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from mpmath.libmp import (fzero, from_int, from_rational,
fone, fhalf, bitcount, to_int, mpf_mul, mpf_div, mpf_sub,
mpf_add, mpf_sqrt, mpf_pi, mpf_cosh_sinh, mpf_cos, mpf_sin)
from .residue_ntheory import _sqrt_mod_prime_power, is_quad_residue
from sympy.utilities.decorator import deprecated
from sympy.utilities.memoization import recurrence_memo
import math
from itertools import count
def _pre():
maxn = 10**5
global _factor, _totient
_factor = [0]*maxn
_totient = [1]*maxn
lim = int(maxn**0.5) + 5
for i in range(2, lim):
if _factor[i] == 0:
for j in range(i*i, maxn, i):
if _factor[j] == 0:
_factor[j] = i
for i in range(2, maxn):
if _factor[i] == 0:
_factor[i] = i
_totient[i] = i-1
continue
x = _factor[i]
y = i//x
if y % x == 0:
_totient[i] = _totient[y]*x
else:
_totient[i] = _totient[y]*(x - 1)
def _a(n, k, prec):
""" Compute the inner sum in HRR formula [1]_
References
==========
.. [1] https://msp.org/pjm/1956/6-1/pjm-v6-n1-p18-p.pdf
"""
if k == 1:
return fone
k1 = k
e = 0
p = _factor[k]
while k1 % p == 0:
k1 //= p
e += 1
k2 = k//k1 # k2 = p^e
v = 1 - 24*n
pi = mpf_pi(prec)
if k1 == 1:
# k = p^e
if p == 2:
mod = 8*k
v = mod + v % mod
v = (v*pow(9, k - 1, mod)) % mod
m = _sqrt_mod_prime_power(v, 2, e + 3)[0]
arg = mpf_div(mpf_mul(
from_int(4*m), pi, prec), from_int(mod), prec)
return mpf_mul(mpf_mul(
from_int((-1)**e*(2 - (m % 4))),
mpf_sqrt(from_int(k), prec), prec),
mpf_sin(arg, prec), prec)
if p == 3:
mod = 3*k
v = mod + v % mod
if e > 1:
v = (v*pow(64, k//3 - 1, mod)) % mod
m = _sqrt_mod_prime_power(v, 3, e + 1)[0]
arg = mpf_div(mpf_mul(from_int(4*m), pi, prec),
from_int(mod), prec)
return mpf_mul(mpf_mul(
from_int(2*(-1)**(e + 1)*(3 - 2*(m % 3))),
mpf_sqrt(from_int(k//3), prec), prec),
mpf_sin(arg, prec), prec)
v = k + v % k
jacobi3 = -1 if k % 12 in [5, 7] else 1
if v % p == 0:
if e == 1:
return mpf_mul(
from_int(jacobi3),
mpf_sqrt(from_int(k), prec), prec)
return fzero
if not is_quad_residue(v, p):
return fzero
_phi = p**(e - 1)*(p - 1)
v = (v*pow(576, _phi - 1, k))
m = _sqrt_mod_prime_power(v, p, e)[0]
arg = mpf_div(
mpf_mul(from_int(4*m), pi, prec),
from_int(k), prec)
return mpf_mul(mpf_mul(
from_int(2*jacobi3),
mpf_sqrt(from_int(k), prec), prec),
mpf_cos(arg, prec), prec)
if p != 2 or e >= 3:
d1, d2 = math.gcd(k1, 24), math.gcd(k2, 24)
e = 24//(d1*d2)
n1 = ((d2*e*n + (k2**2 - 1)//d1)*
pow(e*k2*k2*d2, _totient[k1] - 1, k1)) % k1
n2 = ((d1*e*n + (k1**2 - 1)//d2)*
pow(e*k1*k1*d1, _totient[k2] - 1, k2)) % k2
return mpf_mul(_a(n1, k1, prec), _a(n2, k2, prec), prec)
if e == 2:
n1 = ((8*n + 5)*pow(128, _totient[k1] - 1, k1)) % k1
n2 = (4 + ((n - 2 - (k1**2 - 1)//8)*(k1**2)) % 4) % 4
return mpf_mul(mpf_mul(
from_int(-1),
_a(n1, k1, prec), prec),
_a(n2, k2, prec))
n1 = ((8*n + 1)*pow(32, _totient[k1] - 1, k1)) % k1
n2 = (2 + (n - (k1**2 - 1)//8) % 2) % 2
return mpf_mul(_a(n1, k1, prec), _a(n2, k2, prec), prec)
def _d(n, j, prec, sq23pi, sqrt8):
"""
Compute the sinh term in the outer sum of the HRR formula.
The constants sqrt(2/3*pi) and sqrt(8) must be precomputed.
