Overlapping segments in Bentley-Ottmann Algorithm - computational-geometry

I'm trying to implement Bentley-Ottmann Algorithm according to description found in Mark de Berg, et al. Computational Geometry Algorithms and Applications Third Edition book. There is not clear statement about support of overlapping (collinear and not disjoint) segments, however another degenerate cases described precisely.
From wikipedia I found that there exists implementation that handles all degenerate cases, and it descibed in Bartuschka, U.; Mehlhorn, K.; Näher, S. However it is very hard to understand.
Exactly the same approach provided here. In the subsection 4.3 I found the following:
The basic ingredients of the line sweep algorithm go back to work by Bentley and Ottmann from 1979. The particular formulation discussed here, which takes all possible degeneracies into account, is due to Mehlhorn and Näher.
After thorough investigation of it, I come to conclusion that the only difference is that sweep line going from left to right, but not from top to bottom as in de Berg's book. But I can't understand why this algorithm handles overlapping segments. I want to see some proof why such segments are handled correctly and how are they handled. What should we output as intersection: there is no intersections at all or segments are intersect in infinite number of points?
(I've alrady looked at The Generalization of Bentley-Ottmann Algorithm.)
Plase help me figure out how and why this algorithm handle overlapping segments.

Overlapping (by which I assume you mean, collinear and not disjoint) segments aren't that difficult to handle. You have a couple of options.
You can merge overlapping segments when you encounter them at an event, treating them as one segment for the purposes of the algorithm.
You can decide that overlapping segments don't intersect at all, in which case you don't really have to do anything except filter out those intersections at events. In order to simplify the logic of event handling and locating segments on the sequence, though, it is useful to come up with a canonical order for the segments, such that if two segments are collinear, the order is used to determine which one's on top.

Related

Given an unordered list of n-dimensional points, how can I best find the smallest volume defined by n+1 of those points that encloses a given point?

Note: I'm very much from a programming background, not a mathematics background. This will become obvious very quickly.
Assume I have a bounded n-dimensional space - for example here I'll use n=2. In that space, I have a set of pre-defined points. (As it happens, I'm doing something questionable with genetic algorithms for finding minima in non-mathematically-solvable equations, but that's not directly relevant).
Next, I have defined a new point within that 2D space. I want to know which three (n+1) points form the smallest (or possibly nearest?) triangle that contains that point. Illustration here:
Now, as that illustration shows, I'm not entirely sure what I'm doing, in that I've failed to adequately describe the criteria by which the candidate triangles are judged - points 10, 5, and 8 would enclose the point within a triangle of smaller area, for example. This is because those specifics are somewhat flexible. What I really care about is:
Computational efficiency: I could potentially end up scaling this algorithm up to thousands of points in a hundred dimensions. As such, I need a solution that's better than exhaustively testing every potential convex hull against a specific equation, and ideally one with a decent big-O notation. Presumably I'm going to need a more intelligent structure that a big unordered list to do this.
If I had a preferred criteria, it's proximity of the hull vertices to the test point. However, if it's easier to build an algorithm which judges by area/volume or something else, I can live with that.
I need to be able to handle edge-cases, where the test-point is, for example, on the edge between two vertices.
Where do I even start with this? Some cursory googling suggests Voronoi diagrams might be a step in the right direction, is that correct? What are the right tools for this job? Any help would be greatly appreciated.

How to break a geometry into blocks?

