Given an array of 3D integers, what is the algorithmic complexity of determining which of those integers exist within a cube? I'm assuming the points can be represented in a number of concurrent data structures, each sorted in one or more dimensions.
My intuition tells me given a sorted array of points in 1D one can determine the subset of points between some lower and upper bound in something like O(log(n), but I would be very grateful for any insights others can offer on this notion (and any help others can offer generalizing to the multidimensional case!).
If you're unfamiliar with the math involved, I recommend doing this problem in two dimensions first, with a rectangle. That way, you can get familiar with the math, which is really just a bit of basic trigonometry. After that, stepping up to three dimensions isn't very difficult.
The problem is much simpler if the cube (or rectangle) is axis aligned, so you probably should do that first. For an example of determining the rotation you need, see How to calculate rotation angle from rectangle points?.
Once you've determined the rotation angle, you can translate the rectangle to the origin and rotate it by doing the first two steps in the accepted answer here: Drawing a Rotated Rectangle.
You now have an axis-aligned rectangle that's centered at the origin.
Finally, for each of your points:
Apply the same translation and rotation that you applied to the rectangle.
Test to see if the x and y coordinates in the resulting point are within the rectangle. This is a matter of, at most, four bounds checks.
If the point is in the rectangle, save it.
Once you've done this in two dimensions, you should be able to apply those concepts to three dimensions.
The algorithm is O(n), where n is the number of points.
Contour lines (aka isolines) are curves that trace constant values across a 2D scalar field. For example, in a geographical map you might have contour lines to illustrate the elevation of the terrain by showing where the elevation is constant. In this case, let's store contour lines as lists of points on the map.
Suppose you have map that has several contour lines at known elevations, and otherwise you know nothing about the elevations of the map. What algorithm would you use to fill in additional contour lines to approximate the unknown elevations of the map, assuming the landscape is continuous and doesn't do anything surprising?
It is easy to find advise about interpolating the elevation of an individual point using contour lines. There are also algorithms like Marching Squares for turning point elevations into contour lines, but none of these exactly capture this use case. We don't need the elevation of any particular point; we just want the contour lines. Certainly we could solve this problem by filling an array with estimated elevations and then using Marching Squares to estimate the contour lines based on the array, but the two steps of that process seem unnecessarily expensive and likely to introduce artifacts. Surely there is a better way.
IMO, about all methods will amount to somehow reconstructing the 3D surface by interpolation, even if implicitly.
You may try by flattening the curves (turning them to polylines) and triangulating the resulting polygons thay they will define. (There will be a step of closing the curves that end on the border of the domain.)
By intersection of the triangles with a new level (unsing linear interpolation along the sides), you will obtain new polylines corresponding to new isocurves. Notice that the intersections with the old levels recreates the old polylines, which is sound.
You may apply a post-smoothing to the curves, but you will have no guarantee to retrieve the original old curves and cannot prevent close surves to cross each other.
Beware that increasing the density of points along the curves will give you a false feeling of accuracy, as the error due to the spacing of the isolines will remain (indeed the reconstructed surface will be cone-like, with one of the curvatures being null; the surface inside the bottommost and topmost lines will be flat).
Alternatively to using flat triangles, one may think of a scheme where you compute a gradient vector at every vertex (f.i. from a least square fit of a plane on the vertex and its neighbors), and use this information to generate a bivariate polynomial surface in the triangle. You must do this in such a way that the values along a side will coincide for the two triangles that share it. (Unfortunately, I have no formula to give you.)
The isolines are then obtained by a further subdivision of the triangle in smaller triangles, with a flat approximation.
Actually, this is not very different from getting sample points, (Delaunay) triangulating them and fitting picewise continuous patches to the triangles.
Whatever method you will use, be it 2D or 3D, it is useful to reason on what happens if you sweep the range of z values in a continous way. This thought experiment does reconstruct a 3D surface, which will possess continuity and smoothness properties.
A possible improvement over the crude "flat triangulation" model could be to extend every triangle side between to iso-polylines with sides leading to the next iso-polylines. This way, higher order interpolation (cubic) can be achieved, giving a smoother reconstruction.
Anyway, you can be sure that this will introduce discontinuities or other types of artifacts.
