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).
Related
Okay first I wasn't sure if this was better suited to the MathSO so apologies if it needs migrating.
I have a 3D grid of points (representing the centers of voxels) with pitch varying in each dimension, but regular. For example resolution may be 100 by 50 by 40 for a cube shaped object.
Giving me nVox = 200,000.
For each voxel - I would like to cast (nVox - 1) rays, ending at the center, and originating from each of the other voxels.
Now there is obviously a lot of overlap here but I am having trouble finding how to calculate the minimum set of rays required. This sounds like a problem that has an elegant solution, I am however struggling to find it.
As a start, it is obvious that you only need to compute
[nVox * (nVox - 1)] / 2
of the rays, as the other half will simply be in the opposite directions. It is also easy in the 2D case to combine all of those parallel to one of the grid axes (and the two diagonals).
So how do I find the minimum set of rays I need, to pass from all voxel centers, to all others?
If someone could point me in the right direction that'd be great. Any and all help will be much appreciated.
Your problem really isn't about three dimensions in any specific way. All the conceptual complexity is present in the two dimensional case.
Instead of connecting points individually, think about the set of lines that pass through at least two points on your grid. Thus instead of thinking about points initially, think about directions. For 2-D these directions are slopes of lines. These slopes have to be rational numbers, since they intersect points on an integer lattice. Since you have a finite lattice, the numerator and the denominator of the slope can be bounded by the size of the figure. So your underlying problem is enumerating possible slopes for rational numbers of bounded "height" (math jargon).
There's an algorithm for that. It's the one used to generate the Farey sequence of reduced fractions. If your figure is N pixels wide, there will (in general) be a slope with denominator N in the somewhere, but there can't be a slope in reduced form with denominator >N; it wouldn't fit.
It's easier to deal with slopes between 0 and 1 directly. You get the other directions by two operations: negating the slope and by interchanging axes. For three dimensions, you need two slopes to define a direction.
Given an arbitrary direction (no necessarily a rational one as above), there's a perpendicular linear space of dimension k-1; for 3-D that's a plane. Projecting a 3-D parallelpiped onto this plane yields a hexagon in general; two vertices project onto the interior, six project to the vertices of the hexagon.
For a given discrete direction, there's a minimal bounding box on the integer lattice such that two opposite vertices lie along that direction. As long as that bounding box fits within your original grid, each of the interior points of the projection each correspond to a line that intersects your grid in at least two points.
In summary, enumerate directions, then for each direction enumerate where that direction intersects your grid in at least two points.
I am trying to compute a Voronoi tesselation in 2D with the Manhattan distance in R.
Ideally this would be a function that takes a set of two-dimensional points and outputs a list of polygons that partition the space. I am not certain what representations of Voronoi tesselations are standard.
There are of course many ways to do this with the Euclidean metric (packages like deldir and qhull make this very easy), but I haven't found a way to do this for the manhattan distance. A search using sos's findFn('voronoi') also yielded no results.
Info: taxicabgeometry.net
Interactive: Manhattan-metric Voronoi diagram(Click version)
I've been rolling my own in python, and can sum up the basics here:
Between neighboring centroids is a perpendicular line, in manhattan metric - two rays and a 45 degree diagonal most likely, if the centroids are randomly generated, but a straight horizontal, vertical, or 45 degree diagonal line may also occur. Given a set of such lines for every centroid pair, the edges separating the regions are among them. Collect intersect points of each pair of lines which are equal-distant (within an epsilon), in manhattan metric, to it's 3 nearest centroids. Also collect the two mid points of the 45 degree diagonal which are similarly equal-distant to their nearest two centroids. The outer polies won't be closed. How to deal with them depends on what you need. The poly borders and border verts will need sorting, so your polies aren't a zigzagged mess. Winding order can be determined if they should be clockwise or other. More can be done, just depends on what you need.
Unfortunately, this gets exponentially slower the more points are involved. The intersecting of every bisector to every other bisector is the bottleneck. I've been attempting an insertion method, with some success, but . Now I'm thinking to try first creating a nearest-neighbor linkage between the centroids. If the neighbors are known, the bisectors to intersect will be minimal, and many centroids can be computed quickly.
Anyway, the brute-force approach does work:
The point near the cursor is actually 2 points of a tiny diagonal. It's a precise method, but more complicated than it first seems. The java code from the interactive link above may be faster, but was difficult to get solid and precise geometry from.
Sorry, I dont know R.
Maybe the question is about finding the maximum area of a square that match inside a circumcircle (of a triangle). The equation for such a square abs(x)+abs(y)=r (www.mathematische-basteleien.de/taxicabgeometry.htm). When you have a mesh of triangles the voronoi diagram is the dual.
I have a computer graphics plotting application where we often plot regular convex polygon shapes as symbols for different data points. I'd like to scale the radius (aka circumradius, distance from center to vertex) of the polygons so that polygons with different numbers of sides all have equal area (so presumably similar perceptual impact). i.e. if a circle with radius=1 has area Pi*radius^2, how much do I need to scale the radius to get a square or a triangle with the same area? What would the formula be to compute this for regular polygons with arbitrary numbers of sides?
Seems like this should be a simple geometry/algebra problem, but that was a long time ago... :-)
Using the formula below (taken from this site):
one can derive that:
R = sqrt(2*area / (N*sin(2*pi/N)))
I am trying to find sphere that surly encompasses given list of points.
Points will have x, y and z co-ordinate[Points are in 3D].
Actually I am trying to find new three points based on given list of points by some calculations like find MinX,MaxX ,MinY,MaxY,and MinZ and MaxZ and do some operation and find new three points
And I will draw sphere from these three points.
And I will also taking all these three points on the diameter of sphere so I have a unique sphere.
Is there any standard way for finding encompassing sphere of given list of points?
Yes, the standard algorithm is Welzl's algorithm (assuming you want the minimal sphere around your points). Particularly the improved version of Gaertner is very useful, robust and numerically stable! It handles all the degenerate cases well too.
At its core, the algorithm permutes the points (randomly) to find the 1-4 points that lie on the boundary of the sphere. It's basically a clever trial-and-error algorithm. From these points, you can find the center by finding a point that has the same distance to all those points. Gärtner's version uses an improved numerical device to find the center. Also, it employs an extra pivoting step that presumably makes the algorithm work better for a large number of input points.
If all you want is a sphere around three points, I suggest you still use Gärtners "device" to compute the circumsphere of the triangle. Otherwise, the method will probably degenerate easily (i.e. when the triangle is very flat).
Do you need 3 points, or any number of points?
If you only need the answer for 3 points, each pair of points defines a line segment. Take the longest line segment. Take a sphere centered at the middle of that line segment, whose radius is half the length of the line segment. There are two cases.
The third point is inside of that initial sphere. If so, then you have the smallest sphere.
The third point is outside of that initial sphere. Then the solution at Find Circum Center of Three point of Triangle [Not using Compass] will give you the center of the smallest sphere containing those 3 points.
If you need an arbitrary number of points, I'd do some sort of iterative approximation algorithm. Since you don't seem like you need that, I won't work out the details.
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).