Given the following sample of code:
example(Ls) :-
Ls = [X,Y],
Ls ins 1..2,
Cost #= max((X #= 1)*3 + (Y #= 1)*5,
(X #= 2)*3 + (Y #= 2)*5),
labeling([minimize(Cost)], Ls).
The idea is to find the assignment to variables of Ls that minimizes Cost (in this simple example, it would be either X=1 and Y=2, or X=2 and Y=1).
I am trying to use the fact that the constraint #=/2 is reifiable, which means reflecting their truth values into Boolean values represented by the integers 0 and 1. (taken from the manual http://www.swi-prolog.org/man/clpfd.html).
However, it doesn't work. I get the following error:
ERROR: Domain error: `clpfd_expression' expected, found `_G3154#=1'
What would be an equivalent, correct version?
Reification involves separate constraints (#<==>/2
, #==>/2
etc.), you can use them for example like:
example(Ls) :-
Ls = [X,Y],
Ls ins 1..2,
X #= 1 #<==> B1,
Y #= 1 #<==> B2,
X #= 2 #<==> B3,
Y #= 2 #<==> B4,
Cost #= max(B1*3 + B2*5, B3*3 + B4*5),
labeling([min(Cost)], Ls).
Sample query and its result:
?- example(Ls).
Ls = [1, 2] ;
Ls = [2, 1] ;
Ls = [1, 1] ;
Ls = [2, 2] ;
false.
As an alternative to using reification you could also use additional arithmetic constraints for "capturing" equality in an expression:
example([X,Y]) :-
X in 1..2,
Y in 1..2,
Cost #= max(3*(1-min(1,abs(X-1))) + 5*(1-min(1,abs(Y-1))),
3*(1-min(1,abs(X-2))) + 5*(1-min(1,abs(Y-2)))),
labeling([min(Cost)], [X,Y]).
Note that the expression inside Cost #= max(...)
can be slightly simplified.
Sample use:
?- example(Ls).
Ls = [1,2]
; Ls = [2,1]
; Ls = [1,1]
; Ls = [2,2]
; false.
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