Vehicle-cpp/hkpc/software/sympy/ntheory/continued_fraction.py

from __future__ import annotations
from sympy.core.exprtools import factor_terms
from sympy.core.numbers import Integer, Rational
from sympy.core.singleton import S
from sympy.core.symbol import Dummy
from sympy.core.sympify import _sympify
from sympy.utilities.misc import as_int


def continued_fraction(a) -> list:
    """Return the continued fraction representation of a Rational or
    quadratic irrational.

    Examples
    ========

    >>> from sympy.ntheory.continued_fraction import continued_fraction
    >>> from sympy import sqrt
    >>> continued_fraction((1 + 2*sqrt(3))/5)
    [0, 1, [8, 3, 34, 3]]

    See Also
    ========
    continued_fraction_periodic, continued_fraction_reduce, continued_fraction_convergents
    """
    e = _sympify(a)
    if all(i.is_Rational for i in e.atoms()):
        if e.is_Integer:
            return continued_fraction_periodic(e, 1, 0)
        elif e.is_Rational:
            return continued_fraction_periodic(e.p, e.q, 0)
        elif e.is_Pow and e.exp is S.Half and e.base.is_Integer:
            return continued_fraction_periodic(0, 1, e.base)
        elif e.is_Mul and len(e.args) == 2 and (
                e.args[0].is_Rational and
                e.args[1].is_Pow and
                e.args[1].base.is_Integer and
                e.args[1].exp is S.Half):
            a, b = e.args
            return continued_fraction_periodic(0, a.q, b.base, a.p)
        else:
            # this should not have to work very hard- no
            # simplification, cancel, etc... which should be
            # done by the user.  e.g. This is a fancy 1 but
            # the user should simplify it first:
            # sqrt(2)*(1 + sqrt(2))/(sqrt(2) + 2)
            p, d = e.expand().as_numer_denom()
            if d.is_Integer:
                if p.is_Rational:
                    return continued_fraction_periodic(p, d)
                # look for a + b*c
                # with c = sqrt(s)
                if p.is_Add and len(p.args) == 2:
                    a, bc = p.args
                else:
                    a = S.Zero
                    bc = p
                if a.is_Integer:
                    b = S.NaN
                    if bc.is_Mul and len(bc.args) == 2:
                        b, c = bc.args
                    elif bc.is_Pow:
                        b = Integer(1)
                        c = bc
                    if b.is_Integer and (
                            c.is_Pow and c.exp is S.Half and
                            c.base.is_Integer):
                        # (a + b*sqrt(c))/d
                        c = c.base
                        return continued_fraction_periodic(a, d, c, b)
    raise ValueError(
        'expecting a rational or quadratic irrational, not %s' % e)


def continued_fraction_periodic(p, q, d=0, s=1) -> list:
    r"""
    Find the periodic continued fraction expansion of a quadratic irrational.

    Compute the continued fraction expansion of a rational or a
    quadratic irrational number, i.e. `\frac{p + s\sqrt{d}}{q}`, where
    `p`, `q \ne 0` and `d \ge 0` are integers.

    Returns the continued fraction representation (canonical form) as
    a list of integers, optionally ending (for quadratic irrationals)
    with list of integers representing the repeating digits.

    Parameters
    ==========

    p : int
        the rational part of the number's numerator
    q : int
        the denominator of the number
    d : int, optional
        the irrational part (discriminator) of the number's numerator
    s : int, optional
        the coefficient of the irrational part

    Examples
    ========

    >>> from sympy.ntheory.continued_fraction import continued_fraction_periodic
    >>> continued_fraction_periodic(3, 2, 7)
    [2, [1, 4, 1, 1]]

    Golden ratio has the simplest continued fraction expansion:

    >>> continued_fraction_periodic(1, 2, 5)
    [[1]]

    If the discriminator is zero or a perfect square then the number will be a
    rational number:

    >>> continued_fraction_periodic(4, 3, 0)
    [1, 3]
    >>> continued_fraction_periodic(4, 3, 49)
    [3, 1, 2]

    See Also
    ========

    continued_fraction_iterator, continued_fraction_reduce

    References
    ==========

    .. [1] https://en.wikipedia.org/wiki/Periodic_continued_fraction
    .. [2] K. Rosen. Elementary Number theory and its applications.
           Addison-Wesley, 3 Sub edition, pages 379-381, January 1992.

    """
    from sympy.functions import sqrt, floor

    p, q, d, s = list(map(as_int, [p, q, d, s]))

    if d < 0:
        raise ValueError("expected non-negative for `d` but got %s" % d)

    if q == 0:
        raise ValueError("The denominator cannot be 0.")

    if not s:
        d = 0

    # check for rational case
    sd = sqrt(d)
    if sd.is_Integer:
        return list(continued_fraction_iterator(Rational(p + s*sd, q)))

    # irrational case with sd != Integer
    if q < 0:
        p, q, s = -p, -q, -s

    n = (p + s*sd)/q
    if n < 0:
        w = floor(-n)
        f = -n - w
        one_f = continued_fraction(1 - f)  # 1-f < 1 so cf is [0 ... [...]]
        one_f[0] -= w + 1
        return one_f

    d *= s**2
    sd *= s

    if (d - p**2)%q:
        d *= q**2
        sd *= q
        p *= q
        q *= q

    terms: list[int] = []
    pq = {}

    while (p, q) not in pq:
        pq[(p, q)] = len(terms)
        terms.append((p + sd)//q)
        p = terms[-1]*q - p
        q = (d - p**2)//q

    i = pq[(p, q)]
    return terms[:i] + [terms[i:]]  # type: ignore


def continued_fraction_reduce(cf):
    """
    Reduce a continued fraction to a rational or quadratic irrational.

