SIFT特征描述符的实现

Question

According to Lowe's paper about the original SIFT algorithm, a feature descriptor consisting of 4 x 4 orientation histograms is calculated from a 16 x 16 window. The scale of the descriptor is only used to select the level of gaussian blur for the image.

Looking at the OpenCV implementation, this doesn't seem to be the case. In calcSIFTDescriptor there is the following code to calculate the histograms:

for( k = 0; k < len; k++ )
{
  // histogram update
}

Where len is the number of samples used. According to Lowe's algorithm, this should always be 256 (16 x 16), shouln't be? In OpenCV implementation the len depends on the scale of the descriptor.

Could someone clarify this?

Thanks

Answer 1

You're not looking at the code that builds the histogram but at the code that explores the neighbourhood of a given pixel to compute the gradient orientation and norm.

The relevant code is below in the source file:

for( k = 0; k < len; k++ )
{
  int bin = cvRound((n/360.f)*Ori[k]);
  if( bin >= n )
    bin -= n;
  if( bin < 0 )
    bin += n;
  temphist[bin] += W[k]*Mag[k];
}

In this excerpt, the line cvRound(...) is the one that computes the target bin in the 16-bin histogram.

Normally, in SIFT you should first resample the feature point neighbourhood by recreating a grid of 4x4 blocks of 4x4 pixels. The code that you've shown is used to compute the gradient statistics without having to resample the image. The variable radius is here to take into account the image location in the scale space.