A naive approach is to find, for each edge in the polygon, the point on that edge closest to the given point, and then take the one that's closest. Is there a faster algorithm? My goal is to implement a 2D Super Mario Galaxy-style platformer.
Apparently this can be done with Voronoi regions, as in this video: http://www.youtube.com/watch?v=Ldh2YKobuWo
However, I can't find any Voronoi algorithms that deal with edges as well as points. Ideas?
Calculate the point-line distance for each of the edges, then pick the shortest one. There is no shortcut. This site has a good explanation and even implementations in various languages.
However, finding "the point on that edge closest to the given point" is a computationally unnecessary intermediate result.
If the polygon is convex, then the overhead of the voronoi calculation far exceeds that of the naive approach.
If this is run many times, and each time the point changes slightly, you only need to check 3 segments (think about it: as you move around, assuming many checks, then the closest edge will only change to an adjacent edge)
Related
I would like to find two vertices of two meshes (1 vertex per mesh) that define the closest distance between them. Or the two triangles would be fine I guess.
However I'm not sure how to search for this in CGAL's documentation, I'm sure that this is doable with some existing tool (probably based on a 3d distance field and/or AABBs). Could I please get a hint (keywords/link) on what to look for?
I've been pointed to the Optimal Distances CGAL package, but it's not exactly what I want, since it outputs the distance and the coordinates, so finding the vertex ID is an additional computational overhead.
I've already implemented a collision detection with CGAL to find the triangle-triangle intersection in a triangle-soup, using AABB-trees. I guess that I should be somehow close to this, although now a simple soup with all me object-triangles wouldn't do the job.
The solution found was this:
CGAL's Optimal Distances package can give an approximation of the closest distance between the convex hulls of two meshes, without explicitly computing the hulls. As a result one gets the shortest distance between these hulls, and the coordinates of the 2 points that lie on them and define this distance.
Then these coordinates can be used as a search-query in kd-trees that contains the original vertices of the meshes in order to find the closest vertices.
In case one mesh is non-convex, the hull that CGAL is using is very approximate, so convex decomposition might be necessary. In such a case one would have to check distances for each convex part and then take the shortest distance.
The above would result in something like this:
enter link description here
I am currently wondering if there is a common algorithm to check whether a set of plane polygones, not nescessarily triangles, contruct a watertight polyhedra. Each polygon has an oriantation (normal vector). A simple solution would just be to say yes or no. A more advanced version would be to point out the edges, where the polyhedron is "open". I am not really interesed on how to close to polyhedra.
I would like to point out, that my "holes" are not nescessarily small, e.g., one face of a cube might be missing. Thus, the "undersampling correction" algorithms dont seem to be the correct approach. Furthermore, I am talking of about 100 - 1000, not 1000000 polygons, so computation time should not really be a problem.
Any hints or tips?
kind regards,
curator
I believe you can use a simple topological test -- count the number of times each edge appears in the full list of polygons.
If the set of polygons define the surface of a closed volume, each edge should have count>=2, indicating that each edge is shared by (at least) two adjacent polygons. If the surface is manifold count==2 exactly.
Edges with count==1 indicate open regions of the surface.
The above answer does not cover many cases. A more correct (but not necessarily complete: I wouldn't know) algorithm is to ensure that every edge of every polygon (or of the mesh/polyhedron) has an even number of faces connected to it. Consider the following mesh:
The segment (line) between the closest vertex and the one below is attached to 3 faces (one one of the outer triangle and two of the inner triangle), which is greater than two faces. However this is clearly not closed.
Currently I am looking for an efficient algorithm to compute the intersection of two triangle meshes. I have searched over the internet, but haven't found valuable materials. The book Real-Time Collision Detection is a helpful book but is too complex for my task. I also found the post:Triangle to triangle collision detection in 3D. However I hope to find a detailed description about the algorithm.
Regards
Jogging
Well it depends on meshes size, testing each triangle in each mesh against the other is only valid in small meshes since it has n^2 complexity.
To work around that most algorithms use
Spatial portioning
first to subdivide the space into smaller ones and then tackles each one separately.
For spatial portioning most algorithms use
OcTrees
or BSPTrees however if you don't need to complicate things you can just subdivide the space into n boxes then check triangle triangle intersection in each box
To detect if a point is in a polygon, you project a line from the point, to infinity, and see how many of polygon's vertices it intersects with... simple enough. My problem is that if the ray intersects the polygon on one of the points, then it is counted as intersecting two segments, and considered outside the polygon. I altered my function to make it only count one of the segments when the ray intersects a point of the polygon, but there are cases where a line could intersect the point while still being outside as well. Take this image as an example:
If you assume the point in the top left is "infinity", and cast a ray to either of the other points, both intersect at a point of the polygon, and would count as intersecting the same number of vertices even though one is inside, and one is outside.
Is there a way to compensate for that, or do I just have to assume that those fringe cases won't pop up?
If the ray crosses a side exactly on a vertex, only count that side if the other vertex is above the ray. That will fix your corner case.
For example in the picture you posted, the lower ray crosses two sides of the square at the top-left vertex, but one side is above the ray and the other below, so that contributes 1 and the target point is found to be inside. The upper ray crosses two sides at the top-right vertex, both sides are below the ray, so they contribute 0 to the count and the target point is found to be outside.
Update:
I remembered reading an article which describes a technique for dealing with singular cases in general. Please read my other answer if interested.
While my first answer should do the trick for this simple problem, I can't help but mention that there exist general techniques for dealing with these kinds of special cases.
This article describes a technique for dealing with these kinds of issues in general. And one of the first examples they provide happens to be the algorithm you ask about!
The idea is to apply Automatic differentiation aka Dual numbers to compute symbolic perturbations.
By the way the same technique can also be used to avoid handling 0/0 as a special case in programs!
Here is the blog post I originally learned this from, it gives some great background to the technique, and the author blogs a lot about automatic differentiation (AD).
Despite appearances AD is a very practical technique especially in languages with good support for operator overloading (eg: C++, Haskell, Python ...) and I have used it in "real life" (industrial applications in C++).
Send ray in another direction.
If you try n+1 different directions (n is number of polygon points) one of them surely will not pass through any vertex.
This will simplify the code compared to consideration of corner cases.
Worst case becomes O(n)*CheckComplexity(n) which is likely O(n^2). If it's not acceptable, you can just sort all vertices by direction from the point to them and select middle of some interval. This will give O(n*log n).
I have a set of 2D points inside a finite 2D region of space (let's say a world-aligned rectangle to keep things simple for now). What would be an exceedingly efficient way to insert a new point into the set that has a relatively large distance to its new closest neighbour?
I could slowly build a Delaunay triangulation and limit my search to the largest triangles only, but I was hoping someone has a different (better) idea.
Goodwill,
David
Edit:
Forgot to mention that I need to do this thousands of times, every time taking all the previous points into account. I'm looking for an algorithm that doesn't slow down to a crawl as my point set grows.
Use the Bowyer-Watson or other incremental algorithm to maintain the Voronoi diagram. The vertexes of the Voronoi diagram are candidate points, keep all the candidate points in a priority queue ordered by distance to the source points. That should be pretty fast, and optimal (at least, optimal at each step).
Were you looking for something faster and less optimal?