Question

我对编码很陌生，我一直在寻找在线帮助编写一个将使用黄金分割法（显然是GNU科学图书馆有的C代码）的不可能的时间，虽然我没有任何运气找到它）找到牛顿最小化方法失败的最小函数。

具体来说，我想输入一个x值作为起点，并让代码输出函数的最小值和最小值点的x坐标。我的函数是f（x）= x ²⁰。我也被允许一些错误（＆lt; 10 ^-3）。

我甚至不知道从哪里开始，我一直在互联网上，并没有找到任何有用的东西。我会非常感谢一些帮助，我可以在哪里找到更多信息，或者我如何实现这种方法。

编辑：

这是我现在的代码：

#include <gsl/gsl_errno.h>  /* Defines GSL_SUCCESS, etc. */
#include <gsl/gsl_math.h>
#include <gsl/gsl_min.h>

int minimize_convex(gsl_function *F,double a, double b, double *x_min, double tol)
{
    int status;
    double h = (b - a) * .0000001;   /* Used to test slope at boundaries */
    /* First deal with the special cases */
    if (b - a < tol)
    {
        *x_min = b;
        status = GSL_SUCCESS;
    }
/* If the min is at a, then the derivative at a is >= 0.  Test for
 * this case. */
else if (GSL_FN_EVAL(F, a + h) - GSL_FN_EVAL(F, a) >= 0)
{
    *x_min = a;
    status = GSL_SUCCESS;
}
/* If the min is at b, then the derivative at b is >= 0.  Test for
 * this case. */
else if (GSL_FN_EVAL(F, b - h) - GSL_FN_EVAL(F, b) >= 0)
{
    *x_min = b;
    status = GSL_SUCCESS;
}
else
{
    /* Choose x_guess so that it's value is less than either of the two
     * endpoint values. Since we've got this far, we know that at least
     * of of F(a + h) and F(b - h) has this property. */
    double x_guess;
    x_guess = (GSL_FN_EVAL(F, a + h) < GSL_FN_EVAL(F, b - h)) ? 
        a + h : b - h;
    int iter = 0, max_iter = 200;
    const gsl_min_fminimizer_type *T;
    gsl_min_fminimizer *s;
    T = gsl_min_fminimizer_goldensection;
    s = gsl_min_fminimizer_alloc(T);
    gsl_min_fminimizer_set(s, F, x_guess, a, b);

    do
    {
       iter++;
       status = gsl_min_fminimizer_iterate(s);  /* perform iteration */
       status = 
           gsl_min_test_interval(a, b, tol, 0.0); /* |a - b| < tol? */

       a = gsl_min_fminimizer_x_lower(s);
       b = gsl_min_fminimizer_x_upper(s);

       if (status == GSL_SUCCESS)
       {
           *x_min = gsl_min_fminimizer_x_minimum(s);  /* current est */
       }
    }
    while (status == GSL_CONTINUE && iter < max_iter);

    gsl_min_fminimizer_free(s);
}
return status;
}
double f(double x, void *params)
{
    double *p = (double *) params;
    return (x^(50)) + *p;
}
double C = 0.0;
int main (void)
{
    double m = 0.0, result;
    double a = -1.0, b = 1.0;
    double epsilon = 0.001;
    int exit_val;
    gsl_function F;
    F.function = &f;
    F.params = &C;
    exit_val = minimize_convex(&F, a, b, m, &result, epsilon);
    printf("Minimizer: %g\n", result);
    printf("Function value: %g\n", f(result, &C));
    printf("%d\n", exit_val);
    return 0;
}

我收到以下错误： try.c：69：14：错误：二进制文件的操作数无效表达式（'double'和'double'） return（x ^（50））+ * p;

try.c：81：54：错误：函数的参数太多了打电话，预计5，有6 exit_val = minimize_convex（＆amp; F，a，b，m，＆amp; result，epsilon）;

有什么想法吗？

Answer 1

gsl有一个通用的最小化器，可以使用多种方法来实现最小化。有关如何使用最小化器的说明，请参见documentation。您可以通过将方法显示为gsl_min_fminimizer_goldensection来将其设置为黄金分割方法。

C中的黄金分割方法

1 个答案: