modelparameters.sympy.ntheory package

Submodules

modelparameters.sympy.ntheory.bbp_pi module

This implementation is a heavily modified fixed point implementation of BBP_formula for calculating the nth position of pi. The original hosted at: http://en.literateprograms.org/Pi_with_the_BBP_formula_(Python)

# Permission is hereby granted, free of charge, to any person obtaining # a copy of this software and associated documentation files (the # “Software”), to deal in the Software without restriction, including # without limitation the rights to use, copy, modify, merge, publish, # distribute, sub-license, and/or sell copies of the Software, and to # permit persons to whom the Software is furnished to do so, subject to # the following conditions: # # The above copyright notice and this permission notice shall be # included in all copies or substantial portions of the Software. # # THE SOFTWARE IS PROVIDED “AS IS”, WITHOUT WARRANTY OF ANY KIND, # EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF # MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. # IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY # CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, # TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE # SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.

Modifications:

1.Once the nth digit and desired number of digits is selected, the number of digits of working precision is calculated to ensure that the hexadecimal digits returned are accurate. This is calculated as

int(math.log(start + prec)/math.log(16) + prec + 3) ————————————— ——–

/ /

number of hex digits additional digits

This was checked by the following code which completed without errors (and dig are the digits included in the test_bbp.py file):

for i in range(0,1000):
for j in range(1,1000):

a, b = pi_hex_digits(i, j), dig[i:i+j] if a != b:

print(‘%s

%s’%(a,b))

Deceasing the additional digits by 1 generated errors, so ‘3’ is the smallest additional precision needed to calculate the above loop without errors. The following trailing 10 digits were also checked to be accurate (and the times were slightly faster with some of the constant modifications that were made):

>> from time import time >> t=time();pi_hex_digits(10**2-10 + 1, 10), time()-t (‘e90c6cc0ac’, 0.0) >> t=time();pi_hex_digits(10**4-10 + 1, 10), time()-t (‘26aab49ec6’, 0.17100000381469727) >> t=time();pi_hex_digits(10**5-10 + 1, 10), time()-t (‘a22673c1a5’, 4.7109999656677246) >> t=time();pi_hex_digits(10**6-10 + 1, 10), time()-t (‘9ffd342362’, 59.985999822616577) >> t=time();pi_hex_digits(10**7-10 + 1, 10), time()-t (‘c1a42e06a1’, 689.51800012588501)

2. The while loop to evaluate whether the series has converged quits when the addition amount dt has dropped to zero.

3. the formatting string to convert the decimal to hexidecimal is calculated for the given precision.

4. pi_hex_digits(n) changed to have coefficient to the formula in an array (perhaps just a matter of preference).

modelparameters.sympy.ntheory.bbp_pi.pi_hex_digits(n, prec=14)[source]

Returns a string containing prec (default 14) digits starting at the nth digit of pi in hex. Counting of digits starts at 0 and the decimal is not counted, so for n = 0 the returned value starts with 3; n = 1 corresponds to the first digit past the decimal point (which in hex is 2).

Examples

>>> from .bbp_pi import pi_hex_digits
>>> pi_hex_digits(0)
'3243f6a8885a30'
>>> pi_hex_digits(0, 3)
'324'

References

modelparameters.sympy.ntheory.continued_fraction module

modelparameters.sympy.ntheory.continued_fraction.continued_fraction_convergents(cf)[source]

Return an iterator over the convergents of a continued fraction (cf).

The parameter should be an iterable returning successive partial quotients of the continued fraction, such as might be returned by continued_fraction_iterator. In computing the convergents, the continued fraction need not be strictly in canonical form (all integers, all but the first positive). Rational and negative elements may be present in the expansion.

Examples

>>> from ..core import Rational, pi
>>> from .. import S
>>> from .continued_fraction import             continued_fraction_convergents, continued_fraction_iterator

>>> list(continued_fraction_convergents([0, 2, 1, 2]))
[0, 1/2, 1/3, 3/8]

>>> list(continued_fraction_convergents([1, S('1/2'), -7, S('1/4')]))
[1, 3, 19/5, 7]

>>> it = continued_fraction_convergents(continued_fraction_iterator(pi))
>>> for n in range(7):
...     print(next(it))
3
22/7
333/106
355/113
103993/33102
104348/33215
208341/66317

See also

continued_fraction_iterator

modelparameters.sympy.ntheory.continued_fraction.continued_fraction_iterator(x)[source]

Return continued fraction expansion of x as iterator.

Examples

>>> from ..core import Rational, pi
>>> from .continued_fraction import continued_fraction_iterator

>>> list(continued_fraction_iterator(Rational(3, 8)))
[0, 2, 1, 2]
>>> list(continued_fraction_iterator(Rational(-3, 8)))
[-1, 1, 1, 1, 2]

>>> for i, v in enumerate(continued_fraction_iterator(pi)):
...     if i > 7:
...         break
...     print(v)
3
7
15
1
292
1
1
1

References

modelparameters.sympy.ntheory.continued_fraction.continued_fraction_periodic(p, q, d=0)[source]

Find the periodic continued fraction expansion of a quadratic irrational.

Compute the continued fraction expansion of a rational or a quadratic irrational number, i.e. frac{p + sqrt{d}}{q}, where p, q and d ge 0 are integers.

Returns the continued fraction representation (canonical form) as a list of integers, optionally ending (for quadratic irrationals) with repeating block as the last term of this list.

Parameters:

  • p (int) – the rational part of the number’s numerator

  • q (int) – the denominator of the number

  • d (int, optional) – the irrational part (discriminator) of the number’s numerator

Examples

>>> from .continued_fraction import continued_fraction_periodic
>>> continued_fraction_periodic(3, 2, 7)
[2, [1, 4, 1, 1]]

Golden ratio has the simplest continued fraction expansion:

>>> continued_fraction_periodic(1, 2, 5)
[[1]]

If the discriminator is zero or a perfect square then the number will be a rational number:

>>> continued_fraction_periodic(4, 3, 0)
[1, 3]
>>> continued_fraction_periodic(4, 3, 49)
[3, 1, 2]

See also

continued_fraction_iterator, continued_fraction_reduce

References

modelparameters.sympy.ntheory.continued_fraction.continued_fraction_reduce(cf)[source]

Reduce a continued fraction to a rational or quadratic irrational.

Compute the rational or quadratic irrational number from its terminating or periodic continued fraction expansion. The continued fraction expansion (cf) should be supplied as a terminating iterator supplying the terms of the expansion. For terminating continued fractions, this is equivalent to list(continued_fraction_convergents(cf))[-1], only a little more efficient. If the expansion has a repeating part, a list of the repeating terms should be returned as the last element from the iterator. This is the format returned by continued_fraction_periodic.

For quadratic irrationals, returns the largest solution found, which is generally the one sought, if the fraction is in canonical form (all terms positive except possibly the first).

