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-boun...@gecode.org [mailto:users-boun...@gecode.org] On Behalf Of Max Ostrowski Sent: Tuesday, October 23, 2012 1:31 PM To: users@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 _______________________________________________ Gecode users mailing list users@gecode.org https://www.gecode.org/mailman/listinfo/gecode-users
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