A mixed method:
flatten the isolines to polylines;
triangulate the poygons formed by the polylines and the borders;
on every node, estimate the surface gradient (least-square fit of a plane to the node and its neighborrs);
in every triangle, consider the two sides along which you need to interpolate and compute the derivative at endpoints (from the known gradients and the side directions);
use Hermite interpolation along these sides and solve for the desired iso-levels;
join the points obtained on both sides.
This method should be a good tradeoff between complexity and smoothness. It does reconstruct a continuous surface (except maybe for the remark below).
Note that is some cases, yo will obtain three solutions of the cubic. If there are three on each side, join them in order. Otherwise, make a decision on which to join and use the remaining two to close the curve.
I am interested in using shapes like these:
Usually a tangram is made of 7 shapes(5 triangles, 1 square and 1 parallelogram).
What I want to do is fill a shape only with tangram shapes, so at this point,
the size and repetition of shapes shouldn't matter.
Here's something I manually tried:
I am a bit lost on how to approach this.
Assuming I have a path (an ordered list/array of points of the outline),
I imagine I should try to do some sort of triangulation.
Is there such a thing as Deulanay triangulation with triangles constrained to 45 degrees
right angled triangles ?
A more 'brute' approach would be to add a bunch of triangles(45 degrees) and use SAT
for collision detection to 'fix' overlaps, and hopefully gaps will be avoided.
Since the square and parallelogram can be made of triangles(45 degrees) too, I imagine there
would be a nice clean geometric solution, right ?
How do I pack triangles(45 degrees) inside an arbitrary shape ?
Any ideas are welcome.
A few random thoughts (maybe they help you find a better solution) if you're using only the original sizes of the shapes:
as you point out, all shapes in the tangram can be made composed of e.g. the yellow or pink triangle (d-g-c), so try also thinking of a bottom-up approach such as first trying to place as many yellow triangles into your shape and then combine them into larger shapes if possible. In the worst case, you'll end up with a set of these smallest triangles.
any kind triangulation of non-polygons (such as the half-moon in your example) probably does not work very well...
It looks like you require that the shapes can only have a few discrete orientations. To find the best fit of these triangles into the given shape, I'd propose the following approximate solution: draw a grid of triangles (i.e. a square grid with diagonal lines) across the shape and take those triangles which are fully contained. This most likely will not give you the optimal coverage but then you could repeatedly shift the grid by a tenth of the grid size in horizontal and vertical direction and see whether you'll find something which covers a larger fraction of the original shape (or you could go in steps of 1/2 then 1/4 etc. of the original grid size in the spirit of a binary search).
If you allow any arbitrary scaling of the shapes you could approximate any (reasonably smooth ?) shape to arbitrary precision by adding smaller and smaller shapes. E.g. if you have a raster image, you can e.g. choose the size of the yellow triangle such that two of them make a pixel on the image and then you can represent any such raster image.
Starting with a 3D mesh, how would you give a rounded appearance to the edges and corners between the polygons of that mesh?
Without wishing to discourage other approaches, here's how I'm currently approaching the problem:
Given the mesh for a regular polyhedron, I can give the mesh's edges a rounded appearance by scaling each polygon along its plane and connecting the edges using cylinder segments such that each cylinder is tangent to each polygon where it meets that polygon.
Here's an example involving a cube:
Here's the cube after scaling its polygons:
Here's the cube after connecting the polygons' edges using cylinders:
What I'm having trouble with is figuring out how to deal with the corners between polygons, especially in cases where more than three edges meet at each corner. I'd also like an algorithm that works for all closed polyhedra instead of just those that are regular.
I post this as an answer because I can't put images into comments.
Sattle point
Here's an image of two brothers camping:
They placed their simple tents right beside each other in the middle of a steep walley (that's one bad place for tents, but thats not the point), so one end of each tent points upwards. At the point where the four squares meet you have a sattle point. The two edges on top of each tent can be rounded normally as well as the two downward edges. But at the sattle point you have different curvature in both directions and therefore its not possible to use a sphere. This rules out Svante's solution.
Selfintersection
The following image shows some 3D polygons if viewed from the side. Its some sharp thing with a hole drilled into it from the other side. The left image shows it before, the right after rounding.
