4, 2)
"""
n = int(n)
if n < 3:
return False
logn = math.log(n, 2)
max_possible = int(logn) + 2 # only check values less than this
not_square = n % 10 in [2, 3, 7, 8] # squares cannot end in 2, 3, 7, 8
if not candidates:
candidates = primerange(2 + not_square, max_possible)
afactor = 2 + n % 2
for e in candidates:
if e < 3:
if e == 1 or e == 2 and not_square:
continue
if e > max_possible:
return False
# see if there is a factor present
if factor:
if n % afactor == 0:
# find what the potential power is
if afactor == 2:
e = trailing(n)
else:
e = multiplicity(afactor, n)
# if it's a trivial power we are done
if e == 1:
return False
# maybe the bth root of n is exact
r, exact = integer_nthroot(n, e)
if not exact:
# then remove this factor and check to see if
# any of e's factors are a common exponent; if
# not then it's not a perfect power
n //= afactor**e
m = perfect_power(n, candidates=primefactors(e), big=big)
if m is False:
return False
else:
r, m = m
# adjust the two exponents so the bases can
# be combined
g = igcd(m, e)
if g == 1:
return False
m //= g
e //= g
r, e = r**m*afactor**e, g
if not big:
e0 = primefactors(e)
if len(e0) > 1 or e0[0] != e:
e0 = e0[0]
r, e = r**(e//e0), e0
return r, e
else:
# get the next factor ready for the next pass through the loop
afactor = nextprime(afactor)
# Weed out downright impossible candidates
if logn/e < 40:
b = 2.0**(logn/e)
if abs(int(b + 0.5) - b) > 0.01:
continue
# now see if the plausible e makes a perfect power
r, exact = integer_nthroot(n, e)
if exact:
if big:
m = perfect_power(r, big=big, factor=factor)
if m is not False:
r, e = m[0], e*m[1]
return int(r), e
else:
return False
[docs]def pollard_rho(n, s=2, a=1, retries=5, seed=1234, max_steps=None, F=None):
r"""
Use Pollard's rho method to try to extract a nontrivial factor
of ``n``. The returned factor may be a composite number. If no
factor is found, ``None`` is returned.
The algorithm generates pseudo-random values of x with a generator
function, replacing x with F(x). If F is not supplied then the
function x**2 + ``a`` is used. The first value supplied to F(x) is ``s``.
Upon failure (if ``retries`` is > 0) a new ``a`` and ``s`` will be
supplied; the ``a`` will be ignored if F was supplied.
The sequence of numbers generated by such functions generally have a
a lead-up to some number and then loop around back to that number and
begin to repeat the sequence, e.g. 1, 2, 3, 4, 5, 3, 4, 5 -- this leader
and loop look a bit like the Greek letter rho, and thus the name, 'rho'.
For a given function, very different leader-loop values can be obtained
so it is a good idea to allow for retries:
>>> from sympy.ntheory.generate import cycle_length
>>> n = 16843009
>>> F = lambda x:(2048*pow(x, 2, n) + 32767) % n
>>> for s in range(5):
... print('loop length = %4i; leader length = %3i' % next(cycle_length(F, s)))
...
loop length = 2489; leader length = 42
loop length = 78; leader length = 120
loop length = 1482; leader length = 99
loop length = 1482; leader length = 285
loop length = 1482; leader length = 100
Here is an explicit example where there is a two element leadup to
a sequence of 3 numbers (11, 14, 4) that then repeat:
>>> x=2
>>> for i in range(9):
... x=(x**2+12)%17
... print(x)
...
16
13
11
14
4
11
14
4
11
>>> next(cycle_length(lambda x: (x**2+12)%17, 2))
(3, 2)
>>> list(cycle_length(lambda x: (x**2+12)%17, 2, values=True))
[16, 13, 11, 14, 4]
Instead of checking the differences of all generated values for a gcd
with n, only the kth and 2*kth numbers are checked, e.g. 1st and 2nd,
2nd and 4th, 3rd and 6th until it has been detected that the loop has been
traversed. Loops may be many thousands of steps long before rho finds a
factor or reports failure. If ``max_steps`` is specified, the iteration
is cancelled with a failure after the specified number of steps.
