I am working on image segmentation and I thought the convex hull can provide me with a simple solution to my problem. Currently I have polygons with for sides (see image below). Due to image processing issues, the shape does not have clean straight sides and hence when I use the standard convex hull (in Matlab) I may get more than the four main corners to define it.
My goal is to force the convex hull algorithm to find the best 4 vertices that will enclose my polygons (i.e. 4 best enclosing vertices per polygon). Is this possible? An example code will be appreciated.
Thanks
The problem of the minimum area bounding polygon is briefly mentioned in "Geometric applications of a matrix-searching algorithm" (see Applications section). It is not simple and is probably not the way for you.
For an easier (but approximate) answer to your question, you can consider the four cardinal directions and find the farthest points in these, which define a quadrilateral. (Also consider the four intermediate directions, which are more appropriate for an axis-aligned rectangle.)
If you insist having an enclosing quadrilateral, you can translate the four edges to the farthest points in the respective perpendicular directions, and find the pairwise intersections.
If you insist having a rectangle, compute the convex hull and find the minimum area or minimum perimeter bounding rectangle by the Rotating Calipers method. https://geidav.wordpress.com/tag/rotating-calipers/
Related
Most of the times, I determine contour orientation generating 2D points and computing the closed polygon area. Depending on the area value sign I can understand if the contour is oriented clockwise or not (see How to determine if a list of polygon points are in clockwise order?).
Would it be possible to do the same computations without generating 2D points? I mean, relying only on geometric curve properties?
We are interested in determining the orientation of contours like these ones without sampling them with 2D points.
EDIT: Some interesting solutions can be found here:
https://math.stackexchange.com/questions/423718/general-way-to-find-out-whether-a-curve-is-positively-oriented
Scientific paper: Determining the orientation of closed planar curves, DJ Filip (1990)
How are those geometric curves defined?
Do you have an angle for them? The radius doesn't matter, only the difference between entry-angle and exit-angle of each curve.
In that case, a trivial idea crossing my mind is to just sum up all the angles. If the result is positive, you know you had more curves towards the right meaning it's a clockwise contour. If it was negative, then more curves were leftwards -> anti-clockwise contour. (assuming that positive angels determine a right-curve and vica versa)
After thinking about this for awhile, for polygons that contain arcs I think there are three ways to do this.
One, is to break the arcs into line segments and then use the area formula as described above. The success of this approach seems to be tied to how close the interpolation of the arcs is as this could cause the polygon to intersect itself.
A quicker way than the above would be to do the interpolation of the arcs and then find a vertex in the corner (minimal Y, if tie minimal X) and use the sign of the cross product for that vertex. Positive CCW, negative CW. Again, this is still tied to the accuracy of the interpolation.
I think a better approach would be to find the midpoint of the arc and create two line segments, one from the beginning of the arc to the midpoint and another from the midpoint to the end of the arc and replace the arc with these line segments. Now you have a polygon with only line segments. Then you can add up all the normalized cross products of all the vertices. The sign will tell you the direction. Positive is counter-clockwise, negative is clockwise. In this case it doesn't matter if the polygon self-intersects.
Shown Figure (1) is a typical Delaunay triangulation (blue) and it has a boundary line (black rectangle).
Each vertex in the Delaunay triangulation has a height value. So I can calculate the height inside convex hull. I am figuring out a method to calculate the height up to the boundary line (some sort of extrapolation).
There are two things associated with this task
Triangulate up to the boundary point
Figuring out the height at newly created triangle vertices
Anybody come across this issue?
Figure 1:
I'd project the convex hull points of the triangulation to the visible box segments and then insert the 4 box corners and the projected points into the triangulation.
There is no unique correct way to assign heights to the new points. One easy and stable method would be to assign to each new point the height of the closest visible convex hull vertex. Be careful with extrapolation: Triangles of the convex hull tend to have unstable slopes, see the large triangles in front of the below terrain image. Their projection to the xy plane has almost 0 area but due to the height difference they are large and almost 90 degrees to the xy plane.
I've had some luck with the following approach:
Find the segment on the convex hull that is closet to the extrapolation point
If I can drop a perpendicular onto the segment, interpolate between the two vertices of the segment.
If I can not construct the perpendicular, just use the closest vertex
This approach results in a continuous surface, but does not provide 1st derivative continuity.
You can find some code that might be helpful at TriangularFacetInterpolator.java. Look for the interpolateWithExteriorSupport method.
