Suppose random points P1 to P20 scattered in a plane.
Then is there any way to sort those points in either clock-wise or anti-clock wise.
Here we can’t use degree because you can see from the image many points can have same degree.
E.g, here P4,P5 and P13 acquire the same degree.
If your picture has realistic distance between the points, you might get by with just choosing a point at random, say P1, and then always picking the nearest unvisited neighbour as your next point. Traveling Salesman, kind of.
Are you saying you want an ordered result P1, P2, ... P13?
If that's the case, you need to find the convex hull of the points. Walking around the circumference of the hull will then give you the order of the points that you need.
In a practical sense, have a look at OpenCV's documentation -- calling convexHull with clockwise=true gives you a vector of points in the order that you want. The link is for C++, but there are C and Python APIs there as well. Other packages like Matlab should have a similar function, as this is a common geometrical problem to solve.
EDIT
Once you get your convex hull, you could iteratively collapse it from the outside to get the remaining points. Your iterations would stop when there are no more pixels left inside the hull. You would have to set up your collapse function such that closer points are included first, i.e. such that you get:
and not:
In both diagrams, green is the original convex hull, the other colors are collapsed areas.
Find the right-most of those points (in O(n)) and sort by the angle relative to that point (O(nlog(n))).
It's the first step of graham's convex-hull algorithm, so it's a very common procedure.
Edit: Actually, it's just not possible, since the polygonal representation (i.e. the output-order) of your points is ambiguous. The algorithm above will only work for convex polygons, but it can be extended to work for star-shaped polygons too (you need to pick a different "reference-point").
You need to define the order you actually want more precisely.
Related
I'm looking for an algorithm that provides what I call a "shrunken convex hull" (as distinct from a "reduced convex hull") in 3D. I am defining the shrunken hull, H', as the volume of space that has, no less than D distance from some original convex hull, H.
Analytically, this can be formed by moving each plane of H inwards along its normal by D, then computing the convex hull (if it exists) of the resultant planes. The tricky bit is some planes might be trimmed or dropped, others may move past other planes, and get entirely "snipped" out due to normal reversal (if D is big enough). I’m a bit fuzzy on how to do the algorithm, but have some badly thought out ideas below.
I am doing this to to identify the subset of points in a dataset which are guaranteed to be no less than a given distance from the surface of the original point set (which is assumed to be convex, and I have this). This is to remove surface effects that are disrupting our signal in some calculations we are doing.
I'm really looking for a name, or examples of anyone doing this, or another way to compute this. Ideally some good-old open code would be great, but I think my problem is far too niche.
I found reduced convex hulls, but this seems to be a different idea. The current closest thing I can find is "Hausdorff Cores" - however this seems like the more complicated case of non-convex polygons, and is pretty damned dense.
Do not read beyond here, unless you really really want to.
Current, incomplete/badly thought out algorithm
The slow way (i.e. current way) of identifying the reduced point set it is to compute the signed distance for all points, and reject those that are less than a given distance. However, this is pretty damned slow, as the number of points can be up to 100M. I think operating on the original hull to generate the shrunken hull, and computing its AABB and spherical BB, then retaining only those inside the shrunken hull might be much faster (I hope -willing to accept comments saying this is stupid).
I think it should be possible, as I don't strictly need the full distance information for each point, just D_point > D. So once I know this I should be able to stop.
I can see how the shrunken hull might be done in 2D, where you look at each vertex, then use an analytical solution to a constant velocity Eikonal, then move the vertex along the vector derived from each corner.
However, the situation is more complex for the 3D version, afaics, as there are multiple facets (>2) for each vertex . My current plan is to look at each edge pair individually, then work from there to (somehow - create half spaces and union them?) to build this hull.