Examples
========
>>> from sympy import pollard_rho
>>> n=16843009
>>> F=lambda x:(2048*pow(x,2,n) + 32767) % n
>>> pollard_rho(n, F=F)
257
Use the default setting with a bad value of ``a`` and no retries:
>>> pollard_rho(n, a=n-2, retries=0)
If retries is > 0 then perhaps the problem will correct itself when
new values are generated for a:
>>> pollard_rho(n, a=n-2, retries=1)
257
References
==========
- Richard Crandall & Carl Pomerance (2005), "Prime Numbers:
A Computational Perspective", Springer, 2nd edition, 229-231
"""
n = int(n)
if n < 5:
raise ValueError('pollard_rho should receive n > 4')
prng = random.Random(seed + retries)
V = s
for i in range(retries + 1):
U = V
if not F:
F = lambda x: (pow(x, 2, n) + a) % n
j = 0
while 1:
if max_steps and (j > max_steps):
break
j += 1
U = F(U)
V = F(F(V)) # V is 2x further along than U
g = igcd(U - V, n)
if g == 1:
continue
if g == n:
break
return int(g)
V = prng.randint(0, n - 1)
a = prng.randint(1, n - 3) # for x**2 + a, a%n should not be 0 or -2
F = None
return None
[docs]def pollard_pm1(n, B=10, a=2, retries=0, seed=1234):
"""
Use Pollard's p-1 method to try to extract a nontrivial factor
of ``n``. Either a divisor (perhaps composite) or ``None`` is returned.
The value of ``a`` is the base that is used in the test gcd(a**M - 1, n).
The default is 2. If ``retries`` > 0 then if no factor is found after the
first attempt, a new ``a`` will be generated randomly (using the ``seed``)
and the process repeated.
Note: the value of M is lcm(1..B) = reduce(ilcm, range(2, B + 1)).
A search is made for factors next to even numbers having a power smoothness
less than ``B``. Choosing a larger B increases the likelihood of finding a
larger factor but takes longer. Whether a factor of n is found or not
depends on ``a`` and the power smoothness of the even mumber just less than
the factor p (hence the name p - 1).
Although some discussion of what constitutes a good ``a`` some
descriptions are hard to interpret. At the modular.math site referenced
below it is stated that if gcd(a**M - 1, n) = N then a**M % q**r is 1
for every prime power divisor of N. But consider the following:
>>> from sympy.ntheory.factor_ import smoothness_p, pollard_pm1
>>> n=257*1009
>>> smoothness_p(n)
(-1, [(257, (1, 2, 256)), (1009, (1, 7, 16))])
So we should (and can) find a root with B=16:
>>> pollard_pm1(n, B=16, a=3)
1009
If we attempt to increase B to 256 we find that it doesn't work:
>>> pollard_pm1(n, B=256)
>>>
But if the value of ``a`` is changed we find that only multiples of
257 work, e.g.:
>>> pollard_pm1(n, B=256, a=257)
1009
Checking different ``a`` values shows that all the ones that didn't
work had a gcd value not equal to ``n`` but equal to one of the
factors:
>>> from sympy.core.numbers import ilcm, igcd
>>> from sympy import factorint, Pow
>>> M = 1
>>> for i in range(2, 256):
... M = ilcm(M, i)
...
>>> set([igcd(pow(a, M, n) - 1, n) for a in range(2, 256) if
... igcd(pow(a, M, n) - 1, n) != n])
set([1009])
But does aM % d for every divisor of n give 1?
>>> aM = pow(255, M, n)
>>> [(d, aM%Pow(*d.args)) for d in factorint(n, visual=True).args]
[(257**1, 1), (1009**1, 1)]
No, only one of them. So perhaps the principle is that a root will
be found for a given value of B provided that:
1) the power smoothness of the p - 1 value next to the root
does not exceed B
2) a**M % p != 1 for any of the divisors of n.
