I have arbitrary many lines in 3D space which are all parallel to each other. Now I want to find the convex hull of these lines. To illustrate this, I've drawn a picture:
I know the start- and endpoints of all lines (the blue dots). The lines are not equally long. If a viewer looks in the direction of the lines (marked as the viewer direction in the pic) he sees only the dots. Now I want to find the convex hull of these dots. Hopefully its clear what I mean.
My idea was to project the start or endpoints on a plane which is perpendicular to the line's direction. After that I can apply some kind of convex hull algorithm to these points. But I have no idea how.
Your idea is exactly correct. One way to accomplish this is to define a vector v along your viewing direction, and then rotate v to the z-axis. The same rotation will convert lines to vertical lines. Then drop the z-coordinate of the endpoints to get your projected points. Then compute the convex hull. There are hull algorithms all over the web, including my own here.
Here's a suggestion based on the calculus of variations.
Consider enclosing your collection of parallel line segments in a simple closed curve minimizing the area of the curve given the constraint that it has to enclose all your segments.
Your "curve" is going to be piecewise linear, so there you might be able to use a P.W basis function in the iterations, though it's possible that you could run into some singularities when the algorithm needs to drop a segment.
I'd like to find the largest convex hull which fits in the interior of a set of points. I have a set of points which are roughly circular, with a large number of outlier points outside of the circle I'd like to fit. Imagine a circle with "solar flares"... I want to fit the circle and completely ignore the flares. I've tried various fit and culling strategies, but they aren't working well.
I've searched quite a bit and not found a solution. Thanks in advance.
The notion you need may be alpha shapes. The convex hull is a sub-set of the alpha-shape for an extreme value for alpha. The alpha shape is fitting a set of point closer than the convex hull with some values for alpha.
Theory has been developed by Edelbrunner. This is a good start: http://www.mpi-inf.mpg.de/~jgiesen/tch/sem06/Celikik.pdf
For computation, you must: compute delaunay triangulation and/or voronoi diagram, then select points that observe one condition.
Example alpha shape:
This is in fact a concave hull, and it may disregard outliers.
Standard convex hull algorithms will not work with (longitude, latitude)-points, because standard algorithms assume you want the hull of a set of Cartesian points. Latitude-longitude points are not Cartesian, because longitude "wraps around" at the anti-meridian (+/- 180 degrees). I.e., two degrees east of longitude 179 is -179.
So if your set of points happens to straddle the anti-meridian, you will compute spurious hulls that stretch all the way around the world incorrectly.
Any suggestions for tricks I could apply with a standard convex hull algorithm to correct for this, or pointers to proper "geospherical" hull algorithms?
Now that I think on it, there are more interesting cases to consider than straddling the anti-merdian. Consider a "band" of points that encircle the earth -- its convex hull would have no east/west bounds. Or even further, what is the convex hull of {(0,0), (0, 90), (0, -90), (90, 0), (-90, 0), (180, 0)}? -- it would seem to contain the entire surface of the earth, so which points are on its perimeter?
Standard convex hull algorithms are not defeated by the wrapping-around of the coordinates on the surface of the Earth but by a more fundamental problem. The surface of a sphere (let's forget the not-quite-sphericity of the Earth) is not a Euclidean space so Euclidean geometry doesn't work, and convex hull routines which assume that the underlying space is Euclidean (show me one which doesn't, please) won't work.
The surface of the sphere conforms to the concepts of an elliptic geometry where lines are great circles and antipodal points are considered the same point. You've already started to experience the issues arising from trying to apply a Euclidean concept of convexity to an elliptic space.
One approach open to you would be to adopt the definitions of geodesic convexity and implement a geodesic convex hull routine. That looks quite hairy. And it may not produce results which conform to your (generally Euclidean) expectations. In many cases, for 3 arbitrary points, the convex hull turns out to be the entire surface of the sphere.
Another approach, one adopted by navigators and cartographers through the ages, would be to project part of the surface of the sphere (a part containing all your points) into Euclidean space (which is the subject of map projections and I won't bother you with references to the extensive literature thereon) and to figure out the convex hull of the projected points. Project the area you are interested in onto the plane and adjust the coordinates so that they do not wrap around; for example, if you were interested in France you might adjust all longitudes by adding 30deg so that the whole country was coordinated by +ve numbers.
While I'm writing, the idea proposed in #Li-aung Yip's answer, of using a 3D convex hull algorithm, strikes me as misguided. The 3D convex hull of the set of surface points will include points, edges and faces which lie inside the sphere. These literally do not exist on the 2D surface of the sphere and only change your difficulties from wrestling with the not-quite-right concept in 2D to quite-wrong in 3D. Further, I learned from the Wikipedia article I referenced that a closed hemisphere (ie one which includes its 'equator') is not convex in the geometry of the surface of the sphere.
Instead of considering your data as latitude-longitude data, could you instead consider it in 3D space and apply a 3D convex hull algorithm? You may be able to then find the 2D convex hull you desire by analysing the 3D convex hull.
This returns you to well-travelled algorithms for cartesian convex hulls (albeit in three dimensions) and has no issues with wrap around of the coordinates.
Alternately, there's this paper: Computing the Convex Hull of a Simple Polygon on the Sphere (1996) which seems to deal with some of the same issues that you're dealing with (coordinate wrap-around, etc.)
If all your points are within a hemisphere (that is, if you can find a cut plane through the center of the Earth that puts them all on one side), then you can do a central a.k.a. gnomic a.k.a. gnomonic projection from the center of the Earth to a plane parallel to the cut plane. Then all great circles become straight lines in the projection, and so a convex hull in the projection will map back to a correct convex hull on the Earth. You can see how wrong lat/lon points are by looking at the latitude lines in the "Gnomonic Projection" section here (notice that the longitude lines remain straight).
