Talk:Gaussian blur

message: I spent some time looking through the article to find what the term "Gaussian" means, and didn't find any answers. After some further googling I found that it is named after a man named Gauss(http://www.webmonkey.com/webmonkey/glossary/gaussian.html). I am surprised the introduction doesn't mention this, and assumes the reader knows (or doesn't care) what Gaussian actually means -- I looked up this article to discover that, and the entire Wiki article didn't help. 
Additionally, does Gaussian always necessarily mean it relates in some way to Karl Friedrich Gauss? Perhaps that clarification is beyond the scope of this article, but surely the fact that Gaussion in this case refers to Karl Friedrich Gauss is not?  —Preceding unsigned comment added by Abeall (talk • contribs) 18:55, 3 January 2008 (UTC)
replies: text: "Gaussian" almost always refers to CF Gauss. I cleaned up the intro accordingly. —Ben FrantzDale (talk) 01:36, 26 April 2008 (UTC)
thread_title:  What does "Gaussian" mean? 

message: The page claims most programs use a radius of 6σ+1. Won't this depend entirely on the bit depth? That is, to get "perfect" results you need a radius such that the value of the convolution kernel at that radius would round to zero in that bit depth? Actually, you might need to go further... I think the worst case is bluring a black circle with a white background. To find the result corresponding to the center of the circle you want to make sure your filter is large enough to get the correct value at the center of the circle in the current bit depth. I could work out the math, but either way, the optimum radius of the kernel will certainly be a function of bit depth. —Ben FrantzDale 15:14, 2 May 2006 (UTC)
replies: text: The truncation boundary 6σ+1 is rather arbitrary; see the article on scale-space implementation for a complementary description. Tpl 17:41, 9 July 2006 (UTC)
thread_title: Radius

message: :G(r) = \frac{1}{\sqrt{2\pi} \sigma} e^{-r^2/(2 \sigma^2)}
or
:G(u,v) = \frac{1}{2\pi \sigma^2} e^{-(u^2 + v^2)/(2 \sigma^2)}
Why has the square root disappeared from the second formula, and why has a \sigma^2 appeared?
We're told that r^2 = u^2 + v^2 so why wasn't that substitution the only difference?
I ask not because I think it's wrong, but because I'm trying to understand the formula before I use it in a program I'm writing.
Edited to add: for example see http://www.maths.abdn.ac.uk/~igc/tch/mx4002/notes/node99.html - this page shows this formula:
:G(i,j) = \frac{1}{\sqrt{2\pi \sigma^2}} exp(\frac{-(i^2 + j^2)}{(2 \sigma^2)})
It looks to me as if the square root should be in the second formula above, and that it should extend over the sigma squared (and perhaps the first formula should extend it too and square the sigma for consistency?).
Can anyone clear this up for me?
:It should be clarified that G(r) is a Gaussian distribution in 1 dimension. G(u,v) is simply the result of G(u,v) = G(u) * G(v) which is the correct way to use the Gaussian distribution in 2 dimensions. If you use G(r) with r^2 = u^2 + v^2 then the result of the integral in both dimensions is no longer 1. The formula on the reference link you found can be proved to be incorrect by counter-example: we use i=j=0 (which is the center of the matrix) G(0,0) = 0.398942 according to the given formula. But the value in the center of the matrix is 159/1003 = 0.158524 ... surprisingly this is very close to 1 / (2 * Pi). So in my opinion the forumlas in the article G(r) is correct in 1 dimension and G(u,v) is correct in 2 dimensions. So the statement r^2 = u^2 + v^2 should be removed.
::Okay, how about this as a proposed replacement? I've replaced r with d in the one-dimensional example and taken out the bit about the radius.
:::...The equation of Gaussian distribution in one dimension is
::::G(d) = \frac{1}{\sqrt{2\pi} \sigma} e^{-d^2/(2 \sigma^2)}
:::or, in two dimensions,
::::G(u,v) = \frac{1}{2\pi \sigma^2} e^{-(u^2 + v^2)/(2 \sigma^2)}
:::where σ is the standard deviation of the Gaussian distribution...
::What do you think? David 11:30, 22 June 2006 (UTC)
replies: text: Surely it is approximately equal, and not equal therefore? 129.78.220.7 05:28, 18 July 2007 (UTC)
thread_title:  The formulae 

message: :(Note 12 July 2006 - xShin)
:The matrix dimension would be better calculated in this manner n=ceil(3\sigma) Kernel Size = (2n+1) \times (2n+1).
:(End of Note 12 July 2006 - xShin)
::I removed that note from the article, since it seemed to belong here. Luna Santin 07:01, 12 July 2006 (UTC)
thread_title:  Note removed from article 

message: Something should probably done to connect this article to the gaussian filter one, especially since the subjects are so closely related and there's already some duplication of content. -Roger (talk) 18:36, 24 February 2009 (UTC)
thread_title:  Relation to Gaussian filter 

message: Would anyone address those issues using Gaussaian Filters?
--Royi A (talk) 12:39, 16 October 2009 (UTC)
thread_title:  Halos 

message: "Applying multiple, successive gaussian blurs to an image has the same effect as applying a single, larger gaussian blur, whose radius is the square root of the sum of the squares of the blur radii that were actually applied. For example, applying successive gaussian blurs with radii of 6 and 8 gives the same results as applying a single gaussian blur of radius 10, since . Because of this relationship, processing time cannot be saved by simulating a gaussian blur with successive, smaller blurs — the time required will be at least as great as performing the single large blur."
I don't believe this is accurate, referencing gpu massively parallel applications and smaller mask windows  — Preceding unsigned comment added by 70.208.73.107 (talk) 17:42, 16 November 2013 (UTC)
thread_title:  computation time 

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thread_title:  External links modified 

message: While the Gaussian function is very important in statistics, does the same hold true for optics? The Gaussian blur seems to be the go-to blur method, preferred over more naïve digital methods for its supposedly appealing retention of edges, but this alone doesn't say much about its objective basis. I find it interesting that the similar Cauchy function never gets used for this purpose, despite it also offering seemingly natural-looking blurs with long tails yielding an oneric aesthetic. I guess the question boils down to how close the Gaussian function resembles the behaviour of blurring incurred by the interaction of light with real materials and the human eye. A notable case is how it poorly represents the bokeh blur of out-of-focus camera shots. 212.60.107.203 (talk) 09:21, 13 June 2019 (UTC)
thread_title:  Physical basis for Gaussian blurs