I want to create a set of N random convex disjoint polygons on a plane where each polygon must have at most V vertices, where N and V are parameters for my function, and I'd like to obtain a distribution as close as possible to uniform (every possible set being equally probable). Also I need to randomly select two points on the plane that either match with one of the vertices in the scene or are in empty space (not inside a polygon).
I already implemented for other reasons in the same programming language an AABB tree, Separating Axis Theorem-based collision detection between convex polygons and I can generate a random convex polygon with arbitrary amount of vertices inside a circle of given radius. My best bet thus far:
Generate a random polygon using the function I have available.
Query the AABB tree to test for interception with existing polygons.
If the AABB tree query returns empty set I push the generated polygon into it, otherwise I test with SAT against all the other polygons whose AABB overlaps with the generated one's. If SAT returns "no intersection" I push the polygon, otherwise I discard it.
Repeat from 1 until N polygons are generated.
Generate a random number in {0,1}
If the generated number is 1 I pick a random polygon and a random vertex on it as a point
If the generated number is 0 I generate a random position in (x,y) and test if it falls within some polygon (I might create a tiny AABB around it and exploit the AABB tree to reduce the required number of PiP tests). In case it's not inside any polygon I approve it as a valid point, otherwise I repeat from 5.
Repeat from 5 once more to get the second point.
I think the solution would possibly work, but unfortunately there's no way to guarantee that I can generate N such polygons for very large N, or find two good points in an acceptable time, and I'm programming in React, where long operations run on the main thread blocking the UI till they end. I could circumvent the issue by ejecting from create-react-app and learn Web Workers, which would require probably more time than it's worth for me.
This is definitely non-uniform distribution, but perhaps you could begin by generating N points in the plane and then computing the Voronoi diagram for those points. The Voronoi diagram can be computed in O(n log n) time with Fortune's algorithm. The cells of the Voronoi diagram are convex, so you can then construct a random polygon of the desired number of vertices that lies within each cell of the diagram.
By Balu Ertl - Own work, CC BY-SA 4.0, Link
Ok, here is another proposal. I have little knowledge of js, but could cook up something in Python.
Use Poisson disk sampling with distance parameter d to generate N samples of the centers
For a given center make a circle with R≤d.
Generate V angles using Dirichlet distribution such that sum of angles is equal to 2π. Sort them.
Place vertices on the circle using angles generate at step#3 and connect them. This would be be your polygon
UPDATE
Instead using Poisson disk sampling for step 1, one could use Sobol quasi-random sequences. For N points sampled in the 1x1 box (well, you have to scale it afterwards), least distance between points would be
d = 0.5 * sqrt(D) / N,
where D is dimension of the problem, 2 in your case. So radius of the circle for step 2 would be 0.25 * sqrt(2) / N. This ties nicely N and d.
https://www.sciencedirect.com/science/article/abs/pii/S0378475406002382
Related
I have multiple faces in 3D space creating cells. All these faces lie within a predefined cube (e.g. of size 100x100x100).
Every face is convex and defined by a set of corner points and a normal vector. Every cell is convex. The cells are result of 3d voronoi tessellation, and I know the initial seed points of the cells.
Now for every integer coordinate I want the smallest distance to any face.
My current solution uses this answer https://math.stackexchange.com/questions/544946/determine-if-projection-of-3d-point-onto-plane-is-within-a-triangle/544947 and calculates for every point for every face for every possible triple of this faces points the projection of the point to the triangle created by the triple, checks if the projection is inside the triangle. If this is the case I return the distance between projection and original point. If not I calculate the distance from the point to every possible line segment defined by two points of a face. Then I choose the smallest distance. I repeat this for every point.
This is quite slow and clumsy. I would much rather calculate all points that lie on (or almost lie on) a face and then with these calculate the smallest distance to all neighbour points and repeat this.
I have found this Get all points within a Triangle but am not sure how to apply it to 3D space.
