Are you saying that using "next" in a loop makes no sense with BAB
in that only the first result is the optimal result and the "next"
result is not the next optimal result, but just another possible
result?
Thanks,
Dean
On Wed, 31 Jul 2013 23:19:37 -0400, Guido Tack <t...@gecode.org> wrote:
Hi,
BAB is not meant to produce all solutions, the only guarantee it gives
is that the last solution is going to be optimal, but intermediate
solutions can be skipped (that's in fact the whole point of BAB).
If you need all solutions and still want to know which one is best,
simply use DFS and then sort the solutions afterwards.
Cheers,
Guido
On 31/07/2013, at 11:53 PM, Dean Hiller <hil...@employees.org> wrote:
I was having a problem with finding all my solutions while using BAB.
I finally created a very simple model that demonstrated my problem
and using that model, I was able to find a solution to the problem.
I could not find this issue discussed anywhere, so I am going to
describe it here in case any one else runs into it and is looking
for a solution to it.
This is the scenario:
IntVar X(0,1);
branch(X, INT_VAL_MIN());
IntVar Y(0,2);
branch(Y, INT_VAL_MIN());
IntArgs C(1,2);
IntVarArray V(2, 0, 2);
rel(V[0] == X);
rel(V[1] == Y);
IntVar Cost(0,5);
linear(C, X, V, IRT_EQ, Cost);
If I solve this with DFS, I get 6 solutions.
If I solve this with BAB and in the constrain function say
rel(cost > b.cost);
I only get 4 solutions.
Note: If I run with GIST and expand the nodes in the
correct order, I can cause BAB to display all 6 solutions,
but the normal search order does not find all of them.
To have BAB find all 6 solutions, I removed the above
two branch statements and replaced them with
branch(Cost, INT_VAL_MIN());
So, the general rule I would conclude from this test is
that when using linear to create a cost function for
BAB, always branch on the output of the equation rather
than on the terms making up the equation.
I don't know if this holds true for other uses of linear.
Dean
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