That was a bit tricky, but I found the problem ;-)
The reified dom constraint does not behave as you expect. When you call
dom(home, x, s, b) it means b <-> x in s.
So if !b, then x not in s (all values from s will be removed from x!).
I guess what you want is just the implication b -> x in s, which can be
implemented like this:
BoolVar b(*this, 0,1);
dom(*this, start_times[j], v, b);
rel(*this, m(j,i), BOT_IMPL, b, 1);
or more compactly using minimodel:
rel(*this, m(j,i) >> (singleton(start_times[j]) <= v));
Cheers,
Guido
On 31 May 2011, at 08:52, alin gherman wrote:
> Hallo,
>
> I now understand the reasoning for using the boolean matrix
> so that each unary constraint corresponds to a certain resource
> and to the starting times of the resource.
>
> but thaw it all make sense now I still don't get the result
> I expect. There are no different resources assigned to the task
> though different resources can do the task faster since they
> can work in parallel,
> and another problem is that when I specify constraints
> for what resources can do a certain task I get no solution
> again.
>
> I attached a small sample that behaves like my program
> maybe someone with more experience can help me.
>
> thanks.
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--
Guido Tack, http://people.cs.kuleuven.be/~guido.tack/
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