Written by Aryan Sidhwani on 10/13/2024

Sobel Feldman Operator

Robert Cross

An edge detector should have the following properties: the produced edges should be well-defined, the background should contribute as little noise as possible, and the intensity of edges should correspond as close as possible to what a human would perceive

n order to perform edge detection with the Roberts operator we first convolve the original image, with the following two kernels:

$\begin{bmatrix} +1 & 0 \\ 0 & -1 \end{bmatrix}$ $\begin{bmatrix} 0 & +1 \\ -1 & 0 \end{bmatrix}$

Let $I (x, y)$ be a point in the original image and $G_{x} (x, y)$ be a point in an image formed by convolving with the first kernel and $G_{y} (x, y)$ be a point in an image formed by convolving with the second kernel. The gradient can then be defined as:

$\nabla I(x,y)=G(x,y) = \begin{bmatrix}G_{x}\\G_{y}\end{bmatrix}$
$\lVert \nabla I(x,y) \rVert = \sqrt {G_{x}^{2}+G_{y}^{2}}$

The direction of the gradient can also be defined as follows:

$\Theta (x,y)=\arctan {\left ({\frac {G_{y}(x,y)}{G_{x}(x,y)}} \right)} - \frac{3 \pi}{4}$

Motivation for sobel

The motivation to develop it was to get an efficiently
computable gradient estimate which would be more isotropic than
the then popular Roberts Cross operator

The vector summation $\sqrt {G_{x}^{2}+G_{y}^{2}}$ operation provides
an averaging over directions-of-measurement of the gradient.

Simple Central Gradient estimate is the vector sum of 4 gradients possible for simple central gradient (corresponding to its '4-neighbours') ->

Horizontal Gradient: $G_{x} = I (1, 0) - I (1, 2) $
Vertical Gradient: $G_{y} = I (0, 1) - I (2, 1) $
Diagonal Gradient 1: $G_{d1}=I(0,0)−I(2,2)$
Diagonal Gradient 2: $G d 2 = I (0, 2) - I (2, 0) $

Let the neighbourhood be -

$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$

So,

G(x,y) = \frac{c - g}{4} \cdot \begin{bmatrix} 1 \\ 1 \end{bmatrix} + \frac{a - i}{4} \cdot \begin{bmatrix} -1 \\ 1 \end{bmatrix} + \frac{b - h}{4} \cdot \begin{bmatrix} 0 \\ 1 \end{bmatrix} + \frac{f - d}{4} \cdot \begin{bmatrix} 1 \\ 0 \end{bmatrix}

where the coefficients are the directional derivative estimates for the each of the four unit vectors

Evaluating this further,

$4 \cdot G = \begin{bmatrix} c - g - a + i + 2(f-d) \\ c - g + a - i + 2 (b - h) \end{bmatrix}$

In the x-component, this addition of components can be represented by a dot product -

$\begin{bmatrix} c \\ g \\ a \\ i \\ f \\ d \end{bmatrix} \cdot \begin{bmatrix} 1 \\ -1 \\ -1 \\ 1 \\ 2 \\ -2 \end{bmatrix}$

we can add 3 more zeros to account for the weights for the image positions not in the sum ->

$\implies \lVert G_x * A(\text{neighbourhood of 'e'}) \rVert = \begin{bmatrix} c \\ g \\ a \\ i \\ f \\ d \\ b \\ e \\ h \end{bmatrix} \cdot \begin{bmatrix} 1 \\ -1 \\ -1 \\ 1 \\ 2 \\ -2 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

converting these weights to a matrix form so that the weights are put in place of the original neighbourhood matrix ->

$G_x = \begin{bmatrix} -1 & 0 & 1 \\ -2 & 0 & 2 \\ -1 & 0 & 1 \end{bmatrix}$

Doing the exact same analysis for the y-component, we have ->

$\lVert G_y * A(\text{neighbourhood of 'e'}) \rVert = \begin{bmatrix} c \\ g \\ a \\ i \\ b \\ h \\ f \\ e \\ d \end{bmatrix} \cdot \begin{bmatrix} 1 \\ -1 \\ 1 \\ -1 \\ 2 \\ -2 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

Therefore,

$G_y = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ -1 & -2 & -1 \end{bmatrix}$

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