java – delaunay三角测量的代码如何工作？

我有这个Java代码,它带有一组Point in输入,返回一组代表Delaunay三角剖分的图形边缘.

我想知道使用什么策略,如果存在,使用的算法名称.

在此代码中,GraphEdge包含两个awt Point并表示三角剖分中的边,GraphPoint扩展Awt Point,并在TreeSet对象中返回最终三角剖分的边.

我的目的是了解这种方法的工作原理：

public TreeSet getEdges(int n, int[] x, int[] y, int[] z)

在这个三角测量的完整源代码下面：

import java.awt.Point;
import java.util.Iterator;
import java.util.TreeSet;

public class DelaunayTriangulation
{
   int[][] adjMatrix;

   DelaunayTriangulation(int size)
   {
     this.adjMatrix = new int[size][size];
   }
   public int[][] getAdj() {
     return this.adjMatrix;
   }

   public TreeSet getEdges(int n, int[] x, int[] y, int[] z)
   {
     TreeSet result = new TreeSet();

     if (n == 2)
     {
       this.adjMatrix[0][1] = 1;
       this.adjMatrix[1][0] = 1;
       result.add(new GraphEdge(new GraphPoint(x[0], y[0]), new GraphPoint(x[1], y[1])));

       return result;
     }

     for (int i = 0; i < n - 2; i++) {
       for (int j = i + 1; j < n; j++) {
         for (int k = i + 1; k < n; k++)
         {
           if (j == k) {
             continue;
           }
           int xn = (y[j] - y[i]) * (z[k] - z[i]) - (y[k] - y[i]) * (z[j] - z[i]);

           int yn = (x[k] - x[i]) * (z[j] - z[i]) - (x[j] - x[i]) * (z[k] - z[i]);

           int zn = (x[j] - x[i]) * (y[k] - y[i]) - (x[k] - x[i]) * (y[j] - y[i]);
           boolean flag;
           if (flag = (zn < 0 ? 1 : 0) != 0) {
             for (int m = 0; m < n; m++) {
               flag = (flag) && ((x[m] - x[i]) * xn + (y[m] - y[i]) * yn + (z[m] - z[i]) * zn <= 0);
             }

           }

           if (!flag)
           {
             continue;
           }
           result.add(new GraphEdge(new GraphPoint(x[i], y[i]), new GraphPoint(x[j], y[j])));
           //System.out.println("----------");
           //System.out.println(x[i]+" "+ y[i] +"----"+x[j]+" "+y[j]);

          result.add(new GraphEdge(new GraphPoint(x[j], y[j]), new GraphPoint(x[k], y[k])));
          //System.out.println(x[j]+" "+ y[j] +"----"+x[k]+" "+y[k]);
          result.add(new GraphEdge(new GraphPoint(x[k], y[k]), new GraphPoint(x[i], y[i])));
           //System.out.println(x[k]+" "+ y[k] +"----"+x[i]+" "+y[i]);
           this.adjMatrix[i][j] = 1;
           this.adjMatrix[j][i] = 1;
           this.adjMatrix[k][i] = 1;
           this.adjMatrix[i][k] = 1;
           this.adjMatrix[j][k] = 1;
           this.adjMatrix[k][j] = 1;
         }

       }

     }

     return result;
   }

   public TreeSet getEdges(TreeSet pointsSet)
   {
     if ((pointsSet != null) && (pointsSet.size() > 0))
     {
       int n = pointsSet.size();

       int[] x = new int[n];
       int[] y = new int[n];
       int[] z = new int[n];

       int i = 0;

       Iterator iterator = pointsSet.iterator();
       while (iterator.hasNext())
       {
         Point point = (Point)iterator.next();

         x[i] = (int)point.getX();
         y[i] = (int)point.getY();
         z[i] = (x[i] * x[i] + y[i] * y[i]);

         i++;
       }

       return getEdges(n, x, y, z);
     }

     return null;
   }
 }

解决方法:

看起来像这里描述的http://en.wikipedia.org/wiki/Delaunay_triangulation：

The problem of finding the Delaunay triangulation of a set of points in d-dimensional Euclidean space can be converted to the problem of finding the convex hull of a set of points in (d + 1)-dimensional space, by giving each point p an extra coordinate equal to |p|2, taking the bottom side of the convex hull, and mapping back to d-dimensional space by deleting the last coordinate.

在你的例子中,d是2.

向量(xn,yn,zn)是向量的交叉积(点i – >点j)和(点i – >点k)或换句话说垂直于三角形的向量(点i,点) j,点k).

标志的计算检查该三角形的法线是否指向负z方向,以及所有其他点是否位于与三角形法线相反的一侧(相反,因为其他点需要在三角形的平面上方,因为我们是感兴趣的是凸壳的底部).如果是这种情况,三角形(i,j,k)是3D凸包的一部分,因此x和y分量(3D三角形在x,y平面上的投影)是(2D)的一部分Delaunay三角剖分.

标签：triangulation,java,delaunay
来源： https://codeday.me/bug/20190926/1818311.html