The two operators for detecting and smoothing horizontal and vertical edges are shown below:
[-1 0 1]
[-2 0 2]
[-1 0 1]
and
[-1 -2 -1]
[ 0 0 0]
[ 1 2 1]
But after much Googling, I still have no idea where these operators come from. I would appreciate it if someone can show me how they are derived.
The formulation was proposed by Irwin Sobel a long time ago. I think about 1974. There is a great page on the subject here.
The main advantage of convolving the 9 pixels surrounding one at which gradients are to be detected is that this simple operator is really fast and can be done with shifts and adds in low-cost hardware.
They are not the greatest edge detectors in the world - Google Canny edge detectors for something better, but they are fast and suitable for a lot of simple applications.
So spatial filters, like the Sobel kernels, are applied by "sliding" the kernel over the image (this is called convolution). If we take this kernel:
[-1 0 1]
[-2 0 2]
[-1 0 1]
After applying the Sobel operator, each result pixel gets a:
high (positive) value if the pixels on the right side are bright and pixels on the left are dark
low (negative) value if the pixels on the right side are dark and pixels on the left are bright.
This is because in discrete 2D convolution, the result is the sum of each kernel value multiplied by the corresponding image pixel. Thus a vertical edge causes the value to have a large negative or positive value, depending on the direction of the edge gradient. We can then take the absolute value and scale to interval [0, 1], if we want to display the edges as white and don't care about the edge direction.
This works identically for the other kernel, except it finds horizontal edges.
Related
I have a 8-bit image and I want to filter it with a matrix for edge detection. My kernel matrix is
0 1 0
1 -4 1
0 1 0
For some indices it gives me a negative value. What am I supposed to with them?
Your kernel is a Laplace filter. Applying it to an image yields a finite difference approximation to the Laplacian operator. The Laplace operator is not an edge detector by itself.
But you can use it as a building block for an edge detector: you need to detect the zero crossings to find edges (this is the Marr-Hildreth edge detector). To find zero crossings, you need to have negative values.
You can also use the Laplace filtered image to sharpen your image. If you subtract it from the original image, the result will be an image with sharper edges and a much crisper feel. For this, negative values are important too.
For both these applications, clamping the result of the operation, as suggested in the other answer, is wrong. That clamping sets all negative values to 0. This means there are no more zero crossings to find, so you can't find edges, and for the sharpening it means that one side of each edge will not be sharpened.
So, the best thing to do with the result of the Laplace filter is preserve the values as they are. Use a signed 16-bit integer type to store your results (I actually prefer using floating-point types, it simplifies a lot of things).
On the other hand, if you want to display the result of the Laplace filter to a screen, you will have to do something sensical with the pixel values. Common in this case is to add 128 to each pixel. This shifts the zero to a mid-grey value, shows negative values as darker, and positive values as lighter. After adding 128, values above 255 and below 0 can be clipped. You can also further stretch the values if you want to avoid clipping, for example laplace / 2 + 128.
Out of range values are extremely common in JPEG. One handles them by clamping.
If X < 0 then X := 0 ;
If X > 255 then X := 255 ;
Gx = [-1 0 1
-2 0 2
-1 0 1]
Gy = [-1 -2 -1
0 0 0
1 2 1]
I knew these are the combination of smoothing filter and gradient but how are they combine to get this output ?
The Sobel Kernel is a convolution of the derivation kernel [-1 0 1] with a smoothing kernel [1 2 1]'. The former is straightforward, the later is rather arbitrary - you can see it as some sort of discrete implementation of a 1D Gaussian of a certain sigma if you want.
I think edge detection (ie gradient) influence is obvious - if there is a vertical edge. sobel operator Gx will definitely give big values relative to places where there is no edge because you just subtract two different values (intensity on the one side of an edge differs much from intensity on another side). The same thought on horizontal edges.
About smoothing, if you see e.g. mask for gaussian function for simga=1.0:
which actually does smoothing, you can catch an idea: we actaully set a pixel to a value associated to values of its neighbor. It means we 'average' values respectively to the pixel we are considering. In our case, Gx and Gy, it perforsm slightly smoothing in comparision to gaussian, but still idea remains the same.
I am new to image processing and had to do some edge detection. I understood that there are 2 types of detectors- Gaussian and Laplacian which look for maximas and zero crossings respectively. What I don't understand is how this is implemented by simply convolving the image with 2d kernels. I mean how does convolving equals finding maxima and zero crossing?