Examples

>>> from .continued_fraction import continued_fraction_reduce
>>> continued_fraction_reduce([1, 2, 3, 4, 5])
225/157
>>> continued_fraction_reduce([-2, 1, 9, 7, 1, 2])
-256/233
>>> continued_fraction_reduce([2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8]).n(10)
2.718281835
>>> continued_fraction_reduce([1, 4, 2, [3, 1]])
(sqrt(21) + 287)/238
>>> continued_fraction_reduce([[1]])
1/2 + sqrt(5)/2
>>> from .continued_fraction import continued_fraction_periodic
>>> continued_fraction_reduce(continued_fraction_periodic(8, 5, 13))
(sqrt(13) + 8)/5

See also

continued_fraction_periodic

modelparameters.sympy.ntheory.egyptian_fraction module

modelparameters.sympy.ntheory.egyptian_fraction.egypt_golomb(x, y)[source]
modelparameters.sympy.ntheory.egyptian_fraction.egypt_graham_jewett(x, y)[source]
modelparameters.sympy.ntheory.egyptian_fraction.egypt_greedy(x, y)[source]
modelparameters.sympy.ntheory.egyptian_fraction.egypt_harmonic(r)[source]
modelparameters.sympy.ntheory.egyptian_fraction.egypt_takenouchi(x, y)[source]
modelparameters.sympy.ntheory.egyptian_fraction.egyptian_fraction(r, algorithm='Greedy')[source]

Return the list of denominators of an Egyptian fraction expansion [1]_ of the said rational r.

Parameters:

  • r (Rational) – a positive rational number.

  • algorithm ({ "Greedy", "Graham Jewett", "Takenouchi", "Golomb" }, optional) – Denotes the algorithm to be used (the default is “Greedy”).

Examples

>>> from .. import Rational
>>> from .egyptian_fraction import egyptian_fraction
>>> egyptian_fraction(Rational(3, 7))
[3, 11, 231]
>>> egyptian_fraction(Rational(3, 7), "Graham Jewett")
[7, 8, 9, 56, 57, 72, 3192]
>>> egyptian_fraction(Rational(3, 7), "Takenouchi")
[4, 7, 28]
>>> egyptian_fraction(Rational(3, 7), "Golomb")
[3, 15, 35]
>>> egyptian_fraction(Rational(11, 5), "Golomb")
[1, 2, 3, 4, 9, 234, 1118, 2580]

See also

sympy.core.numbers.Rational

Notes

Currently the following algorithms are supported:

  1. Greedy Algorithm

    Also called the Fibonacci-Sylvester algorithm [2]_. At each step, extract the largest unit fraction less than the target and replace the target with the remainder.

    It has some distinct properties:

    1. Given p/q in lowest terms, generates an expansion of maximum length p. Even as the numerators get large, the number of terms is seldom more than a handful.

    2. Uses minimal memory.

    3. The terms can blow up (standard examples of this are 5/121 and 31/311). The denominator is at most squared at each step (doubly-exponential growth) and typically exhibits singly-exponential growth.

  2. Graham Jewett Algorithm

    The algorithm suggested by the result of Graham and Jewett. Note that this has a tendency to blow up: the length of the resulting expansion is always 2**(x/gcd(x, y)) - 1. See [3].

  3. Takenouchi Algorithm

    The algorithm suggested by Takenouchi (1921). Differs from the Graham-Jewett algorithm only in the handling of duplicates. See [3].

  4. Golomb’s Algorithm

    A method given by Golumb (1962), using modular arithmetic and inverses. It yields the same results as a method using continued fractions proposed by Bleicher (1972). See [4].

If the given rational is greater than or equal to 1, a greedy algorithm of summing the harmonic sequence 1/1 + 1/2 + 1/3 + … is used, taking all the unit fractions of this sequence until adding one more would be greater than the given number. This list of denominators is prefixed to the result from the requested algorithm used on the remainder. For example, if r is 8/3, using the Greedy algorithm, we get [1, 2, 3, 4, 5, 6, 7, 14, 420], where the beginning of the sequence, [1, 2, 3, 4, 5, 6, 7] is part of the harmonic sequence summing to 363/140, leaving a remainder of 31/420, which yields [14, 420] by the Greedy algorithm. The result of egyptian_fraction(Rational(8, 3), “Golomb”) is [1, 2, 3, 4, 5, 6, 7, 14, 574, 2788, 6460, 11590, 33062, 113820], and so on.

References

modelparameters.sympy.ntheory.factor_ module

Integer factorization

modelparameters.sympy.ntheory.factor_.antidivisor_count(n)[source]

Return the number of antidivisors [1]_ of n.

References

Examples

>>> from .factor_ import antidivisor_count
>>> antidivisor_count(13)
4
>>> antidivisor_count(27)
5

See also

factorint, divisors, antidivisors, divisor_count, totient

modelparameters.sympy.ntheory.factor_.antidivisors(n, generator=False)[source]

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

Examples

>>> from .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

modelparameters.sympy.ntheory.factor_.core(n, t=2)[source]

Calculate core(n,t) = core_t(n) of a positive integer n

core_2(n) is equal to the squarefree part of n

If n’s prime factorization is:

\[n = \prod_{i=1}^\omega p_i^{m_i},\]

then

\[core_t(n) = \prod_{i=1}^\omega p_i^{m_i \mod t}.\]
Parameters:

t (core(n,t) calculates the t-th power free part of n) –

core(n, 2) is the squarefree part of n core(n, 3) is the cubefree part of n

Default for t is 2.

References

Examples

>>> from .factor_ import core
>>> core(24, 2)
6
>>> core(9424, 3)
1178
>>> core(379238)
379238
>>> core(15**11, 10)
15

See also

factorint, sympy.solvers.diophantine.square_factor

modelparameters.sympy.ntheory.factor_.digits(n, b=10)[source]

Return a list of the digits of n in base b. The first element in the list is b (or -b if n is negative).

Examples

>>> from .factor_ import digits
>>> digits(35)
[10, 3, 5]
>>> digits(27, 2)
[2, 1, 1, 0, 1, 1]
>>> digits(65536, 256)
[256, 1, 0, 0]
>>> digits(-3958, 27)
[-27, 5, 11, 16]
modelparameters.sympy.ntheory.factor_.divisor_count(n, modulus=1)[source]

Return the number of divisors of n. If modulus is not 1 then only those that are divisible by modulus are counted.

References

>>> from .. import divisor_count
>>> divisor_count(6)
4

See also

factorint, divisors, totient

class modelparameters.sympy.ntheory.factor_.divisor_sigma(n, k=1)[source]

Bases: 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:

\[n = \prod_{i=1}^\omega p_i^{m_i},\]

then

\[\sigma_k(n) = \prod_{i=1}^\omega (1+p_i^k+p_i^{2k}+\cdots + p_i^{m_ik}).\]
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

Examples

>>> from ..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

default_assumptions = {}
classmethod eval(n, k=1)[source]

Returns a canonical form of cls applied to arguments args.

The eval() method is called when the class cls is about to be instantiated and it should return either some simplified instance (possible of some other class), or if the class cls should be unmodified, return None.

