def n_order(a, n):
"""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 sympy.ntheory import n_order
>>> n_order(3, 7)
6
>>> n_order(4, 7)
3
"""
a, n = as_int(a), as_int(n)
if igcd(a, n) != 1:
raise ValueError("The two numbers should be relatively prime")
group_order = totient(n)
factors = factorint(group_order)
order = 1
if a > n:
a = a % n
for p, e in factors.items():
exponent = group_order
for f in xrange(e + 1):
if pow(a, exponent, n) != 1:
order *= p ** (e - f + 1)
break
exponent = exponent // p
return order
def legendre_symbol(a, p):
"""
Returns
=======
1. 0 if a is multiple of p
2. 1 if a is a quadratic residue of p
3. -1 otherwise
p should be an odd prime by definition
Examples
========
>>> from sympy.ntheory import legendre_symbol
>>> [legendre_symbol(i, 7) for i in range(7)]
[0, 1, 1, -1, 1, -1, -1]
>>> list(set([i**2 % 7 for i in range(7)]))
[0, 1, 2, 4]
See Also
========
is_quad_residue, jacobi_symbol
"""
a, p = as_int(a), as_int(p)
if not isprime(p) or p == 2:
raise ValueError("p should be an odd prime")
a = a % p
if not a:
return 0
if is_quad_residue(a, p):
return 1
return -1
def multiplicity(p, n):
"""
Find the greatest integer m such that p**m divides n.
Examples
========
>>> from sympy.ntheory import multiplicity
>>> from sympy.core.numbers import Rational as R
>>> [multiplicity(5, n) for n in [8, 5, 25, 125, 250]]
[0, 1, 2, 3, 3]
>>> multiplicity(3, R(1, 9))
-2
"""
try:
p, n = as_int(p), as_int(n)
except ValueError:
if all(isinstance(i, (SYMPY_INTS, Rational)) for i in (p, n)):
try:
p = Rational(p)
n = Rational(n)
like_min = min(
multiplicity(p.p, n.p) if p.p != 1 else oo,
multiplicity(p.q, n.q) if p.q != 1 else oo)
cross_min = min(
multiplicity(p.p, n.q) if p.p != 1 else oo,
multiplicity(p.q, n.p) if p.q != 1 else oo)
return like_min - cross_min
except AttributeError:
pass
raise ValueError('expecting ints or fractions, got %s and %s' % (p, n))
if p == 2:
return trailing(n)
if p < 2:
raise ValueError('p must be an integer, 2 or larger, but got %s' % p)
if p == n:
return 1
m = 0
n, rem = divmod(n, p)
while not rem:
m += 1
if m > 5:
# The multiplicity could be very large. Better
# to increment in powers of two
e = 2
while 1:
ppow = p**e
if ppow < n:
nnew, rem = divmod(n, ppow)
if not rem:
m += e
e *= 2
n = nnew
continue
return m + multiplicity(p, n)
n, rem = divmod(n, p)
return m
def search(self, n):
"""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 sympy import sieve
>>> sieve.search(25)
(9, 10)
>>> sieve.search(23)
(9, 9)
"""
from sympy.functions.elementary.integers import ceiling
# wrapping ceiling in as_int will raise an error if there was a problem
# determining whether the expression was exactly an integer or not
test = as_int(ceiling(n))
n = as_int(n)
if n < 2:
raise ValueError("n should be >= 2 but got: %s" % n)
if n > self._list[-1]:
self.extend(n)
b = bisect(self._list, n)
if self._list[b - 1] == test:
return b, b
else:
return b, b + 1
def primerange(self, a, b):
"""Generate all prime numbers in the range [a, b).
Examples
========
>>> from sympy import sieve
>>> print([i for i in sieve.primerange(7, 18)])
[7, 11, 13, 17]
"""
from sympy.functions.elementary.integers import ceiling
# wrapping ceiling in as_int will raise an error if there was a problem
# determining whether the expression was exactly an integer or not
a = max(2, as_int(ceiling(a)))
b = as_int(ceiling(b))
if a >= b:
return
self.extend(b)
i = self.search(a)[1]
maxi = len(self._list) + 1
while i < maxi:
p = self._list[i - 1]
if p < b:
yield p
i += 1
else:
return
def ones(r, c=None):
"""Returns a matrix of ones with ``r`` rows and ``c`` columns;
if ``c`` is omitted a square matrix will be returned.
See Also
========
zeros
eye
diag
"""
from dense import Matrix
if is_sequence(r):
SymPyDeprecationWarning(
feature="The syntax ones([%i, %i])" % tuple(r),
useinstead="ones(%i, %i)." % tuple(r),
issue=3381, deprecated_since_version="0.7.2",
).warn()
r, c = r
else:
c = r if c is None else c
r = as_int(r)
c = as_int(c)
return Matrix(r, c, [S.One]*r*c)
def is_quad_residue(a, p):
"""
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 sympy.ntheory import is_quad_residue
>>> list(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
"""
a, p = as_int(a), as_int(p)
if p < 1:
raise ValueError('p must be > 0')
if a >= p or a < 0:
a = a % p
if a < 2 or p < 3:
return True
if not isprime(p):
if p % 2 and jacobi_symbol(a, p) == -1:
return False
r = sqrt_mod(a, p)
if r is None:
return False
else:
return True
return pow(a, (p - 1) // 2, p) == 1
def is_primitive_root(a, p):
"""
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 sympy.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
"""
a, p = as_int(a), as_int(p)
if igcd(a, p) != 1:
raise ValueError("The two numbers should be relatively prime")
if a > p:
a = a % p
return n_order(a, p) == totient(p)
def is_nthpow_residue(a, n, m):
"""
Returns True if ``x**n == a (mod m)`` has solutions.
References
==========
.. [1] P. Hackman "Elementary Number Theory" (2009), page 76
"""
a, n, m = as_int(a), as_int(n), as_int(m)
if m <= 0:
raise ValueError('m must be > 0')
if n < 0:
raise ValueError('n must be >= 0')
if a < 0:
raise ValueError('a must be >= 0')
if n == 0:
if m == 1:
return False
return a == 1
if n == 1:
return True
if n == 2:
return is_quad_residue(a, m)
return _is_nthpow_residue_bign(a, n, m)
def digits(n, b=10):
"""
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 sympy.ntheory.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]
"""
b = as_int(b)
n = as_int(n)
if b <= 1:
raise ValueError("b must be >= 2")
else:
x, y = abs(n), []
while x >= b:
x, r = divmod(x, b)
y.append(r)
y.append(x)
y.append(-b if n < 0 else b)
y.reverse()
return y
def sqrt_mod_iter(a, p, domain=int):
"""
iterate over solutions to ``x**2 = a mod p``
Parameters
==========
a : integer
p : positive integer
domain : integer domain, ``int``, ``ZZ`` or ``Integer``
Examples
========
>>> from sympy.ntheory.residue_ntheory import sqrt_mod_iter
>>> list(sqrt_mod_iter(11, 43))
[21, 22]
"""
from sympy.polys.galoistools import gf_crt1, gf_crt2
from sympy.polys.domains import ZZ
a, p = as_int(a), abs(as_int(p))
if isprime(p):
a = a % p
if a == 0:
res = _sqrt_mod1(a, p, 1)
else:
res = _sqrt_mod_prime_power(a, p, 1)
if res:
if domain is ZZ:
for x in res:
yield x
else:
for x in res:
yield domain(x)
else:
f = factorint(p)
v = []
pv = []
for px, ex in f.items():
if a % px == 0:
rx = _sqrt_mod1(a, px, ex)
if not rx:
raise StopIteration
else:
rx = _sqrt_mod_prime_power(a, px, ex)
if not rx:
raise StopIteration
v.append(rx)
pv.append(px**ex)
mm, e, s = gf_crt1(pv, ZZ)
if domain is ZZ:
for vx in _product(*v):
r = gf_crt2(vx, pv, mm, e, s, ZZ)
yield r
else:
for vx in _product(*v):
r = gf_crt2(vx, pv, mm, e, s, ZZ)
yield domain(r)
def __new__(cls, partition, integer=None):
"""
Generates a new IntegerPartition object from a list or dictionary.