Laplacian zero crossing is a 2nd derivative operation, since the local maxima is equivalent with a zero crossing in a 2nd derivative. So it can be written as f_xx+f_yy. If we use a 1D vector to represent f_xx and f_yy, it is [-1 2 -1] (f(x+1,y)-2*f(x,y)+f(x-1,y)). Since the laplacian is f_xx + f_yy, it can be rephrased in a 2D kernal:
0 -1 0
-1 4 -1
0 -1 0
or if you consider the diagonal elements as well, it is:
-1 -1 -1
-1 8 -1
-1 -1 -1
On the other hand,Gaussian kernal as a low pass filter is used here for scaling. The scaling ratio is controlled by the sigma. This mainly enhances the edges with different widths. Basically the larger the sigma, the thicker edges are enhanced.
Combined Laplacian and Gaussian is mathematically equivalent with G_xx + G_yy where G is the Gaussian kernel. But usually people used Difference of Gaussian instead of Laplacian of Gaussian to reduce the computational cost.
For image derivative computation, Sobel operator looks this way:
[-1 0 1]
[-2 0 2]
[-1 0 1]
I don't quite understand 2 things about it,
1.Why the centre pixel is 0? Can't I just use an operator like below,
[-1 1]
[-1 1]
[-1 1]
2.Why the centre row is 2 times the other rows?
I googled my questions, didn't find any answer which can convince me. Please help me.
In computer vision, there's very often no perfect, universal way of doing something. Most often, we just try an operator, see its results and check whether they fit our needs. It's true for gradient computation too: Sobel operator is one of many ways of computing an image gradient, which has proved its usefulness in many usecases.
In fact, the simpler gradient operator we could think of is even simpler than the one you suggest above:
[-1 1]
Despite its simplicity, this operator has a first problem: when you use it, you compute the gradient between two positions and not at one position. If you apply it to 2 pixels (x,y) and (x+1,y), have you computed the gradient at position (x,y) or (x+1,y)? In fact, what you have computed is the gradient at position (x+0.5,y), and working with half pixels is not very handy. That's why we add a zero in the middle:
[-1 0 1]
Applying this one to pixels (x-1,y), (x,y) and (x+1,y) will clearly give you a gradient for the center pixel (x,y).
This one can also be seen as the convolution of two [-1 1] filters: [-1 1 0] that computes the gradient at position (x-0.5,y), at the left of the pixel, and [0 -1 1] that computes the gradient at the right of the pixel.
Now this filter still has another disadvantage: it's very sensitive to noise. That's why we decide not to apply it on a single row of pixels, but on 3 rows: this allows to get an average gradient on these 3 rows, that will soften possible noise:
[-1 0 1]
[-1 0 1]
[-1 0 1]
But this one tends to average things a little too much: when applied to one specific row, we lose much of what makes the detail of this specific row. To fix that, we want to give a little more weight to the center row, which will allow us to get rid of possible noise by taking into account what happens in the previous and next rows, but still keeping the specificity of that very row. That's what gives the Sobel filter:
[-1 0 1]
[-2 0 2]
[-1 0 1]
Tampering with the coefficients can lead to other gradient operators such as the Scharr operator, which gives just a little more weight to the center row:
[-3 0 3 ]
[-10 0 10]
[-3 0 3 ]
There are also mathematical reasons to this, such as the separability of these filters... but I prefer seeing it as an experimental discovery which proved to have interesting mathematical properties, as experiment is in my opinion at the heart of computer vision.
Only your imagination is the limit to create new ones, as long as it fits your needs...
EDIT The true reason that the Sobel operator looks that way can be be
found by reading an interesting article by Sobel himself. My
quick reading of this article indicates Sobel's idea was to get an
improved estimate of the gradient by averaging the horizontal,
vertical and diagonal central differences. Now when you break the
gradient into vertical and horizontal components, the diagonal central
differences are included in both, while the vertical and horizontal
central differences are only included in one. Two avoid double
counting the diagonals should therefore have half the weights of the
vertical and horizontal. The actual weights of 1 and 2 are just
convenient for fixed point arithmetic (and actually include a scale
factor of 16).
I agree with #mbrenon mostly, but there are a couple points too hard to make in a comment.
Firstly in computer vision, the "Most often, we just try an operator" approach just wastes time and gives poor results compared to what might have been achieved. (That said, I like to experiment too.)
