normalized Laplacian of Gaussian

Question

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?

Answer 1

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 $\sigma^2$.

Answer 2

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.)

Answer 3

I found a detailed explanation here. In short, we use normalization to keep the magnitude of the response $\sigma$-independent.