Examples of eval() for the function “sign”

@classmethod def eval(cls, arg):

if arg is S.NaN:

return S.NaN

if arg is S.Zero: return S.Zero if arg.is_positive: return S.One if arg.is_negative: return S.NegativeOne if isinstance(arg, Mul):

coeff, terms = arg.as_coeff_Mul(rational=True) if coeff is not S.One:

return cls(coeff) * cls(terms)

modelparameters.sympy.ntheory.factor_.divisors(n, generator=False)[source]

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 .. 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

modelparameters.sympy.ntheory.factor_.factorint(n, limit=None, use_trial=True, use_rho=True, use_pm1=True, verbose=False, visual=None, multiple=False)[source]

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 ..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 ..ntheory import isprime
>>> from ..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}

List of Factors:

If multiple is set to True then a list containing the prime factors including multiplicities is returned.

>>> factorint(24, multiple=True)
[2, 2, 2, 3]

Visual Factorization:

If visual is set to True, then it will return a visual factorization of the integer. For example:

>>> from .. 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 .. 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:

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

modelparameters.sympy.ntheory.factor_.factorrat(rat, limit=None, use_trial=True, use_rho=True, use_pm1=True, verbose=False, visual=None, multiple=False)[source]

Given a Rational r, factorrat(r) returns a dict containing the prime factors of r as keys and their respective multiplicities as values. For example:

>>> from ..ntheory import factorrat
>>> from ..core.symbol import S
>>> factorrat(S(8)/9)    # 8/9 = (2**3) * (3**-2)
{2: 3, 3: -2}
>>> factorrat(S(-1)/987)    # -1/789 = -1 * (3**-1) * (7**-1) * (47**-1)
{-1: 1, 3: -1, 7: -1, 47: -1}

Please see the docstring for factorint for detailed explanations and examples of the following keywords:

  • limit: Integer limit up to which trial division is done

  • 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

  • verbose: Toggle detailed printing of progress

  • multiple: Toggle returning a list of factors or dict

  • visual: Toggle product form of output

modelparameters.sympy.ntheory.factor_.multiplicity(p, n)[source]

Find the greatest integer m such that p**m divides n.

Examples

>>> from ..ntheory import multiplicity
>>> from ..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
modelparameters.sympy.ntheory.factor_.perfect_power(n, candidates=None, big=True, factor=True)[source]

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 .. import perfect_power
>>> perfect_power(16)
(2, 4)
>>> perfect_power(16, big = False)
(4, 2)
modelparameters.sympy.ntheory.factor_.pollard_pm1(n, B=10, a=2, retries=0, seed=1234)[source]

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 .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 ..core.numbers import ilcm, igcd
>>> from .. 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])
{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 ..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 ..utilities import flatten
>>> from .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

modelparameters.sympy.ntheory.factor_.pollard_rho(n, s=2, a=1, retries=5, seed=1234, max_steps=None, F=None)[source]

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 .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 .. 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

modelparameters.sympy.ntheory.factor_.primefactors(n, limit=None, verbose=False)[source]

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 ..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

class modelparameters.sympy.ntheory.factor_.primenu(n)[source]

Bases: Function

Calculate the number of distinct prime factors for a positive integer n.

If n’s prime factorization is:

\[n = \prod_{i=1}^k p_i^{m_i},\]

then primenu(n) or nu(n) is:

\[\nu(n) = k.\]

References

Examples

>>> from .factor_ import primenu
>>> primenu(1)
0
>>> primenu(30)
3

See also

factorint

default_assumptions = {}
classmethod eval(n)[source]

Returns a canonical form of cls applied to arguments args.

The eval() method is called when the class cls is about to be instantiated and it should return either some simplified instance (possible of some other class), or if the class cls should be unmodified, return None.

Examples of eval() for the function “sign”

@classmethod def eval(cls, arg):

if arg is S.NaN:

return S.NaN

if arg is S.Zero: return S.Zero if arg.is_positive: return S.One if arg.is_negative: return S.NegativeOne if isinstance(arg, Mul):

coeff, terms = arg.as_coeff_Mul(rational=True) if coeff is not S.One:

return cls(coeff) * cls(terms)

class modelparameters.sympy.ntheory.factor_.primeomega(n)[source]

Bases: Function

Calculate the number of prime factors counting multiplicities for a positive integer n.

If n’s prime factorization is:

\[n = \prod_{i=1}^k p_i^{m_i},\]

then primeomega(n) or Omega(n) is:

\[\Omega(n) = \sum_{i=1}^k m_i.\]

References

Examples

>>> from .factor_ import primeomega
>>> primeomega(1)
0
>>> primeomega(20)
3

See also

factorint

default_assumptions = {}
classmethod eval(n)[source]

Returns a canonical form of cls applied to arguments args.

The eval() method is called when the class cls is about to be instantiated and it should return either some simplified instance (possible of some other class), or if the class cls should be unmodified, return None.

Examples of eval() for the function “sign”

@classmethod def eval(cls, arg):

if arg is S.NaN:

return S.NaN

if arg is S.Zero: return S.Zero if arg.is_positive: return S.One if arg.is_negative: return S.NegativeOne if isinstance(arg, Mul):

coeff, terms = arg.as_coeff_Mul(rational=True) if coeff is not S.One:

return cls(coeff) * cls(terms)

class modelparameters.sympy.ntheory.factor_.reduced_totient(n)[source]

Bases: Function

Calculate the Carmichael reduced totient function lambda(n)

reduced_totient(n) or lambda(n) is the smallest m > 0 such that k^m equiv 1 mod n for all k relatively prime to n.

References

Examples

>>> from ..ntheory import reduced_totient
>>> reduced_totient(1)
1
>>> reduced_totient(8)
2
>>> reduced_totient(30)
4

See also

totient

default_assumptions = {}
classmethod eval(n)[source]

Returns a canonical form of cls applied to arguments args.

The eval() method is called when the class cls is about to be instantiated and it should return either some simplified instance (possible of some other class), or if the class cls should be unmodified, return None.

Examples of eval() for the function “sign”

@classmethod def eval(cls, arg):

if arg is S.NaN:

return S.NaN

if arg is S.Zero: return S.Zero if arg.is_positive: return S.One if arg.is_negative: return S.NegativeOne if isinstance(arg, Mul):

coeff, terms = arg.as_coeff_Mul(rational=True) if coeff is not S.One:

return cls(coeff) * cls(terms)

modelparameters.sympy.ntheory.factor_.smoothness(n)[source]

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 .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

modelparameters.sympy.ntheory.factor_.smoothness_p(n, m=-1, power=0, visual=None)[source]

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 .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

class modelparameters.sympy.ntheory.factor_.totient(n)[source]

Bases: Function

Calculate the Euler totient function phi(n)

totient(n) or phi(n) is the number of positive integers leq n that are relatively prime to n.

References

Examples

>>> from ..ntheory import totient
>>> totient(1)
1
>>> totient(25)
20

See also

divisor_count

default_assumptions = {}
classmethod eval(n)[source]

Returns a canonical form of cls applied to arguments args.

The eval() method is called when the class cls is about to be instantiated and it should return either some simplified instance (possible of some other class), or if the class cls should be unmodified, return None.

