最小二乘法用于直线，多项式，圆，椭圆的拟合及程序实现（资料整理的非常全）
2018-03-14 18:14
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最小二乘法是一种优化算法，最小二乘法名字的缘由有两个：一是要将误差最小化，二是将误差最小化的方法是使误差的平方和最小化。利用最小二乘法可以简便地求得未知的数据，并使得这些求得的数据与实际数据之间误差的平方和为最小。最小二乘法还可用于曲线拟合，所拟合的曲线可以是线性拟合与非线性拟合。
残差平方和：
则通过Q最小确定这条直线，即确定
，以
为变量，把它们看作是Q的函数，就变成了一个求极值的问题，可以通过求导数得到。求Q对两个待估参数的偏导数：
解得：
实例（c++）：[cpp] view plain copy#include <iostream>
#include <algorithm>
#include <valarray>
#include <vector>
using namespace std;
int main()
{
double x[] = { 1, 2, 3, 4, 5, 6 };
double y[] = { 3, 5.5, 6.8, 8.8, 11, 12};
valarray<double> data_x(x, 6);
valarray<double> data_y(y, 6);
float A = 0.0;
float B = 0.0;
float C = 0.0;
float D = 0.0;
A = (data_x*data_x).sum();
B = data_x.sum();
C = (data_x*data_y).sum();
D = data_y.sum();
float tmp = A*data_x.size() - B*B;
float k, b;
k = (C*data_x.size() - B*D) / tmp;
b = (A*D - C*B) / tmp;
cout << "y=" << k << "x+" << b << endl;
return 0;
}
运行结果：
注：valarray类似vector，也是一个模板类，主要用来对一系列元素进行高速的数字计算，其与vector的主要区别在于以下两点：
1、valarray定义了一组在两个相同长度和相同类型的valarray类对象之间的数字计算；
2、通过重载operater[]，可以返回valarray的相关信息（valarray其中某个元素的引用、特定下标的值或者其某个子集）。valarray类构造函数:
valarray( );
explicit valarray(size_t _Count);
valarray( const Type& _Val, size_t _Count);
valarray( const Type* _Ptr, size_t _Count);
valarray( const slice_array<Type>& _SliceArray);
valarray( const gslice_array<Type>& _GsliceArray);
valarray( const mask_array<Type>& _MaskArray);
valarray( const indirect_array<Type>& _IndArray);
valarray 类用法:
1. apply 将 valarray 数组的每一个值都用 apply 所接受到的函数进行计算
2. cshift 将 valarray 数组的数据进行循环移动，参数为正者左移为负就右移
3. max 返回 valarray 数组的最大值
4. min 返回 valarray 数组的最小值
5. resize 重新设置 valarray 数组大小，并对其进行初始化
6. shift 将 valarray 数组移动，参数为正者左移，为负者右移，移动后由 0 填充剩余位
7. size 得到数组的大小
8. sum 数组求和
各点到这条曲线的距离之和，即偏差平方和如下：
对等式右边求ai偏导数，得到：
.......
把这些等式表示成矩阵的形式，就可以得到下面的矩阵：
（3）进行化简计算：
，
上面公式（3）可以写为:
，
[cpp] view plain copy#include "stdio.h"
#include "stdlib.h"
#include "math.h"
#include "vector"
using namespace std;
struct point
{
double x;
double y;
};
typedef vector<double> doubleVector;
vector<point> getFile(char *File); //获取文件数据
doubleVector getCoeff(vector<point> sample, int n); //矩阵方程
void main()
{
int i, n;
char *File = "XY.txt";
vector<point> sample;
doubleVector coefficient;
sample = getFile(File);
printf("拟合多项式阶数n=");
scanf_s("%d", &n);
coefficient = getCoeff(sample, n);
printf("\n拟合矩阵的系数为：\n");
for (i = 0; i < coefficient.size(); i++)
printf("a%d = %lf\n", i, coefficient[i]);
}
//矩阵方程
doubleVector getCoeff(vector<point> sample, int n)
{
vector<doubleVector> matFunX; //公式3左矩阵
vector<doubleVector> matFunY; //公式3右矩阵
doubleVector temp;
double sum;
int i, j, k;
//公式3左矩阵
for (i = 0; i <= n; i++)
{
temp.clear();
for (j = 0; j <= n; j++)
{
sum = 0;
for (k = 0; k < sample.size(); k++)
sum += pow(sample[k].x, j + i);
temp.push_back(sum);
}
matFunX.push_back(temp);
}
//printf("matFunX.size=%d\n", matFunX.size());
//printf("matFunX[3][3]=%f\n", matFunX[3][3]);
//公式3右矩阵
for (i = 0; i <= n; i++)
{
temp.clear();
sum = 0;
for (k = 0; k < sample.size(); k++)
sum += sample[k].y*pow(sample[k].x, i);
temp.push_back(sum);
matFunY.push_back(temp);
}
printf("matFunY.size=%d\n", matFunY.size());
//矩阵行列式变换
double num1, num2, ratio;
for (i = 0; i < matFunX.size() - 1; i++)
{
num1 = matFunX[i][i];
for (j = i + 1; j < matFunX.size(); j++)
{
num2 = matFunX[j][i];
ratio = num2 / num1;
for (k = 0; k < matFunX.size(); k++)
matFunX[j][k] = matFunX[j][k] - matFunX[i][k] * ratio;
matFunY[j][0] = matFunY[j][0] - matFunY[i][0] * ratio;
}
}
//计算拟合曲线的系数
doubleVector coeff(matFunX.size(), 0);
for (i = matFunX.size() - 1; i >= 0; i--)
{
if (i == matFunX.size() - 1)
coeff[i] = matFunY[i][0] / matFunX[i][i];
else
{
for (j = i + 1; j < matFunX.size(); j++)
matFunY[i][0] = matFunY[i][0] - coeff[j] * matFunX[i][j];
coeff[i] = matFunY[i][0] / matFunX[i][i];
}
}
return coeff;
}
//获取文件数据
vector<point> getFile(char *File)
{
int i = 1;
vector<point> dst;
FILE *fp=fopen(File, "r");
if (fp == NULL)
{
printf("Open file error!!!\n");
exit(0);
}
point temp;
double num;
while (fscanf(fp, "%lf", &num) != EOF)
{
if (i % 2 == 0)
{
temp.y = num;
dst.push_back(temp);
}
else
temp.x = num;
i++;
}
fclose(fp);
return dst;
}
XY.txt内容：[cpp] view plain copy0 1.0
0.25 1.28
0.5 1.65
0.75 2.12
1 2.72
另外在http://blog.csdn.net/lsh_2013/article/details/46697625里也有相关程序：[cpp] view plain copy#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
//最小二乘拟合相关函数定义
double sum(vector<double> Vnum, int n);
double MutilSum(vector<double> Vx, vector<double> Vy, int n);
double RelatePow(vector<double> Vx, int n, int ex);
double RelateMutiXY(vector<double> Vx, vector<double> Vy, int n, int ex);
void EMatrix(vector<double> Vx, vector<double> Vy, int n, int ex, double coefficient[]);
void CalEquation(int exp, double coefficient[]);
double F(double c[],int l,int m);
double Em[6][4];
//主函数，这里将数据拟合成二次曲线
int main(int argc, char* argv[])
{
double arry1[5]={0,0.25,0,5,0.75};
double arry2[5]={1,1.283,1.649,2.212,2.178};
double coefficient[5];
memset(coefficient,0,sizeof(double)*5);
vector<double> vx,vy;
for (int i=0; i<5; i++)
{
vx.push_back(arry1[i]);
vy.push_back(arry2[i]);
}
EMatrix(vx,vy,5,3,coefficient);
printf("拟合方程为：y = %lf + %lfx + %lfx^2 \n",coefficient[1],coefficient[2],coefficient[3]);
return 0;
}
//累加
double sum(vector<double> Vnum, int n)
{
double dsum=0;
for (int i=0; i<n; i++)
{
dsum+=Vnum[i];
}
return dsum;
}
//乘积和
double MutilSum(vector<double> Vx, vector<double> Vy, int n)
{
double dMultiSum=0;
for (int i=0; i<n; i++)
{
dMultiSum+=Vx[i]*Vy[i];
}
return dMultiSum;
}
//ex次方和
double RelatePow(vector<double> Vx, int n, int ex)
{
double ReSum=0;
for (int i=0; i<n; i++)
{
ReSum+=pow(Vx[i],ex);
}
return ReSum;
}
//x的ex次方与y的乘积的累加
