I want to automatically find the "knee" point of the eigenvalue plot. I.e. I have a vector of eigenvalues (sorted from highest to lowest) and I want some heuristic to find the "knee" point.
Is there some heuristic for doing that
I've found the two following proposals so far.
Setting a threshold, say 0.99, or 0.95 and keep m of n eigenvalues when T(m-1) < 0.99 *T(n) <= T(m) where T(m) = sum(i=1:m){lambda(i)}
The knee is located at a point where the radius of curvature is a local minimum. For a curve y = f(x) the curvature is k = y''/(1+(y')^2)^(3/2). Just replace the derivatives with finite differences.
What do you think of these two proposals? How can I implement the second one? I don't understand how to replace the derivatives with the differences
Did you read this paper?
Non-Graphical Solutions for Cattell’s Scree Test
https://ppw.kuleuven.be/okp/_pdf/Raiche2013NGSFC.pdf
Related
I have a Nurbs curve (Ctrl Points, Knot Vector, Weights & 3rd or 4th degree). I am able to compute the length by sampling along the curve using a parameter t [0, 1]. Computing the sum of the distances along the curve gives an approximate length of the curve.
Is there a better way to compute the length of the curve?
Linear sampling: I would like to sample the curve linearly such that the distance between the first sample t = t0 = 0 to t = t1 is S1 and the last sample between t n-1 to tn = 1 is S2 and all the samples in between have lengths that interpolate linearly from S1 to S2.
S1 and S2 are fixed.
The curve length follows the equation
ds / dt = √(x'²(t) + y'²(t))
where s is the curvilinear abscissa, t the curve parameter and the derivatives are taken on t.
What you are willing to do amounts to constructing the function t(s) and imposing your values of s. This is conveniently done by writing the differential equation
dt / ds = 1 / √(x'²(t) + y'²(t))
and integrating it numerically, for example with Runge-Kutta.
Yes. Sampling as you do, instead of pairs, work by the triples of points, interpreting them as arcs passing through those three points.
You will get much smaller approximation error with the same spacing of the sampling points, vs. the straight line segments.
To compute the length of a parametrized curve c(u)=(x(u), y(u)) you can use the general formula.
see curvilinear abscissa from wikipedia
You explicitly know x(u) and y(u) since
NURBS from wikipedia
I believe you have the formula of the derivative of the rational basis functions. Therefore you have x'(u) and y'(u). Then you can either integrate using simpson's rule or use gauss points specifically to integrate rational polynomial or better yet use your favorite symbolic calculus tool (maple, wolfram,...) to compute exactly the integral.
As my question states, I want to calculate the Fourier transform F(q) of a radial function f(r) (defined on [0,infinity[ and which decays like an exponential exp(-Ar +b) at large r) as accurately as possible in Fortran. The function values come from a data file (which I can easily interpolate through cubic interpolation for example and extrapolate since the behaviour at large r is known).
I'm using the "physics" definition of the Fourier transform in 3D, which gives (because f is radial) :
I first tried to calculate this integral for some chosen values of q by using Gauss-Legendre quadrature, by generating some 60 or 100 abscissas and weights via the NAG routine D01BCF (D01BCF link). In the case of Gauss Legendre quadrature, the problem is to choose the interval [0,B] on which to integrate. While the function f loses 4 to 5 orders of magnitude from r=10 to r=20 (example), the choice of B as a strong influence on the result of the calculation... When I compared the result I get to a "nearly exact" calculation (made with matlab but with a veeeery long computation time), I saw that in fact this was only valid for small values of q (of the order of 5, when I have to deal with values as large as 150). A Gauss-Laguerre quadrature does not give any better result, probably because of the oscillatory part of the integrand.
I then tried to compute this Fourier transform for some given values of q with the routine D01ASF (D01ASF link). It is a "one-dimensional quadrature, adaptive, semi-infinite interval, weight function cos(ωx) or sin(ωx) ", which is exactly what I need. The results are quite convincing for q up to 80 or 100 if I input absolute error tolerances of 10E-5. Problems are : I would need to go at larger q, and the Fourier transform F(q) oscillates with a magnitude of ~ 10E-6 at such q's. Lowering the tolerance to 10E-5 already takes some time and even makes the whole thing to output some error message from the subroutine so I don't know if 10E-6 would be feasible.
I'm thus currently wondering if trying to calculate this Fourier transform with FFT wouldn't be a good idea ? The problems I face are that I don't know how to calculate radial wave functions with FFT (and also that I don't even know how to use FFT properly either since the definition of the transform is not even the same (exponent sign and argument) and that I never used it before).
Would you have ideas ? :)
EDIT 2 : I tried by FFT (using the routine C06FAF from NAG library). It works quite well up to some large values of q. The problem I face is that there is always some constant normalising factor to account for. I don't get why. This normalising factor evolves with the number N of points used in the mesh. It has the for of a power law : Normalising Factor F = N^(-0.5) x exp(9.9) approximately (see figure where the black line is the "exact" Fourier Transform and the green, magenta, blue, red and yellow lines are the FFT calculated for different values of N)
EDIT3 : I found the factor to be A*N^(-0.5) where A is the length of the integration mesh
The closest thing I've found to help explain what I need is here in this question: Draw equidistant points on a spiral
However, that's not exactly what I want.
The spiral to draw is an archimedean spiral and the points obtained must be equidistant from each other. (Quote: From the question linked above.)
This is precisely what I want given the Archimedean Spiral equation, .
There is a specific set of data a user can input, they are NOT based on spirals but circular figures in general. They are as follows: center point [X,Y,Z], radius, horizontal separation [can be called X separation, depends on figure], and vertical separation [can be called Y separation, depends on figure], and most importantly degrees of rotation. I'd like the horizontal separation to be the distance between consecutive points since they are the ones that need to be the same distance between each other. I'd also like vertical separation to be the distance between the 'parallel' curves.