I am certain there is already some algorithm that does what I need, but I am not sure what phrase to Google, or what is the algorithm category.
Here is my problem: I have a polyhedron made up by several contacting blocks (hyperslabs), i. e. the edges are axis aligned and the angles between edges are 90°. There may be holes inside the polyhedron.
I want to break up this concave polyhedron in as little convex rectangular axis-aligned whole blocks are possible (if the original polyhedron is convex and has no holes, then it is already such a block, and therefore, the solution). To illustrate, some 2-D images I made (but I need the solution for 3-D, and preferably, N-D):
I have this geometry:
One possible breakup into blocks is this:
But the one I want is this (with as few blocks as possible):
I have the impression that an exact algorithm may be too expensive (is this problem NP-hard?), so an approximate algorithm is suitable.
One detail that maybe make the problem easier, so that there could be a more appropriated/specialized algorithm for it is that all edges have sizes multiple of some fixed value (you may think all edges sizes are integer numbers, or that the geometry is made up by uniform tiny squares, or voxels).
Background: this is the structured grid discretization of a PDE domain.
What algorithm can solve this problem? What class of algorithms should I
search for?
Update: Before you upvote that answer, I want to point out that my answer is slightly off-topic. The original poster have a question about the decomposition of a polyhedron with faces that are axis-aligned. Given such kind of polyhedron, the question is to decompose it into convex parts. And the question is in 3D, possibly nD. My answer is about the decomposition of a general polyhedron. So when I give an answer with a given implementation, that answer applies to the special case of polyhedron axis-aligned, but it might be that there exists a better implementation for axis-aligned polyhedron. And when my answer says that a problem for generic polyhedron is NP-complete, it might be that there exists a polynomial solution for the special case of axis-aligned polyhedron. I do not know.
Now here is my (slightly off-topic) answer, below the horizontal rule...
The CGAL C++ library has an algorithm that, given a 2D polygon, can compute the optimal convex decomposition of that polygon. The method is mentioned in the part 2D Polygon Partitioning of the manual. The method is named CGAL::optimal_convex_partition_2. I quote the manual:
This function provides an implementation of Greene's dynamic programming algorithm for optimal partitioning [2]. This algorithm requires O(n4) time and O(n3) space in the worst case.
In the bibliography of that CGAL chapter, the article [2] is:
[2] Daniel H. Greene. The decomposition of polygons into convex parts. In Franco P. Preparata, editor, Computational Geometry, volume 1 of Adv. Comput. Res., pages 235–259. JAI Press, Greenwich, Conn., 1983.
It seems to be exactly what you are looking for.
Note that the same chapter of the CGAL manual also mention an approximation, hence not optimal, that run in O(n): CGAL::approx_convex_partition_2.
Edit, about the 3D case:
In 3D, CGAL has another chapter about Convex Decomposition of Polyhedra. The second paragraph of the chapter says "this problem is known to be NP-hard [1]". The reference [1] is:
[1] Bernard Chazelle. Convex partitions of polyhedra: a lower bound and worst-case optimal algorithm. SIAM J. Comput., 13:488–507, 1984.
CGAL has a method CGAL::convex_decomposition_3 that computes a non-optimal decomposition.
I have the feeling your problem is NP-hard. I suggest a first step might be to break the figure into sub-rectangles along all hyperplanes. So in your example there would be three hyperplanes (lines) and four resulting rectangles. Then the problem becomes one of recombining rectangles into larger rectangles to minimize the final number of rectangles. Maybe 0-1 integer programming?
I think dynamic programming might be your friend.
The first step I see is to divide the polyhedron into a trivial collection of blocks such that every possible face is available (i.e. slice and dice it into the smallest pieces possible). This should be trivial because everything is an axis aligned box, so k-tree like solutions should be sufficient.
This seems reasonable because I can look at its cost. The cost of doing this is that I "forget" the original configuration of hyperslabs, choosing to replace it with a new set of hyperslabs. The only way this could lead me astray is if the original configuration had something to offer for the solution. Given that you want an "optimal" solution for all configurations, we have to assume that the original structure isn't very helpful. I don't know if it can be proven that this original information is useless, but I'm going to make that assumption in this answer.
The problem has now been reduced to a graph problem similar to a constrained spanning forest problem. I think the most natural way to view the problem is to think of it as a graph coloring problem (as long as you can avoid confusing it with the more famous graph coloring problem of trying to color a map without two states of the same color sharing a border). I have a graph of nodes (small blocks), each of which I wish to assign a color (which will eventually be the "hyperslab" which covers that block). I have the constraint that I must assign colors in hyperslab shapes.
