Box filter approximations for the oriented gabor filters

debugcn Published at Dev

AngelCastillo

I'm currently reading this paper: Approximations in the HMAX Model, I'm particularly interested in the approximation of the gabor filter using Box filters.

Last year I developed a synthetic fingerprint generator (you can see some of it here How to convolve an image with different gabor filters adjusted according to the local orientation and density using FFT?).

But the biggest flaw in the generator is the time it takes to apply the gabor filter, I have been researching how to implement a separable version of the filter but have failed horribly since mathematics is not my forte. Now this seem better for my needs than the separable filter, since it seems to be faster and the results seem good enough for what I need. But I don’t understand how to implement the box filtering for gabor from the equation in that paper.

It goes as follow:

In this section we attempt to perform a grosser approximation of the ﬁlter to achieve signiﬁcantly higher eﬃciency. Currently, the S1 layer in the model, consists of simple line detectors that are modeled using oriented Gabor ﬁlters. However, oriented lines can be represented using much simpler represented such as box ﬁlters as shown in Figure 5. These ﬁlters have very sparse derivatives allowing us to compute convolution quite eﬃciently. The convolution of the image with the ﬁlters B(x, y) can be eﬃciently computed using the ﬁlter’s derivatives as follows.

The second derivative of the box ﬁlter consists of just impulses and therefore needs to be evaluated at only four points for every position in the image. The integral image can also be eﬃciently computed using single pass algorithm as outlined in [5, 1]. The cost of convolution using this approximation is O(N2c), c=4. Unlike other approximations, the computational cost does not depend on the size of the ﬁlter at all!.

What are those terms? Is that the integral of the integral of the image? I’m sorry if this is like super basic stuff I’m asking, but I honestly have no clue :(.

Daniel Heilper

The rhs equation in your question is the infinitesimal description of convolution - you have the convolution operator " * " which is applied over a 2D subrange of the image - that 2d subrange is expressed as an 2D integral. The discrete description is similar, but with 2D subrange expressed with summation operator using the sigma notation "Σ". instead of 2D integral, it is applied as "ΣΣ" (I don't know how to put math notation here).

Naive Implementation of the convolution is using 2 loops - in each point in the image you sum the kernel values multiplied with the that point and neighboring points -

P(x+c/2) = P(x)*C(1)+P(x+1)*C(2)...+P(x+c)*C(c) (c is the size of the kernel). The process is similar to long multiplication or polynomial with polynomial multiplication.

Basically convolution can be applied in various ways, one way is straight forward (summation), or by transformation to other spaces (Fast Fourier Transform) - Convolution Theorem.

FFT time is O(NlogN). using FFT for convolution is efficient if the size of the kernel is big. Otherwise, if kernel size is small : O(c) < O(logN) - convolution is faster. Since filter size is 4 - go with straight forward convolution.

Also you can parallel the convolution (via cuda, open-mp, etc...)

Collected from the Internet

Please contact [email protected] to delete if infringement.

edited at2021-06-27

Comments

0 comments

From Dev

Related Related

Article