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问题:
Consider a (meta-logical) predicate var_in_vars(Var, Vars)
which takes a variable Var
and a list of variables Vars
and succeeds if Var
occurs in Vars
. So we do not need to ensure that Var
is a variable, nor that Vars
is a list of variables.
What is the most compact and canonical way to express this in ISO Prolog? Here is an overview of the built-ins in ISO/IEC 13211-1:1995 including Cor.2:2012.
?- var_in_vars(V, [U,V,W]).
true.
?- var_in_vars(V, [X,Y,Z]).
false.
回答1:
One possibility:
var_in_vars(V, Vs) :- \+ unify_with_occurs_check(V, Vs).
and shorter:
var_in_vars(V, Vs) :- \+ subsumes_term(V, Vs).
EDIT: Future readers, please take into account the context of the question, which is a specific compactness challenge involving the expressivity of ISO predicates under given circumstances.
In other circumstances, you will likely benefit more from a definition like:
var_in_vars(V, Vs) :-
must_be(list, Vs),
once((member(X, Vs), V == X)).
回答2:
this definition passes the tests, but... do I miss some subtlety ?
var_in_vars(V, [H|_]) :- V == H, !.
var_in_vars(V, [_|T]) :- var_in_vars(V, T).
回答3:
And here goes another one, although a bit more complex:
var_in_vars(V, Vs) :-
term_variables(Vs+V, Ws),
Ws == Vs.
So this relies on the precise order how variables are visited. And since this is well defined in the standard we can rely that they
... appear according to their first occurrence in left-to-right traversal ...
A drawback of this definition is that it has minimum cost proportional to the length of Vs
. But since an internal traversal is often quite efficiently implemented, this is not such a problem.
It has one big advantage: It only succeeds if Vs
is a list of variables.
回答4:
An alternative solution is:
var_in_vars(V, Vs) :-
\+ (V = Vs, acyclic_term(Vs)).
But the solutions by @mat are better and more elegant and the solution by @CapelliC was for a long time the most portable one (the subsumes_term/2
predicate was only standardized recently and not all systems provided the unify_with_occurs_check/2
predicate).
回答5:
The solution @false can be simplified to:
var_in_vars(V, Vs) :-
term_variables(Vs+V, Vs).
When V
is a member of the Vs
list, the second argument returns Vs
(due to the left-to-right traversal of the Vs+V
term). When V
is not a member of Vs
, the second argument returns a list that have one more element than Vs
and thus cannot unify with it. Although there's an implicit unification in the second argument, in neither case there's a danger of creating a cyclic term. I.e. unification being STO is not a problem in this simplified solution.
But is the simplification worth it w.r.t. performance? The use of equality, (==)/2
have the potential of failing earlier and thus making the original solution faster.
回答6:
Solution Synthesis
Solution : Short
var_in_vars(V, Vs) :- \+ subsumes_term(V, Vs).
Alternative 1 : Simple
var_in_vars(V, Vs) :- \+ unify_with_occurs_check(V, Vs).
Alternative 2 : Depending on circumstances, this could be more suitable
var_in_vars(V, Vs) :-
must_be(list, Vs),
once((member(X, Vs), V == X)).
Alternative 3 : More complex
var_in_vars(V, Vs) :-
term_variables(Vs+V, Ws),
Ws == Vs.
Alternative 4 : Other possibility
var_in_vars(V, [H|_]) :- V == H, !.
var_in_vars(V, [_|T]) :- var_in_vars(V, T).
Links :
Standard
Note : The context of the question, which is a specific compactness challenge involving the expressivity of ISO predicates under given circumstances.