Laplacian kernels of higher order in image processing

杀马特。学长 韩版系。学妹 提交于 2020-01-30 07:54:39

问题


In literature on digital image processing you find examples of Laplace kernels of relatively low orders, typically, 3 or 5. I wonder, is there any general way to build Laplace kernels or arbitrary order? Links or/and references would be appreciated.


回答1:


The Laplace operator is defined as the sum of the second derivatives along each of the axes of the image. (That is, it is the trace of the Hessian matrix):

  • I = ( ∂2/∂x2 + ∂2/∂y2 ) I

There are two common ways to discretize this:

  1. Use finite differences. The derivative operator is the convolution by [1,-1] or [0.5,0,-0.5], the second derivative operator applying the [1,-1] convolution twice, leading to a convolution with [1,-2,1].

  2. Convolve with the derivative of a regularization kernel. The optimal regularization kernel is the Gaussian, leading to a Laplace of Gaussian operator. The result is the exact Laplace of the image smoothed by the Gaussian kernel.


An alternative is to replace the regularization kernel with an interpolating kernel. A former colleague of mine published a paper on this method:

A. Hast, "Simple filter design for first and second order derivatives by a double filtering approach", Pattern Recognition Letters 42(1):65-71, 2014.

He used a "double filter", but with linear filters that can always be simplified to a single convolution.

The idea is simply that, take an interpolating kernel, and compute its derivative at integer locations. The interpolating kernel is always 1 at the origin, and 0 at other integer locations, but it waves through these "knot points", meaning that its derivative is not zero at these integer locations.

In the extreme case, take the ideal interpolator, the sinc function:

  • sinc(x) = sin(πx) / πx

Its second derivative is:

  • d2/dx2(sinc(πx)) = [ (2 - π2x2) sin(πx) - 2πx cos(πx) ] / (πx3)

Which sampled at 11 integer locations leads to:

[ 0.08 -0.125 0.222 -0.5 2 -3 2 -0.5 0.222 -0.125 0.08 ]

But note that the normalization is not correct here, as we're cutting off the infinitely long kernel. Thus, it's better to pick a shorter kernel, such as the cubic spline kernel.


A second alternative is to compute the Laplace operator through the Fourier domain. This simply requires multiplying with -πu2v2, with u and v the frequencies.

This is some MATLAB code that applies this filter to a unit impulse image, leading to an image of the kernel of size 256x256:

[u,v] = meshgrid((-128:127)/256,(-128:127)/256);
Dxx = -4*(pi*u).^2;
Dyy = -4*(pi*v).^2;
L = Dxx + Dyy;
l = fftshift(ifft2(ifftshift(L)));
l = real(l);        % discard insignificant imaginary component (probably not necessary in MATLAB, but Octave leaves these values there)
l(abs(l)<1e-6) = 0; % set near-zero values to zero

l here is the same as the result above for the ideal interpolator, adding the vertical and horizontal ones together, and normalizing for a length of 256.


Finally, I'd like to mention that the Laplace operator is very sensitive to noise (high frequencies are enhanced significantly). The methods discussed here are meaningful only for data without nose (presumably synthetic data?). For any real-world data, I highly recommend that you use the Laplace of Gaussian. This will give you the exact Laplace of the smoothed image. The smoothing is necessary to prevent influence from noise. With little noise, you can use a small Gaussian sigma (e.g. σ=0.8). This will give you much more useful results than any other approach.



来源:https://stackoverflow.com/questions/50793646/laplacian-kernels-of-higher-order-in-image-processing

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