I have a problem with SIFT that I do not understand. Lowe [1] proposed in his work the s=3 levels of scale are enough for one octave. Afterwards, he mentioned that you need to compute s+3 levels.

Why there are 3 and not 2 additional levels required. I understand that you require one additional level above and one additional level below the scales since you search for extrema in neighbored scales.

For what is the third additional level of scale?

Thank you very much in advance!

[1] Distinctive Image Features from Scale-Invariant Keypoints
    D. G. Lowe
    Int. Journal of Computer Vision 60(2) (2004), pp. 91--110

2 Answers 2


We must produce s + 3 images in the stack of blurred images for each octave, so that final extrema detection covers a complete octave.

For $s=3$ this means you will have $s + 3 = 6$ blurred images (the Gaussian images shown in the paper in Figure 1 on the left). Having $6$ Gaussian images will result in $5$ DoG images (shown in Figure 1 on the right). This will allow you to do the extrema detection on $s=3$ scales (using the method shown in Figure 2).

  • $\begingroup$ Ah yes, this makes sense. The whole time I thought that $s+3$ DoG images are required. Thank you! $\endgroup$
    – who9vy
    Commented Aug 1, 2013 at 15:03
  • $\begingroup$ So, we're generating $s$ scales by blurring with $k^p\sigma$ given $p = 1 \dots s$, then generate $3$ additional scales (with $p = (s+1) \dots (s+3)$) "on top" of the existing $s$ ones, build the $(s+3)-1$ DoG images by subtracting and then find the extremas in the $s$ middle DoG images, i.e. leaving out the first one obtained from $p=1,2$ and the last one obtained from $p=(s+2),s=(s+3)$? That is, we do search in $s$ laplacians of the scales, but neither in the ones from the first nor equidistantly spaced scales of the octave? $\endgroup$
    – sunside
    Commented Jan 16, 2016 at 15:28

I can only find this is the paper:

As this graph shows, the highest repeatability is obtained when sampling 3 scales per octave, and this is the number of scale samples used for all other experiments throughout this paper. It might seem surprising that the repeatability does not continue to improve as more scales are sampled. The reason is that this results in many more local extrema being detected, but these extrema are on average less stable and therefore are less likely to be detected in the transformed image.

Where are you referring to when he says s+3?

I guess it also depends on the ratio of DOG filters you are using. I've seen DOGupper/DOGlower = 1.6 as it is pretty stable (D. Marr & E. Hildreth, "Theory of Edge Detection", 1980). This also approach a LOG filter.

  • $\begingroup$ On page 7 he said: We must produce s + 3 images in the stack of blurred images for each octave, so that final extrema detection covers a complete octave. $\endgroup$
    – who9vy
    Commented Jul 27, 2013 at 9:01

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