I'm trying to create a convolution function but I'm having trouble during the access to the kernel data (cv::Mat).
I create the 3x3 kernel:
cv::Mat krn(3, 3, CV_32FC1);
krn.setTo(1);
krn = krn/9;
And I try to loop over it. Next the image Mat will be the image to which I want to apply the convolution operator and output will be the result of convolution:
for (int r = 0; r < image.rows - krn.rows; ++r) {
for (int c = 0; c < image.cols - krn.cols; ++c) {
int sum = 0;
for (int rs = 0; rs < krn.rows; ++rs) {
for (int cs = 0; cs < krn.cols; ++cs) {
sum += krn.data[rs * krn.cols + cs] * image.data[(r + rs) * image.cols + c + cs];
}
}
output.data[(r+1)*src.cols + c + 1]=sum; // assuming 3x3 kernel
}
}
However the output is not as desired (only randomic black and white pixel).
However, if I change my code this way:
for (int r = 0; r < image.rows - krn.rows; ++r) {
for (int c = 0; c < image.cols - krn.cols; ++c) {
int sum = 0;
for (int rs = 0; rs < krn.rows; ++rs) {
for (int cs = 0; cs < krn.cols; ++cs) {
sum += 0.11 * image.data[(r + rs) * image.cols + c + cs]; // CHANGE HERE
}
}
output.data[(r+1)*src.cols + c + 1]=sum; // assuming 3x3 kernel
}
}
Using 0.11 instead of the kernel values seems to give the correct output.
For this reason I think I'm doing something wrong accessing the kernel's data.
P.S: I cannot use krn.at<float>(rs,cs).
Thanks!
Instead of needlessly using memcpy, you can just cast the pointer. I'll use a C-style cast because why not.
cv::Mat krn = 1 / (cv::Mat_<float>(3,3) <<
1, 2, 3,
4, 5, 6,
7, 8, 9);
for (int i = 0; i < krn.rows; i += 1)
{
for (int j = 0; j < krn.cols; j += 1)
{
// to see clearly what's happening
uint8_t *byteptr = krn.data + krn.step[0] * i + krn.step[1] * j;
float *floatptr = (float*) byteptr;
// or in one step:
float *floatptr = (float*) (krn.data + krn.step[0] * i + krn.step[1] * j);
cout << "krn.at<float>(" << i << "," << j << ") = " << (*floatptr) << endl;
endl;
}
}
krn.at<float>(0,0) = 1
krn.at<float>(0,1) = 0.5
krn.at<float>(0,2) = 0.333333
krn.at<float>(1,0) = 0.25
krn.at<float>(1,1) = 0.2
krn.at<float>(1,2) = 0.166667
krn.at<float>(2,0) = 0.142857
krn.at<float>(2,1) = 0.125
krn.at<float>(2,2) = 0.111111
Note that pointer arithmetic may not be obvious. if you have a uint8_t*, adding 1 moves it by one uint8_t, and if you have a float*, adding 1 moves it by one float which is four bytes. The step[] contains offsets expressed in bytes.
Consult the documentation for details, which include information on the step[] array that contains the strides/steps to calculate the offset given a tuple of indices into the matrix.
cv::Mat::data is pointer of type uchar.
By data[y * cols + x] you access some byte of stored float values in krn. To get full float values use at method template:
krn.at<float>(rs,cs)
Consider changing type of sum variable to be real. Without this, you may lose partial results when calculating convolution .
So, if you cannot use at, just read 4 bytes from data pointer:
float v = 0.0;
memcpy(&v, krn.data + (rs * krn.step + cs * sizeof(float)), 4);
step - means total bytes occupied by one line in mat.
Related
I have the raw color data for four images, let's call them 1, 2, 3, and 4. I am storing the data in an unsigned char * with allocated memory. Individually I can manipulate or encode the images but when trying to concatenate or order them into a single image it works but takes more time than I would like.
I would like to create a 2 by 2 of the raw image data to encode as a single image.
1 2
3 4
For my example each image is 400 by 225 with RGBA (360000 bytes). Iim doing a for loop with memcpy where
for (int j = 0; j < 225; j++)
{
std::memcpy(dest + (j * (400 + 400) * 4), src + (j * 400 * 4), 400 * 4); //
}
for each image with an offset for the starting position added in (the example above would only work for the top left of course).
This works but I'm wondering if this is a solved problem with a better solution, either in an algorithm described somewhere or a small library.
#include <iostream>
const int width = 6;
const int height = 4;
constexpr int n = width * height;
int main()
{
unsigned char a[n], b[n], c[n], d[n];
unsigned char dst[n * 4];
int i = 0, j = 0;
/* init data */
for (; i < n; i++) {
a[i] = 'a';
b[i] = 'b';
c[i] = 'c';
d[i] = 'd';
}
/* re-order */
i = 0;
for (int y = 0; y < height; y++) {
for (int x = 0; x < width; x++, i++, j++) {
dst[i ] = a[j];
dst[i + width] = b[j];
dst[i + n * 2 ] = c[j];
dst[i + n * 2 + width] = d[j];
}
i += width;
}
/* print result */
i = 0;
for (int y = 0; y < height * 2; y++) {
for (int x = 0; x < width * 2; x++, i++)
std::cout << dst[i];
std::cout << '\n';
}
return 0;
}
I am trying to implement Laplace sharpening using C++ , here's my code so far:
img = imread("cow.png", 0);
Mat convoSharp() {
//creating new image
Mat res = img.clone();
for (int y = 0; y < res.rows; y++) {
for (int x = 0; x < res.cols; x++) {
res.at<uchar>(y, x) = 0.0;
}
}
//variable declaration
int filter[3][3] = { {0,1,0},{1,-4,1},{0,1,0} };
//int filter[3][3] = { {-1,-2,-1},{0,0,0},{1,2,1} };
int height = img.rows;
int width = img.cols;
int filterHeight = 3;
int filterWidth = 3;
int newImageHeight = height - filterHeight + 1;
int newImageWidth = width - filterWidth + 1;
int i, j, h, w;
//convolution
for (i = 0; i < newImageHeight; i++) {
for (j = 0; j < newImageWidth; j++) {
for (h = i; h < i + filterHeight; h++) {
for (w = j; w < j + filterWidth; w++) {
res.at<uchar>(i,j) += filter[h - i][w - j] * img.at<uchar>(h,w);
}
}
}
}
//img - laplace
for (int y = 0; y < res.rows; y++) {
for (int x = 0; x < res.cols; x++) {
res.at<uchar>(y, x) = img.at<uchar>(y, x) - res.at<uchar>(y, x);
}
}
return res;
}
I don't really know what went wrong, I also tried different filter (1,1,1),(1,-8,1),(1,1,1) and the result is also same (more or less). I don't think that I need to normalize the result because the result is in range of 0 - 255. Can anyone explain what really went wrong in my code?
Problem: uchar is too small to hold partial results of filerting operation.
You should create a temporary variable and add all the filtered positions to this variable then check if value of temp is in range <0,255> if not, you need to clamp the end result to fit <0,255>.
By executing below line
res.at<uchar>(i,j) += filter[h - i][w - j] * img.at<uchar>(h,w);
partial result may be greater than 255 (max value in uchar) or negative (in filter you have -4 or -8). temp has to be singed integer type to handle the case when partial result is negative value.
Fix:
for (i = 0; i < newImageHeight; i++) {
for (j = 0; j < newImageWidth; j++) {
int temp = res.at<uchar>(i,j); // added
for (h = i; h < i + filterHeight; h++) {
for (w = j; w < j + filterWidth; w++) {
temp += filter[h - i][w - j] * img.at<uchar>(h,w); // add to temp
}
}
// clamp temp to <0,255>
res.at<uchar>(i,j) = temp;
}
}
You should also clamp values to <0,255> range when you do the subtraction of images.
The problem is partially that you’re overflowing your uchar, as rafix07 suggested, but that is not the full problem.
