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.
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 want to write a program to draw a picture which covers a plane with tiled irregular quadrangles, just like this one:
However, I don't know the relevant algorithms, for example, in which order should I draw the edges?
Could someone point a direction for me?
Apologies, in my previous answer, I misunderstood the question.
Here is one stab at an algorithm (not necessarily the most optimal way, but a way). All you need is the ability to render a polygon and a basic rotation.
If you don't want the labels to be flipped, draw them separately (the labels can be stored in the vertices, e.g., and rotated with the polygon points, but drawn upright as text).
Edit
I received a question about the "start with an arbitrary polygon" step. I didn't communicate that step very clearly, as I actually intended to merely suggest an arbitrary polygon from the provided diagram, and not any arbitrary polygon in the world.
However, this should work at least for arbitrary quads, including concave ones, like so:
I'm afraid I lack the proper background to provide a proof as to why this works, however. Perhaps more mathematically-savvy people can help there with the proof.
I think one way to tackle the proof is to first start with the notion that all tiled edges are manifold -- this is a given considering that we're generating a neighboring polygon at every edge in order to generate the tiled result. Then we might be able to prove that every 2-valence boundary vertex is going to become a 4-valence vertex as a result of this operation (since each of its two edges are going to become manifold, and that introduces two new vertex edges into the mix -- this seems like the hardest part to prove to me). Last step might be to prove that the sum of the angles at each 4-valence vertex will always add up to 360 degrees.
I am searching for an algorithm which calculates, from a list of lines, the area a given point (P) is in. The lines are stored with x and y coordinates, the resulting area schould be stored as a polygon which may has holes in it.
Here two images to illustrate what I mean:
Many CAD applications use such algorithms to let the user hatch areas. I don't know how to call such a method so I don't really know what to search for.
Edit 1: To clearify what I'm looking for:
I don't need an algorithm to calculate the area of a polygon, I need an algorithm which returns the polygon which is formed by lines around P. If P is outside any possible polygon, the algorithm fails.
(I also edited the images)
Edit 2: I read the accepted answer of this possible duplicate and it seams I'm looking for a point location algorithm.
I also stumpled across this site which does exactly explains what I'm looking for and more important it led me to the open-source CGAL library which provides the functionality to do such things. Point 4.6 of the CGAL manual gives an example how to use this library to form a region from a bunch of line segments.
One way to do it is to imagine a line from P to infinity. For each polygon test each edge to see if it crosses the line. Count up how many lines cross. If its even then the point is outside the polygon, if its odd then the point is inside the polygon.
This is a fairly standard mathematical result. If you want to get a little deeper into the subject have a look at https://en.wikipedia.org/wiki/Winding_number.
I did a fairly similar things earlier this week Detecting Regions Described by Lines on HTML5 Canvas
this calculates a the polygon around a point given a set of lines. You can see a working example at the jsfiddle mentioned in the answer
the difference is that it uses infinite mathematical lines rather than line segments, but it could easily be modified for this example.
An outline algorithm
First construct two data-structures one of line-segments one of intersection-points of line-segments. In each record which two lines give each intersection, so you can flip from
one to the other. You can get this by pair-wise intersection of line-segments.
It might be easiest to cut the segments at the intersection points so each segment only have two solutions on it, one at each end. We assume we have done this.
Prune the data-structures, remove all line-segments with no intersections and only one segment - these cannot contribute.
Find possible starting lines. You could calculate the distance from each line to the point and take the smallest. You could check for intersections with a line from the point to infinity.
Walk around the polygon anti-clockwise. With you starting line find the most anti-clockwise end. At that point find the most anti-clockwise segment. Follow that and repeat. It may happen that closed loop is formed, in which case discard all the segments in the loop. Continue until you get back to the starting point.
Check if the polygon actually encloses the point. If so we are done. If not discard segments in the polygon and start again.
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.