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/*
* Copyright 2012 Google Inc.
*
* Use of this source code is governed by a BSD-style license that can be
* found in the LICENSE file.
*/
// http://metamerist.com/cbrt/CubeRoot.cpp
//
#include <math.h>
#include "CubicUtilities.h"
#define TEST_ALTERNATIVES 0
#if TEST_ALTERNATIVES
typedef float (*cuberootfnf) (float);
typedef double (*cuberootfnd) (double);
// estimate bits of precision (32-bit float case)
inline int bits_of_precision(float a, float b)
{
const double kd = 1.0 / log(2.0);
if (a==b)
return 23;
const double kdmin = pow(2.0, -23.0);
double d = fabs(a-b);
if (d < kdmin)
return 23;
return int(-log(d)*kd);
}
// estiamte bits of precision (64-bit double case)
inline int bits_of_precision(double a, double b)
{
const double kd = 1.0 / log(2.0);
if (a==b)
return 52;
const double kdmin = pow(2.0, -52.0);
double d = fabs(a-b);
if (d < kdmin)
return 52;
return int(-log(d)*kd);
}
// cube root via x^(1/3)
static float pow_cbrtf(float x)
{
return (float) pow(x, 1.0f/3.0f);
}
// cube root via x^(1/3)
static double pow_cbrtd(double x)
{
return pow(x, 1.0/3.0);
}
// cube root approximation using bit hack for 32-bit float
static float cbrt_5f(float f)
{
unsigned int* p = (unsigned int *) &f;
*p = *p/3 + 709921077;
return f;
}
#endif
// cube root approximation using bit hack for 64-bit float
// adapted from Kahan's cbrt
static double cbrt_5d(double d)
{
const unsigned int B1 = 715094163;
double t = 0.0;
unsigned int* pt = (unsigned int*) &t;
unsigned int* px = (unsigned int*) &d;
pt[1]=px[1]/3+B1;
return t;
}
#if TEST_ALTERNATIVES
// cube root approximation using bit hack for 64-bit float
// adapted from Kahan's cbrt
#if 0
static double quint_5d(double d)
{
return sqrt(sqrt(d));
const unsigned int B1 = 71509416*5/3;
double t = 0.0;
unsigned int* pt = (unsigned int*) &t;
unsigned int* px = (unsigned int*) &d;
pt[1]=px[1]/5+B1;
return t;
}
#endif
// iterative cube root approximation using Halley's method (float)
static float cbrta_halleyf(const float a, const float R)
{
const float a3 = a*a*a;
const float b= a * (a3 + R + R) / (a3 + a3 + R);
return b;
}
#endif
// iterative cube root approximation using Halley's method (double)
static double cbrta_halleyd(const double a, const double R)
{
const double a3 = a*a*a;
const double b= a * (a3 + R + R) / (a3 + a3 + R);
return b;
}
#if TEST_ALTERNATIVES
// iterative cube root approximation using Newton's method (float)
static float cbrta_newtonf(const float a, const float x)
{
// return (1.0 / 3.0) * ((a + a) + x / (a * a));
return a - (1.0f / 3.0f) * (a - x / (a*a));
}
// iterative cube root approximation using Newton's method (double)
static double cbrta_newtond(const double a, const double x)
{
return (1.0/3.0) * (x / (a*a) + 2*a);
}
// cube root approximation using 1 iteration of Halley's method (double)
static double halley_cbrt1d(double d)
{
double a = cbrt_5d(d);
return cbrta_halleyd(a, d);
}
// cube root approximation using 1 iteration of Halley's method (float)
static float halley_cbrt1f(float d)
{
float a = cbrt_5f(d);
return cbrta_halleyf(a, d);
}
// cube root approximation using 2 iterations of Halley's method (double)
static double halley_cbrt2d(double d)
{
double a = cbrt_5d(d);
a = cbrta_halleyd(a, d);
return cbrta_halleyd(a, d);
}
#endif
// cube root approximation using 3 iterations of Halley's method (double)
static double halley_cbrt3d(double d)
{
double a = cbrt_5d(d);
a = cbrta_halleyd(a, d);
a = cbrta_halleyd(a, d);
return cbrta_halleyd(a, d);
}
#if TEST_ALTERNATIVES
// cube root approximation using 2 iterations of Halley's method (float)
static float halley_cbrt2f(float d)
{
float a = cbrt_5f(d);
a = cbrta_halleyf(a, d);
return cbrta_halleyf(a, d);
}
// cube root approximation using 1 iteration of Newton's method (double)
static double newton_cbrt1d(double d)
{
double a = cbrt_5d(d);
return cbrta_newtond(a, d);
}
// cube root approximation using 2 iterations of Newton's method (double)
static double newton_cbrt2d(double d)
{
double a = cbrt_5d(d);
a = cbrta_newtond(a, d);
return cbrta_newtond(a, d);
}
// cube root approximation using 3 iterations of Newton's method (double)
static double newton_cbrt3d(double d)
{
double a = cbrt_5d(d);
a = cbrta_newtond(a, d);