Are there any techniques or algorithms to do this efficiently?
Since we're working with a Voronoi tessellation, we can simplify the current algorithm. Given a grid point p, it belongs to the cell of some site q. Take the minimum over each neighboring site r of the distance from p to the plane that is the perpendicular bisector of qr. We don't need to worry whether the closest point s on the plane belongs to the face between q and r; if not, the segment ps intersects some other face of the cell, which is necessarily closer.
Actually it doesn't even matter if we loop r over some sites that are not neighbors. So if you don't have access to a point location subroutine, or it's slow, we can use a fast nearest neighbors algorithm. Given the grid point p, we know that q is the closest site. Find the second closest site r and compute the distance d(p, bisector(qr)) as above. Now we can prune the sites that are too far away from q (for every other site s, we have d(p, bisector(qs)) ≥ d(q, s)/2 − d(p, q), so we can prune s unless d(q, s) ≤ 2 (d(p, bisector(qr)) + d(p, q))) and keep going until we have either considered or pruned every other site. To do pruning in the best possible way requires access to the guts of the nearest neighbor algorithm; I know that it slots right into the best-first depth-first search of a kd-tree or a cover tree.
I'm currently working on a project where I want to draw different mathematical objects onto a 3D cube. It works as it should for Points and Lines given as a vector equation. Now I have a plane given as a parametric equation. This plane can be somewhere in the 3D space and may be visible on the screen, which is this 3D cube. The cube acts as an AABB.
First thing I needed to know was whether the plane intersects with the cube. To do this I made lines who are identical to the edges of this cube and then doing 12 line/plane intersections, calculating whether the line is hit inside the line segment(edge) which is part of the AABB. Doing this I will get a set of Points defining the visible part of the plane in the cube which I have to draw.
I now have up to 6 points A, B, C, D, E and F defining the polygon ABCDEF I would like to draw. To do this I want to split the polygon into triangles for example: ABC, ACD, ADE, AED. I would draw this triangles like described here. The problem I am currently facing is, that I (believe I) need to order the points to get correct triangles and then a correctly drawn polygon. I found out about convex hulls and found QuickHull which works in three dimensional space. There is just one problem with this algorithm: At the beginning I need to create a three dimensional simplex to have a starting point for the algorithm. But as all my points are in the same plane they simply form a two dimensional plane. Thus I think this algorithm won't work.
My question is now: How do I order these 3D points resulting in a polygon that should be a 2D convex hull of these points? And if this is a limitation: I need to do this in C.
Thanks for your help!
One approach is to express the coordinates of the intersection points in the space of the plane, which is 2D, instead of the global 3D space. Depending on how exactly you computed these points, you may already have these (say (U, V)) coordinates. If not, compute two orthonormal vectors that belong to the plane and take the dot products with the (X, Y, Z) intersections. Then you can find the convex hull in 2D.
The 8 corners of the cube can be on either side of the plane, and have a + or - sign when the coordinates are plugged in the implicit equation of the plane (actually the W coordinate of the vertices). This forms a maximum of 2^8=256 configurations (of which not all are possible).
For efficiency, you can solve all these configurations once for all, and for every case list the intersections that form the polygon in the correct order. Then for a given case, compute the 8 sign bits, pack them in a byte and lookup the table of polygons.
Update: direct face construction.
Alternatively, you can proceed by tracking the intersection points from edge to edge.
Start from an edge of the cube known to traverse the plane. This edge belongs to two faces. Choose one arbitrarily. Then the plane cuts this face in a triangle and a pentagon, or two quadrilaterals. Go to the other the intersection with an edge of the face. Take the other face bordered by this new edge. This face is cut in a triangle and a pentagon...
Continuing this process, you will traverse a set of faces and corresponding segments that define the section polygon.
In the figure, you start from the intersection on edge HD, belonging to face DCGH. Then move to the edge GC, also in face CGFB. From there, move to edge FG, also in face EFGH. Move to edge EH, also in face ADHE. And you are back on edge HD.
