问题
library(clpb) is currently available in SICStus (original version), and SWI (by mat). Let me come to the essence quite rapidly:
?- X = 1+1, sat(X), X = 1+1.
X = 1+1.
?- sat(X), X = 1+1.
false.
So this is a similar problem as it exists in the default state of library(clpfd).
What to do in such a situation?
Update: In library(clpfd) of mat, there is now the functor # /1 for this purpose. Ideally, accompanied with an operator declaration op(100,fx,#), we now can write:
?- X = 1+1, #X #= Y.
ERROR: Type error: `integer' expected, found `1+1' (a compound)
To ensure full algebraic properties one has to declare:
:- set_prolog_flag(clpfd_monotonic, true).
Now, variables that are left ambiguous (thus, being either integers only, or expressions) produce instantiation errors:
?- 1 + 1 #= Y.
ERROR: Arguments are not sufficiently instantiated
?- 1 + 1 #= #Y.
Y = 2.
回答1:
Yes, same in CLP(FD) and CLP(B):
Situation
CLP(FD), the (+)/2 acts as addition in finite domain:
?- X = 1+1, X #= 2.
X = 1+1.
?- X #= 2.
X = 2
?- X #= 2, X = 1+1.
false.
CLP(B), the (+)/2 acts as disjunction in boolean domain:
?- X = 1+1, sat(X)
X = 1+1
?- sat(X)
X = 1
?- sat(X), X = 1+1.
false
But now watch out, such things happen already in ordinary prolog:
?- X = 3, X \= 4.
X = 3.
?- X \= 4, X = 3.
false.
Problem
So what should we tell the students? I don't think that CLP(FD) or CLP(B) is to blame here.
Its more ingrained into (=)/2, and the impossibility to hook into (=)/2 so that it does more than a verify_attribute/3.
In an ideal world (=)/2 would work algebraically, and the X=1+1 wouldn't fail, the system would see that 1+1 equals 2 (FD case) respectively 1+1 equals 1 (B case).
But for this the Prolog system would need types as a start, otherwise the (=)/2 cannot interpret the (+)/2, it has two different interpretations in our example.
Solution
The solution is to refrain from using (=)/2 where particular types are needed. Hereby I recommend reading:
Constraint Logic Programming using ECLiPSe
Krzysztof R. Apt and Mark Wallace,
May 16, 2006, Full Text PDF
The above paper highlights a little bit the motivation of the hash sign (#) as a prefix for operators such as (=)/2, (<)/2, etc.. Its already a type annotation. There is also another prefix in this paper, namely the dollar sign ($).
In ECLiPSe Prolog the hash sign (#) implies the constraint integers/1, and the dollar sign ($) implies the constraint reals/1. These constraints can also be used independently as in the following example shows:
?- integers([X,Y]), X = 3, Y = 4.5.
No
?- reals(X), X = [1,2.1,a].
No
Coming back to your problem of non commutativity of (=)/2. Although (=)/2 is non-commutativ, the (#=)/2 and its boolean brother are of course commutative. We have:
CLP(FD), the (+)/2 acts as addition in finite domain, and with (#=)/2 commutativiy is there:
?- X #= 1+1, X #= 2.
X = 2
?- X #= 2, X #= 1+1.
X = 2
CLP(B), the (+)/2 acts as disjunction in boolean domain, and with sat/1 and (=:=)/2 commutativity is there:
?- sat(X=:=1+1), sat(X).
X = 1.
?- sat(X), sat(X=:=1+1).
X = 1
A verify_attribute/3 hook will only wake up constraints which can additionally fail a Herbrand domain unification. But they cannot evaluate terms on the fly before assigning them to a Prolog variable.
Only the woken up constraints might assign values to other variables. But for example the current SICStus Spec of verify_atribute/3 even recommends not doing this directly, and only in the continuation queue. See recent SWI-Prolog group discussion.
So long talk, short explanation: Never forget that (=)/1 only works for the Herbrand domain in Prolog, even if we hook in constraints. I dont say a Prolog system couldnt work differently, but this is the state of the Art of a Herd or Prolog systems.
Bye
来源：https://stackoverflow.com/questions/33697654/undesirable-properties-of-clpb