Is there a known way / algorithm that can give me a 2d curve or set of curves that encircles certain shapes, for example circles, like this:
I know of no analytic way to do this, but if you can discretize your shapes you could use the convex hull algorithm to get a curve that encompasses your original shapes. Like the algorithm name suggests, the resulting shape will be convex and won't bend inward toward the spaces between the circles as in your picture. The algorithm is O(n log n) in 2 and 3 dimensions.
Related
I am looking for an algorithm to distribute a bunch of points (could be anywhere from a few hundred to millions) within a sphere. In this case the sphere is centered at (0,0,0).
For random points a simple method is
repeat
x:=random*diameter-radius;
y:=random*diameter-radius;
z:=random*diameter-radius;
until ((x*x+y*y+z*z)<(radius*radius));
But I want to get the points evenly spaced within the sphere and without bunching at the poles.
Any good tricks/algorithms/formulas/code snippet to accomplish this?
You could do something like this:
Put the center of your sphere at a random position within an infinite volume of evenly-spaced points, like a tetrahedral or cubic lattice.
Enumerate points in order of increasing distance from the center until you have the right number.
Rescale the selected points around the center so that the distance to the furthest point is equal to the desired radius.
If you need evenly spaced points - just place them in grid nodes.
Sphere with radius R has volume
V=4/3*Pi*R^3
so for placing N points every cell of cubic grid (perhaps you might want to use hexagonal close packing) should have volume
v=4/3*Pi*R^3/N
and edge length
l = R * (4*Pi/(3*N))^1/3
Then generate points in coordinates (a*l, b*l, c*l) where a,b,c are integers limited by -R..+R (with appropriate sum of squares).
Proposed approach is quite rough estimation and perhaps some points from N needed ones might run outside of the sphere. In this case one have to diminish cell size or use more exact value - it might be calculated using 3D analog of Gauss circle formula ()
Thanks for the explanation by joriki. The above formula has been explained in proof of the formula mentioned. But the formula has some limitations. The cube length can be 2 (-1 to 1 in all directions). From the above concept, mentioned by joriki or proof of the formula mentioned we can generalize for the cube of any length (i.e 2a (-a to a in all directions)). Here 2a is the side length of the cube.
$a^6-(a^2-x^2 )(a^2-y^2)(a^2-z^2)=x^2 a^2 (a^2-\frac{z^2}{2}-\frac{y^2}{2}+\frac{y^2 z^2}{3a^2})+y^2 a^2 (a^2-\frac{z^2}{2}-\frac{x^2}{2}+\frac{x^2 z^2}{3a^2})+z^2 a^2 (a^2-\frac{x^2}{2}-\frac{y^2}{2}+\frac{y^2 x^2}{3a^2})$
$x'=xa \sqrt{a^2-\frac{z^2}{2}-\frac{y^2}{2}+\frac{y^2 z^2}{3a^2}}$,
$y'=ya \sqrt{a^2-\frac{z^2}{2}-\frac{x^2}{2}+\frac{x^2 z^2}{3a^2}}$,
$z'=za \sqrt{a^2-\frac{x^2}{2}-\frac{y^2}{2}+\frac{y^2 x^2}{3a^2}}$,
From the above equation, we can easily separate the (x',y',z') as explained in 1. The above equation will readily find the mapping of the generalized length of the cube to a sphere. I hope this will be helpful.
Form the above equation we can also find the evenly distribution of points within the sphere by varying the grid size of the cube. By discretizing the cube into different grids and simultaneously mapping into the sphere will give the evenly distribution of points within the sphere.
Kindly ignore some typos.
C++ code for evenly distributing points within the sphere:-
for(x=-a;x<=a;x=x+n)
for(y=-a;y<=a;y=y+n)
{
for(z=-a;z<=a;z=z+n)
{
xnew=x*a*(sqrt((a*a)-(y*y)/(2)-(z*z)/(2)+(y*y*z*z)/(3*a*a)));
ynew=y*a*(sqrt((a*a)-(z*z)/(2)-(x*x)/(2)+(x*x*z*z)/(3*a*a)));
znew=z*a*(sqrt((a*a)-(x*x)/(2)-(y*y)/(2)+(y*y*x*x)/(3*a*a)));
cout<<" "<<xnew<<" "<<ynew<<" "<<znew<<endl;
}
}
}
Note:- Here in code, n is the grid size of a cube.
