image-processingcomputer-visionimage-segmentationsuperpixels

How the spatial extent of a superpixel is a region of size S*S?


I am reading this paper Achanta-SLIC Superpixel segmentation where it says that the every superpixel cluster center is located at a distance of S = root(N/k) and that expected spatial extent of a superpixel is a region of S * S and the search for similar pixels is done in a spatial region of 2S*2S.

Can someone please explain me this point as I am stuck at it?


Solution

  • From the paper:

    Our algorithm takes as input a desired number of approximately equally-sized superpixels K.

    So, let's assume that our SP are approximately squares. You will have K of them.

    For an image with N pixels, the approximate size of each superpixel is therefore N/K pixels

    If you divide the image area N in K SP, every SP has (almost) N/K pixels. I.e., the area of each SP is N/K.

    For roughly equally sized superpixels there would be a superpixel center at every grid interval S = sqrt(N/K).

    Each SP is assumed to be squared, with area N/K. The side of the square will then be sqrt(area) = sqrt(N/K) = S. This means that a SP center is S far from neighbours's centers.

    Since the spatial extent of any superpixel is approximately S^2 (the approximate area of a superpixel)

    Well, the side of each square is S, then its area is S^2 (which is the same as N/K = sqrt(N/K)^2 = S^2).

    we can safely assume that pixels that are associated with this cluster center lie within a 2S × 2S area around the superpixel center

    We mentioned that each side of the square will be S, then each pixels of the SP will lie within the size of half the diagonal from the center sqrt(S/2), which is less than the side sqrt(S/2) < S. But SP are not exactly squares, so we want to be a little more flexible, and say that all pixels lie within the double of this distance: 2S.