d6176b0dca
This CL is part II of IV (I broke down the 1280 files into 4 CLs). Review URL: https://codereview.appspot.com/6474054 git-svn-id: http://skia.googlecode.com/svn/trunk@5263 2bbb7eff-a529-9590-31e7-b0007b416f81
388 lines
9.0 KiB
C++
388 lines
9.0 KiB
C++
// http://metamerist.com/cbrt/CubeRoot.cpp
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//
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#include <math.h>
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#include "CubicUtilities.h"
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#define TEST_ALTERNATIVES 0
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#if TEST_ALTERNATIVES
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typedef float (*cuberootfnf) (float);
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typedef double (*cuberootfnd) (double);
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// estimate bits of precision (32-bit float case)
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inline int bits_of_precision(float a, float b)
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{
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const double kd = 1.0 / log(2.0);
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if (a==b)
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return 23;
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const double kdmin = pow(2.0, -23.0);
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double d = fabs(a-b);
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if (d < kdmin)
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return 23;
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return int(-log(d)*kd);
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}
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// estiamte bits of precision (64-bit double case)
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inline int bits_of_precision(double a, double b)
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{
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const double kd = 1.0 / log(2.0);
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if (a==b)
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return 52;
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const double kdmin = pow(2.0, -52.0);
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double d = fabs(a-b);
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if (d < kdmin)
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return 52;
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return int(-log(d)*kd);
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}
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// cube root via x^(1/3)
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static float pow_cbrtf(float x)
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{
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return (float) pow(x, 1.0f/3.0f);
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}
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// cube root via x^(1/3)
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static double pow_cbrtd(double x)
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{
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return pow(x, 1.0/3.0);
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}
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// cube root approximation using bit hack for 32-bit float
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static float cbrt_5f(float f)
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{
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unsigned int* p = (unsigned int *) &f;
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*p = *p/3 + 709921077;
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return f;
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}
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#endif
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// cube root approximation using bit hack for 64-bit float
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// adapted from Kahan's cbrt
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static double cbrt_5d(double d)
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{
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const unsigned int B1 = 715094163;
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double t = 0.0;
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unsigned int* pt = (unsigned int*) &t;
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unsigned int* px = (unsigned int*) &d;
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pt[1]=px[1]/3+B1;
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return t;
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}
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#if TEST_ALTERNATIVES
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// cube root approximation using bit hack for 64-bit float
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// adapted from Kahan's cbrt
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#if 0
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static double quint_5d(double d)
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{
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return sqrt(sqrt(d));
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const unsigned int B1 = 71509416*5/3;
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double t = 0.0;
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unsigned int* pt = (unsigned int*) &t;
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unsigned int* px = (unsigned int*) &d;
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pt[1]=px[1]/5+B1;
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return t;
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}
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#endif
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// iterative cube root approximation using Halley's method (float)
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static float cbrta_halleyf(const float a, const float R)
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{
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const float a3 = a*a*a;
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const float b= a * (a3 + R + R) / (a3 + a3 + R);
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return b;
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}
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#endif
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// iterative cube root approximation using Halley's method (double)
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static double cbrta_halleyd(const double a, const double R)
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{
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const double a3 = a*a*a;
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const double b= a * (a3 + R + R) / (a3 + a3 + R);
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return b;
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}
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#if TEST_ALTERNATIVES
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// iterative cube root approximation using Newton's method (float)
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static float cbrta_newtonf(const float a, const float x)
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{
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// return (1.0 / 3.0) * ((a + a) + x / (a * a));
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return a - (1.0f / 3.0f) * (a - x / (a*a));
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}
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// iterative cube root approximation using Newton's method (double)
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static double cbrta_newtond(const double a, const double x)
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{
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return (1.0/3.0) * (x / (a*a) + 2*a);
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}
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// cube root approximation using 1 iteration of Halley's method (double)
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static double halley_cbrt1d(double d)
