Suppose I have a six-dimensional point cloud D, it has only one cluster and no noise, and its density is uneven.
Given an examination point C, how to calculate the distance from C to the boundary of D?
This is easy when C is outside the point cloud D; in this case, the distance is the minimum distance from C to all points in D.
But how about the case when C is in the interior of D?
2D-example
Thanks a lot!
I have tried the density-based algorithm DBSCAN to detect the boundary points of the point cloud, but it can not detect enough actual boundary points that enclose the point cloud.
Assumptions
On the basis that the points represent a defined ellipse with a smooth, sharp surface (as per your answer to my question), the boundary position you find will necessarily be a statistical estimate, as I expect you're aware.
I assume the points are uniformly randomly distributed.
My idea
The location on the edge of the ellipse (or on the surface of the cloud) is going to be somewhere on a line that starts at the origin of the point cloud and goes through C.
Project a cylinder from the origin to C (I'll get to the diameter later).
Count the number of points enclosed by the cylinder. I found a bit of C++ code that can check if a point is inside a cylinder here. That code is for 3 dimensions but it's easy to see how it can be adapted for 6, and it's simple enough that you could translate it into any language.
Increase the length of the cylinder and count the points it encloses again.
Keep going until the number of points enclosed by the cylinder hasn't changed for a few iterations.
You can then work back to find the cylinder length where the point count stopped increasing.
The cylinder now essentially represents a vector whose end lies on the boundary of D.
You can also subtract the distance from the origin of D to C from the length of the final cylinder to get the distance from C to the boundary.
You can have an outer iteration that chooses random locations for C and goes through these steps each time. The ends of all those cylinders represent a set of new points that describe the surface of the point cloud, which I imagined is the purpose of the exercise (to generate a shell from the cloud).
Cylinder radius
As for the radius of the cylinder, this needs to be large enough to ensure it's not so thin that it misses too many points near the boundary and so gives you a distance that's too small. You could trial-and-error this, use some judgment based on the density of the point cloud, or carry out the test above a number of times using different radii.
Notes
This method should actually work with any shape, even an amorphous one as long as as the boundary never overlaps itself (i.e. all points on the surface would be visible from the origin).
So it would work with a cube, but not a banana! You could adapt it for abritrary shapes but as that's not in the question I'll leave that for now.
Related
I am given a set of N points in 2D plane represented as (x,y) coordinate pairs. What is a fast algorithm to choose three points so that the triangle formed by these points has maximum perimeter?
This is preemptive in nature
Pick a point farthest from the flock, lets call it point A.
draw an imaginary straight line that cuts through A to rest of the flock.
pick another opposite point, that its deviation(from the imaginary
straight line) is highest to right.
pick another opposite point, that
is deviation(from the imaginary straight line) is highest to the
left.
Check if a triangle can be made?.
if no check another highest point in another axis
Here's a rough idea (I'm not too versed in computational geometry). A triangle with a fixed perimeter and base can generate an ellipse. For example, here B and C are fixed and any point, A, on the ellipse will keep the triangle perimeter the same:
For each segment connecting two points, pick a random third point in our set. Generate the relevant ellipse, then pick another random point from our set that's outside that ellipse. Each ellipse will exclude points that generate triangles of the same or smaller perimeter until we run out of points, having found the largest. Of course, we would need some efficient methods to find relevant points (perhaps using space partitioning?).
I have a set of 3d points that lie in a plane. Somewhere on the plane, there will be a hole (which is represented by the lack of points), as in this picture:
I am trying to find the contour of this hole. Other solutions out there involve finding convex/concave hulls but those apply to the outer boundaries, rather than an inner one.
Is there an algorithm that does this?
If you know the plane (which you could determine by PCA), you can project all points into this plane and continue with the 2D coordinates. Thus, your problem reduces to finding boundary points in a 2D data set.
Your data looks as if it might be uniformly sampled (independently per axis). Then, a very simple check might be sufficient: Calculate the centroid of the - let's say 30 - nearest neighbors of a point. If the centroid is very far away from the original point, you are very likely on a boundary.
A second approach might be recording the directions in which you have neighbors. I.e. keep something like a bit field for the discretized directions (e.g. angles in 10° steps, which will give you 36 entries). Then, for every neighbor, calculate its direction and mark that direction, including a few of the adjacent directions, as occupied. E.g. if your neighbor is in the direction of 27.4°, you could mark the direction bits 1, 2, and 3 as occupied. This additional surrounding space will influence how fine-grained the result will be. You might also want to make it depend on the distance of the neighbor (i.e. treat the neighbors as circles and find the angular range that is spanned by the circle). Finally, check if all directions are occupied. If not, you are on a boundary.