What your thinking of is downscaling the 3D convex hull, it works just like downscaling a 2D image, except for how the angle
Outline for the algorithm (in 2D) looks something like this:
1. Compute the convex hull.
2. For each point, P, in the convex hull:
3. Find the hull points before and after, P
4. Bisect the angle formed to obtain the angle, A, required.
5. Create a new point, P', along the angle A at a distance, D, from `P`.
7. Add P' to the scaled-down (shrunken) convex hull.
The only difference in 3D occurs in lines 3 and 4. In 3D, step 3 obtains 3 points. In step 4, a 3D angle is used. Thus you'll find a fair bit of benefit in using the 3D transforms in a graphics/geometry libary, as the math may be tricky.
If your objective is to remove surface effects, and it's not important that every surface of the convex hull be displaced by the same distance, you could instead
Identify a point known to be inside the hull (e.g. the centroid of the point cloud or the hull)
Scale the hull inward towards that point
Unless you scale infinitely (collapsing everything to a point), this operation should give an inwardly-displaced hull which has the same connectivity - no points added or removed.
Im looking for some fairly easy (I know polygon union is NOT an easy operation but maybe someone could point me in the right direction with a relativly easy one) algorithm on merging two intersecting polygons. Polygons could be concave without holes and also output polygon should not have holes in it. Polygons are represented in counter-clockwise manner. What I mean is presented on a picture. As you can see even if there is a hole in union of polygons I dont need it in the output. Input polygons are for sure without holes. I think without holes it should be easier to do but still I dont have an idea.
Remove all the vertices of the polygons which lie inside the other polygon: http://paulbourke.net/geometry/insidepoly/
Pick a starting point that is guaranteed to be in the union polygon (one of the extremes would work)
Trace through the polygon's edges in counter-clockwise fashion. These are points in your union. Trace until you hit an intersection (note that an edge may intersect with more than one edge of the other polygon).
Find the first intersection (if there are more than one). This is a point in your Union.
Go back to step 3 with the other polygon. The next point should be the point that makes the greatest angle with the previous edge.
You can proceed as below:
First, add to your set of points all the points of intersection of your polygons.
Then I would proceed like graham scan algorithm but with one more constraint.
Instead of selecting the point that makes the highest angle with the previous line (have a look at graham scan to see what I mean (*), chose the one with the highest angle that was part of one of the previous polygon.
You will get an envellope (not convex) that will describe your shape.
Note:
It's similar to finding the convex hull of your points.
For example graham scan algorithm will help you find the convex hull of the set of points in O (N*ln (N) where N is the number of points.
Look up for convex hull algorithms, and you can find some ideas.
Remarques:
(*)From wikipedia:
The first step in this algorithm is to find the point with the lowest
y-coordinate. If the lowest y-coordinate exists in more than one point
in the set, the point with the lowest x-coordinate out of the
candidates should be chosen. Call this point P. This step takes O(n),
where n is the number of points in question.
Next, the set of points must be sorted in increasing order of the
angle they and the point P make with the x-axis. Any general-purpose
sorting algorithm is appropriate for this, for example heapsort (which
is O(n log n)). In order to speed up the calculations, it is not
necessary to calculate the actual angle these points make with the
x-axis; instead, it suffices to calculate the cosine of this angle: it
is a monotonically decreasing function in the domain in question
(which is 0 to 180 degrees, due to the first step) and may be
calculated with simple arithmetic.
In the convex hull algorithm you chose the point of the angle that makes the largest angle with the previous side.
To "stick" with your previous polygon, just add the constraint that you must select a side that previously existed.
And you take off the constraint of having angle less than 180°
I don't have a full answer but I'm about to embark on a similar problem. I think there are two step which are fairly important. First would be to find a point on some polygon which lies on the outside edge. Second would be to make a list of bounding boxes for all the vertices and see which of these overlap. This means when you iterate through vertices, you don't have to do tests for all of them, only those which you know have a chance of intersecting (bounding box problems are lightweight).
