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 [1]_.
The correctness of this implementation has been tested through 10**10.
Examples
========
>>> from sympy.ntheory import npartitions
>>> npartitions(25)
1958
References
==========
.. [1] http://mathworld.wolfram.com/PartitionFunctionP.html
"""
n = int(n)
if n < 0:
return 0
if n <= 5:
return [1, 1, 2, 3, 5, 7][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.log(10, 2)
prec = p = int(pbits*1.1 + 100)
s = fzero
M = max(6, int(0.24*n**0.5 + 4))
if M > 10**5:
raise ValueError("Input too big") # Corresponds to n > 1.7e11
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)
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
return int(to_int(mpf_add(s, fhalf, prec)))
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 [1]_.
The correctness of this implementation has been tested through 10**10.
Examples
========
>>> from sympy.ntheory import npartitions
>>> npartitions(25)
1958
References
==========
.. [1] http://mathworld.wolfram.com/PartitionFunctionP.html
"""
n = int(n)
if n < 0:
return 0
if n <= 5:
return [1, 1, 2, 3, 5, 7][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.log(10, 2)
prec = p = int(pbits * 1.1 + 100)
s = fzero
M = max(6, int(0.24 * n**0.5 + 4))
if M > 10**5:
raise ValueError("Input too big") # Corresponds to n > 1.7e11
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)
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
return int(to_int(mpf_add(s, fhalf, 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)
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)
def npartitions(n):
"""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.
The correctness of this implementation has been tested for 10**n
up to n = 8.
Examples
========
>>> npartitions(25)
1958
References
==========
* https://mathworld.wolfram.com/PartitionFunctionP.html
"""
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 range(1, M):
a = _a(n, q, p)
d = _d(n, q, p, sq23pi, sqrt8)
s = mpf_add(s, mpf_mul(a, d), prec)
debug('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
return int(to_int(mpf_add(s, fhalf, prec)))
def npartitions(n):
"""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.
The correctness of this implementation has been tested for 10**n
up to n = 8.
Examples
========
>>> npartitions(25)
1958
References
==========
* http://mathworld.wolfram.com/PartitionFunctionP.html
"""
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 range(1, M):
a = _a(n, q, p)
d = _d(n, q, p, sq23pi, sqrt8)
s = mpf_add(s, mpf_mul(a, d), prec)
debug("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
return int(to_int(mpf_add(s, fhalf, prec)))
def _make_tol(ctx):
hundred = (0, 25, 2, 5)
eps = (0, MPZ_ONE, 1-ctx.prec, 1)
return mpf_mul(hundred, eps)
def evalf_mul(v, prec, options):
res = pure_complex(v)
if res:
# the only pure complex that is a mul is h*I
_, h = res
im, _, im_acc, _ = evalf(h, prec, options)
return None, im, None, im_acc
args = list(v.args)
# see if any argument is NaN or oo and thus warrants a special return
special = []
from sympy.core.numbers import Float
for arg in args:
arg = evalf(arg, prec, options)
if arg[0] is None:
continue
arg = Float._new(arg[0], 1)
if arg is S.NaN or arg.is_infinite:
special.append(arg)
if special:
from sympy.core.mul import Mul
special = Mul(*special)
return evalf(special, prec + 4, {})
# With guard digits, multiplication in the real case does not destroy
# accuracy. This is also true in the complex case when considering the
# total accuracy; however accuracy for the real or imaginary parts
# separately may be lower.
acc = prec
# XXX: big overestimate
working_prec = prec + len(args) + 5
# Empty product is 1
start = man, exp, bc = MPZ(1), 0, 1
# First, we multiply all pure real or pure imaginary numbers.
# direction tells us that the result should be multiplied by
# I**direction; all other numbers get put into complex_factors
# to be multiplied out after the first phase.
last = len(args)
direction = 0
args.append(S.One)
complex_factors = []
for i, arg in enumerate(args):
if i != last and pure_complex(arg):
args[-1] = (args[-1] * arg).expand()
continue
elif i == last and arg is S.One:
continue
re, im, re_acc, im_acc = evalf(arg, working_prec, options)
if re and im:
complex_factors.append((re, im, re_acc, im_acc))
continue
elif re:
(s, m, e, b), w_acc = re, re_acc
elif im:
(s, m, e, b), w_acc = im, im_acc
direction += 1
else:
return None, None, None, None
direction += 2 * s
man *= m
exp += e
bc += b
if bc > 3 * working_prec:
man >>= working_prec
exp += working_prec
acc = min(acc, w_acc)
sign = (direction & 2) >> 1
if not complex_factors:
v = normalize(sign, man, exp, bitcount(man), prec, rnd)
# multiply by i
if direction & 1:
return None, v, None, acc
else:
return v, None, acc, None
else:
# initialize with the first term
if (man, exp, bc) != start:
# there was a real part; give it an imaginary part
re, im = (sign, man, exp, bitcount(man)), (0, MPZ(0), 0, 0)
i0 = 0
else:
# there is no real part to start (other than the starting 1)
wre, wim, wre_acc, wim_acc = complex_factors[0]
acc = min(acc, complex_accuracy((wre, wim, wre_acc, wim_acc)))
re = wre
im = wim
i0 = 1
for wre, wim, wre_acc, wim_acc in complex_factors[i0:]:
