Given two spatial masks \(S_1=\begin{bmatrix}0&1&0\\1&-4&1\\0&1&0\end{bmatrix}\) and \(S_2=\begin{bmatrix}1&1&1\\1&-8&1\\1&1&1\end{bmatrix}\) The Laplacian of an image at all points (x, y) can be implemented by convolving the image with spatial mask. Which of the following can be used as the spatial mask ?
Both S1 and S2
The Laplacian is the isotropic second-derivative operator \(\nabla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}\). Replacing each second derivative by a finite difference gives a 3 × 3 mask.
Using \(\frac{\partial^2 f}{\partial x^2} \approx f(x+1,y) + f(x-1,y) - 2f(x,y)\) and the same in y, \(\nabla^2 f = f(x+1,y) + f(x-1,y) + f(x,y+1) + f(x,y-1) - 4f(x,y)\). These coefficients are exactly S1 (centre −4, four neighbours 1).
Adding the two diagonal directions as well gives centre −8 with all eight neighbours equal to 1, which is S2. It is isotropic for rotations in steps of 45°.
Both masks have coefficients summing to \(-4+4 = 0\) and \(-8+8 = 0\), so they give zero response in regions of constant intensity, as a second-derivative operator must.
Therefore either mask can be used to compute the Laplacian.
Hence, the answer is Both S1 and S2.
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