How to express a transitive relationship

问题:

I want to express a transitive relationship. If A references B and B references C then A references C. I have this:

proj(A).
proj(B).
proj(C).
ref(A,B).
ref(B,C).

When I query using proj(A) I obtain:

[46] ?-proj(A).
A = _639

What does "_639" mean? I expected a yes or no and got that strangeness. I need to add a rule to say:

ref(A,C). and get YES. Ideally, if possible, I would like to show how this relationship came about: (A => B => C).

回答1:

The _639 is an uninstantiated, anonymous variable. Your "facts" have variables rather than atoms. For example:

proj(A).   % This has a variable A and means "any A is a project"

So if I query:

:- proj(X).
X = _blah    % anonymous variable: anything is a project!

You need atoms:

proj(a).
proj(b).

Which results in the query:

:- proj(X).
X = a ;
X = b 

If you have:

ref(a,b).
ref(b,c).

Then, the simplest way in Prolog to express a transitive property is by declaring a rule (or what's known as a predicate in Prolog):

ref(A,C) :- ref(A,B), ref(B,C).

This says that, ref(A,C) is true if ref(A,B) is true, AND ref(B,C) is true.. Running the query:

:- ref(a,c).
true ;
Out of stack

Or:

:- ref(a,X).
X = b ;
X = c ;
Out of stack

So it sounds logical but has an issue: you can get into a loop due to the self-reference. A simple way around that is to make the rule name different than the fact:

refx(A, B) :- ref(A, B).
refx(A, C) :- ref(A, B), refx(B, C).

Now if I query:

:- refx(a, b).
true ;
no

:- refx(a, c).
yes

:- refx(a, X).
X = b ;
X = c
yes

Etc.

There are still cases where this could have termination issues, however, if the facts contain reflexive or commutative relationships. For example:

ref(a,b).
ref(b,a).
ref(b,c).

In this case, a query to refx(a, b) yields:

| ?- refx(a, b).
true ? ;
true ? ;
true ? ;
...

As @lambda.xy.x points out, this could be resolved by tracking where we've been and avoiding repeat "visits":

refx(X, Y) :-
    refx(X, Y, []).

refx(X, Y, A) :-
    ref(X, Y),
    \+ memberchk((X, Y), A).   % Make sure we haven't visited this case
refx(X, Y, A) :-
    ref(X, Z),
    \+ memberchk((X, Z), A),   % Make sure we haven't visited this case
    refx(Z, Y, [(X,Z)|A]).

Now we terminate with refx(a,b) and succeed once:

| ?- refx(a,b).
true ? ;
no
| ?-

And refx(X, Y) produces all of the solutions (albeit, some repeats due to succeeding more than once):

| ?- refx(X, Y).

X = a
Y = b ? ;

X = b
Y = a ? ;

X = b
Y = c ? ;

X = a
Y = a ? ;

X = a
Y = c ? ;

X = b
Y = b ? ;

X = b
Y = c ? ;

(2 ms) no