Finding if Path2D self-intersects

2019-03-25 13:42发布

I need to find if Path2D intersects itself. For now, I do it by simply extracting an array of lines from path, and finding if any of these intersect. But it has O(n^2) complexity, and so it is very slow. Is there a faster way to do it?

2条回答
Rolldiameter
2楼-- · 2019-03-25 14:05

Here is my Java implementation of this algorithm:

import java.awt.Point;
import java.awt.geom.Line2D;
import java.awt.geom.PathIterator;
import java.util.*;

/**
 * Path2D helper functions.
 * <p/>
 * @author Gili Tzabari
 */
public class Path2Ds
{
    /**
     * Indicates if a Path2D intersects itself.
     * <p/>
     * @return true if a Path2D intersects itself
     */
    public static boolean isSelfIntersecting(PathIterator path)
    {
        SortedSet<Line2D> lines = getLines(path);
        if (lines.size() <= 1)
            return false;

        Set<Line2D> candidates = new HashSet<Line2D>();
        for (Line2D line: lines)
        {
            if (Double.compare(line.getP1().distance(line.getP2()), 0) <= 0)
            {
                // Lines of length 0 do not cause self-intersection
                continue;
            }
            for (Iterator<Line2D> i = candidates.iterator(); i.hasNext();)
            {
                Line2D candidate = i.next();

                // Logic borrowed from Line2D.intersectsLine()
                int lineRelativeToCandidate1 = Line2D.relativeCCW(line.getX1(), line.getY1(), line.
                    getX2(),
                    line.getY2(), candidate.getX1(), candidate.getY1());
                int lineRelativeToCandidate2 = Line2D.relativeCCW(line.getX1(), line.getY1(), line.
                    getX2(),
                    line.getY2(), candidate.getX2(), candidate.getY2());
                int candidateRelativeToLine1 = Line2D.relativeCCW(candidate.getX1(),
                    candidate.getY1(),
                    candidate.getX2(), candidate.getY2(), line.getX1(), line.getY1());
                int candidateRelativeToLine2 = Line2D.relativeCCW(candidate.getX1(),
                    candidate.getY1(),
                    candidate.getX2(), candidate.getY2(), line.getX2(), line.getY2());
                boolean intersection = (lineRelativeToCandidate1 * lineRelativeToCandidate2 <= 0)
                    && (candidateRelativeToLine1 * candidateRelativeToLine2 <= 0);
                if (intersection)
                {
                    // Lines may share a point, so long as they extend in different directions
                    if (lineRelativeToCandidate1 == 0 && lineRelativeToCandidate2 != 0)
                    {
                        // candidate.P1 shares a point with line
                        if (candidateRelativeToLine1 == 0 && candidateRelativeToLine2 != 0)
                        {
                            // line.P1 == candidate.P1
                            continue;
                        }
                        if (candidateRelativeToLine1 != 0 && candidateRelativeToLine2 == 0)
                        {
                            // line.P2 == candidate.P1
                            continue;
                        }
                        // else candidate.P1 intersects line
                    }
                    else if (lineRelativeToCandidate1 != 0 && lineRelativeToCandidate2 == 0)
                    {
                        // candidate.P2 shares a point with line
                        if (candidateRelativeToLine1 == 0 && candidateRelativeToLine2 != 0)
                        {
                            // line.P1 == candidate.P2
                            continue;
                        }
                        if (candidateRelativeToLine1 != 0 && candidateRelativeToLine2 == 0)
                        {
                            // line.P2 == candidate.P2
                            continue;
                        }
                        // else candidate.P2 intersects line
                    }
                    else
                    {
                        // line and candidate overlap
                    }
                    return true;
                }
                if (candidate.getX2() < line.getX1())
                    i.remove();
            }
            candidates.add(line);
        }
        return false;
    }


    /**
     * Returns all lines in a path. The lines are constructed such that the starting point is found
     * on the left (or same x-coordinate) of the ending point.
     * <p/>
     * @param path the path
     * @return the lines, sorted in ascending order of the x-coordinate of the starting point and
     * ending point, respectively
     */
    private static SortedSet<Line2D> getLines(PathIterator path)
    {
        double[] coords = new double[6];
        SortedSet<Line2D> result = new TreeSet<Line2D>(new Comparator<Line2D>()
        {
            @Override
            public int compare(Line2D o1, Line2D o2)
            {
                int result = Double.compare(o1.getX1(), o2.getX1());
                if (result == 0)
                {
                    // Ensure we are consistent with equals()
                    return Double.compare(o1.getX2(), o2.getX2());
                }
                return result;
            }
        });
        if (path.isDone())
            return result;
        int type = path.currentSegment(coords);
        assert (type == PathIterator.SEG_MOVETO): type;
        Point.Double startPoint = new Point.Double(coords[0], coords[1]);
        Point.Double openPoint = startPoint;
        path.next();

        while (!path.isDone())
        {
            type = path.currentSegment(coords);
            assert (type != PathIterator.SEG_CUBICTO && type != PathIterator.SEG_QUADTO): type;
            switch (type)
            {
                case PathIterator.SEG_MOVETO:
                {
                    openPoint = startPoint;
                    break;
                }
                case PathIterator.SEG_CLOSE:
                {
                    coords[0] = openPoint.x;
                    coords[1] = openPoint.y;
                    break;
                }
            }
            Point.Double endPoint = new Point.Double(coords[0], coords[1]);
            if (Double.compare(startPoint.getX(), endPoint.getX()) < 0)
                result.add(new Line2D.Double(startPoint, endPoint));
            else
                result.add(new Line2D.Double(endPoint, startPoint));
            path.next();
            startPoint = endPoint;
        }
        return result;
    }
}
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forever°为你锁心
3楼-- · 2019-03-25 14:10

You can do this faster using the sweep-line algorithm: http://en.wikipedia.org/wiki/Sweep_line_algorithm

Pseudocode:

Each line has a start point and an end point. Say that `start_x` <= `end_x` for all the lines.
Create an empty bucket of lines.
Sort all the points by their x coordinates, and then iterate through the sorted list.
If the current point is a start point, test its line against all the lines in the bucket, and then add its line to the 
bucket.
if the current point is an end point, remove its line from the bucket.

The worst case is still O(N^2), but the average case is O(NlogN)

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