193 lines
5.8 KiB
Python
193 lines
5.8 KiB
Python
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from mpmath.libmp import (fzero, from_int, from_rational,
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fone, fhalf, bitcount, to_int, to_str, mpf_mul, mpf_div, mpf_sub,
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mpf_add, mpf_sqrt, mpf_pi, mpf_cosh_sinh, mpf_cos, mpf_sin)
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from sympy.core.numbers import igcd
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from .residue_ntheory import (_sqrt_mod_prime_power,
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legendre_symbol, jacobi_symbol, is_quad_residue)
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import math
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def _pre():
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maxn = 10**5
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global _factor
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global _totient
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_factor = [0]*maxn
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_totient = [1]*maxn
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lim = int(maxn**0.5) + 5
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for i in range(2, lim):
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if _factor[i] == 0:
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for j in range(i*i, maxn, i):
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if _factor[j] == 0:
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_factor[j] = i
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for i in range(2, maxn):
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if _factor[i] == 0:
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_factor[i] = i
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_totient[i] = i-1
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continue
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x = _factor[i]
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y = i//x
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if y % x == 0:
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_totient[i] = _totient[y]*x
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else:
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_totient[i] = _totient[y]*(x - 1)
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def _a(n, k, prec):
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""" Compute the inner sum in HRR formula [1]_
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References
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==========
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.. [1] https://msp.org/pjm/1956/6-1/pjm-v6-n1-p18-p.pdf
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"""
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if k == 1:
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return fone
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k1 = k
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e = 0
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p = _factor[k]
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while k1 % p == 0:
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k1 //= p
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e += 1
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k2 = k//k1 # k2 = p^e
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v = 1 - 24*n
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pi = mpf_pi(prec)
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if k1 == 1:
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# k = p^e
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if p == 2:
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mod = 8*k
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v = mod + v % mod
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v = (v*pow(9, k - 1, mod)) % mod
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m = _sqrt_mod_prime_power(v, 2, e + 3)[0]
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arg = mpf_div(mpf_mul(
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from_int(4*m), pi, prec), from_int(mod), prec)
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return mpf_mul(mpf_mul(
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from_int((-1)**e*jacobi_symbol(m - 1, m)),
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mpf_sqrt(from_int(k), prec), prec),
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mpf_sin(arg, prec), prec)
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if p == 3:
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mod = 3*k
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v = mod + v % mod
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if e > 1:
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v = (v*pow(64, k//3 - 1, mod)) % mod
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m = _sqrt_mod_prime_power(v, 3, e + 1)[0]
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arg = mpf_div(mpf_mul(from_int(4*m), pi, prec),
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from_int(mod), prec)
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return mpf_mul(mpf_mul(
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from_int(2*(-1)**(e + 1)*legendre_symbol(m, 3)),
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mpf_sqrt(from_int(k//3), prec), prec),
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mpf_sin(arg, prec), prec)
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v = k + v % k
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if v % p == 0:
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if e == 1:
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return mpf_mul(
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from_int(jacobi_symbol(3, k)),
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mpf_sqrt(from_int(k), prec), prec)
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return fzero
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if not is_quad_residue(v, p):
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return fzero
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_phi = p**(e - 1)*(p - 1)
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v = (v*pow(576, _phi - 1, k))
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m = _sqrt_mod_prime_power(v, p, e)[0]
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arg = mpf_div(
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mpf_mul(from_int(4*m), pi, prec),
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from_int(k), prec)
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return mpf_mul(mpf_mul(
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from_int(2*jacobi_symbol(3, k)),
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mpf_sqrt(from_int(k), prec), prec),
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mpf_cos(arg, prec), prec)
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if p != 2 or e >= 3:
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d1, d2 = igcd(k1, 24), igcd(k2, 24)
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e = 24//(d1*d2)
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n1 = ((d2*e*n + (k2**2 - 1)//d1)*
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pow(e*k2*k2*d2, _totient[k1] - 1, k1)) % k1
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n2 = ((d1*e*n + (k1**2 - 1)//d2)*
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pow(e*k1*k1*d1, _totient[k2] - 1, k2)) % k2
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return mpf_mul(_a(n1, k1, prec), _a(n2, k2, prec), prec)
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if e == 2:
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n1 = ((8*n + 5)*pow(128, _totient[k1] - 1, k1)) % k1
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n2 = (4 + ((n - 2 - (k1**2 - 1)//8)*(k1**2)) % 4) % 4
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return mpf_mul(mpf_mul(
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from_int(-1),
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_a(n1, k1, prec), prec),
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_a(n2, k2, prec))
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n1 = ((8*n + 1)*pow(32, _totient[k1] - 1, k1)) % k1
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n2 = (2 + (n - (k1**2 - 1)//8) % 2) % 2
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return mpf_mul(_a(n1, k1, prec), _a(n2, k2, prec), prec)
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def _d(n, j, prec, sq23pi, sqrt8):
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"""
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Compute the sinh term in the outer sum of the HRR formula.
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The constants sqrt(2/3*pi) and sqrt(8) must be precomputed.
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"""
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j = from_int(j)
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pi = mpf_pi(prec)
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a = mpf_div(sq23pi, j, prec)
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b = mpf_sub(from_int(n), from_rational(1, 24, prec), prec)
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c = mpf_sqrt(b, prec)
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ch, sh = mpf_cosh_sinh(mpf_mul(a, c), prec)
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D = mpf_div(
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mpf_sqrt(j, prec),
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mpf_mul(mpf_mul(sqrt8, b), pi), prec)
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E = mpf_sub(mpf_mul(a, ch), mpf_div(sh, c, prec), prec)
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return mpf_mul(D, E)
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def npartitions(n, verbose=False):
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"""
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Calculate the partition function P(n), i.e. the number of ways that
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n can be written as a sum of positive integers.
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P(n) is computed using the Hardy-Ramanujan-Rademacher formula [1]_.
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The correctness of this implementation has been tested through $10^{10}$.
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Examples
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========
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>>> from sympy.ntheory import npartitions
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>>> npartitions(25)
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1958
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References
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==========
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.. [1] https://mathworld.wolfram.com/PartitionFunctionP.html
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"""
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n = int(n)
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if n < 0:
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return 0
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if n <= 5:
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return [1, 1, 2, 3, 5, 7][n]
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if '_factor' not in globals():
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_pre()
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# Estimate number of bits in p(n). This formula could be tidied
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pbits = int((
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math.pi*(2*n/3.)**0.5 -
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math.log(4*n))/math.log(10) + 1) * \
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math.log(10, 2)
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prec = p = int(pbits*1.1 + 100)
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s = fzero
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M = max(6, int(0.24*n**0.5 + 4))
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if M > 10**5:
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raise ValueError("Input too big") # Corresponds to n > 1.7e11
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sq23pi = mpf_mul(mpf_sqrt(from_rational(2, 3, p), p), mpf_pi(p), p)
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sqrt8 = mpf_sqrt(from_int(8), p)
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for q in range(1, M):
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a = _a(n, q, p)
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d = _d(n, q, p, sq23pi, sqrt8)
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s = mpf_add(s, mpf_mul(a, d), prec)
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if verbose:
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print("step", q, "of", M, to_str(a, 10), to_str(d, 10))
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# On average, the terms decrease rapidly in magnitude.
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# Dynamically reducing the precision greatly improves
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# performance.
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p = bitcount(abs(to_int(d))) + 50
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return int(to_int(mpf_add(s, fhalf, prec)))
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__all__ = ['npartitions']
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