Reputation: 95
Many sources indicate that well-known fast inverse square root algorithm can be generalized to calculation arbitrary power inverse root. Unfortunately I have not found such C++ implementation and I'm not so good at math to generalize this method by myself. Could you help me to do this or perhaps provide a ready-made solution? I think this will be useful to many, especially with good explanations.
This is the original algorithm and I do not quite understand what I need to change to get for example 1 /cbrt(x)
:
float Q_rsqrt( float number )
{
long i;
float x2, y;
const float threehalfs = 1.5F;
x2 = number * 0.5F;
y = number;
i = * ( long * ) &y; // evil floating point bit level hacking
i = 0x5f3759df - ( i >> 1 ); // what the...?
y = * ( float * ) &i;
y = y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration
// y = y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed
return y;
}
Upvotes: 4
Views: 1112
Reputation: 2260
The algorithm consists of two steps - a rough solution estimation and the solution improvement using several Newton method steps.
The basic idea is to use relationship between float number logarithm log2(x)
and its integer representation Ix
:
(Image from https://en.wikipedia.org/wiki/Fast_inverse_square_root)
Now use the well-known logarithm identity for root:
.
Combining the identities obtained earlier, we get:
Substituting numerical values L * (B - s) = 0x3F7A3BEA
, so
Iy = 0x3F7A3BEA / c * (c + 1) - Ix / c;
.
For a simple float point number representation as an integer and back it is convenient to use union
type:
union
{
float f; // float representation
uint32_t i; // integer representation
} t;
t.f = x;
t.i = 0x3F7A3BEA / n * (n + 1) - t.i / n; // Work with integer representation
float y = t.f; // back to float representation
Note that for n=2
the expression is simplified to t.i = 0x5f3759df - t.i / 2;
which is identical to original i = 0x5f3759df - ( i >> 1 );
Transform equality
into an equation that should be solved:
Now construct Newton steps:
Programmatically it looks like: y = y * (1 + n - x * pow(y,n)) / n;
. As an initial y
, we use the value obtained at Rough estimation step.
Note for the particular case of square root (n = 2
) we get y = y * (3 - x*y*y) / 2;
which is identically to the original formula y = y * (threehalfs - (x2 * y * y))
;
Final code as template function. Parameter N
determines root power.
template<unsigned N>
float power(float x) {
if (N % 2 == 0) return power<N / 2>(x * x);
else if (N % 3 == 0) return power<N / 3>(x * x * x);
return power<N - 1>(x) * x;
};
template<>
float power<0>(float x){ return 1; }
// fast_inv_nth_root<2>(x) - inverse square root 1/sqrt(x)
// fast_inv_nth_root<3>(x) - inverse cube root 1/cbrt(x)
template <unsigned n>
float fast_inv_nth_root(float x)
{
union { float f; uint32_t i; } t = { x };
// Approximate solution
t.i = 0x3F7A3BEA / n * (n + 1) - t.i / n;
float y = t.f;
// Newton's steps. Copy for more accuracy.
y = y * (n + 1 - x * power<n>(y)) / n;
y = y * (n + 1 - x * power<n>(y)) / n;
return y;
}
Testing code:
int main()
{
std::cout << "|x ""|fast2 "" actual2 "
"|fast3 "" actual3 "
"|fast4 "" actual4 "
"|fast5 "" actual5 ""|" << std::endl;
for (float i = 0.00001; i < 10000; i *= 10)
std::cout << std::setprecision(5) << std::fixed
<< std::scientific << '|'
<< i << '|'
<< fast_inv_nth_root<2>(i) << " " << 1 / sqrt(i) << "|"
<< fast_inv_nth_root<3>(i) << " " << 1 / cbrt(i) << "|"
<< fast_inv_nth_root<4>(i) << " " << pow(i, -0.25) << "|"
<< fast_inv_nth_root<5>(i) << " " << pow(i, -0.2) << "|"
<< std::endl;
}
Results:
|x |fast2 actual2 |fast3 actual3 |fast4 actual4 |fast5 actual5 |
|1.00000e-05|3.16226e+02 3.16228e+02|4.64152e+01 4.64159e+01|1.77828e+01 1.77828e+01|9.99985e+00 1.00000e+01|
|1.00000e-04|9.99996e+01 1.00000e+02|2.15441e+01 2.15443e+01|9.99991e+00 1.00000e+01|6.30949e+00 6.30957e+00|
|1.00000e-03|3.16227e+01 3.16228e+01|1.00000e+01 1.00000e+01|5.62339e+00 5.62341e+00|3.98103e+00 3.98107e+00|
|1.00000e-02|9.99995e+00 1.00000e+01|4.64159e+00 4.64159e+00|3.16225e+00 3.16228e+00|2.51185e+00 2.51189e+00|
|1.00000e-01|3.16227e+00 3.16228e+00|2.15443e+00 2.15443e+00|1.77828e+00 1.77828e+00|1.58487e+00 1.58489e+00|
|1.00000e+00|9.99996e-01 1.00000e+00|9.99994e-01 1.00000e+00|9.99991e-01 1.00000e+00|9.99987e-01 1.00000e+00|
|1.00000e+01|3.16226e-01 3.16228e-01|4.64159e-01 4.64159e-01|5.62341e-01 5.62341e-01|6.30948e-01 6.30957e-01|
|1.00000e+02|9.99997e-02 1.00000e-01|2.15443e-01 2.15443e-01|3.16223e-01 3.16228e-01|3.98102e-01 3.98107e-01|
|1.00000e+03|3.16226e-02 3.16228e-02|1.00000e-01 1.00000e-01|1.77827e-01 1.77828e-01|2.51185e-01 2.51189e-01|
|1.00000e+04|9.99996e-03 1.00000e-02|4.64155e-02 4.64159e-02|9.99995e-02 1.00000e-01|1.58487e-01 1.58489e-01|
Upvotes: 8