OK, say I've got a bunch of discs sitting on a plane in fixed known locations. Each disc is 1 unit in radius. The plane is fully covered by the set of discs, in fact, it is extensively over-covered by the set of discs, by an order of magnitude or two in some areas. I'd like to find a subset of the discs that still fully cover the plane. Optimal is nice, but not necessary.
Here's the before illustration:
And here's the after illustration:
It seems to me that there's a dual problem having to do with Delaunay triangulation, but I'm not quite sure that helps me. I also know that this is similar, but not the same as, the disc covering problem in computational geometry. Is this a standard problem whose name I don't know?
Possible approaches seem to me to include growing a covering set using a local greedy search, and iteratively using a nearest-pair query to remove discs one at a time. I'm not sure if either is guaranteed to work well, and I haven't worked through the details.
Oh, and the application is, if you haven't guessed it, finding a subsample of ZIP code centroids to cover a map when making queries, so n is about 50,000.
Game Plan
The following is basically just a more precise restatement of your problem, but it might help:
Enumerate every connected region in the plane that results when the boundaries of all disks are drawn. By assumption, each of these regions is covered by 1 or more disks.
Each region is a "thing to be covered", and each disk is a "covering thing". Find the minimum set cover on this set of regions. This is NP-hard unfortunately.
This might not be exploiting all the structure available in the problem, but it will definitely give you an optimal answer.
Enumerating Regions
Enumerating the regions and recording which disks cover each in step 1 is the tricky part. Regions are not in general convex which makes intersection tests tricky, and every circle you add potentially doubles the number of regions. Here is how I would approach that:
Forget about the actual location of each region, and define a region only in terms of which disks it is inside and which it is outside. I.e. a region is defined by a length-n vector of 0/1 values, each indicating whether the region inside or outside that disk is to be included in the intersection -- the region in question is formed by intersecting all these n regions. So in principle you could have up to 2^n regions, but in practice some (most) vectors produce empty regions because they entail intersecting two disks that have no intersection -- this is easy to test for, thankfully. It should be straightforward to recursively generate all non-empty regions, except that...
Bad News
Unfortunately I now see that it is necessary to perform full intersection testing, because it's not always possible to tell when a region will be empty. The critical counterexample is that, given two disks A and B that have a small sliver of overlap and another disk C that overlaps each of A and B, depending on the positions of all 3 disks, the intersection of all 3 either may or may not be non-empty. (To see this, draw 3 disks in different colours with 50% opacity in a drawing program, and move them around.)
A Workable Hack
Since generating the exact list of non-empty regions looks like it will be a lot of work and take a long time due to intersection testing, and you claim you don't need optimal solutions, you could try just using a grid of sample points as the set of "things to be covered" instead of the exact list of non-empty regions. It's straightforward to determine which disks cover a given sample point. Then solve maximum set cover as before.
To get confidence that there are no gaps, rerun several times, randomly jittering the sample points' co-ordinates each time. Increase the density of sample points until there is no change in the final result.
Related
I have any number of points on an imaginary 2D surface. I also have a grid on the same surface with points at regular intervals along the X and Y access. My task is to map each point to the nearest grid point.
The code is straight forward enough until there are a shortage of grid points. The code I've been developing finds the closest grid point, displaying an already mapped point if the distance will be shorter for the current point.
I then added a second step that compares each mapped point to another and, if swapping the mapping with another point produces a smaller sum of the total mapped distance of both points, I swap them.
This last step seems important as it reduces the number crossed map lines. (This would be used to map points on a plate to a grid on another plate, with pins connecting the two, and lines that don't cross seem to have a higher chance that the pins would not make contact.)
Questions:
Can anyone comment on my thinking that if the image above were truly optimized, (that is, the mapped points--overall--would have the smallest total distance), then none of the lines were cross?
And has anyone seen any existing algorithms to help with this. I've searched but came up with nothing.
The problem could be approached as a variation of the Assignment Problem, with the "agents" being the grid squares and the points being the "tasks", (or vice versa) with the distance between them being the "cost" for that agent-task combination. You could solve with the Hungarian algorithm.
To handle the fact that there are more grid squares than points, find a bounding box for the possible grid squares you want to consider and add dummy points that have a cost of 0 associated with all grid squares.
The Hungarian algorithm is O(n3), perhaps your approach is already good enough.
See also:
How to find the optimal mapping between two sets?
How to optimize assignment of tasks to agents with these constraints?
