Given two MPFR numbers x and y, compute floor(x / y),
    rounded if necessary using the given rounding mode.
    The result is placed in 'rop'.

def mpfr_floordiv(rop, x, y, rnd):
    """
    Given two MPFR numbers x and y, compute floor(x / y),
    rounded if necessary using the given rounding mode.
    The result is placed in 'rop'.
    """
    # Algorithm notes
    # ---------------
    # A simple and obvious approach is to compute floor(x / y) exactly, and
    # then round to the nearest representable value using the given rounding
    # mode.  This requires computing x / y to a precision sufficient to ensure
    # that floor(x / y) is exactly representable.  If abs(x / y) < 2**r, then
    # abs(floor(x / y)) <= 2**r, and so r bits of precision is enough.
    # However, for large quotients this is impractical, and we need some other
    # method.  For x / y sufficiently large, it's possible to show that x / y
    # and floor(x / y) are indistinguishable, in the sense that both quantities
    # round to the same value.  More precisely, we have the following theorem:
    #
    # Theorem.  Suppose that x and y are nonzero finite binary floats
    # representable with p and q bits of precision, respectively.  Let R be any
    # of the IEEE 754 standard rounding modes, and choose a target precision r.
    # Write rnd for the rounding operation from Q to precision-r binary floats
    # with rounding mode R.  Write bin(x) for the binade of a nonzero float x.
    #
    # If R is a round-to-nearest rounding mode, and either
    #
    # (1) p <= q + r and |x / y| >= 2^(q + r), or
    # (2) p > q + r and bin(x) - bin(y) >= p
    #
    # then
    #
    #    rnd(floor(x / y)) == rnd(x / y)
    #
    # Conversely, if R is a directed rounding mode, and either
    #
    # (1) p < q + r and |x / y| >= 2^(q + r - 1), or
    # (2) p >= q + r and bin(x) - bin(y) >= p
    #
    # then again
    #
    #    rnd(floor(x / y)) == rnd(x / y).
    #
    # Proof.  See separate notes and Coq proof in the float-proofs
    # repository.
    #
    # Rather than distinguish between the various cases (R directed
    # or not, p large versus p small) above, we use a weaker but
    # simpler amalgamation of the above result:
    #
    # Corollary 1. With x, y, p, q, R, r and rnd as above, if
    #
    #     |x / y| >= 2^max(q + r, p)
    #
    # then
    #
    #     rnd(floor(x / y)) == rnd(x / y).
    #
    # Proof. Note that |x / y| >= 2^p implies bin(x) - bin(y) >= p,
    # so it's enough that |x / y| >= 2^max(p, q + r) in the case of
    # a round-to-nearest mode, and that |x / y| >= 2^max(p, q + r - 1)
    # in the case of a directed rounding mode.

    # In special cases, it's safe to defer to mpfr_div: the result in
    # these cases is always 0, infinity, or nan.
    if not mpfr.mpfr_regular_p(x) or not mpfr.mpfr_regular_p(y):
        return mpfr.mpfr_div(rop, x, y, rnd)

    e = _quotient_exponent(x, y)

    p = mpfr.mpfr_get_prec(x)
    q = mpfr.mpfr_get_prec(y)
    r = mpfr.mpfr_get_prec(rop)

    # If e - 1 >= max(p, q+r) then |x / y| >= 2^(e-1) >= 2^max(p, q+r),
    # so by the above theorem, round(floordiv(x, y)) == round(div(x, y)).
    if e - 1 >= max(p, q + r):
        return mpfr.mpfr_div(rop, x, y, rnd)

    # Slow version: compute to sufficient bits to get integer precision.  Given
    # that 2**(e-1) <= x / y < 2**e, need >= e bits of precision.
    z_prec = max(e, 2)
    z = mpfr.Mpfr_t()
    mpfr.mpfr_init2(z, z_prec)

    # Compute the floor exactly. The division may set the
    # inexact flag, so we save its state first.
    old_inexact = mpfr.mpfr_inexflag_p()
    mpfr.mpfr_div(z, x, y, mpfr.MPFR_RNDD)
    if not old_inexact:
        mpfr.mpfr_clear_inexflag()

    # Floor result should be exactly representable, so any rounding mode will
    # do.
    ternary = mpfr.mpfr_rint_floor(z, z, rnd)
    assert ternary == 0

    # ... and round to the given rounding mode.
    return mpfr.mpfr_set(rop, z, rnd)