double RelateMutiXY(vector<double> Vx, vector<double> Vy, int n, int ex)
{
double dReMultiSum=0;
for (int i=0; i<n; i++)
{
dReMultiSum+=pow(Vx[i],ex)*Vy[i];
}
return dReMultiSum;
}
//计算方程组的增广矩阵
void EMatrix(vector<double> Vx, vector<double> Vy, int n, int ex, double coefficient[])
{
for (int i=1; i<=ex; i++)
{
for (int j=1; j<=ex; j++)
{
Em[i][j]=RelatePow(Vx,n,i+j-2);
}
Em[i][ex+1]=RelateMutiXY(Vx,Vy,n,i-1);
}
Em[1][1]=n;
CalEquation(ex,coefficient);
}
//求解方程
void CalEquation(int exp, double coefficient[])
{
for(int k=1;k<exp;k++) //消元过程
{
for(int i=k+1;i<exp+1;i++)
{
double p1=0;
if(Em[k][k]!=0)
p1=Em[i][k]/Em[k][k];
for(int j=k;j<exp+2;j++)
Em[i][j]=Em[i][j]-Em[k][j]*p1;
}
}
coefficient[exp]=Em[exp][exp+1]/Em[exp][exp];
for(int l=exp-1;l>=1;l--) //回代求解
coefficient[l]=(Em[l][exp+1]-F(coefficient,l+1,exp))/Em[l][l];
}
//供CalEquation函数调用
double F(double c[],int l,int m)
{
double sum=0;
for(int i=l;i<=m;i++)
sum+=Em[l-1][i]*c[i];
return sum;
}
memset相关介绍：http://baike.baidu.com/link?url=p7JreiRCj9yPs3r3WAfsXgynjvtGrWoQ_exF9tFGK6fsVP7V6tdm-_13QhCZxqPrfRi0wH0EihhRL_-qVvrewq http://c.biancheng.net/cpp/html/157.html src="https://oscdn.geek-share.com/Uploads/Images/Content/202009/10/406a062c292f40b61ef6e476092f6aa4.jpg" border="0" >
[cpp] view plain copy/*
最小二乘法拟合圆，拟合出的圆以圆心坐标和半径的形式表示
*/
typedef complex<int> POINT;
bool FitCircle(const std::vector<POINT> &points, double &er_x, double &er_y, double &radius)
{
cent_x = 0.0f;
cent_y = 0.0f;
radius = 0.0f;
if (points.size() < 3)
{
return false;
}
double sum_x = 0.0f, sum_y = 0.0f;
double sum_x2 = 0.0f, sum_y2 = 0.0f;
double sum_x3 = 0.0f, sum_y3 = 0.0f;
double sum_xy = 0.0f, sum_x1y2 = 0.0f, sum_x2y1 = 0.0f;
int N = points.size();
for (int i = 0; i < N; i++)
{
double x = points[i].real();
double y = points[i].imag();
double x2 = x * x;
double y2 = y * y;
sum_x += x;
sum_y += y;
sum_x2 += x2;
sum_y2 += y2;
sum_x3 += x2 * x;
sum_y3 += y2 * y;
sum_xy += x * y;
sum_x1y2 += x * y2;
sum_x2y1 += x2 * y;
}
double C, D, E, G, H;
double a, b, c;
C = N * sum_x2 - sum_x * sum_x;
D = N * sum_xy - sum_x * sum_y;
E = N * sum_x3 + N * sum_x1y2 - (sum_x2 + sum_y2) * sum_x;
G = N * sum_y2 - sum_y * sum_y;
H = N * sum_x2y1 + N * sum_y3 - (sum_x2 + sum_y2) * sum_y;
a = (H * D - E * G) / (C * G - D * D);
b = (H * C - E * D) / (D * D - G * C);
c = -(a * sum_x + b * sum_y + sum_x2 + sum_y2) / N;
cent_x = a / (-2);
cent_y = b / (-2);
radius = sqrt(a * a + b * b - 4 * c) / 2;
return true;
}
[cpp] view plain copy/*************************************************************************
功能说明： 对平面上的一些列点给出最小二乘的椭圆拟合，利用奇异值分解法
解得最小二乘解作为椭圆参数。
调用形式： cvFitEllipse2f（arrayx,arrayy,box）；
参数说明： arrayx: arrayx
,每个值为x轴一个点
arrayx: arrayy
,每个值为y轴一个点
n : 点的个数
box : box[5],椭圆的五个参数，分别为center.x,center.y,2a,2b,xtheta
esp: 解精度，通常取1e-6,这个是解方程用的说
***************************************************************************/
#include<cstdlib>
#include<float.h>
#include<vector>
using namespace std;
class LSEllipse
{
public:
LSEllipse(void);
~LSEllipse(void);
vector<double> getEllipseparGauss(vector<CPoint> vec_point);
void cvFitEllipse2f( int *arrayx, int *arrayy,int n,float *box );
private:
int SVD(float *a,int m,int n,float b[],float x[],float esp);
int gmiv(float a[],int m,int n,float b[],float x[],float aa[],float eps,float u[],float v[],int ka);
int ginv(float a[],int m,int n,float aa[],float eps,float u[],float v[],int ka);
int muav(float a[],int m,int n,float u[],float v[],float eps,int ka);
};
//LSEllipse.cpp[cpp] view plain copy#include "LSEllipse.h"
#include <cmath>
LSEllipse::LSEllipse(void)
{
}
LSEllipse::~LSEllipse(void)
{
}
//列主元高斯消去法
//A为系数矩阵，x为解向量，若成功，返回true，否则返回false，并将x清空。
bool RGauss(const vector<vector<double> > & A, vector<double> & x)
{
x.clear();
int n = A.size();
int m = A[0].size();
x.resize(n);
//复制系数矩阵，防止修改原矩阵
vector<vector<double> > Atemp(n);
for (int i = 0; i < n; i++)
{
vector<double> temp(m);
for (int j = 0; j < m; j++)
{
temp[j] = A[i][j];
}
Atemp[i] = temp;
temp.clear();
}
for (int k = 0; k < n; k++)
{
//选主元
double max = -1;
int l = -1;
for (int i = k; i < n; i++)
{
if (abs(Atemp[i][k]) > max)
{
max = abs(Atemp[i][k]);
l = i;
}
}
if (l != k)
{
//交换系数矩阵的l行和k行
for (int i = 0; i < m; i++)
{
double temp = Atemp[l][i];
Atemp[l][i] = Atemp[k][i];
Atemp[k][i] = temp;
}
}
//消元
for (int i = k+1; i < n; i++)
{
double l = Atemp[i][k]/Atemp[k][k];
for (int j = k; j < m; j++)
{
Atemp[i][j] = Atemp[i][j] - l*Atemp[k][j];
}
}
}
//回代
x[n-1] = Atemp[n-1][m-1]/Atemp[n-1][m-2];
for (int k = n-2; k >= 0; k--)
{
double s = 0.0;
for (int j = k+1; j < n; j++)
{
s += Atemp[k][j]*x[j];
}
x[k] = (Atemp[k][m-1] - s)/Atemp[k][k];
}
return true;
}
vector<double> LSEllipse::getEllipseparGauss(vector<CPoint> vec_point)
{
vector<double> vec_result;
double x3y1 = 0,x1y3= 0,x2y2= 0,yyy4= 0, xxx3= 0,xxx2= 0,x2y1= 0,yyy3= 0,x1y2= 0 ,yyy2= 0,x1y1= 0,xxx1= 0,yyy1= 0;
int N = vec_point.size();
for (int m_i = 0;m_i < N ;++m_i )
{
double xi = vec_point[m_i].x ;
double yi = vec_point[m_i].y;
x3y1 += xi*xi*xi*yi ;
x1y3 += xi*yi*yi*yi;
x2y2 += xi*xi*yi*yi; ;
yyy4 +=yi*yi*yi*yi;
xxx3 += xi*xi*xi ;
xxx2 += xi*xi ;
x2y1 += xi*xi*yi;
x1y2 += xi*yi*yi;
yyy2 += yi*yi;
x1y1 += xi*yi;
xxx1 += xi;
yyy1 += yi;
yyy3 += yi*yi*yi;
}
double resul[5];
resul[0] = -(x3y1);
resul[1] = -(x2y2);
resul[2] = -(xxx3);
resul[3] = -(x2y1);
resul[4] = -(xxx2);
long double Bb[5],Cc[5],Dd[5],Ee[5],Aa[5];
Bb[0] = x1y3, Cc[0] = x2y1, Dd[0] = x1y2, Ee[0] = x1y1, Aa[0] = x2y2;
Bb[1] = yyy4, Cc[1] = x1y2, Dd[1] = yyy3, Ee[1] = yyy2, Aa[1] = x1y3;
Bb[2] = x1y2, Cc[2] = xxx2, Dd[2] = x1y1, Ee[2] = xxx1, Aa[2] = x2y1;
Bb[3] = yyy3, Cc[3]= x1y1, Dd[3] = yyy2, Ee[3] = yyy1, Aa[3] = x1y2;
