用于在图中查找人脸的 Java 算法

Question

我有一个我自己创建的平面图。我想找到这个图的面，但我找不到这样做的工作算法。到目前为止我所做的是使用一种算法来查找图中的所有循环，但这给了我所有可能的循环，我已经尝试过但没有找到一种方法来只对面部进行排序。我的一个想法是使用 Path2Ds contains 方法来查看另一个形状是否重叠，但由于面共享节点，这不起作用。下图展示了我想要的内容，之后的代码展示了我的可复制示例。

import java.awt.geom.Point2D;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Set;

public class PolygonFinder {

    //  Graph modeled as list of edges
    static int[][] graph
            = {
                {1, 2}, {1, 6}, {1, 5}, {2, 6},
                {2, 3}, {3, 7}, {7, 4}, {3, 4},
                {5, 4}, {6, 5}
            };

    static List<int[]> cycles = new ArrayList<>();

    /**
     * @param args
     */
    public static void main(String[] args) {

        for (int[] graph1 : graph) {
            for (int j = 0; j < graph1.length; j++) {
                findNewCycles(new int[]{graph1[j]});
            }
        }

        cycles.stream().map(cy -> {
            String s = "" + cy[0];
            for (int i = 1; i < cy.length; i++) {
                s += "," + cy[i];
            }
            return s;
        }).forEachOrdered(s -> {
            System.out.println(s);
        });
    }

    static void findNewCycles(int[] path) {
        int n = path[0];
        int x;
        int[] sub = new int[path.length + 1];

        for (int[] graph1 : graph) {
            for (int y = 0; y <= 1; y++) {
                if (graph1[y] == n) {
                    x = graph1[(y + 1) % 2];
                    if (!visited(x, path)) //  neighbor node not on path yet
                    {
                        sub[0] = x;
                        System.arraycopy(path, 0, sub, 1, path.length);
                        //  explore extended path
                        findNewCycles(sub);
                    } else if ((path.length > 2) && (x == path[path.length - 1])) //  cycle found
                    {
                        int[] p = normalize(path);
                        int[] inv = invert(p);
                        if (isNew(p) && isNew(inv)) {
                            cycles.add(p);
                        }
                    }
                }
            }
        }
    }

    //  check of both arrays have same lengths and contents
    static Boolean equals(int[] a, int[] b) {
        Boolean ret = (a[0] == b[0]) && (a.length == b.length);

        for (int i = 1; ret && (i < a.length); i++) {
            if (a[i] != b[i]) {
                ret = false;
            }
        }

        return ret;
    }

    //  create a path array with reversed order
    static int[] invert(int[] path) {
        int[] p = new int[path.length];

        for (int i = 0; i < path.length; i++) {
            p[i] = path[path.length - 1 - i];
        }

        return normalize(p);
    }

    //  rotate cycle path such that it begins with the smallest node
    static int[] normalize(int[] path) {
        int[] p = new int[path.length];
        int x = smallest(path);
        int n;

        System.arraycopy(path, 0, p, 0, path.length);

        while (p[0] != x) {
            n = p[0];
            System.arraycopy(p, 1, p, 0, p.length - 1);
            p[p.length - 1] = n;
        }

        return p;
    }

    //  compare path against known cycles
    //  return true, iff path is not a known cycle
    static Boolean isNew(int[] path) {
        Boolean ret = true;

        for (int[] p : cycles) {
            if (equals(p, path)) {
                ret = false;
                break;
            }
        }

        return ret;
    }

    //  return the int of the array which is the smallest
    static int smallest(int[] path) {
        int min = path[0];

        for (int p : path) {
            if (p < min) {
                min = p;
            }
        }

        return min;
    }

    //  check if vertex n is contained in path
    static Boolean visited(int n, int[] path) {
        Boolean ret = false;

        for (int p : path) {
            if (p == n) {
                ret = true;
                break;
            }
        }

        return ret;
    }
}

运行上述代码后的结果是:

1,6,2
1,5,6,2
1,5,4,7,3,2
1,6,5,4,7,3,2
1,5,4,3,2
1,6,5,4,3,2
1,5,4,7,3,2,6
1,5,4,3,2,6
1,5,6
2,3,7,4,5,6
2,3,4,5,6
3,4,7

