Is it valid to use OpenCV fitEllipse for circle fitting.
fitEllipse() returns cv::RotatedRect how about averaging width and height to get fitted circle radius?
I think that the "validity" of using cv::fitEllipse for fitting circles depends on the precision you require for the fitting.
For example you can run your algorithm on a test set, fitting points with cv::fitEllipse and logging the length of the two axes of the ellipse, then have a look at the distributions of the ratio of two axes or at the difference between the major and the minor axis; you can find how much your supposed circles differ from a circle and then asses if you can use the cv::fitEllipse.
You can take the average of the width and the height of the cv::RotatedRect returned by cv::fitEllipse to get an approximation of the diameter of the circle (you wrote the radius but I think it was a trivial error).
You can have a look at this very readable article
UMBACH, Dale; JONES, Kerry N. A few methods for fitting circles to data. Instrumentation and Measurement, IEEE Transactions on, 2003, 52.6: 1881-1885. and write your own circle interpolator.
If you want to minimize the geometric error (the sum of the squares of the distances from the points to the circle, as explained in the Introduction of the article) you maybe need a reliable implementation of a non linear minimization algorithm.
Otherwise you can write a simple circle interpolator with the formulae from (II.8) to (II.15) (a closed-form solution wich minimize an error different from the geometric one) with some warning:
from an implementation point of view you have to take care of the usually warnings about roundoff error and truncation error.
the closed form solution cannot be robust enough in case of outlier points, in that case you may need to implement a robust interpolator like RANSAC (random choose three points, interpolate a circle with that three points with formulae from (25) to (34) from Weisstein, Eric W. "Circle." From MathWorld--A Wolfram Web Resource, compute the consensus and iterate). This warning applies also to the circle found with the minimization of the geometric error.
There is a function for circle fitting: minEnclosingCircle
Related
Suppose that the Canny edge detector successfully detects an edge in an image. The edge is then rotated by θ, where the relationship between a point on the original edge (x,y)(x,y) and a point on the rotated edge (x′,y′)(x′,y′) is defined as x′ = xcosθ; y′ = xsinθ;
Will the rotated edge be detected using the same Canny edge detector?
(I think we should find answer considering that the detection of an edge by the Canny edge detector depends only on the magnitude of its derivative.)
The answer is both yes and no, and which one you go for depends on how literally you take the question.
First of all, we're dealing with a rectangular grid, so given an integer location (x,y), the corresponding point (x',y') in a rotated image is highly likely not an integer location. And considering that the output of Canny is a set of points, and not a smooth function that can be interpolated, it would be difficult to establish a correspondence between the set resulting from the rotated and the one resulting from the original image.
Think for example about the number of pixels on a discrete line of a given length at 0 degrees and at 45 degrees. (Hint: the line at 45 degrees has sqrt(2) times fewer pixels.)
But if you take the question more generally and interpret it as "will an edge that is detected in the original image also be detected after rotating the image by θ degrees?" then the answer is yes, in theory.
Of course practice is always a bit different than theory. The details of the implementation matter here. And there is always numerical imprecision to contend with.
Let's start by assuming the rotation is computed correctly, with a precise interpolation scheme (cubic, Lanczos) and not rounded after to uint8 or something (i.e. we're computing using floating-point values).
If you read the original paper by Canny, you'll see he proposes using Gaussian derivatives as the best compromise between compact support and computational precision. I have seen few implementations that actually do. Typically I see a convolution with a Gaussian and then Sobel derivatives. Especially for smaller sigmas (less smoothing) the difference can be quite large. Gaussian derivatives are rotationally invariant, Sobel derivatives are not.
The next step in the algorithm is non-maximum suppression. This is where the continuous gradient is converted to a set of points. For each pixel, it checks to see if it is a local maximum in the direction of the gradient. Because this is done per pixel, a different set of locations are tested in the rotated image compared to the original. Nonetheless, it should detect points along the same ridges in both cases.
Next, a hysteresis threshold is applied. This is a two-threshold operation that keeps pixels above one threshold as long as at least one pixel above a second threshold is present in the same connected component. This is where the differences could occur between rotated and original image. Remember we're dealing with a set of pixels. We have samples the continuous gradient function at discrete points. There could be an edge that has one pixel above the second threshold in one version of the image, but not in the other. This would only occur for edges very close to the chosen threshold, of course.
Next comes a thinning. Because the non-maximum suppression can yield points along a thicker line, a thinning operation is applied that removes pixels from the set that are not needed to maintain connectivity of the lines. Which pixels are selected here will also differ between rotated and original images, but this does not change the geometry of the solution, so we still have the same set of points.
So, the answer is yes and no. :)
Note that the same logic applies to translation.
What is Distance Transform?What is the theory behind it?if I have 2 similar images but in different positions, how does distance transform help in overlapping them?The results that distance transform function produce are like divided in the middle-is it to find the center of one image so that the other is overlapped just half way?I have looked into the documentation of opencv but it's still not clear.
Look at the picture below (you may want to increase you monitor brightness to see it better). The pictures shows the distance from the red contour depicted with pixel intensities, so in the middle of the image where the distance is maximum the intensities are highest. This is a manifestation of the distance transform. Here is an immediate application - a green shape is a so-called active contour or snake that moves according to the gradient of distances from the contour (and also follows some other constraints) curls around the red outline. Thus one application of distance transform is shape processing.
