How to make nodes' removal from a graph as a function of time?

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Hello,
I'm simulating failures of nodes in an interdependent system (two graphs, g and h) and their adjacency matrices are A and B. Some nodes in graph g (layer A) will get attacked randomly, and then this failure will get propagated to graph h (layer B) through the interlinks between the two graphs.
I'm trying to simulate failure or nodes removal based on time, that is,
at: t=1, remove first node in A.
t=2, remove the second node in A
t=3, failure propagates to B and first node in B get removed and so on.
Your help would be highly appreciated.
  1 comentario
Waseem AL Aqqad
Waseem AL Aqqad el 14 de En. de 2021
It would be fine If you can show me how this works for a single layer. I can then apply it to multi-layers or graphs.

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Respuesta aceptada

Christine Tobler
Christine Tobler el 15 de En. de 2021
You may want to look at the graph and digraph classes: Graph and Network Algorithms and their method rmnode specifically.
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Waseem AL Aqqad
Waseem AL Aqqad el 2 de Feb. de 2021
Editada: Waseem AL Aqqad el 2 de Feb. de 2021
Thanks, Christine!
No, G and F will not be changing. But this could be another nice idea.
I wrote another function "Self_Healing" to try to recover as many as possible of those failed nodes. I want to call my "Self_Healing" function at a time instant right after the failure of the first nodes.
So, at t=1 Cascading_Failure starts by removing two nodes, at t=2 Self_Healing will start trying to recover the two removed nodes while at the same time instant the Cascading_Failure will be removing the next nodes, and so on.
Another question please, based on my Cascading_Failure function how can I simulate the time needed for each node to get failed? As I believe, in real world applications, one node might take 30 sec to get removed and another one might take several minutes.
Thanks again!
Waseem AL Aqqad
Waseem AL Aqqad el 5 de Feb. de 2021
Hi Christine, I don’t mean to bother you but I was wondering if you have had a chance to look at my problem yet? Your help is greatly appreciated.

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