The Laplace of an image contains negative values. It has to. And you can’t clamp those to 0, you need to preserve the negative values. Also, it can values up to 4*255 given your version of the filter. What this means is that you need to use a signed 16 bit type to store this output.
But there is a simpler and more efficient approach!
You are computing img - laplace(img). In terms of convolutions (*), this is 1 * img - laplace_kernel * img = (1 - laplace_kernel) * img. That is to say, you can combine both operations into a single convolution. The 1 kernel that doesn’t change the image is [(0,0,0),(0,1,0),(0,0,0)]. Subtract your Laplace kernel from that and you obtain [(0,-1,0),(-1,5,-1),(0,-1,0)].
So, simply compute the convolution with that kernel, and do it using int as intermediate type, which you then clamp to the uchar output range as shown by rafix07.
I have an image, as a cv::Mat. I am getting the raw data from this, with:
uchar* data = (uchar *)pImg.data;
I need to pass this data to a function, then cycle through each pixel of the image. I would have done:
for (int i = 0; i < image.rows; ++i)
{
for (int j = 0; j < image.cols; ++j)
{
//pixel = cv::Point(i,j);
}
}
What is the equivalent of this, using the uchar* data?
It is pretty easy but you need to remember one thing, this image.elemSize() indicates how many bytes there are per pixel (this function is taken from OpenCV mat). So this loop will look little bit different for different image formats. There is a example inside the loop
for (auto i = 0; i < image.rows * image.cols; i+=image.elemSize())
{
//for CV_8UC1
//auto pixel = *(image.data + i)
//for RGB as CV_8UC3
auto r = *(image.data + i)
auto g = *(image.data + i + 1)
auto b = *(image.data + i + 2)
}
The correct pixel value can be accessed from the raw data provided the following parameters are known:
X coordinate of pixel ( column number )
Y coordinate of pixel ( row number )
Image depth (actual data type of a single pixel i.e. uchar, ushort, float etc)
Number of channels of the image
Image step in bytes
Given the above information, the pixel can be accessed as follows (for CV_8UC3 type):
uchar* data = (uchar *)pImg.data;
for (int i = 0; i < image.rows; ++i)
{
for (int j = 0; j < image.cols; ++j)
{
uchar b = data[i * pImg.step + pImg.channels() * j + 0];
uchar g = data[i * pImg.step + pImg.channels() * j + 1];
uchar r = data[i * pImg.step + pImg.channels() * j + 2];
}
}
everyone I am trying to implement patter matching with FFT but I am not sure what the result should be (I think I am missing something even though a read a lot of stuff about the problem and tried a lot of different implementations this one is the best so far). Here is my FFT correlation function.
void fft2d(fftw_complex**& a, int rows, int cols, bool forward = true)
{
fftw_plan p;
for (int i = 0; i < rows; ++i)
{
p = fftw_plan_dft_1d(cols, a[i], a[i], forward ? FFTW_FORWARD : FFTW_BACKWARD, FFTW_ESTIMATE);
fftw_execute(p);
}
fftw_complex* t = (fftw_complex*)fftw_malloc(rows * sizeof(fftw_complex));
for (int j = 0; j < cols; ++j)
{
for (int i = 0; i < rows; ++i)
{
t[i][0] = a[i][j][0];
t[i][1] = a[i][j][1];
}
p = fftw_plan_dft_1d(rows, t, t, forward ? FFTW_FORWARD : FFTW_BACKWARD, FFTW_ESTIMATE);
fftw_execute(p);
for (int i = 0; i < rows; ++i)
{
a[i][j][0] = t[i][0];
a[i][j][1] = t[i][1];
}
}
fftw_free(t);
}
int findCorrelation(int argc, char* argv[])
{
BMP bigImage;
BMP keyImage;
BMP result;
RGBApixel blackPixel = { 0, 0, 0, 1 };
const bool swapQuadrants = (argc == 4);
if (argc < 3 || argc > 4) {
cout << "correlation img1.bmp img2.bmp" << endl;
return 1;
}
if (!keyImage.ReadFromFile(argv[1])) {
return 1;