By trying more than one ``a`` it is possible that one of them
will yield a factor.
Examples
========
With the default smoothness bound, this number can't be cracked:
>>> from sympy.ntheory import pollard_pm1, primefactors
>>> pollard_pm1(21477639576571)
Increasing the smoothness bound helps:
>>> pollard_pm1(21477639576571, B=2000)
4410317
Looking at the smoothness of the factors of this number we find:
>>> from sympy.utilities import flatten
>>> from sympy.ntheory.factor_ import smoothness_p, factorint
>>> print(smoothness_p(21477639576571, visual=1))
p**i=4410317**1 has p-1 B=1787, B-pow=1787
p**i=4869863**1 has p-1 B=2434931, B-pow=2434931
The B and B-pow are the same for the p - 1 factorizations of the divisors
because those factorizations had a very large prime factor:
>>> factorint(4410317 - 1)
{2: 2, 617: 1, 1787: 1}
>>> factorint(4869863-1)
{2: 1, 2434931: 1}
Note that until B reaches the B-pow value of 1787, the number is not cracked;
>>> pollard_pm1(21477639576571, B=1786)
>>> pollard_pm1(21477639576571, B=1787)
4410317
The B value has to do with the factors of the number next to the divisor,
not the divisors themselves. A worst case scenario is that the number next
to the factor p has a large prime divisisor or is a perfect power. If these
conditions apply then the power-smoothness will be about p/2 or p. The more
realistic is that there will be a large prime factor next to p requiring
a B value on the order of p/2. Although primes may have been searched for
up to this level, the p/2 is a factor of p - 1, something that we don't
know. The modular.math reference below states that 15% of numbers in the
range of 10**15 to 15**15 + 10**4 are 10**6 power smooth so a B of 10**6
will fail 85% of the time in that range. From 10**8 to 10**8 + 10**3 the
percentages are nearly reversed...but in that range the simple trial
division is quite fast.
References
==========
- Richard Crandall & Carl Pomerance (2005), "Prime Numbers:
A Computational Perspective", Springer, 2nd edition, 236-238
- http://modular.math.washington.edu/edu/2007/spring/ent/ent-html/node81.html
- http://www.cs.toronto.edu/~yuvalf/Factorization.pdf
"""
n = int(n)
if n < 4 or B < 3:
raise ValueError('pollard_pm1 should receive n > 3 and B > 2')
prng = random.Random(seed + B)
# computing a**lcm(1,2,3,..B) % n for B > 2
# it looks weird, but it's right: primes run [2, B]
# and the answer's not right until the loop is done.
for i in range(retries + 1):
aM = a
for p in sieve.primerange(2, B + 1):
e = int(math.log(B, p))
aM = pow(aM, pow(p, e), n)
g = igcd(aM - 1, n)
if 1 < g < n:
return int(g)
# get a new a:
# since the exponent, lcm(1..B), is even, if we allow 'a' to be 'n-1'
# then (n - 1)**even % n will be 1 which will give a g of 0 and 1 will
# give a zero, too, so we set the range as [2, n-2]. Some references
# say 'a' should be coprime to n, but either will detect factors.
a = prng.randint(2, n - 2)
def _trial(factors, n, candidates, verbose=False):
"""
Helper function for integer factorization. Trial factors ``n`
against all integers given in the sequence ``candidates``
and updates the dict ``factors`` in-place. Returns the reduced
value of ``n`` and a flag indicating whether any factors were found.
"""
if verbose:
factors0 = list(factors.keys())
nfactors = len(factors)
for d in candidates:
if n % d == 0:
m = multiplicity(d, n)
n //= d**m
factors[d] = m
if verbose:
for k in sorted(set(factors).difference(set(factors0))):
print(factor_msg % (k, factors[k]))
return int(n), len(factors) != nfactors
def _check_termination(factors, n, limitp1, use_trial, use_rho, use_pm1,
verbose):
"""
Helper function for integer factorization. Checks if ``n``
is a prime or a perfect power, and in those cases updates
the factorization and raises ``StopIteration``.
