Laplacian of Gaussian formula for 2d case is $$\operatorname{LoG}(x,y) = \frac{1}{\pi\sigma^4}\left(\frac{x^2+y^2}{2\sigma^2} - 1\right)e^{-\frac{x^2+y^2}{2\sigma^2}},$$ in scale-space related processing of digital images, to make the Laplacian of Gaussian operator invariant to scales, it is always said to normalize $LoG$ by multiplying $\sigma^2$, that is $$\operatorname{LoG}_\text{normalized}(x,y) = \sigma^2\cdot \operatorname{LoG}(x,y) = \frac{1}{\pi\sigma^2}\left(\frac{x^2+y^2}{2\sigma^2} - 1\right)e^{-\frac{x^2+y^2}{2\sigma^2}}.$$ I wonder why multiply by $\sigma^2$ not $\sigma^4$ or anything else?


Laplacian response decays as $\sigma$ increases, and it is the second Gaussian derivative so it is multiplied by $\sigma^2$.

LOG is defined as $\bigtriangledown^2G$ so the scale normalized LOG would be $\sigma^2\bigtriangledown^2G$. You need to get rid of the scaling factor of the Gaussian which is $\sigma^2$.

  • $\begingroup$ Could you please give more details? $\endgroup$ – avocado Sep 11 '13 at 13:17

What @bruvel says is right, but there's a much simpler way to understand this. Imagine you have a single white dot convolved with a gaussian with size $\sigma$:

$f(x,y)=\frac{e^{-\frac{x^2+y^2}{2 \sigma }}}{2 \pi \sqrt{\sigma ^2}}$

Now, if you convolve this with a LoG with sigma $scale$, you get at x/y=0:

$\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }f(x,y) \log (x,y)dydx$

$= -\frac{1}{\pi (\sigma +\text{scale})^2}$

Obviously, for a dot convolved with a gaussian with size $\sigma$, we would like to have a scale space maximum at $scale = \sigma$. If you multiply the LoG output with $scale$, you get just that:

$\frac{\partial \frac{-\text{scale}}{\pi (\sigma +\text{scale})^2}}{\partial \text{scale}}=0\Longrightarrow \text{scale}=\sigma$

So if you want to detect "dot-like" features at the right location in scale space (e.g. for keypoint detection), multiplying the LoG result with $\sigma$ gives the right results. (But this is not necessarily true for line or step-edge features! For those, multiplying with $\sigma ^ k$ for some $k>0$ might give more accurate results.)

  • $\begingroup$ Thank you, by now I just can say that I understand your answer to some extent, not fully. Why the convolution of $f(x,y)$ with $LoG$ gives $-\frac{1}{\pi(\sigma + scale)^2}$, how is the computation of double integral done? $\endgroup$ – avocado Sep 11 '13 at 13:12
  • $\begingroup$ I read a paper by Lindeberg, in which he introduced a concept $\gamma$-normalized derivative, and based on this concept, he discussed scale-selection. Is this normalized derivative related to my problem? $\endgroup$ – avocado Sep 11 '13 at 14:12
  • $\begingroup$ Is this all about dimensionless differential operator? Like this en.wikipedia.org/wiki/Nondimensionalization#Substitutions $\endgroup$ – avocado Sep 11 '13 at 14:22
  • $\begingroup$ Convolve with white "dot" has only sense in discrete case. About what convolution we are talking about discrete or continious? Where is a derivation of second formula? $\endgroup$ – bruziuz May 14 '16 at 0:26
  • $\begingroup$ @bruziuz: It's been some time since I wrote this, I guess I meant discrete convolution. But if we called the "dot" a "Dirac delta peak" in a continuous function, wouldn't the math exactly the same? $\endgroup$ – Niki Estner May 14 '16 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.