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{
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double a = cbrt_5d(d);
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return cbrta_halleyd(a, d);
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}
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// cube root approximation using 1 iteration of Halley's method (float)
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static float halley_cbrt1f(float d)
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{
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float a = cbrt_5f(d);
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return cbrta_halleyf(a, d);
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}
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// cube root approximation using 2 iterations of Halley's method (double)
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static double halley_cbrt2d(double d)
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{
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double a = cbrt_5d(d);
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a = cbrta_halleyd(a, d);
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return cbrta_halleyd(a, d);
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}
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#endif
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// cube root approximation using 3 iterations of Halley's method (double)
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static double halley_cbrt3d(double d)
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{
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double a = cbrt_5d(d);
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a = cbrta_halleyd(a, d);
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a = cbrta_halleyd(a, d);
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return cbrta_halleyd(a, d);
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}
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#if TEST_ALTERNATIVES
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// cube root approximation using 2 iterations of Halley's method (float)
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static float halley_cbrt2f(float d)
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{
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float a = cbrt_5f(d);
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a = cbrta_halleyf(a, d);
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return cbrta_halleyf(a, d);
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}
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// cube root approximation using 1 iteration of Newton's method (double)
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static double newton_cbrt1d(double d)
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{
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double a = cbrt_5d(d);
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return cbrta_newtond(a, d);
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}
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// cube root approximation using 2 iterations of Newton's method (double)
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static double newton_cbrt2d(double d)
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{
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double a = cbrt_5d(d);
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a = cbrta_newtond(a, d);
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return cbrta_newtond(a, d);
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}
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// cube root approximation using 3 iterations of Newton's method (double)
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static double newton_cbrt3d(double d)
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{
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double a = cbrt_5d(d);
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a = cbrta_newtond(a, d);
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a = cbrta_newtond(a, d);
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return cbrta_newtond(a, d);
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}
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// cube root approximation using 4 iterations of Newton's method (double)
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static double newton_cbrt4d(double d)
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{
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double a = cbrt_5d(d);
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a = cbrta_newtond(a, d);
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a = cbrta_newtond(a, d);
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a = cbrta_newtond(a, d);
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return cbrta_newtond(a, d);
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}
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// cube root approximation using 2 iterations of Newton's method (float)
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static float newton_cbrt1f(float d)
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{
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float a = cbrt_5f(d);
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return cbrta_newtonf(a, d);
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}
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// cube root approximation using 2 iterations of Newton's method (float)
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static float newton_cbrt2f(float d)
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{
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float a = cbrt_5f(d);
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a = cbrta_newtonf(a, d);
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return cbrta_newtonf(a, d);
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}
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// cube root approximation using 3 iterations of Newton's method (float)
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static float newton_cbrt3f(float d)
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{
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float a = cbrt_5f(d);
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a = cbrta_newtonf(a, d);
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a = cbrta_newtonf(a, d);
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return cbrta_newtonf(a, d);
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}
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// cube root approximation using 4 iterations of Newton's method (float)
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static float newton_cbrt4f(float d)
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{
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float a = cbrt_5f(d);
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a = cbrta_newtonf(a, d);
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a = cbrta_newtonf(a, d);
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a = cbrta_newtonf(a, d);
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return cbrta_newtonf(a, d);
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}
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static double TestCubeRootf(const char* szName, cuberootfnf cbrt, double rA, double rB, int rN)
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{
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const int N = rN;
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float dd = float((rB-rA) / N);
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// calculate 1M numbers
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int i=0;
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float d = (float) rA;
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double s = 0.0;
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for(d=(float) rA, i=0; i<N; i++, d += dd)
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{
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s += cbrt(d);
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}
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double bits = 0.0;
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double worstx=0.0;