"""
if verbose:
print('Check for termination')
# since we've already been factoring there is no need to do
# simultaneous factoring with the power check
p = perfect_power(n, factor=False)
if p is not False:
base, exp = p
if limitp1:
limit = limitp1 - 1
else:
limit = limitp1
facs = factorint(base, limit, use_trial, use_rho, use_pm1,
verbose=False)
for b, e in facs.items():
if verbose:
print(factor_msg % (b, e))
factors[b] = exp*e
raise StopIteration
if isprime(n):
factors[int(n)] = 1
raise StopIteration
if n == 1:
raise StopIteration
trial_int_msg = "Trial division with ints [%i ... %i] and fail_max=%i"
trial_msg = "Trial division with primes [%i ... %i]"
rho_msg = "Pollard's rho with retries %i, max_steps %i and seed %i"
pm1_msg = "Pollard's p-1 with smoothness bound %i and seed %i"
factor_msg = '\t%i ** %i'
fermat_msg = 'Close factors satisying Fermat condition found.'
complete_msg = 'Factorization is complete.'
def _factorint_small(factors, n, limit, fail_max):
"""
Return the value of n and either a 0 (indicating that factorization up
to the limit was complete) or else the next near-prime that would have
been tested.
Factoring stops if there are fail_max unsuccessful tests in a row.
If factors of n were found they will be in the factors dictionary as
{factor: multiplicity} and the returned value of n will have had those
factors removed. The factors dictionary is modified in-place.
"""
def done(n, d):
"""return n, d if the sqrt(n) wasn't reached yet, else
n, 0 indicating that factoring is done.
"""
if d*d <= n:
return n, d
return n, 0
d = 2
m = trailing(n)
if m:
factors[d] = m
n >>= m
d = 3
if limit < d:
if n > 1:
factors[n] = 1
return done(n, d)
# reduce
m = 0
while n % d == 0:
n //= d
m += 1
if m == 20:
mm = multiplicity(d, n)
m += mm
n //= d**mm
break
if m:
factors[d] = m
# when d*d exceeds maxx or n we are done; if limit**2 is greater
# than n then maxx is set to zero so the value of n will flag the finish
if limit*limit > n:
maxx = 0
else:
maxx = limit*limit
dd = maxx or n
d = 5
fails = 0
while fails < fail_max:
if d*d > dd:
break
# d = 6*i - 1
# reduce
m = 0
while n % d == 0:
n //= d
m += 1
if m == 20:
mm = multiplicity(d, n)
m += mm
n //= d**mm
break
if m:
factors[d] = m
dd = maxx or n
fails = 0
else:
fails += 1
d += 2
if d*d > dd:
break
# d = 6*i - 1
# reduce
m = 0
while n % d == 0:
n //= d
m += 1
if m == 20:
mm = multiplicity(d, n)
m += mm
n //= d**mm
break
if m:
factors[d] = m
dd = maxx or n
fails = 0
else:
fails += 1
# d = 6*(i+1) - 1
d += 4
return done(n, d)
[docs]def factorint(n, limit=None, use_trial=True, use_rho=True, use_pm1=True,
verbose=False, visual=None):
r"""
Given a positive integer ``n``, ``factorint(n)`` returns a dict containing
the prime factors of ``n`` as keys and their respective multiplicities
as values. For example:
>>> from sympy.ntheory import factorint
>>> factorint(2000) # 2000 = (2**4) * (5**3)
{2: 4, 5: 3}
>>> factorint(65537) # This number is prime
{65537: 1}
For input less than 2, factorint behaves as follows:
- ``factorint(1)`` returns the empty factorization, ``{}``
- ``factorint(0)`` returns ``{0:1}``
- ``factorint(-n)`` adds ``-1:1`` to the factors and then factors ``n``
Partial Factorization:
If ``limit`` (> 3) is specified, the search is stopped after performing
trial division up to (and including) the limit (or taking a
corresponding number of rho/p-1 steps). This is useful if one has
a large number and only is interested in finding small factors (if
any). Note that setting a limit does not prevent larger factors
from being found early; it simply means that the largest factor may
be composite. Since checking for perfect power is relatively cheap, it is
done regardless of the limit setting.
