Label propagation equation: what are the terms?

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  • Thread starter Master1022
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  • #1
Master1022
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What terms in the equation (from the linked paper) are vectors or scalars?
Hi,

This is a simple question that I just wanted to clarify. I was reading the following paper on label propagation: HERE and I can't understand whether the terms are vectors or scalars in one of the equations - specifically, equation (2.15) shown in the image below.
Screen Shot 2022-05-03 at 11.02.57 AM.png


My understanding:
- ##f## is a vector
- ## S ## is a matrix
- ## \alpha ## is a scalar
- I am not too sure about ##y##: could be a vector or a scalar.
- ##\nu##: I am not too sure, but I think it could be referring to a specific node? That is, ## f(\nu) ## could be the value of the vector ## f## at node ## \nu ##.
- ## y ##: I am not sure, but I think it is a vector (see reasoning below).

Case is ## y ## is a scalar:
- That would make sense mathematically, but does that mean that we are using the same scalar ## y ## the equation for all nodes. That is, it doesn't matter what node ## \nu ## we are considering, we will always have the same ## y ## scalar in the equation? However, there is another equation above (shown below) which uses y as follows. This suggests that ##y## is a vector because then we have matrix-vector multiplication:

Screen Shot 2022-05-03 at 11.03.14 AM.png

Case if ## y ## is a vector:
- It could be a vector (as suggested by image above), but then we are adding a vector ## (1 - \alpha) y ## to a scalar ## \alpha S f ## is a vector, and we are extracted the value at a certain node ## \nu ##, so it is a scalar. Therefore, it seems unlikely that ## y ## is a vector unless my interpretation of ## \nu ## is incorrect.


Apologies if this is sparse with information. I didn't want to rewrite the paper in this post and I am unsure of some of the definitions of variables in there. Any help would be greatly appreciated.
 

Answers and Replies

  • #2
DrClaude
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##f## and ##y## are functions of ##v##, where ##v## designates a vertex, ##S## an operator, ##\alpha## a positive scalar.

For a given set of vertices ##V##, it is possible to write ##f## and ##y## as a vector over the set of vertices, and ##S## as a matrix.

Note that ##y## can't be a scalar otherwise eq. (2.15) would represent the sum of disparate elements.
 
  • #3
Master1022
611
116
##f## and ##y## are functions of ##v##, where ##v## designates a vertex, ##S## an operator, ##\alpha## a positive scalar.

For a given set of vertices ##V##, it is possible to write ##f## and ##y## as a vector over the set of vertices, and ##S## as a matrix.

Note that ##y## can't be a scalar otherwise eq. (2.15) would represent the sum of disparate elements.
Many thanks for the response!

So would ## f( \nu ) ## be a vector instead of just referring to the entry of vector ## f## corresponding to ## \nu ##?
 
  • #4
DrClaude
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So would ## f( \nu ) ## be a vector instead of just referring to the entry of vector ## f## corresponding to ## \nu ##?
I don't understand your question.

##f(v)## is a function, but if you have a discrete set ##V## of vertices ##v##, then ##f(v)## over ##V## can be written as a vector.
 

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