First of all a disclaimer: I'm posting this question here even if I realize it is quite maths heavy, because I have trouble figuring out on what other site it could belong.
I'm writing a 2d spaceships game where the player will have to select the ship's destination and a course will be automatically plotted.
Along with this, I'm offering various options to control the ship's acceleration while it gets there. All these options have to do with the target velocity at the destination.
One option is to select the desired destination and velocity vector, in which case the program will use cubic interpolation, since starting and target coordinates and velocity are available.
Another option is to just select the destination point, but let the game calculate the final velocity vector. This is done through quadratic interpolation (ie. acceleration is constant).
I would like to introduce another option: let the player select the destination and the maximum absolute value of the velocity vector, as in
sqrt( vf_x^2 + vf_y^2 ) <= Vf_max
I take I shall use a 3rd order polynomial to model the course in this case, but I'm having a pretty hard time figuring out the coefficients, since I miss one equation for each coordinate (the one given by velocity at destination). Furthermore, I'm confused as to how I should use the Vf_max constraint to help me figure out the missing coefficients.
I suspect it could be an optimization problem, but I'm quite ignorant about this topic.
Can anybody help me find a solution or point me in the right direction, please?
Related
// Compute the normals
pcl::NormalEstimation<pcl::PointXYZ, pcl::Normal> normalEstimation;
pcl::search::KdTree<pcl::PointXYZ>::Ptr tree(new pcl::search::KdTree<pcl::PointXYZ>);
normalEstimation.setInputCloud(source_cloud);
normalEstimation.setSearchMethod(tree);
Hello everyone,
I am beginner for learning PCL
I don't understand at "normalEstimation.setSearchMethod(tree);"
what does this part mean?
Does it mean there are some methods that we have to choose?
Sometimes I see the code is like this
// Normal estimation
pcl::NormalEstimation<pcl::PointXYZ, pcl::Normal> n;
pcl::search::KdTree<pcl::PointXYZ>::Ptr tree(new pcl::search::KdTree<pcl::PointXYZ>);
tree->setInputCloud(cloud_smoothed)*/; [this part I dont understand too]
n.setInputCloud(cloud_smoothed);
n.setSearchMethod(tree);
Thankyou guys
cheers
you can find some information on how normals are computed here: http://pointclouds.org/documentation/tutorials/normal_estimation.php.
Basically, to compute a normal at a specific point, you need to "analyze" its neighbourhood, so the points that are around it. Usually, you take either the N closest points or all the points that are within a certain radius around the target point.
Finding these neighbour points is not trivial if the pointcloud is completely unstructured. KD-trees are here to speedup this search. In fact, they are an optimized data structure which allows very fast nearest-neighbour search for example. You can find more information on KD-trees here http://pointclouds.org/documentation/tutorials/kdtree_search.php.
So, the line normalEstimation.setSearchMethod(tree); just sets an empty KD-tree that will be used by the normal estimation algorithm.
Problem
I want to find
The first root
The first local minimum/maximum
of a black-box function in a given range.
The function has following properties:
It's continuous and differentiable.
It's combination of constant and periodic functions. All periods are known.
(It's better if it can be done with weaker assumptions)
What is the fastest way to get the root and the extremum?
Do I need more assumptions or bounds of the function?
What I've tried
I know I can use root-finding algorithm. What I don't know is how to find the first root efficiently.
It needs to be fast enough so that it can run within a few miliseconds with precision of 1.0 and range of 1.0e+8, which is the problem.
Since the range could be quite large and it should be precise enough, I can't brute-force it by checking all the possible subranges.
I considered bisection method, but it's too slow to find the first root if the function has only one big root in the range, as every subrange should be checked.
It's preferable if the solution is in java, but any similar language is fine.
Background
I want to calculate when arbitrary celestial object reaches certain height.
It's a configuration-defined virtual object, so I can't assume anything about the object.
It's not easy to get either analytical solution or simple approximation because various coordinates are involved.
I decided to find a numerical solution for this.
For a general black box function, this can't really be done. Any root finding algorithm on a black box function can't guarantee that it has found all the roots or any particular root, even if the function is continuous and differentiable.
The property of being periodic gives a bit more hope, but you can still have periodic functions with infinitely many roots in a bounded domain. Given that your function relates to celestial objects, this isn't likely to happen. Assuming your periodic functions are sinusoidal, I believe you can get away with checking subranges on the order of one-quarter of the shortest period (out of all the periodic components).
Maybe try Brent's Method on the shortest quarter period subranges?
Another approach would be to apply your root finding algorithm iteratively. If your range is (a, b), then apply your algorithm to that range to find a root at say c < b. Then apply your algorithm to the range (a, c) to find a root in that range. Continue until no more roots are found. The last root you found is a good candidate for your minimum root.
