# Revision history [back]

My understanding concerning :

1. "Compute the determinant of the homography, and see if it's too close to zero for comfort." => compute the determinant of the top left 2x2 homography matrix, and check if it's "too close" to zero for comfort...btw you can also check if it's too far from zero because then the invert matrix would have a determinant too close to zero. A determinant of zero would mean the matrix is not inversible, too close to zero would mean singular (like you see the plane object at 90°, which is almost impossible if you use *good matches).

const double det = H.at<double>(0, 0) * H.at<double>(1, 1) - H.at<double>(1, 0) * H.at<double>(0, 1);

One way to do it would be to specify a positive value, say N, and then:

if((fabs(det)>(double)N) && (fabs(det)<(1.0/(double)N)) return false; // bad homography

And while we are at it...if det<0 the homography is not conserving the orientation (clockwise<->anticlockwise), except if you are watching the object in a mirror...it is certainly not good (plus the sift/surf descriptors are not done to be mirror invariants as far as i know, so you would probably don'thave good maches).

if(det<0) return false; // no mirrors in the scene

/* [Detecting Planar Homographies in an Image Pair Etienne Vincent and Robert Laganiere School of Information Technology and Engineering University of Ottawa Canada K1N 6N5] http://www.site.uottawa.ca/research */

My understanding concerning :

1. "Compute the determinant of the homography, and see if it's too close to zero for comfort." => compute the determinant of the top left 2x2 homography matrix, and check if it's "too close" to zero for comfort...btw you can also check if it's too far from zero because then the invert matrix would have a determinant too close to zero. A determinant of zero would mean the matrix is not inversible, too close to zero would mean singular (like you see the plane object at 90°, which is almost impossible if you use *good matches).

const double det = H.at<double>(0, 0) * H.at<double>(1, 1) - H.at<double>(1, 0) * H.at<double>(0, 1);

One way to do it would be to specify a positive value, say N, and then:

if((fabs(det)>(double)N) && (fabs(det)<(1.0/(double)N)) return false; // bad homography

And while we are at it...if det<0 the homography is not conserving the orientation (clockwise<->anticlockwise), except if you are watching the object in a mirror...it is certainly not good (plus the sift/surf descriptors are not done to be mirror invariants as far as i know, so you would probably don'thave good maches).

if(det<0) return false; // no mirrors in the scene

/* [Detecting Planar Homographies in an Image Pair Etienne Vincent and Robert Laganiere School of Information Technology and Engineering University of Ottawa Canada K1N 6N5] http://www.site.uottawa.ca/research */

My understanding concerning :

1. "Compute the determinant of the homography, and see if it's too close to zero for comfort." => compute the determinant of the top left 2x2 homography matrix, and check if it's "too close" to zero for comfort...btw you can also check if it's too far from zero because then the invert matrix would have a determinant too close to zero. A determinant of zero would mean the matrix is not inversible, too close to zero would mean singular (like you see the plane object at 90°, which is almost impossible if you use *good matches).

const double det = H.at<double>(0, 0) * H.at<double>(1, 1) - H.at<double>(1, 0) * H.at<double>(0, 1);

One way to do it would be to specify a positive value, higher than 1, say N, and then:

if((fabs(det)>(double)N) && || (fabs(det)<(1.0/(double)N)) return false; // bad homography

And while we are at it...if det<0 the homography is not conserving the orientation (clockwise<->anticlockwise), except if you are watching the object in a mirror...it is certainly not good (plus the sift/surf descriptors are not done to be mirror invariants as far as i know, so you would probably don'thave good maches).

if(det<0) return false; // no mirrors in the scene

/* /* for reading & reference : [Detecting Planar Homographies in an Image Pair Etienne Vincent and Robert Laganiere School of Information Technology and Engineering University of Ottawa Canada K1N 6N5] */

My understanding concerning :

1. "Compute the determinant of the homography, and see if it's too close to zero for comfort." => compute comfort."

Compute the determinant of the top left 2x2 homography matrix, and check if it's "too close" to zero for comfort...btw you can also check if it's too far *too *far from zero because then the invert matrix would have a determinant too close to zero. A determinant of zero would mean the matrix is not inversible, too close to zero would mean singular *singular (like you see the plane object at 90°, which is almost impossible if you use *good *good matches).

const double det = H.at<double>(0, 0) * H.at<double>(1, 1) - H.at<double>(1, 0) * H.at<double>(0, 1);

One way to do it would be to specify a positive value, higher than 1, say N, and then:

if((fabs(det)>(double)N) || (fabs(det)<(1.0/(double)N)) return false; // bad homography

And while we are at it...if det<0 the homography is not conserving the orientation (clockwise<->anticlockwise), except if you are watching the object in a mirror...it is certainly not good (plus the sift/surf descriptors are not done to be mirror invariants as far as i know, so you would probably don'thave good maches).

if(det<0) return false; // no mirrors in the scene

/* for reading & reference : [Detecting Planar Homographies in an Image Pair Etienne Vincent and Robert Laganiere School of Information Technology and Engineering University of Ottawa Canada K1N 6N5] */