I am implementing Lowe's method, "SIFT", for finding and describing features in an image.
I have found interest points, and now I have to describe them: Using Lowe's method, I have calculated the magnitude and gradient in an area around the keypoint, and created a Gauss weighted histogram, with 36 bins, each corresponding to an orientation of 10 degrees. For each keypoint, there is a histogram. Each bin is the sum of the weighted magnitude, in that direction. An example taken from aishack.in: http://www.aishack.in/static/img/tut/sift-orientation-histogram.jpg
Bins within 80% the size of the maximum bin, is made a new keypoint. After describing, it says in the paper: "Finally, a parabola is fit to the 3 histogram values closest to each peak to interpolate the peak position for better accuracy". I am not sure i get this.
In my understanding, it means the peak, the left, and the right value of that peak, will have a parabola fit, like this(be warned! Drawn free hand)
https://i.sstatic.net/7V8pb.jpg
and the orientation of the keypoint will be where the extremum of the parabola is. For instance: If the parabola fitted at 10-19, 20-29, and 30-39 (with 20-29 being the histogram peak), had extremum at a point, that reached in the 30-39, then this would be the orientation of that keypoint. Am i understanding this correctly? In this way, the orientation of the keypoint, can only be within 36 orientations
Another option: Same idea as above, only the histogram is no longer discrete: the extremum of the parapola will thus be a continuous value, and this value is assigned to the keypoint.
The idea of the parabola fitting is to find the peak with better than bin resolution. As you see in your example, the peak is at 20-29 (average 24.5) but the 10-19 bin is higher than the 30-39 bin. It's therefore likely that the precise peak should be below 24.5.
You can't have a non-discrete histogram, that defeats the point of a histogram. What you can have is overlapping bins: create a bin for 20-29, but also a bin for 21-30 and 22-31 etc. So the value 24 would map to 10 bins, from 15-24 to 24-35.
And when you increment a bin, you don't necessarily need to increment it by 1. You can also increment a bin by a variable amount, e.g. the distance from the given value to the edge of the bin. So 24 would add 1 to bin 16-25 but 4 to bin 20-29.