Since you now have an outside point, you can now iterate through connected points until you detect an intersection. If you know which side is inside and which outside (you may need to do some work on the first vertex to know this), you know which way to go on the intersection. Then it's merely a matter of switching polygons.
This gets a little more interesting if you want to maintain that hole (which I do) in which case, I would probably make sure I had used up all my intersecting bounding boxes. You also didn't specify what should happen if your polygons don't intersect at all. But that's either going to be leave them alone (which could potentially be a problem if you're expecting one polygon out) or return an error.
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How to find largest triangle in convex hull aside from brute force search
I have a set of random points from which i want to find the largest triangle by area who's verticies are each on one of those points.
So far I have figured out that the largest triangle's verticies will only lie on the outside points of the cloud of points (or the convex hull) so i have programmed a function to do just that (using Graham scan in nlogn time).
However that's where I'm stuck. The only way I can figure out how to find the largest triangle from these points is to use brute force at n^3 time which is still acceptable in an average case as the convex hull algorithm usually kicks out the vast majority of points. However in a worst case scenario where points are on a circle, this method would fail miserably.
Dose anyone know an algorithm to do this more efficiently?
Note: I know that CGAL has this algorithm there but they do not go into any details on how its done. I don't want to use libraries, i want to learn this and program it myself (and also allow me to tweak it to exactly the way i want it to operate, just like the graham scan in which other implementations pick up collinear points that i don't want).
Don't know if this help, but if you choose two points from the convex hull and rotate all points of the hull so that the connecting line of the two points is parallel to the x-Axis, either the point with the maximum or the one with the minimum y-coordinate forms the triangle with the largest area together with the two points chosen first.
Of course once you have tested one point for all possible base lines, you can remove it from the list.
Here's a thought on how to get it down to O(n2 log n). I don't really know anything about computational geometry, so I'll mark it community wiki; please feel free to improve on this.
Preprocess the convex hull by finding for each point the range of slopes of lines through that point such that the set lies completely on one side of the line. Then invert this relationship: construct an interval tree for slopes with points in leaf nodes, such that when querying with a slope you find the points such that there is a tangent through those points.
If there are no sets of three or more collinear points on the convex hull, there are at most four points for each slope (two on each side), but in case of collinear points we can just ignore the intermediate points.
Now, iterate through all pairs of points (P,Q) on the convex hull. We want to find the point R such that triangle PQR has maximum area. Taking PQ as the base of the triangle, we want to maximize the height by finding R as far away from the line PQ as possible. The line through R parallel to PQ must be such that all points lie on one side of the line, so we can find a bounded number of candidates in time O(log n) using the preconstructed interval tree.
To improve this further in practice, do branch-and-bound in the set of pairs of points: find an upper bound for the height of any triangle (e.g. the maximum distance between two points), and discard any pair of points whose distance multiplied by this upper bound is less than the largest triangle found so far.
I think the rotating calipers method may apply here.
Off the top of my head, perhaps you could do something involving gridding/splitting the collection of points up into groups? Maybe... separating the points into three groups (not sure what the best way to do that in this case would be, though), doing something to discard those points in each group that are closer to the other two groups than other points in the same group, and then using the remaining points to find the largest triangle that can be made having one vertex in each group? This would actually make the case of all points being on a circle a lot simpler, because you'd just focus on the points that are near the center of the arcs contained within each group, as those would be the ones in each group furthest from the other two groups.
I'm not sure if this would give you the proper result for certain triangles/distributions of points, though. There may be situations where the resultant triangle isn't of optimal area, either because the grouping and/or the vertex choosing aren't/isn't optimal. Something like that.
Anyway, those are my thoughts on the problem. I hope I've at least been able to give you ideas for how to work on it.
How about dropping a point at a time from the convex hull? Starting with the convex hull, calculate the area of the triangle formed by each triple of adjacent points (p1p2p3, p2p3p4, etc.). Find the triangle with minimum area, then drop the middle of the three points that formed that triangle. (In other words, if the smallest area triangle is p3p4p5, drop P4.) Now you have a convex polygon with N-1 points. Repeat the same procedure until you are left with three points. This should take O(N^2) time.
