Algorithm 如何为凹多边形生成回波路径
我需要一个算法来绘制任意多边形的回波路径。如果多边形是凸的,这个问题很容易解决。为了理解我的意思,请看下面的图片,黑色是原始多边形,红色是原始多边形生成的回声多边形。Algorithm 如何为凹多边形生成回波路径,algorithm,path,polygon,Algorithm,Path,Polygon,我需要一个算法来绘制任意多边形的回波路径。如果多边形是凸的,这个问题很容易解决。为了理解我的意思,请看下面的图片,黑色是原始多边形,红色是原始多边形生成的回声多边形。 d是给定的回波路径之间的距离 β角度很容易计算出已知顶点的坐标 正如您所看到的,对于每个顶点,我们可以计算L,从而为下一个回波路径创建新的顶点 问题是当我们在某个点上有凹多边形时,我们会得到一张丑陋的自交多边形的图片。请看这张照片。 我想做的是生成没有自交部分的回声多边形,即没有带虚线的部分。一个算法或java代码将非常有用。
d
是给定的回波路径之间的距离
β
角度很容易计算出已知顶点的坐标
正如您所看到的,对于每个顶点,我们可以计算L
,从而为下一个回波路径创建新的顶点
问题是当我们在某个点上有凹多边形时,我们会得到一张丑陋的自交多边形的图片。请看这张照片。
我想做的是生成没有自交部分的回声多边形,即没有带虚线的部分。一个算法或java
代码将非常有用。谢谢
编辑
只是添加一段代码,按照注释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;
}
public List createEchoCoCentral(列表点原始、浮点编码循环距离、整数贴花循环计数){
列出轮廓点=点或原始点;
List echopath=new ArrayList();
for(整数舍入=0;舍入<贴花计算;舍入++){
列表回显点=新建ArrayList();
int size=contourPoints.size()+1;//+1,因为我们从一端连接到另一端
Point previousPoint=contourPoints.get(contourPoints.size()-1);
对于(int i=0;i
您正在寻找:
(图片来源于。)
好的,找到了一个可以满足我需要的库。它叫 如果有人感兴趣,还可以使用java实现 使用Java库,几行代码就可以做到这一点
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,所以我将深入了解实现细节。感谢所有尝试帮助的人。此问题称为计算多边形偏移。解决此问题有两种常见方法: 1) 最有效的方法是通过计算缠绕数来计算偏移多边形(据我所知,Clipper库使用此算法) 2) 计算直骨架图,帮助您构建偏移多边形 有关此主题的有趣文章: Chen,通过计算缠绕数进行多边形偏移
Felkel计算直线骨架的算法你能给我们展示一下你的凸多边形算法/代码吗?它太开放了,看不到你是如何实现的。骨架可以帮助某人填补空白。@AndrewCheong这是一段有点长的代码,但一般来说,我只需检查所有顶点,根据该角度计算两条相邻边之间的角度,找到一个距离顶点等于L的点,然后将该点添加到下一条路径的点列表中。一旦我靠近我的PCSo,我会在后面添加代码,你不想在下面的示例中的“C”中包含孤立的三角形吗?@samgak我不想包含带虚线的部分,它也会淡出一点。但是,我已经找到了解决方案。请参阅相关QAs:和