0

I am learning this course Stanford CS224N: natural language processing with Deep Learning.

The professor is saying

on the one hand, whenever you see these pictures you should hold onto your wallet. Because there's a huge amount of detail on the original vector space that got completely killed and went away.

is it the vector space got completely killed?

what does the phrase "killed and went away" mean in this context?

|improve this question
  • 1
    It is quite common in programming to refer to the removal of unnecessary data as "killing". So he is saying that the detail was "killed" and is no longer available (it "went away"). – user323578 Apr 15 at 14:48
  • 2
    No, the detail got killed and went away. The author removed detail (killed it) and you can no longer see evidence of it (it went away) by doing whatever was done to paint the pretty picture. – HBruijn Apr 15 at 14:48
  • Please do not answer questions in comments. If you know the answer, please write it as an answer. – DJClayworth Apr 15 at 17:27
  • It's all in the details! – Hot Licks Apr 15 at 22:36
1

In the professor's example, there are 2 following vector spaces,

VS1: original vector space, which is a 100-dimensional space;

VS2: a 2-dimensional space, into which original information are being projected down.

During the projection, some information in original vector space(i.e. VS1) would be lost.

Phrase "killed and went away" here means "lost".

Some knowledge about linear algebra is needed to under this.

|improve this answer
  • Or you could change the 100 to 3 and think of it as the sun and an object (sun+object are in 3D) casting a shadow on the floor (in 2D). You can recreate the shadow if you know all the 3D information but you cannot construct the object and the sun (taking into account size and location) from the shadow alone. – JJJ Apr 15 at 23:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.