.
The mass thats get removed from the sharp edge containts the end of the drill hole.
There is someething else to see here. The drill holes sides might be very large polygons (lets say it's not a hole but a slit). Still you only get small radii at the top. you can't just scale your polygons, you have to take into account the neighboring polygon.
Convexity
You say you're only removing mass, this is only true if your geometry is convex. Look at the image you posted. But now assume that the viewer is inside the volume. The radii turn away from you and therefore add mass.
NURBS
I'm not a nurbs specialist my self. But the constraints would look something like this:
The corners of the nurbs patch must be at the same position as the corners of the scaled-down polygons. The normal vectors of the nurb surface at the corners must be equal to the normal of the polygon. This should be sufficent to gurarantee that the nurb edge will be a straight line following the polygon edge. The normals also ensure that no visible edges will result at the border between polygon and nurbs patch.
I'd just do the math myself. nurbs are just polygons. You'll have some unknown coefficients and your constraints. This gives you a system of equations (often linear) that you can solve.
Is there any upper bound on the number of faces, that meet at that corner?
You might you might employ concepts from CAGD, especially Non-Uniform Rational B-Splines (NURBS) might be of interest for you.
Your current approach - glueing some fixed geometrical primitives might be too inflexible to solve the problem. NURBS require some mathematical work to get used to, but might be more suitable for your needs.
Extrapolating your cylinder-edge approach, the corners should be spheres, resp. sphere segments, that have the same radius as the cylinders meeting there and the centre at the intersection of the cylinders' axes.
Here we have a single C++ header for generating triangulated rounded 3D boxes. The code is in C++ but also easy to transplant to other coding languages. Also it's easy to be modified for other primitives like quads.
https://github.com/nepluno/RoundCornerBox
As #Raymond suggests, I also think that the nepluno repo provides a very good implementation to solve this issue; efficient and simple.
To complete his answer, I just wrote a solution to this issue in JS, based on the BabylonJS 3D engine. This solution can be found here, and can be quite easily replaced by another 3D engine:
https://playground.babylonjs.com/#AY7B23
I've been working on a visualization project for 2-dimensional continuous data. It's the kind of thing you could use to study elevation data or temperature patterns on a 2D map. At its core, it's really a way of flattening 3-dimensions into two-dimensions-plus-color. In my particular field of study, I'm not actually working with geographical elevation data, but it's a good metaphor, so I'll stick with it throughout this post.
Anyhow, at this point, I have a "continuous color" renderer that I'm very pleased with:
The gradient is the standard color-wheel, where red pixels indicate coordinates with high values, and violet pixels indicate low values.
The underlying data structure uses some very clever (if I do say so myself) interpolation algorithms to enable arbitrarily deep zooming into the details of the map.
At this point, I want to draw some topographical contour lines (using quadratic bezier curves), but I haven't been able to find any good literature describing efficient algorithms for finding those curves.
To give you an idea for what I'm thinking about, here's a poor-man's implementation (where the renderer just uses a black RGB value whenever it encounters a pixel that intersects a contour line):
There are several problems with this approach, though:
Areas of the graph with a steeper slope result in thinner (and often broken) topo lines. Ideally, all topo lines should be continuous.
Areas of the graph with a flatter slope result in wider topo lines (and often entire regions of blackness, especially at the outer perimeter of the rendering region).
So I'm looking at a vector-drawing approach for getting those nice, perfect 1-pixel-thick curves. The basic structure of the algorithm will have to include these steps:
At each discrete elevation where I want to draw a topo line, find a set of coordinates where the elevation at that coordinate is extremely close (given an arbitrary epsilon value) to the desired elevation.
Eliminate redundant points. For example, if three points are in a perfectly-straight line, then the center point is redundant, since it can be eliminated without changing the shape of the curve. Likewise, with bezier curves, it is often possible to eliminate cetain anchor points by adjusting the position of adjacent control points.
Assemble the remaining points into a sequence, such that each segment between two points approximates an elevation-neutral trajectory, and such that no two line segments ever cross paths. Each point-sequence must either create a closed polygon, or must intersect the bounding box of the rendering region.
