My purpose is to implement the algorithm to check whether a Delaunay edge is Gabriel.
As the definition, a Delaunay triangulation edge is said to be a Gabriel edge if its diametrical circle is empty. So, to check whether it is a Gabriel edge or not, we need to scan through all finite vertices in Delaunay to check if any is in that diametrical circle OR we just need to check on its 2 adjacent triangles. Which is the accurate option?
You only need to check the two adjacent triangles. Suppose the third vertex on one of the neighboring triangles is not inside the diametral ball of an edge (i.e., it suggests that the edge may have the Gabriel property). The empty (by the Delaunay property) circumcircle of this triangle (dashed below) contains a semi-circle corresponding to the Gabriel ball (in gray below). If you check both Delaunay triangles attached to an edge, you know that both halves of the Gabriel ball are empty.
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Given two convex polygons P, Q separated by a line, how can I find their common tangents?
There should be 4 total.
Geometry isn't my strong side so any help will be appreciated :)
Any tangent will be generated from (at least) one point on each polygon. The tangent forms a half-space for each polygon, containing that entire polygon; different tangents are generated by requiring that each polygon be above or below the tangent line.
Whenever you have a problem involving two convex polygons, the algorithm probably involves "pick a point on each one, then iterate." This algorithm is no different.
What you do, basically, is start with a random guess, and refine. Pick a vertex on each polygon and calculate the line through the two vertices. Look at polygon A and see whether either of the two neighboring vertices is on the wrong side of the line. If one is, replace the current vertex with that vertex. Then do the same check on polygon B. Then, if you updated either vertex, repeat. Eventually the lines will converge to tangents with each vertex the extremal one in the desired direction.
I am working with several convex polygons that overlap each other and I need to combine them back together to form one single polygon that may be convex or concave.
The problem is always as follows:
1) The polygons that I need to merge together are always convex.
2) The vertices of each polygon are defined in clockwise order.
3) The polygons are never in any specific order.
4) The final polygon can only be simple convex or concave polygon, i.e. no self-intersection, no duplicate vertices or holes in the shape.
Here is an example of the kind of polygons that I am working with.
![overlapping convex polygons]"image removed")
My current approach is to start from the first polygon and vertex by vertex I loop through all vertices of all of the polygons to find overlap. If there is no overlap, I store the vertex for the final outline and continue.
Upon finding overlapping vertices, I determine which polygon to continue to by measuring the angles of the possible paths and by choosing the one that leads towards the outside of the shape.
This method works until I encounter polygons that do not have vertices overlapping each other, but instead one polygon's vertex is overlapping another polygon's side, as is the case with the rectangle in the image.
I am currently planning on solving these situations by running line intersect checks for all shapes that I have not yet processed, but I am convinced that this cannot be the easiest or the best method in terms of performance.
Does someone know how I should approach this problem in a more efficient manner and/or universal manner?
I solved this issue and I'm posting the answer here in case someone else runs into this issue as well.
My first step was to implement a pre-processing loop based on trincot's suggestions.
I calculated the minimum and maximum x and y bounds for each individual shape.
I used these values to determine all overlapping shapes and I stored a simple array for each shape that I could later use to only look at shapes that can overlap each other.
Then, for the actual loop that determines the outline of the final polygon:
I start from the first shape and simply compare its vertices to those of the nearby shapes. If there is at least one vertex that isn't shared by another vertex, it must be on the outer edge and the loop starts from there. If there are only overlapping vertices, then I add the first shape to a table for all checked shapes and repeat this process with another shape until I find a vertex that is on the outer edge.
Once the starting vertex is found, the main loop will check the vertices of the starting shape one by one and measure how far from the given vertex is from every nearby shapes' edges. If the distance is zero, then the vertex either overlaps with another shape's vertex or the vertex lies on the side of another shape.
Upon finding the aforementioned type of vertex, I add the previous shape's number to the table of checked shapes so that it isn't checked again. Then, I check if there are other shapes that share this particular vertex. If there are, then I determine the outermost shape and continue from there, starting back from step 2.
Once all shapes have been checked, I check that all non-overlapping vertices from the starting shape were indeed added to the outline. If they weren't, I add them at the end.
There may be computationally faster methods, but I found this one to be simple to write, that it meets all of my requirements and it is fast enough for my needs.
Given a vertex, you could speed up the search of an "overlapping" vertex or edge as follows:
Finding vertices
Assuming that the coordinates are exact, in the sense that if two vertices overlap, they have exactly the same x and y coordinates, without any "error" of imprecision, then it would be good to first create a hash by x-coordinate, and then for each x-entry you would have a hash by y-coordinate. The value of that inner hash would be a list of polygons that have that vertex.
That structure can be built in O(n) time, and will allow you to find a matching vertex in constant time.
Only if that gives no match, you would go to the next algorithm:
Finding edges
In a pre-processing step (only once), create a segment tree for these polygons where a "segment" corresponds to a min/max x-coordinate range for a particular polygon.
Given a vertex, use the segment tree to find the polygons that are in the right x-coordinate range, i.e. where the x-coordinate of the vertex is within the min/max range of x-coordinates of the polygon.
Iterate those polygons, and eliminate those that do not have an y-coordinate range that has the y-coordinate of the vertex.
If no polygons remain, the vertex does not participate in any edge of another polygon.
You cannot get more than one polygon here, since that would mean another polygon shares the vertex, which is a case already covered by the hash-based algorithm.
