Undirected network

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mahya s

May 4, 2023, 10:21:49 AM5/4/23
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Hi ORA team,
I have several networks that when I visualize them and click on "show arrows", they are directed. I want my links to be undirected and just have a link between my nodes (see attached 1). When I want to check the "always symmetric" box, it allows me for the networks including the same nodes (Agent x Agent) but not for networks with different types of nodes (Task x Agent) (see attached 2).


Mihovil Bartulović

May 4, 2023, 11:41:44 AM5/4/23
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ORA reads rows and adds a link from Node1 to Node2 if there is a non-zero value in the row for Node1 towards Node2. Therefore, symmetric networks will have two arrows on the same link - one pointing to Node1 and one pointing to Node2.

If you want the links in the graph to be undirected, simply unselect the "Show Arrows" option.

The "always symmetric" box is a feature of an older version of ORA. Newer versions have replaced it with "Treat as symmetric," which makes the networks symmetric for the purpose of calculations.

mahya s

May 4, 2023, 12:12:31 PM5/4/23
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Thank you for your response. So, that means, if I want to have a regular undirected link between 2 nodes for my calculations, I should consider that node 1 is related to node 2 and also node 2 is related to node 1. So, if I turn on "show arrows", I can see a 2-direction arrow. Then turn off the "show arrow" and continue with my calculation. is that right?

Mihovil Bartulović

May 4, 2023, 1:45:37 PM5/4/23
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Yes, but with a caveat:

For unipartite networks (i.e. Agent x Agent), every link would have to have a double arrow, and those two links need to be of equal strength (have the same value). The adjacency matrix of your unipartite network needs to be a symmetric square matrix.

The image you showed is a bipartite network, which by default is considered undirected as the link between two different nodesets is considered to be a connection (i.e., Agent1 knows how to do Task1). However, given that ORA draws network graphs by reading the lines, ORA will always point from Task -> Agent in a Task x Agent network.

mahya s

May 4, 2023, 2:59:34 PM5/4/23
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Thank you, I think I got it! Just to clarify, here are my networks and the direction between nodes, are they make sense?
Agent x Agent (who works with): Agent <-> Agent
Task x Task (what task is precedent to or dependent on what task): Task <-> Task
Resource x Resource (what resource is used for other resources): Resource <-> Resource
knowledge x knowledge (what knowledge is related to what knowledge): Knowledge <-> knowledge
Task x Agent (who is assigned to what task): Task -> Agent
Task x Resource (what resource is needed to do what task): Task -> Resource
Task x knowledge (what knowledge is needed for what task): Task -> Knowledge
Agent x knowledge (who knows what): Agent -> Knowledge
Resource x Knowledge (what knowledge is needed to use what resource): Resource -> knowledge
Agent x Resource (who can use what resource): Agent -> Resource

I really appreciate your response.

Mihovil Bartulović

May 4, 2023, 8:12:18 PM5/4/23
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All of the networks you mentioned make sense to me !

all of the unimodal networks seem to be undirected/symmetrical, bimodal networks are undirected by default.

for the unimodal networks (i.e. agent x agent) you click on the network in the meta network manager -> on the right hand side select editor tab -> convert links -> symmetrize by method and ORA will make sure to create a symmetric (undirected) network from your data given the chosen method.

This method adds, removes, or modifies links to symmetrize each unimodal network.
A network is symmetric if for each pair of nodes {A,B} the links (A,B) and (B,A) exist and have the same value.

ORA offers different ways to symmetrize a network.

Maximum sets the values of links (A,B) and (B,A) to be the maximum of the two existing values
Minimum sets the values of links (A,B) and (B,A) to be the minimum of the two existing values
Sum sets the values of links (A,B) and (B,A) to be the sum of the two existing values
Average sets the values of links (A,B) and (B,A) to be the average of the two existing values

mahya s

May 12, 2023, 10:34:59 AM5/12/23
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Thank you so much for all your help!
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