- Implementation of `tan` for 16-bit floating point inputs scaled by pi.
i.e,. `tanpif16()`
- Implementation of Tanpi in MPFRWrapper for MPFR versions < 4.2
- Exhaustive tests for `tanpif16()`
Implementation of `cos` for half precision floating point inputs scaled
by pi (i.e., `cospi`), correctly rounded for all rounding modes.
---------
Co-authored-by: OverMighty <its.overmighty@gmail.com>
- added all variations of ffma and fdiv
- will add all new headers into yaml for next patch
- only fsub is left then all basic operations for float is complete
---------
Co-authored-by: OverMighty <its.overmighty@gmail.com>
- fadd removed because I need to add for different input types
- finishing rest of basic operations
- noticed duplicates will remove
---------
Co-authored-by: OverMighty <its.overmighty@gmail.com>
Division-less Newton iterations algorithm for cube roots.
1. **Range reduction**
For `x = (-1)^s * 2^e * (1.m)`, we get 2 reduced arguments `x_r` and `a`
as:
```
x_r = 1.m
a = (-1)^s * 2^(e % 3) * (1.m)
```
Then `cbrt(x) = x^(1/3)` can be computed as:
```
x^(1/3) = 2^(e / 3) * a^(1/3).
```
In order to avoid division, we compute `a^(-2/3)` using Newton method
and then
multiply the results by a:
```
a^(1/3) = a * a^(-2/3).
```
2. **First approximation to a^(-2/3)**
First, we use a degree-7 minimax polynomial generated by Sollya to
approximate `x_r^(-2/3)` for `1 <= x_r < 2`.
```
p = P(x_r) ~ x_r^(-2/3),
```
with relative errors bounded by:
```
| p / x_r^(-2/3) - 1 | < 1.16 * 2^-21.
```
Then we multiply with `2^(e % 3)` from a small lookup table to get:
```
x_0 = 2^(-2*(e % 3)/3) * p
~ 2^(-2*(e % 3)/3) * x_r^(-2/3)
= a^(-2/3)
```
with relative errors:
```
| x_0 / a^(-2/3) - 1 | < 1.16 * 2^-21.
```
This step is done in double precision.
3. **First Newton iteration**
We follow the method described in:
Sibidanov, A. and Zimmermann, P., "Correctly rounded cubic root
evaluation
in double precision", https://core-math.gitlabpages.inria.fr/cbrt64.pdf
to derive multiplicative Newton iterations as below:
Let `x_n` be the nth approximation to `a^(-2/3)`. Define the n^th error
as:
```
h_n = x_n^3 * a^2 - 1
```
Then:
```
a^(-2/3) = x_n / (1 + h_n)^(1/3)
= x_n * (1 - (1/3) * h_n + (2/9) * h_n^2 - (14/81) * h_n^3 + ...)
```
using the Taylor series expansion of `(1 + h_n)^(-1/3)`.
Apply to `x_0` above:
```
h_0 = x_0^3 * a^2 - 1
= a^2 * (x_0 - a^(-2/3)) * (x_0^2 + x_0 * a^(-2/3) + a^(-4/3)),
```
it's bounded by:
```
|h_0| < 4 * 3 * 1.16 * 2^-21 * 4 < 2^-17.
```
So in the first iteration step, we use:
```
x_1 = x_0 * (1 - (1/3) * h_n + (2/9) * h_n^2 - (14/81) * h_n^3)
```
Its relative error is bounded by:
```
| x_1 / a^(-2/3) - 1 | < 35/242 * |h_0|^4 < 2^-70.
```
Then we perform Ziv's rounding test and check if the answer is exact.
This step is done in double-double precision.
4. **Second Newton iteration**
If the Ziv's rounding test from the previous step fails, we define the
error
term:
```
h_1 = x_1^3 * a^2 - 1,
```
And perform another iteration:
```
x_2 = x_1 * (1 - h_1 / 3)
```
with the relative errors exceed the precision of double-double.
We then check the Ziv's accuracy test with relative errors < 2^-102 to
compensate for rounding errors.
5. **Final iteration**
If the Ziv's accuracy test from the previous step fails, we perform
another
iteration in 128-bit precision and check for exact outputs.
Fixes https://github.com/llvm/llvm-project/issues/92874
Algorithm: Let `x = (-1)^s * 2^e * (1 + m)`.
- Step 1: Range reduction: reduce the exponent with:
```
y = cbrt(x) = (-1)^s * 2^(floor(e/3)) * 2^((e % 3)/3) * (1 + m)^(1/3)
```
- Step 2: Use the first 4 bit fractional bits of `m` to look up for a
degree-7 polynomial approximation to:
```
(1 + m)^(1/3) ~ 1 + m * P(m).
```
- Step 3: Perform the multiplication:
```
2^((e % 3)/3) * (1 + m)^(1/3).
```
- Step 4: Check for exact cases to prevent rounding and clear
`FE_INEXACT` floating point exception.
- Step 5: Combine with the exponent and sign before converting down to
`float` and return.
I also fixed a comment in sinpif.cpp in the first commit. Should this be
included in this PR?
All tests were passed, including the exhaustive test.
CC: @lntue