If I understand your main concern correctly, minimising total length of line segments, the algorithm you used does not find the best mapping and it is clear in your image. e.g. when two line segments cross each other, simple mathematic says that if you rearrange their endpoints such that they do not cross, it provides a better total sum. You can use this simple approach (rearranging crossed items) to get better approximation to the optimum, you should apply swapping for more somehow many iterations.
In the following picture you can see why crossing has longer length than non crossing (first question) and also why by swapping once there still exists crossing edges (second question and w.r.t. Comments), I just drew one sample, in fact one may need many iterations of swapping to get non crossed result.
This is a heuristic algorithm certainly not optimum but I expect to be very good and efficient and simple to implement.
I have polygons that define the contour of counties in the UK. These shapes are very detailed (10k to 20k points each), thus rendering the related computations (is point X in polygon P?) quite computationaly expensive.
Thus, I would like to "subsample" my polygons, to obtain a similar shape but with less points. What are the different techniques to do so?
The trivial one would be to take one every N points (thus subsampling by a factor N), but this feels too "crude". I would rather do some averaging of points, or something of that flavor. Any pointer?
Two solutions spring to mind:
1) since the map of the UK is reasonably squarish, you could choose to render a bitmap with the counties. Assign each a specific colour, and then render the borders with a 1 or 2 pixel thick black line. This means you'll only have to perform the expensive interior/exterior calculation if a sample happens to lie on the border. The larger the bitmap, the less often this will happen.
2) simplify the county outlines. You can use a recursive Ramer–Douglas–Peucker algorithm to recursively simplify the boundaries. Just make sure you cache the results. You may also have to solve this not for entire county boundaries but for shared boundaries only, to ensure no gaps. This might be quite tricky.
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!
Polygon triangulation should help here. You'll still have to check many polygons, but these are triangles now, so they are easier to check and you can use some optimizations to determine only a small subset of polygons to check for a given region or point.
As it seems you have all the algorithms you need for polygons, not only for triangles, you can also merge several triangles that are too small after triangulation or if triangle count gets too high.
Problem Statement:
I have the following problem:
There are more than a billion points in 3D space. The goal is to find the top N points which has largest number of neighbors within given distance R. Another condition is that the distance between any two points of those top N points must be greater than R. The distribution of those points are not uniform. It is very common that certain regions of the space contain a lot of points.
Goal:
To find an algorithm that can scale well to many processors and has a small memory requirement.
Thoughts:
Normal spatial decomposition is not sufficient for this kind of problem due to the non-uniform distribution. irregular spatial decomposition that evenly divide the number of points may help us the problem. I will really appreciate that if someone can shed some lights on how to solve this problem.
Use an Octree. For 3D data with a limited value domain that scales very well to huge data sets.
Many of the aforementioned methods such as locality sensitive hashing are approximate versions designed for much higher dimensionality where you can't split sensibly anymore.
Splitting at each level into 8 bins (2^d for d=3) works very well. And since you can stop when there are too few points in a cell, and build a deeper tree where there are a lot of points that should fit your requirements quite well.
For more details, see Wikipedia:
https://en.wikipedia.org/wiki/Octree
Alternatively, you could try to build an R-tree. But the R-tree tries to balance, making it harder to find the most dense areas. For your particular task, this drawback of the Octree is actually helpful! The R-tree puts a lot of effort into keeping the tree depth equal everywhere, so that each point can be found at approximately the same time. However, you are only interested in the dense areas, which will be found on the longest paths in the Octree without even having to look at the actual points yet!
I don't have a definite answer for you, but I have a suggestion for an approach that might yield a solution.
I think it's worth investigating locality-sensitive hashing. I think dividing the points evenly and then applying this kind of LSH to each set should be readily parallelisable. If you design your hashing algorithm such that the bucket size is defined in terms of R, it seems likely that for a given set of points divided into buckets, the points satisfying your criteria are likely to exist in the fullest buckets.
Having performed this locally, perhaps you can apply some kind of map-reduce-style strategy to combine spatial buckets from different parallel runs of the LSH algorithm in a step-wise manner, making use of the fact that you can begin to exclude parts of your problem space by discounting entire buckets. Obviously you'll have to be careful about edge cases that span different buckets, but I suspect that at each stage of merging, you could apply different bucket sizes/offsets such that you remove this effect (e.g. perform merging spatially equivalent buckets, as well as adjacent buckets). I believe this method could be used to keep memory requirements small (i.e. you shouldn't need to store much more than the points themselves at any given moment, and you are always operating on small(ish) subsets).
If you're looking for some kind of heuristic then I think this result will immediately yield something resembling a "good" solution - i.e. it will give you a small number of probable points which you can check satisfy your criteria. If you are looking for an exact answer, then you are going to have to apply some other methods to trim the search space as you begin to merge parallel buckets.
