I am trying to understand the fisheye model by Scaramuzza, which is implemented in Matlab, see https://de.mathworks.com/help/vision/ug/fisheye-calibration-basics.html#mw_8aca38cc-44de-4a26-a5bc-10fb312ae3c5
The backprojection (uv to xyz) seems fairly straightforward according to the following equation:
, where rho=sqrt(u^2 +v^2)
However, how does the projection (from xyz to uv) work?! In my understanding we get a rather complex set of equations. Unfortunately, I don't find any details on that....
Okay, I believe I understand it now fully after analyzing the functions of the (windows) calibration toolbox by Scaramuzza, see https://sites.google.com/site/scarabotix/ocamcalib-toolbox/ocamcalib-toolbox-download-page
Method 1 found in file "world2cam.m"
For the projection, use the same equation above. In the projection case, the equation has three known (x,y,z) and three unknown variables (u,v and lambda). We first substitute lambda with rho by realizing that
u = x/lambda
v = y/lambda
rho=sqrt(u^2+v^2) = 1/lambda * sqrt(x^2+y^2) --> lambda = sqrt(x^2+y^2) / rho
After that, we have the unknown variables (u,v and rho)
u = x/lambda = x / sqrt(x^2+y^2) * rho
v = y/lambda = y / sqrt(x^2+y^2) * rho
z / lambda = z /sqrt(x^2+y^2) * rho = a0 + a2*rho^2 + a3*rho^3 + a4*rho^4
As you can see, the last equation now has only one unknown, namely rho. Thus, we can solve it easily using e.g. the roots function in matlab. However, the result does not always exist nor is it necessarily unique. After solving the unknown variable rho, calculating uv is very simple using the equation above.
This procedure needs to be performed for each point (x,y,z) separately and is thus rather computationally expensive for an image.
Method 2 found in file "world2cam_fast.m"
The last equation has the form rho(x,y,z). However, if we define m = z / sqrt(x^2+y^2) = tan(90°-theta), it only depends on one variable, namely rho(m).
Instead of solving this equation rho(m) for every new m, the authors "plot" the function for several values of m and fit an 8th order polynomial to these points. Using this polynomial they can calculate an approximate value for rho(m) much quicker in the following.
This becomes clear, because "world2cam_fast.m" makes use of ocam_model.pol, which is calculated in "undistort.m". "undistort.m" in turn makes use of "findinvpoly.m".
Related
Using numpy, how can I do an orthogonal projection of, for example, the vector np.array([0.3,0.5,0.2]) into the plane 3x+2y-2z=0 ?
EDIT:
I think one may simply use numpy.linalg.lstsq to find the orthogonal projection?
Your hyperplane is defined by the set of x such that <a,x>=0, where a is a vector orthogonal to the plane. In your example,
a = (3,2,-2).
Then The projection of a point p is in the hyperplane is a point p_proj such that p-p_proj is orthogonal to the plane. This means that it is parallel to a, or in other words p-p_proj=lambda*a. So
p_proj = p- lambda*a (1).
since p_proj is in the hyperplane, <p_proj,a> = 0 so multiplying by a on the equality(1) gives
lambda= <p,a>/<a,a>.
Substituting into (2), you get
Projection(p) = p_proj = p-<p,a>/<a,a>a
which can be done easily in numpy using np.dot(v_1,v_2) wherever we encounter <v_1,v_2>:
def projection(p,a):
lambda_val = np.dot(p,a)/np.dot(a,a)
return p - lambda_val * a
(Note that this is a Gram-Schmidt iteration).
In going through the exercises of SICP, it defines a fixed-point as a function that satisfies the equation F(x)=x. And iterating to find where the function stops changing, for example F(F(F(x))).
The thing I don't understand is how a square root of, say, 9 has anything to do with that.
For example, if I have F(x) = sqrt(9), obviously x=3. Yet, how does that relate to doing:
F(F(F(x))) --> sqrt(sqrt(sqrt(9)))
Which I believe just converges to zero:
>>> math.sqrt(math.sqrt(math.sqrt(math.sqrt(math.sqrt(math.sqrt(9))))))
1.0349277670798647
Since F(x) = sqrt(x) when x=1. In other words, how does finding the square root of a constant have anything to do with finding fixed points of functions?
When calculating the square-root of a number, say a, you essentially have an equation of the form x^2 - a = 0. That is, to find the square-root of a, you have to find an x such that x^2 = a or x^2 - a = 0 -- call the latter equation as (1). The form given in (1) is an equation which is of the form g(x) = 0, where g(x) := x^2 - a.
To use the fixed-point method for calculating the roots of this equation, you have to make some subtle modifications to the existing equation and bring it to the form f(x) = x. One way to do this is to rewrite (1) as x = a/x -- call it (2). Now in (2), you have obtained the form required for solving an equation by the fixed-point method: f(x) is a/x.
Observe that this method requires both sides of the equation to have an 'x' term; an equation of the form sqrt(a) = x doesn't meet the specification and hence can't be solved (iteratively) using the fixed-point method.
The thing I don't understand is how a square root of, say, 9 has anything to do with that.
For example, if I have F(x) = sqrt(9), obviously x=3. Yet, how does that relate to doing: F(F(F(x))) --> sqrt(sqrt(sqrt(9)))
These are standard methods for numerical calculation of roots of non-linear equations, quite a complex topic on its own and one which is usually covered in Engineering courses. So don't worry if you don't get the "hang of it", the authors probably felt it was a good example of iterative problem solving.
You need to convert the problem f(x) = 0 to a fixed point problem g(x) = x that is likely to converge to the root of f(x). In general, the choice of g(x) is tricky.
if f(x) = x² - a = 0, then you should choose g(x) as follows:
g(x) = 1/2*(x + a/x)
(This choice is based on Newton's method, which is a special case of fixed-point iterations).
To find the square root, sqrt(a):
guess an initial value of x0.
Given a tolerance ε, compute xn+1 = 1/2*(xn + a/xn) for n = 0, 1, ... until convergence.
I need to do something very similar to what is detailed in this post. But the way the stencils are done are not obvious to me... well the stencil for _flux is, but the ones for temp_bz & temp_bx are not.
I think the picture would get clearer with variables, instead of numbers (something like stencil = np.array([[a, b], [c, d]]) with a=0.5, b=...
As example, if the recurrence relation is
flux2[i,j] = a*flux2[i-1,j] + b*bz[i-1,j]*dx + c*flux2[i,j-1] - d*bx[i,j-1]*dz
how the code would be changed ?
Having flux2, bz and bx variables, and assuming they are numpy arrays (if they are not, they should), you could write that ecuation in a vectorized form as follows:
flux2[1:,1:] = a * flux2[:-1,1:] + b * bz[:-1,1:] * dx + c * flux2[1:,:-1] - d * bx[1:,:-1] * dz
Note that, since you didn't mention dz, I assumed it is a constant, if it is a matrix of the same shape as flux2, replace with dz[1:, 1:] (same applies to dx).
That line above will vectorize the operation to every i,j of the matrix, and thus, remove the for loop, giving a considerable speedup.
You would have to define the boundary conditions for row and column 0, as your equation doesn't define what to do in those special cases.
So, in short, as your stencil only uses one position for each variable, and only has 4 interactions, I would say is way faster to calculate it in its analytic form, rather than convolving 3 images with almost all-0 stencils (which would be quite a lot of overkill).
In steganography, the least significant bit (LSB) substitution method embeds the secret bits in the place of bits from the cover medium, for example, image pixels. In some methods, the Discrete Wavelet Transform (DWT) of the image is taken and the secret bits are embedded in the DWT coefficients, after which the inverse trasform is used to reconstruct the stego image.
However, the DWT produces float coefficients and for the LSB substitution method integer values are required. Most papers I've read use the 2D Haar Wavelet, yet, they aren't clear on their methodology. I've seen the transform being defined in terms of low and high pass filters (float transforms), or taking the sum and difference of pair values, or the average and mean difference, etc.
More explicitly, either in the forward or the inverse transform (but not necessarily in both depending on the formulas used) eventually float numbers will appear. I can't have them for the coefficients because the substitution won't work and I can't have them for the reconstructed pixels because the image requires integer values for storage.
For example, let's consider a pair of pixels, A and B as a 1D array. The low frequency coefficient is defined by the sum, i.e., s = A + B, and the high frequency coefficient by the difference, i.e., d = A - B. We can then reconstruct the original pixels with B = (s - d) / 2 and A = s - B. However, after any bit twiddling with the coefficients, s - d may not be even anymore and float values will emerge for the reconstructed pixels.
For the 2D case, the 1D transform is applied separately for the rows and the columns, so eventually a division by 4 will occur somewhere. This can result in values with float remainders .00, .25, .50 and .75. I've only come across one paper which addresses this issue. The rest are very vague in their methodology and I struggle to replicate them. Yet, the DWT has been widely implemented for image steganography.
My question is, since some of the literature I've read hasn't been enlightening, how can this be possible? How can one use a transform which introduces float values, yet the whole steganography method requires integers?
One solution that has worked for me is using the Integer Wavelet Transform, which some also refer to as a lifting scheme. For the Haar wavelet, I've seen it defined as:
s = floor((A + B) / 2)
d = A - B
And for inverse:
A = s + floor((d + 1) / 2)
B = s - floor(d / 2)
All the values throughout the whole process are integers. The reason it works is because the formulas contain information about both the even and odd parts of the pixels/coefficients, so there is no loss of information from rounding down. Even if one modifies the coefficients and then takes the inverse transform, the reconstructed pixels will still be integers.
Example implementation in Python:
import numpy as np
def _iwt(array):
output = np.zeros_like(array)
nx, ny = array.shape
x = nx // 2
for j in xrange(ny):
output[0:x,j] = (array[0::2,j] + array[1::2,j])//2
output[x:nx,j] = array[0::2,j] - array[1::2,j]
return output
def _iiwt(array):
output = np.zeros_like(array)
nx, ny = array.shape
x = nx // 2
for j in xrange(ny):
output[0::2,j] = array[0:x,j] + (array[x:nx,j] + 1)//2
output[1::2,j] = output[0::2,j] - array[x:nx,j]
return output
def iwt2(array):
return _iwt(_iwt(array.astype(int)).T).T
def iiwt2(array):
return _iiwt(_iiwt(array.astype(int).T).T)
Some languages already have built-in functions for this purpose. For example, Matlab uses lwt2() and ilwt2() for 2D lifting-scheme wavelet transform.
els = {'p',[-0.125 0.125],0};
lshaarInt = liftwave('haar','int2int');
lsnewInt = addlift(lshaarInt,els);
[cAint,cHint,cVint,cDint] = lwt2(x,lsnewInt) % x is your image
xRecInt = ilwt2(cAint,cHint,cVint,cDint,lsnewInt);
An article example where IWT was used for image steganography is Raja, K.B. et. al (2008) Robust image adaptive steganography using integer wavelets.
edit
So based on the answers so far (thanks for taking your time) I'm getting the sense that I'm probably NOT looking for a Normal Distribution function. Perhaps I'll try to re-describe what I'm looking to do.
Lets say I have an object that returns a number of 0 to 10. And that number controls "speed". However instead of 10 being the top speed, I need 5 to be the top speed, and anything lower or higher would slow down accordingly. (with easing, thus the bell curve)
I hope that's clearer ;/
-original question
These are the times I wish I remembered something from math class.
I'm trying to figure out how to write a function in obj-C where I define the boundries, ex (0 - 10) and then if x = foo y = ? .... where x runs something like 0,1,2,3,4,5,6,7,8,9,10 and y runs 0,1,2,3,4,5,4,3,2,1,0 but only on a curve
Something like the attached image.
I tried googling for Normal Distribution but its way over my head. I was hoping to find some site that lists some useful algorithms like these but wasn't very successful.
So can anyone help me out here ? And if there is some good sites which shows useful mathematical functions, I'd love to check them out.
TIA!!!
-added
I'm not looking for a random number, I'm looking for.. ex: if x=0 y should be 0, if x=5 y should be 5, if x=10 y should be 0.... and all those other not so obvious in between numbers
alt text http://dizy.cc/slider.gif
Okay, your edit really clarifies things. You're not looking for anything to do with the normal distribution, just a nice smooth little ramp function. The one Paul provides will do nicely, but is tricky to modify for other values. It can be made a little more flexible (my code examples are in Python, which should be very easy to translate to any other language):
def quarticRamp(x, b=10, peak=5):
if not 0 <= x <= b:
raise ValueError #or return 0
return peak*x*x*(x-b)*(x-b)*16/(b*b*b*b)
Parameter b is the upper bound for the region you want to have a slope on (10, in your example), and peak is how high you want it to go (5, in the example).
Personally I like a quadratic spline approach, which is marginally cheaper computationally and has a different curve to it (this curve is really nice to use in a couple of special applications that don't happen to matter at all for you):
def quadraticSplineRamp(x, a=0, b=10, peak=5):
if not a <= x <= b:
raise ValueError #or return 0
if x > (b+a)/2:
x = a + b - x
z = 2*(x-a)/b
if z > 0.5:
return peak * (1 - 2*(z-1)*(z-1))
else:
return peak * (2*z*z)
This is similar to the other function, but takes a lower bound a (0 in your example). The logic is a little more complex because it's a somewhat-optimized implementation of a piecewise function.
The two curves have slightly different shapes; you probably don't care what the exact shape is, and so could pick either. There are an infinite number of ramp functions meeting your criteria; these are two simple ones, but they can get as baroque as you want.
The thing you want to plot is the probability density function (pdf) of the normal distribution. You can find it on the mighty Wikipedia.
Luckily, the pdf for a normal distribution is not difficult to implement - some of the other related functions are considerably worse because they require the error function.
To get a plot like you showed, you want a mean of 5 and a standard deviation of about 1.5. The median is obviously the centre, and figuring out an appropriate standard deviation given the left & right boundaries isn't particularly difficult.
A function to calculate the y value of the pdf given the x coordinate, standard deviation and mean might look something like:
double normal_pdf(double x, double mean, double std_dev) {
return( 1.0/(sqrt(2*PI)*std_dev) *
exp(-(x-mean)*(x-mean)/(2*std_dev*std_dev)) );
}
A normal distribution is never equal to 0.
Please make sure that what you want to plot is indeed a
normal distribution.
If you're only looking for this bell shape (with the tangent and everything)
you can use the following formula:
x^2*(x-10)^2 for x between 0 and 10
0 elsewhere
(Divide by 125 if you need to have your peek on 5.)
double bell(double x) {
if ((x < 10) && (x>0))
return x*x*(x-10.)*(x-10.)/125.;
else
return 0.;
}
Well, there's good old Wikipedia, of course. And Mathworld.
What you want is a random number generator for "generating normally distributed random deviates". Since Objective C can call regular C libraries, you either need a C-callable library like the GNU Scientific Library, or for this, you can write it yourself following the description here.
Try simulating rolls of dice by generating random numbers between 1 and 6. If you add up the rolls from 5 independent dice rolls, you'll get a surprisingly good approximation to the normal distribution. You can roll more dice if you'd like and you'll get a better approximation.
Here's an article that explains why this works. It's probably more mathematical detail than you want, but you could show it to someone to justify your approach.
If what you want is the value of the probability density function, p(x), of a normal (Gaussian) distribution of mean mu and standard deviation sigma at x, the formula is
p(x) = exp( ((x-mu)^2)/(2*sigma^2) ) / (sigma * 2 * sqrt(pi))
where pi is the area of a circle divided by the square of its radius (approximately 3.14159...). Using the C standard library math.h, this is:
#include <math>
double normal_pdf(double x, double mu, double sigma) {
double n = sigma * 2 * sqrt(M_PI); //normalization factor
p = exp( -pow(x-mu, 2) / (2 * pow(sigma, 2)) ); // unnormalized pdf
return p / n;
}
Of course, you can do the same in Objective-C.
For reference, see the Wikipedia or MathWorld articles.
It sounds like you want to write a function that yields a curve of a specific shape. Something like y = f(x), for x in [0:10]. You have a constraint on the max value of y, and a general idea of what you want the curve to look like (somewhat bell-shaped, y=0 at the edges of the x range, y=5 when x=5). So roughly, you would call your function iteratively with the x range, with a step that gives you enough points to make your curve look nice.
So you really don't need random numbers, and this has nothing to do with probability unless you want it to (as in, you want your curve to look like a the outline of a normal distribution or something along those lines).
If you have a clear idea of what function will yield your desired curve, the code is trivial - a function to compute f(x) and a for loop to call it the desired number of times for the desired values of x. Plot the x,y pairs and you're done. So that's your algorithm - call a function in a for loop.
The contents of the routine implementing the function will depend on the specifics of what you want the curve to look like. If you need help on functions that might return a curve resembling your sample, I would direct you to the reading material in the other answers. :) However, I suspect that this is actually an assignment of some sort, and that you have been given a function already. If you are actually doing this on your own to learn, then I again echo the other reading suggestions.
y=-1*abs(x-5)+5