Examples of eval() for the function “sign”

@classmethod def eval(cls, arg):

if arg is S.NaN:

return S.NaN

if arg is S.Zero: return S.Zero if arg.is_positive: return S.One if arg.is_negative: return S.NegativeOne if isinstance(arg, Mul):

coeff, terms = arg.as_coeff_Mul(rational=True) if coeff is not S.One:

return cls(coeff) * cls(terms)

modelparameters.sympy.ntheory.factor_.trailing(n)[source]

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 .. import trailing
>>> trailing(128)
7
>>> trailing(63)
0
modelparameters.sympy.ntheory.factor_.udivisor_count(n)[source]

Return the number of unitary divisors of n.

References

>>> from .factor_ import udivisor_count
>>> udivisor_count(120)
8

See also

factorint, divisors, udivisors, divisor_count, totient

class modelparameters.sympy.ntheory.factor_.udivisor_sigma(n, k=1)[source]

Bases: Function

Calculate the unitary divisor function sigma_k^*(n) for positive integer n

udivisor_sigma(n, k) is equal to sum([x**k for x in udivisors(n)])

If n’s prime factorization is:

\[n = \prod_{i=1}^\omega p_i^{m_i},\]

then

\[\sigma_k^*(n) = \prod_{i=1}^\omega (1+ p_i^{m_ik}).\]
Parameters:

k (power of divisors in the sum) –

for k = 0, 1: udivisor_sigma(n, 0) is equal to udivisor_count(n) udivisor_sigma(n, 1) is equal to sum(udivisors(n))

Default for k is 1.

References

Examples

>>> from .factor_ import udivisor_sigma
>>> udivisor_sigma(18, 0)
4
>>> udivisor_sigma(74, 1)
114
>>> udivisor_sigma(36, 3)
47450
>>> udivisor_sigma(111)
152

See also

divisor_count, totient, divisors, udivisors, udivisor_count, divisor_sigma, factorint

default_assumptions = {}
classmethod eval(n, k=1)[source]

Returns a canonical form of cls applied to arguments args.

The eval() method is called when the class cls is about to be instantiated and it should return either some simplified instance (possible of some other class), or if the class cls should be unmodified, return None.

Examples of eval() for the function “sign”

@classmethod def eval(cls, arg):

if arg is S.NaN:

return S.NaN

if arg is S.Zero: return S.Zero if arg.is_positive: return S.One if arg.is_negative: return S.NegativeOne if isinstance(arg, Mul):

coeff, terms = arg.as_coeff_Mul(rational=True) if coeff is not S.One:

return cls(coeff) * cls(terms)

modelparameters.sympy.ntheory.factor_.udivisors(n, generator=False)[source]

Return all unitary divisors of n sorted from 1..n by default. If generator is True an unordered generator is returned.

The number of unitary divisors of n can be quite large if there are many prime factors. If only the number of unitary divisors is desired use udivisor_count(n).

References

Examples

>>> from .factor_ import udivisors, udivisor_count
>>> udivisors(15)
[1, 3, 5, 15]
>>> udivisor_count(15)
4

>>> sorted(udivisors(120, generator=True))
[1, 3, 5, 8, 15, 24, 40, 120]

See also

primefactors, factorint, divisors, divisor_count, udivisor_count

modelparameters.sympy.ntheory.generate module

Generating and counting primes.

class modelparameters.sympy.ntheory.generate.Sieve[source]

Bases: object

An infinite list of prime numbers, implemented as a dynamically growing sieve of Eratosthenes. When a lookup is requested involving an odd number that has not been sieved, the sieve is automatically extended up to that number.

Examples

>>> from .. import sieve
>>> sieve._reset() # this line for doctest only
>>> 25 in sieve
False
>>> sieve._list
array('l', [2, 3, 5, 7, 11, 13, 17, 19, 23])
extend(n)[source]

Grow the sieve to cover all primes <= n (a real number).

Examples

>>> from .. import sieve
>>> sieve._reset() # this line for doctest only
>>> sieve.extend(30)
>>> sieve[10] == 29
True
extend_to_no(i)[source]

Extend to include the ith prime number.

i must be an integer.

The list is extended by 50% if it is too short, so it is likely that it will be longer than requested.

Examples

>>> from .. import sieve
>>> sieve._reset() # this line for doctest only
>>> sieve.extend_to_no(9)
>>> sieve._list
array('l', [2, 3, 5, 7, 11, 13, 17, 19, 23])
primerange(a, b)[source]

Generate all prime numbers in the range [a, b).

Examples

>>> from .. import sieve
>>> print([i for i in sieve.primerange(7, 18)])
[7, 11, 13, 17]
search(n)[source]

Return the indices i, j of the primes that bound n.

If n is prime then i == j.

Although n can be an expression, if ceiling cannot convert it to an integer then an n error will be raised.

Examples

>>> from .. import sieve
>>> sieve.search(25)
(9, 10)
>>> sieve.search(23)
(9, 9)
modelparameters.sympy.ntheory.generate.composite(nth)[source]

Return the nth composite number, with the composite numbers indexed as composite(1) = 4, composite(2) = 6, etc….

Examples

>>> from .. import composite
>>> composite(36)
52
>>> composite(1)
4
>>> composite(17737)
20000

See also

sympy.ntheory.primetest.isprime

Test if n is prime

primerange

Generate all primes in a given range

primepi

Return the number of primes less than or equal to n

prime

Return the nth prime

compositepi

Return the number of positive composite numbers less than or equal to n

modelparameters.sympy.ntheory.generate.compositepi(n)[source]

Return the number of positive composite numbers less than or equal to n. The first positive composite is 4, i.e. compositepi(4) = 1.

Examples

>>> from .. import compositepi
>>> compositepi(25)
15
>>> compositepi(1000)
831

See also

sympy.ntheory.primetest.isprime

Test if n is prime

primerange

Generate all primes in a given range

prime

Return the nth prime

primepi

Return the number of primes less than or equal to n

composite

Return the nth composite number

modelparameters.sympy.ntheory.generate.cycle_length(f, x0, nmax=None, values=False)[source]

For a given iterated sequence, return a generator that gives the length of the iterated cycle (lambda) and the length of terms before the cycle begins (mu); if values is True then the terms of the sequence will be returned instead. The sequence is started with value x0.

Note: more than the first lambda + mu terms may be returned and this is the cost of cycle detection with Brent’s method; there are, however, generally less terms calculated than would have been calculated if the proper ending point were determined, e.g. by using Floyd’s method.

>>> from .generate import cycle_length

This will yield successive values of i <– func(i):

>>> def iter(func, i):
...     while 1:
...         ii = func(i)
...         yield ii
...         i = ii
...

A function is defined:

>>> func = lambda i: (i**2 + 1) % 51

and given a seed of 4 and the mu and lambda terms calculated:

>>> next(cycle_length(func, 4))
(6, 2)

We can see what is meant by looking at the output:

>>> n = cycle_length(func, 4, values=True)
>>> list(ni for ni in n)
[17, 35, 2, 5, 26, 14, 44, 50, 2, 5, 26, 14]

There are 6 repeating values after the first 2.