This number, for example, has two small factors and a huge
semi-prime factor that cannot be reduced easily:
>>> from sympy.ntheory import isprime
>>> from sympy.core.compatibility import long
>>> a = 1407633717262338957430697921446883
>>> f = factorint(a, limit=10000)
>>> f == {991: 1, long(202916782076162456022877024859): 1, 7: 1}
True
>>> isprime(max(f))
False
This number has a small factor and a residual perfect power whose
base is greater than the limit:
>>> factorint(3*101**7, limit=5)
{3: 1, 101: 7}
Visual Factorization:
If ``visual`` is set to ``True``, then it will return a visual
factorization of the integer. For example:
>>> from sympy import pprint
>>> pprint(factorint(4200, visual=True))
3 1 2 1
2 *3 *5 *7
Note that this is achieved by using the evaluate=False flag in Mul
and Pow. If you do other manipulations with an expression where
evaluate=False, it may evaluate. Therefore, you should use the
visual option only for visualization, and use the normal dictionary
returned by visual=False if you want to perform operations on the
factors.
You can easily switch between the two forms by sending them back to
factorint:
>>> from sympy import Mul, Pow
>>> regular = factorint(1764); regular
{2: 2, 3: 2, 7: 2}
>>> pprint(factorint(regular))
2 2 2
2 *3 *7
>>> visual = factorint(1764, visual=True); pprint(visual)
2 2 2
2 *3 *7
>>> print(factorint(visual))
{2: 2, 3: 2, 7: 2}
If you want to send a number to be factored in a partially factored form
you can do so with a dictionary or unevaluated expression:
>>> factorint(factorint({4: 2, 12: 3})) # twice to toggle to dict form
{2: 10, 3: 3}
>>> factorint(Mul(4, 12, evaluate=False))
{2: 4, 3: 1}
The table of the output logic is:
====== ====== ======= =======
Visual
------ ----------------------
Input True False other
====== ====== ======= =======
dict mul dict mul
n mul dict dict
mul mul dict dict
====== ====== ======= =======
Notes
=====
Algorithm:
The function switches between multiple algorithms. Trial division
quickly finds small factors (of the order 1-5 digits), and finds
all large factors if given enough time. The Pollard rho and p-1
algorithms are used to find large factors ahead of time; they
will often find factors of the order of 10 digits within a few
seconds:
>>> factors = factorint(12345678910111213141516)
>>> for base, exp in sorted(factors.items()):
... print('%s %s' % (base, exp))
...
2 2
2507191691 1
1231026625769 1
Any of these methods can optionally be disabled with the following
boolean parameters:
- ``use_trial``: Toggle use of trial division
- ``use_rho``: Toggle use of Pollard's rho method
- ``use_pm1``: Toggle use of Pollard's p-1 method
``factorint`` also periodically checks if the remaining part is
a prime number or a perfect power, and in those cases stops.
If ``verbose`` is set to ``True``, detailed progress is printed.
