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：
1.7320508075688772
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/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：
————
编辑∑Gemini
来源：CSDN