}
if (!bigImage.ReadFromFile(argv[2])) {
return 1;
}
//Preparations
const int maxWidth = std::max(bigImage.TellWidth(), keyImage.TellWidth());
const int maxHeight = std::max(bigImage.TellHeight(), keyImage.TellHeight());
const int rowsCount = maxHeight;
const int colsCount = maxWidth;
BMP bigTemp = bigImage;
BMP keyTemp = keyImage;
keyImage.SetSize(maxWidth, maxHeight);
bigImage.SetSize(maxWidth, maxHeight);
for (int i = 0; i < rowsCount; ++i)
for (int j = 0; j < colsCount; ++j) {
RGBApixel p1;
if (i < bigTemp.TellHeight() && j < bigTemp.TellWidth()) {
p1 = bigTemp.GetPixel(j, i);
} else {
p1 = blackPixel;
}
bigImage.SetPixel(j, i, p1);
RGBApixel p2;
if (i < keyTemp.TellHeight() && j < keyTemp.TellWidth()) {
p2 = keyTemp.GetPixel(j, i);
} else {
p2 = blackPixel;
}
keyImage.SetPixel(j, i, p2);
}
//Here is where the transforms begin
fftw_complex **a = (fftw_complex**)fftw_malloc(rowsCount * sizeof(fftw_complex*));
fftw_complex **b = (fftw_complex**)fftw_malloc(rowsCount * sizeof(fftw_complex*));
fftw_complex **c = (fftw_complex**)fftw_malloc(rowsCount * sizeof(fftw_complex*));
for (int i = 0; i < rowsCount; ++i) {
a[i] = (fftw_complex*)fftw_malloc(colsCount * sizeof(fftw_complex));
b[i] = (fftw_complex*)fftw_malloc(colsCount * sizeof(fftw_complex));
c[i] = (fftw_complex*)fftw_malloc(colsCount * sizeof(fftw_complex));
for (int j = 0; j < colsCount; ++j) {
RGBApixel p1;
p1 = bigImage.GetPixel(j, i);
a[i][j][0] = (0.299*p1.Red + 0.587*p1.Green + 0.114*p1.Blue);
a[i][j][1] = 0.0;
RGBApixel p2;
p2 = keyImage.GetPixel(j, i);
b[i][j][0] = (0.299*p2.Red + 0.587*p2.Green + 0.114*p2.Blue);
b[i][j][1] = 0.0;
}
}
fft2d(a, rowsCount, colsCount);
fft2d(b, rowsCount, colsCount);
result.SetSize(maxWidth, maxHeight);
for (int i = 0; i < rowsCount; ++i)
for (int j = 0; j < colsCount; ++j) {
fftw_complex& y = a[i][j];
fftw_complex& x = b[i][j];
double u = x[0], v = x[1];
double m = y[0], n = y[1];
c[i][j][0] = u*m + n*v;
c[i][j][1] = v*m - u*n;
int fx = j;
if (fx>(colsCount / 2)) fx -= colsCount;
int fy = i;
if (fy>(rowsCount / 2)) fy -= rowsCount;
float r2 = (fx*fx + fy*fy);
const double cuttoffCoef = (maxWidth * maxHeight) / 37992.;
if (r2<128 * 128 * cuttoffCoef)
c[i][j][0] = c[i][j][1] = 0;
}
fft2d(c, rowsCount, colsCount, false);
const int halfCols = colsCount / 2;
const int halfRows = rowsCount / 2;
if (swapQuadrants) {
for (int i = 0; i < halfRows; ++i)
for (int j = 0; j < halfCols; ++j) {
std::swap(c[i][j][0], c[i + halfRows][j + halfCols][0]);
std::swap(c[i][j][1], c[i + halfRows][j + halfCols][1]);
}
for (int i = halfRows; i < rowsCount; ++i)
for (int j = 0; j < halfCols; ++j) {
std::swap(c[i][j][0], c[i - halfRows][j + halfCols][0]);
std::swap(c[i][j][1], c[i - halfRows][j + halfCols][1]);
}
}
for (int i = 0; i < rowsCount; ++i)
for (int j = 0; j < colsCount; ++j) {
const double& g = c[i][j][0];
RGBApixel pixel;
pixel.Alpha = 0;
int gInt = 255 - static_cast<int>(std::floor(g + 0.5));
pixel.Red = gInt;
pixel.Green = gInt;
pixel.Blue = gInt;
result.SetPixel(j, i, pixel);
}
BMP res;
res.SetSize(maxWidth, maxHeight);
result.WriteToFile("result.bmp");
return 0;
}
Sample output
This question would probably be more appropriately posted on another site like cross validated (metaoptimize.com used to also be a good one, but it appears to be gone)
That said:
There's two similar operations you can perform with FFT: convolution and correlation. Convolution is used for determining how two signals interact with each-other, whereas correlation can be used to express how similar two signals are to each-other. Make sure you're doing the right operation as they're both commonly implemented throught a DFT.
