The easiest way of changing the label of one node is to obtain its
Context using `match`, and update the label in the Context and then
put it back in the graph with (&).
Ignoring log factors (which are from the graph implementations, not
the FGL model) this ends up being O(|degree of node|).
On 7 August 2015 at 08:29, Jeffrey Brown
Is changing the label at a node O(N)?
The only way I can think of to do it is with a map, like the following -- which works, but seems like its speed must be linear in the number of nodes of the graph:
:m Data.Graph.Inductive.Example Data.Graph.Inductive.Graph let f c@(adjIn, node, lab, adjOut) = case node of {1 -> (adjIn, node, 'b', adjOut); _ -> c} loop mkGraph [(1,'a')] [(1,1,())] gmap f loop mkGraph [(1,'b')] [(1,1,())]
Is that in fact the right way? Am I somehow missing the point of FGL if I have to do a lot of that?
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