If a sequence is suspected of being longer than you might wish, nmax can be used to exit early (and mu will be returned as None):

>>> next(cycle_length(func, 4, nmax = 4))
(4, None)
>>> [ni for ni in cycle_length(func, 4, nmax = 4, values=True)]
[17, 35, 2, 5]
Code modified from:

http://en.wikipedia.org/wiki/Cycle_detection.

modelparameters.sympy.ntheory.generate.nextprime(n, ith=1)[source]

Return the ith prime greater than n.

i must be an integer.

Notes

Potential primes are located at 6*j +/- 1. This property is used during searching.

>>> from .. import nextprime
>>> [(i, nextprime(i)) for i in range(10, 15)]
[(10, 11), (11, 13), (12, 13), (13, 17), (14, 17)]
>>> nextprime(2, ith=2) # the 2nd prime after 2
5

See also

prevprime

Return the largest prime smaller than n

primerange

Generate all primes in a given range

modelparameters.sympy.ntheory.generate.prevprime(n)[source]

Return the largest prime smaller than n.

Notes

Potential primes are located at 6*j +/- 1. This property is used during searching.

>>> from .. import prevprime
>>> [(i, prevprime(i)) for i in range(10, 15)]
[(10, 7), (11, 7), (12, 11), (13, 11), (14, 13)]

See also

nextprime

Return the ith prime greater than n

primerange

Generates all primes in a given range

modelparameters.sympy.ntheory.generate.prime(nth)[source]

Return the nth prime, with the primes indexed as prime(1) = 2, prime(2) = 3, etc…. The nth prime is approximately n*log(n).

Logarithmic integral of x is a pretty nice approximation for number of primes <= x, i.e. li(x) ~ pi(x) In fact, for the numbers we are concerned about( x<1e11 ), li(x) - pi(x) < 50000

Also, li(x) > pi(x) can be safely assumed for the numbers which can be evaluated by this function.

Here, we find the least integer m such that li(m) > n using binary search. Now pi(m-1) < li(m-1) <= n,

We find pi(m - 1) using primepi function.

Starting from m, we have to find n - pi(m-1) more primes.

For the inputs this implementation can handle, we will have to test primality for at max about 10**5 numbers, to get our answer.

References

Examples

>>> from .. import prime
>>> prime(10)
29
>>> prime(1)
2
>>> prime(100000)
1299709

See also

sympy.ntheory.primetest.isprime

Test if n is prime

primerange

Generate all primes in a given range

primepi

Return the number of primes less than or equal to n

modelparameters.sympy.ntheory.generate.primepi(n)[source]

Return the value of the prime counting function pi(n) = the number of prime numbers less than or equal to n.

Algorithm Description:

In sieve method, we remove all multiples of prime p except p itself.

Let phi(i,j) be the number of integers 2 <= k <= i which remain after sieving from primes less than or equal to j. Clearly, pi(n) = phi(n, sqrt(n))

If j is not a prime, phi(i,j) = phi(i, j - 1)

if j is a prime, We remove all numbers(except j) whose smallest prime factor is j.

Let x= j*a be such a number, where 2 <= a<= i / j Now, after sieving from primes <= j - 1, a must remain (because x, and hence a has no prime factor <= j - 1) Clearly, there are phi(i / j, j - 1) such a which remain on sieving from primes <= j - 1

Now, if a is a prime less than equal to j - 1, x= j*a has smallest prime factor = a, and has already been removed(by sieving from a). So, we don’t need to remove it again. (Note: there will be pi(j - 1) such x)

Thus, number of x, that will be removed are: phi(i / j, j - 1) - phi(j - 1, j - 1) (Note that pi(j - 1) = phi(j - 1, j - 1))

=> phi(i,j) = phi(i, j - 1) - phi(i / j, j - 1) + phi(j - 1, j - 1)

So,following recursion is used and implemented as dp:

phi(a, b) = phi(a, b - 1), if b is not a prime phi(a, b) = phi(a, b-1)-phi(a / b, b-1) + phi(b-1, b-1), if b is prime

Clearly a is always of the form floor(n / k), which can take at most 2*sqrt(n) values. Two arrays arr1,arr2 are maintained arr1[i] = phi(i, j), arr2[i] = phi(n // i, j)

Finally the answer is arr2[1]

Examples

>>> from .. import primepi
>>> primepi(25)
9

See also

sympy.ntheory.primetest.isprime

Test if n is prime

primerange

Generate all primes in a given range

prime

Return the nth prime

modelparameters.sympy.ntheory.generate.primerange(a, b)[source]

Generate a list of all prime numbers in the range [a, b).

If the range exists in the default sieve, the values will be returned from there; otherwise values will be returned but will not modify the sieve.

Notes

Some famous conjectures about the occurence of primes in a given range are [1]:

  • Twin primes: though often not, the following will give 2 primes
    an infinite number of times:

    primerange(6*n - 1, 6*n + 2)

  • Legendre’s: the following always yields at least one prime

    primerange(n**2, (n+1)**2+1)

  • Bertrand’s (proven): there is always a prime in the range

    primerange(n, 2*n)

  • Brocard’s: there are at least four primes in the range

    primerange(prime(n)**2, prime(n+1)**2)

The average gap between primes is log(n) [2]; the gap between primes can be arbitrarily large since sequences of composite numbers are arbitrarily large, e.g. the numbers in the sequence n! + 2, n! + 3 … n! + n are all composite.

References

  1. http://en.wikipedia.org/wiki/Prime_number

  2. http://primes.utm.edu/notes/gaps.html

Examples

>>> from .. import primerange, sieve
>>> print([i for i in primerange(1, 30)])
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]

The Sieve method, primerange, is generally faster but it will occupy more memory as the sieve stores values. The default instance of Sieve, named sieve, can be used:

>>> list(sieve.primerange(1, 30))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]

See also

nextprime

Return the ith prime greater than n

prevprime

Return the largest prime smaller than n

randprime

Returns a random prime in a given range

primorial

Returns the product of primes based on condition

Sieve.primerange

return range from already computed primes or extend the sieve to contain the requested range.

modelparameters.sympy.ntheory.generate.primorial(n, nth=True)[source]

Returns the product of the first n primes (default) or the primes less than or equal to n (when nth=False).

>>> from .generate import primorial, randprime, primerange
>>> from .. import factorint, Mul, primefactors, sqrt
>>> primorial(4) # the first 4 primes are 2, 3, 5, 7
210
>>> primorial(4, nth=False) # primes <= 4 are 2 and 3
6
>>> primorial(1)
2
>>> primorial(1, nth=False)
1
>>> primorial(sqrt(101), nth=False)
210

One can argue that the primes are infinite since if you take a set of primes and multiply them together (e.g. the primorial) and then add or subtract 1, the result cannot be divided by any of the original factors, hence either 1 or more new primes must divide this product of primes.

In this case, the number itself is a new prime:

>>> factorint(primorial(4) + 1)
{211: 1}

In this case two new primes are the factors:

>>> factorint(primorial(4) - 1)
{11: 1, 19: 1}

Here, some primes smaller and larger than the primes multiplied together are obtained:

>>> p = list(primerange(10, 20))
>>> sorted(set(primefactors(Mul(*p) + 1)).difference(set(p)))
[2, 5, 31, 149]

See also

primerange

Generate all primes in a given range

modelparameters.sympy.ntheory.generate.randprime(a, b)[source]

Return a random prime number in the range [a, b).

