开根号的笔算算法图解_开平方的 7 种算法

作者：nash_
链接：https://blog.csdn.net/zmazon/article/details/8217866 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 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 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：

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