Now a key observation is that not all possibilities must be considered. Take the final colored graph we want to see. We can partition this graph in any way we please by breaking any hyperslab which crosses the partition into two pieces. However, not every partition is meaningful. The only partitions that make sense are axis aligned cuts, which always break a hyperslab into two hyperslabs (as opposed to any more complicated shape which could occur if the cut was not axis aligned).
Now this cut is the reverse of the problem we're really trying to solve. That cutting is actually the thing we did in the first step. While we want to find the optimal merging algorithm, undoing those cuts. However, this shows a key feature we will use in dynamic programming: the only features that matter for merging are on the exposed surface of a cut. Once we find the optimal way of forming the central region, it generally doesn't play a part in the algorithm.
So let's start by building a collection of hyperslab-spaces, which can define not just a plain hyperslab, but any configuration of hyperslabs such as those with holes. Each hyperslab-space records:
The number of leaf hyperslabs contained within it (this is the number we are eventually going to try to minimize)
The internal configuration of hyperslabs.
A map of the surface of the hyperslab-space, which can be used for merging.
We then define a "merge" rule to turn two or more adjacent hyperslab-spaces into one:
Hyperslab-spaces may only be combined into new hyperslab-spaces (so you need to combine enough pieces to create a new hyperslab, not some more exotic shape)
Merges are done simply by comparing the surfaces. If there are features with matching dimensionalities, they are merged (because it is trivial to show that, if the features match, it is always better to merge hyperslabs than not to)
Now this is enough to solve the problem with brute force. The solution will be NP-complete for certain. However, we can add an additional rule which will drop this cost dramatically: "One hyperslab-space is deemed 'better' than another if they cover the same space, and have exactly the same features on their surface. In this case, the one with fewer hyperslabs inside it is the better choice."
Now the idea here is that, early on in the algorithm, you will have to keep track of all sorts of combinations, just in case they are the most useful. However, as the merging algorithm makes things bigger and bigger, it will become less likely that internal details will be exposed on the surface of the hyperslab-space. Consider
+===+===+===+---+---+---+---+
| : : A | X : : : :
+---+---+---+---+---+---+---+
| : : B | Y : : : :
+---+---+---+---+---+---+---+
| : : | : : : :
+===+===+===+ +---+---+---+
Take a look at the left side box, which I have taken the liberty of marking in stronger lines. When it comes to merging boxes with the rest of the world, the AB:XY surface is all that matters. As such, there are only a handful of merge patterns which can occur at this surface
No merges possible
A:X allows merging, but B:Y does not
B:Y allows merging, but A:X does not
Both A:X and B:Y allow merging (two independent merges)
We can merge a larger square, AB:XY
There are many ways to cover the 3x3 square (at least a few dozen). However, we only need to remember the best way to achieve each of those merge processes. Thus once we reach this point in the dynamic programming, we can forget about all of the other combinations that can occur, and only focus on the best way to achieve each set of surface features.
In fact, this sets up the problem for an easy greedy algorithm which explores whichever merges provide the best promise for decreasing the number of hyperslabs, always remembering the best way to achieve a given set of surface features. When the algorithm is done merging, whatever that final hyperslab-space contains is the optimal layout.
I don't know if it is provable, but my gut instinct thinks that this will be an O(n^d) algorithm where d is the number of dimensions. I think the worst case solution for this would be a collection of hyperslabs which, when put together, forms one big hyperslab. In this case, I believe the algorithm will eventually work its way into the reverse of a k-tree algorithm. Again, no proof is given... it's just my gut instinct.
You can try a constrained delaunay triangulation. It gives very few triangles.
Are you able to determine the equations for each line?
If so, maybe you can get the intersection (points) between those lines. Then if you take one axis, and start to look for a value which has more than two points (sharing this value) then you should "draw" a line. (At the beginning of the sweep there will be zero points, then two (your first pair) and when you find more than two points, you will be able to determine which points are of the first polygon and which are of the second one.
Eg, if you have those lines:
verticals (red):
x = 0, x = 2, x = 5
horizontals (yellow):
y = 0, y = 2, y = 3, y = 5
and you start to sweep through of X axis, you will get p1 and p2, (and we know to which line-equation they belong ) then you will get p3,p4,p5 and p6 !! So here you can check which of those points share the same line of p1 and p2. In this case p4 and p5. So your first new polygon is p1,p2,p4,p5.
Now we save the 'new' pair of points (p3, p6) and continue with the sweep until the next points. Here we have p7,p8,p9 and p10, looking for the points which share the line of the previous points (p3 and p6) and we get p7 and p10. Those are the points of your second polygon.
When we repeat the exercise for the Y axis, we will get two points (p3,p7) and then just three (p1,p2,p8) ! On this case we should use the farest point (p8) in the same line of the new discovered point.
As we are using lines equations and points 2 or more dimensions, the procedure should be very similar
ps, sorry for my english :S
I hope this helps :)

