Hough Transform for Circles 

You are right. Thanks for pointing it out.

Happy to help 😇

Yes you are right. Thanks for pointing it out 😇

At line 14, 15 you will have to change the xp and yp xp=a*cos(ang); yp=b*sin(ang); where a is the major axis and b is the minor axis of ellipse. You can put the parametric equation of the curve you want to find here.

@@pratikian thanks a lot for replying bro actually I had to do this for lame curve so can you help me with some code for it, as on net very complex codes are there

@@akainu3668 I could try

Am I right? I fixing a: a_min = X - R; a_max = X + R; after that I do for (i = a_min; i

So what do X and Y mean here? Apart from that I feel that the algorithm is fine it should work. here is an algorithm that I had used. first define a circle with radius r. ang=0:0.1:2*pi; xp=r*cos(ang); yp=r*sin(ang); Now we just have to move this circle so that it centers at the edge point (x,y). Now make an accumulator matrix which is of size (m+2*r, n+2*r) (where the image size is m x n ) now run a for loop for all the edge points x,y detected make the circle in the accumulator matrix centered at every edge point (x,y) v = x + xp; w = y + yp; increment the accumulator at v,w position accumulator(v,w) += 1; Finally choose the point which is having the maximum value in the accumulator. That is the required center of the circle. here you can see the matlab code for the above algorithm drive.google.com/drive/u/0/folders/1UTIgcj0Z-dKo67zNZj8qwmueH0g3OPp0

here is an algorithm that I had used. first define a circle with radius r. ang=0:0.1:2*pi; xp=r*cos(ang); yp=r*sin(ang); Now we just have to move this circle so that it centers at the edge point (x,y). Now make an accumulator matrix which is of size (m+2*r, n+2*r) (where the image size is m x n ) now run a for loop for all the edge points x,y detected make the circle in the accumulator matrix centered at every edge point (x,y) v = x + xp; w = y + yp; increment the accumulator at v,w position accumulator(v,w) += 1; Finally choose the point which is having the maximum value in the accumulator. That is the required center of the circle. here you can see the matlab code for the above algorithm drive.google.com/drive/u/0/folders/1UTIgcj0Z-dKo67zNZj8qwmueH0g3OPp0

You can also look at some other implementations my code is not very Robust in.mathworks.com/matlabcentral/fileexchange/4985-circle-detection-via-standard-hough-transform

So for that you can make a vector of the radius of the different circles that is there in the image and run the algo for each radius

@@pratikian can you help me out with it in your implementation to see how it will work I tried using a for loop but so many smaller circles

Okay I will try. Acctually my code is not very robust meanwhile you can also try other implementations in.mathworks.com/matlabcentral/fileexchange/4985-circle-detection-via-standard-hough-transform Hope that helps

All right thank you

I did not get your question. What do you mean by all the co ordinates of (a,b)

@@pratikian yes I want all centes coordinates in a given image for all detected circles

Sure here is the link for the matlab code drive.google.com/drive/mobile/folders/1UTIgcj0Z-dKo67zNZj8qwmueH0g3OPp0?usp=sharing This code is not very robust it is for a constant r , it is good for understanding purpose