I currently have a method for detecting a card in an image and for the most part it works when the lighting is fairly consistent and the background is very calm.
Here is the code I am using to preform this operation:
Mat img = inImg.clone();
outImg = Mat(inImg.size(), CV_8UC1);
inImg.copyTo(outImg);
Mat img_fullRes = img.clone();
pyrDown(img, img);
Mat imgGray;
cvtColor(img, imgGray, CV_RGB2GRAY);
outImg_gray = imgGray.clone();
// Find Edges //
Mat detectedEdges = imgGray.clone();
bilateralFilter(imgGray, detectedEdges, 0, 185, 3, 0);
Canny( detectedEdges, detectedEdges, 20, 65, 3 );
dilate(detectedEdges, detectedEdges, Mat::ones(3,3,CV_8UC1));
Mat cdst = img.clone();
vector<Vec4i> lines;
HoughLinesP(detectedEdges, lines, 1, CV_PI/180, 60, 50, 3 );
for( size_t i = 0; i < lines.size(); i++ )
{
Vec4i l = lines[i];
// For debug
//line( cdst, cv::Point(l[0], l[1]), cv::Point(l[2], l[3]), Scalar(0,0,255), 1);
}
//cdst.copyTo(inImg);
// // Find points of intersection //
cv::Rect imgROI;
int ext = 10;
imgROI.x = ext;
imgROI.y = ext;
imgROI.width = img.size().width - ext;
imgROI.height = img.size().height - ext;
int N = lines.size();
// Creating N amount of points // N == lines.size()
cv::Point** poi = new cv::Point*[N];
for( int i = 0; i < N; i++ )
poi[i] = new cv::Point[N];
vector<cv::Point> poiList;
for( int i = 0; i < N; i++ )
{
poi[i][i] = cv::Point(-1,-1);
Vec4i line1 = lines[i];
for( int j = i + 1; j < N; j++ )
{
Vec4i line2 = lines[j];
cv::Point p = computeIntersect(line1, line2, imgROI);
if( p.x != -1 )
{
//line(cdst, p-cv::Point(2,0), p+cv::Point(2,0), Scalar(0,255,0));
//line(cdst, p-cv::Point(0,2), p+cv::Point(0,2), Scalar(0,255,0));
poiList.push_back(p);
}
poi[i][j] = p;
poi[j][i] = p;
}
}
cdst.copyTo(inImg);
if(poiList.size()==0)
{
outImg = inImg.clone();
//circle(outImg, cv::Point(100,100), 50, Scalar(255,0,0), -1);
return;
}
convexHull(poiList, poiList, false, true);
for( int i=0; i<poiList.size(); i++ )
{
cv::Point p = poiList[i];
//circle(cdst, p, 3, Scalar(255,0,0), 2);
}
//Evaluate all possible quadrilaterals
cv::Point cardCorners[4];
float metric_max = 0;
int Npoi = poiList.size();
for( int p1=0; p1<Npoi; p1++ )
{
cv::Point pts[4];
pts[0] = poiList[p1];
for( int p2=p1+1; p2<Npoi; p2++ )
{
pts[1] = poiList[p2];
if( isCloseBy(pts[1],pts[0]) )
continue;
for( int p3=p2+1; p3<Npoi; p3++ )
{
pts[2] = poiList[p3];
if( isCloseBy(pts[2],pts[1]) || isCloseBy(pts[2],pts[0]) )
continue;
for( int p4=p3+1; p4<Npoi; p4++ )
{
pts[3] = poiList[p4];
if( isCloseBy(pts[3],pts[0]) || isCloseBy(pts[3],pts[1])
|| isCloseBy(pts[3],pts[2]) )
continue;
// get the metrics
float area = getArea(pts);
cv::Point a = pts[0]-pts[1];
cv::Point b = pts[1]-pts[2];
cv::Point c = pts[2]-pts[3];
cv::Point d = pts[3]-pts[0];
float oppLenDiff = abs(a.dot(a)-c.dot(c)) + abs(b.dot(b)-d.dot(d));
float metric = area - 0.35*oppLenDiff;
if( metric > metric_max )
{
metric_max = metric;
cardCorners[0] = pts[0];
cardCorners[1] = pts[1];
cardCorners[2] = pts[2];
cardCorners[3] = pts[3];
}
}
}
}
}
// find the corners corresponding to the 4 corners of the physical card
sortPointsClockwise(cardCorners);
// Calculate Homography //