I would not be at all surprised if there is some pathological case where this doesn't work, but I expect that it would work for the majority of cases. (In other words, I haven't proven this, and I have no source to cite.)
I need to compute the area of the region of overlap between two triangles in the 2D plane. Oddly, I have written up code for the triangle-circle problem, and that works quite well and robustly, but I have trouble with the triangle-triangle problem.
I already first check to see if one entirely contains the other, or if the other contains the first, as well as obtain all the edge-wise intersection points. These intersection points (up to 6, as in the star of David), combined with the triangle vertices that are contained within the other triangle, are the vertices of the intersection region. These points must form a convex polygon.
The solution I seek is the answer to either of these questions:
Given a set of points known to all lie on the convex hull of the point set, compute the area of the convex hull. Note that they are in random order.
Given a set of half-planes, determine the intersecting area. This is equivalent to describing both triangles as the intersection of three half-planes, and computing the solution as the direct intersection of this description.
I have considered for question 1 simply adding up all areas of all possible triangles, and then dividing by the multiplicity in counting, but that seems dumb, and I'm not sure if it is correct. I feel like there is some kind of sweep-line algorithm that would do the trick. However, the solution must also be relatively numerically robust.
I simply have no idea how to solve question 2, but a general answer would be very useful, and providing code would make my day. This would allow for direct computation of intersection areas of convex polygons instead of having to perform a triangle decomposition on them.
Edit: I am aware of this article which describes the general case for finding the intersection polygon of two convex polygons. It seems rather involved for just triangles, and furthermore, I don't really need the resulting polygon itself. So maybe this question is just asked in laziness at this point.
Question 1: why are the points in a random order? If they are, you have to order them so that connecting consecutive points with straight lines yields a convex polygon. How to order them -- for example, by running a convex hull algorithm (though there are probably also simpler methods). Once you have ordered them, compute the area as described here.
--
Question 2 is simpler. Half-plane is defined by a single line having an implicit equation a*x+b*y+c=0 ; all points (x, y) for which a*x+b*y+c <= 0 (note the inequality) are "behind" the half-plane. Now, you need at least three planes so that the intersection of their negative half-spaces is closed (this is necessary, but not sufficient condition). If the intersection is closed, it will be a convex polygon.
I suggest that you maintain a linked list of vertices. The algorithm is initialized with THREE lines. Compute the three points (in general case) where the lines intersect; these are the starting vertices of your region (triangle). You must also check that each vertex is "behind" the half-plane defined by the line going through the other two vertices; this guarantees that the intersection actually IS a closed region.
These three vertices define also the the three edges of a triangle. When you intersect by a new half-plane, simply check for the intersection between the line defining the half-plane and each of the edges of the current region; in general you will get two intersection points, but you must watch out for degenerate cases where the line goes through a vertex of the region. (You can also end up with an empty set!)
The new intersection vertices define a line that splits the current region in TWO regions. Again, use orientation of the new half-plane to decide which of the two new regions to assign to the new "current region", and which one to discard.
The points in the list defining the edges of the current region will be correctly ordered so you can apply the formula in the above link to compute its area.
If this description is not detailed/understandable, the next-best advice I can give you is that you invest in a book on computational geometry and linear algebra.
Having a set of (2D) points from a GIS file (a city map), I need to generate the polygon that defines the 'contour' for that map (its boundary). Its input parameters would be the points set and a 'maximum edge length'. It would then output the corresponding (probably non-convex) polygon.
The best solution I found so far was to generate the Delaunay triangles and then remove the external edges that are longer than the maximum edge length. After all the external edges are shorter than that, I simply remove the internal edges and get the polygon I want. The problem is, this is very time-consuming and I'm wondering if there's a better way.