I found myself in quite a big problem. I am average in math and I need to solve something, which is not very covered on the internet.
My problem: I have 2D space defined by X and Y. This space is just a drawing space. I want to assign to particular Xs,Ys a color with RGB values.
So let says I have 4 points with defined position in XY and color in Z:
[0,0, [255,0,0]]
[0,10, [0,255,0]]
[10,10,[0,0,255]]
[5,5, [0,0,0]]
and my drawing space is xy: 15x15.
And I want to distribute the colors to all empty points
For me its quite a delicate problem, because Z axis is basicly 3D space by itself.
My whole intention is to create a color map in which points 1,2,3,4 have between them smooth transition.
I am able to solve this in 1D where the transition is between 2 points. But I need to create 2D color map in XY drawing space based on fitted surface to these 4 points, which kind of depend both on the space of 3D-RGB and distance between them in XY drawing space.
Thanks in advance for help
You do not show any algorithm or code, so I will just explain a high-level algorithm. If you need more details or code or mathematical formulae, show more of your own work then ask. You do not explain just what you mean by "smooth transition"--there are multiple meanings. This will result in continuous shading but may not be smooth enough for your purposes.
First, given your points in the rectangular drawing space, find the Voronoi diagram for those points. This divides the drawing space into convex polygons, each polygon around one of your points.
For each vertex in the Voronoi diagram, figure which points are closest to the vertex--there will usually be just three of your points but there could be more. Then at that vertex point, assign the color that is the average of the RGB values of the nearby given points. That is, average the R values and the G values and the B values separately.
For any point on a Voronoi polygon edge, its color is the weighted average of the two colors at the endpoints. I.e. If the point is one-third of the distance from one end, its RGB value is one-third of the distance from the values at the endpoints.
Finally, for any point inside a Voronoi polygon, calculate the ray from the point that defined that polygon (the "center point") through the current point you are looking at. Find where that ray intersects the polygon. The RGB value is then the weighted average of the values of the center point and the polygon-intersection point.
The hardest part of all that is finding the Voronoi diagram. Fortune's algorithm can do this in a reasonable time. You can probably find a library to do that for you in your chosen programming language.
Another algorithm is to start with a triangulation of your given points and the corners of the drawing region. Then the color of any point in a triangle is the weighted average of the colors of the vertices. This will be automatically consistent for points on the vertices or edges of the triangles, so this is probably simpler than my previous algorithm. The difficulty here is finding a triangulation (any will do).
I'm writing a data analysis program and part of it requires finding the volume of a shape. The shape information comes in the form of a lost of points, giving the radius and the angular coordinates of the point.
If the data points were uniformly distributed in coordinate space I would be able to perform the integral, but unfortunately the data points are basically randomly distributed.
My inefficient approach would be to find the nearest neighbours to each point and stitch the shape together like that, finding the volume of the stitched together parts.
Does anyone have a better approach to take?
Thanks.
IF those are surface points, one good way to do it would be to discretize the surface as triangles and convert the volume integral to a surface integral using Green's Theorem. Then you can use simple Gauss quadrature over the triangles.
Ok, here it is, along duffymo's lines I think.
First, triangulate the surface, and make sure you have consistent orientation of the triangles. Meaning that orientation of neighbouring triangle is such that the common edge is traversed in opposite directions.
Second, for each triangle ABC compute this expression: H*cross2D(B-A,C-A), where cross2D computes cross product using coordinates X and Y only, ignoring the Z coordinates, and H is the Z-coordinate of any convenient point in the triangle (although the barycentre would improve precision).
Third, sum up all the above expressions. The result would be the signed volume inside the surface (plus or minus depending on the choice of orientation).
Sounds like you want the convex hull of a point cloud. Fortunately, there are efficient ways of getting you there. Check out scipy.spatial.ConvexHull.