vector<Point2f> srcPts(4);
srcPts[0] = cardCorners[0]*2;
srcPts[1] = cardCorners[1]*2;
srcPts[2] = cardCorners[2]*2;
srcPts[3] = cardCorners[3]*2;
vector<Point2f> dstPts(4);
cv::Size outImgSize(1400,800);
dstPts[0] = Point2f(0,0);
dstPts[1] = Point2f(outImgSize.width-1,0);
dstPts[2] = Point2f(outImgSize.width-1,outImgSize.height-1);
dstPts[3] = Point2f(0,outImgSize.height-1);
Mat Homography = findHomography(srcPts, dstPts);
// Apply Homography
warpPerspective( img_fullRes, outImg, Homography, outImgSize, INTER_CUBIC );
outImg.copyTo(inImg);
Where computeIntersect
is defined as:
cv::Point computeIntersect(cv::Vec4i a, cv::Vec4i b, cv::Rect ROI)
{
int x1 = a[0], y1 = a[1], x2 = a[2], y2 = a[3];
int x3 = b[0], y3 = b[1], x4 = b[2], y4 = b[3];
cv::Point p1 = cv::Point (x1,y1);
cv::Point p2 = cv::Point (x2,y2);
cv::Point p3 = cv::Point (x3,y3);
cv::Point p4 = cv::Point (x4,y4);
// Check to make sure all points are within the image boundrys, if not reject them.
if( !ROI.contains(p1) || !ROI.contains(p2)
|| !ROI.contains(p3) || !ROI.contains(p4) )
return cv::Point (-1,-1);
cv::Point vec1 = p1-p2;
cv::Point vec2 = p3-p4;
float vec1_norm2 = vec1.x*vec1.x + vec1.y*vec1.y;
float vec2_norm2 = vec2.x*vec2.x + vec2.y*vec2.y;
float cosTheta = (vec1.dot(vec2))/sqrt(vec1_norm2*vec2_norm2);
float den = ((float)(x1-x2) * (y3-y4)) - ((y1-y2) * (x3-x4));
if(den != 0)
{
cv::Point2f pt;
pt.x = ((x1*y2 - y1*x2) * (x3-x4) - (x1-x2) * (x3*y4 - y3*x4)) / den;
pt.y = ((x1*y2 - y1*x2) * (y3-y4) - (y1-y2) * (x3*y4 - y3*x4)) / den;
if( !ROI.contains(pt) )
return cv::Point (-1,-1);
// no-confidence metric
float d1 = MIN( dist2(p1,pt), dist2(p2,pt) )/vec1_norm2;
float d2 = MIN( dist2(p3,pt), dist2(p4,pt) )/vec2_norm2;
float no_confidence_metric = MAX(sqrt(d1),sqrt(d2));
// If end point ratios are greater than .5 reject
if( no_confidence_metric < 0.5 && cosTheta < 0.707 )
return cv::Point (int(pt.x+0.5), int(pt.y+0.5));
}
return cv::Point(-1, -1);
}
sortPointsClockWise
is defined as:
void sortPointsClockwise(cv::Point a[])
{
cv::Point b[4];
cv::Point ctr = (a[0]+a[1]+a[2]+a[3]);
ctr.x /= 4;
ctr.y /= 4;
b[0] = a[0]-ctr;
b[1] = a[1]-ctr;
b[2] = a[2]-ctr;
b[3] = a[3]-ctr;
for( int i=0; i<4; i++ )
{
if( b[i].x < 0 )
{
if( b[i].y < 0 )
a[0] = b[i]+ctr;
else
a[3] = b[i]+ctr;
}
else
{
if( b[i].y < 0 )
a[1] = b[i]+ctr;
else
a[2] = b[i]+ctr;
}
}
}
getArea
is defined as:
float getArea(cv::Point arr[])
{
cv::Point diag1 = arr[0]-arr[2];
cv::Point diag2 = arr[1]-arr[3];
return 0.5*(diag1.cross(diag2));
}
isCloseBy
is defined as:
bool isCloseBy( cv::Point p1, cv::Point p2 )
{
int D = 10;
// Checking that X values are within 10, same for Y values.
return ( abs(p1.x-p2.x)<=D && abs(p1.y-p2.y)<=D );
}
And finally dist2
:
float dist2( cv::Point p1, cv::Point p2 )
{
return float((p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y));
}
Here are several test images and their results:
Sorry for the very lengthy post, however I am hoping someone can suggest a way I can make my method for extracting the card from the image more robust. One that can better handle disruptive backgrounds along with inconsistent lighting.