Bb[4]= yyy2, Cc[4]= xxx1, Dd[4] = yyy1, Ee[4] = N, Aa[4]= x1y1;
vector<vector<double>>Ma(5);
vector<double>Md(5);
for(int i=0;i<5;i++)
{
Ma[i].push_back(Aa[i]);
Ma[i].push_back(Bb[i]);
Ma[i].push_back(Cc[i]);
Ma[i].push_back(Dd[i]);
Ma[i].push_back(Ee[i]);
Ma[i].push_back(resul[i]);
}
RGauss(Ma,Md);
long double A=Md[0];
long double B=Md[1];
long double C=Md[2];
long double D=Md[3];
long double E=Md[4];
double XC=(2*B*C-A*D)/(A*A-4*B);
double YC=(2*D-A*C)/(A*A-4*B);
long double fenzi=2*(A*C*D-B*C*C-D*D+4*E*B-A*A*E);
long double fenmu=(A*A-4*B)*(B-sqrt(A*A+(1-B)*(1-B))+1);
long double fenmu2=(A*A-4*B)*(B+sqrt(A*A+(1-B)*(1-B))+1);
double XA=sqrt(fabs(fenzi/fenmu));
double XB=sqrt(fabs(fenzi/fenmu2));
double Xtheta=0.5*atan(A/(1-B))*180/3.1415926;
if(B<1)
Xtheta+=90;
vec_result.push_back(XC);
vec_result.push_back(YC);
vec_result.push_back(XA);
vec_result.push_back(XB);
vec_result.push_back(Xtheta);
return vec_result;
}
void LSEllipse::cvFitEllipse2f( int *arrayx, int *arrayy,int n,float *box )
{
float cx=0,cy=0;
double rp[5], t;
float *A1=new float[n*5];
float *A2=new float[2*2];
float *A3=new float[n*3];
float *B1=new float
,*B2=new float[2],*B3=new float
;
const double min_eps = 1e-6;
int i;
for( i = 0; i < n; i++ )
{
cx += arrayx[i]*1.0;
cy += arrayy[i]*1.0;
}
cx /= n;
cy /= n;
for( i = 0; i < n; i++ )
{
int step=i*5;
float px,py;
px = arrayx[i]*1.0;
py = arrayy[i]*1.0;
px -= cx;
py -= cy;
B1[i] = 10000.0;
A1[step] = -px * px;
A1[step + 1] = -py * py;
A1[step + 2] = -px * py;
A1[step + 3] = px;
A1[step + 4] = py;
}
float *x1=new float[5];
//解出Ax^2+By^2+Cxy+Dx+Ey=10000的最小二乘解！
SVD(A1,n,5,B1,x1,min_eps);
A2[0]=2*x1[0],A2[1]=A2[2]=x1[2],A2[3]=2*x1[1];
B2[0]=x1[3],B2[1]=x1[4];
float *x2=new float[2];
//标准化，将一次项消掉，求出center.x和center.y;
SVD(A2,2,2,B2,x2,min_eps);
rp[0]=x2[0],rp[1]=x2[1];
for( i = 0; i < n; i++ )
{
float px,py;
px = arrayx[i]*1.0;
py = arrayy[i]*1.0;
px -= cx;
py -= cy;
B3[i] = 1.0;
int step=i*3;
A3[step] = (px - rp[0]) * (px - rp[0]);
A3[step+1] = (py - rp[1]) * (py - rp[1]);
A3[step+2] = (px - rp[0]) * (py - rp[1]);
}
//求出A(x-center.x)^2+B(y-center.y)^2+C(x-center.x)(y-center.y)的最小二乘解
SVD(A3,n,3,B3,x1,min_eps);
rp[4] = -0.5 * atan2(x1[2], x1[1] - x1[0]);
t = sin(-2.0 * rp[4]);
if( fabs(t) > fabs(x1[2])*min_eps )
t = x1[2]/t;
else
t = x1[1] - x1[0];
rp[2] = fabs(x1[0] + x1[1] - t);
if( rp[2] > min_eps )
rp[2] = sqrt(2.0 / rp[2]);
rp[3] = fabs(x1[0] + x1[1] + t);
if( rp[3] > min_eps )
rp[3] = sqrt(2.0 / rp[3]);
box[0] = (float)rp[0] + cx;
box[1]= (float)rp[1] + cy;
box[2]= (float)(rp[2]*2);
box[3] = (float)(rp[3]*2);
if( box[2] > box[3] )
{
double tmp=box[2];
box[2]=box[3];
box[3]=tmp;
}
box[4] = (float)(90 + rp[4]*180/3.1415926);
if( box[4] < -180 )
box[4] += 360;
if( box[4] > 360 )
box[4] -= 360;
delete []A1;
delete []A2;
delete []A3;
delete []B1;
delete []B2;
delete []B3;
delete []x1;
delete []x2;
}
int LSEllipse::SVD(float *a,int m,int n,float b[],float x[],float esp)
{
float *aa;
float *u;
float *v;
aa=new float[n*m];
u=new float[m*m];
v=new float[n*n];
int ka;
int flag;
if(m>n)
{
ka=m+1;
}else
{
ka=n+1;
}
flag=gmiv(a,m,n,b,x,aa,esp,u,v,ka);
delete []aa;
delete []u;
delete []v;
return(flag);
}
int LSEllipse::gmiv( float a[],int m,int n,float b[],float x[],float aa[],float eps,float u[],float v[],int ka)
{
int i,j;
i=ginv(a,m,n,aa,eps,u,v,ka);
if (i<0) return(-1);
for (i=0; i<=n-1; i++)
{ x[i]=0.0;
for (j=0; j<=m-1; j++)
x[i]=x[i]+aa[i*m+j]*b[j];
}
return(1);
}
int LSEllipse::ginv(float a[],int m,int n,float aa[],float eps,float u[],float v[],int ka)
{
// int muav(float a[],int m,int n,float u[],float v[],float eps,int ka);
int i,j,k,l,t,p,q,f;
i=muav(a,m,n,u,v,eps,ka);
if (i<0) return(-1);
j=n;
if (m<n) j=m;
j=j-1;
k=0;
while ((k<=j)&&(a[k*n+k]!=0.0)) k=k+1;
k=k-1;
for (i=0; i<=n-1; i++)
for (j=0; j<=m-1; j++)
{ t=i*m+j; aa[t]=0.0;
for (l=0; l<=k; l++)
{ f=l*n+i; p=j*m+l; q=l*n+l;
aa[t]=aa[t]+v[f]*u[p]/a[q];
}
}
return(1);
}
int LSEllipse::muav(float a[],int m,int n,float u[],float v[],float eps,int ka)
{ int i,j,k,l,it,ll,kk,ix,iy,mm,nn,iz,m1,ks;
float d,dd,t,sm,sm1,em1,sk,ek,b,c,shh,fg[2],cs[2];
float *s,*e,*w;
//void ppp();
// void sss();
void ppp(float a[],float e[],float s[],float v[],int m,int n);
void sss(float fg[],float cs[]);
s=(float *) malloc(ka*sizeof(float));
e=(float *) malloc(ka*sizeof(float));
w=(float *) malloc(ka*sizeof(float));
it=60; k=n;
if (m-1<n) k=m-1;
l=m;
if (n-2<m) l=n-2;
if (l<0) l=0;
ll=k;
if (l>k) ll=l;
if (ll>=1)
{ for (kk=1; kk<=ll; kk++)
{ if (kk<=k)
{ d=0.0;
for (i=kk; i<=m; i++)
{ ix=(i-1)*n+kk-1; d=d+a[ix]*a[ix];}
s[kk-1]=(float)sqrt(d);
if (s[kk-1]!=0.0)
{ ix=(kk-1)*n+kk-1;
if (a[ix]!=0.0)
{ s[kk-1]=(float)fabs(s[kk-1]);
if (a[ix]<0.0) s[kk-1]=-s[kk-1];
}
for (i=kk; i<=m; i++)
{ iy=(i-1)*n+kk-1;
a[iy]=a[iy]/s[kk-1];
}
a[ix]=1.0f+a[ix];
}
s[kk-1]=-s[kk-1];
}
if (n>=kk+1)
{ for (j=kk+1; j<=n; j++)
{ if ((kk<=k)&&(s[kk-1]!=0.0))
{ d=0.0;
for (i=kk; i<=m; i++)
{ ix=(i-1)*n+kk-1;
iy=(i-1)*n+j-1;
d=d+a[ix]*a[iy];
}
d=-d/a[(kk-1)*n+kk-1];
for (i=kk; i<=m; i++)
{ ix=(i-1)*n+j-1;
iy=(i-1)*n+kk-1;
a[ix]=a[ix]+d*a[iy];
}
}
e[j-1]=a[(kk-1)*n+j-1];
}
}
if (kk<=k)
{ for (i=kk; i<=m; i++)
{ ix=(i-1)*m+kk-1; iy=(i-1)*n+kk-1;
u[ix]=a[iy];
}
}
if (kk<=l)
{ d=0.0;
for (i=kk+1; i<=n; i++)
d=d+e[i-1]*e[i-1];
e[kk-1]=(float)sqrt(d);
if (e[kk-1]!=0.0)
{ if (e[kk]!=0.0)
{ e[kk-1]=(float)fabs(e[kk-1]);
if (e[kk]<0.0) e[kk-1]=-e[kk-1];
}
for (i=kk+1; i<=n; i++)
e[i-1]=e[i-1]/e[kk-1];
e[kk]=1.0f+e[kk];
}
e[kk-1]=-e[kk-1];
if ((kk+1<=m)&&(e[kk-1]!=0.0))
{ for (i=kk+1; i<=m; i++) w[i-1]=0.0;
for (j=kk+1; j<=n; j++)
for (i=kk+1; i<=m; i++)
w[i-1]=w[i-1]+e[j-1]*a[(i-1)*n+j-1];
for (j=kk+1; j<=n; j++)
for (i=kk+1; i<=m; i++)
{ ix=(i-1)*n+j-1;