Bertrand’s postulate assures that randprime(a, 2*a) will always succeed for a > 1.

References

Examples

>>> from .. import randprime, isprime
>>> randprime(1, 30) 
13
>>> isprime(randprime(1, 30))
True

See also

primerange

Generate all primes in a given range

modelparameters.sympy.ntheory.modular module

modelparameters.sympy.ntheory.modular.crt(m, v, symmetric=False, check=True)[source]

Chinese Remainder Theorem.

The moduli in m are assumed to be pairwise coprime. The output is then an integer f, such that f = v_i mod m_i for each pair out of v and m. If symmetric is False a positive integer will be returned, else |f| will be less than or equal to the LCM of the moduli, and thus f may be negative.

If the moduli are not co-prime the correct result will be returned if/when the test of the result is found to be incorrect. This result will be None if there is no solution.

The keyword check can be set to False if it is known that the moduli are coprime.

As an example consider a set of residues U = [49, 76, 65] and a set of moduli M = [99, 97, 95]. Then we have:

>>> from .modular import crt, solve_congruence

>>> crt([99, 97, 95], [49, 76, 65])
(639985, 912285)

This is the correct result because:

>>> [639985 % m for m in [99, 97, 95]]
[49, 76, 65]

If the moduli are not co-prime, you may receive an incorrect result if you use check=False:

>>> crt([12, 6, 17], [3, 4, 2], check=False)
(954, 1224)
>>> [954 % m for m in [12, 6, 17]]
[6, 0, 2]
>>> crt([12, 6, 17], [3, 4, 2]) is None
True
>>> crt([3, 6], [2, 5])
(5, 6)

Note: the order of gf_crt’s arguments is reversed relative to crt, and that solve_congruence takes residue, modulus pairs.

Programmer’s note: rather than checking that all pairs of moduli share no GCD (an O(n**2) test) and rather than factoring all moduli and seeing that there is no factor in common, a check that the result gives the indicated residuals is performed – an O(n) operation.

See also

solve_congruence

sympy.polys.galoistools.gf_crt

low level crt routine used by this routine

modelparameters.sympy.ntheory.modular.crt1(m)[source]

First part of Chinese Remainder Theorem, for multiple application.

Examples

>>> from .modular import crt1
>>> crt1([18, 42, 6])
(4536, [252, 108, 756], [0, 2, 0])
modelparameters.sympy.ntheory.modular.crt2(m, v, mm, e, s, symmetric=False)[source]

Second part of Chinese Remainder Theorem, for multiple application.

Examples

>>> from .modular import crt1, crt2
>>> mm, e, s = crt1([18, 42, 6])
>>> crt2([18, 42, 6], [0, 0, 0], mm, e, s)
(0, 4536)
modelparameters.sympy.ntheory.modular.solve_congruence(*remainder_modulus_pairs, **hint)[source]

Compute the integer n that has the residual ai when it is divided by mi where the ai and mi are given as pairs to this function: ((a1, m1), (a2, m2), …). If there is no solution, return None. Otherwise return n and its modulus.

The mi values need not be co-prime. If it is known that the moduli are not co-prime then the hint check can be set to False (default=True) and the check for a quicker solution via crt() (valid when the moduli are co-prime) will be skipped.

If the hint symmetric is True (default is False), the value of n will be within 1/2 of the modulus, possibly negative.

Examples

>>> from .modular import solve_congruence

What number is 2 mod 3, 3 mod 5 and 2 mod 7?

>>> solve_congruence((2, 3), (3, 5), (2, 7))
(23, 105)
>>> [23 % m for m in [3, 5, 7]]
[2, 3, 2]

If you prefer to work with all remainder in one list and all moduli in another, send the arguments like this:

>>> solve_congruence(*zip((2, 3, 2), (3, 5, 7)))
(23, 105)

The moduli need not be co-prime; in this case there may or may not be a solution:

>>> solve_congruence((2, 3), (4, 6)) is None
True

>>> solve_congruence((2, 3), (5, 6))
(5, 6)

The symmetric flag will make the result be within 1/2 of the modulus:

>>> solve_congruence((2, 3), (5, 6), symmetric=True)
(-1, 6)

See also

crt

high level routine implementing the Chinese Remainder Theorem

modelparameters.sympy.ntheory.modular.symmetric_residue(a, m)[source]

Return the residual mod m such that it is within half of the modulus.

>>> from .modular import symmetric_residue
>>> symmetric_residue(1, 6)
1
>>> symmetric_residue(4, 6)
-2

modelparameters.sympy.ntheory.multinomial module

modelparameters.sympy.ntheory.multinomial.binomial_coefficients(n)[source]

Return a dictionary containing pairs \({(k1,k2) : C_kn}\) where \(C_kn\) are binomial coefficients and \(n=k1+k2\). .. rubric:: Examples

>>> from ..ntheory import binomial_coefficients
>>> binomial_coefficients(9)
{(0, 9): 1, (1, 8): 9, (2, 7): 36, (3, 6): 84,
 (4, 5): 126, (5, 4): 126, (6, 3): 84, (7, 2): 36, (8, 1): 9, (9, 0): 1}

See also

binomial_coefficients_list, multinomial_coefficients

modelparameters.sympy.ntheory.multinomial.binomial_coefficients_list(n)[source]

Return a list of binomial coefficients as rows of the Pascal’s triangle.

Examples

>>> from ..ntheory import binomial_coefficients_list
>>> binomial_coefficients_list(9)
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]

See also

binomial_coefficients, multinomial_coefficients

modelparameters.sympy.ntheory.multinomial.multinomial_coefficients(m, n)[source]

Return a dictionary containing pairs {(k1,k2,..,km) : C_kn} where C_kn are multinomial coefficients such that n=k1+k2+..+km.

For example:

>>> from ..ntheory import multinomial_coefficients
>>> multinomial_coefficients(2, 5) # indirect doctest
{(0, 5): 1, (1, 4): 5, (2, 3): 10, (3, 2): 10, (4, 1): 5, (5, 0): 1}

The algorithm is based on the following result:

\[\binom{n}{k_1, \ldots, k_m} = \frac{k_1 + 1}{n - k_1} \sum_{i=2}^m \binom{n}{k_1 + 1, \ldots, k_i - 1, \ldots}\]

Code contributed to Sage by Yann Laigle-Chapuy, copied with permission of the author.

See also

binomial_coefficients_list, binomial_coefficients

modelparameters.sympy.ntheory.multinomial.multinomial_coefficients0(m, n, _tuple=<class 'tuple'>, _zip=<class 'zip'>)[source]

Return a dictionary containing pairs {(k1,k2,..,km) : C_kn} where C_kn are multinomial coefficients such that n=k1+k2+..+km.

