def _quotient_exponent(x, y):
"""
Given two positive finite MPFR instances x and y,
find the exponent of x / y; that is, the unique
integer e such that 2**(e-1) <= x / y < 2**e.
"""
assert mpfr.mpfr_regular_p(x)
assert mpfr.mpfr_regular_p(y)
# Make copy of x with the exponent of y.
x2 = mpfr.Mpfr_t()
mpfr.mpfr_init2(x2, mpfr.mpfr_get_prec(x))
mpfr.mpfr_set(x2, x, mpfr.MPFR_RNDN)
mpfr.mpfr_set_exp(x2, mpfr.mpfr_get_exp(y))
# Compare x2 and y, disregarding the sign.
extra = mpfr.mpfr_cmpabs(x2, y) >= 0
return extra + mpfr.mpfr_get_exp(x) - mpfr.mpfr_get_exp(y)
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)
def mpfr_mod(rop, x, y, rnd):
"""
Given two MPRF numbers x and y, compute
x - floor(x / y) * y, rounded if necessary using the given
rounding mode. The result is placed in 'rop'.
This is the 'remainder' operation, with sign convention
compatible with Python's % operator (where x % y has
the same sign as y).
"""
# There are various cases:
#
# 0. If either argument is a NaN, the result is NaN.
#
# 1. If x is infinite or y is zero, the result is NaN.
#
# 2. If y is infinite, return 0 with the sign of y if x is zero, x if x has
# the same sign as y, and infinity with the sign of y if it has the
# opposite sign.
#
# 3. If none of the above cases apply then both x and y are finite,
# and y is nonzero. If x and y have the same sign, simply
# return the result of fmod(x, y).
#
# 4. Now both x and y are finite, y is nonzero, and x and y have
# differing signs. Compute r = fmod(x, y) with sufficient precision
# to get an exact result. If r == 0, return 0 with the sign of y
# (which will be the opposite of the sign of x). If r != 0,
# return r + y, rounded appropriately.
if not mpfr.mpfr_number_p(x) or mpfr.mpfr_nan_p(y) or mpfr.mpfr_zero_p(y):
return mpfr.mpfr_fmod(rop, x, y, rnd)
elif mpfr.mpfr_inf_p(y):
x_negative = mpfr.mpfr_signbit(x)
y_negative = mpfr.mpfr_signbit(y)
if mpfr.mpfr_zero_p(x):
mpfr.mpfr_set_zero(rop, -y_negative)
return 0
elif x_negative == y_negative:
return mpfr.mpfr_set(rop, x, rnd)
else:
mpfr.mpfr_set_inf(rop, -y_negative)
return 0
x_negative = mpfr.mpfr_signbit(x)
y_negative = mpfr.mpfr_signbit(y)
if x_negative == y_negative:
return mpfr.mpfr_fmod(rop, x, y, rnd)
else:
p = max(mpfr.mpfr_get_prec(x), mpfr.mpfr_get_prec(y))
z = mpfr.Mpfr_t()
mpfr.mpfr_init2(z, p)
# Doesn't matter what rounding mode we use here; the result
# should be exact.
ternary = mpfr.mpfr_fmod(z, x, y, rnd)
assert ternary == 0
if mpfr.mpfr_zero_p(z):
mpfr.mpfr_set_zero(rop, -y_negative)
return 0
else:
return mpfr.mpfr_add(rop, y, z, rnd)
