Prolog, Determine if graph is acyclic

问题

I need to define a predicate acyclic/1 that takes a graph in as input and determine if that graph is acyclic. So from my understanding

graph1(a,b).
graph1(b,c). 
graph1(c,a). 

Will return no and

graph2(a,b).
graph2(b,c). 

will return yes

I made a predicate to determine if 2 nodes in a graph are connected and if so they will return yes.

isConnected(X,Y) :- a(X,Z), isConnected(Z,Y).

is there a way that I can use this to determine if a graph is acyclic?

I do not want to use any predefined predicates.

回答1:

Using closure0/3:

:- meta_predicate acyclic(2).
:- meta_predicate cyclic(2).

acyclic(R_2) :-
   \+cyclic(R_2).

cyclic(R_2) :-
  closure0(R_2, X0,X),
  call(R_2, X,X0).

?- acyclic(graph2).
true.

?- acyclic(graph1).
false.

cyclic/1 succeeds if the following exists:

1. an acyclic connexion from X0 to X, thus:

   closure0(R_2, X0,X) or more verbosely:

   call(R_2, X0,X1), call(R_2, X1,X2), call(R_2, X2,X3), ..., call(R_2, Xn,X) with X0,X1,...,Xn all pairwise different

2. one edge back

   call(R_2, X,X0).

so that is a cycle. In other words, a cyclic graph is a graph that contains at least one cycle. And that cycle consists of an acyclic part plus one edge back. And it is only this edge back that makes this a cycle.

回答2:

Your recursive predicate isConnected/2 misses the base case:

isConnected(X,Y) :- graph1(X,Y).

(assuming we are checking graph1, of course).

Anyway, you cannot use isConnected/2, since Prolog will loop on cyclic graphs.

来源：https://stackoverflow.com/questions/26962312/prolog-determine-if-graph-is-acyclic