See Also
========
smoothness, smoothness_p, divisors
"""
factordict = {}
if visual and not isinstance(n, Mul) and not isinstance(n, dict):
factordict = factorint(n, limit=limit, use_trial=use_trial,
use_rho=use_rho, use_pm1=use_pm1,
verbose=verbose, visual=False)
elif isinstance(n, Mul):
factordict = dict([(int(k), int(v)) for k, v in
list(n.as_powers_dict().items())])
elif isinstance(n, dict):
factordict = n
if factordict and (isinstance(n, Mul) or isinstance(n, dict)):
# check it
for k in list(factordict.keys()):
if isprime(k):
continue
e = factordict.pop(k)
d = factorint(k, limit=limit, use_trial=use_trial, use_rho=use_rho,
use_pm1=use_pm1, verbose=verbose, visual=False)
for k, v in d.items():
if k in factordict:
factordict[k] += v*e
else:
factordict[k] = v*e
if visual or (type(n) is dict and
visual is not True and
visual is not False):
if factordict == {}:
return S.One
if -1 in factordict:
factordict.pop(-1)
args = [S.NegativeOne]
else:
args = []
args.extend([Pow(*i, evaluate=False)
for i in sorted(factordict.items())])
return Mul(*args, evaluate=False)
elif isinstance(n, dict) or isinstance(n, Mul):
return factordict
assert use_trial or use_rho or use_pm1
n = as_int(n)
if limit:
limit = int(limit)
# special cases
if n < 0:
factors = factorint(
-n, limit=limit, use_trial=use_trial, use_rho=use_rho,
use_pm1=use_pm1, verbose=verbose, visual=False)
factors[-1] = 1
return factors
if limit and limit < 2:
if n == 1:
return {}
return {n: 1}
elif n < 10:
# doing this we are assured of getting a limit > 2
# when we have to compute it later
return [{0: 1}, {}, {2: 1}, {3: 1}, {2: 2}, {5: 1},
{2: 1, 3: 1}, {7: 1}, {2: 3}, {3: 2}][n]
factors = {}
# do simplistic factorization
if verbose:
sn = str(n)
if len(sn) > 50:
print('Factoring %s' % sn[:5] + \
'..(%i other digits)..' % (len(sn) - 10) + sn[-5:])
else:
print('Factoring', n)
if use_trial:
# this is the preliminary factorization for small factors
small = 2**15
fail_max = 600
small = min(small, limit or small)
if verbose:
print(trial_int_msg % (2, small, fail_max))
n, next_p = _factorint_small(factors, n, small, fail_max)
else:
next_p = 2
if factors and verbose:
for k in sorted(factors):
print(factor_msg % (k, factors[k]))
if next_p == 0:
if n > 1:
factors[int(n)] = 1
if verbose:
print(complete_msg)
return factors
# continue with more advanced factorization methods
# first check if the simplistic run didn't finish
# because of the limit and check for a perfect
# power before exiting
try:
if limit and next_p > limit:
if verbose:
print('Exceeded limit:', limit)
_check_termination(factors, n, limit, use_trial, use_rho, use_pm1,
verbose)
if n > 1:
factors[int(n)] = 1
return factors
else:
# Before quitting (or continuing on)...
# ...do a Fermat test since it's so easy and we need the
# square root anyway. Finding 2 factors is easy if they are
# "close enough." This is the big root equivalent of dividing by
# 2, 3, 5.
sqrt_n = integer_nthroot(n, 2)[0]
a = sqrt_n + 1
a2 = a**2
b2 = a2 - n
for i in range(3):
b, fermat = integer_nthroot(b2, 2)
if fermat:
break
b2 += 2*a + 1 # equiv to (a+1)**2 - n
a += 1
if fermat:
if verbose:
print(fermat_msg)
if limit:
limit -= 1
for r in [a - b, a + b]:
facs = factorint(r, limit=limit, use_trial=use_trial,
use_rho=use_rho, use_pm1=use_pm1,
verbose=verbose)
factors.update(facs)
raise StopIteration
# ...see if factorization can be terminated
_check_termination(factors, n, limit, use_trial, use_rho, use_pm1,
verbose)
except StopIteration:
if verbose:
print(complete_msg)
return factors
# these are the limits for trial division which will
# be attempted in parallel with pollard methods
low, high = next_p, 2*next_p
limit = limit or sqrt_n
# add 1 to make sure limit is reached in primerange calls
limit += 1
while 1:
try:
high_ = high
if limit < high_:
