[Python-ideas] Fwd: Trigonometry in degrees

Tim Peters tim.peters at gmail.com
Thu Jun 14 14:44:34 EDT 2018

I should note that numeric code "that works" is often much subtler than it
appears at first glance.  So, for educational purposes, I'll point out some
of what _wasn't_ said about this crucial function:


>     import mpmath
>     from math import fmod
>     # Return (n, x) such that:
>     # 1. d degrees is equivalent to x + n*(pi/2) radians.
>     # 2. x is an mpmath float in [-pi/4, pi/4].
>     # 3. n is an integer in range(4).
>     # There is one potential rounding error, when mpmath.radians() is
>     # used to convert a number of degrees between -45 and 45.  This is
>     # done using the current mpmath precision.
>     def treduce(d):
>         d = fmod(d, 360.0)
>         n = round(d / 90.0)
>         assert -4 <= n <= 4
>         d -= n * 90.0
>         assert -45.0 <= d <= 45.0
>         return n & 3, mpmath.radians(d)
> How do we know there is at most one rounding error in that?  No, it's not

That `fmod` is exact is guaranteed by the relevant standards, but most
people who write a libm don't get it right at first.  There is no "cheap"
way to implement it correctly.  It requires getting the effect of doing
exact integer division on, potentially, multi-thousand bit integers.
Assuming x > y > 0, a correct implementation of fmod(x, y) ends up in a
loop that goes around a number of times roughly equal to log2(x/y),
simulating one-bit-at-a-time long division  For example, here's glibc's


Don't expect that to be easy to follow either ;-)

Then how do we know that `d -= n * 90.0" is exact?  That's not obvious
either.  It follows from the "Sterbenz lemma", one version of which:  if x
and y are non-zero floats of the same sign within a factor of 2 of each

    1/2 <= x/y <= 2  (mathematically)

then x-y is exactly representable as a float too.  This is true regardless
of the floating-point base, or of rounding mode in use.  In IEEE-754, it
doesn't even need weasel words to exempt underflowing cases.

That lemma needs to be applied by cases, for each of the possible values of
(the integer) `n`.  It gets closest to failing for |n| = 1.  For example,
if d is a tiny bit larger than 45, n is 1, and then d/90 is
(mathematically) very close to 1/2.

Which is another thing that needs to be shown:  "if d is a tiny bit larger
than 45, n is 1".  Why?  It's certainly true if we were using infinite
precision, but we're not.  The smallest representable double > 45 is 45 +

>>> d = 45 + 2**-47
>>> d
>>> _.hex()

Then d/90.0 (the argument to round()) is, with infinite precision,

(45 + 2**-47)/90 =
0.5 + 2**-47/90

1 ULP with respect to 0.5 is 2**-53, so that in turn is equal to

0.5 + 2**-53/(90/64) =
0.5 + (64/90)*2**-53 =
0.5 + 0.71111111111... * 2**-53

Because the tail (0.711...) is greater than 0.5 ULP, it rounds up under
nearest-even rounding, to

0.5 + 2**-53

>>> d / 90
>>> 0.5 + 2**-53

and so Python's round() rounds it up to 1:

>>> round(_)

Note that it would _not_ be true if truncating "rounding" were done, so
round-nearest is a hidden assumption in the code.

Similar analysis needs to be done at values near the boundaries around all
possible values of `n`.

That `assert -45.0 <= d <= 45.0` can't fall then follows from all of that.

In all, a proof that the code is correct is much longer than the code
itself.  That's typical.  Alas, it's also typical that math library sources
rarely point out the subtleties.
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