如何为凹多边形生成回声路径

Question

我需要一种算法来绘制任意多边形的回声路径。如果多边形是凸的，问题就很容易解决。要理解我的意思，请看下面的图片，黑色是原始多边形，红色是从原始多边形生成的回声多边形。

d 是给定的回声路径之间的距离

β角度很容易计算知道我们有顶点的坐标

因此，正如您所见，对于每个顶点，我们可以计算 L，从而为下一个回声路径提供新的顶点。

问题是当我们在某个点上有凹多边形时，我们会得到一张难看的自交叉多边形图片。请看看这张照片。

我想要做的是生成没有自交叉部分的回声多边形，即没有带有虚线的部分。算法或 java 代码会很有帮助。谢谢。

编辑

只需添加一段代码，按照评论中的要求为凸多边形生成回声路径。

public List<MyPath> createEchoCoCentral( List<Point> pointsOriginal, float encoderEchoDistance, int appliqueEchoCount){

    List<Point> contourPoints = pointsOriginal;
    List<MyPath> echoPaths = new ArrayList<>();
    for (int round = 0; round < appliqueEchoCount; round++) {

        List<Point> echoedPoints = new ArrayList<>();
        int size = contourPoints.size()+1;//+1 because we connect end to start

        Point previousPoint = contourPoints.get(contourPoints.size() - 1);
        for (int i = 0; i < size; i++) {
            Point currentPoint;
            if (i == contourPoints.size()) {
                currentPoint = new Point(contourPoints.get(0));
            } else {
                currentPoint = contourPoints.get(i);
            }
            final Point nextPoint;
            if (i + 1 == contourPoints.size()) {
                nextPoint = contourPoints.get(0);
            } else if (i == contourPoints.size()) {
                nextPoint = contourPoints.get(1);
            } else {
                nextPoint = contourPoints.get(i + 1);
            }
            if (currentPoint.x == previousPoint.x && currentPoint.y == previousPoint.y) continue;
            if (currentPoint.x == nextPoint.x && currentPoint.y == nextPoint.y) continue;

            // signs needed o determine to which side of polygon new point will go
            float currentSlope = (float) (Math.atan((previousPoint.y - currentPoint.y) / (previousPoint.x - currentPoint.x)));
            float signX = Math.signum((previousPoint.x - currentPoint.x));
            float signY = Math.signum((previousPoint.y - currentPoint.y));
            signX = signX == 0 ? 1 : signX;
            signY = signY == 0 ? 1 : signY;

            float nextSignX = Math.signum((currentPoint.x - nextPoint.x));
            float nextSignY = Math.signum((currentPoint.y - nextPoint.y));
            nextSignX = nextSignX == 0 ? 1 : nextSignX;
            nextSignY = nextSignY == 0 ? 1 : nextSignY;

            float nextSlope = (float) (Math.atan((currentPoint.y - nextPoint.y) / (currentPoint.x - nextPoint.x)));
            float nextSlopeD = (float) Math.toDegrees(nextSlope);

            //calculateMidAngle - is a bit tricky function that calculates angle between two adjacent edges
            double S = calculateMidAngle(currentSlope, nextSlope, signX, signY, nextSignX, nextSignY);
            Point p2 = new Point();

            double ew = encoderEchoDistance / Math.cos(S - (Math.PI / 2));
            p2.x = (int) (currentPoint.x + (Math.cos(currentSlope - S)) * ew * signX);
            p2.y = (int) (currentPoint.y + (Math.sin(currentSlope - S)) * ew * signX);

            echoedPoints.add(p2);
            previousPoint = currentPoint;


        }

        //createPathFromPoints just creates MyPath objects from given Poins set
        echoPaths.add(createPathFromPoints(echoedPoints));
        //remove last point since it was just to connect end to first point
        echoedPoints.remove(echoedPoints.size() - 1);
        contourPoints = echoedPoints;
    }
    return echoPaths;
}

Answer 1

您正在寻找 straight skeleton：

^{（图片来自Wikipedia。）}

Answer 2

好的，找到了一个可以满足我需要的库。它叫 Clipper

如果有人感兴趣，还有 java 实现 here。

用 Java 库几行代码就可以了

    Path originalPath = new Path();
    for (PointF areaPoint:pointsOriginal){
        originalPath.add(new LongPoint((long)areaPoint.x, (long)areaPoint.y));
    }
    final ClipperOffset clo = new ClipperOffset();
    Paths clips = new Paths();
    Paths solution = new Paths();
    clips.add(originalPath);
    clo.addPaths( clips, Clipper.JoinType.SQUARE, Clipper.EndType.CLOSED_LINE );
    float encoderEchoDistance = (float) UnitUtils.convertInchOrMmUnitsToEncoderUnits(this, inchOrMm, appliqueEchoDistance);
    clo.execute( solution, encoderEchoDistance );
    // Now solution.get(0) will contain path that has offset from original path
    // and what is most important it will not have self intersections.

它是开源的，因此我将深入研究实现细节。感谢所有试图提供帮助的人。

Answer 3

这个问题叫做计算多边形偏移。有两种常见的方法可以解决这个问题：

1）最有效的方法是通过计算绕组数来计算偏移多边形（据我了解，该算法被 Clipper 库使用）

2) 计算直骨架图，帮助您构建偏移多边形

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