When a card is placed on a contrasting background with good lighting my method works nearly 90% of the time. But it is clear I need a more robust approach.
Does anyone have any suggestions?
Thanks.
ATTEMPT of dhanushka's soloution
Mat gray, bw; pyrDown(inImg, inImg);
cvtColor(inImg, gray, CV_RGB2GRAY);
int morph_size = 3;
Mat element = getStructuringElement( MORPH_ELLIPSE, cv::Size( 4*morph_size + 1, 2*morph_size+1 ), cv::Point( morph_size, morph_size ) );
morphologyEx(gray, gray, 2, element);
threshold(gray, bw, 160, 255, CV_THRESH_BINARY);
vector<vector<cv::Point> > contours;
vector<Vec4i> hierarchy;
findContours( bw, contours, hierarchy, CV_RETR_TREE, CV_CHAIN_APPROX_SIMPLE, cv::Point(0, 0) );
int largest_area=0;
int largest_contour_index=0;
cv::Rect bounding_rect;
for( int i = 0; i< contours.size(); i++ )
{
double a=contourArea( contours[i],false); // Find the area of contour
if(a>largest_area){
largest_area=a;
largest_contour_index=i; //Store the index of largest contour
bounding_rect=boundingRect(contours[i]);
}
}
//Scalar color( 255,255,255);
rectangle(inImg, bounding_rect, Scalar(0,255,0),1, 8,0);
Mat biggestRect = inImg(bounding_rect);
Mat card1 = biggestRect.clone();
The art of image processing is (in my 10+ years experience) just that: an art. No single answer exists, and there is always more than one way to do it. And it will definitely fail in some cases.
In my experience of working on automatically detecting features in medical images, it takes a long time to build to reliable algorithm, but in hindsight the best result is obtained with a relative simple algorithm. However, it takes a lot of time to get to this simple algorithm.
To get to this, the general approach is always the same:
That having said, I think that you need to answer these questions during the tuning of the algorithm:
These will make the requirements and assumptions on the image to acquire. Assumptions on which you can rely are very strong: they make the algorithm fast, robust and simple if you choose the right ones. Also let these requirements and assumptions be part of the testing database.
So what would I choose? Based on the three images you provided I would start with something like:
The algorithm then would look like:
cornerHarris
.Try to build parallelograms with one corner per each quadrant by combining points from each quadrant. Create a fitness function which gives higher score to:
This fitness function gives a lot of tuning possibilities later on.
Return the parallelogram with the highest score.
So why using corner-detection instead of a hough-transform to do line detection? In my opinion the hough-transform is (next to being slow) quite sensitive to patterns in the background (which is what you see in your first image -- it detects a stronger line in the background then of the card), and it cannot handle a little curved lines that well, unless you use a larger bin size which will worsen the detection.
Good luck!
A more general approach would definitely be something like Rutger Nijlunsing suggested in his answer. However, in your case, at least for the provided sample images, a very simple approach like morphological opening followed by thresholding, contour processing and convexhull would yield the result you want. Use a scaled down version of the images for processing so that you don't have to use a large kernel for morphological operations. Below are the images processed this way.
pyrDown(large, rgb0);
pyrDown(rgb0, rgb0);
pyrDown(rgb0, rgb0);
Mat small;
cvtColor(rgb0, small, CV_BGR2GRAY);
Mat morph;
Mat kernel = getStructuringElement(MORPH_ELLIPSE, Size(11, 11));
morphologyEx(small, morph, MORPH_OPEN, kernel);
Mat bw;
threshold(morph, bw, 0, 255.0, CV_THRESH_BINARY | CV_THRESH_OTSU);
Mat bdry;
kernel = getStructuringElement(MORPH_ELLIPSE, Size(3, 3));
erode(bw, bdry, kernel);
subtract(bw, bdry, bdry);
// do contour processing on bdry
This approach will not work in general, so I would strongly recommend something like Rutger suggested.
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