我无法计算<1的0<=数的平方

Question

我编写了一个计算sqrt(X)估计值的程序：

对下面代码的解释

我试图找到最近的数字(S)到(x)，它可以表示为S=sqrt*sqrt，一直循环到fabs(_sqrt - sqrt) >= 1e-8不正确为止，这意味着sqrt和_sqrt非常接近。

#include <stdio.h>
#include <stdlib.h>

int main (int argc, char *argv[]) {
    double val, min, max; 
    double sqrt;
    double S, _sqrt;
    int cnt = 0;

    // enter an argument,

    if (argc < 2) {
        printf ("Usage: sqrt < number >\n");
        return 1;
    }
     val = fabs (atof (argv[1]));
        min = 0;//minimum 
        max = sqrt = val;//maximum 
        _sqrt = 1;
        //1e-8 = 10^(-8) = 0.00000001
        while (fabs(_sqrt - sqrt) >= 1e-8) {
            _sqrt = sqrt;//keep the old sqrt
            sqrt = (max + min) / 2;
            S = sqrt * sqrt;
            printf("test n(%d)\tsqrt(%lf) = %.6lf\n",++cnt,val, sqrt);
                if (S>val){
                    max = sqrt;//setting max to the current sqrt
                } else{
                    min = sqrt;//setting min to the current sqrt
                }//end else
        }//end while
        puts("");
        puts("\t====================================================");
        printf ("\tfinal value.\tsqrt(%lf) = %.6lf\n", val, sqrt);
        puts("\t====================================================");
    return 0;
}//end main()

输出

[ar.lnx@host square-root] $ ./sqrt 2        
test n(1)       sqrt(2.000000) = 1.000000
test n(2)       sqrt(2.000000) = 1.500000
test n(3)       sqrt(2.000000) = 1.250000
test n(4)       sqrt(2.000000) = 1.375000
test n(5)       sqrt(2.000000) = 1.437500
test n(6)       sqrt(2.000000) = 1.406250
test n(7)       sqrt(2.000000) = 1.421875
test n(8)       sqrt(2.000000) = 1.414062
test n(9)       sqrt(2.000000) = 1.417969
test n(10)      sqrt(2.000000) = 1.416016
test n(11)      sqrt(2.000000) = 1.415039
test n(12)      sqrt(2.000000) = 1.414551
test n(13)      sqrt(2.000000) = 1.414307
test n(14)      sqrt(2.000000) = 1.414185
test n(15)      sqrt(2.000000) = 1.414246
test n(16)      sqrt(2.000000) = 1.414215
test n(17)      sqrt(2.000000) = 1.414200
test n(18)      sqrt(2.000000) = 1.414207
test n(19)      sqrt(2.000000) = 1.414211
test n(20)      sqrt(2.000000) = 1.414213
test n(21)      sqrt(2.000000) = 1.414214
test n(22)      sqrt(2.000000) = 1.414214
test n(23)      sqrt(2.000000) = 1.414213
test n(24)      sqrt(2.000000) = 1.414214
test n(25)      sqrt(2.000000) = 1.414214
test n(26)      sqrt(2.000000) = 1.414214
test n(27)      sqrt(2.000000) = 1.414214
test n(28)      sqrt(2.000000) = 1.414214

        ====================================================
        final value.    sqrt(2.000000) = 1.414214
        ====================================================

但是它只适用于每一个x>=1，如果我想计算0.5或0.254或0.1的平方.这不管用！

[ar.lnx@host square-root] $ ./sqrt1 0.5
test n(1)       sqrt(0.500000) = 0.250000
test n(2)       sqrt(0.500000) = 0.375000
test n(3)       sqrt(0.500000) = 0.437500
test n(4)       sqrt(0.500000) = 0.468750
test n(5)       sqrt(0.500000) = 0.484375
test n(6)       sqrt(0.500000) = 0.492188
test n(7)       sqrt(0.500000) = 0.496094
test n(8)       sqrt(0.500000) = 0.498047
test n(9)       sqrt(0.500000) = 0.499023
test n(10)      sqrt(0.500000) = 0.499512
test n(11)      sqrt(0.500000) = 0.499756
test n(12)      sqrt(0.500000) = 0.499878
test n(13)      sqrt(0.500000) = 0.499939
test n(14)      sqrt(0.500000) = 0.499969
test n(15)      sqrt(0.500000) = 0.499985
test n(16)      sqrt(0.500000) = 0.499992
test n(17)      sqrt(0.500000) = 0.499996
test n(18)      sqrt(0.500000) = 0.499998
test n(19)      sqrt(0.500000) = 0.499999
test n(20)      sqrt(0.500000) = 0.500000
test n(21)      sqrt(0.500000) = 0.500000
test n(22)      sqrt(0.500000) = 0.500000
test n(23)      sqrt(0.500000) = 0.500000
test n(24)      sqrt(0.500000) = 0.500000
test n(25)      sqrt(0.500000) = 0.500000
test n(26)      sqrt(0.500000) = 0.500000

        ====================================================
        final value.    sqrt(0.500000) = 0.500000
        ====================================================

有人能帮我弄清楚问题是什么以及如何解决吗？

Answer 1

原因是您的转换逻辑只有在val > 1时才能工作。

注意测试sqrt(.5)的方式，因为val == 0.5获得了S = val * val，即0.25。这将min重置为0.25，因为sqrt = (max + min) / 2;和初始max是val (0.5)，您永远无法通过0.5。

您必须确定，由于您的初始val小于1，所以您希望初始max为1，因此您的平均值为val < sqrt < 1。

如果您这样做，那么您应该收敛到正确的值。

此时我无法访问C编译器。但是，通过初步检验，第一个平均值是1.5/2 == .75，它重置了最大值。新的平均值是1.25/2 == .625，它重置最小值，然后继续，直到收敛到正确的值为止。

FYI您也可以使用while循环的内容作为一个不同的函数来教自己递归。也就是说，如果S> lim {调用新的估计函数}一旦它通过递归，它将传递最后的估计值回初始调用函数，而不是循环遍历时间。

Answer 2

问题在于您的max值，它必须大于或等于预期的结果。如果您的输入小于1.0，则结果大于输入，小于1.0。您对max的价值至少必须是1.0。像这样调整你的代码。

#include <stdio.h>
#include <stdlib.h>

int main (int argc, char *argv[]) {
    double val, min, max; 
    double sqrt;
    double S, _sqrt;
    int cnt = 0;

    // enter an argument,

    if (argc < 2) {
        printf ("Usage: sqrt < number >\n");
        return 1;
    }
    val = fabs (atof (argv[1]));       

    if ( val < 1.0 )
    {
        min = val;        
        max = sqrt = 1.0; // if val < 1.0 result is greater than val but less than 1.0
    }
    else
    {
        min = 0;          //minimum 
        max = sqrt = val; //maximum
    } 

    do
    {
        _sqrt = sqrt;//keep the old sqrt
        sqrt = (max + min) / 2;
        S = sqrt * sqrt;
        printf("test n(%d)\tsqrt(%lf) = %.6lf\n",++cnt,val, sqrt);
        if (S>val){
            max = sqrt;//setting max to the current sqrt
        } else{
            min = sqrt;//setting min to the current sqrt
        }//end else
    }
    while ( fabs(_sqrt - sqrt) >= 1e-8); //1e-8 = 10^(-8) = 0.00000001

    puts("");
    puts("\t====================================================");
    printf ("\tfinal value.\tsqrt(%lf) = %.6lf\n", val, sqrt);
    puts("\t====================================================");
    return 0;
}