Hopefully there are some computational geometry folks here who can help me out with the following problem -
Please imagine that I take a freely moving ball in 3-space and create a 'cage' around it by defining a set of impassible coordinates, Sc (i.e. points in 3-space that no part of the diffusing ball is allowed to overlap). These points reside within the volume, V(cage), of some larger sphere, where V(cage) >> V(ball).
Provided the set of impassible coordinates, Sc, is there a computationally efficient and/or nice way to determine if the ball can ever escape the cage?
Please see my earlier post at MathOverflow - https://mathoverflow.net/questions/21911/when-can-a-freely-moving-sphere-escape-from-a-cage-defined-by-a-set-of-impassib
Say the ball has radius R. As your friends over at MathOverflow have noted, the problem is equivalent to replacing the impassible points with balls of radius R, and the ball with a point P. So the question is whether P is in a chamber sealed off by the balls.
To answer this question, you need to compute the union of the balls, then take the surface S of this union. If P is inside any of the components of S, it will be trapped, otherwise it can escape.
The are well established algorithms for computing the union of balls, dating back to at least this paper by EdelsBrunner. There's quite a bit of learning curve to understanding and implementing that algorithm, but fortunately it is implemented in the CGAL library.
Once CGAL has calculated the skin,you can look at the connected components of the skin (all of which should be closed meshes) and check whether P is inside any of them (if you need help with that, shout).
Since the tessellated skin is an approximation of the actual sphere surfaces, you may ask whether there are precision problems. I would expect the mesh faces to actually be inside the original balls, and assuming P was originally outside of all the balls, I think that checking whether P is inside any of the skin components will give you accurate results.
Related
I have a turtle-graphics-based algorithm for generating a space-filling Hilbert curve in two dimensions. It is recursive and goes like this:
Wa want to draw a curve of order n, in direction x (where x ∈ {L, R}), and let y be the direction opposite to x. We do as follows:
turn in the direction y
draw a Hilbert curve of order n-1, direction y
move one step forward
turn in the direction x
draw a Hilbert curve of order n-1, direction x
move one step forward
draw a Hilbert curve of order n-1, direction x
turn in the direction x
move one step forward
draw a Hilbert curve of order n-1, direction y
I understand this and was able to implement a working solution. However, I'm now trying to "upgrade" this to 3D, and here's where I basically hit a wall; in 3D, when we reach a vertex, we can turn not in two, but four directions (going straight or backing up is obviously not an option, hence four and not six). Intuitively, I think I should store the plane on which the turtle is "walking" and its general direction in the world, represented by an enum with six values:
Up
Down
Left
Right
In (from the camera's perspective, it goes "inside" the world)
Out (same as above, outside)
The turtle, like in 2D, has a state containing the information outlined above, and when it reaches as vertex (which can be thought of as a "crossing") has to make a decision where to go next, based on that state. Whereas in two dimensions it is rather simple, in three, I'm stumped.
Is my approach correct? (i.e., is this what I should store in the turtle's state?)
If it is, how can I use that information to make a decision where to go next?
Because there are many variants of 3D space filling Hilbert curves, I should specify that this is what I'm using as reference and to aid my imagination:
I'm aware that a similar question has already been asked, but the accepted answer links to a website there this problem is solved using a different approach (i.e., not turtle graphics).
Your 2d algorithm can be summarized as “LRFL” or “RLFR” (with “F” being “forward”). Each letter means “turn that direction, draw a (n-1)-curve in that direction, and take a step forward”. (This assumes the x in step 8 should be a y.)
In 3d, you can summarize the algorithm as the 7 turns you would need to go along your reference. This will depend on how you visualize the turtle starting. If it starts at the empty circle, facing the filled circle, and being right-side-up (with its back facing up), then your reference would be “DLLUULL”.
I am solving a fourth order non-linear partial differential equation in time and space (t, x) on a square domain with periodic or free boundary conditions with MATHEMATICA.
WITHOUT using conformal mapping, what boundary conditions at the edge or corner could I use to make the square domain "seem" like a circular domain for my non-linear partial differential equation which is cartesian?
The options I would NOT like to use are:
Conformal mapping
changing my equation to polar/cylindrical coordinates?
This is something I am pursuing purely out of interest just in case someone screams bloody murder if misconstrued as a homework problem! :P
That question was asked on the time people found out that the world was spherical. They wanted to make rectangular maps of the surface of the world...
It is not possible.
The reason why is not possible is because the sphere has an intrinsic curvature, while the cube/parallelepiped has not. It can be shown that for two elements with different intrinsic curvatures, their surfaces cannot be mapped while either keeping constant infinitesimal distances, either the distance between two points is given by the euclidean distance.
The easiest way to understand this problem is to pick some rectangular piece of paper and try to make a sphere of it without locally stretch it or compress it (you can fold). You can't. On the other hand, you can make a cylinder surface, because the cylinder has also no intrinsic curvature.
In maps, normally people use one of the two options:
approximate the local surface of the sphere by a tangent plane and make a rectangle out of it. (a local map of some region)
make world maps but implement some curved lines everywhere identifying that the measuring distances must be made according to those lines.
