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The isotropic TV is defined as the estimation of 2-norm of gradients $\sqrt{(y_{i+1,j}-y_{i,j})^2+(y_{i,j+1}-y_{i,j})^2}$, while the anisotropic TV is defined as the estimation of 1-norm of gradients $|y_{i+1,j}-y_{i,j}|+|y_{i,j+1}-y_{i,j}|$.

Now I am wondering why the second one will be called anisotropic. Since from my perspective, the isotropic TV bounded the 2-norm of the gradient, and thus it can bound the norm of gradients for every directions. But for the second one, since the absolute values are bounded, it means that the 2-norm of every directions can also be bounded. Isn't it also an isotropic way to bound the gradient?

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  • $\begingroup$ One of my guess is that when we bound the 1-norm, we actually don't know how much we bound each terms. For example, $a+b<3$ can be $a<2$ and $b<1$. And thus when we take directional derivatives, it will be $\sqrt{a^2\cos^2(\theta)+b^2\sin^2(\theta)}$. And thus it's anisotropic. Is my intuition correct? $\endgroup$ – 07216 Apr 27 at 23:05
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In the Total Variation framework we define 2 flavors:

$$ \text{Isotropic TV} \; {TV}_{ {L}_{2} } \left( X \right) = \sum_{ij} \sqrt{ { \left( {D}_{h} X \right) }_{ij}^{2} + { \left( {D}_{v} X \right) }_{ij}^{2} } $$

$$ \text{Anisotropic TV} \; {TV}_{ {L}_{1} } \left( X \right) = \sum_{ij} \sqrt{ { \left( {D}_{h} X \right) }_{ij}^{2} } + \sqrt{{ \left( {D}_{v} X \right) }_{ij}^{2} } $$

Where $ {D}_{h} $ and $ {D}_{v} $ are the horizontal / vertical finite differences operators.

Quoting Isotropic and Anisotropic Total Variation Regularization in Electrical Impedance Tomography:

It is worth noticing that referring to functional as ”anisotropic” is slightly misleading, because the weights corresponding to all coordinate directions are equal.

The term Anisotropic is used because this flavor separate Vertical and Horizontal components as done in Anisotropic Diffusion. Hence it doesn't have rotational invariant term as the Isotropic Flavor. As you know, sum of quadratic terms is connected to Ellipse and with equal weights it is a Sphere / Circle which is Isotropic.

Great reading about the subject is Laurent Condat - Discrete Total Variation - New Definition and Minimization (DOI Address). It analyze the properties of different TV definitions regarding their spatial smoothing.

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  • $\begingroup$ Could the one who -1 this answer say why? $\endgroup$ – Royi May 21 at 19:04

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