Equations for a "fast" method.
This repository contains a set of procedures to compute numerical methods in the vein of the fast inverse root method. In particular, we will generate code that
- Computes rational powers () to an arbitrary precision.
- Computes irrational powers () to within 10% relative error.
- Computes to within 10% relative error.
- Computes to within 10% relative error.
- Computes the geometric mean of an
std::array
quickly to within 10% error.
Additionally, we will do so using mostly just integer arithmetic.
You can use everything in floathacks
by including hacks.h
#include <floathacks/hacks.h>
using namespace floathacks; // Comment this out if you don't want your top-level namespace to be polluted
This document is compiled from READOTHER.md
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can run
python -m readme2tex --output README.md --branch svgs
to recompile these docs.
To generate an estimation for , where is any floating point number, you can run
float approximate_root = fpow<FLOAT(0.12345)>::estimate(x);
Since estimates of pow
can be refined into better iterates (as long as c
is "rational enough"), you can also
compute a more exact result via
float root = pow<FLOAT(0.12345), n>(x);
where n
is the number of newton iterations to perform. The code generated by this template will unroll itself, so it's
relatively efficient.
However, the optimized code does not let you use it as a constexpr
or where the exponent is not constant. In those cases,
you can use consts::fpow(x, c)
and consts::pow(x, c, iterations = 2)
instead:
float root = consts::pow(x, 0.12345, n);
Note that the compiler isn't able to deduce the optimal constants in these cases, so you'll incur additional penalties computing the constants of the method.
You can also compute an approximation of with
float guess = fexp(x);
Unfortunately, since there are no refinement methods available for exponentials, we can't do much
with this result if it's too coarse for your needs. In addition, due to overflow, this method breaks down
when x
approaches 90.
Similarly, you can also compute an approximation of with
float guess = flog(x);
Again, as is with the case of fexp
, there are no refinement methods available for logarithms either.
All of the f***
methods above have bounded relative errors of at most 10%. The refined pow
method
can be made to give arbitrary precision by increasing the number of refinement iterations. Each refinement
iteration takes time proportional to the number of digits in the floating point representation of the exponent.
Note that since floats are finite, this is bounded above by 32 (and more tightly, 23).
You can compute the geometric mean () of a std::array<float, n>
with
float guess = fgmean<3>({ 1, 2, 3 });
This can be refined, but you typically do not care about the absolute precision of a mean-like statistic. To refine this, you can run Newton's method on . As far as I am aware, this is also an original method.
The key ingredient of these types of methods is the pair of transformations and .
-
takes a
IEEE 754
single precision floating point number and outputs its "machine" representation. In essence, it acts likeunsigned long f2l(float x) { union {float fl; unsigned long lg;} lens = { x }; return lens.lg; }
-
takes an unsigned long representing a float and returns a
IEEE 754
single precision floating point number. It acts likefloat l2f(unsigned long z) { union {float fl; unsigned long lg;} lens = { z }; return lens.fl; }
So for example, the fast inverse root method:
union {float fl; unsigned long lg;} lens = { x };
lens.lg = 0x5f3759df - lens.lg / 2;
float y = lens.fl;
can be equivalently expressed as
In a similar vein, a fast inverse cube-root method is presented at the start of this page.
We will justify this in the next section.
We can approximate any using just integer arithmetic on the machine representation of . To do so, compute
where . In general, any value of bias
, as long
as it is reasonably small, will work. At bias = 0
, the method computes a value whose error is completely positive.
Therefore, by increasing the bias, we can shift some of the error down into the negative plane and
halve the error.
As seen in the fast-inverse-root method, a bias of -0x5c416
tend to work well for pretty much every case that I've
tried, as long as we tack on at least one Newton refinement stage at the end. It works well without refinement as well,
but an even bias of -0x5c000
works even better.
Why does this work? See these slides for the derivation. In particular, the fast inverse square-root is a subclass of this method.
We can approximate up to using a similar set of bit tricks. I'll first give its equation, and then give its derivations. As far as I am aware, these are original. However, since there are no refinement methods for the computation of , there is practically no reason to ever resort to this approximation unless you're okay with 10% error.
Here, is the machine epsilon for single precision, and it is computed by .
To give a derivation of this equation, we'll need to borrow a few mathematical tools from analysis. In particular, while
l2f
and f2l
have many discontinuities ( of them to be exact), it is mostly smooth. This
carries over to its "rate-of-change" as well, so we will just pretend that it has mostly smooth derivatives
everywhere.
Consider the function
where the equality is a consequence of the chain-rule, assuming that f2l
is differentiable at the particular value of
. Now, this raises an interesting question: What does it mean to take a derivative of ?
Well, it's not all that mysterious. The derivative of f2l
is just the rate at which a number's IEEE 754 machine
representation changes as we make small perturbations to a number. Unfortunately, while it might be easy to compute
this derivative as a numerical approximation, we still don't have an approximate form for algebraic manipulation.
While might be difficult to construct, we can fair much better with its sibling, . Now, the derivative is the rate that a float will change given that we make small perturbations to its machine representation. However, since its machine representations are all bit-vectors, it doesn't make sense to take a derivative here since we can't make these perturbations arbitrarily small. The smallest change we can make is to either add or subtract one. However, if we just accept our fate, then we can define the "derivative" as the finite difference
where
Here, equality holds when is a perfect power of (including fractions of the form ).
Therefore,
From here, we also have
Given , antidifferentiating both sides gives
Similarly, since satisfies , we have
This makes sense, since we'd like these two functions to be inverses of each other.
Consider
Since we would like , we can impose the boundary condition
which gives . However, while this method gives bounded relative error, in its unbiased form this is pretty off the mark for general purposes (it approximates some other ). Instead, we can add in an unbiased form:
where, empirically, gives a good approximation. Notice that the we've chosen is close to , which is what we need to transform to . In particular, for all , the , , and relative error is always below 10%.
In a similar spirit, we can use the approximation
to derive
Imposing a boundary condition at gives , so we should expect
However, this actually computes some other logarithm , and we'll have to, again, unbias this term
where the term came from the fact that the base computation approximates the 2-logarithm. Empirically, I've found that a bias of works well. In particular, for all , the , , and relative error is always below 10%.
There's a straightforward derivation of the geometric mean. Consider the approximations of and , we can refine them as
Therefore, a bit of algebra will show that
which reduces to the equation for the geometric mean.
Notice that we just add a series of integers, followed by an integer divide, which is pretty efficient.
For more information on how the constant () is derived for the cube-root, visit http://www.bullshitmath.lol/.
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