C++ 中圆的最小二乘法

Question

我应该找到这个 çember 方程，我为此写了四个代码，但他们有一半代码没有完成（我无法完成它:() 首先我写了一个代码来找到这些黑点坐标：

  Mat img=imread("n.jpg");
Mat thresh;
threshold(img,thresh,150,500,THRESH_BINARY_INV);
//  int point[thresh.rows][thresh.cols];
vector<int> x,y;
    for(int i = 0; i < thresh.rows; ++i) {
        for(int j = 0; j < thresh.cols; ++j) {
            int b = int(thresh.at<cv::Vec3b>(i,j)[0]);//burada biz b g r yi kısaca yazabilmek için tanımladık
            int g = int(thresh.at<cv::Vec3b>(i,j)[1]);
            int r = int(thresh.at<cv::Vec3b>(i,j)[2]);
        //cout<<b<<" "<<g<<" "<<r<<endl;
    //  cout<<i<<" "<<j<<endl;
        if(b==255&&g==255&&r==255)
        {
            x.push_back(i);
            y.push_back(j);

        cout<<i<<"  "<<j<<endl;
    }
                //cout<<b<<"  "<<g<<"  "<<r<<endl;
    }

}

而且比 1）我写了一个行列式代码来找到给定 3 点的圆

   float determinant(float arr[][3])
{
    float det=0.0;
    int i;
    for(i=0;i<3;i++)
      det = det + (arr[0][i]*(arr[1][(i+1)%3]*arr[2][(i+2)%3] - arr[1][(i+2)%3]*arr[2][(i+1)%3]));
      return det;
}


int main()
{
    float matris1[3][3],matris2[3][3],matris3[3][3];
    float x[3],y[3];//x1,x2,x3/y1,y2,y3
    int i,j;

    printf("x1 x2 ve x3 u giriniz\n");
    scanf("%f%f%f",&x[0],&x[1],&x[2]);

    printf("y1 y2 ve y3 u giriniz\n");
    scanf("%f%f%f",&y[0],&y[1],&y[2]);
    //matris1 i oluşturma
    for(i=0;i<3;i++)
    for(j=0;j<3;j++)
    if(j==0)
    matris1[i][j]=pow(x[i],2)+pow(y[i],2);
    else if(j==1)
    matris1[i][j]=y[i];
    else 
    matris1[i][j]=1;

    //matris2 yi oluşturma
    for(i=0;i<3;i++)
    for(j=0;j<3;j++)
    if(j==0)
    matris2[i][j]=x[i];
    else if(j==1)
    matris2[i][j]=y[i];
    else 
    matris2[i][j]=1;

    //matris3 yi oluşturma
    for(i=0;i<3;i++)
    for(j=0;j<3;j++)
    if(j==0)
    matris3[i][j]=x[i];
    else if(j==1)
    matris3[i][j]=pow(x[i],2)+pow(y[i],2);
    else 
    matris3[i][j]=1;

    //matris1 i ve determinantını yazdırma
    for(i=0;i<3;i++)
    {
    for(j=0;j<3;j++)
    {
        printf("%.2f ",matris1[i][j]);
    }
    printf("\n");
    }
    printf("Determinanti1 =%.2f \n\n",determinant(matris1));
    for(i=0;i<3;i++)
    {
    for(j=0;j<3;j++)
    {
        printf("%.2f ",matris2[i][j]);
    }
    printf("\n");
    }
    printf("Determinanti2 =%.2f \n\n",determinant(matris2));

    //matris3 i ve determinantını yazdırma
    for(i=0;i<3;i++)
    {
    for(j=0;j<3;j++)
    {
        printf("%.2f ",matris3[i][j]);
    }
    printf("\n");
    }
    printf("Determinanti3 =%.2f \n\n",determinant(matris3));
    float a,b;
    a=determinant(matris1)/(2*determinant(matris2));
    b=determinant(matris3)/(2*determinant(matris2));
    printf("%.2f ",a);
    printf("%.2f ",b);
    float yaricap;


    yaricap=sqrt((x[0]-a)*(x[0]-a)+(y[0]-b)*(y[0]-b));

       printf("yaricap =%.2f \n\n",yaricap);
      }

但是这段代码有 3 点，但我有很多点，我无法实现 2）最小二乘的代码：

typedef struct {
  double x, y;
} Point2;
int CircleFit(int N, Point2 * P, double *pa, double *pb, double *pr)
{

  const int maxIterations = 256;
  const double tolerance = 1e-06;

  double a, b, r;


  int i, j;
  double xAvr = 0.0;
  double yAvr = 0.0;

  for (i = 0; i < N; i++) {
    xAvr += P[i].x;
    yAvr += P[i].y;
  }
  xAvr /= N;
  yAvr /= N;


  a = xAvr;
  b = yAvr;

  for (j = 0; j < maxIterations; j++) {

    double a0 = a;
    double b0 = b;


    double LAvr = 0.0;
    double LaAvr = 0.0;
    double LbAvr = 0.0;

    for (i = 0; i < N; i++) {
      double dx = P[i].x - a;
      double dy = P[i].y - b;
      double L = sqrt(dx * dx + dy * dy);
      if (fabs(L) > tolerance) {
        LAvr += L;
        LaAvr -= dx / L;
        LbAvr -= dy / L;
      }
    }
    LAvr /= N;
    LaAvr /= N;
    LbAvr /= N;

    a = xAvr + LAvr * LaAvr;
    b = yAvr + LAvr * LbAvr;
    r = LAvr;

    if (fabs(a - a0) <= tolerance && fabs(b - b0) <= tolerance)
      break;
  }

  *pa = a;
  *pb = b;
  *pr = r;

  return (j < maxIterations ? j : -1);
}

enum {
  M_SHOW_CIRCLE, M_CIRCLE_INFO, M_RESET_POINTS, M_QUIT
};

但我不知道如何将我的观点和这个结合起来？？感谢您的进阶

Answer 1

如果一组点是对称的，最好的方法是对 x 和 y 求和，然后您将找到中心，然后计算中心和一组点之间的平均距离，使其具有半径

Answer 2

这是一个圆回归问题我确定你可以问https://math.stackexchange.com/

连本站都有例子

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