Black box function for any range? You cannot even be sure it has the continuous domain over that range. What kind of solutions are you looking for? Natural numbers, integers, real numbers, complex? These are all the question that greatly impact the answer.
So 1st thing should be determining what kind of number you accept as the result.
Second is having some kind of protection against limes of function that will try to explode your calculations as it goes for plus or minus infinity.
Since we are touching the limes topics you could have your solution edge towards zero and look like a solution but never touch 0 and become a solution. This depends on your margin of error, how close something has to be to be considered ok, it's good enough.
I think for this your SIMPLEST TO IMPLEMENT bet for real number solutions (I assume those) is to take an interval and this divide and conquer algorithm:
Take lower and upper border and middle value (or approx middle value for infinity decimals border/borders)
Try to calculate solution with all 3 and have some kind of protection against infinities
remember all 3 values in an array with results from them (3 pair of values)
remember the current best value (one its closest to solution) in seperate variable (a pair of value and result for that value)
STEP FORWARD - repeat above with 1st -2nd value range and 2nd -3rd value range
have a new pair of value and result to be closest to solution.
clear the old value-result pairs, replace them with new ones gotten from this iteration while remembering the best value solution pair (total)
Repeat above for how precise you wish to get and look at that memory explode with each iteration, keep in mind you are gonna to have exponential growth of values there. It can be further improved if you lets say take one interval and go as deep as you wanna, remember best value-result pair and then delete all other memory and go for next interval and dig deep.
How big should epsilon be when checking if dot product is close to 0?
I am working on raytracing project, and i need to check if the dot product is 0. But
that will probably never happen, so I want to take it as 0 if its value is in a small
area [-eps, +eps], but I am not sure how big should eps be?
Thanks
Since you describe this as part of a ray-tracing project, your accuracy needed is likely dictated by the "world coordinates" of a scene, or perhaps even the screen coordinates to which those are translated. Either of these would give an acceptable target for absolute error in your calculation.
It might be possible to backtrack from that to the accuracy required in the intermediate calculations you are doing, such as forming an inner product which is supposed "theoretically" to be zero. For example, you might be trying to find a shortest path (reflected light) between two smooth bodies, and the disappearance of the inner product (perpendicularity) gives the location of a point.
In a case like that the inner product may be a quadratic in the unknowns (location of a point) that you seek. It's possible that the unknowns form a "double root" (zero of multiplicity 2), making the location of that root extra sensitive to the computation of the inner product being zero.
For such cases you would want to get roughly twice the number of digits "zero" in the inner product as needed in the accuracy of the location. Essentially the inner product changes very slowly with location in the neighborhood of a double root.
But your application might not be so sensitive; analysis of the algorithm involved is necessary to give you a good answer. As a general rule I do the inner product in double precision to get answers that may be reliable as far as single precision, but this may be too costly if the ray-tracing is to be done in real time.
There is no definitive answer. I use two approaches.
If you are only concerned by floating point error then you can use a pretty small value, comparable to the smallest floating number that the compiler can handle. In c/c++ you can use the definitions provided in float.h, such as DBL_MIN to check for these numbers. I'd use a small multiple of the number, e.g., 10. * DBL_MIN as the value for eps.
If the problem is not floating point math rounding error, then I use a value that is small (say 1%) compared to the modulus of the smallest vector.
I have a linear regression equation from school , which gives a value between 1 and -1 indicative of whether or not a set of data points are close enough to a linear function
and the equation given here
http://people.hofstra.edu/stefan_waner/realworld/calctopic1/regression.html
under best fit of a line. I would like to use these to do simple gesture detection based on a point in 3-space (x,y,z) - forward, back, left, right, up, down. First I would see if they fall on a line in 2 of the 3 dimensions, then I would see if that line's slope approached zero or infinity.
Is this fast enough for functional gesture recognition? If not, could someone propose an alternative algorithm?
If I've understood your question correctly then (1) the calculation you describe here would probably be plenty fast enough, (2) it may not actually do what you want, and (3) the stuff that'll be slow in an actual implementation would lie elsewhere.
So, I think you're proposing to do this. (1) Identify the positions of ... something ... (the user's hand, perhaps) in three-dimensional space, at several successive times. (2) For (say) each of {x,y} and {x,z}, look at those two coordinates of each point, compute the correlation coefficient (which is what your formula describes) and see whether it's close to +-1. (3) If both correlation coefficients are close to +-1 then the points lie approximately on a straight line; calculate the gradient of that line (using a formula similar to that of the correlation coefficient). (4) If the gradients are both very close to 0 or +- infinity, then your line is approximately parallel to one axis, which is the case you're trying to recognize.