I'm working on a science project. I have a list of xyz-coordinates of voronoi diagram vertices, and a list of points that create my protein's convex-hull (from triangulation). Now some of the vertices lie quite far from the hull and I'd like to filter them out. How can I do it in c++ ? For now it doesn't have to be super optimised, I'm only focused on removing those points.
visualization
I was also thinking to check if a line from voronoi vertex(red crosses) to center of protein(pink sphere) is intersecting with hull face at any point, but I'm not sure how to achive that.
I've read that you can check if a point is inside a polygon by counting the times an infinite line from the point is crossing the hull, but that was for two dimensions. Can similar approach be implemented to suit my needs ?
https://www.geeksforgeeks.org/how-to-check-if-a-given-point-lies-inside-a-polygon/
Let's start with the two-dimensional case. You can find the solution to that on CS.SX:
Determine whether a point lies in a convex hull of points in O(logn)
The idea is that each two consecutive points on the convex hull define a triangular slice of the plane (to some internal point of the hull); you can find the slice in which your point is located, and check whether it's closer to the internal point than the line segment bounding the slide.
For the three-dimensional case, I'm guessing you should be able to do something similar, but the search for the 3 points forming the relevant triangle might be a little tricky. In particular, it would also depend on how the convex hull is represented - as in the 2-d case the convex hull is just a sequence of consecutive points on a cycle.
I know this doesn't quite solve your problem but it's the best I've got given what you've written...
I have a set of points(clusters) in higher dimension (30d to 100d). I need to identify concave hull of these points in an efficient manner.
Is there a way to do get the exact concave hull or atleast approximate concave hull of these set of points?
Further, if we have a set of points identified as a border point, is there a way to verify whether the points are actually border points?
In 100d, almost every point will be on the convex hull.
Just remember that a rectangle in 2d has 4 corners, but in 100d, it has 2^100 corners.
As an extremely rough approximation, take the minimum and maximum along each axis. If it is unique, the point is on the hull. For additional points, you can sample some random projections.
But again, the expected behaviour is that almost every point is on the hull, because it is the smallest or largest in some linear combination of features.
I have a 3D point cloud that I've transformed into a Delaunay triangulation using Matlab's function DelaunayTri. Now I have a test point in 3D and want to compute, in Matlab, the smallest distance between this point and the triangulation.
So far, I've thought of using the nearestNeighbor(...) member function of the DelaunayTri class in Matlab to find the point in the triangulation closest to my test point and then computing the distance between them. That is something but it is not what I really want.
The closest point on the triangulation to my test point is, in general, not a vertex of the triangulation but somewhere on the face of a triangle. How can I find this point?
Thank you!!!
I've written codes for these things, but they are not on the File Exchange. I can be convinced to give them out by direct mail though.
It is relatively easy to find the distance to a convex hull, but not trivial. A delaunay tessellation is bounded by the convex hull anyway. So you can convert that tessellation into the convex hull easily enough, or just use the convex hull. Note that a convex hull is often a terribly poor approximation for many purposes, especially if you are using this for color mapping, which is perhaps the most common thing I see this used for. In that case, an alpha shape is a far better choice. Alpha shapes also will have a triangulated boundary surface, though in general it will not be convex.
So, to find the nearest point on a convex triangulation:
Convert to the convex boundary surface, i.e., the convex hull. This reduces to finding those triangles which are not shared between pairs of tetrahedra. Internal facets will always appear exactly twice in the list of all facets. This trick also works for non-convex tessellations of course, so for alpha shapes.
Compute a bounding circumcircle for each triangular surface facet. This allows you to know when to stop checking facets.
Get the distances to each point on the surface. Sort each facet by the distance to the nearest point in that facet. Start by looking at the nearest facet in that list.
Compute the distance to the apparently nearest facet found in step 3. A simple solution is found by least distance programming (LDP), which can be converted to a constrained linear least squares. Lawson & Hanson have an algorithm for this.
Repeat step 4 until the current best distance found is less than the distance, comparing it to any of the circumcircles from step 2. This loop will be quite short really, at least for a convex hull. For a more general non-convex hull from an alpha shape, it may take more time.
You can also reduce the search space a bit by excluding the facets from your search that point AWAY from the point in question. Use those facet normals for this test.
I wrote the tool point2trimesh for this problem. It's kind of a "brute force" solution which works also for non-convex surfaces.