[Haskell-cafe] FGL: The cost and propoer way of changing the label at a node.

[Jeffrey Brown]

What about changing an edge's label? Again all I can think of is to replace
the edge with a new one, but that seems likely inefficient, particularly if
I wanted to handle a lot of edges as a batch.

On Thu, Aug 6, 2015 at 3:41 PM, Ivan Lazar Miljenovic <
ivan.miljenovic at gmail.com> wrote:

> 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 <jeffbrown.the at gmail.com> wrote:
> > 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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>
>
>
> --
> Ivan Lazar Miljenovic
> Ivan.Miljenovic at gmail.com
> http://IvanMiljenovic.wordpress.com
>
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