The partition can be given as a list of positive integers or a
dictionary of (integer, multiplicity) items. If the partition is
preceded by an integer an error will be raised if the partition
does not sum to that given integer.
Examples
========
>>> from sympy.combinatorics.partitions import IntegerPartition
>>> a = IntegerPartition([5, 4, 3, 1, 1])
>>> a
IntegerPartition(14, (5, 4, 3, 1, 1))
>>> print(a)
[5, 4, 3, 1, 1]
>>> IntegerPartition({1:3, 2:1})
IntegerPartition(5, (2, 1, 1, 1))
If the value that the partition should sum to is given first, a check
will be made to see n error will be raised if there is a discrepancy:
>>> IntegerPartition(10, [5, 4, 3, 1])
Traceback (most recent call last):
...
ValueError: The partition is not valid
"""
if integer is not None:
integer, partition = partition, integer
if isinstance(partition, (dict, Dict)):
_ = []
for k, v in sorted(list(partition.items()), reverse=True):
if not v:
continue
k, v = as_int(k), as_int(v)
_.extend([k]*v)
partition = tuple(_)
else:
partition = tuple(sorted(map(as_int, partition), reverse=True))
sum_ok = False
if integer is None:
integer = sum(partition)
sum_ok = True
else:
integer = as_int(integer)
if not sum_ok and sum(partition) != integer:
raise ValueError("Partition did not add to %s" % integer)
if any(i < 1 for i in partition):
raise ValueError("The summands must all be positive.")
obj = Basic.__new__(cls, integer, partition)
obj.partition = list(partition)
obj.integer = integer
return obj
def _number_theoretic_transform(seq, prime, inverse=False):
"""Utility function for the Number Theoretic Transform"""
if not iterable(seq):
raise TypeError("Expected a sequence of integer coefficients "
"for Number Theoretic Transform")
p = as_int(prime)
if isprime(p) == False:
raise ValueError("Expected prime modulus for "
"Number Theoretic Transform")
a = [as_int(x) % p for x in seq]
n = len(a)
if n < 1:
return a
b = n.bit_length() - 1
if n&(n - 1):
b += 1
n = 2**b
if (p - 1) % n:
raise ValueError("Expected prime modulus of the form (m*2**k + 1)")
a += [0]*(n - len(a))
for i in range(1, n):
j = int(ibin(i, b, str=True)[::-1], 2)
if i < j:
a[i], a[j] = a[j], a[i]
pr = primitive_root(p)
rt = pow(pr, (p - 1) // n, p)
if inverse:
rt = pow(rt, p - 2, p)
w = [1]*(n // 2)
for i in range(1, n // 2):
w[i] = w[i - 1]*rt % p
h = 2
while h <= n:
hf, ut = h // 2, n // h
for i in range(0, n, h):
for j in range(hf):
u, v = a[i + j], a[i + j + hf]*w[ut * j]
a[i + j], a[i + j + hf] = (u + v) % p, (u - v) % p
h *= 2
if inverse:
rv = pow(n, p - 2, p)
a = [x*rv % p for x in a]
return a
def as_dict(e):
# creates a dictionary of the expression using either
# as_coefficients_dict or as_powers_dict, depending on Func
# SHARES meth
d = getattr(e, meth, lambda: {a: S.One for a in e.args})()
for k in list(d.keys()):
try:
as_int(d[k])
except ValueError:
d[F(k, d.pop(k))] = S.One
return d
def give(a, b, seq=seed):
a, b = as_int(a), as_int(b)
w = b - a
if w < 0:
raise ValueError('_randint got empty range')
try:
x = seq.pop()
except IndexError:
raise ValueError('_randint sequence was too short')
if a <= x <= b:
return x
else:
return give(a, b, seq)
def zeros(cls, r, c=None):
"""Return an r x c matrix of zeros, square if c is omitted."""
if is_sequence(r):
SymPyDeprecationWarning(
feature="The syntax zeros([%i, %i])" % tuple(r),
useinstead="zeros(%i, %i)." % tuple(r),
issue=3381, deprecated_since_version="0.7.2",
).warn()
r, c = r
else:
c = r if c is None else c
r = as_int(r)
c = as_int(c)
return cls._new(r, c, [S.Zero]*r*c)
def __new__(cls, *args, **kwargs):
self = object.__new__(cls)
if len(args) == 1 and isinstance(args[0], SparseMatrix):
self.rows = args[0].rows
self.cols = args[0].cols
self._smat = dict(args[0]._smat)
return self
self._smat = {}
if len(args) == 3:
self.rows = as_int(args[0])
self.cols = as_int(args[1])
if isinstance(args[2], Callable):
op = args[2]
for i in range(self.rows):
for j in range(self.cols):
value = self._sympify(
op(self._sympify(i), self._sympify(j)))
if value:
self._smat[(i, j)] = value
elif isinstance(args[2], (dict, Dict)):
# manual copy, copy.deepcopy() doesn't work
for key in args[2].keys():
v = args[2][key]
if v:
self._smat[key] = self._sympify(v)
elif is_sequence(args[2]):
if len(args[2]) != self.rows*self.cols:
raise ValueError(
'List length (%s) != rows*columns (%s)' %
(len(args[2]), self.rows*self.cols))
flat_list = args[2]
for i in range(self.rows):
for j in range(self.cols):
value = self._sympify(flat_list[i*self.cols + j])
if value:
self._smat[(i, j)] = value
else:
# handle full matrix forms with _handle_creation_inputs
r, c, _list = Matrix._handle_creation_inputs(*args)
self.rows = r
self.cols = c
for i in range(self.rows):
for j in range(self.cols):
value = _list[self.cols*i + j]
if value:
self._smat[(i, j)] = value
return self
def give(a, b=None, seq=seed):
if b is None:
a, b = 0, a
a, b = as_int(a), as_int(b)
w = b - a
if w < 1:
raise ValueError('_randrange got empty range')
try:
x = seq.pop()
except IndexError as e:
raise ValueError('_randrange sequence was too short')
if a <= x < b:
return x
else:
return give(a, b, seq)
def give(a, b, seq=seed):
a, b = as_int(a), as_int(b)
w = b - a
if w < 0:
raise ValueError("_randint got empty range")
try:
x = seq.pop()
except AttributeError:
raise ValueError("_randint expects a list-like sequence")
except IndexError:
raise ValueError("_randint sequence was too short")
if a <= x <= b:
return x
else:
return give(a, b, seq)
def ordinal(num):
"""Return ordinal number string of num, e.g. 1 becomes 1st.
"""
# modified from https://codereview.stackexchange.com/questions/41298/producing-ordinal-numbers
n = as_int(num)
if n < 0:
return '-%s' % ordinal(-n)
if n == 0 or 4 <= n <= 20:
suffix = 'th'
elif n == 1 or (n % 10) == 1:
suffix = 'st'
elif n == 2 or (n % 10) == 2:
suffix = 'nd'
elif n == 3 or (n % 10) == 3:
suffix = 'rd'
elif n < 101:
suffix = 'th'
else:
suffix = ordinal(n % 100)
if len(suffix) == 3:
# e.g. 103 -> 3rd
# so add other 0 back
suffix = '0' + suffix
n = str(n)[:-2]
return str(n) + suffix
def antidivisor_count(n):
"""
Return the number of antidivisors [1]_ of ``n``.
References
==========
.. [1] formula from https://oeis.org/A066272
Examples
========
>>> from sympy.ntheory.factor_ import antidivisor_count
>>> antidivisor_count(13)
4
>>> antidivisor_count(27)
5
See Also
========
factorint, divisors, antidivisors, divisor_count, totient
"""
n = as_int(abs(n))
if n <= 2:
return 0
return divisor_count(2*n-1) + divisor_count(2*n+1) + \
divisor_count(n) - divisor_count(n, 2) - 5
def antidivisors(n, generator=False):
r"""
Return all antidivisors of n sorted from 1..n by default.
Antidivisors [1]_ of n are numbers that do not divide n by the largest
possible margin. If generator is True an unordered generator is returned.
