sqrt()函数，是绝大部分语言支持的常用函数，它实现的是开方运算；开方运算最早是在我国魏晋时数学家刘徽所著的《九章算术》被提及。今天写了几个函数加上国外大神的几个神级程序带大家领略sqrt的神奇之处。
1、古人算法（暴力法）
原理：从0开始0.00001,000002...一个一个试，直到找到x的平方根，代码如下：
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); }
测试结果：
1、7320509999476947
2、牛顿迭代法
计算机科班出身的童鞋可能首先会想到的是《数值分析》中的牛顿迭代法求平方根。原理是：随意选一个数比如说8，要求根号3，我们可以这么算：
(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
做了4步基本算出了近似值了，这种迭代的方式就是传说中的牛顿迭代法了，代码如下：
public class APIsqrt { static double newtonSqrt(double x) { if (x < 0) { System.out.println("负数没事开什么方"); 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); } }
测试结果：
17320508075688772
3、暴力-牛顿综合法
原理：还是以根号3为例，先用暴力法讲根号3逼近到1.7，然后再利用上述的牛顿迭代法。虽然没有用牛顿迭代好，但是也为我们提供一种思路。代码如下：
public class APIsqrt { static double baoliAndNewTonSqrt(double x) { if (x < 0) { System.out.println("负数没事开什么方"); 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); } }
测试结果：
1、7320508075689423
4、二分开方法
原理：还是以3举例：
(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;
直到前后两次平均值只差小于自定义精度为止，代码如下：
public class APIsqrt { static double erfenSqrt(double x) { if (x < 0) { System.out.println("负数没事开什么方"); 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); } }
测试结果：
1、732051134109497
5、计算 (int)(sqrt(x))算法
PS:此算法非博主所写
原理：空间换时间，细节请大家自行探究，代码如下：
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)); } }
测试结果：8
6、最快的sqrt算法
PS:此算法非博主所写
这个算法很有名，大家可能也见过，作者是开发游戏的，图形算法中经常用到sqrt，作者才写了一个神级算法，和他那神秘的0x5f3759df，代码如下
#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; }
测试结果：
感兴趣的朋友可以参考http://wenku.baidu.com/view/a0174fa20029bd64783e2cc0.html 是作者解释这个算法的14页论文《Fast Inverse Square Root》
7、一个与算法6相似的算法
PS:此算法非博主所写
代码如下：
#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; }
测试结果：
