Detect semicircle in OpenCV

Question

I am trying to detect full circles and semicircles in an image.

I am following the below mentioned procedure: Process image (including Canny edge detection). Find contours and draw them on an empty image, so that I can eliminate unwanted components (The processed image is exactly what I want). Detect circles using HoughCircles. And, this is what I get:

I tried varying the parameters in HoughCircles but the results are not consistent as it varies based on lighting and the position of circles in the image. I accept or reject a circle based on its size. So, the result is not acceptable. Also, I have a long list of "acceptable" circles. So, I need some allowance in the HoughCircle params. As for the full circles, it's easy - I can simply find the "roundness" of the contour. The problem is semicircles!

Please find the edited image before Hough transform

Answer 1

Use houghCircle directly on your image, don't extract edges first. Then test for each detected circle, how much percentage is really present in the image:

int main() {     cv::Mat color = cv::imread("../houghCircles.png");     cv::namedWindow("input"); cv::imshow("input", color);      cv::Mat canny;      cv::Mat gray;     /// Convert it to gray     cv::cvtColor( color, gray, CV_BGR2GRAY );      // compute canny (don't blur with that image quality!!)     cv::Canny(gray, canny, 200,20);     cv::namedWindow("canny2"); cv::imshow("canny2", canny>0);      std::vector<cv::Vec3f> circles;      /// Apply the Hough Transform to find the circles     cv::HoughCircles( gray, circles, CV_HOUGH_GRADIENT, 1, 60, 200, 20, 0, 0 );      /// Draw the circles detected     for( size_t i = 0; i < circles.size(); i++ )      {         Point center(cvRound(circles[i][0]), cvRound(circles[i][1]));         int radius = cvRound(circles[i][2]);         cv::circle( color, center, 3, Scalar(0,255,255), -1);         cv::circle( color, center, radius, Scalar(0,0,255), 1 );     }      //compute distance transform:     cv::Mat dt;     cv::distanceTransform(255-(canny>0), dt, CV_DIST_L2 ,3);     cv::namedWindow("distance transform"); cv::imshow("distance transform", dt/255.0f);      // test for semi-circles:     float minInlierDist = 2.0f;     for( size_t i = 0; i < circles.size(); i++ )      {         // test inlier percentage:         // sample the circle and check for distance to the next edge         unsigned int counter = 0;         unsigned int inlier = 0;          cv::Point2f center((circles[i][0]), (circles[i][1]));         float radius = (circles[i][2]);          // maximal distance of inlier might depend on the size of the circle         float maxInlierDist = radius/25.0f;         if(maxInlierDist<minInlierDist) maxInlierDist = minInlierDist;          //TODO: maybe paramter incrementation might depend on circle size!         for(float t =0; t<2*3.14159265359f; t+= 0.1f)          {             counter++;             float cX = radius*cos(t) + circles[i][0];             float cY = radius*sin(t) + circles[i][1];              if(dt.at<float>(cY,cX) < maxInlierDist)              {                 inlier++;                 cv::circle(color, cv::Point2i(cX,cY),3, cv::Scalar(0,255,0));             }             else                 cv::circle(color, cv::Point2i(cX,cY),3, cv::Scalar(255,0,0));         }         std::cout << 100.0f*(float)inlier/(float)counter << " % of a circle with radius " << radius << " detected" << std::endl;     }      cv::namedWindow("output"); cv::imshow("output", color);     cv::imwrite("houghLinesComputed.png", color);      cv::waitKey(-1);     return 0; } 

For this input:

It gives this output:

The red circles are Hough results.

The green sampled dots on the circle are inliers.

The blue dots are outliers.

Console output:

100 % of a circle with radius 27.5045 detected 100 % of a circle with radius 25.3476 detected 58.7302 % of a circle with radius 194.639 detected 50.7937 % of a circle with radius 23.1625 detected 79.3651 % of a circle with radius 7.64853 detected 

If you want to test RANSAC instead of Hough, have a look at this.