For example:

>>> from .. import multinomial_coefficients
>>> multinomial_coefficients(2, 5) # indirect doctest
{(0, 5): 1, (1, 4): 5, (2, 3): 10, (3, 2): 10, (4, 1): 5, (5, 0): 1}

The algorithm is based on the following result:

Consider a polynomial and its n-th exponent:

P(x) = sum_{i=0}^m p_i x^i
P(x)^n = sum_{k=0}^{m n} a(n,k) x^k

The coefficients a(n,k) can be computed using the J.C.P. Miller Pure Recurrence [see D.E.Knuth, Seminumerical Algorithms, The art of Computer Programming v.2, Addison Wesley, Reading, 1981;]:

a(n,k) = 1/(k p_0) sum_{i=1}^m p_i ((n+1)i-k) a(n,k-i),

where a(n,0) = p_0^n.

modelparameters.sympy.ntheory.multinomial.multinomial_coefficients_iterator(m, n, _tuple=<class 'tuple'>)[source]

multinomial coefficient iterator

This routine has been optimized for m large with respect to n by taking advantage of the fact that when the monomial tuples t are stripped of zeros, their coefficient is the same as that of the monomial tuples from multinomial_coefficients(n, n). Therefore, the latter coefficients are precomputed to save memory and time.

>>> from .multinomial import multinomial_coefficients
>>> m53, m33 = multinomial_coefficients(5,3), multinomial_coefficients(3,3)
>>> m53[(0,0,0,1,2)] == m53[(0,0,1,0,2)] == m53[(1,0,2,0,0)] == m33[(0,1,2)]
True

Examples

>>> from .multinomial import multinomial_coefficients_iterator
>>> it = multinomial_coefficients_iterator(20,3)
>>> next(it)
((3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), 1)

modelparameters.sympy.ntheory.partitions_ module

modelparameters.sympy.ntheory.partitions_.npartitions(n, verbose=False)[source]

Calculate the partition function P(n), i.e. the number of ways that n can be written as a sum of positive integers.

P(n) is computed using the Hardy-Ramanujan-Rademacher formula [1]_.

The correctness of this implementation has been tested through 10**10.

Examples

>>> from ..ntheory import npartitions
>>> npartitions(25)
1958

References

modelparameters.sympy.ntheory.primetest module

Primality testing

modelparameters.sympy.ntheory.primetest.is_extra_strong_lucas_prp(n)[source]

Extra Strong Lucas compositeness test. Returns False if n is definitely composite, and True if n is a “extra strong” Lucas probable prime.

The parameters are selected using P = 3, Q = 1, then incrementing P until (D|n) == -1. The test itself is as defined in Grantham 2000, from the Mo and Jones preprint. The parameter selection and test are the same as used in OEIS A217719, Perl’s Math::Prime::Util, and the Lucas pseudoprime page on Wikipedia.

With these parameters, there are no counterexamples below 2^64 nor any known above that range. It is 20-50% faster than the strong test.

Because of the different parameters selected, there is no relationship between the strong Lucas pseudoprimes and extra strong Lucas pseudoprimes. In particular, one is not a subset of the other.

References

Examples

>>> from .primetest import isprime, is_extra_strong_lucas_prp
>>> for i in range(20000):
...     if is_extra_strong_lucas_prp(i) and not isprime(i):
...        print(i)
989
3239
5777
10877
modelparameters.sympy.ntheory.primetest.is_lucas_prp(n)[source]

Standard Lucas compositeness test with Selfridge parameters. Returns False if n is definitely composite, and True if n is a Lucas probable prime.

This is typically used in combination with the Miller-Rabin test.

References

Examples

>>> from .primetest import isprime, is_lucas_prp
>>> for i in range(10000):
...     if is_lucas_prp(i) and not isprime(i):
...        print(i)
323
377
1159
1829
3827
5459
5777
9071
9179
modelparameters.sympy.ntheory.primetest.is_square(n, prep=True)[source]

Return True if n == a * a for some integer a, else False. If n is suspected of not being a square then this is a quick method of confirming that it is not.

References

[1] http://mersenneforum.org/showpost.php?p=110896

See also

sympy.core.power.integer_nthroot

modelparameters.sympy.ntheory.primetest.is_strong_lucas_prp(n)[source]

Strong Lucas compositeness test with Selfridge parameters. Returns False if n is definitely composite, and True if n is a strong Lucas probable prime.

This is often used in combination with the Miller-Rabin test, and in particular, when combined with M-R base 2 creates the strong BPSW test.

References

Examples

>>> from .primetest import isprime, is_strong_lucas_prp
>>> for i in range(20000):
...     if is_strong_lucas_prp(i) and not isprime(i):
...        print(i)
5459
5777
10877
16109
18971
modelparameters.sympy.ntheory.primetest.isprime(n)[source]

Test if n is a prime number (True) or not (False). For n < 2^64 the answer is definitive; larger n values have a small probability of actually being pseudoprimes.

Negative numbers (e.g. -2) are not considered prime.

The first step is looking for trivial factors, which if found enables a quick return. Next, if the sieve is large enough, use bisection search on the sieve. For small numbers, a set of deterministic Miller-Rabin tests are performed with bases that are known to have no counterexamples in their range. Finally if the number is larger than 2^64, a strong BPSW test is performed. While this is a probable prime test and we believe counterexamples exist, there are no known counterexamples.

Examples

>>> from ..ntheory import isprime
>>> isprime(13)
True
>>> isprime(15)
False

See also

sympy.ntheory.generate.primerange

Generates all primes in a given range

sympy.ntheory.generate.primepi

Return the number of primes less than or equal to n

sympy.ntheory.generate.prime

Return the nth prime

References

modelparameters.sympy.ntheory.primetest.mr(n, bases)[source]

Perform a Miller-Rabin strong pseudoprime test on n using a given list of bases/witnesses.

References

  • Richard Crandall & Carl Pomerance (2005), “Prime Numbers: A Computational Perspective”, Springer, 2nd edition, 135-138

A list of thresholds and the bases they require are here: http://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test#Deterministic_variants_of_the_test

Examples

>>> from .primetest import mr
>>> mr(1373651, [2, 3])
False
>>> mr(479001599, [31, 73])
True

modelparameters.sympy.ntheory.residue_ntheory module

modelparameters.sympy.ntheory.residue_ntheory.discrete_log(n, a, b, order=None, prime_order=None)[source]

Compute the discrete logarithm of a to the base b modulo n.

This is a recursive function to reduce the discrete logarithm problem in cyclic groups of composite order to the problem in cyclic groups of prime order.

It employs different algorithms depending on the problem (subgroup order size, prime order or not):

  • Trial multiplication

  • Baby-step giant-step

  • Pollard’s Rho

  • Pohlig-Hellman

References

Examples

>>> from ..ntheory import discrete_log
>>> discrete_log(41, 15, 7)
3
modelparameters.sympy.ntheory.residue_ntheory.is_nthpow_residue(a, n, m)[source]

Returns True if x**n == a (mod m) has solutions.