a = cbrta_newtond(a, d);
return cbrta_newtond(a, d);
}
// cube root approximation using 4 iterations of Newton's method (double)
static double newton_cbrt4d(double d)
{
double a = cbrt_5d(d);
a = cbrta_newtond(a, d);
a = cbrta_newtond(a, d);
a = cbrta_newtond(a, d);
return cbrta_newtond(a, d);
}
// cube root approximation using 2 iterations of Newton's method (float)
static float newton_cbrt1f(float d)
{
float a = cbrt_5f(d);
return cbrta_newtonf(a, d);
}
// cube root approximation using 2 iterations of Newton's method (float)
static float newton_cbrt2f(float d)
{
float a = cbrt_5f(d);
a = cbrta_newtonf(a, d);
return cbrta_newtonf(a, d);
}
// cube root approximation using 3 iterations of Newton's method (float)
static float newton_cbrt3f(float d)
{
float a = cbrt_5f(d);
a = cbrta_newtonf(a, d);
a = cbrta_newtonf(a, d);
return cbrta_newtonf(a, d);
}
// cube root approximation using 4 iterations of Newton's method (float)
static float newton_cbrt4f(float d)
{
float a = cbrt_5f(d);
a = cbrta_newtonf(a, d);
a = cbrta_newtonf(a, d);
a = cbrta_newtonf(a, d);
return cbrta_newtonf(a, d);
}
static double TestCubeRootf(const char* szName, cuberootfnf cbrt, double rA, double rB, int rN)
{
const int N = rN;
float dd = float((rB-rA) / N);
// calculate 1M numbers
int i=0;
float d = (float) rA;
double s = 0.0;
for(d=(float) rA, i=0; i<N; i++, d += dd)
{
s += cbrt(d);
}
double bits = 0.0;
double worstx=0.0;
double worsty=0.0;
int minbits=64;
for(d=(float) rA, i=0; i<N; i++, d += dd)
{
float a = cbrt((float) d);
float b = (float) pow((double) d, 1.0/3.0);
int bc = bits_of_precision(a, b);
bits += bc;
if (b > 1.0e-6)
{
if (bc < minbits)
{
minbits = bc;
worstx = d;
worsty = a;
}
}
}
bits /= N;
printf(" %3d mbp %6.3f abp\n", minbits, bits);
return s;
}
static double TestCubeRootd(const char* szName, cuberootfnd cbrt, double rA, double rB, int rN)
{
const int N = rN;
double dd = (rB-rA) / N;
int i=0;
double s = 0.0;
double d = 0.0;
for(d=rA, i=0; i<N; i++, d += dd)
{
s += cbrt(d);
}
double bits = 0.0;
double worstx = 0.0;
double worsty = 0.0;
int minbits = 64;
for(d=rA, i=0; i<N; i++, d += dd)
{
double a = cbrt(d);
double b = pow(d, 1.0/3.0);
int bc = bits_of_precision(a, b); // min(53, count_matching_bitsd(a, b) - 12);
bits += bc;
if (b > 1.0e-6)
{
if (bc < minbits)
{
bits_of_precision(a, b);
minbits = bc;
worstx = d;
worsty = a;
}
}
}
bits /= N;
printf(" %3d mbp %6.3f abp\n", minbits, bits);
return s;
}
static int _tmain()
{
// a million uniform steps through the range from 0.0 to 1.0
// (doing uniform steps in the log scale would be better)
double a = 0.0;
double b = 1.0;
int n = 1000000;
printf("32-bit float tests\n");
printf("----------------------------------------\n");
TestCubeRootf("cbrt_5f", cbrt_5f, a, b, n);
TestCubeRootf("pow", pow_cbrtf, a, b, n);
TestCubeRootf("halley x 1", halley_cbrt1f, a, b, n);
TestCubeRootf("halley x 2", halley_cbrt2f, a, b, n);
TestCubeRootf("newton x 1", newton_cbrt1f, a, b, n);
TestCubeRootf("newton x 2", newton_cbrt2f, a, b, n);
TestCubeRootf("newton x 3", newton_cbrt3f, a, b, n);
TestCubeRootf("newton x 4", newton_cbrt4f, a, b, n);
printf("\n\n");
printf("64-bit double tests\n");
printf("----------------------------------------\n");
TestCubeRootd("cbrt_5d", cbrt_5d, a, b, n);
TestCubeRootd("pow", pow_cbrtd, a, b, n);
TestCubeRootd("halley x 1", halley_cbrt1d, a, b, n);
TestCubeRootd("halley x 2", halley_cbrt2d, a, b, n);
TestCubeRootd("halley x 3", halley_cbrt3d, a, b, n);
TestCubeRootd("newton x 1", newton_cbrt1d, a, b, n);
TestCubeRootd("newton x 2", newton_cbrt2d, a, b, n);
TestCubeRootd("newton x 3", newton_cbrt3d, a, b, n);
TestCubeRootd("newton x 4", newton_cbrt4d, a, b, n);
printf("\n\n");
return 0;
}
#endif
double cube_root(double x) {
if (approximately_zero(x)) {
return 0;
}
double result = halley_cbrt3d(fabs(x));
if (x < 0) {
result = -result;
}
return result;
}
#if TEST_ALTERNATIVES
// http://bytes.com/topic/c/answers/754588-tips-find-cube-root-program-using-c
/* cube root */
int icbrt(int n) {
int t=0, x=(n+2)/3; /* works for n=0 and n>=1 */
for(; t!=x;) {
int x3=x*x*x;
t=x;
x*=(2*n + x3);
x/=(2*x3 + n);
}
return x ; /* always(?) equal to floor(n^(1/3)) */
}
#endif
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