a[ix]=a[ix]-w[i-1]*e[j-1]/e[kk];
}
}
for (i=kk+1; i<=n; i++)
v[(i-1)*n+kk-1]=e[i-1];
}
}
}
mm=n;
if (m+1<n) mm=m+1;
if (k<n) s[k]=a[k*n+k];
if (m<mm) s[mm-1]=0.0;
if (l+1<mm) e[l]=a[l*n+mm-1];
e[mm-1]=0.0;
nn=m;
if (m>n) nn=n;
if (nn>=k+1)
{ for (j=k+1; j<=nn; j++)
{ for (i=1; i<=m; i++)
u[(i-1)*m+j-1]=0.0;
u[(j-1)*m+j-1]=1.0;
}
}
if (k>=1)
{ for (ll=1; ll<=k; ll++)
{ kk=k-ll+1; iz=(kk-1)*m+kk-1;
if (s[kk-1]!=0.0)
{ if (nn>=kk+1)
for (j=kk+1; j<=nn; j++)
{ d=0.0;
for (i=kk; i<=m; i++)
{ ix=(i-1)*m+kk-1;
iy=(i-1)*m+j-1;
d=d+u[ix]*u[iy]/u[iz];
}
d=-d;
for (i=kk; i<=m; i++)
{ ix=(i-1)*m+j-1;
iy=(i-1)*m+kk-1;
u[ix]=u[ix]+d*u[iy];
}
}
for (i=kk; i<=m; i++)
{ ix=(i-1)*m+kk-1; u[ix]=-u[ix];}
u[iz]=1.0f+u[iz];
if (kk-1>=1)
for (i=1; i<=kk-1; i++)
u[(i-1)*m+kk-1]=0.0;
}
else
{ for (i=1; i<=m; i++)
u[(i-1)*m+kk-1]=0.0;
u[(kk-1)*m+kk-1]=1.0;
}
}
}
for (ll=1; ll<=n; ll++)
{ kk=n-ll+1; iz=kk*n+kk-1;
if ((kk<=l)&&(e[kk-1]!=0.0))
{ for (j=kk+1; j<=n; j++)
{ d=0.0;
for (i=kk+1; i<=n; i++)
{ ix=(i-1)*n+kk-1; iy=(i-1)*n+j-1;
d=d+v[ix]*v[iy]/v[iz];
}
d=-d;
for (i=kk+1; i<=n; i++)
{ ix=(i-1)*n+j-1; iy=(i-1)*n+kk-1;
v[ix]=v[ix]+d*v[iy];
}
}
}
for (i=1; i<=n; i++)
v[(i-1)*n+kk-1]=0.0;
v[iz-n]=1.0;
}
for (i=1; i<=m; i++)
for (j=1; j<=n; j++)
a[(i-1)*n+j-1]=0.0;
m1=mm; it=60;
while (1==1)
{ if (mm==0)
{ ppp(a,e,s,v,m,n);
free(s); free(e); free(w); return(1);
}
if (it==0)
{ ppp(a,e,s,v,m,n);
free(s); free(e); free(w); return(-1);
}
kk=mm-1;
while ((kk!=0)&&(fabs(e[kk-1])!=0.0))
{ d=(float)(fabs(s[kk-1])+fabs(s[kk]));
dd=(float)fabs(e[kk-1]);
if (dd>eps*d) kk=kk-1;
else e[kk-1]=0.0;
}
if (kk==mm-1)
{ kk=kk+1;
if (s[kk-1]<0.0)
{ s[kk-1]=-s[kk-1];
for (i=1; i<=n; i++)
{ ix=(i-1)*n+kk-1; v[ix]=-v[ix];}
}
while ((kk!=m1)&&(s[kk-1]<s[kk]))
{ d=s[kk-1]; s[kk-1]=s[kk]; s[kk]=d;
if (kk<n)
for (i=1; i<=n; i++)
{ ix=(i-1)*n+kk-1; iy=(i-1)*n+kk;
d=v[ix]; v[ix]=v[iy]; v[iy]=d;
}
if (kk<m)
for (i=1; i<=m; i++)
{ ix=(i-1)*m+kk-1; iy=(i-1)*m+kk;
d=u[ix]; u[ix]=u[iy]; u[iy]=d;
}
kk=kk+1;
}
it=60;
mm=mm-1;
}
else
{ ks=mm;
while ((ks>kk)&&(fabs(s[ks-1])!=0.0))
{ d=0.0;
if (ks!=mm) d=d+(float)fabs(e[ks-1]);
if (ks!=kk+1) d=d+(float)fabs(e[ks-2]);
dd=(float)fabs(s[ks-1]);
if (dd>eps*d) ks=ks-1;
else s[ks-1]=0.0;
}
if (ks==kk)
{ kk=kk+1;
d=(float)fabs(s[mm-1]);
t=(float)fabs(s[mm-2]);
if (t>d) d=t;
t=(float)fabs(e[mm-2]);
if (t>d) d=t;
t=(float)fabs(s[kk-1]);
if (t>d) d=t;
t=(float)fabs(e[kk-1]);
if (t>d) d=t;
sm=s[mm-1]/d; sm1=s[mm-2]/d;
em1=e[mm-2]/d;
sk=s[kk-1]/d; ek=e[kk-1]/d;
b=((sm1+sm)*(sm1-sm)+em1*em1)/2.0f;
c=sm*em1; c=c*c; shh=0.0;
if ((b!=0.0)||(c!=0.0))
{ shh=(float)sqrt(b*b+c);
if (b<0.0) shh=-shh;
shh=c/(b+shh);
}
fg[0]=(sk+sm)*(sk-sm)-shh;
fg[1]=sk*ek;
for (i=kk; i<=mm-1; i++)
{ sss(fg,cs);
if (i!=kk) e[i-2]=fg[0];
fg[0]=cs[0]*s[i-1]+cs[1]*e[i-1];
e[i-1]=cs[0]*e[i-1]-cs[1]*s[i-1];
fg[1]=cs[1]*s[i];
s[i]=cs[0]*s[i];
if ((cs[0]!=1.0)||(cs[1]!=0.0))
for (j=1; j<=n; j++)
{ ix=(j-1)*n+i-1;
iy=(j-1)*n+i;
d=cs[0]*v[ix]+cs[1]*v[iy];
v[iy]=-cs[1]*v[ix]+cs[0]*v[iy];
v[ix]=d;
}
sss(fg,cs);
s[i-1]=fg[0];
fg[0]=cs[0]*e[i-1]+cs[1]*s[i];
s[i]=-cs[1]*e[i-1]+cs[0]*s[i];
fg[1]=cs[1]*e[i];
e[i]=cs[0]*e[i];
if (i<m)
if ((cs[0]!=1.0)||(cs[1]!=0.0))
for (j=1; j<=m; j++)
{ ix=(j-1)*m+i-1;
iy=(j-1)*m+i;
d=cs[0]*u[ix]+cs[1]*u[iy];
u[iy]=-cs[1]*u[ix]+cs[0]*u[iy];
u[ix]=d;
}
}
e[mm-2]=fg[0];
it=it-1;
}
else
{ if (ks==mm)
{ kk=kk+1;
fg[1]=e[mm-2]; e[mm-2]=0.0;
for (ll=kk; ll<=mm-1; ll++)
{ i=mm+kk-ll-1;
fg[0]=s[i-1];
sss(fg,cs);
s[i-1]=fg[0];
if (i!=kk)
{ fg[1]=-cs[1]*e[i-2];
e[i-2]=cs[0]*e[i-2];
}
if ((cs[0]!=1.0)||(cs[1]!=0.0))
for (j=1; j<=n; j++)
{ ix=(j-1)*n+i-1;
iy=(j-1)*n+mm-1;
d=cs[0]*v[ix]+cs[1]*v[iy];
v[iy]=-cs[1]*v[ix]+cs[0]*v[iy];
v[ix]=d;
}
}
}
else
{ kk=ks+1;
fg[1]=e[kk-2];
e[kk-2]=0.0;
for (i=kk; i<=mm; i++)
{ fg[0]=s[i-1];
sss(fg,cs);
s[i-1]=fg[0];
fg[1]=-cs[1]*e[i-1];
e[i-1]=cs[0]*e[i-1];
if ((cs[0]!=1.0)||(cs[1]!=0.0))
for (j=1; j<=m; j++)
{ ix=(j-1)*m+i-1;
iy=(j-1)*m+kk-2;
d=cs[0]*u[ix]+cs[1]*u[iy];
u[iy]=-cs[1]*u[ix]+cs[0]*u[iy];
u[ix]=d;
}
}
}
}
}
}
free(s);free(e);free(w);
return(1);
}
void ppp(float a[],float e[],float s[],float v[],int m,int n)
{ int i,j,p,q;
float d;
if (m>=n) i=n;
else i=m;
for (j=1; j<=i-1; j++)
{ a[(j-1)*n+j-1]=s[j-1];
a[(j-1)*n+j]=e[j-1];
}
a[(i-1)*n+i-1]=s[i-1];
if (m<n) a[(i-1)*n+i]=e[i-1];
for (i=1; i<=n-1; i++)
for (j=i+1; j<=n; j++)
{ p=(i-1)*n+j-1; q=(j-1)*n+i-1;
d=v[p]; v[p]=v[q]; v[q]=d;
}
return;
}
void sss(float fg[],float cs[])
{ float r,d;
if ((fabs(fg[0])+fabs(fg[1]))==0.0)
{ cs[0]=1.0; cs[1]=0.0; d=0.0;}
else
{ d=(float)sqrt(fg[0]*fg[0]+fg[1]*fg[1]);
if (fabs(fg[0])>fabs(fg[1]))
{ d=(float)fabs(d);
if (fg[0]<0.0) d=-d;
}
if (fabs(fg[1])>=fabs(fg[0]))
{ d=(float)fabs(d);
if (fg[1]<0.0) d=-d;
}
cs[0]=fg[0]/d; cs[1]=fg[1]/d;
}
r=1.0;
if (fabs(fg[0])>fabs(fg[1])) r=cs[1];
else
if (cs[0]!=0.0) r=1.0f/cs[0];
fg[0]=d; fg[1]=r;
return;
}
参考:线性，非线性多项式：http://blog.csdn.net/qll125596718/article/details/8248249 http://blog.csdn.net/poxiaozhuimeng/article/details/41117947 http://blog.csdn.net/ouyangying123/article/details/53996403http://blog.csdn.net/jairuschan/article/details/7517773/http://blog.csdn.net/zang141588761/article/details/50523036 http://www.cnblogs.com/gnuhpc/archive/2012/12/09/2809997.html http://download.csdn.net/download/biaobiao11/9755119
圆拟合：http://blog.sina.com.cn/s/blog_b27f71160101gxun.html http://www.cnblogs.com/dotLive/archive/2007/04/06/524633.htmlhttp://blog.csdn.net/andylao62/article/details/24522365 http://blog.csdn.net/liyuanbhu/article/details/50889951 http://blog.csdn.net/liyuanbhu/article/details/50890587
椭圆拟合：http://doc.okbase.net/u013708970/archive/121532.html
转：http://blog.csdn.net/piaoxuezhong/article/details/54973750