I want to create a Voronoi diagram on several pairs of
latitudes/longitudes, but want to use the great circle distance
between them, not the (inaccurate) Pythagorean distance.
Can I make qhull/qvoronoi or some other Linux program do this?
I considered mapping the points to 3D, having qvoronoi create a 3D
Voronoi diagram[1], and intersecting the result with the unit sphere, but
I'm not sure that's easy.
[1] I realize the 3D distance between two latitudes/longitudes (the
"through the Earth" path) isn't the same as the great circle distance,
but it's easy to prove that this transformation preserves relative
distances, which is all that matters for a Voronoi diagram.
I assume you've found this article. From that, it seems like you have the right idea by using a 3D embedding. Your question is then how to intersect the result with the sphere.
First of all you need to consider how you're going to represent the voronoi diagram. If you want to work in lat/long coordinates in a 2D plane, then your voronoi diagram will contain curved edges, so maybe it is best to just use a 3D representation.
If you use a program like qvoronoi, you should in theory only need the inifinite hyperplane data (generated by Fo). This gives you the equation of the plane and the two points it corresponds to. Usually you only need to use the voronoi diagram to test for inclusion within regions, and the hyperplanes should be enough for that.
See also this question: Algorithm to compute a Voronoi diagram on a sphere?
I have two 2d circles in 3d space (defined by a center, normal, and radius) and I'm trying to come up with a pair of points that is one of the set of closest pairs of points. I know that there are anywhere from 1 to an infinite number of point pairs, I just need a single matching pair.
Is there a simple way to do that? Precision is not essential. The radius of both circles are the same, non-zero value.
In case the background is helpful, my overall algorithm takes in a NURBS curve in space and extrudes a 2d polygon along the curve, yielding a deformed cylinder. I just sample several points along the curve. The normal of each circle is the NURBS curve tangent, and I'm trying to figure out how to align adjacent samples, so I don't get weird twisting. It seems that the closest points on adjacent samples should be aligned.
Thanks for all the responses here.. this part of the project got a little delayed, which is why I haven't tested all the answers yet. I'll be sure to toss up some images here and mark an answer when I get to work on this again.
What you are really trying to compute is the pair of points that minimizes the distance between points that lie on 2 different circles in 3 dimensions. The method that you should be employing to find the exact solution (as in almost all optimization problems) is to represent the distance as a function of all possible points and to take its derivate with respect to the independent variables and set the resulting expressions to 0. Since you have 2 circles, you will have 2 independent variables (ie. the angle of a point on one circle and one on the other circle). Once you have solved the minimization equations you would have also found the points on the circles that will satisfy your constraint. (Basically you will find the angles on the circles for the pair of points you are looking for.)
I have found a paper online (at this site) that rigorously goes through with the calculations but the end result is solving an 8th order polynomial equation. You might try to simplify the equations and come up with a less exact solution that satisfies your needs.
There is also an paper that claims to have a much faster algorithm for finding the distance between two circles in 3d; however, I cannot view the contents and, thus, cannot tell if it also gives you the pair of points that satisfy that condition.
UPDATE: Having re-read your question, I see that even though you are asking for a way to find the closest pair of points on two circles in 3 dimensions, I think, you should pay more attention to the properties of the NURBS curve that you are trying to extrude the 2D polygon along. You mention that the orientation of the circle at a given point on the curve is specified by the tangent vector at that point. However, there is more to 3D curves than just the tangent vector; there is the normal (or curvature) vector that points towards the center of curvature of the curve at a given point and then there is the torsion vector that basically specifies the amount of "lift" of the curve from the plane given by the tangent and the normal vectors. All of these define a (what is called) Frenet frame. You can read up more on these at the Wikipedia article.
My suspicion is that you can achieve the effect you desire by joining the points of consecutive circles that each lie along the the normal vector direction of the underlying 3D curve. That way, you will have twisting only when the curve is actually twisting, ie when the torsion vector is non-zero and the normal vector is changing direction as well. In other circumstances, this should satisfy your actual need.
