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zwt233 / NDLS

Node Dependent Local Smoothing for Scalable Graph Learning (NeurIPS'21, Spotlight)
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Some Questions #1

Closed LIUkhan closed 2 years ago

LIUkhan commented 2 years ago

Hi wentao! Excellent work!!! Due to my poor capability of math , I raise the following questions for the paper: when a connected graph with the adjacent matrix A without self-loop and using r=0.5(Acap = D{^0.5}AD{^0.5} ) is used, will the Eq. 2 in the paper become : Acap-infinity = sqrt(di)*sqrt(dj)/ (2m + n) ? will the Theorem 3.1. become: (di +1) in the Eq. 6 to di ? will the Theorem 3.2. still be corret? and are there some tips to choose a proper epsilon for a specific task? thank you for reading this issue!

zwt233 commented 2 years ago

when a connected graph with the adjacent matrix A without self-loop and using r=0.5(Acap = D{^0.5}AD{^0.5} ) is used, will the Eq. 2 in the paper become : Acap-infinity = sqrt(di)*sqrt(dj)/ (2m + n) ? ANSWER: Yes.

will the Theorem 3.1. become: (di +1) in the Eq. 6 to di ? ANSWER: In the Theorem 3.1 \tilde{di} denotes (di+1), but when you set r=0.5, Theorem 3.1 may not hold.

will the Theorem 3.2. still be corret? ANSWER: Yes, Theorem 3.2 is not limited to r.

and are there some tips to choose a proper epsilon for a specific task? ANSWER: Maybe you could calculate the average distance from the inital state to the final state dis, and set epsilon as dis/10 or others.

LIUkhan commented 2 years ago

will the Eq. 2 in the paper become : Acap-infinity = sqrt(di)*sqrt(dj)/ (2m + n) ? ANSWER: Yes. Thanks for your immediate and kind reply!!!!

  • An extra check for this question when I rethink my quesiton:
  • [√ ] Acap-infinity = sqrt(di)*sqrt(dj)/ (2m + n) ？
  • [ ×] Acap-infinity = sqrt(di)*sqrt(dj)/ (2m) ？ without self-loop, the A mat only has 2m non-zero elements
zwt233 commented 2 years ago

will the Eq. 2 in the paper become : Acap-infinity = sqrt(di)*sqrt(dj)/ (2m + n) ? ANSWER: Yes. Thanks for your immediate and kind reply!!!!

  • An extra check for this question when I rethink my quesiton:
  • [√ ] Acap-infinity = sqrt(di)*sqrt(dj)/ (2m + n) ？
  • [ ×] Acap-infinity = sqrt(di)*sqrt(dj)/ (2m) ？ without self-loop, the A mat only has 2m non-zero elements

Mathematically, sqrt(di)*sqrt(dj)/ (2m) is correct without self loop. But in many real application (e.g., node classification and node clustering), the graph information of node itself cannot be ultilized if we ignore self loop, leading to non-optimal model performance.