"""
j = from_int(j)
pi = mpf_pi(prec)
a = mpf_div(sq23pi, j, prec)
b = mpf_sub(from_int(n), from_rational(1, 24, prec), prec)
c = mpf_sqrt(b, prec)
ch, sh = mpf_cosh_sinh(mpf_mul(a, c), prec)
D = mpf_div(
mpf_sqrt(j, prec),
mpf_mul(mpf_mul(sqrt8, b), pi), prec)
E = mpf_sub(mpf_mul(a, ch), mpf_div(sh, c, prec), prec)
return mpf_mul(D, E)
@recurrence_memo([1, 1])
def _partition_rec(n: int, prev) -> int:
""" Calculate the partition function P(n)
Parameters
==========
n : int
nonnegative integer
"""
v = 0
penta = 0 # pentagonal number: 1, 5, 12, ...
for i in count():
penta += 3*i + 1
np = n - penta
if np < 0:
break
s = prev[np]
np -= i + 1
# np = n - gp where gp = generalized pentagonal: 2, 7, 15, ...
if 0 <= np:
s += prev[np]
v += -s if i % 2 else s
return v
def _partition(n: int) -> int:
""" Calculate the partition function P(n)
Parameters
==========
n : int
"""
if n < 0:
return 0
if (n <= 200_000 and n - _partition_rec.cache_length() < 70 or
_partition_rec.cache_length() == 2 and n < 14_400):
# There will be 2*10**5 elements created here
# and n elements created by partition, so in case we
# are going to be working with small n, we just
# use partition to calculate (and cache) the values
# since lookup is used there while summation, using
# _factor and _totient, will be used below. But we
# only do so if n is relatively close to the length
# of the cache since doing 1 calculation here is about
# the same as adding 70 elements to the cache. In addition,
# the startup here costs about the same as calculating the first
# 14,400 values via partition, so we delay startup here unless n
# is smaller than that.
return _partition_rec(n)
if '_factor' not in globals():
_pre()
# Estimate number of bits in p(n). This formula could be tidied
pbits = int((
math.pi*(2*n/3.)**0.5 -
math.log(4*n))/math.log(10) + 1) * \
math.log2(10)
prec = p = int(pbits*1.1 + 100)
# find the number of terms needed so rounded sum will be accurate
# using Rademacher's bound M(n, N) for the remainder after a partial
# sum of N terms (https://arxiv.org/pdf/1205.5991.pdf, (1.8))
c1 = 44*math.pi**2/(225*math.sqrt(3))
c2 = math.pi*math.sqrt(2)/75
c3 = math.pi*math.sqrt(2/3)
def _M(n, N):
sqrt = math.sqrt
return c1/sqrt(N) + c2*sqrt(N/(n - 1))*math.sinh(c3*sqrt(n)/N)
big = max(9, math.ceil(n**0.5)) # should be too large (for n > 65, ceil should work)
assert _M(n, big) < 0.5 # else double big until too large
while big > 40 and _M(n, big) < 0.5:
big //= 2
small = big
big = small*2
while big - small > 1:
N = (big + small)//2
if (er := _M(n, N)) < 0.5:
big = N
elif er >= 0.5:
small = N
M = big # done with function M; now have value
# sanity check for expected size of answer
if M > 10**5: # i.e. M > maxn
raise ValueError("Input too big") # i.e. n > 149832547102
# calculate it
s = fzero
sq23pi = mpf_mul(mpf_sqrt(from_rational(2, 3, p), p), mpf_pi(p), p)
sqrt8 = mpf_sqrt(from_int(8), p)
for q in range(1, M):
a = _a(n, q, p)
d = _d(n, q, p, sq23pi, sqrt8)
s = mpf_add(s, mpf_mul(a, d), prec)
# On average, the terms decrease rapidly in magnitude.
# Dynamically reducing the precision greatly improves
# performance.
p = bitcount(abs(to_int(d))) + 50
return int(to_int(mpf_add(s, fhalf, prec)))
@deprecated("""\
The `sympy.ntheory.partitions_.npartitions` has been moved to `sympy.functions.combinatorial.numbers.partition`.""",
deprecated_since_version="1.13",
active_deprecations_target='deprecated-ntheory-symbolic-functions')
def npartitions(n, verbose=False):
"""
Calculate the partition function P(n), i.e. the number of ways that
n can be written as a sum of positive integers.
.. deprecated:: 1.13
The ``npartitions`` function is deprecated. Use :class:`sympy.functions.combinatorial.numbers.partition`
instead. See its documentation for more information. See
:ref:`deprecated-ntheory-symbolic-functions` for details.
P(n) is computed using the Hardy-Ramanujan-Rademacher formula [1]_.
The correctness of this implementation has been tested through $10^{10}$.
Examples
========
>>> from sympy.functions.combinatorial.numbers import partition
>>> partition(25)
1958
References
==========
.. [1] https://mathworld.wolfram.com/PartitionFunctionP.html
"""
from sympy.functions.combinatorial.numbers import partition as func_partition
return func_partition(n)