In Gonzalez' book Digital Image Processing, third edition, he writes on page 182 that

It can be shown that the simplest isotropic derivative operator is the laplacian operator

where the Laplacian is given by $\nabla^2f = {\frac {\partial^2 f }{\partial x^2}} + {\frac {\partial^2 f }{\partial y^2}}$, and the particular paper he links in reference is not available for free.

It is : Rosenfeld A, Kak A.C. Digital Picture Processing, vols. 1 and 2, 2nd ed., Academic Press, New York.

I was looking for insights to why:

a)The Laplacian is isotropic

b) What he means by "simplest". I am assuming this means in terms of computational complexity of things like how many differentiation operations, additions ops, multiplication ops, etc.

Any help appreciated.


2 Answers 2


Isotropic can be rephrased as "invariant" to the direction. Its value should not depends on the angle under which you see the image.

In other words: if you have a regular function $g(x,y)$ (a continuous 2D image) expressed in the standard $(X,Y)$ referential, or in a $\theta$-rotated referential $(X \cos \theta + Y \sin \theta,-X \sin \theta + Y \cos \theta)$, their Laplacians should be equal (at least close enough). The Laplace operator (in its continuous expression) is rotationally invariant (and more generally, invariant under orthogonal transformations). This is a classical result of rotational invariance for operators, and answered in SE.maths Show Laplace operator is rotationally invariant:

$$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial v^2}$$

where $$u = x \cos \theta + y \sin \theta$$ $$v = -x \sin \theta + y \cos \theta$$

Now, discrete images are sampled along a cartesian grid that is not really compatible with arbitrary rotations. There are many discrete approximation candidates. Let us focus on the $3\times 3$ kernels. The two most standard ways of linking pixels for 2D images is the 4-connectivity (or Von Neumann neighborhood: north, east, south, west) and the 8-connectivity (or Moore neighborhood, adding corners NE, NW, SE, SW). From these connectivities you get the two most classical discrete Laplacians:

$$\begin{bmatrix} 0 & -1 & 0\\-1 & 4 & -1\\0 & -1 & 0 \end{bmatrix}$$ and $$\begin{bmatrix} -1 & -1 & -1\\-1 & 8 & -1\\-1 & -1 & -1 \end{bmatrix}$$

The first one is invariant to $90°$ rotations, the second one is less sensitive to $45°$ angles. Yet, many people are tried to derive better versions, see form instance Constructing an "isotropic" Laplacian operator. Back to the works of A. Rosenfeld, in his paper Optimally isotropic Laplacian operator, 1999:

Laplacian operators used in the literature for digital image processing are not rotationally invariant. We examine the anisotropy of $3\times 3$ Laplacian operators for images quantized in square pixels, and find the operator which has the minimum overall anisotropy.

he finds an optimum: $$-\frac{1}{4}\begin{bmatrix} 1 & 2 & 1\\2 & -12 & 2\\1 & 2 & 1 \end{bmatrix}$$ "by minimizing the anisotropy over all angles and all edge distances". And there are many other works with larger kernels, under different penalties.

Additional information: Role of the rank of the filter mask matrix in image processing?

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    $\begingroup$ This is a fantastic answer and precisely what I was looking for. Thank you so much for your help. $\endgroup$ Commented May 12, 2020 at 20:03
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    $\begingroup$ Happy to be helpful here! $\endgroup$ Commented May 12, 2020 at 20:21

Isotropic operator is an operator that has no bias to any particular direction in the image. If you see the below Laplacian mask, Edge detection is taking place in every direction.

If you see the below image, three filter kernels are applied. The first kernel detects edges in x direction, the second in y direction and the third is Laplacian kernel detecting edges in all directions, hence it is isotropic. enter image description here

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    $\begingroup$ I guess I am looking for a mathematical reason why this is the case. This picture is instructive tho, thank you. $\endgroup$ Commented May 11, 2020 at 22:27
  • $\begingroup$ I suppose just looking at the operator itself, the second filter you have shown has the same value for each 8-neighbour (4 connected neighbours and 4 connected diagonals), where this makes it obvious why it is isotropic. Whereas the first filter you have written is less obvious to me. $\endgroup$ Commented May 11, 2020 at 23:43
  • $\begingroup$ Isotropy shows no bias in all directions. The first one is not isotropic. My bad. $\endgroup$
    – DSP Novice
    Commented May 12, 2020 at 2:31

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