我在执行一个算法时正在摸索我的头脑,我相信会计算一个方程的系数,这个方程会给我一个将限定一组点的椭圆。鉴于我不知道该算法如何实际执行它应该做的事情,我很难理解为什么该算法实际上并不像我认为的那样工作。我确信我已经准确地实现了该算法。我意识到我可能已经塞满了某个地方。在Java中实现边界椭圆
我的实现是从this implementation in C++建模的,因为我发现它比伪代码given here更容易使用。 C++实现的OP说它基于相同的伪代码。
这是我实现:
// double tolerance = 0.2;
// int n = 10;
// int d = 2;
double tolerance=0.02;
int n=10;
int d=2;
// MatrixXd p = MatrixXd::Random(d,n);
RealMatrix p=new BlockRealMatrix(d,n,new double[][]{{70,56,44,93,77,12,30,51,35,82,74,38,92,49,22,69,71,91,39,13}},false);
// MatrixXd q = p;
// q.conservativeResize(p.rows() + 1, p.cols());
RealMatrix q=p.createMatrix(d+1,n);
q.setSubMatrix(p.getData(),0,0);
// for(size_t i = 0; i < q.cols(); i++)
// {
// q(q.rows() - 1, i) = 1;
// }
//
// const double init_u = 1.0/(double) n;
// MatrixXd u = MatrixXd::Constant(n, 1, init_u);
double[]ones=new double[n];
double[]uData=new double[n];
for(int i=0;i<n;i++){
ones[i]=1;
uData[i]=((double)1)/((double)n);
}
q.setRow(d,ones);
// int count = 1;
// double err = 1;
int count=0;
double err=1;
while(err>tolerance){
// MatrixXd Q_tr = q.transpose();
RealMatrix qTr=q.transpose();
// MatrixXd X = q * u.asDiagonal() * Q_tr;
RealMatrix uDiag=MatrixUtils.createRealDiagonalMatrix(uData);
RealMatrix qByuDiag=q.multiply(uDiag);
RealMatrix x=qByuDiag.multiply(qTr);
// MatrixXd M = (Q_tr * X.inverse() * q).diagonal();
RealMatrix qTrByxInverse=qTr.multiply(MatrixUtils.inverse(x));
RealMatrix qTrByxInverseByq=qTrByxInverse.multiply(q);
int r=qTrByxInverseByq.getRowDimension();
double mData[]=new double[r];
for(int i=0;i<r;i++){
mData[i]=qTrByxInverseByq.getEntry(i,i);
}
// double maximum = M.maxCoeff(&j_x, &j_y);
// As M is a matrix formed from mData where only cells on the
// diagonal are populated with values greater than zero, the row
// and column values will be identical, and will be equal to the
// place where the maximum value occurs in mData. The matrix M
// is never used again in the algorithm, and hence creation of
// the matrix M is unnecessary.
int iMax=0;
double dMax=0;
for(int i=0;i<mData.length;i++){
if(mData[i]>dMax){
dMax=mData[i];
iMax=i;
}
}
// double step_size = (maximum - d - 1)/((d + 1) * (maximum + 1));
double stepSize=(dMax-d-1)/((d+1)*(dMax+1));
// MatrixXd new_u = (1 - step_size) * u;
double[]uDataNew=new double[n];
for(int i=0;i<n;i++){
uDataNew[i]=(((double)1)-stepSize)*uData[i];
}
// new_u(j_x, 0) += step_size;
uDataNew[iMax]+=stepSize;
// 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());
double sum=0;
for(int i=1;i<n;i++){
double cell=uDataNew[i]-uData[i];
sum+=(cell*cell);
}
err=Math.sqrt(sum);
// count++
// u = new_u;
count++;
uData=uDataNew;
}
// MatrixXd U = u.asDiagonal();
RealMatrix uFinal=MatrixUtils.createRealDiagonalMatrix(uData);
// MatrixXd A = (1.0/(double) d) * (p * U * p.transpose() - (p * u) * (p * u).transpose()).inverse();
// Broken down into the following 9 sub-steps:
// 1 p * u
double[][]uMatrixData=new double[1][];
uMatrixData[0]=uData;
RealMatrix u=new BlockRealMatrix(n,1,uMatrixData,false);
RealMatrix cFinal=p.multiply(u);
// 2 (p * u).transpose()
RealMatrix two=cFinal.transpose();
// 3 (p * u) * (p * u).transpose()
RealMatrix three=cFinal.multiply(two);
// 4 p * U
RealMatrix four=p.multiply(uFinal);
// 5 p * U * p.transpose()
RealMatrix five=four.multiply(p.transpose());
// 6 p * U * p.transpose() - (p * u) * (p * u).transpose()
RealMatrix six=five.subtract(three);
// 7 (p * U * p.transpose() - (p * u) * (p * u).transpose()).inverse()
RealMatrix seven=MatrixUtils.inverse(six);
// 8 1.0/(double) d
double eight=((double)1)/d;
// 9 MatrixXd A = (1.0/(double) d) * (p * U * p.transpose() - (p * u) * (p * u).transpose()).inverse()
RealMatrix aFinal=seven.scalarMultiply(eight);
// MatrixXd c = p * u; This has been calculated in sub-step (1) above and stored as cFinal.