high_ = limit
# Trial division
if use_trial:
if verbose:
print(trial_msg % (low, high_))
ps = sieve.primerange(low, high_)
n, found_trial = _trial(factors, n, ps, verbose)
if found_trial:
_check_termination(factors, n, limit, use_trial, use_rho,
use_pm1, verbose)
else:
found_trial = False
if high > limit:
if verbose:
print('Exceeded limit:', limit)
if n > 1:
factors[int(n)] = 1
raise StopIteration
# Only used advanced methods when no small factors were found
if not found_trial:
if (use_pm1 or use_rho):
high_root = max(int(math.log(high_**0.7)), low, 3)
# Pollard p-1
if use_pm1:
if verbose:
print(pm1_msg % (high_root, high_))
c = pollard_pm1(n, B=high_root, seed=high_)
if c:
# factor it and let _trial do the update
ps = factorint(c, limit=limit - 1,
use_trial=use_trial,
use_rho=use_rho,
use_pm1=use_pm1,
verbose=verbose)
n, _ = _trial(factors, n, ps, verbose=False)
_check_termination(factors, n, limit, use_trial,
use_rho, use_pm1, verbose)
# Pollard rho
if use_rho:
max_steps = high_root
if verbose:
print(rho_msg % (1, max_steps, high_))
c = pollard_rho(n, retries=1, max_steps=max_steps,
seed=high_)
if c:
# factor it and let _trial do the update
ps = factorint(c, limit=limit - 1,
use_trial=use_trial,
use_rho=use_rho,
use_pm1=use_pm1,
verbose=verbose)
n, _ = _trial(factors, n, ps, verbose=False)
_check_termination(factors, n, limit, use_trial,
use_rho, use_pm1, verbose)
except StopIteration:
if verbose:
print(complete_msg)
return factors
low, high = high, high*2
[docs]def primefactors(n, limit=None, verbose=False):
"""Return a sorted list of n's prime factors, ignoring multiplicity
and any composite factor that remains if the limit was set too low
for complete factorization. Unlike factorint(), primefactors() does
not return -1 or 0.
Examples
========
>>> from sympy.ntheory import primefactors, factorint, isprime
>>> primefactors(6)
[2, 3]
>>> primefactors(-5)
[5]
>>> sorted(factorint(123456).items())
[(2, 6), (3, 1), (643, 1)]
>>> primefactors(123456)
[2, 3, 643]
>>> sorted(factorint(10000000001, limit=200).items())
[(101, 1), (99009901, 1)]
>>> isprime(99009901)
False
>>> primefactors(10000000001, limit=300)
[101]
See Also
========
divisors
"""
n = int(n)
factors = sorted(factorint(n, limit=limit, verbose=verbose).keys())
s = [f for f in factors[:-1:] if f not in [-1, 0, 1]]
if factors and isprime(factors[-1]):
s += [factors[-1]]
return s
def _divisors(n):
"""Helper function for divisors which generates the divisors."""
factordict = factorint(n)
ps = sorted(factordict.keys())
def rec_gen(n=0):
if n == len(ps):
yield 1
else:
pows = [1]
for j in xrange(factordict[ps[n]]):
pows.append(pows[-1] * ps[n])
for q in rec_gen(n + 1):
for p in pows:
yield p * q
for p in rec_gen():
yield p
[docs]def divisors(n, generator=False):
r"""
Return all divisors of n sorted from 1..n by default.
If generator is True an unordered generator is returned.
The number of divisors of n can be quite large if there are many
prime factors (counting repeated factors). If only the number of
factors is desired use divisor_count(n).
Examples
========
>>> from sympy import divisors, divisor_count
>>> divisors(24)
[1, 2, 3, 4, 6, 8, 12, 24]
>>> divisor_count(24)
8
>>> list(divisors(120, generator=True))
[1, 2, 4, 8, 3, 6, 12, 24, 5, 10, 20, 40, 15, 30, 60, 120]
This is a slightly modified version of Tim Peters referenced at:
http://stackoverflow.com/questions/1010381/python-factorization
See Also
========
primefactors, factorint, divisor_count
"""
n = as_int(abs(n))
if isprime(n):
return [1, n]
if n == 1:
return [1]
if n == 0:
return []
rv = _divisors(n)
if not generator:
return sorted(rv)
return rv
[docs]def divisor_count(n, modulus=1):
"""
Return the number of divisors of ``n``. If ``modulus`` is not 1 then only
those that are divisible by ``modulus`` are counted.