--------------------一元线性函数--------------------
形式1：借用案例（http://blog.csdn.net/qll125596718/article/details/8248249）首先以一元线性方程参数估计为例，样本回归模型：残差平方和：
则通过Q最小确定这条直线，即确定
，以
为变量，把它们看作是Q的函数，就变成了一个求极值的问题，可以通过求导数得到。求Q对两个待估参数的偏导数：
解得：
实例（c++）：[cpp] view plain copy#include <iostream>
#include <algorithm>
#include <valarray>
#include <vector>
using namespace std;
int main()
{
double x[] = { 1, 2, 3, 4, 5, 6 };
double y[] = { 3, 5.5, 6.8, 8.8, 11, 12};
valarray<double> data_x(x, 6);
valarray<double> data_y(y, 6);
float A = 0.0;
float B = 0.0;
float C = 0.0;
float D = 0.0;
A = (data_x*data_x).sum();
B = data_x.sum();
C = (data_x*data_y).sum();
D = data_y.sum();
float tmp = A*data_x.size() - B*B;
float k, b;
k = (C*data_x.size() - B*D) / tmp;
b = (A*D - C*B) / tmp;
cout << "y=" << k << "x+" << b << endl;
return 0;
}
运行结果：
注：valarray类似vector，也是一个模板类，主要用来对一系列元素进行高速的数字计算，其与vector的主要区别在于以下两点：
1、valarray定义了一组在两个相同长度和相同类型的valarray类对象之间的数字计算；
2、通过重载operater[]，可以返回valarray的相关信息（valarray其中某个元素的引用、特定下标的值或者其某个子集）。valarray类构造函数:
valarray( );
explicit valarray(size_t _Count);
valarray( const Type& _Val, size_t _Count);
valarray( const Type* _Ptr, size_t _Count);
valarray( const slice_array<Type>& _SliceArray);
valarray( const gslice_array<Type>& _GsliceArray);
valarray( const mask_array<Type>& _MaskArray);
valarray( const indirect_array<Type>& _IndArray);
valarray 类用法:
1. apply 将 valarray 数组的每一个值都用 apply 所接受到的函数进行计算
2. cshift 将 valarray 数组的数据进行循环移动，参数为正者左移为负就右移
3. max 返回 valarray 数组的最大值
4. min 返回 valarray 数组的最小值
5. resize 重新设置 valarray 数组大小，并对其进行初始化
6. shift 将 valarray 数组移动，参数为正者左移，为负者右移，移动后由 0 填充剩余位
7. size 得到数组的大小
8. sum 数组求和
--------------------N元线性函数--------------------
一元线性方程可以看做多元函数的特例，现在用矩阵形式表述多元函数情况下，最小二乘的一般形式： 设拟合多项式为：各点到这条曲线的距离之和，即偏差平方和如下：
对等式右边求ai偏导数，得到：
.......
把这些等式表示成矩阵的形式，就可以得到下面的矩阵：
（3）进行化简计算：
，
上面公式（3）可以写为:
，
[cpp] view plain copy#include "stdio.h"
#include "stdlib.h"
#include "math.h"
#include "vector"
using namespace std;
struct point
{
double x;
double y;
};
typedef vector<double> doubleVector;
vector<point> getFile(char *File); //获取文件数据
doubleVector getCoeff(vector<point> sample, int n); //矩阵方程
void main()
{
int i, n;
char *File = "XY.txt";
vector<point> sample;
doubleVector coefficient;
sample = getFile(File);
printf("拟合多项式阶数n=");
scanf_s("%d", &n);
coefficient = getCoeff(sample, n);
printf("\n拟合矩阵的系数为：\n");
for (i = 0; i < coefficient.size(); i++)
printf("a%d = %lf\n", i, coefficient[i]);
}
//矩阵方程
doubleVector getCoeff(vector<point> sample, int n)
{
vector<doubleVector> matFunX; //公式3左矩阵
vector<doubleVector> matFunY; //公式3右矩阵
doubleVector temp;
double sum;
int i, j, k;
//公式3左矩阵
for (i = 0; i <= n; i++)
{
temp.clear();
for (j = 0; j <= n; j++)
{
sum = 0;
for (k = 0; k < sample.size(); k++)
sum += pow(sample[k].x, j + i);
temp.push_back(sum);
}
matFunX.push_back(temp);
}
//printf("matFunX.size=%d\n", matFunX.size());
//printf("matFunX[3][3]=%f\n", matFunX[3][3]);
//公式3右矩阵
for (i = 0; i <= n; i++)
{
temp.clear();
sum = 0;
for (k = 0; k < sample.size(); k++)
sum += sample[k].y*pow(sample[k].x, i);
temp.push_back(sum);
matFunY.push_back(temp);
}
printf("matFunY.size=%d\n", matFunY.size());
//矩阵行列式变换
double num1, num2, ratio;
for (i = 0; i < matFunX.size() - 1; i++)
{
num1 = matFunX[i][i];
for (j = i + 1; j < matFunX.size(); j++)
{
num2 = matFunX[j][i];
ratio = num2 / num1;
for (k = 0; k < matFunX.size(); k++)
matFunX[j][k] = matFunX[j][k] - matFunX[i][k] * ratio;
matFunY[j][0] = matFunY[j][0] - matFunY[i][0] * ratio;
}
}
//计算拟合曲线的系数
doubleVector coeff(matFunX.size(), 0);
for (i = matFunX.size() - 1; i >= 0; i--)
{
if (i == matFunX.size() - 1)
coeff[i] = matFunY[i][0] / matFunX[i][i];
else
{
for (j = i + 1; j < matFunX.size(); j++)
matFunY[i][0] = matFunY[i][0] - coeff[j] * matFunX[i][j];
coeff[i] = matFunY[i][0] / matFunX[i][i];
}
}
return coeff;
}
//获取文件数据
vector<point> getFile(char *File)
{
int i = 1;
vector<point> dst;
FILE *fp=fopen(File, "r");
if (fp == NULL)
{
printf("Open file error!!!\n");
exit(0);
}
point temp;
double num;
while (fscanf(fp, "%lf", &num) != EOF)
{
if (i % 2 == 0)
{
temp.y = num;
dst.push_back(temp);
}
else
temp.x = num;
i++;
}
fclose(fp);
return dst;
}
XY.txt内容：[cpp] view plain copy0 1.0
0.25 1.28
0.5 1.65
0.75 2.12
1 2.72
另外在http://blog.csdn.net/lsh_2013/article/details/46697625里也有相关程序：[cpp] view plain copy#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
//最小二乘拟合相关函数定义
double sum(vector<double> Vnum, int n);
double MutilSum(vector<double> Vx, vector<double> Vy, int n);
double RelatePow(vector<double> Vx, int n, int ex);
double RelateMutiXY(vector<double> Vx, vector<double> Vy, int n, int ex);
void EMatrix(vector<double> Vx, vector<double> Vy, int n, int ex, double coefficient[]);
void CalEquation(int exp, double coefficient[]);
double F(double c[],int l,int m);
double Em[6][4];
//主函数，这里将数据拟合成二次曲线
int main(int argc, char* argv[])
{
double arry1[5]={0,0.25,0,5,0.75};
double arry2[5]={1,1.283,1.649,2.212,2.178};
double coefficient[5];
memset(coefficient,0,sizeof(double)*5);
vector<double> vx,vy;
for (int i=0; i<5; i++)
{
vx.push_back(arry1[i]);
vy.push_back(arry2[i]);
}
EMatrix(vx,vy,5,3,coefficient);
printf("拟合方程为：y = %lf + %lfx + %lfx^2 \n",coefficient[1],coefficient[2],coefficient[3]);
return 0;
}
//累加
double sum(vector<double> Vnum, int n)
{
double dsum=0;
for (int i=0; i<n; i++)
{
dsum+=Vnum[i];
}
return dsum;
}
//乘积和
double MutilSum(vector<double> Vx, vector<double> Vy, int n)
{
double dMultiSum=0;
for (int i=0; i<n; i++)
{
dMultiSum+=Vx[i]*Vy[i];
}
return dMultiSum;
}
//ex次方和
double RelatePow(vector<double> Vx, int n, int ex)
{
double ReSum=0;
for (int i=0; i<n; i++)
{
ReSum+=pow(Vx[i],ex);
}
return ReSum;
}
//x的ex次方与y的乘积的累加