For each vertex, find a pair of control points such that the resultant curve exhibits a minimum error, with respect to the redundant points eliminated in step #2.
Ensure that all features of the topography visible at the current rendering scale are represented by appropriate topo lines. For example, if the data contains a spike with high altitude, but with extremely small diameter, the topo lines should still be drawn. Vertical features should only be ignored if their feature diameter is smaller than the overall rendering granularity of the image.
But even under those constraints, I can still think of several different heuristics for finding the lines:
Find the high-point within the rendering bounding-box. From that high point, travel downhill along several different trajectories. Any time the traversal line crossest an elevation threshold, add that point to an elevation-specific bucket. When the traversal path reaches a local minimum, change course and travel uphill.
Perform a high-resolution traversal along the rectangular bounding-box of the rendering region. At each elevation threshold (and at inflection points, wherever the slope reverses direction), add those points to an elevation-specific bucket. After finishing the boundary traversal, start tracing inward from the boundary points in those buckets.
Scan the entire rendering region, taking an elevation measurement at a sparse regular interval. For each measurement, use it's proximity to an elevation threshold as a mechanism to decide whether or not to take an interpolated measurement of its neighbors. Using this technique would provide better guarantees of coverage across the whole rendering region, but it'd be difficult to assemble the resultant points into a sensible order for constructing paths.
So, those are some of my thoughts...
Before diving deep into an implementation, I wanted to see whether anyone else on StackOverflow has experience with this sort of problem and could provide pointers for an accurate and efficient implementation.
Edit:
I'm especially interested in the "Gradient" suggestion made by ellisbben. And my core data structure (ignoring some of the optimizing interpolation shortcuts) can be represented as the summation of a set of 2D gaussian functions, which is totally differentiable.
I suppose I'll need a data structure to represent a three-dimensional slope, and a function for calculating that slope vector for at arbitrary point. Off the top of my head, I don't know how to do that (though it seems like it ought to be easy), but if you have a link explaining the math, I'd be much obliged!
UPDATE:
Thanks to the excellent contributions by ellisbben and Azim, I can now calculate the contour angle for any arbitrary point in the field. Drawing the real topo lines will follow shortly!
Here are updated renderings, with and without the ghetto raster-based topo-renderer that I've been using. Each image includes a thousand random sample points, represented by red dots. The angle-of-contour at that point is represented by a white line. In certain cases, no slope could be measured at the given point (based on the granularity of interpolation), so the red dot occurs without a corresponding angle-of-contour line.
Enjoy!
(NOTE: These renderings use a different surface topography than the previous renderings -- since I randomly generate the data structures on each iteration, while I'm prototyping -- but the core rendering method is the same, so I'm sure you get the idea.)
Here's a fun fact: over on the right-hand-side of these renderings, you'll see a bunch of weird contour lines at perfect horizontal and vertical angles. These are artifacts of the interpolation process, which uses a grid of interpolators to reduce the number of computations (by about 500%) necessary to perform the core rendering operations. All of those weird contour lines occur on the boundary between two interpolator grid cells.
Luckily, those artifacts don't actually matter. Although the artifacts are detectable during slope calculation, the final renderer won't notice them, since it operates at a different bit depth.
UPDATE AGAIN:
Aaaaaaaand, as one final indulgence before I go to sleep, here's another pair of renderings, one in the old-school "continuous color" style, and one with 20,000 gradient samples. In this set of renderings, I've eliminated the red dot for point-samples, since it unnecessarily clutters the image.
Here, you can really see those interpolation artifacts that I referred to earlier, thanks to the grid-structure of the interpolator collection. I should emphasize that those artifacts will be completely invisible on the final contour rendering (since the difference in magnitude between any two adjacent interpolator cells is less than the bit depth of the rendered image).
Bon appetit!!
The gradient is a mathematical operator that may help you.
If you can turn your interpolation into a differentiable function, the gradient of the height will always point in the direction of steepest ascent. All curves of equal height are perpendicular to the gradient of height evaluated at that point.
Your idea about starting from the highest point is sensible, but might miss features if there is more than one local maximum.
