Is there anybody can help me interpret
"Edge points may be located by the maxima of the
module of the gradient, and the direction of edge contour is orthogonal to the direction of the gradient."
Paul R has given you an answer, so I'll just add some images to help make the point.
In image processing, when we refer to a "gradient" we usually mean the change in brightness over a series of pixels. You can create gradient images using software such as GIMP or Photoshop.
Here's an example of a linear gradient from black (left) to white (right):
The gradient is "linear" meaning that the change in intensity is directly proportional to the distance between pixels. This particular gradient is smooth, and we wouldn't say there is an "edge" in this image.
If we plot the brightness of the gradient vs. X-position (left to right), we get a plot that looks like this:
Here's an example of an object on a background. The edges are a bit fuzzy, but this is common in images of real objects. The pixel brightness does not change from black to white from one pixel to the next: there is a gradient that includes shades of gray. This is not obvious since you typically have to zoom into a photo to see the fuzzy edge.
In image processing we can find those edges by looking at sharp transitions (sharp gradients) from one brightness to another. If we zoom into the upper left corner of that box, we can see that there is a transition from white to black over just a few pixels. This transition is a gradient, too. The difference is that the gradient is located between two regions of constant color: white on the left, black on the right.
The red arrow shows the direction of the gradient from background to foreground: pixels are light on the left, and as we move in the +x direction the pixels become darker. If we plot the brightness sampled along the arrow, we'll get something like the following plot, with red squares representing the brightness for a specific pixel. The change isn't linear, but instead will look like one side of a bell curve:
The blue line segment is a rough approximation of the slope of the curve at its steepest. The "true" edge point is the point at which slope is steepest along the gradient corresponding to the edge of an object.
Gradient magnitude and direction can be calculated using horizontal and vertical Sobel filters. You can then calculate the direction of the gradient as:
gradientAngle = arctan(gradientY / gradientX)
The gradient will be steepest when it is perpendicular to the edge of the object.
If you look at some black and white images of real scenes, you can zoom in, look at individual pixel values, and develop a good sense of how these principles apply.
Object edges typically result in a step change in intensity. So if you take the derivative of intensity it will have a large (positive or negative) value at edges and a smaller value elsewhere. If you can identify the direction of steepest gradient then this will be at right angles to (orthogonal to) the object edge.
Related
I have images such as below (I am pasting only a sketch here) where I want to calculate the center of symmetry and the displacement between the 2 marked zones in the image(Marked in Red and blue). Could anyone suggest a simple algorithm for this? (Please note the signal is symmetric to 180-degree rotation).
The idea is to calculate the center of symmetry between the red and blue zones
Here is my algorithm approach:
Distinguish the red and blue pixels in the image by thresholding: Lets say set the blue pixels to wihte(255) and set the red pixels(0) and set the rest of image to the (125) in a gray channel image.
Check all of the pixels horizontally in the middle row of the image.
While you are checking, if you hit the red pixel, then try to search blue one. However, if you hit red pixel again, then start searching blue again.
During the search after red pixel if you hit a blue pixel then you can easiy calculate the middle of it. Then re-start the process.
Here is a demonstration of search line:
Note: Since they are dashed lines, luckily you may pass them with gaps. To fix, this you can try couple of random different rows.
I want to find sharp edges in a heightmap image, while ignoring shallow edges.
OpenCV offers multiple approaches to finding edges in a 2d Image: Canny, Sobel, etc.
However, all these approaches work by comparing the intensity values on both sides of the edge.
If the 2D Image represents a height map of a 3D object, then this results in some weird behaviour.
In a height map, the height of a 3D object at a given X/Y coordinate is represented as the intensity of the 2D Pixel at that X/Y coordinate:
In the above picture, at the edge B the intensity changes only slightly between the left and right side, even though it is a sharp corner.
At the edge A, there is a bigchange in the intensity between pixels on the left side of the edge and the right, even though it is only a shallow angle.
So there is no threshold for Canny or Sobel that will preserve the sharp edge but filter the shallow edge.
(In the above example, the edge B has one side with an ascending slope, and one side with a descending slope. I could filter for this feature; but that would remove the edges C and D as well)
How can I get a binary edge image, containing only edges above a certain angle? (e.g. edge B, C, and D, but not A)
Or alternatively, how can I get a gradient derivative image, where the intensity of each pixel is proportional to the angle of the edge at that pixel?
Probably you'll want to use second derivative instead of first for this task.
Here's my intuition: taking derivative of height (intensity in your case) at each position on an evenly spaced grid would be proportional to arctan of the surface slope between sampling points (or at sampling points if you use a 2-sided derivative approximation). But since you want to detect sharp edges - you are looking for a derivative of slope at the sampling points. This means that you can set a threshold on a derivative of arctan of derivative of intensity to achieve your goal (luckily there's no "need to go deeper" :) )
You will have to be extra careful with taking a derivative of "slope angles" that you'll get - depending on the coordinate system you may come across ambiguity of angle difference (there are 2 ways to get from one angle to another, which are different in general case; you're looking for the "shorter" one). You can look for possible solution here
I have a rather simple approach that I came across wile reading a blog post.