References
==========
- http://www.mayer.dial.pipex.com/maths/formulae.htm
>>> from sympy import divisor_count
>>> divisor_count(6)
4
See Also
========
factorint, divisors, totient
"""
if not modulus:
return 0
elif modulus != 1:
n, r = divmod(n, modulus)
if r:
return 0
if n == 0:
return 0
return Mul(*[v + 1 for k, v in factorint(n).items() if k > 1])
def _antidivisors(n):
"""Helper function for antidivisors which generates the antidivisors."""
for d in _divisors(n):
y = 2*d
if n > y and n % y:
yield y
for d in _divisors(2*n-1):
if n > d >= 2 and n % d:
yield d
for d in _divisors(2*n+1):
if n > d >= 2 and n % d:
yield d
def antidivisors(n, generator=False):
r"""
Return all antidivisors of n sorted from 1..n by default.
Antidivisors [1]_ of n are numbers that do not divide n by the largest
possible margin. If generator is True an unordered generator is returned.
References
==========
.. [1] definition is described in http://oeis.org/A066272/a066272a.html
Examples
========
>>> from sympy.ntheory.factor_ import antidivisors
>>> antidivisors(24)
[7, 16]
>>> sorted(antidivisors(128, generator=True))
[3, 5, 15, 17, 51, 85]
See Also
========
primefactors, factorint, divisors, divisor_count, antidivisor_count
"""
n = as_int(abs(n))
if n <= 2:
return []
rv = _antidivisors(n)
if not generator:
return sorted(rv)
return rv
def antidivisor_count(n):
"""
Return the number of antidivisors [1]_ of ``n``.
References
==========
.. [1] formula from https://oeis.org/A066272
Examples
========
>>> from sympy.ntheory.factor_ import antidivisor_count
>>> antidivisor_count(13)
4
>>> antidivisor_count(27)
5
See Also
========
factorint, divisors, antidivisors, divisor_count, totient
"""
n = as_int(abs(n))
if n <= 2:
return 0
return divisor_count(2*n-1) + divisor_count(2*n+1) + \
divisor_count(n) - divisor_count(n, 2) - 5
[docs]class totient(Function):
"""
Calculate the Euler totient function phi(n)
>>> from sympy.ntheory import totient
>>> totient(1)
1
>>> totient(25)
20
See Also
========
divisor_count
"""
@classmethod
def eval(cls, n):
n = sympify(n)
if n.is_Integer:
if n < 1:
raise ValueError("n must be a positive integer")
factors = factorint(n)
t = 1
for p, k in factors.items():
t *= (p - 1) * p**(k - 1)
return t
class divisor_sigma(Function):
"""
Calculate the divisor function `\sigma_k(n)` for positive integer n
``divisor_sigma(n, k)`` is equal to ``sum([x**k for x in divisors(n)])``
If n's prime factorization is:
.. math ::
n = \prod_{i=1}^\omega p_i^{m_i},
then
.. math ::
\sigma_k(n) = \prod_{i=1}^\omega (1+p_i^k+p_i^{2k}+\cdots
+ p_i^{m_ik}).
Parameters
==========
k : power of divisors in the sum
for k = 0, 1:
``divisor_sigma(n, 0)`` is equal to ``divisor_count(n)``
``divisor_sigma(n, 1)`` is equal to ``sum(divisors(n))``
Default for k is 1.
References
==========
.. [1] http://en.wikipedia.org/wiki/Divisor_function
Examples
========
>>> from sympy.ntheory import divisor_sigma
>>> divisor_sigma(18, 0)
6
>>> divisor_sigma(39, 1)
56
>>> divisor_sigma(12, 2)
210
>>> divisor_sigma(37)
38
See Also
========
divisor_count, totient, divisors, factorint
"""
@classmethod
def eval(cls, n, k=1):
n = sympify(n)
k = sympify(k)
if n.is_prime:
return 1 + n**k
if n.is_Integer:
if n <= 0:
raise ValueError("n must be a positive integer")
else:
return Mul(*[(p**(k*(e + 1)) - 1)/(p**k - 1) if k != 0
else e + 1 for p, e in factorint(n).items()])
```