sorry if this question might seem simple but I couldn't figure it out.
Imagine two arbitrary Rectangles that are randomly overlaped so that their outlines intersect at only two locations. Now you cut the area of overlap out of Rectangle 1.
This "bitten Rectangle 1" has now the following points (vertices): (1) All points of Rectangle 1 that lie outside of Rectangle 2, (2) All points of Rectangle 2 that lie inside Rectangle 1 and (3) the two intersection points.
The problem know is the following: How can I get the order of the new points so that the functions plot(...) or fill(...) would draw the right "Bitten Rectangle 1"?
What I did so far:
I determined the convex hull of all the points that lie outside of Rectangle 2 + the intersection points. Then one has to add the new points that lie inside rectangle 1 (due to overlap with Rectangle 2) between the indices of the first and second intersection point also in the right order.
The problem with this convex hull approach is that it only works if the intersection points lie on different lines of Rectangle 1, because then they are part of the convex hull.
If they lie on the same line they are no longer treated as part of the convex hull.
What I'd need is a method to get the order of all possible points that lie on the convex hull and not only the outer most ones.
Hope anybody can help me...
Thank you in advance,
Patrick
In Matlab, there is an amazing function called polybool that let you do any set operation on polygons: http://www.mathworks.com/help/toolbox/map/ref/polybool.html
All you have to do is to define four arrays describing the rectangles (rect1x, rect1y, rect2x and rect2y) and to call [resultx resulty] = polybool('subtraction', rect1x, rect1y, rect2x, rect2y). The resulting arrays will be describing the "Bitten Rectangle 1".
Related
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'm using canny edge detection to detect the edges of a rope and eliminate the background, then using morphological filters I'm filling these edges then thinning them to be in pixel size. The plot indicates the xy coordinates of the rope. What I need to do is to get the intersection of the scattered data (blue *) with the red circle (get the coordinates of Points (1,2,3,4).
Then I need to get the whole points coordinates from the point of intersection (point1,2,3,4) to the center,grouped as A,B,C,D.
The center of the circle, the origin of the axes, and the radius are all known.
I've tried some Matlab functions to get the four intersections points like
InterX,intersections
and also I tried to manually find the intersections
idx=find(y1-y2<eps);
but no method gives me the 4 intersection points.
Thank you in Advance
You'll need a thick circle. I presume (from your previous question) that the rope points are at contiguous integer coordinates. Using a thick circle (donut) with a width of 1 ensures you'll find at least one point in each rope end. Connected component analysis will then tell you which of these points belong to the same rope end.
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/
Given an binary mask with an object in Matlab. I am going to find the concavity point of the object boundary. The concavity point I mean here is the deepest concavity point with respect to the Euclidean distance to the convex hull chords K_1, K_2 ,and K_3 in the concavity regions B_1, B_2, B_3, respectively. The red dot indicates the concavity point I want to find, where in concavity region B_1 I draw three lines perpendicular to the chord K_1, the deepest concavity point is the middle one since it has the largest length.
Anyone have efficient way/code to do that? Thanks.
Another figure below gives an example with the convex hull, where the red dot indicates the valid concavity point.
Efficient is relative...
How about computing the convex hull (there are standard algorithms for it) and then shrinking it until it is completely inside the object boundaries. The last point touching is your desired concavity point.
Alternative strategy:
calculate convex hull
find all differences between convex hull and object boundary (have to be straight lines, K1 K2 K3 in your case)
for every line, rotate image such that line is horizontal
take the lowest pixel of the object boundary below the line