I'd suggest
pick height values at which you will draw lines
create a bunch of points on a fine, regularly spaced grid, then walk each point in small steps in the gradient direction towards the nearest height at which you want to draw a line
create curves by stepping each point perpendicular to the gradient; eliminate excess points by killing a point when another curve comes too close to it-- but to avoid destroying the center of hourglass like figures, you might need to check the angle between the oriented vector perpendicular to the gradient for both of the points. (When I say oriented, I mean make sure that the angle between the gradient and the perpendicular value you calculate is always 90 degrees in the same direction.)
In response to your comment to #erickson and to answer the point about calculating the gradient of your function. Instead of calculating the derivatives of your 300 term function you could do a numeric differentiation as follows.
Given a point [x,y] in your image you could calculate the gradient (direction of steepest decent)
g={ ( f(x+dx,y)-f(x-dx,y) )/(2*dx),
{ ( f(x,y+dy)-f(x,y-dy) )/(2*dy)
where dx and dy could be the spacing in your grid. The contour line will run perpendicular to the gradient. So, to get the contour direction, c, we can multiply g=[v,w] by matrix, A=[0 -1, 1 0] giving
c = [-w,v]
Alternately, there is the marching squares algorithm which seems appropriate to your problem, although you may want to smooth the results if you use a coarse grid.
The topo curves you want to draw are isosurfaces of a scalar field over 2 dimensions. For isosurfaces in 3 dimensions, there is the marching cubes algorithm.
I've wanted something like this myself, but haven't found a vector-based solution.
A raster-based solution isn't that bad, though, especially if your data is raster-based. If your data is vector-based too (in other words, you have a 3D model of your surface), you should be able to do some real math to find the intersection curves with horizontal planes at varying elevations.
For a raster-based approach, I look at each pair of neighboring pixels. If one is above a contour level, and one is below, obviously a contour line runs between them. The trick I used to anti-alias the contour line is to mix the contour line color into both pixels, proportional to their closeness to the idealized contour line.
Maybe some examples will help. Suppose that the current pixel is at an "elevation" of 12 ft, a neighbor is at an elevation of 8 ft, and contour lines are every 10 ft. Then, there is a contour line half way between; paint the current pixel with the contour line color at 50% opacity. Another pixel is at 11 feet and has a neighbor at 6 feet. Color the current pixel at 80% opacity.
alpha = (contour - neighbor) / (current - neighbor)
Unfortunately, I don't have the code handy, and there might have been a bit more to it (I vaguely recall looking at diagonal neighbors too, and adjusting by sqrt(2) / 2). I hope this enough to give you the gist.
It occurred to me that what you're trying to do would be pretty easy to do in MATLAB, using the contour function. Doing things like making low-density approximations to your contours can probably be done with some fairly simple post-processing of the contours.
Fortunately, GNU Octave, a MATLAB clone, has implementations of the various contour plotting functions. You could look at that code for an algorithm and implementation that's almost certainly mathematically sound. Or, you might just be able to offload the processing to Octave. Check out the page on interfacing with other languages to see if that would be easier.
Disclosure: I haven't used Octave very much, and I haven't actually tested it's contour plotting. However, from my experience with MATLAB, I can say that it will give you almost everything you're asking for in just a few lines of code, provided you get your data into MATLAB.
Also, congratulations on making a very VanGough-esque slopefield plot.
I always check places like http://mathworld.wolfram.com before going to deep on my own :)
Maybe their curves section would help? Or maybe the entry on maps.
compare what you have rendered with a real-world topo map - they look identical to me! i wouldn't change a thing...
Write the data out as an HGT file (very simple digital elevation data format used by USGS) and use the free and open-source gdal_contour tool to create contours. That works very well for terrestrial maps, the constraint being that the data points are signed 16-bit numbers, which fits the earthly range of heights in metres very well, but may not be enough for your data, which I assume not to be a map of actual terrain - although you do mention terrain maps.
I recommend the CONREC approach:
Create an empty line segment list
Split your data into regular grid squares
For each grid square, split the square into 4 component triangles:
For each triangle, handle the cases (a through j):
If a line segment crosses one of the cases:
Calculate its endpoints
Store the line segment in the list
Draw each line segment in the line segment list
If the lines are too jagged, use a smaller grid. If the lines are smooth enough and the algorithm is taking too long, use a larger grid.