It involves computing the median value of the gray scale image. Using this value we can now set two threshold values:
lower: max(0, (1.0 - 0.33) * v)
upper: min(255, (1.0 + 0.33) * v)
Now pass these two values as parameters into the cv2.Canny() function.
You will now be able to perform an optimized edge detection given any image. The crux of this answer depends on the median value of the image which varies for different images.
If i understand your question correctly, "what you need is basically a corner with high intensity values".
If that is so then look for Harris corner detector which would help you to find points with high gradient change in both direction.
http://docs.opencv.org/2.4/doc/tutorials/features2d/trackingmotion/harris_detector/harris_detector.html
Once you detect the corners you can filter the corners which have high intensity by using a suitable threshold.
I know, that the Meanshift-Algorithm calculates the mean of a pixel density and checks, if the center of the roi is equal with this point. If not, it moves the new ROI center to the mean center and checks again... . Like in this picture:
For density, it is clear how to find the mean point. But it can't simply calculate the mean of a histogram and get the new position by this point. How can this algorithm work using color histogram?
The feature space in your image is 2D.
Say you have an intensity image (so it's 1D) then you would just have a line (e.g. from 0 to 255) on which the points are located. The circles shown above would just be line segments on that [0,255] line. Depending on their means, these line segments would then shift, just like the circles do in 2D.
You talked about color histograms, so I assume you are talking about RGB.
In that case your feature space is 3D, so you have a sphere instead of a line segment or circle. Your axes are R,G,B and pixels from your image are points in that 3D feature space. You then still look where the mean of a sphere is, to then shift the center towards that mean.
I am trying to find circles in images and warp them back to a canonical view (i.e. as if looking into the center). However, circles in general project to ellipses under perspective transformations. So I am first detecting ellipses, roughly doing the following (in OpenCV):
1. Find contours in the image
2. Estimate area of contour
3. Fitting a bounded box to contour and estimating area by width/2 * height/2 * PI (area of ellipse)
4. checking if area of contour and estimated area of ellipse is < a threhsold
Assuming I have found an ellipse by this method, how can I rectify it back to a circle such that I "undo" the perspective transform (although not in plane rotation as this cannot be done I guess). For example, if it was a rectangle I would just compute the homography from the 4 corners of an uprigh rectangle to the detected projected one.
I have no idea how to do this with an ellipse, any help is much appreciated.
Thanks
A circle is indeed transformed into an ellipse by a perspective transformation, however its axes are not the same as the axes of the initial circle, as shown in this illustration:
(source: brian-curtis.com)
You can refer to this link for a detailled demonstration. As a consequence, the bounding rectangle of the ellipse is not the image of the initial square by the perspective tranformation.
EDIT:
This means that the center and the axes of the ellipse you observe are not the images, by the perspective mapping, of the center and axes of the original circle. I tried to make a clearer illustration:
On this image, I drew in green the axes and center of the original circle, after perspective transformation, and the axes and center of the ellipse in red. On this specific example, the vertical axis is not deformed by the perspective mapping, but it would be deformed in the general case. Hence, deforming a circle by a perspective transformation gives an ellipse, but the axes and center that you see are not the axes and center of the original circle.
As a consequence, you cannot simply use the top, bottom, left and right points on the ellipse (the red points, which can easily be detected from the ellipse) to map these onto the top, bottom, left and right points of the circle because they do not correspond under the perspective mapping (the green points do, but they cannot be detected easily from the ellipse).
In the end, I don't think that it is at all possible to estimate the perspective mapping from a single detected ellipse.
This looks like an indeterminate problem.
The projection of a rectangle supplies 8 equations in 8 unknowns (homography coefficients).
With an ellipse, you can only retrieve the center coordinates (2 DOF), the axis (2 DOF) and the axis orientation (1 DOF).
I'm going to find the most look-like rectangles among shapes. The first image is the original image with shapes which possibly be rectangles but they are not. The green rectangles in the second image is what I want. So is there a way to do this with opencv? I've tried hough lines but the result's not good
The source image:
And what I want is to find out the most look-like rectangle among these shapes, like the rectangles in green.
What I want:
A very simple approach is, after you have a rectangle bounding box around your shape, count the percentage of pixels inside the box which are white.
The higher the percentage of white pixels, the closest to a rectangle it is.
To get the bounding boxes you should take a look at either findContours from opencv, or some Blob extracting algorithm, you will find plenty of questions regarding those.
Edit:
Maybe you should first get the Minimum bounding rectangles of the shapes and then do this kind of heuristic:
Shrink the rectangle dimensions until the white-pixel percentage inside the rectangle reaches some threshold defined by you (like 90% of white pixels inside the rectangle).
To get the Minimum bounding rectangle (the smallest rectangle which contains the whole shape), you might check this tutorial:
http://docs.opencv.org/doc/tutorials/imgproc/shapedescriptors/bounding_rects_circles/bounding_rects_circles.html
One thing that might also help is doing the difference of sizes from the minimum bounding rectangle and the maximum inner rectangle (the biggest rectangle you can fit inside the white shape). The less difference there is between those rectangle's properties (width, height, area, center coordinates) the closest is the shape to a rectangle.