Re: Labeling outside of fd_labeling?

Hi Daniel,

Thank you very much for your explanation. BTW: found
another "looping" example:

    | ?- X + 1 #= Y, Y + 1 #= Z, Z #< X.
    (4555 ms) no

So basically, when one for example formulates
scheduling problems etc.., one has to be careful
with inconsistent problems?

Bye

Daniel Diaz schrieb:
>
> Le 16 juin 2012 à 16:44, Jan Burse a écrit :
>
>> Dear All,
>>
>> Small question. I just posed the following query:
>>
>> ?- X #< Y, Y #< Z, Z #< X.
>>
>> The interpreter then was busy for around ~4 secs
>> and then responded with "no".
>>
>> The "no" is actually correct, by transitivity we have
>> X #< Z which conflicts with Z #< X.
>>
>> What I wonder is what keeps the interpreter busy for
>> ~4 secs and whether the "no" is reliable.
>
> It is due to the arc-consistency filtering algorithm. Consider X@<Y, Y#<X. Basically for X#<Y the max(X) is set to max(Y)-1 each time the max(Y) is updated. Conversely for Y#<X, the max(Y) is set to max(X)-1 each time max(X) is updated.
> This results on a decrement of the max(X) and max(Y) from FD_MAX_INT (e.g. 268435455) to 0 (1 by 1). Once X or Y reaches a domain 0..0 the inconsistency is discovered on the other variable (domain 0..-1, i.e. empty).
>
> This takes 4 sec on your machine.
>
> About the "no": yes it is reliable.
>
> Daniel
>
>>
>> Bye
>>
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>
>