Seven mathematical algorithms

The SQRT () function is a common function supported by most languages. It implements open-side operations. The open-side operations were first mentioned in chapter 9 arithmetic by Liu Hui, a mathematician in China's Wei and Jin Dynasties. Today, I wrote several functions and several God-class programs from other countries to help you understand the magic of SQRT.

1. Ancients algorithm (Violence Law)

Principle: from 0 to 0.00001, 000002... try one by one until the square root of X is found. The Code is as follows:

public class APIsqrt {static double baoliSqrt(double x) {final double _JINGDU = 1e-6;double i;for (i = 0; Math.abs(x - i * i) > _JINGDU; i += _JINGDU);return i;}public static void main(String[] args) {double x = 3;double root = baoliSqrt(x);System.out.println(root);}}

Test results:

1.7320509999476947

2. Newton Iteration Method

The first thing that comes to mind in computer science is the Newton Iteration Method in Numerical Analysis for the square root. The principle is: select a random number, for example, 8, with the root number 3. We can calculate it as follows:

(8 + 3/8) = 4.1875

(4.1875 + 3/4.1875)
= 2.4519

(2.4519
+ 3/2. 4519) = 1.837

(1.837 +
3/1. 837) = 1.735

After four steps, the approximate values are basically calculated. This iteration method is the legendary Newton iteration method. The Code is as follows:

Public class apisqrt {static double newtonsqrt (Double X) {If (x <0) {system. out. println ("no way to enable negative numbers"); Return-1 ;}if (x = 0) return 0; double _ AVG = x; double last_avg = double. max_value; Final double _ jingdu = 1e-6; while (math. ABS (_ AVG-last_avg)> _ jingdu) {last_avg = _ AVG; _ AVG = (_ AVG + x/_ avg)/2;} return _ AVG ;} public static void main (string [] ARGs) {Double X = 3; double root = newtonsqrt (x); system. out. println (Root );}}

Test results:

1.7320508075688772

3. Violence-Newton Synthesis Method

Principle: take root 3 as an example. Use the violence law to tell root 3 to approach 1.7, and then use the above Newton Iteration Method. Although Newton's iteration is not good, it also provides us with an idea. The Code is as follows:

Public class apisqrt {static double baoliandnewtonsqrt (Double X) {If (x <0) {system. out. println ("no way to enable negative numbers"); Return-1 ;}if (x = 0) return 0; double I = 0; double _ AVG; double last_avg = double. max_value; for (I = 0; I * I <X; I ++ = 0.1); _ AVG = I; Final double _ jingdu = 1e-6; while (math. ABS (_ AVG-last_avg)> _ jingdu) {last_avg = _ AVG; _ AVG = (_ AVG + x/_ avg)/2;} return _ AVG ;} public static void main (string [] ARGs) {Double X = 3; double root = baoliandnewtonsqrt (x); system. out. println (Root );}}

Test results:

1.7320508075689423

4. Two separate methods

Principle: Take 3 For example:

(0 + 3)/2 = 1.5, 1.5 ^ 2 = 2.25, 2.25 <3;

(1.5 + 3)/2 = 2.25, 2.25 ^ 2 = 5.0625, 5.0625> 3;

(1.5 + 2.25)/2 = 1.875, 1.875 ^ 2 = 3.515625; 3.515625> 3;

.

.

.

The Code is as follows:

Public class apisqrt {static double erfensqrt (Double X) {If (x <0) {system. out. println ("no way to enable negative numbers"); Return-1;} If (x = 0) return 0; Final double _ jingdu = 1e-6; double _ low = 0; double _ high = x; double _ mid = double. max_value; double last_mid = double. min_value; while (math. ABS (_ mid-last_mid)> _ jingdu) {last_mid = _ mid; _ mid = (_ low + _ high)/2; If (_ mid * _ mid> X) _ high = _ mid; If (_ mid * _ mid <X) _ low = _ mid;} return _ mid;} public static void main (string [] ARGs) {Double X = 3; double root = erfensqrt (x); system. out. println (Root );}}

Test results:

1.732051134109497

5. calculation (INT) (SQRT (x) algorithm PS: this algorithm is not written by bloggers

