Prolog - Connectivity Graph Beginner - prolog

Question

I am a beginner in Prolog and I have a task to do.
I need to check if the graph is connected.
For now I have that...
graph(
[arc(a,b)],
[arc(a,f)],
[arc(b,c)],
[arc(c,d)],
[arc(c,e)],
[arc(e,d)],
[arc(f,c)],
[arc(f,e)],
[arc(f,g)],
[arc(g,c)],
[arc(c,a)]).
edge(X,Y):-arc(X,Y);arc(Y,X).
path(X,Y):-edge(X,Y).
path(X,Y):-edge(X,Z),path(Z,Y).
triangle(X,Y,Z):-arc(X,Y),arc(Y,Z),arc(Z,X).
cycle(X):-arc(X,Y),path(Y,X).
connectivity([]):-forall(member(edge(X,Y)),path(X,Y)).
Check:
connectivity(graph).
upper I have arc(x,y) and I need check if every pair is connected.
Could u help me ?

Answer 1

Since you changed the question after I was almost done I will post what would solve the question before the change and you can figure out how to change it to meet your update.
arc(a,b).
arc(a,f).
arc(b,c).
arc(c,d).
arc(c,e).
arc(e,d).
arc(f,c).
arc(f,e).
arc(f,g).
arc(g,c).
arc(c,a).
edge(X,Y) :-
arc(X,Y), !.
edge(X,Y) :-
arc(Y,X).
path_prime(Visited,X,Y) :-
\+ member(X,Visited),
edge(X,Y), !.
path_prime(Visited,X,Y) :-
\+ member(X,Visited),
edge(X,Z),
path_prime([X|Visited],Z,Y).
path(X,X) :-
ground(X), !.
path(X,Y) :-
path_prime([],X,Y).
nodes(Nodes) :-
setof(A,B^arc(A,B),Starts),
setof(B,A^arc(A,B),Ends),
union(Starts,Ends,Nodes).
connected(X,Y) :-
nodes(Nodes),
member(X,Nodes),
member(Y,Nodes),
path(X,Y).
The first thing that has to be done is to get a list of the unique nodes which will be a set.
This can be done using
nodes(Nodes) :-
setof(A,B^arc(A,B),Starts),
setof(B,A^arc(A,B),Ends),
union(Starts,Ends,Nodes).
Notice that both the start and the end node of an arc are done separately. In particular notice that the node d is only in the destination of an arc.
Since you included edge(X,Y):-arc(X,Y);arc(Y,X). in your question, this indicated that the arcs should not be directional and so it is possible to get cycles. To avoid the cycles the list of visited nodes is added to the argument list and checked before proceeding.
As no test cases or examples of a correct solution were given, some times a node connected to itself is valid and so the clause
path(X,X) :-
ground(X), !.
was added.
This is by no means an optimal or best way to do this, just to give you something that works.
Partial run
?- connected(X,Y).
X = Y, Y = a ;
X = a,
Y = b ;
X = a,
Y = c ;
X = a,
Y = d ;
X = a,
Y = e ;
X = a,
Y = f ;
X = a,
Y = g ;
X = b,
Y = a ;
X = Y, Y = b ;
X = b,
Y = c ;
...
As I often comment, you should do problems with pen an paper first before writing code. If you don't know exactly what the code will be before you start typing the first line of code then why are you typing in code?
Questions from comments:
And setof ,union ,whats mean? Im rly beigneer and I don't understand that language and predicates.
setof/3 collects all of the values from arc/2. Since only one of the two values is needed, ^ tells setup/3 not to bind the variable in the Goal, or in beginner terms to just ignore the values from the variable.
union/3 just combines the to sets into one set; remember that a set will only have unique values.