Complete discussion must take into account the case of the plane through one or more vertices of the cube. (But you can cheat by slightly translating the plane, constructing the intersection polygon and removing the tiny edges that may have been artificially created this way.)
Given one large cube (axis aligned and on integer coordinates), and many smaller cubes (also axis aligned and on integer coordinates). How can we check that the large cube is perfectly filled by the smaller cubes.
Currently we check that:
For each small cube it is fully contained by the large cube.
That it doesn't intersect any other small cube.
The sum of the volumes of the small cubes equals the volume of the large cube.
This is ok for small numbers of cubes but we need to support this test of cubes with dimensions greater than 2^32. Even at 2^16 the number of small cubes required to fill the large cube is large enough that step 2 takes a while (O(n^2) checking each cube intersects no other).
Is there a better algorithm?
EDIT:
There seems to be some confusion over this. I am not trying to split a cube into smaller cubes. That's already done. Part of our program splits large OpenCL ranges (axis aligned cubes on integer coordinates) into lots of smaller ranges that fit into a hardware job.
What I'm doing is hooking into this system and checking that the jobs it produces correctly cover the large initial range. My algorithm above works, but it's slow and given the amount of tests we have to run I'd like to keep these tests as fast as possible.
We are talking about 3D right?
For 2D one can do a similar (but simpler) process (with, I believe, an O(n log n) running time algorithm).
The basic idea of the below is the sweep-line algorithm.
Note that rectangle intersection can done by checking whether any corner of any cube is contained in any other cube.
You can improve on (2) as follows:
Split each cube into 2 rectangles on the y-z plane (so you'd have 2 rectangles defined by the same set of 4 (y,z) coordinates, but the x coordinates will be different between the rectangles).
Define the rectangle with the smaller x-coordinate as the start of a cube and the other rectangle as the end of a cube.
Sort the rectangles by x-coordinate
Have an initially empty interval tree
(each interval should also store a reference to the rectangle to which it belongs)
For each rectangle:
Look up the y-coordinate of each point of the rectangle in the interval tree.
For each matching interval, look up its rectangle and check whether the point is also contained within the z-coordinates (this is all that's required because the tree only contains x-coordinates in the correct range and we check the y-coordinates by doing the interval lookup).
If it is, we have overlap.
If the rectangle is the start of a cube, insert the 2 y-coordinates of the rectangle as an interval into the interval tree.
Otherwise, remove the interval defined by the 2 y-coordinates from the tree.
The running time is between O(n) (best case) and O(n2) (worst case), depending on how much overlap there is in the x- and y-coordinates (more overlap is worse).
order your insert cubes
insert the biggest insert cube in one of the corners of your cube and split up the remaining cube into subcubes
insert the second biggest insert cube in the first of the sub cubes that will fit and add the remaining subcubes of this subcube to the set of subcubes
etc.
Another go, again only addressing step 2 in the original question:
Define a space-filling curve with good spatial locality, such as a 3D Hilbert Curve.
For each cube calculate the pair of coordinates on the curve for the points at which the curve both enters and leaves the cube. The space-filling curve will enter and leave some cubes more than once, calculate more than one pair of coordinates for these cases.
You've now got I don't know how many pairs of coordinates, but I'd guess no more than 2^18. These coordinates define intervals along the space-filling curve, so sort them and look for overlaps.
Time complexity is probably dominated by the sort, space complexity is probably quite big.
I know this may be a duplicate, but it seems like a variation on the 'Closest pair of Points' algorithm.
Given a Set of N points (x, y) in the unit square and a distance d, find all pair of points such that the distance between them is at most d.
For large N the brute force method is not an option. Besides the 'sweep line' and 'divide and conquer' methods, is there a simpler solution? These pair of points are the edges of an undirected graph, that i need to traverse it and say if it's connected or not (which i already did using DFS, but when N = 1 million it never finishes!).
