I have some case on linear algebra and I have not found solutions for this case. Additional i have not methematica background. I have three players. I know velocity and position each od them. Also I have some point (another Vector). I want to calculate how many time take to reach this point by each od these players. Could somebody help me ? I have read: http://blog.wolfire.com/2009/07/linear-algebra-for-game-developers-part-1/
But it is not my case :(
Try this in 3 dimensions, or shorten the vectors for 2 dimensions
player1position={1,2,7};
player1velocity=5;
pointposition={2,4,5};
time=Norm[player1position-pointposition]/player1velocity
Related
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?
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.
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I have a system of linear equations that make up an NxM matrix (i.e. Non-square) which I need to solve - or at least attempt to solve in order to show that there is no solution to the system. (more likely than not, there will be no solution)
As I understand it, if my matrix is not square (over or under-determined), then no exact solution can be found - am I correct in thinking this? Is there a way to transform my matrix into a square matrix in order to calculate the determinate, apply Gaussian Elimination, Cramer's rule, etc?
It may be worth mentioning that the coefficients of my unknowns may be zero, so in certain, rare cases it would be possible to have a zero-column or zero-row.
Whether or not your matrix is square is not what determines the solution space. It is the rank of the matrix compared to the number of columns that determines that (see the rank-nullity theorem). In general you can have zero, one or an infinite number of solutions to a linear system of equations, depending on its rank and nullity relationship.
To answer your question, however, you can use Gaussian elimination to find the rank of the matrix and, if this indicates that solutions exist, find a particular solution x0 and the nullspace Null(A) of the matrix. Then, you can describe all your solutions as x = x0 + xn, where xn represents any element of Null(A). For example, if a matrix is full rank its nullspace will be empty and the linear system will have at most one solution. If its rank is also equal to the number of rows, then you have one unique solution. If the nullspace is of dimension one, then your solution will be a line that passes through x0, any point on that line satisfying the linear equations.
Ok, first off: a non-square system of equations can have an exact solution
[ 1 0 0 ][x] = [1]
[ 0 0 1 ][y] [1]
[z]
clearly has a solution (actually, it has an 1-dimensional family of solutions: x=z=1). Even if the system is overdetermined instead of underdetermined it may still have a solution:
[ 1 0 ][x] = [1]
[ 0 1 ][y] [1]
[ 1 1 ] [2]
(x=y=1). You may want to start by looking at least squares solution methods, which find the exact solution if one exists, and "the best" approximate solution (in some sense) if one does not.
Taking Ax = b, with A having m columns and n rows. We are not guaranteed to have one and only one solution, which in many cases is because we have more equations than unknowns (m bigger n). This could be because of repeated measurements, that we actually want because we are cautious about influence of noise.
If we observe that we can not find a solution that actually means, that there is no way to find b travelling the column space spanned by A. (As x is only taking a combination of the columns).
We can however ask for the point in the space spanned by A that is nearest to b. How can we find such a point? Walking on a plane the closest one can get to a point outside it, is to walk until you are right below. Geometrically speaking this is when our axis of sight is perpendicular to the plane.
Now that is something we can have a mathematical formulation of. A perpendicular vector reminds us of orthogonal projections. And that is what we are going to do. The simplest case tells us to do a.T b. But we can take the whole matrix A.T b.
For our equation let us apply the transformation to both sides: A.T Ax = A.T b.
Last step is to solve for x by taking the inverse of A.T A:
x = (A.T A)^-1 * A.T b
The least squares recommendation is a very good one.
I'll add that you can try a singular value decomposition (SVD) that will give you the best answer possible and provide information about the null space for free.
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
Disclaimer
This is not strictly a programming question, but most programmers soon or later have to deal with math (especially algebra), so I think that the answer could turn out to be useful to someone else in the future.
Now the problem
I'm trying to check if m vectors of dimension n are linearly independent. If m == n you can just build a matrix using the vectors and check if the determinant is != 0. But what if m < n?
Any hints?
See also this video lecture.
Construct a matrix of the vectors (one row per vector), and perform a Gaussian elimination on this matrix. If any of the matrix rows cancels out, they are not linearly independent.
The trivial case is when m > n, in this case, they cannot be linearly independent.
Construct a matrix M whose rows are the vectors and determine the rank of M. If the rank of M is less than m (the number of vectors) then there is a linear dependence. In the algorithm to determine the rank of M you can stop the procedure as soon as you obtain one row of zeros, but running the algorithm to completion has the added bonanza of providing the dimension of the spanning set of the vectors. Oh, and the algorithm to determine the rank of M is merely Gaussian elimination.
Take care for numerical instability. See the warning at the beginning of chapter two in Numerical Recipes.
If m<n, you will have to do some operation on them (there are multiple possibilities: Gaussian elimination, orthogonalization, etc., almost any transformation which can be used for solving equations will do) and check the result (eg. Gaussian elimination => zero row or column, orthogonalization => zero vector, SVD => zero singular number)
However, note that this question is a bad question for a programmer to ask, and this problem is a bad problem for a program to solve. That's because every linearly dependent set of n<m vectors has a different set of linearly independent vectors nearby (eg. the problem is numerically unstable)
I have been working on this problem these days.
Previously, I have found some algorithms regarding Gaussian or Gaussian-Jordan elimination, but most of those algorithms only apply to square matrix, not general matrix.
To apply for general matrix, one of the best answers might be this:
http://rosettacode.org/wiki/Reduced_row_echelon_form#MATLAB
You can find both pseudo-code and source code in various languages.
As for me, I transformed the Python source code to C++, causes the C++ code provided in the above link is somehow complex and inappropriate to implement in my simulation.
Hope this will help you, and good luck ^^
If computing power is not a problem, probably the best way is to find singular values of the matrix. Basically you need to find eigenvalues of M'*M and look at the ratio of the largest to the smallest. If the ratio is not very big, the vectors are independent.
Another way to check that m row vectors are linearly independent, when put in a matrix M of size mxn, is to compute
det(M * M^T)
i.e. the determinant of a mxm square matrix. It will be zero if and only if M has some dependent rows. However Gaussian elimination should be in general faster.
Sorry man, my mistake...
The source code provided in the above link turns out to be incorrect, at least the python code I have tested and the C++ code I have transformed does not generates the right answer all the time. (while for the exmample in the above link, the result is correct :) -- )
To test the python code, simply replace the mtx with
[30,10,20,0],[60,20,40,0]
and the returned result would be like:
[1,0,0,0],[0,1,2,0]
Nevertheless, I have got a way out of this. It's just this time I transformed the matalb source code of rref function to C++. You can run matlab and use the type rref command to get the source code of rref.
Just notice that if you are working with some really large value or really small value, make sure use the long double datatype in c++. Otherwise, the result will be truncated and inconsistent with the matlab result.
I have been conducting large simulations in ns2, and all the observed results are sound.
hope this will help you and any other who have encontered the problem...
A very simple way, that is not the most computationally efficient, is to simply remove random rows until m=n and then apply the determinant trick.
m < n: remove rows (make the vectors shorter) until the matrix is square, and then
m = n: check if the determinant is 0 (as you said)
m < n (the number of vectors is greater than their length): they are linearly dependent (always).
The reason, in short, is that any solution to the system of m x n equations is also a solution to the n x n system of equations (you're trying to solve Av=0). For a better explanation, see Wikipedia, which explains it better than I can.