Minimize Polygon Vertices - graphics

What is a good algorithm for reducing the number of vertices in a polygon without changing the way it looks very much?
Input: A polygon, represented as a list of points, with way too many verticies: raw input from the mouse, for example.
Output: A polygon with much fewer verticies that still looks a lot like the original: something usable for collision detection, for example (not necessarily convex).
Edit: The solution to this would be similar to finding a multi-segmented line of best fit on a graph. It's called Segmented Least Squares in my algorithms book.
Edit2: The Douglas Peucker Algorithm is what I really want.

Edit: Oh look, Simplifying Polygons
You mentioned collision detection. You could go really simple and calculate a bounding convex hull around it.
If you cared about the concave areas, you can calculate a concave hull by taking the centroid of your polygon, and choosing a point to start. From the starting point rotate around the centroid, finding each vertex you want to keep, and assigning that as the next vertex in the bounding hull. The complexity of the algorithm would come in how you determined which vertices to keep, but I'm sure you thought of that already. You can throw all your vertices into buckets based on their location relative to the centroid. When a bucket gets more than an arbitrary number of vertices full, you can split it. Then take the mean of the vertices in that bucket as the vertex to use in your bounding hull. Or, forget the buckets, and when you're moving around the centroid, only choose a point if its more than a given distance from the last point.
Actually, you could probably just use all the vertices in your polygon as "cloud of points" and calculate the concave hull around that. I'll look for an algorithm link. Worst case on this would be a completely convex polygon.
Another alternative is to start with a bounding rectangle. For each vertex on the rectangle, find the distance from the point to the polygon. For the farthest vertex, split it into two more vertices and move them in some. Repeat until some proportion of either vertices or area is met. I'd have to think about the details of this one a little more.
If you care about the polygon actually looking similar, even in the case of a self-intersecting polygon, then another approach would be required, but it doesn't sound like thats necessary since you asked about collision detection.
This post has some details about the convex hull part.

There's a lot of material out there. Just google for things like "mesh reduction", "mesh simplification", "mesh optimization", etc.

Related

Polygon triangulation with collinear points?

I have a polygon with collinear points. I want to triangulate the polygon while retaining all the collinear points, since I require those vertices on the generated meshes. Currently I tried to use poly2tri, but it doesn't support collinear points. Is there a polygon triangulation algorithm which support collinear points?
Try moving the points slightly, so they are not colinear. Doing the meshing and then moving the points back.
I tried running the meshing algorithm, then perturbing all the points it missed, then running it again.
It can be quite slow but it does work.
A simple polygon with collinear vertices presents a serious problem for the "ear splitting" triangulation. The best example to see this problem - sketch yourself a 5-point star. Pick any of the points as the convex vertex for a candidate ear.
The diagonal for this ear lies completely within the polygon, and doesn't form "X" intersections with any of the other edges, so by the criteria for a "valid diagonal", it should work. But, notice that once you slice off the ear, the remaining polygon is no longer a simple polygon - because it contains two vertices where the edges meet at a straight (180 deg.) angle. That alone disqualifies the new sub-polygon as "simple", and it will crash any attempt to continue triangulation on it using ear-splitting. I will post a separate question on how best to triangulate these ill-conditioned polys.

Closest distance + vertices of two meshes

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

Convex hull of parallel lines

I have arbitrary many lines in 3D space which are all parallel to each other. Now I want to find the convex hull of these lines. To illustrate this, I've drawn a picture:
I know the start- and endpoints of all lines (the blue dots). The lines are not equally long. If a viewer looks in the direction of the lines (marked as the viewer direction in the pic) he sees only the dots. Now I want to find the convex hull of these dots. Hopefully its clear what I mean.
My idea was to project the start or endpoints on a plane which is perpendicular to the line's direction. After that I can apply some kind of convex hull algorithm to these points. But I have no idea how.
Your idea is exactly correct. One way to accomplish this is to define a vector v along your viewing direction, and then rotate v to the z-axis. The same rotation will convert lines to vertical lines. Then drop the z-coordinate of the endpoints to get your projected points. Then compute the convex hull. There are hull algorithms all over the web, including my own here.
Here's a suggestion based on the calculus of variations.
Consider enclosing your collection of parallel line segments in a simple closed curve minimizing the area of the curve given the constraint that it has to enclose all your segments.
Your "curve" is going to be piecewise linear, so there you might be able to use a P.W basis function in the iterations, though it's possible that you could run into some singularities when the algorithm needs to drop a segment.

Given an irregular polygon's vertex list, how to create internal triangles to build a flat 3D mesh efficiently?

