# Source code for sympy.ntheory.partitions_

from __future__ import print_function, division

from sympy.mpmath.libmp import (fzero,
from_man_exp, from_int, from_rational,
fone, fhalf, bitcount, to_int, to_str, mpf_mul, mpf_div, mpf_sub,
mpf_add, mpf_sqrt, mpf_pi, mpf_cosh_sinh, pi_fixed, mpf_cos)
from sympy.core.numbers import igcd
import math
from sympy.core.compatibility import xrange

def _a(n, j, prec):
"""Compute the inner sum in the HRR formula."""
if j == 1:
return fone
s = fzero
pi = pi_fixed(prec)
for h in xrange(1, j):
if igcd(h, j) != 1:
continue
# & with mask to compute fractional part of fixed-point number
one = 1 << prec
half = one >> 1
g = 0
if j >= 3:
for k in xrange(1, j):
t = h*k*one//j
if t > 0:
else:
g += k*(frac - half)
g = ((g - 2*h*n*one)*pi//j) >> prec
s = mpf_add(s, mpf_cos(from_man_exp(g, -prec), prec), prec)
return s

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)

[docs]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.

P(n) is computed using the Hardy-Ramanujan-Rademacher formula,
described e.g. at http://mathworld.wolfram.com/PartitionFunctionP.html

The correctness of this implementation has been tested for 10**n
up to n = 8.

Examples
========

>>> from sympy.ntheory import npartitions
>>> npartitions(25)
1958
"""
n = int(n)
if n < 0:
return 0
if n <= 5:
return [1, 1, 2, 3, 5, 7][n]
# 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.log(10, 2)
prec = p = int(pbits*1.1 + 100)
s = fzero
M = max(6, int(0.24*n**0.5 + 4))
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 xrange(1, M):
a = _a(n, q, p)
d = _d(n, q, p, sq23pi, sqrt8)
s = mpf_add(s, mpf_mul(a, d), prec)
if verbose:
print("step", q, "of", M, to_str(a, 10), to_str(d, 10))
# On average, the terms decrease rapidly in magnitude. Dynamically
# reducing the precision greatly improves performance.
p = bitcount(abs(to_int(d))) + 50