我解决这个问题的最佳尝试之一是使用以下代码。坐标来自顶部的图片。

    List<Polygon> polys = new LinkedList<>();
    Polygon p1 = new Polygon();
    p1.addPoint(new Point2D.Double(-4, 4));
    p1.addPoint(new Point2D.Double(-1, 3));
    p1.addPoint(new Point2D.Double(-1, 5));
    Polygon p2 = new Polygon();
    p2.addPoint(new Point2D.Double(-4, 4));
    p2.addPoint(new Point2D.Double(0, -2));
    p2.addPoint(new Point2D.Double(-1, 3));
    p2.addPoint(new Point2D.Double(-1, 5));
    Polygon p3 = new Polygon();
    p3.addPoint(new Point2D.Double(-4, 4));
    p3.addPoint(new Point2D.Double(0, -2));
    p3.addPoint(new Point2D.Double(4, 1));
    p3.addPoint(new Point2D.Double(2, 2));
    p3.addPoint(new Point2D.Double(3, 4));
    p3.addPoint(new Point2D.Double(-1, 5));
    Polygon p4 = new Polygon();
    p4.addPoint(new Point2D.Double(-4, 4));
    p4.addPoint(new Point2D.Double(-1, 3));
    p4.addPoint(new Point2D.Double(0, -2));
    p4.addPoint(new Point2D.Double(4, 1));
    p4.addPoint(new Point2D.Double(2, 2));
    p4.addPoint(new Point2D.Double(3, 4));
    p4.addPoint(new Point2D.Double(-1, 5));
    Polygon p5 = new Polygon();
    p5.addPoint(new Point2D.Double(-4, 4));
    p5.addPoint(new Point2D.Double(0, -2));
    p5.addPoint(new Point2D.Double(4, 1));
    p5.addPoint(new Point2D.Double(3, 4));
    p5.addPoint(new Point2D.Double(-1, 5));
    Polygon p6 = new Polygon();
    p6.addPoint(new Point2D.Double(-4, 4));
    p6.addPoint(new Point2D.Double(-1, 3));
    p6.addPoint(new Point2D.Double(0, -2));
    p6.addPoint(new Point2D.Double(4, 1));
    p6.addPoint(new Point2D.Double(3, 4));
    p6.addPoint(new Point2D.Double(-1, 5));
    Polygon p7 = new Polygon();
    p7.addPoint(new Point2D.Double(-4, 4));
    p7.addPoint(new Point2D.Double(0, -2));
    p7.addPoint(new Point2D.Double(4, 1));
    p7.addPoint(new Point2D.Double(2, 2));
    p7.addPoint(new Point2D.Double(3, 4));
    p7.addPoint(new Point2D.Double(-1, 5));
    p7.addPoint(new Point2D.Double(-1, 3));
    Polygon p8 = new Polygon();
    p8.addPoint(new Point2D.Double(-4, 4));
    p8.addPoint(new Point2D.Double(0, -2));
    p8.addPoint(new Point2D.Double(4, 1));
    p8.addPoint(new Point2D.Double(3, 4));
    p8.addPoint(new Point2D.Double(-1, 5));
    p8.addPoint(new Point2D.Double(-1, 3));
    Polygon p9 = new Polygon();
    p9.addPoint(new Point2D.Double(-4, 4));
    p9.addPoint(new Point2D.Double(0, -2));
    p9.addPoint(new Point2D.Double(-1, 3));
    Polygon p10 = new Polygon();
    p10.addPoint(new Point2D.Double(-1, 5));
    p10.addPoint(new Point2D.Double(3, 4));
    p10.addPoint(new Point2D.Double(2, 2));
    p10.addPoint(new Point2D.Double(4, 1));
    p10.addPoint(new Point2D.Double(0, -2));
    p10.addPoint(new Point2D.Double(-1, 3));
    Polygon p11 = new Polygon();
    p11.addPoint(new Point2D.Double(-1, 5));
    p11.addPoint(new Point2D.Double(3, 4));
    p11.addPoint(new Point2D.Double(4, 1));
    p11.addPoint(new Point2D.Double(0, -2));
    p11.addPoint(new Point2D.Double(-1, 3));
    Polygon p12 = new Polygon();
    p12.addPoint(new Point2D.Double(3, 4));
    p12.addPoint(new Point2D.Double(4, 1));
    p12.addPoint(new Point2D.Double(2, 2));
    polys.add(p1);
    polys.add(p2);
    polys.add(p3);
    polys.add(p4);
    polys.add(p5);
    polys.add(p6);
    polys.add(p7);
    polys.add(p8);
    polys.add(p9);
    polys.add(p10);
    polys.add(p11);
    polys.add(p12);
    Set<Integer> toRemove = new HashSet<>();
    for (Polygon polyI : polys) {
        for (Polygon polyJ : polys) {
            if (polyI.equals(polyJ)) {
                continue;
            }
            if (polyI.contains(polyJ)) {
                toRemove.add(polys.indexOf(polyI));
            }
        }
    }
    List<Integer> list = new LinkedList<>(toRemove);
    Collections.sort(list);
    Collections.reverse(list);
    list.forEach((t) -> {
        polys.remove(t.intValue());
    });