References
==========
.. [1] definition is described in http://oeis.org/A066272/a066272a.html
Examples
========
>>> from sympy.ntheory.factor_ import antidivisors
>>> antidivisors(24)
[7, 16]
>>> sorted(antidivisors(128, generator=True))
[3, 5, 15, 17, 51, 85]
See Also
========
primefactors, factorint, divisors, divisor_count, antidivisor_count
"""
n = as_int(abs(n))
if n <= 2:
return []
rv = _antidivisors(n)
if not generator:
return sorted(rv)
return rv
def real_root(arg, n=None):
"""Return the real nth-root of arg if possible. If n is omitted then
all instances of (-n)**(1/odd) will be changed to -n**(1/odd); this
will only create a real root of a principle root -- the presence of
other factors may cause the result to not be real.
Examples
========
>>> from sympy import root, real_root, Rational
>>> from sympy.abc import x, n
>>> real_root(-8, 3)
-2
>>> root(-8, 3)
2*(-1)**(1/3)
>>> real_root(_)
-2
If one creates a non-principle root and applies real_root, the
result will not be real (so use with caution):
>>> root(-8, 3, 2)
-2*(-1)**(2/3)
>>> real_root(_)
-2*(-1)**(2/3)
See Also
========
sympy.polys.rootoftools.RootOf
sympy.core.power.integer_nthroot
root, sqrt
"""
from sympy import im, Piecewise
if n is not None:
try:
n = as_int(n)
arg = sympify(arg)
if arg.is_positive or arg.is_negative:
rv = root(arg, n)
else:
raise ValueError
except ValueError:
return root(arg, n)*Piecewise(
(S.One, ~Equality(im(arg), 0)),
(Pow(S.NegativeOne, S.One/n)**(2*floor(n/2)), And(
Equality(n % 2, 1),
arg < 0)),
(S.One, True))
else:
rv = sympify(arg)
n1pow = Transform(lambda x: -(-x.base)**x.exp,
lambda x:
x.is_Pow and
x.base.is_negative and
x.exp.is_Rational and
x.exp.p == 1 and x.exp.q % 2)
return rv.xreplace(n1pow)
def is_square(n, prep=True):
"""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
"""
if prep:
n = as_int(n)
if n < 0:
return False
if n in [0, 1]:
return True
m = n & 127
if not ((m*0x8bc40d7d) & (m*0xa1e2f5d1) & 0x14020a):
m = n % 63
if not ((m*0x3d491df7) & (m*0xc824a9f9) & 0x10f14008):
from sympy.ntheory import perfect_power
if perfect_power(n, [2]):
return True
return False
def prime(nth):
""" Return the nth prime, with the primes indexed as prime(1) = 2,
prime(2) = 3, etc.... The nth prime is approximately n*log(n) and
can never be larger than 2**n.
References
==========
- http://primes.utm.edu/glossary/xpage/BertrandsPostulate.html
Examples
========
>>> from sympy import prime
>>> prime(10)
29
>>> prime(1)
2
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
"""
n = as_int(nth)
if n < 1:
raise ValueError("nth must be a positive integer; prime(1) == 2")
return sieve[n]
def _eval_subs(self, old, new):
if old.func is self.func and self.base == old.base:
coeff1, terms1 = self.exp.as_independent(C.Symbol, as_Add=False)
coeff2, terms2 = old.exp.as_independent(C.Symbol, as_Add=False)
if terms1 == terms2:
pow = coeff1/coeff2
ok = False # True if int(pow) == pow OR self.base.is_positive
try:
pow = as_int(pow)
ok = True
except ValueError:
ok = self.base.is_positive
if ok:
# issue 2081
return self.func(new, pow) # (x**(6*y)).subs(x**(3*y),z)->z**2
if old.func is C.exp and self.exp.is_real and self.base.is_positive:
coeff1, terms1 = old.args[0].as_independent(C.Symbol, as_Add=False)
# we can only do this when the base is positive AND the exponent
# is real
coeff2, terms2 = (self.exp*C.log(self.base)).as_independent(
C.Symbol, as_Add=False)
if terms1 == terms2:
pow = coeff1/coeff2
if pow == int(pow) or self.base.is_positive:
return self.func(new, pow) # (2**x).subs(exp(x*log(2)), z) -> z
def extend_to_no(self, i):
"""Extend to include the ith prime number.
Parameters
==========
i : integer
Examples
========
>>> from sympy 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])
Notes
=====
The list is extended by 50% if it is too short, so it is
likely that it will be longer than requested.
"""
i = as_int(i)
while len(self._list) < i:
self.extend(int(self._list[-1] * 1.5))
def __new__(cls, *args):
from sympy.functions.elementary.integers import ceiling
# expand range
slc = slice(*args)
start, stop, step = slc.start or 0, slc.stop, slc.step or 1
try:
start, stop, step = [w if w in [S.NegativeInfinity, S.Infinity] else S(as_int(w))
for w in (start, stop, step)]
except ValueError:
raise ValueError("Inputs to Range must be Integer Valued\n" +
"Use ImageSets of Ranges for other cases")
if not step.is_finite:
raise ValueError("Infinite step is not allowed")
if start == stop:
return S.EmptySet
n = ceiling((stop - start)/step)
if n <= 0:
return S.EmptySet
# normalize args: regardless of how they are entered they will show
# canonically as Range(inf, sup, step) with step > 0
if n.is_finite:
start, stop = sorted((start, start + (n - 1)*step))
else:
start, stop = sorted((start, stop - step))
step = abs(step)
if (start, stop) == (S.NegativeInfinity, S.Infinity):
raise ValueError("Both the start and end value of "
"Range cannot be unbounded")
else:
return Basic.__new__(cls, start, stop + step, step)
def _primitive_root_prime_iter(p):
"""
Generates the primitive roots for a prime ``p``
References
==========
[1] W. Stein "Elementary Number Theory" (2011), page 44
Examples
========
>>> from sympy.ntheory.residue_ntheory import _primitive_root_prime_iter
>>> list(_primitive_root_prime_iter(19))
[2, 3, 10, 13, 14, 15]
"""
p = as_int(p)
v = [(p - 1) // i for i in factorint(p - 1).keys()]
a = 2
while a < p:
for pw in v:
if pow(a, pw, p) == 1:
break
else:
yield a
a += 1
def nextprime(n, ith=1):
""" 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 sympy 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
"""
n = int(n)
i = as_int(ith)
if i > 1:
pr = n
j = 1
while 1:
pr = nextprime(pr)
j += 1
if j > i:
break
return pr
if n < 2:
return 2
if n < 7:
return {2: 3, 3: 5, 4: 5, 5: 7, 6: 7}[n]
nn = 6*(n//6)
if nn == n:
n += 1
if isprime(n):
return n
n += 4
elif n - nn == 5:
n += 2
if isprime(n):
return n
n += 4
else:
n = nn + 5
while 1:
if isprime(n):
return n
n += 2
if isprime(n):
return n
n += 4
def __add__(self, other):
"""
Return permutation whose rank is ``other`` greater than current rank,
(mod the maximum rank for the set).
Examples
========
>>> from sympy.combinatorics.partitions import Partition
>>> a = Partition([1, 2], [3])
>>> a.rank
1
>>> (a + 1).rank
2
>>> (a + 100).rank
1
"""
other = as_int(other)
offset = self.rank + other
result = RGS_unrank((offset) % RGS_enum(self.size), self.size)
return Partition.from_rgs(result, self.members)
def __new__(cls, *args):
from sympy.functions.elementary.integers import ceiling
if len(args) == 1:
if isinstance(args[0], range if PY3 else xrange):
args = args[0].__reduce__()[1] # use pickle method
# expand range
slc = slice(*args)
start, stop, step = slc.start or 0, slc.stop, slc.step or 1
try:
start, stop, step = [
w if w in [S.NegativeInfinity, S.Infinity] else sympify(
as_int(w)) for w in (start, stop, step)
]
except ValueError:
raise ValueError("Inputs to Range must be Integer Valued\n" +
"Use ImageSets of Ranges for other cases")
if not step.is_finite:
raise ValueError("Infinite step is not allowed")
if start == stop:
return S.EmptySet
n = ceiling((stop - start) / step)
if n <= 0:
return S.EmptySet
# normalize args: regardless of how they are entered they will show
# canonically as Range(inf, sup, step) with step > 0
if n.is_finite:
start, stop = sorted((start, start + (n - 1) * step))
else:
start, stop = sorted((start, stop - step))
step = abs(step)
if (start, stop) == (S.NegativeInfinity, S.Infinity):
raise ValueError("Both the start and end value of "
"Range cannot be unbounded")
else:
return Basic.__new__(cls, start, stop + step, step)
def _nth_root1(p, n, x, prec):
"""
Univariate series expansion of the nth root of p
While passing p, make sure that it is of the form `1 + f(x)`.
n (integer): compute p**(1/n)
x: name of the series variable
prec: precision of the series
The Newton method is used.
