Recover image from its X and Y gradients - image-processing

I have a single channel image from which I can compute its vertical and horizontal gradients. I would like to make some operations in the gradient domain and subsequently recover back the scalar field (image) which results after the gradient modification. Any idea how to do this? I know if I integrate the modified gradient I can get back the function up to a constant but I would have two different constants C_x and C_y from the partial X and Y derivatives. Also, I don't have an intuition of how to "integrate" a vector field as the gradient.
Thanks!

Related

Hessian matrix, how to combine Ixx & Iyy together?

"Before extracting the lines, you need to detect potential points on them. Apply a Gaussian filter first and use the Sobel filters as derivative operators. Threshold the determinant of the Hessian and then apply non-maximum suppression in 3 × 3 neighborhoods. Ignore pixels for which any of the filters falls even partially out of the image boundaries."
I understand to gaussian an image first to eliminate noise, then take twice with Sobel_x and Sobel_y, respectively, which became Ixx and Iyy in Hessian that would show horizontal line and vertical line in image.But how am I suppose to get Ixxyy? But how could I combine these two image together to make Ixxyy as the right bottom in Hessian matrix?
The two off-diagonal elements of the Hessian matrix are d^2/dxdy. That is, they are the first derivative along y applied to the first derivative along x.
If the top-left element is obtained by Sobel_x( Sobel_x( image )), and the bottom-right element is Sobel_y( Sobel_y( image )), then the two other elements are both Sobel_y( Sobel_x( image )) or, equivalently, Sobel_x( Sobel_y( image )) (note that these two should be identical).
Do take into account that negative values are important here, and you should thus be careful to compute the Sobel filter in a way that preserves those negative values—don't store them in an unsigned integer array!

Finding edges in a height map

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.

How to apply box filter on integral image? (SURF)

Assuming that I have a grayscale (8-bit) image and assume that I have an integral image created from that same image.
Image resolution is 720x576. According to SURF algorithm, each octave is composed of 4 box filters, which are defined by the number of pixels on their side. The
first octave uses filters with 9x9, 15x15, 21x21 and 27x27 pixels. The
second octave uses filters with 15x15, 27x27, 39x39 and 51x51 pixels.The third octave uses filters with 27x27, 51x51, 75x75 and 99x99 pixels. If the image is sufficiently large and I guess 720x576 is big enough (right??!!), a fourth octave is added, 51x51, 99x99, 147x147 and 195x195. These
octaves partially overlap one another to improve the quality of the interpolated results.
// so, we have:
//
// 9x9 15x15 21x21 27x27
// 15x15 27x27 39x39 51x51
// 27x27 51x51 75x75 99x99
// 51x51 99x99 147x147 195x195
The questions are:What are the values in each of these filters? Should I hardcode these values, or should I calculate them? How exactly (numerically) to apply filters to the integral image?
Also, for calculating the Hessian determinant I found two approximations:
det(HessianApprox) = DxxDyy − (0.9Dxy)^2 anddet(HessianApprox) = DxxDyy − (0.81Dxy)^2Which one is correct?
(Dxx, Dyy, and Dxy are Gaussian second order derivatives).
I had to go back to the original paper to find the precise answers to your questions.
Some background first
SURF leverages a common Image Analysis approach for regions-of-interest detection that is called blob detection.
The typical approach for blob detection is a difference of Gaussians.
There are several reasons for this, the first one being to mimic what happens in the visual cortex of the human brains.
The drawback to difference of Gaussians (DoG) is the computation time that is too expensive to be applied to large image areas.
In order to bypass this issue, SURF takes a simple approach. A DoG is simply the computation of two Gaussian averages (or equivalently, apply a Gaussian blur) followed by taking their difference.
A quick-and-dirty approximation (not so dirty for small regions) is to approximate the Gaussian blur by a box blur.
A box blur is the average value of all the images values in a given rectangle. It can be computed efficiently via integral images.
Using integral images
Inside an integral image, each pixel value is the sum of all the pixels that were above it and on its left in the original image.
The top-left pixel value in the integral image is thus 0, and the bottom-rightmost pixel of the integral image has thus the sum of all the original pixels for value.
Then, you just need to remark that the box blur is equal to the sum of all the pixels inside a given rectangle (not originating in the top-lefmost pixel of the image) and apply the following simple geometric reasoning.
If you have a rectangle with corners ABCD (top left, top right, bottom left, bottom right), then the value of the box filter is given by:
boxFilter(ABCD) = A + D - B - C,
where A, B, C, D is a shortcut for IntegralImagePixelAt(A) (B, C, D respectively).
Integral images in SURF
SURF is not using box blurs of sizes 9x9, etc. directly.
What it uses instead is several orders of Gaussian derivatives, or Haar-like features.
Let's take an example. Suppose you are to compute the 9x9 filters output. This corresponds to a given sigma, hence a fixed scale/octave.
The sigma being fixed, you center your 9x9 window on the pixel of interest. Then, you compute the output of the 2nd order Gaussian derivative in each direction (horizontal, vertical, diagonal). The Fig. 1 in the paper gives you an illustration of the vertical and diagonal filters.
The Hessian determinant
There is a factor to take into account the scale differences. Let's believe the paper that the determinant is equal to:
Det = DxxDyy - (0.9 * Dxy)^2.
Finally, the determinant is given by: Det = DxxDyy - 0.81*Dxy^2.
Look at page 17 of this document
http://www.sci.utah.edu/~fletcher/CS7960/slides/Scott.pdf
If you made a code for normal Gaussian 2D convolution, just use the box filter as a Gaussian kernel and the input image will be the same original image not integral image. The results from this method will be same with the one you asked.