# acc is the overall accuracy of the product; we aren't
# computing exact accuracies of the product.
acc = min(acc, complex_accuracy((wre, wim, wre_acc, wim_acc)))
use_prec = working_prec
A = mpf_mul(re, wre, use_prec)
B = mpf_mul(mpf_neg(im), wim, use_prec)
C = mpf_mul(re, wim, use_prec)
D = mpf_mul(im, wre, use_prec)
re = mpf_add(A, B, use_prec)
im = mpf_add(C, D, use_prec)
if options.get('verbose'):
print("MUL: wanted", prec, "accurate bits, got", acc)
# multiply by I
if direction & 1:
re, im = mpf_neg(im), re
return re, im, acc, acc
def _a(n, k, prec):
""" Compute the inner sum in HRR formula [1]_
References
==========
.. [1] http://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*jacobi_symbol(m - 1, m)),
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)*legendre_symbol(m, 3)),
mpf_sqrt(from_int(k//3), prec), prec),
mpf_sin(arg, prec), prec)
v = k + v % k
if v % p == 0:
if e == 1:
return mpf_mul(
from_int(jacobi_symbol(3, k)),
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*jacobi_symbol(3, k)),
mpf_sqrt(from_int(k), prec), prec),
mpf_cos(arg, prec), prec)
if p != 2 or e >= 3:
d1, d2 = igcd(k1, 24), igcd(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 _make_tol(self):
hundred = (0, 25, 2, 5)
eps = (0, MPZ_ONE, 1 - self.prec, 1)
return mpf_mul(hundred, eps)
def _a(n, k, prec):
""" Compute the inner sum in HRR formula [1]_
References
==========
.. [1] http://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 * jacobi_symbol(m - 1, m)),
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) * legendre_symbol(m, 3)),
mpf_sqrt(from_int(k // 3), prec), prec),
mpf_sin(arg, prec), prec)
v = k + v % k
if v % p == 0:
if e == 1:
return mpf_mul(from_int(jacobi_symbol(3, k)),
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 * jacobi_symbol(3, k)),
mpf_sqrt(from_int(k), prec), prec), mpf_cos(arg, prec),
prec)
if p != 2 or e >= 3:
d1, d2 = igcd(k1, 24), igcd(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 evalf_mul(v, prec, options):
res = pure_complex(v)
if res:
# the only pure complex that is a mul is h*I
_, h = res
im, _, im_acc, _ = evalf(h, prec, options)
return None, im, None, im_acc
args = list(v.args)
# see if any argument is NaN or oo and thus warrants a special return
special = []
from sympy.core.numbers import Float
for arg in args:
arg = evalf(arg, prec, options)
if arg[0] is None:
continue
arg = Float._new(arg[0], 1)
if arg is S.NaN or arg.is_infinite:
special.append(arg)
if special:
from sympy.core.mul import Mul
special = Mul(*special)
return evalf(special, prec + 4, {})
# With guard digits, multiplication in the real case does not destroy
# accuracy. This is also true in the complex case when considering the
# total accuracy; however accuracy for the real or imaginary parts
# separately may be lower.
acc = prec
# XXX: big overestimate
working_prec = prec + len(args) + 5
# Empty product is 1
start = man, exp, bc = MPZ(1), 0, 1
# First, we multiply all pure real or pure imaginary numbers.
# direction tells us that the result should be multiplied by
# I**direction; all other numbers get put into complex_factors
# to be multiplied out after the first phase.
last = len(args)
direction = 0
args.append(S.One)
complex_factors = []
for i, arg in enumerate(args):
if i != last and pure_complex(arg):
args[-1] = (args[-1]*arg).expand()
continue
elif i == last and arg is S.One:
continue
re, im, re_acc, im_acc = evalf(arg, working_prec, options)
if re and im:
complex_factors.append((re, im, re_acc, im_acc))
continue
elif re:
(s, m, e, b), w_acc = re, re_acc
elif im:
(s, m, e, b), w_acc = im, im_acc
direction += 1
else:
return None, None, None, None
direction += 2*s
man *= m
exp += e
bc += b
if bc > 3*working_prec:
man >>= working_prec
exp += working_prec
acc = min(acc, w_acc)
sign = (direction & 2) >> 1
if not complex_factors:
v = normalize(sign, man, exp, bitcount(man), prec, rnd)
# multiply by i
if direction & 1:
return None, v, None, acc
else:
return v, None, acc, None
else:
# initialize with the first term
if (man, exp, bc) != start:
# there was a real part; give it an imaginary part
re, im = (sign, man, exp, bitcount(man)), (0, MPZ(0), 0, 0)
i0 = 0
else:
# there is no real part to start (other than the starting 1)
wre, wim, wre_acc, wim_acc = complex_factors[0]
acc = min(acc,
complex_accuracy((wre, wim, wre_acc, wim_acc)))
re = wre
im = wim
i0 = 1
for wre, wim, wre_acc, wim_acc in complex_factors[i0:]:
# acc is the overall accuracy of the product; we aren't
# computing exact accuracies of the product.
acc = min(acc,
complex_accuracy((wre, wim, wre_acc, wim_acc)))
use_prec = working_prec
A = mpf_mul(re, wre, use_prec)
B = mpf_mul(mpf_neg(im), wim, use_prec)
C = mpf_mul(re, wim, use_prec)
D = mpf_mul(im, wre, use_prec)
re = mpf_add(A, B, use_prec)
im = mpf_add(C, D, use_prec)
if options.get('verbose'):
print("MUL: wanted", prec, "accurate bits, got", acc)
# multiply by I
if direction & 1:
re, im = mpf_neg(im), re
return re, im, acc, acc
def _make_tol(ctx):
hundred = (0, 25, 2, 5)
eps = (0, MPZ_ONE, 1-ctx.prec, 1)
return mpf_mul(hundred, eps)
def _make_tol(self):
hundred = (0, 25, 2, 5)
eps = (0, MPZ_ONE, 1 - self.prec, 1)
return mpf_mul(hundred, eps)