    System.out.println("");
    polys.forEach((t) -> {
        System.out.println(t.getPoints());
    });

此处列出了使用的多边形方法。

@Override
public boolean contains(Point2D point) {
    return getPath().contains(point);
}

@Override
public boolean contains(IPolygon polygon) {
    List<Point2D> p2Points = polygon.getPoints();
    for (Point2D point : p2Points) {
        if (getPath().contains(point)) {
            if (!points.contains(point)) {
                return true;
            }
        }
    }
    return false;
}

private Path2D getPath() {
    Path2D path = new Path2D.Double();
    path.moveTo(points.get(0).getX(), points.get(0).getY());
    for (int i = 1; i < points.size(); i++) {
        path.lineTo(points.get(i).getX(), points.get(i).getY());
    }
    path.closePath();
    return path;
}

这段代码给了我下面的结果，不需要第 2-4 个。

[Point2D.Double[-4.0, 4.0], Point2D.Double[-1.0, 3.0], Point2D.Double[-1.0, 5.0]]
[Point2D.Double[-4.0, 4.0], Point2D.Double[0.0, -2.0], Point2D.Double[-1.0, 3.0], Point2D.Double[-1.0, 5.0]]
[Point2D.Double[-4.0, 4.0], Point2D.Double[-1.0, 3.0], Point2D.Double[0.0, -2.0], Point2D.Double[4.0, 1.0], Point2D.Double[2.0, 2.0], Point2D.Double[3.0, 4.0], Point2D.Double[-1.0, 5.0]]
[Point2D.Double[-4.0, 4.0], Point2D.Double[0.0, -2.0], Point2D.Double[4.0, 1.0], Point2D.Double[2.0, 2.0], Point2D.Double[3.0, 4.0], Point2D.Double[-1.0, 5.0], Point2D.Double[-1.0, 3.0]]
[Point2D.Double[-4.0, 4.0], Point2D.Double[0.0, -2.0], Point2D.Double[-1.0, 3.0]]
[Point2D.Double[-1.0, 5.0], Point2D.Double[3.0, 4.0], Point2D.Double[2.0, 2.0], Point2D.Double[4.0, 1.0], Point2D.Double[0.0, -2.0], Point2D.Double[-1.0, 3.0]]
[Point2D.Double[3.0, 4.0], Point2D.Double[4.0, 1.0], Point2D.Double[2.0, 2.0]]

Answer 1

对于每条边，获取边顶点嵌入中的坐标，并使用它们使用三角法计算边的角度。
例如，从 (x1, y1) 到 (x2, y2) 从正 x 轴逆时针测量的角度由 Math.atan2(y2-y1,x2-x1) 给出。

对于每个顶点，通过按角度对边进行排序来创建循环边排序。这可以存储为数组，也可以使用循环列表数据结构。

选择一条边，跟随它到一个相邻的顶点，然后跟随下一个相邻的顺时针边，重复跟随边到下一个顶点，然后下一个顺时针边，直到你回到起始边；那么你已经找到了图形的一个面。

重复步骤 3，选择一条未访问的边或与前一条相反方向的已访问边，然后沿相同的顺时针方向沿着它找到下一个面。重复此操作，直到所有边都被访问了两次(每个方向一次)，然后您就找到了所有的面。