Deforming an image so that curved lines become straight lines

I have an image with free-form curved lines (actually lists of small line-segments) overlayed onto it, and I want to generate some kind of image-warp that will deform the image in such a way that these curves are deformed into horizontal straight lines.
I already have the coordinates of all the line-segment points stored separately so they don't have to be extracted from the image. What I'm looking for is an appropriate method of warping the image such that these lines are warped into straight ones.
thanks
You can use methods similar to those developed here:
http://www-ui.is.s.u-tokyo.ac.jp/~takeo/research/rigid/
What you do, is you define an MxN grid of control points which covers your source image.
You then need to determine how to modify each of your control points so that the final image will minimize some energy function (minimum curvature or something of this sort).
The final image is a linear warp determined by your control points (think of it as a 2D mesh whose texture is your source image and whose vertices' positions you're about to modify).
As long as your energy function can be expressed using linear equations, you can globally solve your problem (figuring out where to send each control point) using linear equations solver.
You express each of your source points (those which lie on your curved lines) using bi-linear interpolation weights of their surrounding grid points, then you express your restriction on the target by writing equations for these points.
After solving these linear equations you end up with destination grid points, then you just render your 2D mesh with the new vertices' positions.
You need to start out with a mapping formula that given an output coordinate will provide the corresponding coordinate from the input image. Depending on the distortion you're trying to correct for, this can get exceedingly complex; your question doesn't specify the problem in enough detail. For example, are the curves at the top of the image the same as the curves on the bottom and the same as those in the middle? Do horizontal distances compress based on the angle of the line? Let's assume the simplest case where the horizontal coordinate doesn't need any correction at all, and the vertical simply needs a constant correction based on the horizontal. Here x,y are the coordinates on the input image, x',y' are the coordinates on the output image, and f() is the difference between the drawn line segment and your ideal straight line.
x = x'
y = y' + f(x')
Now you simply go through all the pixels of your output image, calculate the corresponding point in the input image, and copy the pixel. The wrinkle here is that your formula is likely to give you points that lie between input pixels, such as y=4.37. In that case you'll need to interpolate to get an intermediate value from the input; there are many interpolation methods for images and I won't try to get into that here. The simplest would be "nearest neighbor", where you simply round the coordinate to the nearest integer.

wind filter in opencv

Could someone suggest me how to go about getting the wind filter effect in opencv similar to the one available in photoshop and gimp?
Here is an image of text with wind styled filter applied on it.
Thanks
I suggest the following steps:
Use the original text image as a mask. White pixels are '1', blacks are '0'.
Smooth the image in X direction (like in the example image you added)
You can do the smoothing by
horizontal vector filter
or use distance transform where
distance is calculated only along x
axis.
I think that distance transform will
run faster
Multiply the result by (1-mask) so smoothing will occur only outside the text.
Multiply each row of the result by random number in range [0.1 ..1]. This will make smoothing uneven.
Add to the result the original image of the text to get the final image

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