"""
if rs_is_puiseux(p, x):
return rs_puiseux2(_nth_root1, p, n, x, prec)
R = p.ring
zm = R.zero_monom
if zm not in p:
raise NotImplementedError('No constant term in series')
n = as_int(n)
assert p[zm] == 1
p1 = R(1)
if p == 1:
return p
if n == 0:
return R(1)
if n == 1:
return p
if n < 0:
n = -n
sign = 1
else:
sign = 0
for precx in _giant_steps(prec):
tmp = rs_pow(p1, n + 1, x, precx)
tmp = rs_mul(tmp, p, x, precx)
p1 += p1 / n - tmp / n
if sign:
return p1
else:
return _series_inversion1(p1, x, prec)
def jordan_cell(eigenval, n):
"""
Create matrix of Jordan cell kind:
Examples
========
>>> from sympy.matrices import jordan_cell
>>> from sympy.abc import x
>>> jordan_cell(x, 4)
[x, 1, 0, 0]
[0, x, 1, 0]
[0, 0, x, 1]
[0, 0, 0, x]
"""
n = as_int(n)
out = zeros(n)
for i in range(n - 1):
out[i, i] = eigenval
out[i, i + 1] = S.One
out[n - 1, n - 1] = eigenval
return out
def extend_to_no(self, i):
"""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 sympy import sieve
>>> from array import array # this line and next for doctest only
>>> sieve._list = array('l', [2, 3, 5, 7, 11, 13])
>>> sieve.extend_to_no(9)
>>> sieve._list
array('l', [2, 3, 5, 7, 11, 13, 17, 19, 23])
"""
i = as_int(i)
while len(self._list) < i:
self.extend(int(self._list[-1] * 1.5))
def real_root(arg, n=None):
"""Return the real nth-root of arg if possible. If n is omitted then
all instances of -1**(1/odd) will be changed to -1.
Examples
========
>>> from sympy import root, real_root, Rational
>>> from sympy.abc import x, n
>>> real_root(-8, 3)
-2
>>> root(-8, 3)
2*(-1)**(1/3)
>>> real_root(_)
-2
See Also
========
sympy.polys.rootoftools.RootOf
sympy.core.power.integer_nthroot
root, sqrt
"""
if n is not None:
n = as_int(n)
rv = C.Pow(arg, Rational(1, n))
if n % 2 == 0:
return rv
else:
rv = sympify(arg)
n1pow = Transform(lambda x: S.NegativeOne,
lambda x:
x.is_Pow and
x.base is S.NegativeOne and
x.exp.is_Rational and
x.exp.p == 1 and x.exp.q % 2)
return rv.xreplace(n1pow)
def is_square(n, prep=True):
"""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.
Examples
========
>>> from sympy.ntheory.primetest import is_square
>>> is_square(25)
True
>>> is_square(2)
False
References
==========
[1] http://mersenneforum.org/showpost.php?p=110896
See Also
========
sympy.core.power.integer_nthroot
"""
if prep:
n = as_int(n)
if n < 0:
return False
if n in [0, 1]:
return True
m = n & 127
if not ((m * 0x8bc40d7d) & (m * 0xa1e2f5d1) & 0x14020a):
m = n % 63
if not ((m * 0x3d491df7) & (m * 0xc824a9f9) & 0x10f14008):
from sympy.ntheory import perfect_power
pp = perfect_power(n, [2], big=False)
if pp:
return pp[1] == 2
return False
def binomial_coefficients_list(n):
""" Return a list of binomial coefficients as rows of the Pascal's
triangle.
Examples
========
>>> from sympy.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
"""
n = as_int(n)
d = [1] * (n + 1)
a = 1
for k in range(1, n // 2 + 1):
a = (a * (n - k + 1)) // k
d[k] = d[n - k] = a
return d
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.iteritems():
t *= (p - 1) * p**(k - 1)
return t
def binomial_coefficients(n):
"""Return a dictionary containing pairs :math:`{(k1,k2) : C_kn}` where
:math:`C_kn` are binomial coefficients and :math:`n=k1+k2`.
Examples
========
>>> from sympy.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
"""
n = as_int(n)
d = {(0, n): 1, (n, 0): 1}
a = 1
for k in range(1, n // 2 + 1):
a = (a * (n - k + 1)) // k
d[k, n - k] = d[n - k, k] = a
return d
def is_square(n, prep=True):
"""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.
Examples
========
>>> from sympy.ntheory.primetest import is_square
>>> is_square(25)
True
>>> is_square(2)
False
References
==========
[1] http://mersenneforum.org/showpost.php?p=110896
See Also
========
sympy.core.power.integer_nthroot
"""
if prep:
n = as_int(n)
if n < 0:
return False
if n in [0, 1]:
return True
m = n & 127
if not ((m * 0x8BC40D7D) & (m * 0xA1E2F5D1) & 0x14020A):
m = n % 63
if not ((m * 0x3D491DF7) & (m * 0xC824A9F9) & 0x10F14008):
from sympy.core.power import integer_nthroot
return integer_nthroot(n, 2)[1]
return False
def random_integer_partition(n, seed=None):
"""
Generates a random integer partition summing to ``n`` as a list
of reverse-sorted integers.
Examples
========
>>> from sympy.combinatorics.partitions import random_integer_partition
For the following, a seed is given so a known value can be shown; in
practice, the seed would not be given.
>>> random_integer_partition(100, seed=[1, 1, 12, 1, 2, 1, 85, 1])
[85, 12, 2, 1]
>>> random_integer_partition(10, seed=[1, 2, 3, 1, 5, 1])
[5, 3, 1, 1]
>>> random_integer_partition(1)
[1]
"""
from sympy.testing.randtest import _randint
n = as_int(n)
if n < 1:
raise ValueError("n must be a positive integer")
randint = _randint(seed)
partition = []
while n > 0:
k = randint(1, n)
mult = randint(1, n // k)
partition.append((k, mult))
n -= k * mult
partition.sort(reverse=True)
partition = flatten([[k] * m for k, m in partition])
return partition
def _check(ct1, ct2, old):
"""Return bool, pow where, if bool is True, then the exponent of
Pow `old` will combine with `pow` so the substitution is valid,
otherwise bool will be False,
cti are the coefficient and terms of an exponent of self or old
In this _eval_subs routine a change like (b**(2*x)).subs(b**x, y)
will give y**2 since (b**x)**2 == b**(2*x); if that equality does
not hold then the substitution should not occur so `bool` will be
False.
"""
coeff1, terms1 = ct1
coeff2, terms2 = ct2
if terms1 == terms2:
pow = coeff1 / coeff2
try:
pow = as_int(pow)
combines = True
except ValueError:
combines = Pow._eval_power(
Pow(*old.as_base_exp(), evaluate=False),
pow) is not None
return combines, pow
return False, None
def primerange(a, b):
""" 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 occurrence 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. https://en.wikipedia.org/wiki/Prime_number
2. http://primes.utm.edu/notes/gaps.html
Examples
========
>>> from sympy 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.
"""
from sympy.functions.elementary.integers import ceiling
if a >= b:
return
# if we already have the range, return it
if b <= sieve._list[-1]:
for i in sieve.primerange(a, b):
yield i
return
# otherwise compute, without storing, the desired range.
# wrapping ceiling in as_int will raise an error if there was a problem
# determining whether the expression was exactly an integer or not
a = as_int(ceiling(a)) - 1
b = as_int(ceiling(b))
while 1:
a = nextprime(a)
if a < b:
yield a
else:
return
def primorial(n, nth=True):
"""
Returns the product of the first n primes (default) or
the primes less than or equal to n (when ``nth=False``).
