I'm trying to define the inheritance-check predicate is_a/2
in Prolog, but so far all my trials failed.
The is_a(X, Y)
predicate should return true whenever Y is a superclass of X. For example:
object(bare).
object(mammal).
object(animal).
object(bird).
is_a(bare, mammal).
is_a(mammal, animal).
is_a(bird, animal).
is_a(X, Y):- <definition goes here>.
The definition should go such that the following query will return true:
?- is_a(bare, animal).
true.
I tried to define it the obvious way, but I got stuck in infinite loops:
开发者_开发技巧is_a(X, Y):- X\==Y, object(X), object(Y), object(Z), is_a(X, Z), is_a(Z, Y).
Any suggestions?
One way to avoid the infinite loop, is to add a predicate, which shows "direct" inheritance (no transitive), namely direct/2
. Then you could write something like this:
object(bare).
object(mammal).
object(animal).
object(bird).
direct(bare, mammal).
direct(mammal, animal).
direct(bird, animal).
isa(X, Y) :- object(X), object(Y), direct(X, Y).
isa(X, Y) :- object(X), object(Y), object(Z), direct(X, Z), isa(Z, Y).
Then you get:
?- findall(X, isa(X, animal), L).
L = [mammal,bird,bare] ? ;
no
I'm not sure this is exactly what you ask for though.
Something like
is_a(X, X).
is_a(X, Y) :- X \== Y, is_a_1(X, Z), is_a(Z, Y).
is_a_1(bear, mammal).
is_a_1(mammal, animal).
is_a_1(bird, animal).
Edit: Same idea as electrologos3's answer, who tried harder to keep it like your original code.
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