SIFT matching features with euclidean distance

23 views (last 30 days)
Hai,
I don't understand how exactly euclidean distance can help us to match features from 1 image to another. can someone help me?
  1 Comment
Santhana Raj
Santhana Raj on 31 May 2017
The assumption is that the feature of the image segment being compared is is similar. The feature is not the x,y position in the image, but any feature like intensity etc.
IF you images have segments with same features, the euclidean distance between the features will be less, i.e., they will be closed in the feature graph. IT may not be and mostly never closer in the normal image seen,i.,e in the x,y position.
Hope this helps

Sign in to comment.

Answers (1)

Vinoth Venkatesan
Vinoth Venkatesan on 6 Jul 2017
Each feature point that you obtain using SIFT on an image is usually associated with a 128-dimensional vector that acts as a descriptor for that specific feature. The SIFT algorithm ensures that these descriptors are mostly invariant to in-plane rotation, illumination and position.
Please refer to the MATLAB documentation on Feature Extraction here: Local Feature Extraction
So if a feature from one image is to be matched with the corresponding feature in another image, their descriptor needs to be matched to find the closest matching feature. This can be done in various ways, but the most accepted way is to use the euclidean distance (or the euclidean norm of the difference) between these descriptors.
MATLAB provides a way to do this mathcing using the matchfeatures function. You can refer to the documentation here for the complete functionalities: Feature Matching
The features that are closest to each other (least distanced) are matched, provided the distance is greater than a threshold value. In MATLAB, this can be done using the norm function.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by