What to do with negative rho values in hough transform? - c++

Question

Here is my code for creating the hough accumulator for lines in image :
void hough_lines_acc(cv::Mat img_a_edges, std::vector<std::vector<int> > &hough_acc) {
for (size_t r = 0; r < img_a_edges.rows; r++) {
for (size_t c = 0; c < img_a_edges.cols; c++) {
int theta = static_cast<int> (std::atan2(r, c) * 180 / M_PI);
int rho = static_cast<int> ((c * cos(theta)) + (r * sin(theta)));
if (theta < -90) theta = -90;
if (theta > 89) theta = 89;
++hough_acc[abs(rho)][theta];
}
}
cv::Mat img_mat(hough_acc.size(), hough_acc[0].size(), CV_8U);
std::cout << hough_acc.size() << " " << hough_acc[0].size() << std::endl;
for (size_t i = 0; i < hough_acc.size(); i++) {
for (size_t j = 0; j < hough_acc[0].size(); j++) {
img_mat.at<int> (i,j) = hough_acc[i][j];
}
}
imwrite("../output/ps1-­2-­b-­1.png", img_mat);
}
theta varies from -90 to 89. I am getting negative rho values. Right now I am just replacing the negative who with a positive one but am not getting a correct answer. What do I do to the negative rho? Please explain the answer.
theta = arctan (y / x)
rho = x * cos(theta) + y * sin(theta)
Edited code :
bool hough_lines_acc(cv::Mat img_a_edges, std::vector<std::vector<int> > &hough_acc,\
std::vector<double> thetas, std::vector<double> rhos, int rho_resolution, int theta_resolution) {
int img_w = img_a_edges.cols;
int img_h = img_a_edges.rows;
int max_votes = 0;
int min_votes = INT_MAX;
for (size_t r = 0; r < img_h; r++) {
for (size_t c = 0; c < img_w; c++) {
if(img_a_edges.at<int>(r, c) == 255) {
for (size_t i = 0; i < thetas.size(); i++) {
thetas[i] = (thetas[i] * M_PI / 180);
double rho = ( (c * cos(thetas[i])) + (r * sin(thetas[i])) );
int buff = ++hough_acc[static_cast<int>(abs(rho))][static_cast<int>(i)];
if (buff > max_votes) {
max_votes = buff;
}
if (buff < min_votes) {
min_votes = buff;
}
}
}
}
}
double div = static_cast<double>(max_votes) / 255;
int threshold = 10;
int possible_edge = round(static_cast<double>(max_votes) / div) - threshold;
props({
{"max votes", max_votes},
{"min votes", min_votes},
{"scale", div}
});
// needed for scaling intensity for contrast
// not sure if I am doing it correctly
for (size_t r = 0; r < hough_acc.size(); r++) {
for (size_t c = 0; c < hough_acc[0].size(); c++) {
double val = hough_acc[r][c] / div;
if (val < 0) {
val = 0;
}
hough_acc[r][c] = static_cast<int>(val);
}
}
cv::Mat img_mat = cv::Mat(hough_acc.size(), hough_acc[0].size(), CV_8UC1, cv::Scalar(0));
for (size_t i = 0; i < hough_acc.size(); i++) {
for (size_t j = 0; j < hough_acc[0].size(); j++) {
img_mat.at<uint8_t> (i,j) = static_cast<uint8_t>(hough_acc[i][j]);
}
}
imwrite("../output/ps1-­2-­b-­1.png", img_mat);
return true;
}
Still not correct output. What is the error here?

Answer 1

atan2 of two positive numbers... should not be giving you negative angles, it should only be giving you a range of 0-90
also for the hough transform, I think you want everything relative to one point (ie 0,0 in this case). I think for that you would actually want to make theta=90-atan2(r,c)
Admittedly though, I am a bit confused as I thought you had to encode line direction, rather than just "edge pt". ie I thought at each edge point you had to provide a discrete array of guessed line trajectories and calculate rho and theta for each one and throw all of those into your accumulator. As is... I am not sure what you are calculating.