开根号的笔算算法图解_数学算法开平方的七种算法

阳光穿透心脏的1/2处 2023-01-02 15:26 397阅读 0赞

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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

![8a9db889270e93c7a8d7f1f63add2e8c.gif][]

![9da733d5a3fc5b410f729c4dc6c0b5b0.gif][]

**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

![8a9db889270e93c7a8d7f1f63add2e8c.gif][]

![9da733d5a3fc5b410f729c4dc6c0b5b0.gif][]

**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

![8a9db889270e93c7a8d7f1f63add2e8c.gif][]

![9da733d5a3fc5b410f729c4dc6c0b5b0.gif][]

**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

![8a9db889270e93c7a8d7f1f63add2e8c.gif][]

![9da733d5a3fc5b410f729c4dc6c0b5b0.gif][]

**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;

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

![8a9db889270e93c7a8d7f1f63add2e8c.gif][]

![9da733d5a3fc5b410f729c4dc6c0b5b0.gif][]

**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

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》

![8a9db889270e93c7a8d7f1f63add2e8c.gif][]

![9da733d5a3fc5b410f729c4dc6c0b5b0.gif][]

**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

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 ) );

return number \* y;

int main()

printf("%f",SquareRootFloat(3));

return 0;

测试结果：

![3f72c3bd9af2d3aa8af1f2cc6c1ac1ef.png][]

![8a9db889270e93c7a8d7f1f63add2e8c.gif][]

完

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