>>> from sympy.ntheory.generate import primorial, randprime, primerange
>>> from sympy 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
"""
if nth:
n = as_int(n)
else:
n = int(n)
if n < 1:
raise ValueError("primorial argument must be >= 1")
p = 1
if nth:
for i in range(1, n + 1):
p *= prime(i)
else:
for i in primerange(2, n + 1):
p *= i
return p
def prime(nth):
""" 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
==========
- https://en.wikipedia.org/wiki/Prime_number_theorem#Table_of_.CF.80.28x.29.2C_x_.2F_log_x.2C_and_li.28x.29
- https://en.wikipedia.org/wiki/Prime_number_theorem#Approximations_for_the_nth_prime_number
- https://en.wikipedia.org/wiki/Skewes%27_number
Examples
========
>>> from sympy 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
"""
n = as_int(nth)
if n < 1:
raise ValueError("nth must be a positive integer; prime(1) == 2")
if n <= len(sieve._list):
return sieve[n]
from sympy.functions.special.error_functions import li
from sympy.functions.elementary.exponential import log
a = 2 # Lower bound for binary search
b = int(n * (log(n) + log(log(n)))) # Upper bound for the search.
while a < b:
mid = (a + b) >> 1
if li(mid) > n:
b = mid
else:
a = mid + 1
n_primes = primepi(a - 1)
while n_primes < n:
if isprime(a):
n_primes += 1
a += 1
return a - 1
def nextprime(n, ith=1):
""" 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 sympy 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
"""
n = int(n)
i = as_int(ith)
if i > 1:
pr = n
j = 1
while 1:
pr = nextprime(pr)
j += 1
if j > i:
break
return pr
if n < 2:
return 2
if n < 7:
return {2: 3, 3: 5, 4: 5, 5: 7, 6: 7}[n]
if n <= sieve._list[-2]:
l, u = sieve.search(n)
if l == u:
return sieve[u + 1]
else:
return sieve[u]
nn = 6 * (n // 6)
if nn == n:
n += 1
if isprime(n):
return n
n += 4
elif n - nn == 5:
n += 2
if isprime(n):
return n
n += 4
else:
n = nn + 5
while 1:
if isprime(n):
return n
n += 2
if isprime(n):
return n
n += 4
def multiplicity(p, n):
"""
Find the greatest integer m such that p**m divides n.
Examples
========
>>> from sympy.ntheory import multiplicity
>>> from sympy.core.numbers import Rational as R
>>> [multiplicity(5, n) for n in [8, 5, 25, 125, 250]]
[0, 1, 2, 3, 3]
>>> multiplicity(3, R(1, 9))
-2
"""
try:
p, n = as_int(p), as_int(n)
except ValueError:
if all(isinstance(i, (SYMPY_INTS, Rational)) for i in (p, n)):
try:
p = Rational(p)
n = Rational(n)
if p.q == 1:
if n.p == 1:
return -multiplicity(p.p, n.q)
return S.Zero
elif p.p == 1:
return multiplicity(p.q, n.q)
else:
like = min(multiplicity(p.p, n.p), multiplicity(p.q, n.q))
cross = min(multiplicity(p.q, n.p), multiplicity(p.p, n.q))
return like - cross
except AttributeError:
pass
raise ValueError('expecting ints or fractions, got %s and %s' % (p, n))
if p == 2:
return trailing(n)
if p < 2:
raise ValueError('p must be an integer, 2 or larger, but got %s' % p)
if p == n:
return 1
m = 0
n, rem = divmod(n, p)
while not rem:
m += 1
if m > 5:
# The multiplicity could be very large. Better
# to increment in powers of two
e = 2
while 1:
ppow = p**e
if ppow < n:
nnew, rem = divmod(n, ppow)
if not rem:
m += e
e *= 2
n = nnew
continue
return m + multiplicity(p, n)
n, rem = divmod(n, p)
return m
def eye(cls, n):
"""Return an n x n identity matrix."""
n = as_int(n)
return cls(n, n, {(i, i): S.One for i in range(n)})
def eye(cls, n):
"""Return an n x n identity matrix."""
n = as_int(n)
mat = [cls._sympify(0)]*n*n
mat[::n + 1] = [cls._sympify(1)]*n
return cls._new(n, n, mat)
def zeros(cls, r, c=None):
"""Return an r x c matrix of zeros, square if c is omitted."""
c = r if c is None else c
r = as_int(r)
c = as_int(c)
return cls._new(r, c, [cls._sympify(0)]*r*c)
def composite(nth):
""" Return the nth composite number, with the composite numbers indexed as
composite(1) = 4, composite(2) = 6, etc....
Examples
========
>>> from sympy 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
"""
n = as_int(nth)
if n < 1:
raise ValueError("nth must be a positive integer; composite(1) == 4")
composite_arr = [4, 6, 8, 9, 10, 12, 14, 15, 16, 18]
if n <= 10:
return composite_arr[n - 1]
a, b = 4, sieve._list[-1]
if n <= b - primepi(b) - 1:
while a < b - 1:
mid = (a + b) >> 1
if mid - primepi(mid) - 1 > n:
b = mid
else:
a = mid
if isprime(a):
a -= 1
return a
from sympy.functions.special.error_functions import li
from sympy.functions.elementary.exponential import log
a = 4 # Lower bound for binary search
b = int(n * (log(n) + log(log(n)))) # Upper bound for the search.
while a < b:
mid = (a + b) >> 1
if mid - li(mid) - 1 > n:
b = mid
else:
a = mid + 1
n_composites = a - primepi(a) - 1
while n_composites > n:
if not isprime(a):
n_composites -= 1
a -= 1
if isprime(a):
a -= 1
return a
def is_extra_strong_lucas_prp(n):
"""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
==========
- "Frobenius Pseudoprimes", Jon Grantham, 2000.
http://www.ams.org/journals/mcom/2001-70-234/S0025-5718-00-01197-2/
- OEIS A217719: Extra Strong Lucas Pseudoprimes
https://oeis.org/A217719
- https://en.wikipedia.org/wiki/Lucas_pseudoprime
Examples
========
>>> from sympy.ntheory.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
"""
# Implementation notes:
# 1) the parameters differ from Thomas R. Nicely's. His parameter
# selection leads to pseudoprimes that overlap M-R tests, and
# contradict Baillie and Wagstaff's suggestion of (D|n) = -1.
# 2) The MathWorld page as of June 2013 specifies Q=-1. The Lucas
# sequence must have Q=1. See Grantham theorem 2.3, any of the
# references on the MathWorld page, or run it and see Q=-1 is wrong.
from sympy.ntheory.factor_ import trailing
n = as_int(n)
if n == 2:
return True
if n < 2 or (n % 2) == 0:
return False
if is_square(n, False):
return False
D, P, Q = _lucas_extrastrong_params(n)
if D == 0:
return False
# remove powers of 2 from n+1 (= k * 2**s)
s = trailing(n + 1)
k = (n + 1) >> s
U, V, Qk = _lucas_sequence(n, P, Q, k)
if U == 0 and (V == 2 or V == n - 2):
return True
if V == 0:
return True
for r in range(1, s):
V = (V * V - 2) % n
if V == 0:
return True
return False
def solve_congruence(*remainder_modulus_pairs, **hint):
"""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 sympy.ntheory.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
"""
def combine(c1, c2):
"""Return the tuple (a, m) which satisfies the requirement
that n = a + i*m satisfy n = a1 + j*m1 and n = a2 = k*m2.
