'''
This implementation is a heavily modified fixed point implementation of
BBP_formula for calculating the nth position of pi. The original hosted
at: http://en.literateprograms.org/Pi_with_the_BBP_formula_(Python)

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Modifications:

1.Once the nth digit and desired number of digits is selected, the
number of digits of working precision is calculated to ensure that
the hexadecimal digits returned are accurate. This is calculated as

    int(math.log(start + prec)/math.log(16) + prec + 3)
    ---------------------------------------   --------
                      /                          /
    number of hex digits           additional digits

This was checked by the following code which completed without
errors (and dig are the digits included in the test_bbp.py file):

    for i in range(0,1000):
     for j in range(1,1000):
      a, b = pi_hex_digits(i, j), dig[i:i+j]
      if a != b:
        print('%s\n%s'%(a,b))

Deceasing the additional digits by 1 generated errors, so '3' is
the smallest additional precision needed to calculate the above
loop without errors. The following trailing 10 digits were also
checked to be accurate (and the times were slightly faster with
some of the constant modifications that were made):

    >> from time import time
    >> t=time();pi_hex_digits(10**2-10 + 1, 10), time()-t
    ('e90c6cc0ac', 0.0)
    >> t=time();pi_hex_digits(10**4-10 + 1, 10), time()-t
    ('26aab49ec6', 0.17100000381469727)
    >> t=time();pi_hex_digits(10**5-10 + 1, 10), time()-t
    ('a22673c1a5', 4.7109999656677246)
    >> t=time();pi_hex_digits(10**6-10 + 1, 10), time()-t
    ('9ffd342362', 59.985999822616577)
    >> t=time();pi_hex_digits(10**7-10 + 1, 10), time()-t
    ('c1a42e06a1', 689.51800012588501)

2. The while loop to evaluate whether the series has converged quits
when the addition amount `dt` has dropped to zero.

3. the formatting string to convert the decimal to hexadecimal is
calculated for the given precision.

4. pi_hex_digits(n) changed to have coefficient to the formula in an
array (perhaps just a matter of preference).

'''

import math
from sympy.core.compatibility import as_int


def _series(j, n, prec=14):

    # Left sum from the bbp algorithm
    s = 0
    D = _dn(n, prec)
    D4 = 4 * D
    k = 0
    d = 8 * k + j
    for k in range(n + 1):
        s += (pow(16, n - k, d) << D4) // d
        d += 8

    # Right sum iterates to infinity for full precision, but we
    # stop at the point where one iteration is beyond the precision
    # specified.

    t = 0
    k = n + 1
    e = 4*(D + n - k)
    d = 8 * k + j
    while True:
        dt = (1 << e) // d
        if not dt:
            break
        t += dt
        # k += 1
        e -= 4
        d += 8
    total = s + t

    return total


def pi_hex_digits(n, prec=14):
    """Returns a string containing ``prec`` (default 14) digits
    starting at the nth digit of pi in hex. Counting of digits
    starts at 0 and the decimal is not counted, so for n = 0 the
    returned value starts with 3; n = 1 corresponds to the first
    digit past the decimal point (which in hex is 2).

    Examples
    ========

    >>> from sympy.ntheory.bbp_pi import pi_hex_digits
    >>> pi_hex_digits(0)
    '3243f6a8885a30'
    >>> pi_hex_digits(0, 3)
    '324'

    References
    ==========

    .. [1] http://www.numberworld.org/digits/Pi/
    """
    n, prec = as_int(n), as_int(prec)
    if n < 0:
        raise ValueError('n cannot be negative')
    if prec == 0:
        return ''

    # main of implementation arrays holding formulae coefficients
    n -= 1
    a = [4, 2, 1, 1]
    j = [1, 4, 5, 6]

    #formulae
    D = _dn(n, prec)
    x = + (a[0]*_series(j[0], n, prec)
         - a[1]*_series(j[1], n, prec)
         - a[2]*_series(j[2], n, prec)
         - a[3]*_series(j[3], n, prec)) & (16**D - 1)

    s = ("%0" + "%ix" % prec) % (x // 16**(D - prec))
    return s


def _dn(n, prec):
    # controller for n dependence on precision
    # n = starting digit index
    # prec = the number of total digits to compute
    n += 1  # because we subtract 1 for _series
    return int(math.log(n + prec)/math.log(16) + prec + 3)