One of the former students in our lab used some applicable techniques for his PhD thesis. I believe one of them is called "alpha shapes" and is referenced in the following paper:
http://www.cis.rit.edu/people/faculty/kerekes/pdfs/AIPR_2007_Gurram.pdf
That paper gives some further references you can follow.
This paper discusses the Efficient generation of simple polygons for characterizing the shape of a set of points in the plane and provides the algorithm. There's also a Java applet utilizing the same algorithm here.
The guys here claim to have developed a k nearest neighbors approach to determining the concave hull of a set of points which behaves "almost linearly on the number of points". Sadly their paper seems to be very well guarded and you'll have to ask them for it.
Here's a good set of references that includes the above and might lead you to find a better approach.
The answer may still be interesting for somebody else: One may apply a variation of the marching square algorithm, applied (1) within the concave hull, and (2) then on (e.g. 3) different scales that my depend on the average density of points. The scales need to be int multiples of each other, such you build a grid you can use for efficient sampling. This allows to quickly find empty samples=squares, samples that are completely within a "cluster/cloud" of points, and those, which are in between. The latter category then can be used to determine easily the poly-line that represents a part of the concave hull.
Everything is linear in this approach, no triangulation is needed, it does not use alpha shapes and it is different from the commercial/patented offering as described here ( http://www.concavehull.com/ )
A quick approximate solution (also useful for convex hulls) is to find the north and south bounds for each small element east-west.
Based on how much detail you want, create a fixed sized array of upper/lower bounds.
For each point calculate which E-W column it is in and then update the upper/lower bounds for that column. After you processed all the points you can interpolate the upper/lower points for those columns that missed.
It's also worth doing a quick check beforehand for very long thin shapes and deciding wether to bin NS or Ew.
A simple solution is to walk around the edge of the polygon. Given a current edge om the boundary connecting points P0 and P1, the next point on the boundary P2 will be the point with the smallest possible A, where
H01 = bearing from P0 to P1
H12 = bearing from P1 to P2
A = fmod( H12-H01+360, 360 )
|P2-P1| <= MaxEdgeLength
Then you set
P0 <- P1
P1 <- P2
and repeat until you get back where you started.
This is still O(N^2) so you'll want to sort your pointlist a little. You can limit the set of points you need to consider at each iteration if you sort points on, say, their bearing from the city's centroid.
Good question! I haven't tried this out at all, but my first shot would be this iterative method:
Create a set N ("not contained"), and add all points in your set to N.
Pick 3 points from N at random to form an initial polygon P. Remove them from N.
Use some point-in-polygon algorithm and look at points in N. For each point in N, if it is now contained by P, remove it from N. As soon as you find a point in N that is still not contained in P, continue to step 4. If N becomes empty, you're done.
Call the point you found A. Find the line in P closest to A, and add A in the middle of it.
Go back to step 3
I think it would work as long as it performs well enough — a good heuristic for your initial 3 points might help.
Good luck!
You can do it in QGIS with this plug in;
https://github.com/detlevn/QGIS-ConcaveHull-Plugin
Depending on how you need it to interact with your data, probably worth checking out how it was done here.
As a wildly adopted reference, PostGIS starts with a convexhull and then caves it in, you can see it here.
https://github.com/postgis/postgis/blob/380583da73227ca1a52da0e0b3413b92ae69af9d/postgis/postgis.sql.in#L5819
The Bing Maps V8 interactive SDK has a concave hull option within the advanced shape operations.
https://www.bing.com/mapspreview/sdkrelease/mapcontrol/isdk/advancedshapeoperations?toWww=1&redig=D53FACBB1A00423195C53D841EA0D14E#JS
Within ArcGIS 10.5.1, the 3D Analyst extension has a Minimum Bounding Volume tool with the geometry types of concave hull, sphere, envelope, or convex hull. It can be used at any license level.
There is a concave hull algorithm here: https://github.com/mapbox/concaveman