double RelateMutiXY(vector<double> Vx, vector<double> Vy, int n, int ex)
{
double dReMultiSum=0;
for (int i=0; i<n; i++)
{
dReMultiSum+=pow(Vx[i],ex)*Vy[i];
}
return dReMultiSum;
}
//计算方程组的增广矩阵
void EMatrix(vector<double> Vx, vector<double> Vy, int n, int ex, double coefficient[])
{
for (int i=1; i<=ex; i++)
{
for (int j=1; j<=ex; j++)
{
Em[i][j]=RelatePow(Vx,n,i+j-2);
}
Em[i][ex+1]=RelateMutiXY(Vx,Vy,n,i-1);
}
Em[1][1]=n;
CalEquation(ex,coefficient);
}
//求解方程
void CalEquation(int exp, double coefficient[])
{
for(int k=1;k<exp;k++) //消元过程
{
for(int i=k+1;i<exp+1;i++)
{
double p1=0;
if(Em[k][k]!=0)
p1=Em[i][k]/Em[k][k];
for(int j=k;j<exp+2;j++)
Em[i][j]=Em[i][j]-Em[k][j]*p1;
}
}
coefficient[exp]=Em[exp][exp+1]/Em[exp][exp];
for(int l=exp-1;l>=1;l--) //回代求解
coefficient[l]=(Em[l][exp+1]-F(coefficient,l+1,exp))/Em[l][l];
}
//供CalEquation函数调用
double F(double c[],int l,int m)
{
double sum=0;
for(int i=l;i<=m;i++)
sum+=Em[l-1][i]*c[i];
return sum;
}
memset相关介绍：http://baike.baidu.com/link?url=p7JreiRCj9yPs3r3WAfsXgynjvtGrWoQ_exF9tFGK6fsVP7V6tdm-_13QhCZxqPrfRi0wH0EihhRL_-qVvrewq http://c.biancheng.net/cpp/html/157.html
--------------------拟合圆的方程--------------------
[cpp] view plain copy/*
最小二乘法拟合圆，拟合出的圆以圆心坐标和半径的形式表示
*/
typedef complex<int> POINT;
bool FitCircle(const std::vector<POINT> &points, double &er_x, double &er_y, double &radius)
{
cent_x = 0.0f;
cent_y = 0.0f;
radius = 0.0f;
if (points.size() < 3)
{
return false;
}
double sum_x = 0.0f, sum_y = 0.0f;
double sum_x2 = 0.0f, sum_y2 = 0.0f;
double sum_x3 = 0.0f, sum_y3 = 0.0f;
double sum_xy = 0.0f, sum_x1y2 = 0.0f, sum_x2y1 = 0.0f;
int N = points.size();
for (int i = 0; i < N; i++)
{
double x = points[i].real();
double y = points[i].imag();
double x2 = x * x;
double y2 = y * y;
sum_x += x;
sum_y += y;
sum_x2 += x2;
sum_y2 += y2;
sum_x3 += x2 * x;
sum_y3 += y2 * y;
sum_xy += x * y;
sum_x1y2 += x * y2;
sum_x2y1 += x2 * y;
}
double C, D, E, G, H;
double a, b, c;
C = N * sum_x2 - sum_x * sum_x;
D = N * sum_xy - sum_x * sum_y;
E = N * sum_x3 + N * sum_x1y2 - (sum_x2 + sum_y2) * sum_x;
G = N * sum_y2 - sum_y * sum_y;
H = N * sum_x2y1 + N * sum_y3 - (sum_x2 + sum_y2) * sum_y;
a = (H * D - E * G) / (C * G - D * D);
b = (H * C - E * D) / (D * D - G * C);
c = -(a * sum_x + b * sum_y + sum_x2 + sum_y2) / N;
cent_x = a / (-2);
cent_y = b / (-2);
radius = sqrt(a * a + b * b - 4 * c) / 2;
return true;
}
--------------------拟合椭圆方程--------------------
//LSEllipse.h[cpp] view plain copy/*************************************************************************
功能说明： 对平面上的一些列点给出最小二乘的椭圆拟合，利用奇异值分解法
解得最小二乘解作为椭圆参数。
调用形式： cvFitEllipse2f（arrayx,arrayy,box）；
参数说明： arrayx: arrayx
,每个值为x轴一个点
arrayx: arrayy
,每个值为y轴一个点
n : 点的个数
box : box[5],椭圆的五个参数，分别为center.x,center.y,2a,2b,xtheta
esp: 解精度，通常取1e-6,这个是解方程用的说
***************************************************************************/
#include<cstdlib>
#include<float.h>
#include<vector>
using namespace std;
class LSEllipse
{
public:
LSEllipse(void);
~LSEllipse(void);
vector<double> getEllipseparGauss(vector<CPoint> vec_point);
void cvFitEllipse2f( int *arrayx, int *arrayy,int n,float *box );
private:
int SVD(float *a,int m,int n,float b[],float x[],float esp);
int gmiv(float a[],int m,int n,float b[],float x[],float aa[],float eps,float u[],float v[],int ka);
int ginv(float a[],int m,int n,float aa[],float eps,float u[],float v[],int ka);
int muav(float a[],int m,int n,float u[],float v[],float eps,int ka);
};
//LSEllipse.cpp[cpp] view plain copy#include "LSEllipse.h"
#include <cmath>
LSEllipse::LSEllipse(void)
{
}
LSEllipse::~LSEllipse(void)
{
}
//列主元高斯消去法
//A为系数矩阵，x为解向量，若成功，返回true，否则返回false，并将x清空。
bool RGauss(const vector<vector<double> > & A, vector<double> & x)
{
x.clear();
int n = A.size();
int m = A[0].size();
x.resize(n);
//复制系数矩阵，防止修改原矩阵
vector<vector<double> > Atemp(n);
for (int i = 0; i < n; i++)
{
vector<double> temp(m);
for (int j = 0; j < m; j++)
{
temp[j] = A[i][j];
}
Atemp[i] = temp;
temp.clear();
}
for (int k = 0; k < n; k++)
{
//选主元
double max = -1;
int l = -1;
for (int i = k; i < n; i++)
{
if (abs(Atemp[i][k]) > max)
{
max = abs(Atemp[i][k]);
l = i;
}
}
if (l != k)
{
//交换系数矩阵的l行和k行
for (int i = 0; i < m; i++)
{
double temp = Atemp[l][i];
Atemp[l][i] = Atemp[k][i];
Atemp[k][i] = temp;
}
}
//消元
for (int i = k+1; i < n; i++)
{
double l = Atemp[i][k]/Atemp[k][k];
for (int j = k; j < m; j++)
{
Atemp[i][j] = Atemp[i][j] - l*Atemp[k][j];
}
}
}
//回代
x[n-1] = Atemp[n-1][m-1]/Atemp[n-1][m-2];
for (int k = n-2; k >= 0; k--)
{
double s = 0.0;
for (int j = k+1; j < n; j++)
{
s += Atemp[k][j]*x[j];
}
x[k] = (Atemp[k][m-1] - s)/Atemp[k][k];
}
return true;
}
vector<double> LSEllipse::getEllipseparGauss(vector<CPoint> vec_point)
{
vector<double> vec_result;
double x3y1 = 0,x1y3= 0,x2y2= 0,yyy4= 0, xxx3= 0,xxx2= 0,x2y1= 0,yyy3= 0,x1y2= 0 ,yyy2= 0,x1y1= 0,xxx1= 0,yyy1= 0;
int N = vec_point.size();
for (int m_i = 0;m_i < N ;++m_i )
{
double xi = vec_point[m_i].x ;
double yi = vec_point[m_i].y;
x3y1 += xi*xi*xi*yi ;
x1y3 += xi*yi*yi*yi;
x2y2 += xi*xi*yi*yi; ;
yyy4 +=yi*yi*yi*yi;
xxx3 += xi*xi*xi ;
xxx2 += xi*xi ;
x2y1 += xi*xi*yi;
x1y2 += xi*yi*yi;
yyy2 += yi*yi;
x1y1 += xi*yi;
xxx1 += xi;
yyy1 += yi;
yyy3 += yi*yi*yi;
}
double resul[5];
resul[0] = -(x3y1);
resul[1] = -(x2y2);
resul[2] = -(xxx3);
resul[3] = -(x2y1);
resul[4] = -(xxx2);
long double Bb[5],Cc[5],Dd[5],Ee[5],Aa[5];
Bb[0] = x1y3, Cc[0] = x2y1, Dd[0] = x1y2, Ee[0] = x1y1, Aa[0] = x2y2;
Bb[1] = yyy4, Cc[1] = x1y2, Dd[1] = yyy3, Ee[1] = yyy2, Aa[1] = x1y3;
Bb[2] = x1y2, Cc[2] = xxx2, Dd[2] = x1y1, Ee[2] = xxx1, Aa[2] = x2y1;
Bb[3] = yyy3, Cc[3]= x1y1, Dd[3] = yyy2, Ee[3] = yyy1, Aa[3] = x1y2;
Bb[4]= yyy2, Cc[4]= xxx1, Dd[4] = yyy1, Ee[4] = N, Aa[4]= x1y1;
