3 ms·
That makes slightly more sense, but then why the heck do you end up needing to negate something while computing a dot product?
by eigenket 3y ago
That makes slightly more sense, but then why the heck do you end up needing to negate something while computing a dot product?
- ashvardanian 3y agoSorry, I meant the dot product distance. For vectors A and B it's equal to 1-AB. ...ripped it out of my vector search library :)
- eigenket 3y agoThat's a pretty weird "distance" unless you have some additional constraint like assuming the vectors are 2-norm normalised, in which case you get |a-b|^2 = 2(1-a.b). It is natural to assume vectors have distance 0 from themselves at least, which this thing doesn't (in general). E.g. if you compute this "distance" for a = b = (2,0) then you get that the distance from a to itself is -3 which seems pretty weird.