Sorting Geographical non-contiguous line segments along an implied curve

Given:
A Set (for the sake of discussion we will call it S), which is an unordered collection of line segments. Each line segment is defined as two Longitude-Latitude end-points. While all of the line segments follow an implied curve, there are "gaps" between each of the segments, of various sizes. We refer to this curve as "implied" because it is not explicitly defined anywhere. The only information that we have available are the line segments contained within S.
Desired Result:
A sequence (for the sake of discussion we will call it R), which is an ordered collection of line segments. Each line segment is defined just as before, following the same implied curve as before but are now sorted by their position along the implied curve.
Context (i.e. "Why in the heck do I need this?"):
Basically I have incomplete geographical data that needs to be normalized and "completed" by doing some very simple interpolation to form a complete curve with no gaps. You might ask "why not just fit a curve to all the line segment end-points and be done with it?" -- well, that's not quite what I am after. The line segments are precisely where they should be located, and there is no need for the final curve to be "smooth". In fact, I intend to connect each of the segments with a straight-line (the crudest form of interpolation imaginable). But, connecting the segments is easy; the hard part is sorting them.
So In Summary: What would be a performant algorithm for going from S to R?
You can use a k-d tree or a cover tree to find nearby points quickly.
If you need one continuous curve, I would suggest that a short traveling salesman path that incorporates the given edges would be a reasonable reconstruction. You could use 2-opt together with a k-d tree the way Bentley described (paywalled, sorry; I think there's also a description in this chapter on TSP local search by Johnson and McGeoch). The one modification needed would be to ensure that the initial path includes the given edges and that 2-opt moves do not remove those edges.
I guess the implied curve has two properties. One is it is continious which means there is no segments. Second, its first derivative is continious which means there is no corners.
From second property we can say that if the angle between two line is closer to each other, they are more related. But i guess it is not enough. You can define a cost function which depends on the angle between lines and distance of lines.
C = A*angle + B*distance (where A,B should be tested and tuned)
Form this function you can find how much each line is related to another one. Than you can just simply connect the line with the strongest relations. Though i guess greedy algorithm does not mean you will always get the optimal solution.

Reverse Rectangle Packing

I have a connected shape that consists of squares put together, e.g. take a squared paper and draw a line along the existing lines that ends at its beginning and does not cross itself.
The goal is now to find an algorithm (not brute-force) that fills this shape with as few, non-overlapping rectangles as possible.
I'm looking for the optimal solution. As can be seen in the images, the naive greedy approach (take the largest rectangle) does not work.
(Optimal)
(Greedy)
My scenario is vertices reduction, but I'm sure there are other use-cases as well.
Note: This problem seems basic, but I was not able to find a solution elsewhere. Also, is this problem NP-hard?
Edit: I just realized that, in my scenario, filling the shape with as few non-overlapping triangles as possible, would give an even better result.
I've spend a lot of time researching this, since I asked the initial question. For the first problem (optimally filling the shape with rectangles), I've written the solution here under the header "Optimal Greedy Meshing":
http://blackflux.wordpress.com/2014/03/01/meshing-in-voxel-engines-part-2/
The complexity is actually better (faster) than for optimally triangulating a polygon without holes. The slowest part is the Hopcroft-Karp algorithm.
Treating the shape as a polygon is also discussed in the linked blog post. Note that I'm also considering holes.
The first problem is harder than the one with triangles; for triangles, see the algorithms in
http://en.wikipedia.org/wiki/Polygon_triangulation
which can do it without any extra vertices.