vector<vector<double>>Ma(5);
vector<double>Md(5);
for(int i=0;i<5;i++)
{
Ma[i].push_back(Aa[i]);
Ma[i].push_back(Bb[i]);
Ma[i].push_back(Cc[i]);
Ma[i].push_back(Dd[i]);
Ma[i].push_back(Ee[i]);
Ma[i].push_back(resul[i]);
}
RGauss(Ma,Md);
long double A=Md[0];
long double B=Md[1];
long double C=Md[2];
long double D=Md[3];
long double E=Md[4];
double XC=(2*B*C-A*D)/(A*A-4*B);
double YC=(2*D-A*C)/(A*A-4*B);
long double fenzi=2*(A*C*D-B*C*C-D*D+4*E*B-A*A*E);
long double fenmu=(A*A-4*B)*(B-sqrt(A*A+(1-B)*(1-B))+1);
long double fenmu2=(A*A-4*B)*(B+sqrt(A*A+(1-B)*(1-B))+1);
double XA=sqrt(fabs(fenzi/fenmu));
double XB=sqrt(fabs(fenzi/fenmu2));
double Xtheta=0.5*atan(A/(1-B))*180/3.1415926;
if(B<1)
Xtheta+=90;
vec_result.push_back(XC);
vec_result.push_back(YC);
vec_result.push_back(XA);
vec_result.push_back(XB);
vec_result.push_back(Xtheta);
return vec_result;
}
void LSEllipse::cvFitEllipse2f( int *arrayx, int *arrayy,int n,float *box )
{
float cx=0,cy=0;
double rp[5], t;
float *A1=new float[n*5];
float *A2=new float[2*2];
float *A3=new float[n*3];
float *B1=new float
,*B2=new float[2],*B3=new float
;
const double min_eps = 1e-6;
int i;
for( i = 0; i < n; i++ )
{
cx += arrayx[i]*1.0;
cy += arrayy[i]*1.0;
}
cx /= n;
cy /= n;
for( i = 0; i < n; i++ )
{
int step=i*5;
float px,py;
px = arrayx[i]*1.0;
py = arrayy[i]*1.0;
px -= cx;
py -= cy;
B1[i] = 10000.0;
A1[step] = -px * px;
A1[step + 1] = -py * py;
A1[step + 2] = -px * py;
A1[step + 3] = px;
A1[step + 4] = py;
}
float *x1=new float[5];
//解出Ax^2+By^2+Cxy+Dx+Ey=10000的最小二乘解！
SVD(A1,n,5,B1,x1,min_eps);
A2[0]=2*x1[0],A2[1]=A2[2]=x1[2],A2[3]=2*x1[1];
B2[0]=x1[3],B2[1]=x1[4];
float *x2=new float[2];
//标准化，将一次项消掉，求出center.x和center.y;
SVD(A2,2,2,B2,x2,min_eps);
rp[0]=x2[0],rp[1]=x2[1];
for( i = 0; i < n; i++ )
{
float px,py;
px = arrayx[i]*1.0;
py = arrayy[i]*1.0;
px -= cx;
py -= cy;
B3[i] = 1.0;
int step=i*3;
A3[step] = (px - rp[0]) * (px - rp[0]);
A3[step+1] = (py - rp[1]) * (py - rp[1]);
A3[step+2] = (px - rp[0]) * (py - rp[1]);
}
//求出A(x-center.x)^2+B(y-center.y)^2+C(x-center.x)(y-center.y)的最小二乘解
SVD(A3,n,3,B3,x1,min_eps);
rp[4] = -0.5 * atan2(x1[2], x1[1] - x1[0]);
t = sin(-2.0 * rp[4]);
if( fabs(t) > fabs(x1[2])*min_eps )
t = x1[2]/t;
else
t = x1[1] - x1[0];
rp[2] = fabs(x1[0] + x1[1] - t);
if( rp[2] > min_eps )
rp[2] = sqrt(2.0 / rp[2]);
rp[3] = fabs(x1[0] + x1[1] + t);
if( rp[3] > min_eps )
rp[3] = sqrt(2.0 / rp[3]);
box[0] = (float)rp[0] + cx;
box[1]= (float)rp[1] + cy;
box[2]= (float)(rp[2]*2);
box[3] = (float)(rp[3]*2);
if( box[2] > box[3] )
{
double tmp=box[2];
box[2]=box[3];
box[3]=tmp;
}
box[4] = (float)(90 + rp[4]*180/3.1415926);
if( box[4] < -180 )
box[4] += 360;
if( box[4] > 360 )
box[4] -= 360;
delete []A1;
delete []A2;
delete []A3;
delete []B1;
delete []B2;
delete []B3;
delete []x1;
delete []x2;
}
int LSEllipse::SVD(float *a,int m,int n,float b[],float x[],float esp)
{
float *aa;
float *u;
float *v;
aa=new float[n*m];
u=new float[m*m];
v=new float[n*n];
int ka;
int flag;
if(m>n)
{
ka=m+1;
}else
{
ka=n+1;
}
flag=gmiv(a,m,n,b,x,aa,esp,u,v,ka);
delete []aa;
delete []u;
delete []v;
return(flag);
}
int LSEllipse::gmiv( float a[],int m,int n,float b[],float x[],float aa[],float eps,float u[],float v[],int ka)
{
int i,j;
i=ginv(a,m,n,aa,eps,u,v,ka);
if (i<0) return(-1);
for (i=0; i<=n-1; i++)
{ x[i]=0.0;
for (j=0; j<=m-1; j++)
x[i]=x[i]+aa[i*m+j]*b[j];
}
return(1);
}
int LSEllipse::ginv(float a[],int m,int n,float aa[],float eps,float u[],float v[],int ka)
{
// int muav(float a[],int m,int n,float u[],float v[],float eps,int ka);
int i,j,k,l,t,p,q,f;
i=muav(a,m,n,u,v,eps,ka);
if (i<0) return(-1);
j=n;
if (m<n) j=m;
j=j-1;
k=0;
while ((k<=j)&&(a[k*n+k]!=0.0)) k=k+1;
k=k-1;
for (i=0; i<=n-1; i++)
for (j=0; j<=m-1; j++)
{ t=i*m+j; aa[t]=0.0;
for (l=0; l<=k; l++)
{ f=l*n+i; p=j*m+l; q=l*n+l;
aa[t]=aa[t]+v[f]*u[p]/a[q];
}
}
return(1);
}
int LSEllipse::muav(float a[],int m,int n,float u[],float v[],float eps,int ka)
{ int i,j,k,l,it,ll,kk,ix,iy,mm,nn,iz,m1,ks;
float d,dd,t,sm,sm1,em1,sk,ek,b,c,shh,fg[2],cs[2];
float *s,*e,*w;
//void ppp();
// void sss();
void ppp(float a[],float e[],float s[],float v[],int m,int n);
void sss(float fg[],float cs[]);
s=(float *) malloc(ka*sizeof(float));
e=(float *) malloc(ka*sizeof(float));
w=(float *) malloc(ka*sizeof(float));
it=60; k=n;
if (m-1<n) k=m-1;
l=m;
if (n-2<m) l=n-2;
if (l<0) l=0;
ll=k;
if (l>k) ll=l;
if (ll>=1)
{ for (kk=1; kk<=ll; kk++)
{ if (kk<=k)
{ d=0.0;
for (i=kk; i<=m; i++)
{ ix=(i-1)*n+kk-1; d=d+a[ix]*a[ix];}
s[kk-1]=(float)sqrt(d);
if (s[kk-1]!=0.0)
{ ix=(kk-1)*n+kk-1;
if (a[ix]!=0.0)
{ s[kk-1]=(float)fabs(s[kk-1]);
if (a[ix]<0.0) s[kk-1]=-s[kk-1];
}
for (i=kk; i<=m; i++)
{ iy=(i-1)*n+kk-1;
a[iy]=a[iy]/s[kk-1];
}
a[ix]=1.0f+a[ix];
}
s[kk-1]=-s[kk-1];
}
if (n>=kk+1)
{ for (j=kk+1; j<=n; j++)
{ if ((kk<=k)&&(s[kk-1]!=0.0))
{ d=0.0;
for (i=kk; i<=m; i++)
{ ix=(i-1)*n+kk-1;
iy=(i-1)*n+j-1;
d=d+a[ix]*a[iy];
}
d=-d/a[(kk-1)*n+kk-1];
for (i=kk; i<=m; i++)
{ ix=(i-1)*n+j-1;
iy=(i-1)*n+kk-1;
a[ix]=a[ix]+d*a[iy];
}
}
e[j-1]=a[(kk-1)*n+j-1];
}
}
if (kk<=k)
{ for (i=kk; i<=m; i++)
{ ix=(i-1)*m+kk-1; iy=(i-1)*n+kk-1;
u[ix]=a[iy];
}
}
if (kk<=l)
{ d=0.0;
for (i=kk+1; i<=n; i++)
d=d+e[i-1]*e[i-1];
e[kk-1]=(float)sqrt(d);
if (e[kk-1]!=0.0)
{ if (e[kk]!=0.0)
{ e[kk-1]=(float)fabs(e[kk-1]);
if (e[kk]<0.0) e[kk-1]=-e[kk-1];
}
for (i=kk+1; i<=n; i++)
e[i-1]=e[i-1]/e[kk-1];
e[kk]=1.0f+e[kk];
}
e[kk-1]=-e[kk-1];
if ((kk+1<=m)&&(e[kk-1]!=0.0))
{ for (i=kk+1; i<=m; i++) w[i-1]=0.0;
for (j=kk+1; j<=n; j++)
for (i=kk+1; i<=m; i++)
w[i-1]=w[i-1]+e[j-1]*a[(i-1)*n+j-1];
for (j=kk+1; j<=n; j++)
for (i=kk+1; i<=m; i++)
{ ix=(i-1)*n+j-1;
a[ix]=a[ix]-w[i-1]*e[j-1]/e[kk];
}
}
