给定一组2d点(笛卡尔形式),我需要找到最小面积椭圆,使得集合中的每个点都位于椭圆上或内部。
我在这个网站上以伪代码的形式拥有found the solution,但我尝试用C ++实现解决方案是不成功的。
下图以图形方式说明了我的问题的解决方案:
在我的尝试中,我使用Eigen库进行矩阵的各种操作。
//The tolerance for error in fitting the ellipse
double tolerance = 0.2;
int n = 10; // number of points
int d = 2; // dimension
MatrixXd p = MatrixXd::Random(d,n); //Fill matrix with random points
MatrixXd q = p;
q.conservativeResize(p.rows() + 1, p.cols());
for(size_t i = 0; i < q.cols(); i++)
{
q(q.rows() - 1, i) = 1;
}
int count = 1;
double err = 1;
const double init_u = 1.0 / (double) n;
MatrixXd u = MatrixXd::Constant(n, 1, init_u);
while(err > tolerance)
{
MatrixXd Q_tr = q.transpose();
cout << "1 " << endl;
MatrixXd X = q * u.asDiagonal() * Q_tr;
cout << "1a " << endl;
MatrixXd M = (Q_tr * X.inverse() * q).asDiagonal();
cout << "1b " << endl;
int j_x, j_y;
double maximum = M.maxCoeff(&j_x, &j_y);
double step_size = (maximum - d - 1) / ((d + 1) * (maximum + 1));
MatrixXd new_u = (1 - step_size) * u;
new_u(j_x, 0) += step_size;
cout << "2 " << endl;
//Find err
MatrixXd u_diff = new_u - u;
for(size_t i = 0; i < u_diff.rows(); i++)
{
for(size_t j = 0; j < u_diff.cols(); j++)
u_diff(i, j) *= u_diff(i, j); // Square each element of the matrix
}
err = sqrt(u_diff.sum());
count++;
u = new_u;
}
cout << "3 " << endl;
MatrixXd U = u.asDiagonal();
MatrixXd A = (1.0 / (double) d) * (p * U * p.transpose() - (p * u) * (p * u).transpose()).inverse();
MatrixXd c = p * u;
错误发生在以下行:
MatrixXd M = (Q_tr * X.inverse() * q).asDiagonal();
,内容如下:
run: /usr/include/eigen3/Eigen/src/Core/DenseBase.h:261: void Eigen::DenseBase<Derived>::resize(Eigen::Index, Eigen::Index) [with Derived = Eigen::Diagonal<Eigen::Matrix<double, -1, -1>, 0>; Eigen::Index = long int]: Assertion `rows == this->rows() && cols == this->cols() && "DenseBase::resize() does not actually allow to resize."' failed.
Aborted (core dumped)
有人可以指出为什么会发生这种错误甚至更好吗,请教我如何使用C ++将椭圆拟合到一组点?
答案 0 :(得分:1)
使用Eigen,您可以从.diagonal()
的矩阵中获得对角矢量;您可以使用.asDiagonal()
将矢量视为对角矩阵;但你不能将密集矩阵视为对角矩阵。那条线应该是
MatrixXd M = (Q_tr * X.inverse() * q).diagonal();