Any pseudocode, comments or ideas are welcome,
Thanks!
EDIT: I found this on Sedgewick book (i'm looking at the code right now):
Program 3.18 uses a two-dimensional array of linked lists to improve the running time of Program 3.7 by a factor of about 1/d2 when N is sufficiently large. It divides the unit
square up into a grid of equal-sized smaller squares. Then, for each square, it builds a linked list of all the
points that fall into that square. The two-dimensional array provides the capability to access immediately
the set of points close to a given point; the linked lists provide the flexibility to store the points where
they may fall without our having to know ahead of time how many points fall into each grid square.
We are really looking for points that are inside of a circle of center (x,y) and radius d.
The square that encloses circle is a square of center (x,y) and sides 2d. Any point out of this square does not need to be checked, it's out. So, a point a (xa, ya) is out if abs(xa - x) > d or abs (ya -yb) > d.
Same for the square that is enclosed by that circle is a square of center (x,y) and diagonals 2d. Any point out of this square does not need to be checked, it's in. So, a point a (xa, ya) is in if abs(xa - x) < (d * 1.412) or abs(ya -yb) < (d * 1.412).
Those two easy rules combined reduce a lot the number of points to be checked. If we sort the point by their x, filter the possible points, sort by their y, we come to the ones we really need to calculate the full distance.
For any given point, you can use a Manhattan distance (x-delta plus y-delta) heuristic to filter out most of the points that are not within the distance "d" - filter out any points whose Manhattan distance is greater than (sqrt(2) * d), then run the expensive-and-precise distance test on the remaining points.
I do work in theoretical chemistry on a high performance cluster, often involving molecular dynamics simulations. One of the problems my work addresses involves a static field of N-dimensional (typically N = 2-5) hyper-spheres, that a test particle may collide with. I'm looking to optimize (read: overhaul) the the data structure I use for representing the field of spheres so I can do rapid collision detection. Currently I use a dead simple array of pointers to an N-membered struct (doubles for each coordinate of the center) and a nearest-neighbor list. I've heard of oct- and quad- trees but haven't found a clear explanation of how they work, how to efficiently implement one, or how to then do fast collision detection with one. Given the size of my simulations, memory is (almost) no object, but cycles are.
How best to approach this for your problem depends on several factors that you have not described:
- Will the same hypersphere arrangement be used for many particle collision calculations?
- Are the hyperspheres uniform size?
- What is the movement of the particle (e.g. straight line/curve) and is that movement affected by the spheres?
- Do you consider the particle to have zero volume?
I assume that the particle does not have simple straight line movement as that would be the relatively fast calculation of finding the closest point between a line and a point, which is likely going to be about the same speed as finding which of the boxes the line intersects with (to determine where in the n-tree to examine).
If your hypersphere positions are fixed for a lot of particle collisions then computing a voronoi decomposition/Dirichlet tessellation would give you a fast way of later finding exactly which sphere is closest to your particle for any given point in the space.
However to answer your original question about octrees/quadtrees/2^n-trees, in n dimensions you start with a (hyper)-cube that contains the area of space that you are interested in. This will be subdivided into 2^n hypercubes if you deem the contents to be too complicated. This continues recursively until you have only simple elements (e.g. one hypersphere centroid) in the leaf nodes.
Now that the n-tree is built you use it for collision detection by taking the path of your particle and intersecting it with the outer hypercube. The intersection position will tell you which hypercube in the next level down of the tree to visit next, and you determine the position of intersection with all 2^n hypercubes at that level, following downwards until you reach a leaf node. Once you reach the leaf you can examine interactions between your particle path and the hypersphere stored at that leaf. If you have collision you have finished, otherwise you have to find the exit point of the particle path from the current hypercube leaf and determine which hypercube it moves to next. Continue until you find a collision or entirely leave the overall bounding hypercube.
