算法的C来计算多项式用拉格朗日插值系数 [英] Algorithm in C to calculate coefficients of polynomial using Lagrange interpolation
问题描述
我一直停留在此有一段时间了。我用拉格朗日插值方法写一个算法的C拉出一个多项式的系数。
I've been stuck on this for a while now. I'm writing an algorithm in C to pull out the coefficients of a polynomial using Lagrange's interpolation method.
我的code部分的工作，例如，如果我们在这里做的第一个例子 HTTP：/ /en.wikipedia.org/wiki/Lagrange_polynomial#Example_1 的那么code可以打印出第2系数（0和4.834848）
My code partially works, for instance if we do the first example here http://en.wikipedia.org/wiki/Lagrange_polynomial#Example_1 then the code can print out the first 2 coefficients (0 and 4.834848)
同样以实例3这篇文章中，将打印的2系数6和-11。
Similarly with example 3 on that article, it will print the 2 coefficients 6 and -11.
我需要能够准确地获得所有的系数从任何点集。请告知需要到code中的变化。
I need to be able to accurately get all the coefficients from any set of points. Please advise on the alterations required to the code.
在此先感谢！
更新了最新的code，下午7点57，北京时间8月5日。 9系数现在的工作，越来越难看。将调查迭代过程的n度的明天！
Updated with latest code, 7:57PM, GMT on August 5th. 9 coefficients now working, getting ugly looking. Will investigate iterative process for n degrees tomorrow!
#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);
}
推荐答案
您代数是完全错误的，这其实是隐藏的选择不当的变量名。
在计算我日基础多项式的贡献（别介意是现在）有什么变化再presents系数的 X 2 的任期？这是 coeff3 。而你没有正确计算。
When you calculate the contribution of the ith basis polynomial (never mind y for now) what variable represents the coefficient of the x2 term? It's coeff3. And you don't calculate it correctly.
以一个简单的情况。假设你想工作了（X + A）（X + B）（X + C）（X + D）。第一项为x 4 ，轻松。接下来是（A + B + C + D）x 3 ，不算太糟糕。接下来的是（AB + AC +广告+ BC + BD + CD）x 2 ，而现在很明显，一个循环就不会做这个工作。 这是值得花时间，以确保你可以写code正确处理简单的问题，你试试更复杂的前您需要的是这样的：
Take a simpler case. Suppose you want to work out (x+a)(x+b)(x+c)(x+d). The first term is x4, easy. The next is (a+b+c+d)x3, not too bad. The next is (ab + ac + ad + bc + bd + cd)x2, and now it's clear that a single loop won't do the job. It's worth taking the time to make sure you can write code that handles the simple problem correctly, before you try the more complex one. You need something like this:
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。
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