Alpha shapes can give you both the inner and outer boundaries.
convert to 2D by projecting the points onto your plane
see related QA dealing with this:
C++ plane interpolation from a set of points
find holes in 2D point set
simply apply this related QA:
Finding holes in 2d point sets?
project found holes back to 3D
again see the link in #1
Sorry for almost link only answer but booth links are here on SO/SE and deals exactly with your issue when combined. I was struggling first to flag your question as duplicate and leave this in a comment but this is more readable.
I have a list of coordinates (latitude, longitude) that define a polygon. Its edges are created by connecting two points with the arc that is the shortest path between those points.
My problem is to determine whether another point (let's call it U) lays in or out of the polygon. I've been searching web for hours looking for an algorithm that will be complete and won't have any flaws. Here's what I want my algorithm to support and what to accept (in terms of possible weaknesses):
The Earth may be treated as a perfect sphere (from what I've read it results in 0.3% precision loss that I'm fine with).
It must correctly handle polygons that cross International Date Line.
It must correctly handle polygons that span over the North Pole and South Pole.
I've decided to implement the following approach (as a modification of ray casting algorithm that works for 2D scenario).
I want to pick the point S (latitude, longitude) that is outside of the polygon.
For each pair of vertices that define a single edge, I want to calculate the great circle (let's call it G).
I want to calculate the great circle for pair of points S and U.
For each great circle defined in point 2, I want to calculate whether this great circle intersects with G. If so, I'll check if the intersection point lays on the edge of the polygon.
I will count how many intersections there are, and based on that (even/odd) I'll decide if point U is inside/outside of the polygon.
I know how to implement the calculations from points 2 to 5, but I don't have a clue how to pick a starting point S. It's not that obvious as on 2D plane, since I can't just pick a point that is to the left of the leftmost point.
Any ideas on how can I pick this point (S) and if my approach makes sense and is optimal?
Thanks for any input!
If your polygons are local, you can just take the plane tangent to the earth sphere at the point B, and then calculate the projection of the polygon vertices on that plane, so that the problem becomes reduced to a 2D one.
This method introduces a small error as you are approximating the spherical arcs with straight lines in the projection. If your polygons are small it would probably be insignificant, otherwise, you can add intermediate points along the arcs when doing the projection.
You should also take into account the polygons on the antipodes of B, but those could be discarded taking into account the polygons orientation, or checking the distance between B and some polygon vertex.
Finally, if you have to query too many points for that, you may like to pick some fixed projection planes (for instance, those forming an octahedron wrapping the sphere) and precalculate the projection of the polygons on then. You could even create some 2d indexing structure as a quadtree for every one in order to speed up the lookup.
The biggest issue is to define what we mean by 'inside the polygon'.
On a sphere, every polygon (as long as the lines are not intersecting) defines two regions of the sphere. Both regions are equally qualified to be called the inside of the polygon.
Consider a simple, 1-meter on a side, yellow square around the south pole.
You can think of the yellow area to be the inside of the square OR you can think of the square enclosing everything north of each line (the rest of the earth).
So, technically, any point on the sphere 'validly' inside the polygon.
The only way to disambiguate is to select which side of the polygon you want. For example, define the interior to always be the area to the right of each edge.
I am trying to create an algorithm for 'fleeing' and would like to first find points which are 'safe'. That is to say, points where they are relatively distant from other points.
This is 2D (not that it matters much) and occurs within a fixed sized circle.
I'm guessing the sum of the squared distances would produce a good starting equation, whereby the highest score is the furthest away.
As for picking the points, I do not think it would be possible to solve for X,Y but approximation is sufficient.
I did some reading and determined that in order to cover the area of a circle, you would need 7 half-sized circles (with centers forming a hex, and a seventh at the center)
I could iterate through these, all of which are within the circle to begin with. As I choose the best scoring sphere, I could continue to divide them into 7 spheres. Of course, excluding any points which fall outside the original circle.
I could then iterate to a desired precision or a desired level.
To expand on the approach, the assumption is that it takes time to arrive at a location and while the location may be safe, the trip in between may not. How should I incorporate the distance in the equation so that I arrive at a good solution.
I suppose I could square the distance to the new point and multiply it by the score, and iterate from there. It would strongly favor a local spot, but I imagine that is a good behavior. It would try to resolve a safe spot close by and then upon re-calculating it could find 'outs' and continue to sneak to safety.
Any thoughts on this, or has this problem been done before? I wasn't able to find this problem specifically when I looked.
EDIT:
I've brought in the C# implementation of Fortune's Algorithm, and also added a few points around my points to create a pseudo circular constraint, as I don't understand the algorithm well enough to adjust it manually.
I realize now that the blue lines create a path between nodes. I can use the length of these and the distance between the surrounding points to compute a path (time to traverse and danger) and weigh that against the safety (the empty circle it is trying to get to) to determine what is the best course of action. By playing with how these interact, I can eliminate most of the work I would have had to do, simply by using the voronoi. Also my spawning algorithm will use this now, to determine the LEC and spawn at that spot.
