I am developing an Android app that calculates the sum of all points of the being-seen dominoes pieces -shown in picture- using OpenCV for Android.
The problem is, I can't find a way to filtering other contours and counting only dots I see in the dominoes, I tried to use Canny edge finding then use HoughCircles, but with no result, as I don't have an absolute top view of the rocks and HoughCircles detect perfect circles only :)
Here is my code:
public Mat onCameraFrame(Mat inputFrame) {
inputFrame.copyTo(mRgba);
Mat grey = new Mat();
// Make it greyscale
Imgproc.cvtColor(mRgba, grey, Imgproc.COLOR_RGBA2GRAY);
// init contours arraylist
List<MatOfPoint> contours = new ArrayList<MatOfPoint>(200);
//blur
Imgproc.GaussianBlur(grey, grey, new Size(9,9), 10);
Imgproc.threshold(grey, grey, 80, 150, Imgproc.THRESH_BINARY);
// because findContours modifies the image I back it up
Mat greyCopy = new Mat();
grey.copyTo(greyCopy);
Imgproc.findContours(greyCopy, contours, new Mat(), Imgproc.RETR_TREE, Imgproc.CHAIN_APPROX_NONE);
// Now I have my controus pefectly
MatOfPoint2f mMOP2f1 = new MatOfPoint2f();
//a list for only selected contours
List<MatOfPoint> SelectedContours = new ArrayList<MatOfPoint>(400);
for(int i=0;i<contours.size();i++)
{
if(here I should put a condition that distinguishes my spots, eg: if contour inside is black and is a black disk)
{
SelectedContours.add(contours.get(i));
}
}
Imgproc.drawContours(mRgba, SelectedContours, -1, new Scalar(255,0,0,255), 1);
return mRgba;
}
EDIT:
One unique feature of my contours after threshold is they're totally black from inside, is there anyway I could calculate the mean color/intensity for a given contour ?
There is a similiar problem and possible solution on SO, titled Detection of coins (and fit ellipses) on an image. Here you will find some recomendations about opencv's function fitEllipse
.
You should take a look at this for more info on opencv's function fitEllipse
.
Also, to detect only black elements in an image, you can use HSV color model, to find only black colors. You can find an explanation here.
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