It is true that a good reason to use [-1 0 1] is that it centres the derivative estimate at the pixel. But another good reason is that it is the central difference formula, and you can prove mathematically that it gives a lower error in its estmate of the true derivate than [-1 1].
[1 2 1] is used to filter noise as mbrenon, said. The reason these particular numbers work well is that they are an approximation of a Gaussian which is the only filter that does not introduce artifacts (although from Sobel's article, this seems to be coincidence). Now if you want to reduce noise and you are finding a horizontal derivative you want to filter in the vertical direction so as to least affect the derivate estimate. Convolving transpose([1 2 1]) with [-1 0 1] we get the Sobel operator. i.e.:
[1] [-1 0 1]
[2]*[-1 0 1] = [-2 0 2]
[1] [-1 0 1]
For 2D image you need a mask. Say this mask is:
[ a11 a12 a13;
a21 a22 a23;
a31 a32 a33 ]
Df_x (gradient along x) should be produced from Df_y (gradient along y) by a rotation of 90o, i.e. the mask should be:
[ a11 a12 a11;
a21 a22 a21;
a31 a32 a31 ]
Now if we want to subtract the signal in front of the middle pixel (thats what differentiation is in discrete - subtraction) we want to allocate same weights to both sides of subtraction, i.e. our mask becomes:
[ a11 a12 a11;
a21 a22 a21;
-a11 -a12 -a11 ]
Next, the sum of the weight should be zero, because when we have a smooth image (e.g. all 255s) we want to have a zero response, i.e. we get:
[ a11 a12 a11;
a21 -2a21 a21;
-a31 -a12 -a31 ]
In case of a smooth image we expect the differentiation along X-axis to produce zero, i.e.:
[ a11 a12 a11;
0 0 0;
-a31 -a12 -a31 ]
Finally if we normalize we get:
[ 1 A 1;
0 0 0;
-1 -A -1 ]
and you can set A to anything you want experimentally. A factor of 2 gives the original Sobel filter.
I want to smooth the contour of binarized images and think that erode is the best way to do it. I know that normal way of work is use cvDilate(src, dst, 0, iter); where 0 is a 3x3 matrix.
Problem is 3x3 matrix makes a deep erode in my images. How can I do a erode with a 2x2 matrix or anything smaller than the default 3x3 matrix.
Here you have for your reference the results of using different kernels:
Saludos!
If your goals is to have a binarized image with smooth edges then, if you have the original, it is better to use something like a gaussian blur with cvSmooth() on that before performing the binarization.
That said, you are not restricted to 3x3 kernels. cvDilate() takes an IplConvKernel produced by CreateStructuringElementEx and you can make a structuring element with any (rectangular) shape with that function.
However, a structuring element works relative to an anchor point that must have integer coordinates, so if you use a 2x2 matrix the matrix can not be centered around the pixel. so in most cases it is best to use structuring elements with an odd number of rows and collumns.
What you could do is create a 3x3 structuring element where only the center value and the values directly above, below, left and to the right of that are 1 like such:
0 1 0
1 1 1
0 1 0
rather than the default:
1 1 1
1 1 1
1 1 1
The first kernel will make for some slightly smoother edges.
Here's a quick and dirty approach to tell you whether dilation/erosion will work for you:
Upsample your image.
Erode (dilate, open, close, whatever) with the smallest filter you can use (typically 3x3)
Downsample back to the original image size
With the C API, you can create a dedicated IplConvKernel object of any kind and size with the function CreateStructuringElementEx(). If you to use the C++ API (function dilate()), the structuring element used for dilation is any matrix (Mat) you want.
A kernel with all 1's is a low pass convolution filter. A dilation filter replaces each pixel in the 3X3 region with the darkest pixel in that 3x3 region. An erosion filter replaces each pixel in the 3X3 region with the lightest pixel in that 3x3 region. That is if your background is light and your foreground object is dark. If you flip your background and foreground, then you would also flip your dilation and erosion filter.
Also if you want to perform an 'open' operation, you perform an erosion followed by a dilation. Conversely a 'close' operation is a dilation followed by an erosion. Open tends to remove isolated clumps of pixels and close tends to fill in holes.
Errosion and dilation matrices should be odd order
-- a 2*2 matrix cannot be used
convolution matrices should be of the order 1*1 3*3 5*5 7*7 ... but ODD
try to apply close - Erode then dilate the image operation - use the cvMorpologyEx() function