Another thought I had was that this could relate to finding the metric k-center. It's definitely not the exact same problem, but perhaps some of the methods used in solving that are applicable in this case. The problem is that this assumes you have a metric space in which computing the distance metric is possible - in your case, however, the presence of a billion points makes it undesirable and difficult to perform any kind of global traversal (e.g. sorting of the distances between points). As I said, just a thought, and perhaps a source of further inspiration.
Here are some possible parts of a solution.
There are various choices at each stage,
which will depend on Ncluster, on how fast the data changes,
and on what you want to do with the means.
3 steps: quantize, box, K-means.
1) quantize: reduce the input XYZ coordinates to say 8 bits each,
by taking 2^8 percentiles of X,Y,Z separately.
This will speed up the whole flow without much loss of detail.
You could sort all 1G points, or just a random 1M,
to get 8-bit x0 < x1 < ... x256, y0 < y1 < ... y256, z0 < z1 < ... z256
with 2^(30-8) points in each range.
To map float X -> 8 bit x, unrolled binary search is fast —
see Bentley, Pearls p. 95.
Added: Kd trees
split any point cloud into different-sized boxes, each with ~ Leafsize points —
much better than splitting X Y Z as above.
But afaik you'd have to roll your own Kd tree code
to split only the first say 16M boxes, and keep counts only, not the points.
2) box: count the number of points in each 3d box,
[xj .. xj+1, yj .. yj+1, zj .. zj+1].
The average box will have 2^(30-3*8) points;
the distribution will depend on how clumpy the data is.
If some boxes are too big or get too many points, you could
a) split them into 8,
b) track the centre of the points in each box,
otherwide just take box midpoints.
3)
K-means clustering
on the 2^(3*8) box centres.
(Google parallel "k means" -> 121k hits.)
This depends strongly on K aka Ncluster, also on your radius R.
A rough approach would be to grow a
heap
of the say 27*Ncluster boxes with the most points,
then take the biggest ones subject to your Radius constraint.
(I like to start with a
Minimum spanning tree,
then remove the K-1 longest links to get K clusters.)
See also
Color quantization .
I'd make Nbit, here 8, a parameter from the beginning.
What is your Ncluster ?
Added: if your points are moving in time, see
collision-detection-of-huge-number-of-circles on SO.
I would also suggest to use an octree. The OctoMap framework is very good at dealing with huge 3D point clouds. It does not store all the points directly, but updates the occupancy density of every node (aka 3D box).
After the tree is built, you can use a simple iterator to find the node with the highest density. If you would like to model the point density or distribution inside the nodes, the OctoMap is very easy to adopt.
Here you can see how it was extended to model the point distribution using a planar model.
Just an idea. Create a graph with given points and edges between points when distance < R.
Creation of this kind of graph is similar to spatial decomposition. Your questions can be answered with local search in graph. First are vertices with max degree, second is finding of maximal unconnected set of max degree vertices.
I think creation of graph and search can be made parallel. This approach can have large memory requirement. Splitting domain and working with graphs for smaller volumes can reduce memory need.
I have a set of points which are contained within the rectangle. I'd like to split the rectangles into subrectangles based on point density (giving a number of subrectangles or desired density, whichever is easiest).
The partitioning doesn't have to be exact (almost any approximation better than regular grid would do), but the algorithm has to cope with the large number of points - approx. 200 millions. The desired number of subrectangles however is substantially lower (around 1000).
Does anyone know any algorithm which may help me with this particular task?
Just to understand the problem.
The following is crude and perform badly, but I want to know if the result is what you want>
Assumption> Number of rectangles is even
Assumption> Point distribution is markedly 2D (no big accumulation in one line)
Procedure>
Bisect n/2 times in either axis, looping from one end to the other of each previously determined rectangle counting "passed" points and storing the number of passed points at each iteration. Once counted, bisect the rectangle selecting by the points counted in each loop.
Is that what you want to achieve?
I think I'd start with the following, which is close to what #belisarius already proposed. If you have any additional requirements, such as preferring 'nearly square' rectangles to 'long and thin' ones you'll need to modify this naive approach. I'll assume, for the sake of simplicity, that the points are approximately randomly distributed.
Split your initial rectangle in 2 with a line parallel to the short side of the rectangle and running exactly through the mid-point.
Count the number of points in both half-rectangles. If they are equal (enough) then go to step 4. Otherwise, go to step 3.
