OpenCV C++/Obj-C: Advanced square detection
Solution 1:
You might try using HoughLines to detect the four sides of the square. Next, locate the four resulting line intersections to detect the corners. The Hough transform is fairly robust to noise and occlusions, so it could be useful here. Also, here is an interactive demo showing how the Hough transform works (I thought it was cool at least :). Here is one of my previous answers that detects a laser cross center showing most of the same math (except it just finds a single corner).
You will probably have multiple lines on each side, but locating the intersections should help to determine the inliers vs. outliers. Once you've located candidate corners, you can also filter these candidates by area or how "square-like" the polygon is.
EDIT : All these answers with code and images were making me think my answer was a bit lacking :) So, here is an implementation of how you could do this:
#include <opencv2/core/core.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include <iostream>
#include <vector>
using namespace cv;
using namespace std;
Point2f computeIntersect(Vec2f line1, Vec2f line2);
vector<Point2f> lineToPointPair(Vec2f line);
bool acceptLinePair(Vec2f line1, Vec2f line2, float minTheta);
int main(int argc, char* argv[])
{
Mat occludedSquare = imread("Square.jpg");
resize(occludedSquare, occludedSquare, Size(0, 0), 0.25, 0.25);
Mat occludedSquare8u;
cvtColor(occludedSquare, occludedSquare8u, CV_BGR2GRAY);
Mat thresh;
threshold(occludedSquare8u, thresh, 200.0, 255.0, THRESH_BINARY);
GaussianBlur(thresh, thresh, Size(7, 7), 2.0, 2.0);
Mat edges;
Canny(thresh, edges, 66.0, 133.0, 3);
vector<Vec2f> lines;
HoughLines( edges, lines, 1, CV_PI/180, 50, 0, 0 );
cout << "Detected " << lines.size() << " lines." << endl;
// compute the intersection from the lines detected...
vector<Point2f> intersections;
for( size_t i = 0; i < lines.size(); i++ )
{
for(size_t j = 0; j < lines.size(); j++)
{
Vec2f line1 = lines[i];
Vec2f line2 = lines[j];
if(acceptLinePair(line1, line2, CV_PI / 32))
{
Point2f intersection = computeIntersect(line1, line2);
intersections.push_back(intersection);
}
}
}
if(intersections.size() > 0)
{
vector<Point2f>::iterator i;
for(i = intersections.begin(); i != intersections.end(); ++i)
{
cout << "Intersection is " << i->x << ", " << i->y << endl;
circle(occludedSquare, *i, 1, Scalar(0, 255, 0), 3);
}
}
imshow("intersect", occludedSquare);
waitKey();
return 0;
}
bool acceptLinePair(Vec2f line1, Vec2f line2, float minTheta)
{
float theta1 = line1[1], theta2 = line2[1];
if(theta1 < minTheta)
{
theta1 += CV_PI; // dealing with 0 and 180 ambiguities...
}
if(theta2 < minTheta)
{
theta2 += CV_PI; // dealing with 0 and 180 ambiguities...
}
return abs(theta1 - theta2) > minTheta;
}
// the long nasty wikipedia line-intersection equation...bleh...
Point2f computeIntersect(Vec2f line1, Vec2f line2)
{
vector<Point2f> p1 = lineToPointPair(line1);
vector<Point2f> p2 = lineToPointPair(line2);
float denom = (p1[0].x - p1[1].x)*(p2[0].y - p2[1].y) - (p1[0].y - p1[1].y)*(p2[0].x - p2[1].x);
Point2f intersect(((p1[0].x*p1[1].y - p1[0].y*p1[1].x)*(p2[0].x - p2[1].x) -
(p1[0].x - p1[1].x)*(p2[0].x*p2[1].y - p2[0].y*p2[1].x)) / denom,
((p1[0].x*p1[1].y - p1[0].y*p1[1].x)*(p2[0].y - p2[1].y) -
(p1[0].y - p1[1].y)*(p2[0].x*p2[1].y - p2[0].y*p2[1].x)) / denom);
return intersect;
}
vector<Point2f> lineToPointPair(Vec2f line)
{
vector<Point2f> points;
float r = line[0], t = line[1];
double cos_t = cos(t), sin_t = sin(t);
double x0 = r*cos_t, y0 = r*sin_t;
double alpha = 1000;
points.push_back(Point2f(x0 + alpha*(-sin_t), y0 + alpha*cos_t));
points.push_back(Point2f(x0 - alpha*(-sin_t), y0 - alpha*cos_t));
return points;
}
NOTE : The main reason I resized the image was so I could see it on my screen, and speed-up processing.
Canny
This uses Canny edge detection to help greatly reduce the number of lines detected after thresholding.
Hough transform
Then the Hough transform is used to detect the sides of the square.
Intersections
Finally, we compute the intersections of all the line pairs.
Hope that helps!
Solution 2:
I tried to use convex hull method
which is pretty simple.
Here you find convex hull of the contour detected. It removes the convexity defects at the bottom of paper.
Below is the code (in OpenCV-Python):
import cv2
import numpy as np
img = cv2.imread('sof.jpg')
img = cv2.resize(img,(500,500))
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
ret,thresh = cv2.threshold(gray,127,255,0)
contours,hier = cv2.findContours(thresh,cv2.RETR_LIST,cv2.CHAIN_APPROX_SIMPLE)
for cnt in contours:
if cv2.contourArea(cnt)>5000: # remove small areas like noise etc
hull = cv2.convexHull(cnt) # find the convex hull of contour
hull = cv2.approxPolyDP(hull,0.1*cv2.arcLength(hull,True),True)
if len(hull)==4:
cv2.drawContours(img,[hull],0,(0,255,0),2)
cv2.imshow('img',img)
cv2.waitKey(0)
cv2.destroyAllWindows()
(Here, i haven't found square in all planes. Do it yourself if you want.)
Below is the result i got:
I hope this is what you needed.
Solution 3:
1st: start experimenting with threshold techniques to isolate the white paper sheet from the rest of the image. This is a simple way:
Mat new_img = imread(argv[1]);
double thres = 200;
double color = 255;
threshold(new_img, new_img, thres, color, CV_THRESH_BINARY);
imwrite("thres.png", new_img);
but there are other alternatives that could provide better result. One is to investigate inRange()
, and another is to detect through color by converting the image to the HSV color space.
This thread also provides an interest discussion on the subject.
2nd: after you execute one of this procedures, you could try to feed the result directly into find_squares()
:
An alternative to find_squares()
is to implement the bounding box technique, which has the potential to provide a more accurate detection of the rectangular area (provided that you have a perfect result of threshold). I've used it here and here. It's worth noting that OpenCV has it's own bounding box tutorial.
Another approach besides find_squares()
, as pointed by Abid on his answer, is to use the convexHull method. Check OpenCV's C++ tutorial on this method for code.