There is very clear explanation on how a Summed Area Table/Integral Image works here: https://uk.mathworks.com/help/vision/ref/integralimage.html enter image description here

How to you scale this SAT/Integral Image?

For example, I wish to downscale it from 16x16 to 8x8. How would I go about calculating that?


1 Answer 1


Calculation of the integral image is a linear transform on the original image.

Now, assuming you wanting to downscale the integral image means that you want the result to be the same whether you:

  • downscale the original image, integrate after
  • integrate the original image, downscale after

and remember that "downscaling" by a factor of $N$ the original image means

  1. applying a low-pass filter to avoid aliasing and
  2. throwing away $N-1$ of every $N$ pixels.

since low-pass filtering is also a linear operation, we can just switch orders of summations and down-scale the result image. Nothing special applies.

  • $\begingroup$ Thank you for the response. I am not sure if I fully understand your post: I understand that we want the result to be the same whether we perform the down-scaling followed by the integrate after, or integrate followed by the down-scaling after. I also understand the process on how to downscale the original image. However, I do not understand the process on down-scaling an integral image. As I understand, the process for down-scaling an integral image is not the same as the process for an original image. $\endgroup$ Dec 18, 2018 at 14:41
  • $\begingroup$ But it should be! (aside from potentially a constant factor). $\endgroup$ Dec 18, 2018 at 15:36
  • $\begingroup$ Ah! OK, my apologies, I understand now. Your point "aside from potentially a constant factor" was my clue! :-) If we convert an image from the image domain to the integral domain, downscale it, and convert it back, it will have pixel values much larger than if we were to just downscale the image in the image domain. Would you have any idea on how to calculate the scaling constant factor by any chance? $\endgroup$ Dec 19, 2018 at 12:12
  • $\begingroup$ try with the scaling factor² $\endgroup$ Dec 19, 2018 at 12:51

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