System.out.println();
System.out.println("The coefficients of ellipse's equation are given as follows:");
for(int i=0;i<aFinal.getRowDimension();i++){
for(int j=0;j<aFinal.getColumnDimension();j++){
System.out.printf(" %3.8f",aFinal.getEntry(i,j));
}
System.out.println();
}
System.out.println();
System.out.println("The the axis shifts are given as follows:");
for(int i=0;i<cFinal.getRowDimension();i++){
for(int j=0;j<cFinal.getColumnDimension();j++){
System.out.printf(" %3.8f",cFinal.getEntry(i,j));
}
System.out.println();
}
// Get the centre of the set of points, which will be the centre of the
// ellipse. This part was not actually included in the C++
// implementation. I guess the OP considered it too trivial.
double xmin=0;
double xmax=0;
double ymin=0;
double ymax=0;
for(int i=0;i<p.getRowDimension();i++){
double x=p.getEntry(i,0);
double y=p.getEntry(i,1);
if(i==0){
xmin=xmax=x;
ymin=ymax=y;
}else{
if(x<xmin){
xmin=x;
}else if(x>xmax){
xmax=x;
}
if(y<ymin){
ymin=y;
}else if(y>ymax){
ymax=y;
}
}
}
double x=(xmax-xmin)/2+xmin;
double y=(ymax-ymin)/2+ymin;
System.out.println();
System.out.println("The centre of the ellipse is given as follows:");
System.out.printf(" The x axis is %3.8f.\n",x);
System.out.printf(" The y axis is %3.8f.\n",y);
System.out.println();
System.out.println("The algorithm completed ["+count+"] iterations of its while loop.");
// This code constructs and displays a yellow ellipse with a black border.
ArrayList<Integer>pointsx=new ArrayList<>();
ArrayList<Integer>pointsy=new ArrayList<>();
for (double t=0;t<2*Math.PI;t+=0.02){ // <- or different step
pointsx.add(this.getWidth()/2+(int)(cFinal.getEntry(0,0)*Math.cos(t)*aFinal.getEntry(0,0)-cFinal.getEntry(1,0)*Math.sin(t)*aFinal.getEntry(0,1)));
pointsy.add(this.getHeight()/2+(int)(cFinal.getEntry(0,0)*Math.cos(t)*aFinal.getEntry(1,0)+cFinal.getEntry(1,0)*Math.sin(t)*aFinal.getEntry(1,1)));
}
int[]xpoints=new int[pointsx.size()];
Iterator<Integer>xpit=pointsx.iterator();
for(int i=0;xpit.hasNext();i++){
xpoints[i]=xpit.next();
}
int[]ypoints=new int[pointsy.size()];
Iterator<Integer>ypit=pointsy.iterator();
for(int i=0;ypit.hasNext();i++){
ypoints[i]=ypit.next();
}
g.setColor(Color.yellow);
g.fillPolygon(xpoints,ypoints,pointsx.size());
g.setColor(Color.black);
g.drawPolygon(xpoints,ypoints,pointsx.size());
该程序产生的输出如下:
The coefficients of ellipse's equation are given as follows:
0.00085538 0.00050693
0.00050693 0.00093474
The axis shifts are given as follows:
54.31114965
55.60647648
The centre of the ellipse is given as follows:
The x axis is 72.00000000.
The y axis is 47.00000000.
The algorithm completed [23] iterations of its while loop.
我觉得有些奇怪的是,2×2矩阵的输入是非常小的。我被引导认为这些条目是用于描述椭圆的方向的系数,而第二个2×1矩阵描述了椭圆的x和y轴的大小。
我知道用于获得点的方程式称为参数方程。他们有一个表格,引用here。
椭圆的中心位置和这些值的计算已经被我添加了。它们没有出现在C++实现中,并且在将此算法的输出结合到用于描述椭圆的参数方程之后,我导致相信C++实现的OP给了错误的印象,即这个2×1矩阵描述了椭圆的中心。我承认我形成的印象可能是错误的,因为如果一个人认为我是对的,那么这个中心(两个轴的最低值和最高值之间的中间值)似乎是错误的;它小于y轴的大小,我采取的是径向测量。
当我将这些值插入椭圆的参数方程来生成点I然后使用创建一个Polygon
时,生成的形状占据一个像素。考虑到描述方向的2×2矩阵中给出的值,这就是我所期望的。
因此,在我看来,我如何生成产生方向的2x2矩阵存在一些问题。
我已尽力简洁,并提供所有相关事实,以及我形成的任何相关印象,无论它们是对还是错。我希望有人能提供这个问题的答案。
请:编辑你的问题和工作,最后一段。关注这个问题。请注意:当人们告诉你“问题不好”时,那么你的反应不应该是“无论如何,请继续前进”。因为那你就忽略了有价值的反馈。 – GhostCat
听起来像一个好主意。感谢您的忠告,我会编辑我的问题并继续前进。 – 000