What is the difference between dense SIFT and HoG?

Question

I am new to Computer Vision. I am studying Dense SIFT and HOG. For dense SIFT, the algorithm just considers every point as an interesting point and computes its gradient vector. HOG is another way to describe an image with a gradient vector.

I think Dense SIFT is a special case for HOG. In HoG, if we set the bin size to 8, for each window there are 4 blocks, for each block, there are 4 cells and the block stride is the same as the block size, we can still get a 128 dim vector for this window. And we can set any window stride to slide the window to detect the whole image. If the window stride of both these two algorithms is the same, they can get identical results.

I am not sure whether I am correct. Can anyone help me?

Answer 1

SIFT descriptor chooses a 16x16 and then divides it into 4x4 windows. Over each of these 4 windows it computes a Histogram of Oriented gradients. While computing this histogram, it also performs an interpolation between neighboring angles. Once you have all the 4x4 windows, it uses a gaussian of half the window size, centered at the center of the 16x16 block to weight the values in the whole 16x16 descriptor.

HoG on the other hand only computes a simple histogram of oriented gradients as the name says.

I feel that SIFT is more suited in describing the importance of a point, due to the gaussian weighting involved, while HoG does not have such a bias. Due to this reason, (ideally) HoG should be better suited at classification of images over dense SIFT, if all feature vectors are concatenated into one huge vector (this is my opinion, may not be true)