[gecode-users] Number of branches/choices

[Max Ostrowski]

Thanks for this, so a solution and a failure is also a node.
This formula seems to fit.
Thx

On 10/23/2012 04:50 PM, Christian Schulte wrote:
> Hi,
>
> So we could say that every edge (corresponding to an alternative) in the
> search tree is a decision made during search (even though the branching in
> the parent node defines all alternatives directly below).
>
> Then, if we are talking binary branchings only, we all know that the number
> of edges #e is the number of nodes #n minus 1. The total number of nodes #n
> is the number of nodes reported plus the number of solutions and failures.
> Voila!
>
> Cheers
> Christian
>
> --
> Christian Schulte, www.ict.kth.se/~cschulte/
>
>
> -----Original Message-----
> From: users-bounces at gecode.org [mailto:users-bounces at gecode.org] On Behalf
> Of Max Ostrowski
> Sent: Tuesday, October 23, 2012 1:31 PM
> To: users at gecode.org
> Subject: [gecode-users] Number of branches/choices
>
> Hi,
>
> i want to compare some statistics,
> and want to find out how many choices (non-deterministic decisions) gecode
> has made during branching.
> (So, not talking about non-deterministic propagation, just simple
> select/propagate).
>
> The statistic object
> dfsSearchEngine_->statistics().node
> tells me something about the used nodes, but this not seem to be the right
> thing to use.
>
> So, for a simple CSP, with variable x = 0..9 without any constraint, i get
> 19 nodes (enumerating all solutions).
> Can i simply divide this by 2 to get my 9 decisions?
>
> Best,
> Max
>
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