References
==========
- http://en.wikipedia.org/wiki/Method_of_successive_substitution
"""
a1, m1 = c1
a2, m2 = c2
a, b, c = m1, a2 - a1, m2
g = reduce(igcd, [a, b, c])
a, b, c = [i // g for i in [a, b, c]]
if a != 1:
inv_a, _, g = igcdex(a, c)
if g != 1:
return None
b *= inv_a
a, m = a1 + m1 * b, m1 * c
return a, m
rm = remainder_modulus_pairs
symmetric = hint.get('symmetric', False)
if hint.get('check', True):
rm = [(as_int(r), as_int(m)) for r, m in rm]
# ignore redundant pairs but raise an error otherwise; also
# make sure that a unique set of bases is sent to gf_crt if
# they are all prime.
#
# The routine will work out less-trivial violations and
# return None, e.g. for the pairs (1,3) and (14,42) there
# is no answer because 14 mod 42 (having a gcd of 14) implies
# (14/2) mod (42/2), (14/7) mod (42/7) and (14/14) mod (42/14)
# which, being 0 mod 3, is inconsistent with 1 mod 3. But to
# preprocess the input beyond checking of another pair with 42
# or 3 as the modulus (for this example) is not necessary.
uniq = {}
for r, m in rm:
r %= m
if m in uniq:
if r != uniq[m]:
return None
continue
uniq[m] = r
rm = [(r, m) for m, r in uniq.items()]
del uniq
# if the moduli are co-prime, the crt will be significantly faster;
# checking all pairs for being co-prime gets to be slow but a prime
# test is a good trade-off
if all(isprime(m) for r, m in rm):
r, m = list(zip(*rm))
return crt(m, r, symmetric=symmetric, check=False)
rv = (0, 1)
for rmi in rm:
rv = combine(rv, rmi)
if rv is None:
break
n, m = rv
n = n % m
else:
if symmetric:
return symmetric_residue(n, m), m
return n, m
def __mul__(self, other):
try:
n = as_int(other)
except ValueError:
raise TypeError("Can't multiply sequence by non-integer of type '%s'" % type(other))
return self.func(*(self.args*n))
def intlike(n):
try:
as_int(n, strict=False)
return True
except ValueError:
return False
def linrec(coeffs, init, n):
r"""
Evaluation of univariate linear recurrences of homogeneous type
having coefficients independent of the recurrence variable.
Parameters
==========
coeffs : iterable
Coefficients of the recurrence
init : iterable
Initial values of the recurrence
n : Integer
Point of evaluation for the recurrence
Notes
=====
Let `y(n)` be the recurrence of given type, ``c`` be the sequence
of coefficients, ``b`` be the sequence of initial/base values of the
recurrence and ``k`` (equal to ``len(c)``) be the order of recurrence.
Then,
.. math :: y(n) = \begin{cases} b_n & 0 \le n < k \\
c_0 y(n-1) + c_1 y(n-2) + \cdots + c_{k-1} y(n-k) & n \ge k
\end{cases}
Let `x_0, x_1, \ldots, x_n` be a sequence and consider the transformation
that maps each polynomial `f(x)` to `T(f(x))` where each power `x^i` is
replaced by the corresponding value `x_i`. The sequence is then a solution
of the recurrence if and only if `T(x^i p(x)) = 0` for each `i \ge 0` where
`p(x) = x^k - c_0 x^(k-1) - \cdots - c_{k-1}` is the characteristic
polynomial.
Then `T(f(x)p(x)) = 0` for each polynomial `f(x)` (as it is a linear
combination of powers `x^i`). Now, if `x^n` is congruent to
`g(x) = a_0 x^0 + a_1 x^1 + \cdots + a_{k-1} x^{k-1}` modulo `p(x)`, then
`T(x^n) = x_n` is equal to
`T(g(x)) = a_0 x_0 + a_1 x_1 + \cdots + a_{k-1} x_{k-1}`.
Computation of `x^n`,
given `x^k = c_0 x^{k-1} + c_1 x^{k-2} + \cdots + c_{k-1}`
is performed using exponentiation by squaring (refer to [1]_) with
an additional reduction step performed to retain only first `k` powers
of `x` in the representation of `x^n`.
Examples
========
>>> from sympy.discrete.recurrences import linrec
>>> from sympy.abc import x, y, z
>>> linrec(coeffs=[1, 1], init=[0, 1], n=10)
55
>>> linrec(coeffs=[1, 1], init=[x, y], n=10)
34*x + 55*y
>>> linrec(coeffs=[x, y], init=[0, 1], n=5)
x**2*y + x*(x**3 + 2*x*y) + y**2
>>> linrec(coeffs=[1, 2, 3, 0, 0, 4], init=[x, y, z], n=16)
13576*x + 5676*y + 2356*z
References
==========
.. [1] https://en.wikipedia.org/wiki/Exponentiation_by_squaring
.. [2] https://en.wikipedia.org/w/index.php?title=Modular_exponentiation§ion=6#Matrices
See Also
========
sympy.polys.agca.extensions.ExtensionElement.__pow__
"""
if not coeffs:
return S.Zero
if not iterable(coeffs):
raise TypeError("Expected a sequence of coefficients for"
" the recurrence")
if not iterable(init):
raise TypeError("Expected a sequence of values for the initialization"
" of the recurrence")
n = as_int(n)
if n < 0:
raise ValueError("Point of evaluation of recurrence must be a "
"non-negative integer")
c = [sympify(arg) for arg in coeffs]
b = [sympify(arg) for arg in init]
k = len(c)
if len(b) > k:
raise TypeError("Count of initial values should not exceed the "
"order of the recurrence")
else:
b += [S.Zero]*(k - len(b)) # remaining initial values default to zero
def _square_and_reduce(u, offset):
# squares `(u_0 + u_1 x + u_2 x^2 + \cdots + u_{k-1} x^k)` (and
# multiplies by `x` if offset is 1) and reduces the above result of
# length upto `2k` to `k` using the characteristic equation of the
# recurrence given by, `x^k = c_0 x^{k-1} + c_1 x^{k-2} + \cdots + c_{k-1}`
w = [S.Zero]*(2*len(u) - 1 + offset)
for i, p in enumerate(u):
for j, q in enumerate(u):
w[offset + i + j] += p*q
for j in range(len(w) - 1, k - 1, -1):
for i in range(k):
w[j - i - 1] += w[j]*c[i]
return w[:k]
def _final_coeffs(n):
# computes the final coefficient list - `cf` corresponding to the
# point at which recurrence is to be evalauted - `n`, such that,
# `y(n) = cf_0 y(k-1) + cf_1 y(k-2) + \cdots + cf_{k-1} y(0)`
if n < k:
return [S.Zero]*n + [S.One] + [S.Zero]*(k - n - 1)
else:
return _square_and_reduce(_final_coeffs(n//2), n%2)
return b[n] if n < k else sum(u*v for u, v in zip(_final_coeffs(n), b))
def convolution(a, b, cycle=0, dps=None, prime=None, dyadic=None, subset=None):
"""
Performs convolution by determining the type of desired
convolution using hints.
Exactly one of ``dps``, ``prime``, ``dyadic``, ``subset`` arguments
should be specified explicitly for identifying the type of convolution,
and the argument ``cycle`` can be specified optionally.
For the default arguments, linear convolution is performed using **FFT**.
Parameters
==========
a, b : iterables
The sequences for which convolution is performed.
cycle : Integer
Specifies the length for doing cyclic convolution.
dps : Integer
Specifies the number of decimal digits for precision for
performing **FFT** on the sequence.
prime : Integer
Prime modulus of the form `(m 2^k + 1)` to be used for
performing **NTT** on the sequence.
dyadic : bool
Identifies the convolution type as dyadic (*bitwise-XOR*)
convolution, which is performed using **FWHT**.
subset : bool
Identifies the convolution type as subset convolution.