Based on the distribution of points between the half-rectangles, move the line to even things up again. So if, perchance, the first cut split the points 1/3, 2/3, move the line half-way into the heavy half of the rectangle. Go to step 2. (Be careful not to get trapped here, moving the line in ever decreasing steps first in one direction, then the other.)
Now, pass each of the half-rectangles in to a recursive call to this function, at step 1.
I hope that outlines the proposal well enough. It has limitations: it will produce a number of rectangles equal to some power of 2, so adjust it if that's not good enough. I've phrased it recursively, but it's ideal for parallelisation. Each split creates two tasks, each of which splits a rectangle and creates two more tasks.
If you don't like that approach, perhaps you could start with a regular grid with some multiple (10 - 100 perhaps) of the number of rectangles you want. Count the number of points in each of these tiny rectangles. Then start gluing the tiny rectangles together until the less-tiny rectangle contains (approximately) the right number of points. Or, if it satisfies your requirements well enough, you could use this as a discretisation method and integrate it with my first approach, but only place the cutting lines along the boundaries of the tiny rectangles. This would probably be much quicker as you'd only have to count the points in each tiny rectangle once.
I haven't really thought about the running time of either of these; I have a preference for the former approach 'cos I do a fair amount of parallel programming and have oodles of processors.
You're after a standard Kd-tree or binary space partitioning tree, I think. (You can look it up on Wikipedia.)
Since you have very many points, you may wish to only approximately partition the first few levels. In this case, you should take a random sample of your 200M points--maybe 200k of them--and split the full data set at the midpoint of the subsample (along whichever axis is longer). If you actually choose the points at random, the probability that you'll miss a huge cluster of points that need to be subdivided will be approximately zero.
Now you have two problems of about 100M points each. Divide each along the longer axis. Repeat until you stop taking subsamples and split along the whole data set. After ten breadth-first iterations you'll be done.
If you have a different problem--you must provide tick marks along the X and Y axis and fill in a grid along those as best you can, rather than having the irregular decomposition of a Kd-tree--take your subsample of points and find the 0/32, 1/32, ..., 32/32 percentiles along each axis. Draw your grid lines there, then fill the resulting 1024-element grid with your points.
R-tree
Good question.
I think the area you need to investigate is "computational geometry" and the "k-partitioning" problem. There's a link that might help get you started here
You might find that the problem itself is NP-hard which means a good approximation algorithm is the best you're going to get.
Would K-means clustering or a Voronoi diagram be a good fit for the problem you are trying to solve?
That's looks like Cluster analysis.
Would a QuadTree work?
A quadtree is a tree data structure in which each internal node has exactly four children. Quadtrees are most often used to partition a two dimensional space by recursively subdividing it into four quadrants or regions. The regions may be square or rectangular, or may have arbitrary shapes. This data structure was named a quadtree by Raphael Finkel and J.L. Bentley in 1974. A similar partitioning is also known as a Q-tree. All forms of Quadtrees share some common features:
They decompose space into adaptable cells
Each cell (or bucket) has a maximum capacity. When maximum capacity is reached, the bucket splits
The tree directory follows the spatial decomposition of the Quadtree
I need to evaluate if two sets of 3d points are the same (ignoring translations and rotations) by finding and comparing a proper geometric hash. I did some paper research on geometric hashing techniques, and I found a couple of algorithms, that however tend to be complicated by "vision requirements" (eg. 2d to 3d, occlusions, shadows, etc).
Moreover, I would love that, if the two geometries are slightly different, the hashes are also not very different.
Does anybody know some algorithm that fits my need, and can provide some link for further study?
Thanks
Your first thought may be trying to find the rotation that maps one object to another but this a very very complex topic... and is not actually necessary! You're not asking how to best match the two, you're just asking if they are the same or not.
Characterize your model by a list of all interpoint distances. Sort the list by that distance. Now compare the list for each object. They should be identical, since interpoint distances are not affected by translation or rotation.
Three issues:
1) What if the number of points is large, that's a large list of pairs (N*(N-1)/2). In this case you may elect to keep only the longest ones, or even better, keep the 1 or 2 longest ones for each vertex so that every part of your model has some contribution. Dropping information like this however changes the problem to be probabilistic and not deterministic.
2) This only uses vertices to define the shape, not edges. This may be fine (and in practice will be) but if you expect to have figures with identical vertices but different connecting edges. If so, test for the vertex-similarity first. If that passes, then assign a unique labeling to each vertex by using that sorted distance. The longest edge has two vertices. For each of THOSE vertices, find the vertex with the longest (remaining) edge. Label the first vertex 0 and the next vertex 1. Repeat for other vertices in order, and you'll have assigned tags which are shift and rotation independent. Now you can compare edge topologies exactly (check that for every edge in object 1 between two vertices, there's a corresponding edge between the same two vertices in object 2) Note: this starts getting really complex if you have multiple identical interpoint distances and therefore you need tiebreaker comparisons to make the assignments stable and unique.
