Mark Dickinson <dicki...@gmail.com> added the comment:
Here's the float-and-Fraction-based code that I'm using to compare the
integer-based code against:
def sqrt_frac2(n, m):
"""
Square root of n/m as a float, correctly rounded.
"""
f = fractions.Fraction(n, m)
# First approximation.
x = math.sqrt(n / m)
# Use the approximation to find a pair of floats bracketing the actual sqrt
if fractions.Fraction(x)**2 >= f:
x_lo, x_hi = math.nextafter(x, 0.0), x
else:
x_lo, x_hi = x, math.nextafter(x, math.inf)
# Check the bracketing. If math.sqrt is correctly rounded (as it will be on
a
# typical machine), then the assert can't fail. But we can't rely on
math.sqrt being
# correctly rounded in general, so would need some fallback.
fx_lo, fx_hi = fractions.Fraction(x_lo), fractions.Fraction(x_hi)
assert fx_lo**2 <= f <= fx_hi**2
# Compare true square root with the value halfway between the two floats.
mid = (fx_lo + fx_hi) / 2
if mid**2 < f:
return x_hi
elif mid**2 > f:
return x_lo
else:
# Tricky case: mid**2 == f, so we need to choose the "even" endpoint.
# Cheap trick: the addition in 0.5 * (x_lo + x_hi) will round to even.
return 0.5 * (x_lo + x_hi)
----------
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Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue45876>
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