References

modelparameters.sympy.ntheory.residue_ntheory.is_primitive_root(a, p)[source]

Returns True if a is a primitive root of p

a is said to be the primitive root of p if gcd(a, p) == 1 and totient(p) is the smallest positive number s.t.

a**totient(p) cong 1 mod(p)

Examples

>>> from ..ntheory import is_primitive_root, n_order, totient
>>> is_primitive_root(3, 10)
True
>>> is_primitive_root(9, 10)
False
>>> n_order(3, 10) == totient(10)
True
>>> n_order(9, 10) == totient(10)
False
modelparameters.sympy.ntheory.residue_ntheory.is_quad_residue(a, p)[source]

Returns True if a (mod p) is in the set of squares mod p, i.e a % p in set([i**2 % p for i in range(p)]). If p is an odd prime, an iterative method is used to make the determination:

>>> from ..ntheory import is_quad_residue
>>> sorted(set([i**2 % 7 for i in range(7)]))
[0, 1, 2, 4]
>>> [j for j in range(7) if is_quad_residue(j, 7)]
[0, 1, 2, 4]

See also

legendre_symbol, jacobi_symbol

modelparameters.sympy.ntheory.residue_ntheory.jacobi_symbol(m, n)[source]

Returns the Jacobi symbol (m / n).

For any integer m and any positive odd integer n the Jacobi symbol is defined as the product of the Legendre symbols corresponding to the prime factors of n:

\[\genfrac(){}{}{m}{n} = \genfrac(){}{}{m}{p^{1}}^{\alpha_1} \genfrac(){}{}{m}{p^{2}}^{\alpha_2} ... \genfrac(){}{}{m}{p^{k}}^{\alpha_k} \text{ where } n = p_1^{\alpha_1} p_2^{\alpha_2} ... p_k^{\alpha_k}\]

Like the Legendre symbol, if the Jacobi symbol genfrac(){}{}{m}{n} = -1 then m is a quadratic nonresidue modulo n.

But, unlike the Legendre symbol, if the Jacobi symbol genfrac(){}{}{m}{n} = 1 then m may or may not be a quadratic residue modulo n.

Parameters:

  • m (integer) –

  • n (odd positive integer) –

Examples

>>> from ..ntheory import jacobi_symbol, legendre_symbol
>>> from .. import Mul, S
>>> jacobi_symbol(45, 77)
-1
>>> jacobi_symbol(60, 121)
1

The relationship between the jacobi_symbol and legendre_symbol can be demonstrated as follows:

>>> L = legendre_symbol
>>> S(45).factors()
{3: 2, 5: 1}
>>> jacobi_symbol(7, 45) == L(7, 3)**2 * L(7, 5)**1
True

See also

is_quad_residue, legendre_symbol

modelparameters.sympy.ntheory.residue_ntheory.legendre_symbol(a, p)[source]

Returns the Legendre symbol (a / p).

For an integer a and an odd prime p, the Legendre symbol is defined as

\[\begin{split}\genfrac(){}{}{a}{p} = \begin{cases} 0 & \text{if } p \text{ divides } a\\ 1 & \text{if } a \text{ is a quadratic residue modulo } p\\ -1 & \text{if } a \text{ is a quadratic nonresidue modulo } p \end{cases}\end{split}\]
Parameters:

  • a (integer) –

  • p (odd prime) –

Examples

>>> from ..ntheory import legendre_symbol
>>> [legendre_symbol(i, 7) for i in range(7)]
[0, 1, 1, -1, 1, -1, -1]
>>> sorted(set([i**2 % 7 for i in range(7)]))
[0, 1, 2, 4]

See also

is_quad_residue, jacobi_symbol

class modelparameters.sympy.ntheory.residue_ntheory.mobius(n)[source]

Bases: Function

Möbius function maps natural number to {-1, 0, 1}

It is defined as follows:

  1. 1 if n = 1.

  2. 0 if n has a squared prime factor.

  3. (-1)^k if n is a square-free positive integer with k number of prime factors.

It is an important multiplicative function in number theory and combinatorics. It has applications in mathematical series, algebraic number theory and also physics (Fermion operator has very concrete realization with Möbius Function model).

Parameters:

n (positive integer) –

Examples

>>> from ..ntheory import mobius
>>> mobius(13*7)
1
>>> mobius(1)
1
>>> mobius(13*7*5)
-1
>>> mobius(13**2)
0

References

default_assumptions = {}
classmethod eval(n)[source]

Returns a canonical form of cls applied to arguments args.

The eval() method is called when the class cls is about to be instantiated and it should return either some simplified instance (possible of some other class), or if the class cls should be unmodified, return None.

Examples of eval() for the function “sign”

@classmethod def eval(cls, arg):

if arg is S.NaN:

return S.NaN

if arg is S.Zero: return S.Zero if arg.is_positive: return S.One if arg.is_negative: return S.NegativeOne if isinstance(arg, Mul):

coeff, terms = arg.as_coeff_Mul(rational=True) if coeff is not S.One:

return cls(coeff) * cls(terms)

modelparameters.sympy.ntheory.residue_ntheory.n_order(a, n)[source]

Returns the order of a modulo n.

The order of a modulo n is the smallest integer k such that a**k leaves a remainder of 1 with n.

Examples

>>> from ..ntheory import n_order
>>> n_order(3, 7)
6
>>> n_order(4, 7)
3
modelparameters.sympy.ntheory.residue_ntheory.nthroot_mod(a, n, p, all_roots=False)[source]

Find the solutions to x**n = a mod p

Parameters:

  • a (integer) –

  • n (positive integer) –

  • p (positive integer) –

  • all_roots (if False returns the smallest root, else the list of roots) –

Examples

>>> from .residue_ntheory import nthroot_mod
>>> nthroot_mod(11, 4, 19)
8
>>> nthroot_mod(11, 4, 19, True)
[8, 11]
>>> nthroot_mod(68, 3, 109)
23
modelparameters.sympy.ntheory.residue_ntheory.primitive_root(p)[source]

Returns the smallest primitive root or None

References

Parameters:

p (positive integer) –

Examples

>>> from .residue_ntheory import primitive_root
>>> primitive_root(19)
2
modelparameters.sympy.ntheory.residue_ntheory.quadratic_residues(p)[source]

Returns the list of quadratic residues.

Examples

>>> from .residue_ntheory import quadratic_residues
>>> quadratic_residues(7)
[0, 1, 2, 4]
modelparameters.sympy.ntheory.residue_ntheory.sqrt_mod(a, p, all_roots=False)[source]

Find a root of x**2 = a mod p

Parameters:

  • a (integer) –

  • p (positive integer) –

  • all_roots (if True the list of roots is returned or None) –

Notes

If there is no root it is returned None; else the returned root is less or equal to p // 2; in general is not the smallest one. It is returned p // 2 only if it is the only root.

Use all_roots only when it is expected that all the roots fit in memory; otherwise use sqrt_mod_iter.

Examples

>>> from ..ntheory import sqrt_mod
>>> sqrt_mod(11, 43)
21
>>> sqrt_mod(17, 32, True)
[7, 9, 23, 25]
modelparameters.sympy.ntheory.residue_ntheory.sqrt_mod_iter(a, p, domain=<class 'int'>)[source]

Iterate over solutions to x**2 = a mod p

Parameters:

  • a (integer) –

  • p (positive integer) –

  • domain (integer domain, int, ZZ or Integer) –

Examples

>>> from .residue_ntheory import sqrt_mod_iter
>>> list(sqrt_mod_iter(11, 43))
[21, 22]

Module contents

Number theory module (primes, etc)