Efficiently finding the neighbouring hypercube when exiting a hypercube is one of the most challenging parts of this approach. For 2^n trees Samet's approaches {1, 2} can be adapted. For kd-trees (binary trees) an approach is suggested in {3} section 4.3.3.
Efficient implementation can be as simple as storing a list of 8 pointers from each hypercube to its children hypercubes, and marking the hypercube in a special way if it is a leaf (e.g. make all pointers NULL).
A description of dividing space to create a quadtree (which you can generalise to n-tree) can be found in Klinger & Dyer {4}
As others have mentioned kd-trees may be more suited than 2^n-trees as extension to an arbitrary number of dimensions is more straightforward, however they will result in a deeper tree. It is also easier to adapt the split positions to match the geometry of your
hyperspheres with a kd-tree. The description above of collision detection in a 2^n tree is equally applicable to a kd-tree.
{1} Connected Component Labeling, Hanan Samet, Using Quadtrees Journal of the ACM Volume 28 , Issue 3 (July 1981)
{2} Neighbor finding in images represented by octrees, Hanan Samet, Computer Vision, Graphics, and Image Processing Volume 46 , Issue 3 (June 1989)
{3} Convex hull generation, connected component labelling, and minimum distance
calculation for set-theoretically defined models, Dan Pidcock, 2000
{4} Experiments in picture representation using regular decomposition, Klinger, A., and Dyer, C.R. E, Comptr. Graphics and Image Processing 5 (1976), 68-105.
It sounds like you'd want to implement a kd-tree, which would allow you to more quickly search the N-dimensional space. There's some more information and links to implementations at the Stony Brook Algorithm Repository.
Since your field is static (by which I'm assuming you mean that the hyper spheres don't move), then the fastest solution I know of is a Kdtree.
You can either make your own, or use someone else's, like this one:
http://libkdtree.alioth.debian.org/
A Quad tree is a 2 dimensional tree, in which at each level a node has 4 children, each of which covers 1/4 of the area of the parent node.
An Oct tree is a 3 dimensional tree, in which at each level a node has 8 children, each of which contains 1/8th of the volume of the parent node. Here is picture to help you visualize it: http://en.wikipedia.org/wiki/Octree
If you're doing N dimensional intersection tests, you could generalize this to an N tree.
Intersection algorithms work by starting at the top of the tree and recursively traversing into any child nodes that intersect the object being tested, at some point you get to leaf nodes, which contain the actual objects.
An octree will work as long as you can specify the spheres by their centres - it hierarchically bins points into cubic regions with eight children. Working out neighbours in an octree data structure will require you to do sphere-intersecting-cube calculations (to some extent easier than they look) to work out which cubic regions in an octree are within the sphere.
Finding the nearest neighbours means walking back up the tree until you get a node with more than one populated child and all surrounding nodes included (this ensures the query gets all sides).
From memory, this is the (somewhat naive) basic algorithm for sphere-cube intersection:
i. Is the centre within the cube (this gets the eponymous situation)
ii. Are any of the corners of the cube within radius r of the centre (corners within the sphere)
iii. For each surface of the cube (you can eliminate some of the surfaces by working out which side of the surface the centre lies on) work out (this is all first-year vector arithmetic):
a. A normal of the surface that goes to the centre of the sphere
b. The distance from the centre of the sphere to the intersection of the normal with the plane of the surface (chord intersets plane the surface of the cube)
c. Intersection of the plane lies within the side of the cube (one condition of chord intersection to the cube)
iv. Calculate the size of the chord (Sin of Cos^-1 of ratio of normal length to radius of sphere)
v. If the nearest point on the line is less than the distance of the chord and the point lies between the ends of the line the chord intersects one of the edges of the cube (chord intersects cube surface somewhere along one of the edges).
Slightly dimly remembered but this is something I did for a situation involving spherical regions using an octee data structure (many years ago). You may also wish to check out KD-trees as some of the other posters suggest but your initial question sounds very similar to what I did.