Dear Chris,
thanks for your answer. Think the solution you provided wont do the trick due to the following reason: say I have X=(x1,x2) where x1.x2 are somehow specially dependent (also via constraints). So a symmetric assignment of x1 directly defines an assignment of x2 and vice versa. It is thus no simple value symmetry that can be independently applied to any variable. I guess constraint propagation would ensure the symmetry propagation from one to the other, but not sure if it does not actually make the solver loose solutions.
But I built a workaround via branch and wait statements and special checks. As soon as this is running smooth I will give LDSB a second try. Furthermore, I have some fixed order constraints on X that break some symmetries. Will have to check how this influences the LDSB results.
So long, Martin Am 26.04.2013 03:00, schrieb Chris Mears:
On Tue, 23 Apr 2013 10:55:08 +0200 Martin Mann <mm...@informatik.uni-freiburg.de> wrote:I have a symmetry function "s" that can generate for each solution "a" the set of symmetric assignments on X1 and on X2 (independently). The problem is defined in a way that all symmetries from X1 are theoretically combineable with symmetric assignments on X2. Thus I would like to do something like that: - for each assignment a1 on X1 + find assignment a2 on X2 + ensure during remaining search implication if(a1) --> [X2 not s(a2)] for all symmetries of a2 + ensure during remaining search [X1 not s(a1)] for all symmetries of a1 + continue searchHi Martin, I'm not sure I understand exactly what you are trying to do. Say you have a model like this: class TwoPhase : public Script { protected: IntVarArray x; IntVarArray y; public: TwoPhase(const Options&) : x(*this,2,1,2), y(*this,3,1,3) { Symmetries sx; sx << ValueSymmetry(IntArgs(2,1,2)); Symmetries sy; sy << ValueSymmetry(IntArgs(3,1,2,3)); branch(*this, x, INT_VAR_NONE(), INT_VAL_MIN(), sx); branch(*this, y, INT_VAR_NONE(), INT_VAL_MIN(), sy); } // ... } You would get these solutions: x: 1 1 y: 1 1 1 x: 1 1 y: 1 1 2 x: 1 1 y: 1 2 1 x: 1 1 y: 1 2 2 x: 1 1 y: 1 2 3 x: 1 2 y: 1 1 1 x: 1 2 y: 1 1 2 x: 1 2 y: 1 2 1 x: 1 2 y: 1 2 2 x: 1 2 y: 1 2 3 i.e. the symmetry is broken within the "x" and "y" independently. If this is not what you are looking for, could you please explain a bit more? Cheers, Chris _______________________________________________ Gecode users mailing list users@gecode.org https://www.gecode.org/mailman/listinfo/gecode-users
-- Dr. Martin Mann, Postdoc Bioinformatics - Inst. of Computer Science Albert-Ludwigs-University Freiburg Tel: ++49-761-203-8254 Fax: ++49-761-203-7462 http://www.bioinf.uni-freiburg.de/~mmann/ _______________________________________________ Gecode users mailing list users@gecode.org https://www.gecode.org/mailman/listinfo/gecode-users