This is also the main reason why when traveling from Europe to North America the airplanes seems to make a curve always trying to pass near canada. If we measured the distance from the rectangular map, we see that they should go on a strait line to minimize the distance. However, because we are mapping two different intrinsic curvatures, the real distance must be measured in a different way (and not via a strait line).
For 2D (in fact for nD) the same reasoning applies.
Imagine we have a 2D sky (10000x10000 coordinates). Anywhere on this sky we can have an aircraft, identified by its position (x, y). Any aircraft can start moving to another coordinates (in straight line).
There is a single component that manages all this positioning and movement. When a aircraft wants to move, it send it a message in the form of (start_pos, speed, end_pos). How can I tell in the component, when one aircraft will move in the line of sight of another (each aircraft has this as a property as radius of sight) in order to notify it. Note that many aircrafts can be moving at the same time. Also, this algorithm is good to be effective sa it can handle ~1000 planes.
If there is some constraint, that is limiting your solution - it can probably be removed. The problem is not fixed.
Use a line to represent the flight path.
Convert each line to a rectangle embracing it. The width of the rectangle is determined by your definition of "close" (The bigger the safety distance is, the wider the rectangle should be).
For each new flight plan:
Check if the new rectangle intersects with another rectangle.
If so, calculate when will each plane reach the collision point. If the time difference is too small (and you should define too small according to the scenario), refuse the new flight plan.
If you want to deal with the temporal aspect (i.e. dealing with the fact that the aircraft move), then I think a potentially simplification is lifting the problem by the time dimension (adding one more dimension - hence, the original problem, being 2D, becomes a 3D problem).
Then, the problem becomes a matter of finding the point where a line intersects a (tilted) cylinder. Finding all possible intersections would then be n^2; not too sure if that is efficient enough.
See Wikipedia:Quadtree for a data structure that will make it easy to find which airplanes are close to a given airplane. It will save you from doing O(N^2) tests for closeness.
You have good answers, I'll comment only on one aspect and probably not correctly
you say that you aircrafts move in form (start_pos, speed, end_pos)
if all aircrafts have such, let's call them, flightplans then you should be able to calculate directly when and where they will be within certain distance from each other, or when will they be at closest point from each other or if the will collide/get too near
So, if they indeed move according to the flightplans and do not deviate from them your problem is deterministic - it boils down to solving a set of equations, which for ~1000 planes is not such a big task.
If you do need to solve these equations faster you can employ the techniques described in other answers
using efficient structures that can speedup calculating distances (quadtree, octree, kd-trees),
splitting the problem to solve the equations only for some relevant future timeslice
prioritize solving equations for pairs for which the distance changes most rapidly
Of course converting time to a third dimension turns the aircrafts from points into lines and you end up searching for the closest points between two 3d lines (here's some math)
I actually found an answer to this question.
It is in the book Real-Time Collision Detection, p. 223. It's better named, as well: Intersecting Moving Sphere Against Sphere, where a 2D sphere is a circle. It's not so simple (and I may also violate some rights) to explain it here, but the basic idea is to fix one of the circles as a point, adding its radius to the radius of the moving one. The new direction for the moving one is the sum of the two original vectors.
I found some solutions, but they're too messy.
Yes. The Chebyshev center, x*, of a set C is the center of the largest ball that lies inside C. [Boyd, p. 416] When C is a convex set, then this problem is a convex optimization problem.
Better yet, when C is a polyhedron, then this problem becomes a linear program.
Suppose the m-sided polyhedron C is defined by a set of linear inequalities: ai^T x <= bi, for i in {1, 2, ..., m}. Then the problem becomes
maximize R
such that ai^T x + R||a|| <= bi, i in {1, 2, ..., m}
R >= 0
where the variables of minimization are R and x, and ||a|| is the Euclidean norm of a.
Perhaps these "too messy" solutions are what you actually looking for, and there are no simplier ones?
I can suggest a simple, but potentially imprecise solution, which uses numerical analysis. Assume you have a resilient ball, and you inflate it, starting from radius zero. If its center is not in the center you're looking for, then it will move, because the walls would "push" it in the proper direction, until it reaches the point, from where he can't move anywhere else. I guess, for a convex polygon, the ball will eventually move to the point where it has maximum radius.
You can write a program that emulates the process of circle inflation. Start with an arbitrary point, and "inflate" the circle until it reaches a wall. If you keep inflating it, it will move in one of the directions that don't make it any closer to the walls it already encounters. You can determine the possible ways where it could move by drawing the lines that are parallel to the walls through the center you're currently at.
In this example, the ball would move in one of the directions marked with green:
(source: coldattic.info)
Then, move your ball slightly in one of these directions (a good choice might be moving along the bisection of the angle), and repeat the step. If the new radius would be less than the one you have, retreat and decrease the pace you move it. When you'll have to make your pace less than a value of, say, 1 inch, then you've found the centre with precision of 1 in. (If you're going to draw it on a screen, precision of 0.5 pixel would be good enough, I guess).
If an imprecise solution is enough for you, this is simple enough, I guess.