(Treating the Earth as a sphere still isn't quite right, but it's a good second approximation. I don't think points on a true least-distance path across a more realistic Earth (say WGS84) generally lie on a plane through the center. Maybe pretending they do gives you a better approximation than what you get with a sphere.)
FutureNerd:
You are absolutely correct. I had to solve the exact same problem as Maxy-B for my application. As a first iteration, I just treated (lng,lat) as (x,y) and ran a standard 2D algorithm. This worked fine as long as nobody looked too close, because all my data were in the contiguous U.S. As a second iteration, though, I used your approach and proved the concept.
The points MUST be in the same hemisphere. As it turns out, choosing this hemisphere is non-trivial (it's not just the points' center, as I had initially guessed.) To illustrate, consider the following four points: (0,0), (-60,0), (+60,0) along the equator, and (0,90) the north pole. However you choose to define "center", their center lies on the north pole by symmetry and all four points are in the Northern Hemisphere. However, consider replacing the fourth point with, say (-19, 64) iceland. Now their center is NOT on the north pole, but asymmetrically drawn toward iceland. However, all four points are still in the Northern Hemisphere. Further, the Northern Hemisphere, as uniquely defined by the North Pole, is the ONLY hemisphere they share. So calculating this "pole" becomes algorithmic, not algebraic.
See my repository for the Python code:
https://github.com/VictorDavis/GeoConvexHull
This question has been answered a while ago, but I would like to sum up the results of my research.
The spherical convex hull is basically defined only for non-antipodal points. Supposing all the points are on the same hemisphere, you can compute their convex hull in two main ways:
Project the points to a plane using gnomonic/central projection and apply a planar convex hull algorithm. See Lin-Lin Chen, T. C. Woo, "Computational Geometry on the Sphere With Application to Automated Machining" (1992). If the points are on a known hemisphere, you can hard-code on which plane to project the points unto.
Adapt planar convex hull algorithms to the sphere. See C. Grima and A. Marquez, "Computational Geometry on Surfaces: Performing Computational Geometry on the Cylinder, the Sphere, the Torus, and the Cone", Springer (2002). This reference seems to give a similar method to the abstract referenced by Li-aung Yip above.
For reference, in Python I am working on an implementation of my own, which currently works only for points on the Northern hemisphere.
See also this question on Math Overflow.
All edges of a spherical convex hull can be viewed/treated as great circles (seminally, all edges of a convex hull in euclidean space can be treated as lines (rather than a line segment)). Each one of these great circles cuts the sphere into two hemispheres. You could thus conceive each great circle as a constraint. A point that is within the convex hull will be on each of the hemispheres defined by each constraint.
Each edge of the original polygon is a candidate edge of the convex hull. To verify if it is indeed an edge of the convex hull, you'd simply need to verify if all nodes of the polygon are on the hemisphere defined by the great circle that goes through the two nodes of the edge in question. However, we'd still need to create new edges that surpass the concave nodes of the polygon.
But lets rather shortcut / brute-forces this:
Draw a great circle between every pair of nodes in the polygon. Do this in both directions (i.e. the great circle connecting A to B and the great circle connecting B to A). For a polygon with N nodes, you will thus end up with N^2 great circle. Each one of these great circles is a candidate constraint (i.e. a candidate edge of the convex polygon). Some of these great circles will overlap with the edges of the original polygon, but most won't. Now, remember again: each great circle is a constraint that constrains the sphere to one hemisphere. Now verify if all nodes of the original polygon satisfy the constraint (i.e. if all nodes are on the hemisphere defined by the great circle). If yes, then this great circle is an edge of the convex hull. If, however a single node of the original polygon does not satisfy the constraint, then it isn't and you can discard this great circle.
The beauty of this is that once you converted your latitudes and longitudes into cartesian vectors pointing onto the unit sphere, it really just requires dot products and cross products
- You find the great circle that passes through two points on a sphere by its cross product
- A point is on the hemisphere defined by a great circle if the dot product of the great circle and the point is greater (or equal) to 0.
So even for polygons with a large number of edges, this brute force method should work just fine.
A planar Voronoi tesselation will always have unbounded polygons
(infinite regions).
How do I tell qhull/qvoronoi to intersect the tesselation with either
the original points' convex hull, or with a polygon I explicitly
specify (probably a rectangle).
Specific example ( https://github.com/barrycarter/bcapps/blob/master/bc-temperature-voronoi.pl ):
I map latitudes/longitudes (inaccurately) to the 2-D rectangle
(-180,180)*(-90,90), and apply qvoronoi to the result.
I can't handle unbounded polygons and Google Maps API dislikes
polygons that wander outside (-180,180)(-90,90) [actually, it's more like (-180,180)(-85,85) because Google uses Mercator]
Can I make qhull/qvoronoi intersect the polygons with the points'
convex hull and/or the (-180,180)*(-90,90) rectangle? Of course, the
convex hull is entirely contained within the rectangle.
If not, can another program do it? I suspect I could write this
myself, but would prefer an existing solution.
I have a group of points (x,y) and I need to find out the distance between the two that are farthest apart.
What is the most efficient way to find this?
Thanks
Well, compairing every point against every other point is certainly not efficient.
The most efficient way involves finding the convex hull, which is the convex polygon (no angles > 180) surrounding all points.
After that, you find the farthest points on the hull, using antipodal pairs.
Algorithm described here:
http://www.seas.gwu.edu/~simhaweb/cs153/lectures/module1/module1.html