For this type of application of DFTs you usually wouldn't extract any useful information in the fourier spectrum unless you were looking for frequencies common to both data sources or whatever (eg, if you were comparing two bridges to see if their supports are spaced similarly).
Your 3rd image looks a lot like the power domain; normally I see the correlation output entirely grey except where overlap occurred. Your code definitely appears to be computing the inverse DFT, so unless I'm missing something the only other explanation I've come up with for the fuzzy look could be some of the "fudge factor" code in there like:
if (r2<128 * 128 * cuttoffCoef)
c[i][j][0] = c[i][j][1] = 0;
As for what you should expect: wherever there are common elements between the two images you'll see a peak. The larger the peak, the more similar the two images are near that region.
Some comments and/or recommended changes:
1) Convolution & correlation are not scale invariant operations. In other words, the size of your pattern image can make a significant difference in your output.
2) Normalize your images before correlation.
When you get the image data ready for the forward DFT pass:
a[i][j][0] = (0.299*p1.Red + 0.587*p1.Green + 0.114*p1.Blue);
a[i][j][1] = 0.0;
/* ... */
How you grayscale the image is your business (though I would've picked something like sqrt( r*r + b*b + g*g )). However, I don't see you doing anything to normalize the image.
The word "normalize" can take on a few different meanings in this context. Two common types:
normalize the range of values between 0.0 and 1.0
normalize the "whiteness" of the images
3) Run your pattern image through an edge enhancement filter. I've personally made use of canny, sobel, and I think I messed with a few others. As I recall, canny was "quick'n dirty", sobel was more expensive, but I got comparable results when it came time to do correlation. See chapter 24 of the "dsp guide" book that's freely available online. The whole book is worth your time, but if you're low on time then at a minimum chapter 24 will help a lot.
4) Re-scale the output image between [0, 255]; if you want to implement thresholds, do it after this step because the thresholding step is lossy.
My memory on this one is hazy, but as I recall (edited for clarity):
You can scale the final image pixels (before rescaling) between [-1.0, 1.0] by dividing off the largest power spectrum value from the entire power spectrum
The largest power spectrum value is, conveniently enough, the center-most value in the power spectrum (corresponding to the lowest frequency)
If you divide it off the power spectrum, you'll end up doing twice the work; since FFTs are linear, you can delay the division until after the inverse DFT pass to when you're re-scaling the pixels between [0..255].
If after rescaling most of your values end up so black you can't see them, you can use a solution to the ODE y' = y(1 - y) (one example is the sigmoid f(x) = 1 / (1 + exp(-c*x) ), for some scaling factor c that gives better gradations). This has more to do with improving your ability to interpret the results visually than anything you might use to programmatically find peaks.
edit I said [0, 255] above. I suggest you rescale to [128, 255] or some other lower bound that is gray rather than black.
I'm new using cplex and I try to find some information on internet but didn't find clear stuff to help me in my problem.