3) There's a possibility that two figures have identical edge length populations but they aren't identical.. this is true when one object is the mirror image of the other. This is quite annoying to detect! One way to do it is to use four non-coplanar points (perhaps the ones labeled 0 to 3 from the previous step) and compare the "handedness" of the coordinate system they define. If the handedness doesn't match, the objects are mirror images.
Note the list-of-distances gives you easy rejection of non-identical objects. It also allows you to add "fuzzy" acceptance by allowing a certain amount of error in the orderings. Perhaps taking the root-mean-squared difference between the two lists as a "similarity measure" would work well.
Edit: Looks like your problem is a point cloud with no edges. Then the annoying problem of edge correspondence (#2) doesn't even apply and can be ignored! You still have to be careful of the mirror-image problem #3 though.
There a bunch of SIGGRAPH publications which may prove helpful to you.
e.g. "Global Non-Rigid Alignment of 3-D Scans" by Brown and Rusinkiewicz:
http://portal.acm.org/citation.cfm?id=1276404
A general search that can get you started:
http://scholar.google.com/scholar?q=siggraph+point+cloud+registration
spin images are one way to go about it.
Seems like a numerical optimisation problem to me. You want to find the parameters of the transform which transforms one set of points to as close as possible by the other. Define some sort of residual or "energy" which is minimised when the points are coincident, and chuck it at some least-squares optimiser or similar. If it manages to optimise the score to zero (or as near as can be expected given floating point error) then the points are the same.
Googling
least squares rotation translation
turns up quite a few papers building on this technique (e.g "Least-Squares Estimation of Transformation Parameters Between Two Point Patterns").
Update following comment below: If a one-to-one correspondence between the points isn't known (as assumed by the paper above), then you just need to make sure the score being minimised is independent of point ordering. For example, if you treat the points as small masses (finite radius spheres to avoid zero-distance blowup) and set out to minimise the total gravitational energy of the system by optimising the translation & rotation parameters, that should work.
If you want to estimate the rigid
transform between two similar
point clouds you can use the
well-established
Iterative Closest Point method. This method starts with a rough
estimate of the transformation and
then iteratively optimizes for the
transformation, by computing nearest
neighbors and minimizing an
associated cost function. It can be
efficiently implemented (even
realtime) and there are available
implementations available for
matlab, c++... This method has been
extended and has several variants,
including estimating non-rigid
deformations, if you are interested
in extensions you should look at
Computer graphics papers solving
scan registration problem, where
your problem is a crucial step. For
a starting point see the Wikipedia
page on Iterative Closest Point
which has several good external
links. Just a teaser image from a matlab implementation which was designed to match to point clouds:
(source: mathworks.com)
After aligning you could the final
error measure to say how similar the
two point clouds are, but this is
very much an adhoc solution, there
should be better one.
Using shape descriptors one can
compute fingerprints of shapes which
are often invariant under
translations/rotations. In most cases they are defined for meshes, and not point clouds, nevertheless there is a multitude of shape descriptors, so depending on your input and requirements you might find something useful. For this, you would want to look into the field of shape analysis, and probably this 2004 SIGGRAPH course presentation can give a feel of what people do to compute shape descriptors.
This is how I would do it:
Position the sets at the center of mass
Compute the inertia tensor. This gives you three coordinate axes. Rotate to them. [*]
Write down the list of points in a given order (for example, top to bottom, left to right) with your required precision.
Apply any algorithm you'd like for a resulting array.
To compare two sets, unless you need to store the hash results in advance, just apply your favorite comparison algorithm to the sets of points of step 3. This could be, for example, computing a distance between two sets.
I'm not sure if I can recommend you the algorithm for the step 4 since it appears that your requirements are contradictory. Anything called hashing usually has the property that a small change in input results in very different output. Anyway, now I've reduced the problem to an array of numbers, so you should be able to figure things out.
[*] If two or three of your axis coincide select coordinates by some other means, e.g. as the longest distance. But this is extremely rare for random points.
Maybe you should also read up on the RANSAC algorithm. It's commonly used for stitching together panorama images, which seems to be a bit similar to your problem, only in 2 dimensions. Just google for RANSAC, panorama and/or stitching to get a starting point.