Re: [help-GIFT] Histograms and Normalization

[David Squire]

Sailesh Suvarna wrote:
> Hi all,
> Thanks for the answers to my last question.
>
> I have some more...........
>
> I am using just 4 images in my collection and am using only feature (1) - 
> Colour Histograms.

Thats's not one feature, that's one *feature group*.

> In  CWeightingFunction::subApply if I change-
>
> lRetVal=copysign(min(inDocumentFrequency,fabs(getTermFrequency())),getTermFrequency());

This is doing histogram intersection. In fact it is calculating the 
intersection for just one bin of the histogram, corresponding a 
particular feature.

> lRetVal=fabs((inDocumentFrequency-getTermFrequency()));
> to find out the difference rather than similarity 

The system expects scores to be maximal for an image with identical 
features to the query. This version will be zero when the features are 
identical. This is likely to cause problems later...

> I get the following results: 
> Similarity: inf
> Similarity: nan
> Similarity: inf
> Similarity: inf

Yep. Problems. If I remember correctly, the similarity scores are 
normalized by dividing by the similarity of the query with itself, which 
you have just set up to be zero. Hence problems caused by divide-by-zero 
errors.

NB. This interpretation is not based on detailed knowledge of this code. 
Wolfgang: jump all over me if I'm wrong!

> Also the lQueryScore always computes to 0
>   The following is what I get 
>   
>  Pruning used!
>  Pruning: I will evaluate 39 Features.

Note: 39 features, not one.

Cheers,

David



--
Dr. David McG. Squire, Postgraduate Research Coordinator (Caulfield),
Computer Science and Software Engineering, Monash University, Australia
Monash Provider No. 00008C       http://www.csse.monash.edu.au/~davids/