I have P[k] k will be equal to 1 to 4
and I have a decision variable x[i][k] must be equal to 0 or 1 (also p[k])
the i is between 1 to 5
For now I do like this
IloEnv env;
IloModel model(env);
IloNumVarArray p(env);
p.add(IloNumVar(env, 0, 1));
p.add(IloNumVar(env, 0, 1));
p.add(IloNumVar(env, 0, 1));
IloIntVar x(env, 0, 1);
model.add(IloMaximize(env, 1000 * p[1] + 2000 * p[2] + 500 * p[3] + 1500 * p[4]));
for(int k = 1; k <= 4; k++){
for(int i = 1; i <= 5; i++){
model.add(x[i][k] + x[i][k] + x[i][k] + x[i][k] + x[i][k] => 2 * p[k]; );
}}
The loop should do something like this:
x[1][1] + x[2][1] + x[3][1] + x[4][1] + x[5][1] => 2 * p[1];
x[1][2] + x[2][2] + x[3][2] + x[4][2] + x[5][2] => 2 * p[2];
x[1][3] + x[2][3] + x[3][3] + x[4][3] + x[5][3] => 2 * p[3];
x[1][4] + x[2][4] + x[3][4] + x[4][4] + x[5][4] => 3 * p[4];
but I'm far away from this result.
Does anyone have an idea?
Thanks
You probably want to use an IloNumExpr
for(int k = 0; k < 4; k++){
IloNumExpr sum_over_i(env);
for(int i = 0; i < 5; i++){
sum_over_i += x[i][k];
}
model.add(sum_over_i >= 2 * p[k]; );
}
You also need to declare x as a 2-dimensional array.
IloArray x(env, 4);
for (int k = 0; k < 4; ++k)
x[k] = IloIntVarArray(env, 5, 0, 1);
Also, in c++, array indices are from 0 to size-1, not 1 to size. Your objective should be written
model.add(IloMaximize(env, 1000 * p[0] + 2000 * p[1] + 500 * p[2] + 1500 * p[3]));
Usertfwr already gave a good answer, but I would like to give another version of solution which might help you to code CPLEX applications in a more generic way. First, I would suggest you to use a text file to hold all the data (objective function coefficients) which will be fed into the program. In your case, you only have to copy literally the following matrix like data to notepad and name it as “coef.dat”:
[1000, 2000, 500, 1500]
Now comes the full code, let me know if have difficulties understanding any statement:
#include <ilcplex/ilocplex.h>
#include <fstream>
#include <iostream>
ILOSTLBEGIN
int main(int argc, char **argv) {
IloEnv env;
try {
const char* inputData = "coef.dat";
ifstream inFile(inputData); // put your data in the same directory as your executable
if(!inFile) {
cerr << "Cannot open the file " << inputData << " successfully! " <<endl;
throw(-1);
}
// Define parameters (coef of objective function)
IloNumArray a(env);
// Read in data
inFile >> a;
// Define variables
IloBoolVarArray p(env, a.getSize()); // note that a.getSize() = 4
IloArray<IloBoolVarArray> X(env, 5); // note that you need a 5x4 X variables, not 4x5
for(int i = 0; i < 5; i++) {
X[i] = IloBoolVarArray(env,4);
}
// Build model
IloModel model(env);
// Add objective function
IloExpr objFun (env);
for(int i = 0; i < a.getSize(); i++){
objFun += a[i]*p[i];
}
model.add(IloMaximize(env, objFun));
objFun.end();
// Add constraints -- similar to usertfwr’s answer
for(int i = 0; i < 4; k++){
IloExpr sumConst (env);
for(int j = 0; j < 5; i++){
sumConst += x[j][i];
}
// before clearing sumConst expr, add it to model
model.add(sumConst >= 2*p[i]);
sumConst.end(); // very important to end after having been added to the model
}
// Extract the model to CPLEX
IloCplex cplex(mod);
// Export the LP model to a txt file to check correctness
//cplex.exportModel("model.lp");
// Solve model
cplex.solve();
}
catch (IloException& e) {
cerr << "Concert exception caught: " << e << endl;
}
catch (...) {
cerr << "Unknown exception caught" << endl;
}
env.end();
}