在 Java 中，这将是:

import java.awt.geom.Point2D;
import java.awt.Polygon;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.stream.Collectors;
import java.text.MessageFormat;

public class GraphFaces
{
  static class Vertex
  {
    final int index;
    final Point2D point;
    final ArrayList<Edge> outboundEdges = new ArrayList<>();
    
    
    public Vertex( final int index, final Point2D point )
    {
      this.index = index;
      this.point = point;
    }
    
    public void addEdge( final Edge edge )
    {
      this.outboundEdges.add( edge );
    }
    
    public void sortEdges()
    {
      this.outboundEdges.sort((e1,e2)->Double.compare(e1.angle,e2.angle));
      
      Edge prev = this.outboundEdges.get(this.outboundEdges.size() - 1);
      for ( final Edge edge: this.outboundEdges )
      {
        edge.setNextEdge( prev );
        prev = edge;
      }
    }
    
    @Override
    public String toString()
    {
      return Integer.toString(this.index);
      // return MessageFormat.format("({0},{1})",this.point.getX(),this.point.getY());
    }
  }
  
  static class Edge
  {
    final Vertex from;
    final Vertex to;
    final double angle;
    boolean visited = false;
    Edge next = null;
    Edge reverse = null;
    
    public Edge( final Vertex from, final Vertex to )
    {
      this.from = from;
      this.to = to;
      this.angle = Math.atan2(to.point.getY() - from.point.getY(), to.point.getX() - from.point.getX());
      from.addEdge( this );
    }
    
    public Vertex getFrom()
    {
      return this.from;
    }

    public Vertex getTo()
    {
      return this.to;
    }

    public void setNextEdge( final Edge edge )
    {
      this.next = edge;
    }

    public void setReverseEdge( final Edge edge )
    {
      this.reverse = edge;
    }

    @Override
    public String toString()
    {
      return MessageFormat.format("{0} -> {1}", this.from, this.to);
    }
  }

  public static void main(final String[] args)
  {
    final Vertex[] vertices = {
      new Vertex( 1, new Point2D.Double(-4,+4) ),
      new Vertex( 2, new Point2D.Double(-1,+5) ),
      new Vertex( 3, new Point2D.Double(+3,+4) ),
      new Vertex( 4, new Point2D.Double(+4,+1) ),
      new Vertex( 5, new Point2D.Double(+0,-2) ),
      new Vertex( 6, new Point2D.Double(-1,+3) ),
      new Vertex( 7, new Point2D.Double(+2,+2) )
    };
     
    final int[][] graph = {
      {1, 2}, {1, 6}, {1, 5}, {2, 6}, {2, 3}, {3, 7}, {7, 4}, {3, 4}, {5, 4}, {6, 5}
    };
    
    final Edge[] edges = new Edge[2 * graph.length];

    for ( int i = 0; i < graph.length; i++ )
    {
      final Vertex from = vertices[graph[i][0]-1];
      final Vertex to = vertices[graph[i][1]-1];
      edges[2*i] = new Edge( from, to );
      edges[2*i+1] = new Edge( to, from );
      
      edges[2*i].setReverseEdge(edges[2*i+1]);
      edges[2*i+1].setReverseEdge(edges[2*i]);
    }
    
    
    for ( final Vertex vertex: vertices )
    {
      vertex.sortEdges();
    }
    
    final ArrayList<ArrayList<Edge>> faces = new ArrayList<>();
    for ( final Edge edge: edges )
    {
      if ( edge.visited )
      {
        continue;
      }
      final ArrayList<Edge> face = new ArrayList<>();
      faces.add( face );
      Edge e = edge;
      do
      {
        face.add(e);
        e.visited = true;
        e = e.reverse.next;
      }
      while (e != edge);
      
      System.out.println( face.stream().map(Edge::getFrom).collect(Collectors.toList()) );
    }
  }
}

哪些输出:

[1, 2, 3, 4, 5]
[2, 1, 6]
[6, 1, 5]
[2, 6, 5, 4, 7, 3]
[3, 7, 4]

注意:这包括图形的外表面。
或者，如果您想: 测试图形的平面性；生成(双连通)图的所有可能嵌入；并为这些嵌入中的一个(或多个)生成循环边排序，然后您可以使用博士论文 Planarity Testing by Path Addition ，它在附录中包含完整的 Java 源代码。