Initialisation du repository de Beta

venv/lib/python3.12/site-packages/sympy/ntheory/bbp_pi.py

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+'''
+This implementation is a heavily modified fixed point implementation of
+BBP_formula for calculating the nth position of pi. The original hosted
+at: https://web.archive.org/web/20151116045029/http://en.literateprograms.org/Pi_with_the_BBP_formula_(Python)
+
+# Permission is hereby granted, free of charge, to any person obtaining
+# a copy of this software and associated documentation files (the
+# "Software"), to deal in the Software without restriction, including
+# without limitation the rights to use, copy, modify, merge, publish,
+# distribute, sub-license, and/or sell copies of the Software, and to
+# permit persons to whom the Software is furnished to do so, subject to
+# the following conditions:
+#
+# The above copyright notice and this permission notice shall be
+# included in all copies or substantial portions of the Software.
+#
+# THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
+# EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
+# MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.
+# IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY
+# CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT,
+# TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE
+# SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
+
+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).
+
+'''
+
+from sympy.utilities.misc import as_int
+
+
+def _series(j, n, prec=14):
+
+    # Left sum from the bbp algorithm
+    s = 0
+    D = _dn(n, prec)
+    D4 = 4 * D
+    d = 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 = D4 - 4 # 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).
+
+    Parameters
+    ==========
+
+    n : non-negative integer
+    prec : non-negative integer. default = 14
+
+    Returns
+    =======
+
+    str : Returns a string containing ``prec`` digits
+          starting at the nth digit of pi in hex.
+          If ``prec`` = 0, returns empty string.
+
+    Raises
+    ======
+
+    ValueError
+        If ``n`` < 0 or ``prec`` < 0.
+        Or ``n`` or ``prec`` is not an integer.
+
+    Examples
+    ========
+
+    >>> from sympy.ntheory.bbp_pi import pi_hex_digits
+    >>> pi_hex_digits(0)
+    '3243f6a8885a30'
+    >>> pi_hex_digits(0, 3)
+    '324'
+
+    These are consistent with the following results
+
+    >>> import math
+    >>> hex(int(math.pi * 2**((14-1)*4)))
+    '0x3243f6a8885a30'
+    >>> hex(int(math.pi * 2**((3-1)*4)))
+    '0x324'
+
+    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:
+        raise ValueError('prec 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
+
+    # assert int(math.log(n + prec)/math.log(16)) ==\
+    #  ((n + prec).bit_length() - 1) // 4
+    return ((n + prec).bit_length() - 1) // 4 + prec + 3