I am looking for an algorithm that can connect points together with a continuous curve line. Imagine drawing from point a to b to c until the last point, and when you draw from point to point, the line must be a curve and is continuous with respect to the previous point and next point, as if the given points are just samples of a closed loop. Please see figure below for illustration.
Are there such algorithm for something like this?
*The circles in the figure are my list of points.
Given that your points are ordered, spline interpolation is definitely the best way to go here. (As indicated by by bo1024's comment) I highly recommend the following notes:
http://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/
And specifically the section here would be most relevant to getting a closed loop like you asked for:
http://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/spline/B-spline/bspline-curve-closed.html
EDIT: If the curve has to pass through the points, then the unique degree n solution is the Lagrange interpolating polynomial. You can just make one polynomial for each component of your points vectors using the formula on the wiki page:
http://en.wikipedia.org/wiki/Lagrange_polynomial
Unfortunately Lagrange interpolation can be pretty noisy if you have too many points. As a result, I would still recommend using some fixed degree spline interpolation. Instead of B-splines, another option are Hermite polynomials:
http://en.wikipedia.org/wiki/Cubic_Hermite_spline
These will guarantee that the curve passes through the points. To get a closed curve, you need to repeat the the first d points of your curve when solving for the coefficients, where d is the degree of the Hermite spline you are using to approximate your points.
The problem is very similar to the travelling salesman problem, you may be able to extend some of the algorithms used to solve it to suit your case.
For instance, evolutionary algorithms are easy to adapt and you will find lot of references about using them to solve the TSP.
Related
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 :)
I am trying to use B spline curve fitting. The order of B spline curve is 4. When I have many control points, it works well. However if the number of control points is small such as two, my program will crash. I realize that the number of control points is related to number of knots and the order. Can anyone help me clarify the relationship or give some links on it?
Sounds like you're simply reading out of bounds, which is not a specific issue of calculating splines. To calculate a b-spline of degree n, you'll need at least n + 1 points.
To simplify and show the issue:
The easiest way of interpolation is linear interpolation - just draw a line between two points.
If you've got only one point, you can't interpolate anything, simply due to the fact that you don't know where to draw.
For a quadratic interpolation, you'll need at least three points, etc.
In a similar way, you'll need at least 5 points for a b-spline of 4th degree.
A really nice online demo can be found here:
Pick any b-spline demo on the lower left side, I'd just go for the linear one.
On the right you're now able to set the number of control points as well as the degree of the curve.
Feel free to try around, also by moving the points around with your mouse.
Two control points is not sufficient to define a B-spline of order 4. For B-splines, the number of knots needs to equal the sum of number of control points and order. A single segment degree 3 B-spline will require 4 control points and 8 knot values. So, to calculate a B-spline with order N, you at least need N points. That will give you a B-spline with single segment. If you have more points, then the resulting B-spline will have more segments.
As others stated, the number of control points is equal to the number of knots minus the order of the bspline basis. Thus you cannot have an arbitrary combination of order, say k, and knot vector for your bspline function/curve once you fix the control points.
A very useful link for theory on b-splines and nurbs curves is the following:
http://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/index.html
There you can find the relationship of number of control points with dimensionality of knot vector etc as well as detailed examples and some algorithms.
Depending on your needs, you may also wish to check "The NURBS book" by Piegl and Tiller
http://www.amazon.com/NURBS-Book-Monographs-Visual-Communication/dp/3540615458
they have done an amazing job and in their book they include working algorithms.
The curve fitting problem of a b-spline to data is a rather large subject since you have to take care to avoid over/under fitting. There are several approaches, and most involve including a curvature penalty term. The literature is vast, but you can find a lot of information and a great starting point in the book by Hastie et. al. "The elements of statistical learning" which you can legally download from the authors site:
http://statweb.stanford.edu/~tibs/ElemStatLearn/
The curve fitting problem is covered to some extent in all references I gave. Good luck.
Hi people i have 2 question related to Decasteljau algorithm,they are more of a general questions,but if im right it could help solving many problems.Here it is:
We have some sufrace: Ʃ(i=0,n) Ʃ(j=o,m) Bi,n(U),Bi,m(v) Pi,j analysis that i have found says that first we take some value for one parameter u=uo,then we itterate other parametar v -> 1 get a set of points,then increment u by one etc....for loop inside for loop in code language.My question is can we fix one parameter U=Uo for what ever value,and then just compute points on for parameter v?Because all points that are on one curve are also on the surface,and if distance between curves approaches to zero (which itteration really is) we can apply DeCasteljau algorithm only to one set of curves itterating only one parameter.Or i got something wrong?:)
Second question is i havent really figured out what do we really need DeCasteljau algorithm for,unless we are drawing curves by hand?If we know order of the curve we can easily form Bernstain polynoms for that curve order and compute point for given value of parametar.Because when you unwrap Decasteljau what you get is Bernstain polynom?
So like i said,please help have i got i wrong?
Yes you can fix one parameter (say U) and change the other (V) to generate an iso-U curve.
You can see the things as if you had an NxM array of control points. If you perform a first interpolation on U (actually M interpolations involving N control points), you get M new control points that define a Bezier curve. and by varying U, the curve moves in space.
The De Casteljau's algorithm is used for convenience: it computes the interpolant by using a cascade of linear interpolations between the control points. Direct evaluation of the Bernstein polynomials would require the precomputation of the coefficients, and would not be faster, even when implemented by Horner's scheme, and can be numerically less stable.
The De Casteljau's algorithm is also appreciated for its geometrical interpretation, and for its connection with the subdivision process: if you want to build the control points for just a part of a Bezier curve, De Calsteljau's provides them.
I have spent time looking for information on the best algorithm to create a nesting of irregular polygons in 2D using manual and automatic positioning. I need to use such an algorithm in the context of CAD/CAM software. Here are the real possibilities I've found so far:
Separating Axis Theorem: is a fairly quick and simple algorithm to implement, but the drawback I find with it is that it only works with convex polygons. To work with concave polygons, a convex decomposition would need to be done first. This implies an increase in the run-time and the implementation of a new algorithm that decomposes the concave polygon into convex polygons.
Nesting by a power function: calculating the partial derivatives in the X and Y axes, you could get the escape direction you should take a polygon so that there is a collision between the two polygons. This function of energy and I tested and the three major problems that I have encountered are: first obtaining local minima , second nesting when the collision occurs over a piece and finally the execution time is very high.
Using no-fit polygon: use the no-fit polygon to the nesting can be somewhat interesting. I have read several papers on the subject although there are very few online documentation on it. Not sure if it can really be a useful choice. I still have several doubts on the details of this approach.
Any idea which of these algorithms to choose? Or if you know any other options that can be used? I'm a little confused :-) .
Thank you very much.
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.