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Please refer to Section 2 in this paper for notations, and specifically Section 2.3 for definitions of $d_\square$ and $\delta_\square$.

My question is as follows:

If $G$ and $G^\prime$ are graphs with the same set of nodes and all node weights are the same, then will $\delta_\square(G, G^\prime) = d_\square(G, G^\prime)?$

I believe it should be confirmative, since $\delta_\square$ is a generalization of $d_\square$. However, I cannot prove it based on definitions of $d_\square$ and $\delta_\square$ given in the paper.

Thanks a lot for any help.

Tien Kha Pham
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    It is more reasonable to ask whether $\delta_\square = \widehat \delta_\square$, since both of these allow relabeling nodes. Since $d_\square$ does not allow relabeling nodes, then for a case where $G, G'$ are isomorphic but unequal graphs on the same set of vertices, $\delta_\square(G,G')$ will be positive while $\widehat \delta_\square(G,G')$ and its generalization $\delta_\square$ will be $0$. – Misha Lavrov Aug 22 '22 at 19:21
  • @MishaLavrov Thanks. Just a small issue: I think there's a typo here. "$\delta_\square(G, G^\prime)$ will be positive..." --> it should be $d_\square(G, G^\prime)$, right? – Tien Kha Pham Aug 23 '22 at 00:43
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    Yes, sorry, that's what I meant. – Misha Lavrov Aug 23 '22 at 00:44

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