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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