You probably don't need the overkill of finding closest points on consecutive circles.
For what you describe, it is sufficient to select a point on the perimeter of the first circle and find the point on the perimeter of each circle along that is closest to the one selected for the previous circle; this will completely constrain the polygonization, with no twisting, and should be much easier to solve than the general case - simply find the point on the plane containing the second circle that is closest to that selected in the first, and intersect the line passing through that point and the second circle's center with the second circle's perimeter.
However, this might not yield as pleasing a polygonisation for the extruded cylinder as keeping the polygon area constant as possible, and to do that will require some twisting between adjacent circles.
Yikes, unless the circles happen to be on the same plane or parallel planes I think the only way to do it is to find a minimum on the equation of the distance between two points on the circle.
http://www.physicsforums.com/showthread.php?t=123168
That link shows how to get the equation of each circle in 3D space, then minimize for the distance formula between those equations. Not pretty though, hopefully someone will come up with something more clever.
I think with the two closest points you might still get weird twisting... An extreme example: Let's assume both circles have the R=1. If the first circle's centre is O, and it is sitting on X-Y plane, and the second circle's centre is sitting at X=1,Y=0,Z=0.01, and it just slightly tilted in the growing direction of X, the closest points on the two circles will for sure get the "weird twist" you are trying to avoid. Since the closest points would not get you the weird twist in case the second circle is at X=0,Y=0,Z=0.01 and is equally tilted, then at some point the statements "aligned to two closest points on two circles" and "no weird twisting seen" no longer correspond to each other.
Assuming this can happen within the constraint of NURBS, here's another idea. In the start, take the three points on the NURBS curve - two that belong to the centers of your circles, and the third one precisely inbetween. Draw a plane between the three. This plane will cross the two circles at 4 points. Two of these points will be on the same "side" of the line that connects the centers of the circles - they are your alignment points.
For the next alignment points you would take the alignment point of the "previous circle", and draw the plane between the center of the "previous circle", this alignment point, and the center of the "new circle". From this you get the "next alignment point" based on the intersection with the other circle.
Next step - "previous circle" = "new circle", and the "new circle" - your next one according to the NURBS curve.
If the radii from the centers of the circles to the selected alignment points cross, you know you the picture will look a bit ugly - that's the scenario where with the "closest point" algorithm you'd still get the weird twisting.
I think the coordinates of the point on the circle that is intersection with the plane going via its center should be easy to calculate (it's a point on the line made by intersection of the two planes, one of the circle and the target plane; at the distance R from the center).
I don't have the rigorous proof to fully assert or deny the above - but hopefully it helps at all, and I think it should be quick enough to verify, compared to calculating the closet points on the two circles... (If there are any flaws in my logic, the corrections in the comments are very welcome).
The thread here, mentioned in another answer gives the parameterization formula for a 3D circle: P = R cos(t) u + R sin(t) nxu + c, where u is a unit vector from the centre of the circle to any point on the circumference; R is the radius; n is a unit vector perpendicular to the plane and c is the centre of the circle, t goes from 0 to 2pi, and by nxu I mean "n cross u". Parameterize one circle this way, and another similarly with a different parameter, say s. Then each point Pt on the first circle will have coordinates in the variable t, and each point Ps on the second circle will have coordinates in the variable s.
Write the distance function d(s,t) between Ps and Pt in the usual way (or better, the square of the Euclidean distance so you don't have to mess with the square root when you take derivatives). The graph of this function d of two variables is a surface over a 2pi by 2pi square in the s,t plane, and it's minimum is what you're after. You can determine it with the standard calculus methods, e.g. as explained here.
Extend the circles to planes (using the center points and normals). If the planes are parallel, then any points will do. If the planes are not parallel, then they intersect in a line. Construct the plane through the two centers of the circles perpendicular to the line. The two circles intersect this new plane in four points. These four points are the two nearest points and the two farthest points on the circles.
Isn't this just a matter of constructing the line between the two centers of the circles/spheres and finding the intersection of the line and the circles? The solutions that are closest are it (unless the circle intersect, then the answer depends on how you want to interpret that case).