Examples
========
>>> from sympy import convolution, symbols, S, I
>>> u, v, w, x, y, z = symbols('u v w x y z')
>>> convolution([1 + 2*I, 4 + 3*I], [S(5)/4, 6], dps=3)
[1.25 + 2.5*I, 11.0 + 15.8*I, 24.0 + 18.0*I]
>>> convolution([1, 2, 3], [4, 5, 6], cycle=3)
[31, 31, 28]
>>> convolution([111, 777], [888, 444], prime=19*2**10 + 1)
[1283, 19351, 14219]
>>> convolution([111, 777], [888, 444], prime=19*2**10 + 1, cycle=2)
[15502, 19351]
>>> convolution([u, v], [x, y, z], dyadic=True)
[u*x + v*y, u*y + v*x, u*z, v*z]
>>> convolution([u, v], [x, y, z], dyadic=True, cycle=2)
[u*x + u*z + v*y, u*y + v*x + v*z]
>>> convolution([u, v, w], [x, y, z], subset=True)
[u*x, u*y + v*x, u*z + w*x, v*z + w*y]
>>> convolution([u, v, w], [x, y, z], subset=True, cycle=3)
[u*x + v*z + w*y, u*y + v*x, u*z + w*x]
"""
c = as_int(cycle)
if c < 0:
raise ValueError("The length for cyclic convolution "
"must be non-negative")
dyadic = True if dyadic else None
subset = True if subset else None
if sum(x is not None for x in (prime, dps, dyadic, subset)) > 1:
raise TypeError("Ambiguity in determining the type of convolution")
if prime is not None:
ls = convolution_ntt(a, b, prime=prime)
return ls if not c else [sum(ls[i::c]) % prime for i in range(c)]
if dyadic:
ls = convolution_fwht(a, b)
elif subset:
ls = convolution_subset(a, b)
else:
ls = convolution_fft(a, b, dps=dps)
return ls if not c else [sum(ls[i::c]) for i in range(c)]
def isprime(n):
"""
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 sympy.ntheory import isprime
>>> isprime(13)
True
>>> isprime(13.0) # limited precision
False
>>> isprime(15)
False
Notes
=====
This routine is intended only for integer input, not numerical
expressions which may represent numbers. Floats are also
rejected as input because they represent numbers of limited
precision. While it is tempting to permit 7.0 to represent an
integer there are errors that may "pass silently" if this is
allowed:
>>> from sympy import Float, S
>>> int(1e3) == 1e3 == 10**3
True
>>> int(1e23) == 1e23
True
>>> int(1e23) == 10**23
False
>>> near_int = 1 + S(1)/10**19
>>> near_int == int(near_int)
False
>>> n = Float(near_int, 10) # truncated by precision
>>> n == int(n)
True
>>> n = Float(near_int, 20)
>>> n == int(n)
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
==========
- https://en.wikipedia.org/wiki/Strong_pseudoprime
- "Lucas Pseudoprimes", Baillie and Wagstaff, 1980.
http://mpqs.free.fr/LucasPseudoprimes.pdf
- https://en.wikipedia.org/wiki/Baillie-PSW_primality_test
"""
try:
n = as_int(n)
except ValueError:
return False
# Step 1, do quick composite testing via trial division. The individual
# modulo tests benchmark faster than one or two primorial igcds for me.
# The point here is just to speedily handle small numbers and many
# composites. Step 2 only requires that n <= 2 get handled here.
if n in [2, 3, 5]:
return True
if n < 2 or (n % 2) == 0 or (n % 3) == 0 or (n % 5) == 0:
return False
if n < 49:
return True
if (n % 7) == 0 or (n % 11) == 0 or (n % 13) == 0 or (n % 17) == 0 or \
(n % 19) == 0 or (n % 23) == 0 or (n % 29) == 0 or (n % 31) == 0 or \
(n % 37) == 0 or (n % 41) == 0 or (n % 43) == 0 or (n % 47) == 0:
return False
if n < 2809:
return True
if n <= 23001:
return pow(2, n, n) == 2 and n not in [7957, 8321, 13747, 18721, 19951]
# bisection search on the sieve if the sieve is large enough
from sympy.ntheory.generate import sieve as s
if n <= s._list[-1]:
l, u = s.search(n)
return l == u
# If we have GMPY2, skip straight to step 3 and do a strong BPSW test.
# This should be a bit faster than our step 2, and for large values will
# be a lot faster than our step 3 (C+GMP vs. Python).
from sympy.core.compatibility import HAS_GMPY
if HAS_GMPY == 2:
from gmpy2 import is_strong_prp, is_strong_selfridge_prp
return is_strong_prp(n, 2) and is_strong_selfridge_prp(n)
# Step 2: deterministic Miller-Rabin testing for numbers < 2^64. See:
# https://miller-rabin.appspot.com/
# for lists. We have made sure the M-R routine will successfully handle
# bases larger than n, so we can use the minimal set.
if n < 341531:
return mr(n, [9345883071009581737])
if n < 885594169:
return mr(n, [725270293939359937, 3569819667048198375])
if n < 350269456337:
return mr(
n,
[4230279247111683200, 14694767155120705706, 16641139526367750375])
if n < 55245642489451:
return mr(
n, [2, 141889084524735, 1199124725622454117, 11096072698276303650])
if n < 7999252175582851:
return mr(n, [
2, 4130806001517, 149795463772692060, 186635894390467037,
3967304179347715805
])
if n < 585226005592931977:
return mr(n, [
2, 123635709730000, 9233062284813009, 43835965440333360,
761179012939631437, 1263739024124850375
])
if n < 18446744073709551616:
return mr(n, [2, 325, 9375, 28178, 450775, 9780504, 1795265022])
# We could do this instead at any point:
#if n < 18446744073709551616:
# return mr(n, [2]) and is_extra_strong_lucas_prp(n)
# Here are tests that are safe for MR routines that don't understand
# large bases.
#if n < 9080191:
# return mr(n, [31, 73])
#if n < 19471033:
# return mr(n, [2, 299417])
#if n < 38010307:
# return mr(n, [2, 9332593])
#if n < 316349281:
# return mr(n, [11000544, 31481107])
#if n < 4759123141:
# return mr(n, [2, 7, 61])
#if n < 105936894253:
# return mr(n, [2, 1005905886, 1340600841])
#if n < 31858317218647:
# return mr(n, [2, 642735, 553174392, 3046413974])
#if n < 3071837692357849:
# return mr(n, [2, 75088, 642735, 203659041, 3613982119])
#if n < 18446744073709551616:
# return mr(n, [2, 325, 9375, 28178, 450775, 9780504, 1795265022])
# Step 3: BPSW.
#
# Time for isprime(10**2000 + 4561), no gmpy or gmpy2 installed
# 44.0s old isprime using 46 bases
# 5.3s strong BPSW + one random base
# 4.3s extra strong BPSW + one random base
# 4.1s strong BPSW
# 3.2s extra strong BPSW
# Classic BPSW from page 1401 of the paper. See alternate ideas below.
return mr(n, [2]) and is_strong_lucas_prp(n)
def factorint(n,
limit=None,
use_trial=True,
use_rho=True,
use_pm1=True,
verbose=False,
visual=None):
r"""
Given a positive integer ``n``, ``factorint(n)`` returns a dict containing
the prime factors of ``n`` as keys and their respective multiplicities
as values. For example:
>>> from sympy.ntheory import factorint
>>> factorint(2000) # 2000 = (2**4) * (5**3)
{2: 4, 5: 3}
>>> factorint(65537) # This number is prime
{65537: 1}
For input less than 2, factorint behaves as follows:
- ``factorint(1)`` returns the empty factorization, ``{}``
- ``factorint(0)`` returns ``{0:1}``
- ``factorint(-n)`` adds ``-1:1`` to the factors and then factors ``n``
Partial Factorization:
If ``limit`` (> 3) is specified, the search is stopped after performing
trial division up to (and including) the limit (or taking a
corresponding number of rho/p-1 steps). This is useful if one has
a large number and only is interested in finding small factors (if
any). Note that setting a limit does not prevent larger factors
from being found early; it simply means that the largest factor may
be composite. Since checking for perfect power is relatively cheap, it is
done regardless of the limit setting.
This number, for example, has two small factors and a huge
semi-prime factor that cannot be reduced easily:
>>> from sympy.ntheory import isprime
>>> from sympy.core.compatibility import long
>>> a = 1407633717262338957430697921446883
>>> f = factorint(a, limit=10000)
>>> f == {991: 1, long(202916782076162456022877024859): 1, 7: 1}
True
>>> isprime(max(f))
False
This number has a small factor and a residual perfect power whose
base is greater than the limit:
>>> factorint(3*101**7, limit=5)
{3: 1, 101: 7}
Visual Factorization:
If ``visual`` is set to ``True``, then it will return a visual
factorization of the integer. For example:
>>> from sympy import pprint
>>> pprint(factorint(4200, visual=True))
3 1 2 1
2 *3 *5 *7
Note that this is achieved by using the evaluate=False flag in Mul
and Pow. If you do other manipulations with an expression where
evaluate=False, it may evaluate. Therefore, you should use the
visual option only for visualization, and use the normal dictionary
returned by visual=False if you want to perform operations on the
factors.