Summary: It is not trivial. So it is very unlikely that it will not get messy. But there are some lecture slides which you may find useful.
Source: http://www.eggheadcafe.com/software/aspnet/30304481/finding-the-maximum-inscribed-circle-in-c.aspx
Your problem is not trivial, and there
is no C# code that does this straight
out of the box. You will have to write
your own. I found the problem
intriguing, and did some research, so
here are a few clues that may help.
First, here's an answer in "plain
English" from mathforum.org:
Link
The answer references Voronoi Diagrams
as a methodology for making the
process more efficient. In researching
Voronoi diagrams, in conjunction with
the "maximum empty circle" problem
(same problem, different name), I came
across this informative paper:
http://www.cosy.sbg.ac.at/~held/teaching/compgeo/slides/vd_slides.pdf
It was written by Martin Held, a
Computational Geometry professor at
the University of Salzberg in Austria.
Further investigation of Dr. Held's
writings yielded a couple of good
articles:
http://www.cosy.sbg.ac.at/~held/projects/vroni/vroni.html
http://www.cosy.sbg.ac.at/~held/projects/triang/triang.html
Further research into Vornoi Diagrams
yielded the following site:
http://www.voronoi.com/
This site has lots of information,
code in various languages, and links
to other resources.
Finally, here is the URL to the
Mathematics and Computational Sciences
Division of the National Institute of
Standards and Technology (U.S.), a
wealth of information and links
regarding mathematics of all sorts:
http://math.nist.gov/mcsd/
-- HTH,
Kevin Spencer Microsoft MVP
The largest inscribed circle (I'm assuming it's unique) will intersect some of the faces tangentially, and may fail to intersect others. Let's call a face "relevant" if the largest inscribed circle intersects it, and "irrelevant" otherwise.
If your convex polygon is in fact a triangle, then the problem can be solved by calculating the triangle's incenter, by intersecting angle bisectors. This may seem a trivial case, but even when
your convex polygon is complicated, the inscribed circle will always be tangent to at least three faces (proof? seems geometrically obvious), and so its center can be calculated as the incenter of three relevant faces (extended outwards to make a triangle which circumscribes the original polygon).
Here we assume that no two such faces are parallel. If two are parallel, we have to interpret the "angle bisector" of two parallel lines to mean that third parallel line between them.
This immediately suggests a rather terrible algorithm: Consider all n-choose-3 subsets of faces, find the incenters of all triangles as above, and test each circle for containment in the original polygon. Maximize among those that are legal. But this is cubic in n and we can do much better.
But it's possible instead to identify faces that are irrelevant upfront: If a face is tangent
to some inscribed circle, then there is a region of points bounded by that face and by the two angle bisectors at its endpoints, wherein the circle's center must lie. If even the circle whose center lies at the farthest tip of that triangular region is "legal" (entirely contained in the polygon), then the face itself is irrelevant, and can be removed. The two faces touching it should be extended beyond it so that they meet.
By iteratively removing faces which are irrelevant in this sense, you should be able to reduce the
polygon to a triangle, or perhaps a trapezoid, at which point the problem will be easily solved, and its solution will still lie within the original polygon.
Here is a problem I am trying to solve:
I have an irregular shape. How would I go about evenly distributing 5 points on this shape so that the distance between each point is equal to each other?
David says this is impossible, but in fact there is an answer out of left field: just put all your points on top of each other! They'll all have the same distance to all the other points: zero.
In fact, that's the only algorithm that has a solution (i.e. all pairwise distances are the same) regardless of the input shape.
I know the question asks to put the points "evenly", but since that's not formally defined, I expect that was just an attempt to explain "all pairwise distances are the same", in which case my answer is "even".
this is mathematically impossible. It will only work for a small subset of base shapes.
There are however some solutions you might try:
Analytic approach. Start with a point P0, create a sphere around P0 and intersect it with the base shape, giving you a set of curves C0. Then create another point P1 somewhere on C0. Again, create a sphere around P1 and intersect it with C0, giving you a set of points C1, your third point P2 will be one of the points in C1. And so on and so forth. This approach guarantees distance constraints, but it also heavily depends on initial conditions.
Iterative approach. Essentially form-finding. You create some points on the object and you also create springs between the ones that share a distance constraint. Then you solve the spring forces and move your points accordingly. This will most likely push them away from the base shape, so you need to pull them back onto the base shape. Repeat until your points are no longer moving or until the distance constraint has been satisfied within tolerance.
Sampling approach. Convert your base geometry into a voxel space, and start scooping out all the voxels that are too close to a newly inserted point. This makes sure you never get two points too close together, but it also suffers from tolerance (and probably performance) issues.
If you can supply more information regarding the nature of your geometry and your constraints, a more specific answer becomes possible.
For folks stumbling across here in the future, check out Lloyd's algorithm.
The only way to position 5 points equally distant from one another (other than the trivial solution of putting them through the origin) is in the 4+ dimensional space. It is mathematically impossible to have 5 equally distanced object in 3D.
Four is the most you can have in 3D and that shape is a tetrahedron.