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From: Vladimir Prus (ghost_at_[hidden])
Date: 2002-01-14 09:48:19


David Abrahams wrote:

> > In this case we don't know if (t-->d) is due to (t-->s) plus (s-->d) or
> > there's independent path. Observe that (t-->s) means t and s belong to
> > the same SCC. Again, the interesting case is when d belongs to a
> > different
>
> SCC.
>
> I'm not sure of that:
>
> +---+____ +---+____ +---+
> | d |<- `->| s |<- `->| t |
> +---+ `----+---+ `----+---+

> +---+____ +---+____ +---+
> | d | `->| s | `->| t |
> +---+ +---+ ---+---+
> ^ |
> `----------------------/
>
> Isn't that enough to make the case where they're all in the same SCC
> interesting?

Absolutely true! I'll think more about this. Can you tell me any details
about the type of graph? Is it tree or DAG or general graph? Any information
can help.

- Volodya


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