Source code for sympy.ntheory.factor_

"""
Integer factorization
"""
import random
import math

from sympy.core import Mul
from sympy.core.evalf import bitcount
from sympy.core.numbers import igcd
from sympy.core.power import integer_nthroot, Pow
from sympy.core.mul import Mul
from sympy.core.compatibility import as_int
from .primetest import isprime
from .generate import sieve, primerange, nextprime
from sympy.core.singleton import S

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 + n) 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 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 list(facs.items())]], key=lambda x: (x[1][k], x[0]))) else: rv = n if visual is False or (visual != 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 type(z) is int: 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 >>> [multiplicity(5, n) for n in [8, 5, 25, 125, 250]] [0, 1, 2, 3, 3] """ p, n = as_int(p), as_int(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, end=' ') ... 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 - http://www.csh.rit.edu/~pat/math/quickies/rho/ """ 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.math.mcgill.ca/darmon/courses/05-06/usra/charest.pdf - 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 list(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 >>> a = 1407633717262338957430697921446883 >>> f = factorint(a, limit=10000) >>> f == {991: 1, 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, **dict(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(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 list(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 = 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: if 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) s = [] 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 range(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 = abs(n) if isprime(n): return [1, n] elif n == 1: return [1] elif n == 0: return [] else: 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 list(factorint(n).items()) if k > 1])
[docs]def totient(n): """ Calculate the Euler totient function phi(n) >>> from sympy.ntheory import totient >>> totient(1) 1 >>> totient(25) 20 See Also ======== divisor_count """ n = as_int(n) 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