You can easily switch between the two forms by sending them back to
factorint:
>>> from sympy import Mul, Pow
>>> regular = factorint(1764); regular
{2: 2, 3: 2, 7: 2}
>>> pprint(factorint(regular))
2 2 2
2 *3 *7
>>> visual = factorint(1764, visual=True); pprint(visual)
2 2 2
2 *3 *7
>>> print(factorint(visual))
{2: 2, 3: 2, 7: 2}
If you want to send a number to be factored in a partially factored form
you can do so with a dictionary or unevaluated expression:
>>> factorint(factorint({4: 2, 12: 3})) # twice to toggle to dict form
{2: 10, 3: 3}
>>> factorint(Mul(4, 12, evaluate=False))
{2: 4, 3: 1}
The table of the output logic is:
====== ====== ======= =======
Visual
------ ----------------------
Input True False other
====== ====== ======= =======
dict mul dict mul
n mul dict dict
mul mul dict dict
====== ====== ======= =======
Notes
=====
Algorithm:
The function switches between multiple algorithms. Trial division
quickly finds small factors (of the order 1-5 digits), and finds
all large factors if given enough time. The Pollard rho and p-1
algorithms are used to find large factors ahead of time; they
will often find factors of the order of 10 digits within a few
seconds:
>>> factors = factorint(12345678910111213141516)
>>> for base, exp in sorted(factors.items()):
... print('%s %s' % (base, exp))
...
2 2
2507191691 1
1231026625769 1
Any of these methods can optionally be disabled with the following
boolean parameters:
- ``use_trial``: Toggle use of trial division
- ``use_rho``: Toggle use of Pollard's rho method
- ``use_pm1``: Toggle use of Pollard's p-1 method
``factorint`` also periodically checks if the remaining part is
a prime number or a perfect power, and in those cases stops.
If ``verbose`` is set to ``True``, detailed progress is printed.
See Also
========
smoothness, smoothness_p, divisors
"""
factordict = {}
if visual and not isinstance(n, Mul) and not isinstance(n, dict):
factordict = factorint(n,
limit=limit,
use_trial=use_trial,
use_rho=use_rho,
use_pm1=use_pm1,
verbose=verbose,
visual=False)
elif isinstance(n, Mul):
factordict = dict([(int(k), int(v))
for k, v in list(n.as_powers_dict().items())])
elif isinstance(n, dict):
factordict = n
if factordict and (isinstance(n, Mul) or isinstance(n, dict)):
# check it
for k in list(factordict.keys()):
if isprime(k):
continue
e = factordict.pop(k)
d = factorint(k,
limit=limit,
use_trial=use_trial,
use_rho=use_rho,
use_pm1=use_pm1,
verbose=verbose,
visual=False)
for k, v in d.items():
if k in factordict:
factordict[k] += v * e
else:
factordict[k] = v * e
if visual or (type(n) is dict and visual is not True
and visual is not False):
if factordict == {}:
return S.One
if -1 in factordict:
factordict.pop(-1)
args = [S.NegativeOne]
else:
args = []
args.extend(
[Pow(*i, evaluate=False) for i in sorted(factordict.items())])
return Mul(*args, evaluate=False)
elif isinstance(n, dict) or isinstance(n, Mul):
return factordict
assert use_trial or use_rho or use_pm1
n = as_int(n)
if limit:
limit = int(limit)
# special cases
if n < 0:
factors = factorint(-n,
limit=limit,
use_trial=use_trial,
use_rho=use_rho,
use_pm1=use_pm1,
verbose=verbose,
visual=False)
factors[-1] = 1
return factors
if limit and limit < 2:
if n == 1:
return {}
return {n: 1}
elif n < 10:
# doing this we are assured of getting a limit > 2
# when we have to compute it later
return [{
0: 1
}, {}, {
2: 1
}, {
3: 1
}, {
2: 2
}, {
5: 1
}, {
2: 1,
3: 1
}, {
7: 1
}, {
2: 3
}, {
3: 2
}][n]
factors = {}
# do simplistic factorization
if verbose:
sn = str(n)
if len(sn) > 50:
print('Factoring %s' % sn[:5] + \
'..(%i other digits)..' % (len(sn) - 10) + sn[-5:])
else:
print('Factoring', n)
if use_trial:
# this is the preliminary factorization for small factors
small = 2**15
fail_max = 600
small = min(small, limit or small)
if verbose:
print(trial_int_msg % (2, small, fail_max))
n, next_p = _factorint_small(factors, n, small, fail_max)
else:
next_p = 2
if factors and verbose:
for k in sorted(factors):
print(factor_msg % (k, factors[k]))
if next_p == 0:
if n > 1:
factors[int(n)] = 1
if verbose:
print(complete_msg)
return factors
# continue with more advanced factorization methods
# first check if the simplistic run didn't finish
# because of the limit and check for a perfect
# power before exiting
try:
if limit and next_p > limit:
if verbose:
print('Exceeded limit:', limit)
_check_termination(factors, n, limit, use_trial, use_rho, use_pm1,
verbose)
if n > 1:
factors[int(n)] = 1
return factors
else:
# Before quitting (or continuing on)...
# ...do a Fermat test since it's so easy and we need the
# square root anyway. Finding 2 factors is easy if they are
# "close enough." This is the big root equivalent of dividing by
# 2, 3, 5.
sqrt_n = integer_nthroot(n, 2)[0]
a = sqrt_n + 1
a2 = a**2
b2 = a2 - n
for i in range(3):
b, fermat = integer_nthroot(b2, 2)
if fermat:
break
b2 += 2 * a + 1 # equiv to (a+1)**2 - n
a += 1
if fermat:
if verbose:
print(fermat_msg)
if limit:
limit -= 1
for r in [a - b, a + b]:
facs = factorint(r,
limit=limit,
use_trial=use_trial,
use_rho=use_rho,
use_pm1=use_pm1,
verbose=verbose)
factors.update(facs)
raise StopIteration
# ...see if factorization can be terminated
_check_termination(factors, n, limit, use_trial, use_rho, use_pm1,
verbose)
except StopIteration:
if verbose:
print(complete_msg)
return factors
# these are the limits for trial division which will
# be attempted in parallel with pollard methods
low, high = next_p, 2 * next_p
limit = limit or sqrt_n
# add 1 to make sure limit is reached in primerange calls
limit += 1
while 1:
try:
high_ = high
if limit < high_:
high_ = limit
# Trial division
if use_trial:
if verbose:
print(trial_msg % (low, high_))
ps = sieve.primerange(low, high_)
n, found_trial = _trial(factors, n, ps, verbose)
if found_trial:
_check_termination(factors, n, limit, use_trial, use_rho,
use_pm1, verbose)
else:
found_trial = False
if high > limit:
if verbose:
print('Exceeded limit:', limit)
if n > 1:
factors[int(n)] = 1
raise StopIteration
# Only used advanced methods when no small factors were found
if not found_trial:
if (use_pm1 or use_rho):
high_root = max(int(math.log(high_**0.7)), low, 3)
# Pollard p-1
if use_pm1:
if verbose:
print(pm1_msg % (high_root, high_))
c = pollard_pm1(n, B=high_root, seed=high_)
if c:
# factor it and let _trial do the update
ps = factorint(c,
limit=limit - 1,
use_trial=use_trial,
use_rho=use_rho,
use_pm1=use_pm1,
verbose=verbose)
n, _ = _trial(factors, n, ps, verbose=False)
_check_termination(factors, n, limit, use_trial,
use_rho, use_pm1, verbose)
# Pollard rho
if use_rho:
max_steps = high_root
if verbose:
print(rho_msg % (1, max_steps, high_))
c = pollard_rho(n,
retries=1,
max_steps=max_steps,
seed=high_)
if c:
# factor it and let _trial do the update
ps = factorint(c,
limit=limit - 1,
use_trial=use_trial,
use_rho=use_rho,
use_pm1=use_pm1,
verbose=verbose)
n, _ = _trial(factors, n, ps, verbose=False)
_check_termination(factors, n, limit, use_trial,
use_rho, use_pm1, verbose)
except StopIteration:
if verbose:
print(complete_msg)
return factors
low, high = high, high * 2