Re: [eclipse-clp-users] ECLiPSe - Real variables domain propagation problem

Thank you both Kish and Joachim for your answers.

Wayne

On Fri, Jul 24, 2009 at 2:38 AM, Joachim Schimpf <
joachim.schimpf_at_...44...> wrote:

> Wayne Mac Adams wrote:
> > Or even just a simpler way to represent it is to just focus on the
> > following query(assuming the ic librarby is loaded)
> >
> > ?- A :: 0.0 .. 0.5, [B, C] :: 0.0 .. 1.0, A $= eval(B + C - B * C).
> > A = A{0.0 .. 0.5}
> > B = B{0.0 .. 1.0}
> > C = C{0.0 .. 1.0}
> > There are 4 delayed goals.
> > Yes (0.00s cpu)
> >
> > If i had
> > ?- A :: 0.0 .. 0.5, [B, C] :: 0.0 .. 1.0, A $= eval(B + C).
> > A = A{0.0 .. 0.5}
> > B = B{0.0 .. 0.5}
> > C = C{0.0 .. 0.5}
> > There is 1 delayed goal.
> > Yes (0.00s cpu)
> >
> > it propagates correctly.
> >
> > I have been told( by Helmut Simonis, and I hope I paraphrase this
> > correctly!) it is because ECLiPSe treats B+C and B*C as seperate and not
> > foesn't see the it as as the same terms being used.
>
> There are two aspects here:
> (1)  you seem to expect too much from propagation generally
> (2)  A $= B+C-B*C could propagate more than it actually does
>
>
> As for (1), all that constraint propagation does is to remove
> "impossible" values from the domains, i.e. values that do not
> occur in any solution to the problem.  But there is no guarantee
> that propagation removes _all_ impossible values.  In fact, as
> soon as you have more than a single constraint, you can be almost
> certain that this is not the case.  Even if the individual constraints
> have this property (i.e. achieve "generalised arc consistency"),
> an arbitrary combination of them will not (in general).
> To prove that a value is part of a solution, you have to find
> a concrete solution with this value, you cannot rely on propagation
> alone.
>
>
> As for (2): individual constraints can often be implemented such
> that they achieve "generalised arc consistency", i.e. remove all
> impossible domain values.  But even that is not always done,
> simply because it may be too computationally expensive, or,
> like in this particular case, because what looks like a single
> constraint in your model is actually implemented by decomposition:
> The source constraint
>
>  A $= B+C-B*C
>
> is broken up into two constraints, representing the linear
> and the nonlinear part of your equation separately:
>
>  A $= B+C-Aux,  Aux $= B*C
>
> If you look at these two constraints separately, you see that none
> of them (individually) can achieve more propagation than it does:
>
>  Aux{0.0 .. 1.0} - C{0.0 .. 1.0} - B{0.0 .. 1.0} + A{0.0 .. 0.5} $= 0,
>  Aux{0.0 .. 1.0} $= B{0.0 .. 1.0} * C{0.0 .. 1.0}
>
> Although it would be possible to make an implementation of A $= B+C-B*C
> without decomposition that achieves full arc consistency, the example
> shows that as soon as you add other constraints, you lose the property
> again.
>
>
> >...
> >
> >     Does anyone know how to solve this problem?
> >     It forms part of a larger problem in which I am getting an incorrect
> >     result because of the domains staying between 0.0..1.0.
>
> The issue is that you are not interpreting the "result" correctly: the fact
> that there are "delayed goals" left at the end of the computation implies
> that the problem is not actually solved, and the domains may or may not
> contain solutions.  The only guarantee you have is that any solutions
> (should they exist) are within the computed domains.
>
> In an integer problem you would at this point use labeling/indomain to
> find concrete integer solution values.  Since you are working with reals,
> it is more tricky, because of numerical inaccuracies - some of it is
> discussed in http://eclipse-clp.org/doc/tutorial/tutorial063.html
>
> To get more restricted domains in your example, you could use
>
> ?- A::0.0..0.5, [B,C]::0.0..1.0, A $= B+C-B*C, squash([A,B,C], 0.001, log).
> A = A{0.0 .. 0.5}
> B = B{0.0 .. 0.50091762095689774}
> C = C{0.0 .. 0.50091762095689774}
> There are 4 delayed goals.
> Yes (0.00s cpu)
>
> squash/3 is a search procedure that splits domains and excludes ranges
> if they can be shown not to contain any solutions.
>
>
> -- Joachim
>
>
>
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