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double worsty=0.0;
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int minbits=64;
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for(d=(float) rA, i=0; i<N; i++, d += dd)
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{
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float a = cbrt((float) d);
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float b = (float) pow((double) d, 1.0/3.0);
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int bc = bits_of_precision(a, b);
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bits += bc;
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if (b > 1.0e-6)
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{
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if (bc < minbits)
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{
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minbits = bc;
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worstx = d;
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worsty = a;
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}
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}
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}
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bits /= N;
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printf(" %3d mbp %6.3f abp\n", minbits, bits);
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return s;
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}
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static double TestCubeRootd(const char* szName, cuberootfnd cbrt, double rA, double rB, int rN)
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{
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const int N = rN;
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double dd = (rB-rA) / N;
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int i=0;
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double s = 0.0;
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double d = 0.0;
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for(d=rA, i=0; i<N; i++, d += dd)
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{
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s += cbrt(d);
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}
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double bits = 0.0;
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double worstx = 0.0;
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double worsty = 0.0;
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int minbits = 64;
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for(d=rA, i=0; i<N; i++, d += dd)
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{
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double a = cbrt(d);
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double b = pow(d, 1.0/3.0);
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int bc = bits_of_precision(a, b); // min(53, count_matching_bitsd(a, b) - 12);
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bits += bc;
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if (b > 1.0e-6)
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{
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if (bc < minbits)
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{
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bits_of_precision(a, b);
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minbits = bc;
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worstx = d;
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worsty = a;
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}
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}
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}
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bits /= N;
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printf(" %3d mbp %6.3f abp\n", minbits, bits);
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return s;
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}
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static int _tmain()
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{
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// a million uniform steps through the range from 0.0 to 1.0
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// (doing uniform steps in the log scale would be better)
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double a = 0.0;
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double b = 1.0;
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int n = 1000000;
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printf("32-bit float tests\n");
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printf("----------------------------------------\n");
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TestCubeRootf("cbrt_5f", cbrt_5f, a, b, n);
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TestCubeRootf("pow", pow_cbrtf, a, b, n);
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TestCubeRootf("halley x 1", halley_cbrt1f, a, b, n);
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TestCubeRootf("halley x 2", halley_cbrt2f, a, b, n);
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TestCubeRootf("newton x 1", newton_cbrt1f, a, b, n);
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TestCubeRootf("newton x 2", newton_cbrt2f, a, b, n);
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TestCubeRootf("newton x 3", newton_cbrt3f, a, b, n);
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TestCubeRootf("newton x 4", newton_cbrt4f, a, b, n);
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printf("\n\n");
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printf("64-bit double tests\n");
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printf("----------------------------------------\n");
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TestCubeRootd("cbrt_5d", cbrt_5d, a, b, n);
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TestCubeRootd("pow", pow_cbrtd, a, b, n);
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TestCubeRootd("halley x 1", halley_cbrt1d, a, b, n);
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TestCubeRootd("halley x 2", halley_cbrt2d, a, b, n);
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TestCubeRootd("halley x 3", halley_cbrt3d, a, b, n);
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TestCubeRootd("newton x 1", newton_cbrt1d, a, b, n);
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TestCubeRootd("newton x 2", newton_cbrt2d, a, b, n);
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TestCubeRootd("newton x 3", newton_cbrt3d, a, b, n);
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TestCubeRootd("newton x 4", newton_cbrt4d, a, b, n);
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printf("\n\n");
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return 0;
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}
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#endif
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double cube_root(double x) {
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return halley_cbrt3d(x);
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}
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#if TEST_ALTERNATIVES
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// http://bytes.com/topic/c/answers/754588-tips-find-cube-root-program-using-c
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/* cube root */
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int icbrt(int n) {
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int t=0, x=(n+2)/3; /* works for n=0 and n>=1 */
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for(; t!=x;) {
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int x3=x*x*x;
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t=x;
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x*=(2*n + x3);
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x/=(2*x3 + n);
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}
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return x ; /* always(?) equal to floor(n^(1/3)) */
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}
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#endif
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