I'm using Unity, but the solution should be generic.
I will get user input from mouse clicks, which define the vertex list of a closed irregular polygon.
That vertices will define the outer edges of a flat 3D mesh.
To procedurally generate a mesh in Unity, I have to specify all the vertices and how they are connected to form triangles.
So, for convex polygons it's trivial, I'd just make triangles with vertices 1,2,3 then 1,3,4 etc. forming something like a Peacock tail.
But for concave polygons it's not so simple.
Is there an efficient algorithm to find the internal triangles?
You could make use of a constrained Delaunay triangulation (which is not trivial to implement!). Good library implementations are available within Triangle and CGAL, providing efficient O(n*log(n)) implementations.
If the vertex set is small, the ear-clipping algorithm is also a possibility, although it wont necessarily give you a Delaunay triangulation (it will typically produce sub-optimal triangles) and runs in O(n^2). It is pretty easy to implement yourself though.
Since the input vertices exist on a flat plane in 3d space, you could obtain a 2d problem by projecting onto the plane, computing the triangulation in 2d and then applying the same mesh topology to your 3d vertex set.
I've implemented the ear clipping algorithm as follows:
Iterate over the vertices until a convex vertex, v is found
Check whether any point on the polygon lies within the triangle (v-1,v,v+1). If there are, then you need to partition the polygon along the vertices v, and the point which is farthest away from the line (v-1, v+1). Recursively evaluate both partitions.
If the triangle around vertex v contains no other vertices, add the triangle to your output list and remove vertex v, repeat until done.
Notes:
This is inherently a 2D operation even when working on 3D faces. To consider the problem in 2D, simply ignore the vector coordinate of the face's normal which has the largest absolute value. (This is how you "project" the 3D face into 2D coordinates). For example, if the face had normal (0,1,0), you would ignore the y coordinate and work in the x,z plane.
To determine which vertices are convex, you first need to know the polygon's winding. You can determine this by finding the leftmost (smallest x coordinate) vertex in the polygon (break ties by finding the smallest y). Such a vertex is always convex, so the winding of this vertex gives you the winding of the polygon.
You determine winding and/or convexity with the signed triangle area equation. See: http://softsurfer.com/Archive/algorithm_0101/algorithm_0101.htm. Depending on your polygon's winding, all convex triangles with either have positive area (counterclockwise winding), or negative area (clockwise winding).
The point-in-triangle formula is constructed from the signed-triangle-area formula. See: How to determine if a point is in a 2D triangle?.
In step 2 where you need to determine which vertex (v) is farthest away from the line, you can do so by forming the triangles (L0, v, L1), and checking which one has the largest area (absolute value, unless you're assuming a specific winding direction)
This algorithm is not well defined for self-intersecting polygons, and due to the nature of floating point precision, you will likely encounter such a case. Some safeguards can be implemented for stability: - A point should not be considered to be inside your triangle unless it is a concave point. (Such a case indicates self-intersection and you should not partition your set along this vertex). You may encounter a situation where a partition is entirely concave (i.e. it's wound differently to the original polygon's winding). This partition should be discarded.
Because the algorithm is cyclic and involves partitioning the sets, it is highly efficient to use a bidirectional link list structure with an array for storage. You can then partition the sets in 0(1), however the algorithm still has an average O(n^2) runtime. The best case running time is actually a set where you need to partition many times, as this rapidly reduces the number of comparisons.
There is a community script for triangulating concave polygons but I've not personally used it. The author claims it works on 3D points as well as 2D.
One hack I've used in the past if I want to constrain the problem to 2D is to use principal component analysis to find the 2 axes of greatest change in my 3D data and making these my "X" and "Y".

triangle points around a point

I have given the coordinates of 1000 triangles on a plane (triangle number (T0001-T1000) and its coordinates (x1,y1) (x2,y2),(x3,y3)). Now, for a given point P(x,y), I need to find a triangle which contains the point P.
One option might be to check all the triangles and find the triangle that contain P. But, I am looking for efficient solution for this problem.
You are going to have to check every triangle at some point during the execution of your program. That's obvious right? If you want to maximize the efficiency of this calculation then you are going to create some kind of cache data structure. The details of the data structure depend on your application. For example: How often do the triangles change? How often do you need to calculate where a point is?
One way to make the cache would be this: Divide your plane in to a finite grid of boxes. For each box in the grid, store a list of the triangles that might intersect with the box.
Then when you need to find out which triangles your point is inside of, you would first figure out which box it is in (this would be O(1) time because you just look at the coordinates) and then look at the triangles in the triangle list for that box.
Several different ways you could search through your triangles. I would start by eliminating impossibilities.
Find a lowest left corner for each triangle and eliminate any that lie above and/or to the right of your point. continue search with the other triangles and you should eliminate the vast majority of the original triangles.
Take what you have left and use the polar coordinate system to gather the rest of the needed information based on angles between the corners and the point (java does have some tools for this, I do not know about other languages).
Some things to look at would be convex hull (different but somewhat helpful), Bernoullies triangles, and some methods for sorting would probably be helpful.

Resources