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To make Gaussian blurring a 2d image faster, I know that you can do one axis and then the other.

I'm wondering though, if I did two Gaussian blurs of size $N$, would that be the same mathematically as doing one Gaussian blur of size $2N$?

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Yes, applying two Gaussian blurs is equivalent to doing one Gaussian blur, but with a slightly different size calculation.

Applying a Gaussian blur to an image means doing a convolution of the Gaussian with the image. Convolution is associative: Applying two Gaussian blurs to an image is equivalent to convolving the two Gaussians with each other, and convolving the resulting function with the image.

As it happens, the convolution of two Gaussians with each other is another Gaussian, whose variance is the sum of variances of the two original Gaussians. A Gaussian of N pixel width has variance $N^2$; applying it twice is equivalent to a Gaussian with variance $2\cdot N^2$, which corresponds to a width of $\sqrt{2}\cdot N$ pixels.

In the same vein, applying a Gaussian of width $2\cdot N$ is equivalent to applying a Gaussian of width $N$ four times.

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  • $\begingroup$ Does this mean that applying a Gaussian blur 4 times would give twice the size, as required? $\endgroup$ – trichoplax Aug 15 '15 at 20:08
  • $\begingroup$ Yes, absolutely! $\endgroup$ – Benedikt Bitterli Aug 15 '15 at 20:12
  • $\begingroup$ That covers the question entirely then - would it be worth mentioning how to get twice the size in the question so we can tidy away the comments? $\endgroup$ – trichoplax Aug 15 '15 at 20:19
  • $\begingroup$ If it takes 4 to double the size, what size would 3 represent? $\endgroup$ – Alan Wolfe Aug 16 '15 at 1:01
  • $\begingroup$ I found the answer is N*sqrt(3). It turns out that the total Blur amount of multiple blurs is equal to the square root of the sum of the sizes squared. Wikipedia says that: en.wikipedia.org/wiki/Gaussian_blur $\endgroup$ – Alan Wolfe Aug 16 '15 at 1:08

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