    Compute the rational or quadratic irrational number from its
    terminating or periodic continued fraction expansion.  The
    continued fraction expansion (cf) should be supplied as a
    terminating iterator supplying the terms of the expansion.  For
    terminating continued fractions, this is equivalent to
    ``list(continued_fraction_convergents(cf))[-1]``, only a little more
    efficient.  If the expansion has a repeating part, a list of the
    repeating terms should be returned as the last element from the
    iterator.  This is the format returned by
    continued_fraction_periodic.

    For quadratic irrationals, returns the largest solution found,
    which is generally the one sought, if the fraction is in canonical
    form (all terms positive except possibly the first).

    Examples
    ========

    >>> from sympy.ntheory.continued_fraction import continued_fraction_reduce
    >>> continued_fraction_reduce([1, 2, 3, 4, 5])
    225/157
    >>> continued_fraction_reduce([-2, 1, 9, 7, 1, 2])
    -256/233
    >>> continued_fraction_reduce([2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8]).n(10)
    2.718281835
    >>> continued_fraction_reduce([1, 4, 2, [3, 1]])
    (sqrt(21) + 287)/238
    >>> continued_fraction_reduce([[1]])
    (1 + sqrt(5))/2
    >>> from sympy.ntheory.continued_fraction import continued_fraction_periodic
    >>> continued_fraction_reduce(continued_fraction_periodic(8, 5, 13))
    (sqrt(13) + 8)/5

    See Also
    ========

    continued_fraction_periodic

    """
    from sympy.solvers import solve

    period = []
    x = Dummy('x')

    def untillist(cf):
        for nxt in cf:
            if isinstance(nxt, list):
                period.extend(nxt)
                yield x
                break
            yield nxt

    a = S.Zero
    for a in continued_fraction_convergents(untillist(cf)):
        pass

    if period:
        y = Dummy('y')
        solns = solve(continued_fraction_reduce(period + [y]) - y, y)
        solns.sort()
        pure = solns[-1]
        rv = a.subs(x, pure).radsimp()
    else:
        rv = a
    if rv.is_Add:
        rv = factor_terms(rv)
        if rv.is_Mul and rv.args[0] == -1:
            rv = rv.func(*rv.args)
    return rv


def continued_fraction_iterator(x):
    """
    Return continued fraction expansion of x as iterator.

    Examples
    ========

    >>> from sympy import Rational, pi
    >>> from sympy.ntheory.continued_fraction import continued_fraction_iterator

    >>> list(continued_fraction_iterator(Rational(3, 8)))
    [0, 2, 1, 2]
    >>> list(continued_fraction_iterator(Rational(-3, 8)))
    [-1, 1, 1, 1, 2]

    >>> for i, v in enumerate(continued_fraction_iterator(pi)):
    ...     if i > 7:
    ...         break
    ...     print(v)
    3
    7
    15
    1
    292
    1
    1
    1

    References
    ==========

    .. [1] https://en.wikipedia.org/wiki/Continued_fraction

    """
    from sympy.functions import floor
    while True:
        i = floor(x)
        yield i
        x -= i
        if not x:
            break
        x = 1/x


def continued_fraction_convergents(cf):
    """
    Return an iterator over the convergents of a continued fraction (cf).

    The parameter should be an iterable returning successive
    partial quotients of the continued fraction, such as might be
    returned by continued_fraction_iterator.  In computing the
    convergents, the continued fraction need not be strictly in
    canonical form (all integers, all but the first positive).
    Rational and negative elements may be present in the expansion.

    Examples
    ========

    >>> from sympy.core import pi
    >>> from sympy import S
    >>> from sympy.ntheory.continued_fraction import \
            continued_fraction_convergents, continued_fraction_iterator

    >>> list(continued_fraction_convergents([0, 2, 1, 2]))
    [0, 1/2, 1/3, 3/8]

    >>> list(continued_fraction_convergents([1, S('1/2'), -7, S('1/4')]))
    [1, 3, 19/5, 7]

    >>> it = continued_fraction_convergents(continued_fraction_iterator(pi))
    >>> for n in range(7):
    ...     print(next(it))
    3
    22/7
    333/106
    355/113
    103993/33102
    104348/33215
    208341/66317

    See Also
    ========

    continued_fraction_iterator

    """
    p_2, q_2 = S.Zero, S.One
    p_1, q_1 = S.One, S.Zero
    for a in cf:
        p, q = a*p_1 + p_2, a*q_1 + q_2
        p_2, q_2 = p_1, q_1
        p_1, q_1 = p, q
        yield p/q