Simplified (or smooth) polygons that contain the original detailed polygon

I have a detailed 2D polygon (representing a geographic area) that is defined by a very large set of vertices. I'm looking for an algorithm that will simplify and smooth the polygon, (reducing the number of vertices) with the constraint that the area of the resulting polygon must contain all the vertices of the detailed polygon.
For context, here's an example of the edge of one complex polygon:
My research:
I found the Ramer–Douglas–Peucker algorithm which will reduce the number of vertices - but the resulting polygon will not contain all of the original polygon's vertices. See this article Ramer-Douglas-Peucker on Wikipedia
I considered expanding the polygon (I believe this is also known as outward polygon offsetting). I found these questions: Expanding a polygon (convex only) and Inflating a polygon. But I don't think this will substantially reduce the detail of my polygon.
Thanks for any advice you can give me!
Edit
As of 2013, most links below are not functional anymore. However, I've found the cited paper, algorithm included, still available at this (very slow) server.
Here you can find a project dealing exactly with your issues. Although it works primarily with an area "filled" by points, you can set it to work with a "perimeter" type definition as yours.
It uses a k-nearest neighbors approach for calculating the region.
Samples:
Here you can request a copy of the paper.
Seemingly they planned to offer an online service for requesting calculations, but I didn't test it, and probably it isn't running.
HTH!
I think Visvalingam’s algorithm can be adapted for this purpose - by skipping removal of triangles that would reduce the area.
I had a very similar problem : I needed an inflating simplification of polygons.
I did a simple algorithm, by removing concav point (this will increase the polygon size) or removing convex edge (between 2 convex points) and prolongating adjacent edges. In any case, doing one of those 2 possibilities will remove one point on the polygon.
I choosed to removed the point or the edge that leads to smallest area variation. You can repeat this process, until the simplification is ok for you (for example no more than 200 points).
The 2 main difficulties were to obtain fast algorithm (by avoiding to compute vertex/edge removal variation twice and maintaining possibilities sorted) and to avoid inserting self-intersection in the process (not very easy to do and to explain but possible with limited computational complexity).
In fact, after looking more closely it is a similar idea than the one of Visvalingam with adaptation for edge removal.
That's an interesting problem! I never tried anything like this, but here's an idea off the top of my head... apologies if it makes no sense or wouldn't work :)
Calculate a convex hull, that might be way too big / imprecise
Divide the hull into N slices, for example joining each one of the hull's vertices to the center
Calculate the intersection of your object with each slice
Repeat recursively for each intersection (calculating the intersection's hull, etc)
Each level of recursion should give a better approximation.... when you reached a satisfying level, merge all the hulls from that level to get the final polygon.
Does that sound like it could do the job?
To some degree I'm not sure what you are trying to do but it seems you have two very good answers. One is Ramer–Douglas–Peucker (DP) and the other is computing the alpha shape (also called a Concave Hull, non-convex hull, etc.). I found a more recent paper describing alpha shapes and linked it below.
I personally think DP with polygon expansion is the way to go. I'm not sure why you think it won't substantially reduce the number of vertices. With DP you supply a factor and you can make it anything you want to the point where you end up with a triangle no matter what your input. Picking this factor can be hard but in your case I think it's the best method. You should be able to determine the factor based on the size of the largest bit of detail you want to go away. You can do this with direct testing or by calculating it from your source data.
http://www.it.uu.se/edu/course/homepage/projektTDB/ht13/project10/Project-10-report.pdf
I've written a simple modification of Douglas-Peucker that might be helpful to anyone having this problem in the future: https://github.com/prakol16/rdp-expansion-only
It's identical to DP except that it pushes a line segment outwards a bit if the points that it would remove are outside the polygon. This guarantees that the resulting simplified polygon contains all the original polygon, but it has almost the same number of line segments as the original DP algorithm and is usually reasonably good at approximating the original shape.

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