Principle: Change the space time. For details, please explore the Code as follows:

public class APIsqrt2 {final static int[] table = { 0, 16, 22, 27, 32, 35, 39, 42, 45, 48, 50, 53,55, 57, 59, 61, 64, 65, 67, 69, 71, 73, 75, 76, 78, 80, 81, 83, 84,86, 87, 89, 90, 91, 93, 94, 96, 97, 98, 99, 101, 102, 103, 104,106, 107, 108, 109, 110, 112, 113, 114, 115, 116, 117, 118, 119,120, 121, 122, 123, 124, 125, 126, 128, 128, 129, 130, 131, 132,133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 144,145, 146, 147, 148, 149, 150, 150, 151, 152, 153, 154, 155, 155,156, 157, 158, 159, 160, 160, 161, 162, 163, 163, 164, 165, 166,167, 167, 168, 169, 170, 170, 171, 172, 173, 173, 174, 175, 176,176, 177, 178, 178, 179, 180, 181, 181, 182, 183, 183, 184, 185,185, 186, 187, 187, 188, 189, 189, 190, 191, 192, 192, 193, 193,194, 195, 195, 196, 197, 197, 198, 199, 199, 200, 201, 201, 202,203, 203, 204, 204, 205, 206, 206, 207, 208, 208, 209, 209, 210,211, 211, 212, 212, 213, 214, 214, 215, 215, 216, 217, 217, 218,218, 219, 219, 220, 221, 221, 222, 222, 223, 224, 224, 225, 225,226, 226, 227, 227, 228, 229, 229, 230, 230, 231, 231, 232, 232,233, 234, 234, 235, 235, 236, 236, 237, 237, 238, 238, 239, 240,240, 241, 241, 242, 242, 243, 243, 244, 244, 245, 245, 246, 246,247, 247, 248, 248, 249, 249, 250, 250, 251, 251, 252, 252, 253,253, 254, 254, 255 };/** * A faster replacement for (int)(java.lang.Math.sqrt(x)). Completely * accurate for x < 2147483648 (i.e. 2^31)... */static int sqrt(int x) {int xn;if (x >= 0x10000) {if (x >= 0x1000000) {if (x >= 0x10000000) {if (x >= 0x40000000) {xn = table[x >> 24] << 8;} else {xn = table[x >> 22] << 7;}} else {if (x >= 0x4000000) {xn = table[x >> 20] << 6;} else {xn = table[x >> 18] << 5;}}xn = (xn + 1 + (x / xn)) >> 1;xn = (xn + 1 + (x / xn)) >> 1;return ((xn * xn) > x) ? --xn : xn;} else {if (x >= 0x100000) {if (x >= 0x400000) {xn = table[x >> 16] << 4;} else {xn = table[x >> 14] << 3;}} else {if (x >= 0x40000) {xn = table[x >> 12] << 2;} else {xn = table[x >> 10] << 1;}}xn = (xn + 1 + (x / xn)) >> 1;return ((xn * xn) > x) ? --xn : xn;}} else {if (x >= 0x100) {if (x >= 0x1000) {if (x >= 0x4000) {xn = (table[x >> 8]) + 1;} else {xn = (table[x >> 6] >> 1) + 1;}} else {if (x >= 0x400) {xn = (table[x >> 4] >> 2) + 1;} else {xn = (table[x >> 2] >> 3) + 1;}}return ((xn * xn) > x) ? --xn : xn;} else {if (x >= 0) {return table[x] >> 4;}}}return -1;}public static void main(String[] args){System.out.println(sqrt(65));}}

Test result: 8

6. Fastest SQRT algorithm PS: this algorithm is not written by bloggers

This algorithm is very famous, and you may have seen it. The author developed a game and SQRT is often used in graphics algorithms. The author wrote a God-Level Algorithm and his mysterious 0x5f3759df. The Code is as follows:

#include <math.h>float InvSqrt(float x){ float xhalf = 0.5f*x; int i = *(int*)&x; // get bits for floating VALUE i = 0x5f375a86- (i>>1); // gives initial guess y0 x = *(float*)&i; // convert bits BACK to float x = x*(1.5f-xhalf*x*x); // Newton step, repeating increases accuracy return x;}int main(){  printf("%lf",1/InvSqrt(3));   return 0;}

Test results:

Interested friends can refer to http://wenku.baidu.com/view/a0174fa20029bd64783e2cc0.html
The author explains this algorithm's 14-page essay fast inverse square root

7. An algorithm similar to algorithm 6 Ps: this algorithm is not written by bloggers

The Code is as follows:

#include <math.h>float SquareRootFloat(float number) {    long i;    float x, y;    const float f = 1.5F;    x = number * 0.5F;    y  = number;    i  = * ( long * ) &y;    i  = 0x5f3759df - ( i >> 1 );    y  = * ( float * ) &i;    y  = y * ( f - ( x * y * y ) );    y  = y * ( f - ( x * y * y ) );    return number * y;}int main(){  printf("%f",SquareRootFloat(3));   return 0;}

Test results:

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