for (i=kk+1; i<=n; i++)
v[(i-1)*n+kk-1]=e[i-1];
}
}
}
mm=n;
if (m+1<n) mm=m+1;
if (k<n) s[k]=a[k*n+k];
if (m<mm) s[mm-1]=0.0;
if (l+1<mm) e[l]=a[l*n+mm-1];
e[mm-1]=0.0;
nn=m;
if (m>n) nn=n;
if (nn>=k+1)
{ for (j=k+1; j<=nn; j++)
{ for (i=1; i<=m; i++)
u[(i-1)*m+j-1]=0.0;
u[(j-1)*m+j-1]=1.0;
}
}
if (k>=1)
{ for (ll=1; ll<=k; ll++)
{ kk=k-ll+1; iz=(kk-1)*m+kk-1;
if (s[kk-1]!=0.0)
{ if (nn>=kk+1)
for (j=kk+1; j<=nn; j++)
{ d=0.0;
for (i=kk; i<=m; i++)
{ ix=(i-1)*m+kk-1;
iy=(i-1)*m+j-1;
d=d+u[ix]*u[iy]/u[iz];
}
d=-d;
for (i=kk; i<=m; i++)
{ ix=(i-1)*m+j-1;
iy=(i-1)*m+kk-1;
u[ix]=u[ix]+d*u[iy];
}
}
for (i=kk; i<=m; i++)
{ ix=(i-1)*m+kk-1; u[ix]=-u[ix];}
u[iz]=1.0f+u[iz];
if (kk-1>=1)
for (i=1; i<=kk-1; i++)
u[(i-1)*m+kk-1]=0.0;
}
else
{ for (i=1; i<=m; i++)
u[(i-1)*m+kk-1]=0.0;
u[(kk-1)*m+kk-1]=1.0;
}
}
}
for (ll=1; ll<=n; ll++)
{ kk=n-ll+1; iz=kk*n+kk-1;
if ((kk<=l)&&(e[kk-1]!=0.0))
{ for (j=kk+1; j<=n; j++)
{ d=0.0;
for (i=kk+1; i<=n; i++)
{ ix=(i-1)*n+kk-1; iy=(i-1)*n+j-1;
d=d+v[ix]*v[iy]/v[iz];
}
d=-d;
for (i=kk+1; i<=n; i++)
{ ix=(i-1)*n+j-1; iy=(i-1)*n+kk-1;
v[ix]=v[ix]+d*v[iy];
}
}
}
for (i=1; i<=n; i++)
v[(i-1)*n+kk-1]=0.0;
v[iz-n]=1.0;
}
for (i=1; i<=m; i++)
for (j=1; j<=n; j++)
a[(i-1)*n+j-1]=0.0;
m1=mm; it=60;
while (1==1)
{ if (mm==0)
{ ppp(a,e,s,v,m,n);
free(s); free(e); free(w); return(1);
}
if (it==0)
{ ppp(a,e,s,v,m,n);
free(s); free(e); free(w); return(-1);
}
kk=mm-1;
while ((kk!=0)&&(fabs(e[kk-1])!=0.0))
{ d=(float)(fabs(s[kk-1])+fabs(s[kk]));
dd=(float)fabs(e[kk-1]);
if (dd>eps*d) kk=kk-1;
else e[kk-1]=0.0;
}
if (kk==mm-1)
{ kk=kk+1;
if (s[kk-1]<0.0)
{ s[kk-1]=-s[kk-1];
for (i=1; i<=n; i++)
{ ix=(i-1)*n+kk-1; v[ix]=-v[ix];}
}
while ((kk!=m1)&&(s[kk-1]<s[kk]))
{ d=s[kk-1]; s[kk-1]=s[kk]; s[kk]=d;
if (kk<n)
for (i=1; i<=n; i++)
{ ix=(i-1)*n+kk-1; iy=(i-1)*n+kk;
d=v[ix]; v[ix]=v[iy]; v[iy]=d;
}
if (kk<m)
for (i=1; i<=m; i++)
{ ix=(i-1)*m+kk-1; iy=(i-1)*m+kk;
d=u[ix]; u[ix]=u[iy]; u[iy]=d;
}
kk=kk+1;
}
it=60;
mm=mm-1;
}
else
{ ks=mm;
while ((ks>kk)&&(fabs(s[ks-1])!=0.0))
{ d=0.0;
if (ks!=mm) d=d+(float)fabs(e[ks-1]);
if (ks!=kk+1) d=d+(float)fabs(e[ks-2]);
dd=(float)fabs(s[ks-1]);
if (dd>eps*d) ks=ks-1;
else s[ks-1]=0.0;
}
if (ks==kk)
{ kk=kk+1;
d=(float)fabs(s[mm-1]);
t=(float)fabs(s[mm-2]);
if (t>d) d=t;
t=(float)fabs(e[mm-2]);
if (t>d) d=t;
t=(float)fabs(s[kk-1]);
if (t>d) d=t;
t=(float)fabs(e[kk-1]);
if (t>d) d=t;
sm=s[mm-1]/d; sm1=s[mm-2]/d;
em1=e[mm-2]/d;
sk=s[kk-1]/d; ek=e[kk-1]/d;
b=((sm1+sm)*(sm1-sm)+em1*em1)/2.0f;
c=sm*em1; c=c*c; shh=0.0;
if ((b!=0.0)||(c!=0.0))
{ shh=(float)sqrt(b*b+c);
if (b<0.0) shh=-shh;
shh=c/(b+shh);
}
fg[0]=(sk+sm)*(sk-sm)-shh;
fg[1]=sk*ek;
for (i=kk; i<=mm-1; i++)
{ sss(fg,cs);
if (i!=kk) e[i-2]=fg[0];
fg[0]=cs[0]*s[i-1]+cs[1]*e[i-1];
e[i-1]=cs[0]*e[i-1]-cs[1]*s[i-1];
fg[1]=cs[1]*s[i];
s[i]=cs[0]*s[i];
if ((cs[0]!=1.0)||(cs[1]!=0.0))
for (j=1; j<=n; j++)
{ ix=(j-1)*n+i-1;
iy=(j-1)*n+i;
d=cs[0]*v[ix]+cs[1]*v[iy];
v[iy]=-cs[1]*v[ix]+cs[0]*v[iy];
v[ix]=d;
}
sss(fg,cs);
s[i-1]=fg[0];
fg[0]=cs[0]*e[i-1]+cs[1]*s[i];
s[i]=-cs[1]*e[i-1]+cs[0]*s[i];
fg[1]=cs[1]*e[i];
e[i]=cs[0]*e[i];
if (i<m)
if ((cs[0]!=1.0)||(cs[1]!=0.0))
for (j=1; j<=m; j++)
{ ix=(j-1)*m+i-1;
iy=(j-1)*m+i;
d=cs[0]*u[ix]+cs[1]*u[iy];
u[iy]=-cs[1]*u[ix]+cs[0]*u[iy];
u[ix]=d;
}
}
e[mm-2]=fg[0];
it=it-1;
}
else
{ if (ks==mm)
{ kk=kk+1;
fg[1]=e[mm-2]; e[mm-2]=0.0;
for (ll=kk; ll<=mm-1; ll++)
{ i=mm+kk-ll-1;
fg[0]=s[i-1];
sss(fg,cs);
s[i-1]=fg[0];
if (i!=kk)
{ fg[1]=-cs[1]*e[i-2];
e[i-2]=cs[0]*e[i-2];
}
if ((cs[0]!=1.0)||(cs[1]!=0.0))
for (j=1; j<=n; j++)
{ ix=(j-1)*n+i-1;
iy=(j-1)*n+mm-1;
d=cs[0]*v[ix]+cs[1]*v[iy];
v[iy]=-cs[1]*v[ix]+cs[0]*v[iy];
v[ix]=d;
}
}
}
else
{ kk=ks+1;
fg[1]=e[kk-2];
e[kk-2]=0.0;
for (i=kk; i<=mm; i++)
{ fg[0]=s[i-1];
sss(fg,cs);
s[i-1]=fg[0];
fg[1]=-cs[1]*e[i-1];
e[i-1]=cs[0]*e[i-1];
if ((cs[0]!=1.0)||(cs[1]!=0.0))
for (j=1; j<=m; j++)
{ ix=(j-1)*m+i-1;
iy=(j-1)*m+kk-2;
d=cs[0]*u[ix]+cs[1]*u[iy];
u[iy]=-cs[1]*u[ix]+cs[0]*u[iy];
u[ix]=d;
}
}
}
}
}
}
free(s);free(e);free(w);
return(1);
}
void ppp(float a[],float e[],float s[],float v[],int m,int n)
{ int i,j,p,q;
float d;
if (m>=n) i=n;
else i=m;
for (j=1; j<=i-1; j++)
{ a[(j-1)*n+j-1]=s[j-1];
a[(j-1)*n+j]=e[j-1];
}
a[(i-1)*n+i-1]=s[i-1];
if (m<n) a[(i-1)*n+i]=e[i-1];
for (i=1; i<=n-1; i++)
for (j=i+1; j<=n; j++)
{ p=(i-1)*n+j-1; q=(j-1)*n+i-1;
d=v[p]; v[p]=v[q]; v[q]=d;
}
return;
}
void sss(float fg[],float cs[])
{ float r,d;
if ((fabs(fg[0])+fabs(fg[1]))==0.0)
{ cs[0]=1.0; cs[1]=0.0; d=0.0;}
else
{ d=(float)sqrt(fg[0]*fg[0]+fg[1]*fg[1]);
if (fabs(fg[0])>fabs(fg[1]))
{ d=(float)fabs(d);
if (fg[0]<0.0) d=-d;
}
if (fabs(fg[1])>=fabs(fg[0]))
{ d=(float)fabs(d);
if (fg[1]<0.0) d=-d;
}
cs[0]=fg[0]/d; cs[1]=fg[1]/d;
}
r=1.0;
if (fabs(fg[0])>fabs(fg[1])) r=cs[1];
else
if (cs[0]!=0.0) r=1.0f/cs[0];
fg[0]=d; fg[1]=r;
return;
}
参考:线性，非线性多项式：http://blog.csdn.net/qll125596718/article/details/8248249 http://blog.csdn.net/poxiaozhuimeng/article/details/41117947 http://blog.csdn.net/ouyangying123/article/details/53996403http://blog.csdn.net/jairuschan/article/details/7517773/http://blog.csdn.net/zang141588761/article/details/50523036 http://www.cnblogs.com/gnuhpc/archive/2012/12/09/2809997.html http://download.csdn.net/download/biaobiao11/9755119
圆拟合：http://blog.sina.com.cn/s/blog_b27f71160101gxun.html http://www.cnblogs.com/dotLive/archive/2007/04/06/524633.htmlhttp://blog.csdn.net/andylao62/article/details/24522365 http://blog.csdn.net/liyuanbhu/article/details/50889951 http://blog.csdn.net/liyuanbhu/article/details/50890587
椭圆拟合：http://doc.okbase.net/u013708970/archive/121532.html
转：http://blog.csdn.net/piaoxuezhong/article/details/54973750
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