You can take the convex hull of your set of locations - the vertices of the convex hull will give you the set of "most distant" points. Next, take the centroid of the points you're fleeing from, then determine which vertex of the convex hull is the most distant from the centroid. You may be able to speed this up by, for example, dividing the playing field into quadrants - you only need to test the vertices that are in the furthermost quadrant (e.g., if the centroid is in the positive-x positive-y quadrant, then you only need to check the vertices in the negative-x negative-y quadrant); if the playing field is an irregular shape then this may not be an option.
As an alternative to fleeing to the most distant point, if you have a starting point that you're fleeing from (e.g. the points you're fleeing from are enemies, and the player character is currently at point X which denotes its starting point), then rather than have the player flee to the most distant point you can instead have the player follow the trajectory that most quickly takes them from the centroid of the enemies - draw a ray from the enemies' centroid through the player's location, and that ray gives you the direction that the player should flee.
If the player character is surrounded then both of these algorithms will give nonsense results, but in that case the player character doesn't really have any viable options anyway.
I am programming an algorithm where I have broken up the surface of a sphere into grid points (for simplicity I have the grid lines parallel and perpendicular to the meridians). Given a point A, I would like to be able to efficiently take any grid "square" and determine the point B in the square with the least spherical coordinate distance AB. In the degenerate case the "squares" are actually "triangles".
I am actually only using it to bound which squares I am searching, so I can also accept a lower bound if it is only a tiny bit off. For this reason, I need the algorithm to be extremely quick otherwise it would be better to just take the loss of accuracy and search a few more squares.
I decided to repost this question to Math Overflow: https://mathoverflow.net/questions/854/closest-grid-square-to-a-point-in-spherical-coordinates. More progress has been made here
For points on a sphere, the points closest in the full 3D space will also be closest when measured along the surface of the sphere. The actual distances will be different, but if you're just after the closest point it's probably easiest to minimize the 3D distance rather than worry about great circle arcs, etc.
To find the actual great-circle distance between two (latitidude, longitude) points on the sphere, you can use the first formula in this link.
A few points, for clarity.
Unless you specifically wish these squares to be square (and hence to not fit exactly in this parallel and perpendicular layout with regards to the meridians), these are not exactly squares. This is particularly visible if the dimensions of the square are big.
The question speaks of a [perfect] sphere. Matters would be somewhat different if we were considering the Earth (or other planets) with its flattened poles.
Following is a "algorithm" that would fit the bill, I doubt it is optimal, but could offer a good basis. EDIT: see Tom10's suggestion to work with the plain 3D distance between the points rather than the corresponding great cirle distance (i.e. that of the cord rather than the arc), as this will greatly reduce the complexity of the formulas.
Problem layout: (A, B and Sq as defined in the OP's question)
A : a given point the the surface of the sphere
Sq : a given "square" from the grid
B : solution to problem : point located within Sq which has the shortest
distance to A.
C : point at the center of Sq
Tentative algorithm:
Using the formulas associated with [Great Circle][1], we can:
- find the equation of the circle that includes A and C
- find the distance between A and C. See the [formula here][2] (kindly lifted
from Tom10's reply).
- find the intersect of the Great Circle arc between these points, with the
arcs of parallel or meridian defining the Sq.
There should be only one such point, unless this finds a "corner" of Sq,
or -a rarer case- if the two points are on the same diameter (see
'antipodes' below).
Then comes the more algorithmic part of this procedure (so far formulas...):
- find, by dichotomy, the point on Sq's arc/seqment which is the closest from
point A. We're at B! QED.
Optimization:
It is probably possible make a good "guess" as to the location
of B, based on the relative position of A and C, hence cutting the number of
iterations for the binary search.
Also, if the distance A and C is past a certain threshold the intersection
of the cicles' arcs is probably a good enough estimate of B. Only when A
and C are relatively close will B be found a bit further on the median or
parallel arc in these cases, projection errors between A and C (or B) are
smaller and it may be ok to work with orthogonal coordinates and their
simpler formulas.
Another approach is to calculate the distance between A and each of the 4
corners of the square and to work the dichotomic search from two of these
points (not quite sure which; could be on the meridian or parallel...)
( * ) *Antipodes case*: When points A and C happen to be diametrically
opposite to one another, all great circle lines between A and C have the same
length, that of 1/2 the circonference of the sphere, which is the maximum any
two points on the surface of a sphere may be. In this case, the point B will
be the "square"'s corner that is the furthest from C.
I hope this helps...
The lazy lower bound method is to find the distance to the center of the square, then subtract the half diagonal distance and bound using the triangle inequality. Given these aren't real squares, there will actually be two diagonal distances - we will use the greater. I suppose that it will be reasonably accurate as well.
See Math Overflow: https://mathoverflow.net/questions/854/closest-grid-square-to-a-point-in-spherical-coordinates for an exact solution