Another application is text recognition - one of the powerful cues for text is a stable width of a stroke. The distance transform run on segmented text can confirm this. A corresponding method is called stroke width transform (SWT)
As for aligning two rotated shapes, I am not sure how you can use DT. You can find a center of a shape to rotate the shape but you can also rotate it about any point as well. The difference will be just in translation which is irrelevant if you run matchTemplate to match them in correct orientation.
Perhaps if you upload your images it will be more clear what to do. In general you can match them as a whole or by features (which is more robust to various deformations or perspective distortions) or even using outlines/silhouettes if they there are only a few features. Finally you can figure out the orientation of your object (if it has a dominant orientation) by running PCA or fitting an ellipse (as rotated rectangle).
cv::RotatedRect rect = cv::fitEllipse(points2D);
float angle_to_rotate = rect.angle;
The distance transform is an operation that works on a single binary image that fundamentally seeks to measure a value from every empty point (zero pixel) to the nearest boundary point (non-zero pixel).
An example is provided here and here.
The measurement can be based on various definitions, calculated discretely or precisely: e.g. Euclidean, Manhattan, or Chessboard. Indeed, the parameters in the OpenCV implementation allow some of these, and control their accuracy via the mask size.
The function can return the output measurement image (floating point) - as well as a labelled connected components image (a Voronoi diagram). There is an example of it in operation here.
I see from another question you have asked recently you are looking to register two images together. I don't think the distance transform is really what you are looking for here. If you are looking to align a set of points I would instead suggest you look at techniques like Procrustes, Iterative Closest Point, or Ransac.
I do have two sets of points and I want to find the best transformation between them.
In OpenCV, you have the following function:
Mat H = Calib3d.findHomography(src_points, dest_points);
that returns you a 3x3 Homography matrix, using RANSAC. My problem is now, that I only need translation and rotation (& maybe scale), I don't need affine and perspective.
The thing is, my points are only in 2D.
(1) Is there a function to compute something like a homography but with less degrees of freedom?
(2) If there is none, is it possible to extract a 3x3 matrix that does only translation and rotation from the 3x3 homography matrix?
Thanks in advance for any help!
Isa
OpenCV estimateRigidTransform function is exactly what you need: it returns Translation, Rotation and Scale (use false value for fullAffine flag). And it DOES use RANSAC (see source code to be sure of it).
Homography is for 2D points, the third dimension is just for casting points in 3 dim homogeneous coordinates and performing perspective effects. You can always cast points back:
homogeneous [x, y, w]
cartesian [x/w, y/w]
However since you calculate 6DOF instead of 4DOF (similarity) you result is pretty different from what you expect with 4DOF. More flexible transformation will fit more points in RANSAC at the expense of distortions in transformations you care about. Bottom line - don’t try to decompose H, instead fit similarity or isometry (also called rigid or euclidean). The reason why they are absent in the library - they are expressed in closed form even with correct least squared metric in point coordinates and thus don't require non-linear optimization. In other words, they are very simple.
If you only have rotation and translation, I wrote a quick functions to find them (no RANSAC though). It is probably similar to a rigidTransform but more understandable (hopefully)
https://stackoverflow.com/a/18091472/457687
With scale there is still a closed form solution, but slightly different formulas for translation and scaling. See Learning similarity parameters, p. 25
I'm validating an image segmentation algorithm applied to 2D images. The algorithm generates a contour segment, i.e. a set of connected pixels that form a freecurve in 2D space. The idea is to compare this set of pixels with a ground-truth, in my case another contour segment manually traced by an expert. An image showing what would be a segmentation result and the corresponding manual (ground-truth) segmentation is shown below:
I'm trying to think of an adequate comparison metric to validate the segmentation results. Ideally the best metric would be the point-to-point euclidean distance between corresponding pairs of pixels on each segment, however (as seen in previous figure) the segments don't have the same length (i.e. differ by the total number of pixels) so pixel-to-pixel comparisons have to be discarded.
Can you suggest me an adequate metric for validating my algorithm? Thanks for any suggestion!
For each pixel in the ground truth, take the distance to the nearest pixel in the segmentation result. Then take the sum of that for all ground truth pixels as the total error.
That's basically recall weighted by distance. If you start with the pixels in the result, it would resemble precision instead.
If the curves are closed, you can compute the area between the curves. If you can tell which pixels belong to a segment, that is as easy as computing XOR set of the 2 pixel sets.
Here is an example using that I've created using Matlab:
You could divide each line into n segments of equal length, then compute the euclidean distance between each segment and its pair on the other line.
I want algorithm for recognizing multiple no of shapes(Specially rectangle and squares) in a picture.Preferably I am using C# so, I am looking forward for solutions in C#.
check aforgenet....
http://www.aforgenet.com/forum/
If you are looking for a library that does a lot of image processing for you there is always OpenCV. I think it is is c++ though.
You can use the Circularity algorithm as a first approach, which is very easy to compute:
C = p2/a where p is the perimeter (border area) and a is shape area.
To know how to read/write pixels quickly, take a look here
Alternatively look for shape signature algorithm available at Rafael Gonzales book. In this algorithm you compute the center of the object using central momentum, the you compute the distance between the center and each border pixel. You'll end up with a 1D signal where peaks represent bigger distance from the center. In a square, you have 4 symmetric peaks while in a rectangle 2 big peaks and 2 smaller ones.