So given that specific input selection (and yes, some can be ignored), how can I iterate through all of the consecutive, equidistant points it would take to reach the input degrees (which can be very large but is finite) and return the X and Y point of each point of those points?
Basically what I'm try to achieve is a loop from zero to the number of degrees in the input, given all of the rest of the input and my preferences noted above, and drawing a point for all of the equidistant, consecutive points (if you decide to represent using code, just represent the drawing using a 'print').
I'm having a hard time explaining, but I think I got it pretty much covered. The points on this graph are exactly what I need:
Assuming a 2D case and an archimedean spiral centered around zero (a=0), so with equation . Successive lines are then apart, so to obtain a 'vertical spacing' of , set .
The length of the arc from the centre to a point at given angle is given by Wolfram, but his solution is difficult to working with. Instead, we can approximate the length of the arc (using a very rough for-large-theta approximation) to . Rearranging, , allowing us to determine what angles correspond to the desired 'horizontal spacing'. If this approximation is not good enough, I would look at using something like Newton-Raphson. The question you link to uses also uses an approximation, although not the same one.
Finally, recognising that polar coordinates translate to cartesian as follows: ; .
I get the following:
This is generated by the following MATLAB code, but it should be straight-forward enough to translate to C++ if this is what you actually need.
% Entered by user
vertspacing = 1;
horzspacing = 1;
thetamax = 10*pi;
% Calculation of (x,y) - underlying archimedean spiral.
b = vertspacing/2/pi;
theta = 0:0.01:thetamax;
x = b*theta.*cos(theta);
y = b*theta.*sin(theta);
% Calculation of equidistant (xi,yi) points on spiral.
smax = 0.5*b*thetamax.*thetamax;
s = 0:horzspacing:smax;
thetai = sqrt(2*s/b);
xi = b*thetai.*cos(thetai);
yi = b*thetai.*sin(thetai);
I'm working on an application were I have a set of Contours(each one representing a Potential Line) and I wanna check "How straight" is that contour/shape.
The article I am using as a refrence uses the following technique:
It Matches a "segmented" line crossing the shape like so-
Then grading how "straight" is the line.
Heres an example of the Contours I am working on:
How would you go about implementing this technique?
Is there any other way of checking "How Straight" is a contour\shape?
Regards!
My first guess would be to use a coefficient of determination. That would be, fit a linear line to all your point assuming some reasonable origin where you won't receive rounding errors and calculate R^2.
A more advanced approach, if all contours are disconnected components, would be to calculate the structure model index (the link is for bone morphometry, but they explain the concept and cite the original paper.) This gives you a number that tells you how much your segment is "like a rod". This is just an idea, though. Anything that forms curves or has branches will be less and less like a rod.
I would say that it also depends on what you are using the metric for and if your contours are always generally carrying left to right.
An additional method would be to create the covariance matrix of your points, calculate the eigenvalues from that matrix, and take their ratio (where the ratio is greater than or equal to 1; otherwise, invert the ratio.) This is the basic principle behind a PCA besides the final ratio. If you have a rather linear data set (the data set varies in only one direction) then you will have a very large ratio. As the data set becomes less and less linear (or more uncorrelated) you would see the ratio approach one. A perfectly linear data set would be infinity and a perfect circle one (I believe, but I would appreciate if someone could verify this for me.) Also, working in two dimensions would mean the calculation would be computationally cheap and straight forward.
This would handle outliers very well and would be invariant to the rotation and shape of your contour. You also have a number which is always positive. The only issue would be preventing overflow when dividing the two eigenvalues. Then again you could always divide the smaller eigenvalue by the larger and your metric would be bound between zero and one, one being a circle and zero being a straight line.
Either way, you would need to test if this parameter is sensitive enough for your application.
One example for a simple algorithm is using the dot product between two segments to determine the angle between them. The formula for dot product is:
A * B = ||A|| ||B|| cos(theta)
Solving the equation for cos(theta) yields
cos(theta) = (A * B / (||A|| ||B||))
Since cos(0) = 1, cos(pi) = -1.0 and you're checking for the "straightness" of the lines, a line whose normalization of cos(theta) angles is closest to -1.0 is the straightest.
straightness = SUM(cos(theta))/(number of line segments)
where a straight line is close to -1.0, and a non-straight line approaches 1.0. Keep in mind this is a cursory evaluation of this algorithm and it obviously has edge cases and caveats that would need to be addressed in an implementation.
The trick is to use image moments. In short, you calculate the minimum inertia around an axis, the inertia around an axis perpendicular to this, and the ratio between them (which is always between 0 and 1; since inertia is non-negative)
For a straight line, the inertia along the line is zero, so the ratio is also zero. For a circle, the inertia is the same along all axis so the ratio is one. Your segmented line will be 0.01 or so as it's a fairly good match.
A simpler method is to compare the circumference of the the convex polygon containing the shape with the circumference of the shape itself. For a line, they're trivially equal, and for a not too crooked shape it's still comparable.
I have two discs that can move separately with the help of keyboard. The two discs represent two players and I want to code :
If disc1 touches disc2 then the size of disc2 reduces a little bit
Both the discs should not go out of the screen
Given that they're discs, collision detection is actually fairly simple and straightforward. Given two discs with radii R1 and R2, if the distance between the centers of the two objects is less than or equal to R1+R2, then they've collided.
You can compute the distance between the two center points using the Pythagorean theorem: the distance equals the square root of the sum of the delta X squared and delta Y squared.
If you're doing this very often, you probably want to avoid that square root. Fortunately that's pretty easy: square the sum of the two radii, and compare that to the sum of the squares of delta X and delta Y.