1: Is it fast enough? You might perhaps be sampling at 50 frames per second or thereabouts, and your gestures might take a second to execute. So you'll have somewhere on the order of 50 positions. So, the total number of arithmetic operations you'll need is maybe a few hundred (including a modest number of square roots). In the worst case, you might be doing this in emulated floating-point on a slow ARM processor or something; in that case, each arithmetic operation might take a couple of hundred cycles, so the whole thing might be 100k cycles, which for a really slow processor running at 100MHz would be about a millisecond. You're not going to have any problem with the time taken to do this calculation.
2: Is it the right thing? It's not clear that it's the right calculation. For instance, suppose your user's hand moves back and forth rapidly several times along the x-axis; that will give you a positive result; is that what you want? Suppose the user attempts the gesture you want but moves at slightly the wrong angle; you may get a negative result. Suppose they move exactly along the x-axis for a bit and then along the y-axis for a bit; then the projections onto the {x,y}, {x,z} and {y,z} planes will all pass your test. These all seem like results you might not want.
3: Is it where the real cost will lie? This all assumes you've already got (x,y,z) coordinates. Getting those is probably going to be more expensive than processing them. For instance, if you have a camera-based system of some kind then there'll be some nontrivial image processing for every frame. Or perhaps you're integrating up data from accelerometers (which, by the way, is likely to give nasty inaccurate position results); the chances are that you're doing some filtering and other calculations to get position data. I bet that the cost of performing a calculation like this one will be substantially less than the cost of getting the coordinates in the first place.
Someone somewhere has had to solve this problem. I can find many a great website explaining this problem and how to solve it. While I'm sure they are well written and make sense to math whizzes, that isn't me. And while I might understand in a vague sort of way, I do not understand how to turn that math into a function that I can use.
So I beg of you, if you have a function that can do this, in any language, (sure even fortran or heck 6502 assembler) - please help me out.
prefer an analytical to iterative solution
EDIT: Meant to specify that its a cubic bezier I'm trying to work with.
What you're asking for is the inverse of the arc length function. So, given a curve B, you want a function Linv(len) that returns a t between 0 and 1 such that the arc length of the curve between 0 and t is len.
If you had this function your problem is really easy to solve. Let B(0) be the first point. To find the next point, you'd simply compute B(Linv(w)) where w is the "equal arclength" that you refer to. To get the next point, just evaluate B(Linv(2*w)) and so on, until Linv(n*w) becomes greater than 1.
I've had to deal with this problem recently. I've come up with, or come across a few solutions, none of which are satisfactory to me (but maybe they will be for you).
Now, this is a bit complicated, so let me just give you the link to the source code first:
http://icedtea.classpath.org/~dlila/webrevs/perfWebrev/webrev/raw_files/new/src/share/classes/sun/java2d/pisces/Dasher.java. What you want is in the LengthIterator class. You shouldn't have to look at any other parts of the file. There are a bunch of methods that are defined in another file. To get to them just cut out everything from /raw_files/ to the end of the URL. This is how you use it. Initialize the object on a curve. Then to get the parameter of a point with arc length L from the beginning of the curve just call next(L) (to get the actual point just evaluate your curve at this parameter, using deCasteljau's algorithm, or zneak's suggestion). Every subsequent call of next(x) moves you a distance of x along the curve compared to your last position. next returns a negative number when you run out of curve.
Explanation of code: so, I needed a t value such that B(0) to B(t) would have length LEN (where LEN is known). I simply flattened the curve. So, just subdivide the curve recursively until each curve is close enough to a line (you can test for this by comparing the length of the control polygon to the length of the line joining the end points). You can compute the length of this sub-curve as (controlPolyLength + endPointsSegmentLen)/2. Add all these lengths to an accumulator, and stop the recursion when the accumulator value is >= LEN. Now, call the last subcurve C and let [t0, t1] be its domain. You know that the t you want is t0 <= t < t1, and you know the length from B(0) to B(t0) - call this value L0t0. So, now you need to find a t such that C(0) to C(t) has length LEN-L0t0. This is exactly the problem we started with, but on a smaller scale. We could use recursion, but that would be horribly slow, so instead we just use the fact that C is a very flat curve. We pretend C is a line, and compute the point at t using P=C(0)+((LEN-L0t0)/length(C))*(C(1)-C(0)). This point doesn't actually lie on the curve because it is on the line C(0)->C(1), but it's very close to the point we want. So, we just solve Bx(t)=Px and By(t)=Py. This is just finding cubic roots, which has a closed source solution, but I just used Newton's method. Now we have the t we want, and we can just compute C(t), which is the actual point.
I should mention that a few months ago I skimmed through a paper that had another solution to this that found an approximation to the natural parameterization of the curve. The author has posted a link to it here: Equidistant points across Bezier curves