Question

我已经坚持了一段时间了。我在C中编写一个算法，用拉格朗日插值方法得出多项式的系数。

我的代码部分有效，例如，如果我们在这里做第一个例子http://en.wikipedia.org/wiki/Lagrange_polynomial#Example_1，那么代码可以打印出前两个系数（0和4.834848）

与该文章的例子3类似，它将打印2个系数6和-11。

我需要能够准确地从任何一组点中获取所有系数。请告知代码所需的更改。

提前致谢！

最新代码更新，格林威治标准时间8月5日下午7:57。 9个系数正在运行，看起来很难看。将在明天调查n度的迭代过程！

#include<ncursesw/ncurses.h>
#include<math.h>
#include <stdio.h>
#include <string.h>
#include <errno.h>
#include <stdlib.h>
#define MAX 200
float coeff[MAX], coefftwo[MAX], coeffthree[MAX], coefffour[MAX];
int count;
void main()
{
int n,i,j ;
char ch;
float x[MAX],y[MAX],fp2, coeff1, coeff2;
printf("\n\nn = ");
scanf("%i", &count);

    for(i=0; i < count; i++)
    {
        printf("\n\n The value of x%i= ", i);
        scanf("%f",&x[i]);
        printf("\n The value of f(x%i)= ", i);
        scanf("%f",&y[i]);
    }
    for(i=0;i<count;i++)
    {
        coeff1 = 1.0;
        coeff2 = 0.0;
        coeff3 = 0.0;
        coeff4 = 0.0;
        coeff5 = 0.0;
        coeff6 = 0.0;
        coeff7 = 0.0;
        coeff8 = 0.0;
        coeff9 = 0.0;
        for(j=0; j<count; j++)
        {
            if(i!=j) {
                coeff1 = coeff1 * (array[i]-array[j]);
                coeff2 -= array[j];
                for (int k=j; k < count; k++) {
                if ((j!=k) && (k!=i)) {
                coeff3 += array[j] * array[k];
                    for(int l=k; l < count; l++) {
                    if ((l!=j) && (l!=k) && (l!=i))     {
    coeff4 -= array[j] * array[k] * array[l];

    for (int m = l; m < count; m++) {
        if ((m!=l) && (m!=k) && (m!=j) && (m!=i)) {                 coeff5 += array[j] * array[k] * array[l] * array[m];
            for (int n = m; n < count; n++) {
            if ((n!=m) && (n!=l) && (n!=k) && (n!=j) && (n!=i)) {
            coeff6 -= array[j] * array[k] * array[l] * array[m] * array[n];
            for (int o = n; o < count; o++) {
            if ((o!=n) && (o!=m) && (o!=l) && (o!=k) && (o!=j) && (o!=i)) {
            coeff7 += array[j] * array[k] * array[l] * array[m] * array[n] * array[o];
            for (int p = o; p < count; p++) {
            if ((p!=o) && (p!=n) && (p!=m) && (p!=l) && (p!=k) && (p!=j) && (p!=i)) {
            coeff8 -= array[j] * array[k] * array[l] *array[m] *array[n] * array[o] * array[p];
            for (int q = p; q < count; q++) {
            if ((q!=p) && (q!=o) && (q!=n) && (q!=m) && (q!=l) && (q!=k) && (q!=j) && (q!=i)) {
            coeff9 += array[j] * array[k] * array[l] * array[m] * array[n] * array[o] * array[p] * array[q];
            }
            }
            }
            }
            }
            }
            }
            }
                    }               
        }
                    }
                }
            }
        }    
    }
}       
        coeff[i] = y[i] / coeff1;
        coefftwo[i] = y[i] * coeff2 / coeff1;   
        coeffthree[i] = y[i] * coeff3 / coeff1;
        coefffour[i] = y[i] * coeff4 / coeff1;
        coefffive[i] = y[i] * coeff5 / coeff1;
        coeffsix[i] = y[i] * coeff6 / coeff1;
        coeffseven[i] = y[i] * coeff7 / coeff1;
        coeffeight[i] = y[i] * coeff8 / coeff1;
        coeffnine[i] = y[i] * coeff9 / coeff1;
    }
float coefficientone = 0.0;
float coefficienttwo = 0.0;
float coefficientthree = 0.0;
float coefficientfour = 0.0;
float coefficientfive = 0.0;
float coefficientsix = 0.0;
float coefficientseven = 0.0;
float coefficienteight = 0.0;
float coefficientnine = 0.0;
for (int i = 0; i< count; i++){
        coefficientone = coefficientone + coeff[i];
        coefficienttwo = coefficienttwo + coefftwo[i];
        coefficientthree = coefficientthree + coeffthree[i];
        coefficientfour = coefficientfour + coefffour[i];
        coefficientfive = coefficientfive + coefffive[i];
        coefficientsix = coefficientsix + coeffsix[i];
        coefficientseven = coefficientseven + coeffseven[i];
        coefficienteight = coefficienteight + coeffeight[i];
        coefficientnine = coefficientnine + coeffnine[i];
}
printf("coefficient 1 = %f\n", coefficientone);
printf("coefficient 2 = %f\n", coefficienttwo);
printf("coefficient 3 = %f\n", coefficientthree);
printf("coefficient 4 = %f\n", coefficientfour);
printf("coefficient 5 = %f\n", coefficientfive);
printf("coefficient 6 = %f\n", coefficientsix);
printf("coefficient 7 = %f\n", coefficientseven);
printf("coefficient 8 = %f\n", coefficienteight);
printf("coefficient 9 = %f\n", coefficientnine);

}

Answer 1

你的代数完全错了，选择不当的变量名隐藏了这个事实。

当你计算i ^th基多项式的贡献时（暂不考虑y），什么变量代表 x ²一词？它是coeff3。而且你没有正确计算它。

采取更简单的案例。假设你想算出（x + a）（x + b）（x + c）（x + d）。第一项是x ⁴，很容易。接下来是（a + b + c + d）x ³，也不错。接下来是（ab + ac + ad + bc + bd + cd）x ²，现在很明显单个循环不能完成这项工作。 值得花时间确保在尝试更复杂的问题之前，可以编写能够正确处理简单问题的代码。你需要这样的东西：

for(unsigned int j=0 ; j<count ; ++j)
{
  ...
  coeff2 -= x[j];
  for(unsigned int k=j ; k<count ; ++k)
  {
    if(j!=k && k!=i)
      coeff3 += x[j] * x[k];
    ...
  }
}

这应该足以让你开始。

用C算法用拉格朗日插值计算多项式系数

1 个答案: