I am trying to figure out how to port the del2() function in matlab to C++.
I have a couple of masks that I am working with that are ones and zeros, so I wrote code liket his:
for(size_t i = 1 ; i < nmax-1 ; i++)
{
for(size_t j = 1 ; j < nmax-1 ; j++)
{
transmask[i*nmax+j] = .25*(posmask[(i+1)*nmax + j]+posmask[(i-1)*nmax+j]+posmask[i*nmax+(j+1)]+posmask[i*nmax+(j-1)]);
}
}
to compute the interior points of the laplacians. I think according to some info in "doc del2" in matlab, the border conditions just use the available info to compute, right? SO i guess I just need to write cases for the border conditions at i,j = 0 and nmax
However, i would think these values from the code I have posted here would be correct for the interior points as is, but it seems like the del2 results are different!
I dug through the del2 source, and I guess I am not enough of a matlab wizard to figure out what is going on with some of the code for the interior computation
You can see the code of del2 by edit del2 or type del2.
Note that del2 does cubic interpolation on the boundaries.
The problem is that the line you have there:
transmask[i*nmax+j] = .25*(posmask[(i+1)*nmax + j]+posmask[(i-1)*nmax+j]+posmask[i*nmax+(j+1)]+posmask[i*nmax+(j-1)]);
isn't the discrete Laplacian at all.
What you have is (I(i+1,j) + I(i-1,j) + I(i,j+1) + I(i,j-1) ) / 4
I dont' know what this mask is, but the discrete Laplacian (assuming the spacing between each pixel in each dimension is 1) is:
(-4 * I(i,j) + I(i+1,j) + I(i-1,j) + I(i,j+1) + I(i,j-1) )
So basically, you missed a term, and you don't need to divide by 4. I suggest going back and rederiving the discrete Laplacian from its definition, which is the second x derivative of the image plus the second y derivative of the image.
Edit: I see where you got the /4 from, as Matlab uses this definition for some reason (even though this isn't standard mathematically).
I think that with the Matlab compiler you can convert the m code into C code. Have you tried that?
I found this link where another methot to convert to C is explained.
http://www.kluid.com/mlib/viewtopic.php?t=337
Good luck.
Related
I am using the OpenCV method solve (https://docs.opencv.org/2.4/modules/core/doc/operations_on_arrays.html#solve) in C++ to fit a curve (grade 3, ax^3+bx^2+cx+d) through a set of points. I am solving A * x = B, A contain the powers of the points x-coordinates (so x^3, x^2, x^1, 1), and B contains the y coordinates of the points, x (Matrix) contains the parameters a, b, c and d.
I am using the flag DECOMP_QR on cv::solve to fit the curve.
The problem I am facing is that the set of points do not neccessarily follow a mathematical function (e.g. the function changes it's equation, see picture). So, in order to fit an accurate curve, I need to split the set of points where the curvature changes. In case of the picture below, I would split the regression at the index where the curve starts. So I need to detect where the curvature changes.
So, if I don't split, I'll get the yellow curve as a result, which is inaccurate. What I want is the blue curve.
Finding curvature changes:
To find out where the curvature changes, I want to use the solution accuracy.
So basically:
int splitIndex = 0;
for(int pointIndex = 0; pointIndex < numberOfPoints; pointIndex += 5) {
cv::Range rowR = Range(0, pointIndex); //Selected rows to index
cv::Range colR = Range(0,3); //Grade: 3 (x^3)
cv::Mat x;
bool res = cv::solve(A(rowR, colR), B(rowR, Range(0,1)),x , DECOMP_QR);
if(res == true) {
//Check for accuracy
if (accuracy too bad) {
splitIndex = pointIndex;
return splitIndex;
}
}
}
My questions are:
- is there a way of getting the accuracy / standard deviation from the solve command (efficiently & fast, because of real-time application (around 1ms compute time left))
- is this a good way of finding the curvature change / does anyone know a better way?
Thanks :)
Given a set of points P I need to find a line L that best approximates these points. I have tried to use the function gsl_fit_linear from the GNU scientific library. However my data set often contains points that have a line of best fit with undefined slope (x=c), thus gsl_fit_linear returns NaN. It is my understanding that it is best to use total least squares for this sort of thing because it is fast, robust and it gives the equation in terms of r and theta (so x=c can still be represented). I can't seem to find any C/C++ code out there currently for this problem. Does anyone know of a library or something that I can use? I've read a few research papers on this but the topic is still a little fizzy so I don't feel confident implementing my own.
Update:
I made a first attempt at programming my own with armadillo using the given code on this wikipedia page. Alas I have so far been unsuccessful.
This is what I have so far:
void pointsToLine(vector<Point> P)
{
Row<double> x(P.size());
Row<double> y(P.size());
for (int i = 0; i < P.size(); i++)
{
x << P[i].x;
y << P[i].y;
}
int m = P.size();
int n = x.n_cols;
mat Z = join_rows(x, y);
mat U;
vec s;
mat V;
svd(U, s, V, Z);
mat VXY = V(span(0, (n-1)), span(n, (V.n_cols-1)));
mat VYY = V(span(n, (V.n_rows-1)) , span(n, (V.n_cols-1)));
mat B = (-1*VXY) / VYY;
cout << B << endl;
}
the output from B is always 0.5504, Even when my data set changes. As well I thought that the output should be two values, so I'm definitely doing something very wrong.
Thanks!
To find the line that minimises the sum of the squares of the (orthogonal) distances from the line, you can proceed as follows:
The line is the set of points p+r*t where p and t are vectors to be found, and r varies along the line. We restrict t to be unit length. While there is another, simpler, description in two dimensions, this one works with any dimension.
The steps are
1/ compute the mean p of the points
2/ accumulate the covariance matrix C
C = Sum{ i | (q[i]-p)*(q[i]-p)' } / N
(where you have N points and ' denotes transpose)
3/ diagonalise C and take as t the eigenvector corresponding to the largest eigenvalue.
All this can be justified, starting from the (orthogonal) distance squared of a point q from a line represented as above, which is
d2(q) = q'*q - ((q-p)'*t)^2
Quick method to quickly compute Fibonacci, using Matrix property
Divide_Conquer_Fib(n) {
i = h = 1;
j = k = 0;
while (n > 0) {
if (n%2 == 1) { // if n is odd
t = j*h;
j = i*h + j*k + t;
i = i*k + t;
}
t = h*h;
h = 2*k*h + t;
k = k*k + t;
n = (int) n/2;
}
return j;
}
How do i understand this code? What would your strategy be? Would you put lots of print statements to see how states of variables change?
It is important to see how various developers' minds would go about understanding this code.
I would start off by running it against a few vales of n to check that it actually appears to give the correct answers. Then I'd read up on the mathematical theory to understand how it is likely to be working, and finally use that knowledge to take it to bits…
The Wikipedia entry section on the Matrix form explains the basis for this algorithm.
Well, the proper way to look at this code is to know what it does: Fibonacci numbers are coming up as an interesting exercise frequently, plus there is quite a bit of context saying what it does: it uses a matrix property together with divide and conquer. It turns out that you can compute the vector (Fibn, Fibn-1) as a product of some matrix and (Fibn-1, Fibn-2). Let's assume two rows in the code below are just two rows of the same matrix:
(Fib[n] ) (1 1) (Fib[n-1])
( ) = ( ) * ( )
(Fib[n-1]) (1 0) (Fib[n-2])
Now, matrix multiplication of quadratic matrices is associative, i.e., if the matrix above is M you can compute Fibn as Mn times (1, 0).
The next step is to compute Mn using divide and conquer. The basic trick here is that Mn can be decomposed according to the bits of n: Instead of computing the power by n multiplication you decompose the computation into computing squares and multiplying an extra term if the value is odd.
This is the basic underlying approach. The computation of the powers is done in the other direction, however, which works - I think - because the matrix is symmetric. I don't think you can derive the algorithm from the code easily if you are unaware of the basic approach.
I'm studying about the CSS algorithm and I don't get the hang of the concept of 'Arc Length Parameter'.
According to the literature, planar curve Gamma(u)=(x(u),y(u)) and they say this u is the arc length parameter and apparently, Gaussian Kernel g is also parameterized by this u here.
Stop me if I got something wrong but, aren't x and y location of the pixel? How is it represented by another parameter?
I had no idea when I first saw it on the literature so, I looked up the code. and apparently, I got puzzled even more.
here is the portion of the code
void getGaussianDerivs(double sigma, int M, vector<double>& gaussian,
vector<double>& dg, vector<double>& d2g) {
int L = (M - 1) / 2;
double sigma_sq = sigma * sigma;
double sigma_quad = sigma_sq*sigma_sq;
dg.resize(M); d2g.resize(M); gaussian.resize(M);
Mat_<double> g = getGaussianKernel(M, sigma, CV_64F);
for (double i = -L; i < L+1.0; i += 1.0) {
int idx = (int)(i+L);
gaussian[idx] = g(idx);
// from http://www.cedar.buffalo.edu/~srihari/CSE555/Normal2.pdf
dg[idx] = (-i/sigma_sq) * g(idx);
d2g[idx] = (-sigma_sq + i*i)/sigma_quad * g(idx);
}
}
so, it seems the code uses simple 1D Gaussian Kernel Aperture size of M and it is trying to compute its 1st and 2nd derivatives. As far as I know, 1D Gaussian kernel has parameter of x which is a horizontal coordinate and sigma which is scale. it seems like that 'arc length parameter u' is equivalent to the variable of x. That doesn't make any sense because later in the code, it directly convolutes the set of x and y on the contour.
what is this u?
PS. since I replied to the fellow who tried to answer my question, I think I should modify my question, so, here we go.
What I'm confusing is, how is this parameter 'u' implemented in codes? I think I understood the full code above -of course, I inserted only a portion of the code- but the problem is, I have no idea about what it would be for the 'improved' version of the algorithm. It says it's using 'affine length parameter' instead of this 'arc length parameter' and I'm not so sure how I implement the concept into the code.
According to the literature, the main difference between arc length parameter and affine length parameter is it's sampling interval and arc length parameter uses 1 for the vertical and horizontal direction and root of 2 for the diagonal direction. It makes sense since the portion of the code above is using for loop to compute 1st and 2nd derivatives of the 1d Gaussian and it directly inserts the value of interval 1 but, how is it gonna be with different interval with different variable? Is it possible that I'm not able to use 'for loop' for it?
so i have this image processing program where i am using a linear regression algorithm to find a plane that best fits all of the points (x,y,z: z being the pixel color intensity (0-255)
Simply speaking i have this picture of ? x ? dimension. I run this algorithm and i get these A, B, C values. (3 float values)
then i go every pixel in the program and minus the pixel value with mod_val where
mod_val = (-A * x -B * y ) / C
A,B,C are constants while x,y is the pixel location in a x,y plane.
When the dimension of the picture is divisible by 100 its perfect but when its not the picture fractures. The picture itself is the same as the original but there is a diagonal line with color contrast that goes across the picture. The program is supposed to make the pixel color uniform from the center.
I tried running the picture where mod_val = 0 for not divisble by 100 dimension pictures and it copies a new picture perfectly. So i doubt there is a problem with storing and writing the read data in terms of alignment. (fyi this picture is a grey scale 8 bit.bmp)
I have tried changing the A,B,C values but the diagonal remains the same. The color of the image fragments within the diagonals change.
when i run 1400 x 1100 picture it works perfectly with the mod_val equation written above which is the most baffling part.
I spent a lot of time looking for rounding errors. They are virtually all floats. The dimension i used for breaking picture is 1490 x 1170.
here is a gragment of the code where i think a error is occuring:
int img_row = row_length;
int img_col = col_length;
int i = 0;
float *pfAmultX = new float[img_row];
for (int x = 0; x < img_row; x++)
{
pfAmultX[x] = (A * x)/C;
}
for (int y = 0; y < img_col; y++)
{
float BmultY = B*y/C;
for (int x = 0; x < img_row; x++, i++)
{
modify_val = pfAmultX[x] + BmultY;
int temp = (int) data.data[i];
data.data[i] += (unsigned char) modify_val;
if(temp >= 250){
data.data[i] = 255;
}
else if(temp < 0){
data.data[i] = 0;
}
}
}
delete[] pfAmultX;
The img_row, img_col is correct according to VS debugger mode
Any help would be greatly appreciated. I've been trying to find this bug for many hours now and my boss is telling me that i can't go back home until i find this bug.....
before algorithm (1400 x 1100, works)
after
before (1490 x 1170, demonstrates the problem)
after
UPDATE:
well i have boiled down the problem as something with the x coordinate after extensive testing.
This is because when i use large A or B values or both (C value is always ~.999) for 1400x1100 it does not create diagonals.
However, for the other image, large B values do not create diagonals but a fairly small - avg A value creates diagonals.
Whats even more, when i test a picture where x is disivible by 100 but y is divisible by 10. the answer is correct.
well in the end i found the solution. It was a problem due to the padding the the bitmap. When the dimension on the x was not divisible by 4 it would use padding which would throw off all of the x coordinates. This also meant that the row_value i received from the bmp header was the same as the dimension but not really the same in reality. I had to make a edit where i had to do: 4 * (row_value_from_bmp_header + 3)/ 4.