If you get just one polygon, then continue your search by going through the edges of that polygon to find a match -- which is what you already planned on doing (line intersect check), but now you would only need to do it for one polygon.
You could speed that line intersect check up a little bit by first filtering the edges to those that have the right x-range. For convex polygons you would end up with at most two edges. At most one of those two will have the right y-range. If you get such an edge, check whether the vertex is really on that edge.
How can one determine the exact Voronoi sites (cells/regions) from a Delaunay triangulation?
If one has an already constructed delaunay triangulation it is easy to calculate the edges of a voronoi by simply connecting adjacent circum-circle centers of every triangle.
It is also easy to determine the Voronoi points/sites because they are represented by every point of every triangle in the Delaunay triangulation.
However how do you determine that a specific voronoi site goes with a specific list of edges from a delaunay triangulation?
It seems it is simple to get one and the other as separate entities but putting them together is another challenge?
Looking at the diagram below, you can see the Delaunay triangulation along with the dual Voronoi diagram. All that I described can be pictured below for an easy reference. Ignore the green circle as that is just an artifact of this particular reference i took from the web.
If you want polygons from edges pick the midpoint of each edge and the distance to each site then sort the result and pick the first and second (when they are equal) and save them into polygons. For the borders there is of course only 1 edge. Maybe a dupe:Getting polygons from voronoi edges.
It's a bit tricky and hard to visualize. I am little stuck with the borders. Here is the original answer from Alink:How can I get a dictionary of cells from this Voronoi Diagram data?.
Each vertex in the Delaunay triangulation represents a Voronoi site. So to create the cell of a site you take one such triangle t and a vertex v in t. Now compute the Voronoi edges between v and the two remaining vertices of t. Repeat this process by traversing the triangles around v one by one. Assuming you can store the neighbourhood relation between the triangles this should at most take O(k) time, k being the number of adjacent triangles of v.
This converts the Delaunay triangulation into the Voronoi Diagram in O(n) time/space, for n sites. No sorting is required, otherwise what is the point in having the Delaunay triangulation in the first place.
Say I have a polygon. It can be a convex one or not, it doesn't matter, but it doesn't have holes. It also has "inner" vertices and edges, meaning that it is partitioned.
Is there any kind of popular/known algorithm or standard procedures for when I want to check if a point is inside that kind of polygon?
I'm asking because Winding Number and Ray Casting aren't accurate in this case
Thanks in advance
You need to clarify what you mean by 'inner vertices and edges'. Let's take a very general case and hope that you find relevance.
The ray casting (point in polygon) algorithm shoots off a ray counting the intersections with the sides of the POLYGON (Odd intersections = inside, Even = outside).
Hence it accurately gives the correct result regardless of whether you start from inside the disjoint trapezoidal hole or the triangular hole (inner edges?) or even if a part of the polygon is completely seperated and/or self intersecting.
However, in what order do you feed the vertices of the polygon such that all the points are evaluated correctly?
Though this is code specific, if you're using an implementation that is counting every intersection with the sides of the polygon then this approach will work -
- Break the master polygon into polygonal components. eg - trapezoidal hole is a polygonal component.
- Start with (0,0) vertex (doesn't matter whether (0,0) actually lies wrt your polygon) followed by the first component' vertices, repeating its first vertex after the last vertex.
- Include another (0,0) vertex.
- Include the next component , repeating its first vertex after the last vertex.
- Repeat the above two steps for each component.
- End with a final (0,0) vertex.
2 component eg- Let the vertices of the two components be (1x,1y), (2x,2y), (3x,3y) and (Ax,Ay), (Bx,By), (Cx,Cy). Where (Ax,Ay), (Bx,By), (Cx,Cy) could be anything from a disjoint triangular hole, intersecting triangle or separated triangle.
Hence , the vertices of a singular continous polygon which is mathematically equivalent to the 2 components is -
(0,0),(1x,1y),(2x,2y),(3x,3y),(1x,1y),(0,0),(Ax,Ay),(Bx,By),(Cx,Cy),(Ax,Ay),(0,0)
To understand how it works, try drawing this mathematically equivalent polygon on a scratch pad.-
1. Mark all the vertices but don't join them yet.
2. Mark the repeated vertices separately also. Do this by marking them close to the original points, but not on them. (at a distance e, where e->0 (tends to/approaches) ) (to help visualize)
3. Now join all the vertices in the right order (as in the example above)
You will notice that this forms a continuous polygon and only becomes disjoint at the e=0 limit.
You can now send this mathematically equivalent polygon to your ray casting function (and maybe even winding number function?) without any issues.
Is there a computationally efficient way to determine the points of intersection of a straight line with all the edges of a given Voronoi tessellation in a rectangular plane area?
Thanks
Once you have your first intersection point, the rest is easy.
Prepare a database of edges: for each edge, list both polygons it belongs to, or say it is the outer edge (and so belongs to one polygon only). In your picture, the lower side of the rectangle would contain 4 edges of 4 different polygons.
Draw your line, find your first intersection point ([0, 0.25] in your picture, not circled). Say it's polygon A. Then the next intersection point (the lowest one circled in your picture) also belongs to A. You find the relevant edge with a binary search through the list of edges of A.
Now you have found the second edge of A, find out which other polygon it belongs to. Then use binary search to find which other edge of that other polygon the line intersects. And so on until you exit your rectangle.