The known path is three steps long (i.e. contains four vertices) while the 
paths returned by all_shortest_paths are only two steps long. Is there any way 
of extending the cut off criterion for all_shortest_paths so that it would 
return the known path too?

My problem with the graph.edge(u,v) function for me in this case is the fact 
that in many cases I have multiple edges connecting two given vertices. Each 
edge has a unique ID allowing me to link it back to an entry in an external 
database. Though it is of interest to me which vertices lie along the path when 
I run all_shortest_paths what I really am after are the IDs associated to the 
edges along the path. It seems to me as if graph.edge(u,v) only returns one 
edge when called. So when I have twenty edges that each link u and v I am not 
sure yet how I can access the other edges to obtain their IDs. I could 
presumably note down the edge and then delete it and then rerun graph.edge but 
that probably is not the most ideal way.

Best,

Philipp

-----Original Message-----
From: graph-tool [mailto:[email protected]] On Behalf Of Tiago de 
Paula Peixoto
Sent: 09 March 2016 21:22
To: [email protected]
Subject: Re: [graph-tool] How is all_shortest_paths defined?

On 09.03.2016 22:17, Philipp-Maximilian Jacob wrote:
> When I however run all_shortest_paths(g,source,target) this path is 
> not part of the results set. Why is that?

The function finds all _shortest_ paths, not all paths. Are you sure the known 
path belong to the shortest set?

> Finally, is it possible to also obtain an edge descriptor as a result?
> Currently I think the documentation states that the result is an 
> iterator over the sequence of vertices from source to target.

You can get the edge between any two vertices u and v by calling the 
Graph.edge(u, v) function.

Best,
Tiago

--
Tiago de Paula Peixoto <[email protected]> 
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