detecting 2 lines opencv - c++

Question

I have an image on which I run a dilation, and works fine, now I want to detect two dick lines on it :
and when run on it the part of code:
cv::Canny(dilationResult,canny,50,200,3);
cv::cvtColor(dilationResult,dilationResult,CV_BGR2GRAY);
cv::HoughLines(canny,lines,30,CV_PI/180,500,0);
cv::cvtColor(mask,mask,CV_GRAY2BGR);
if(lines.size()!=0){
std::cout << " line Size " << lines.size()<< std::endl;
for( size_t i = 0; i < lines.size(); i++ )
{
float rho = lines[i][0], theta = lines[i][2];
cv::Point pt1, pt2;
double a = cos(theta), b = sin(theta);
double x0 = a*rho, y0 = b*rho;
pt1.x = cvRound(x0 + 1000*(-b));
pt1.y = cvRound(y0 + 1000*(a));
pt2.x = cvRound(x0 - 1000*(-b));
pt2.y = cvRound(y0 - 1000*(a));
angle = atan2f((pt2.y-pt1.y),(pt2.x-pt1.x))*180.0/CV_PI;
std::cout << "angle " << angle<< std::endl;
line( mask, pt1, pt2, cv::Scalar(0,0,255), 3, CV_AA);
}
}
cv::imshow("mask " ,mask);
here's the result:
what I would like to get is something like this :
getting only 2 lines that have the same width, and by the way I don't want to use findcontour function
any idea how can do this !

Answer 1

I don't get it to work with hough transform, but with the probabilistic version cv::HoughLinesP
with lineDetection_Input.jpg being your linked image
int main()
{
cv::Mat color = cv::imread("../lineDetection_Input.jpg");
cv::Mat gray;
cv::cvtColor(color, gray, CV_RGB2GRAY);
std::vector<cv::Vec4i> lines;
cv::HoughLinesP( gray, lines, 1, 2*CV_PI/180, 100, 100, 50 );
for( size_t i = 0; i < lines.size(); i++ )
{
cv::line( color, cv::Point(lines[i][0], lines[i][1]),
cv::Point(lines[i][2], lines[i][3]), cv::Scalar(0,0,255), 1);
}
cv::imwrite("lineDetection_Output.jpg", color);
cv::namedWindow("output"); cv::imshow("output", color); cv::waitKey(-1);
return 0;
}
lineDetection_Output.jpg:
for rotated image:
and for some different intersection angle:
there you can see some lines detected with a slightly false angle that start in the top-right and end near the intersection (close behind), but these might be easily filtered by length or something.