How to check if the paths are connected between rooms

Question

I have the following facts that build ne of the wings of my dungeon.

path(room1,e,room2).
path(room2,w,room1).
path(room2,e,room5).
path(room5,w,room2).
path(room5,e,room6).
path(room6,w,room5).
path(room2,n,room3).
path(room3,s,room2).
path(room3,e,room4).
path(room4,w,room3).
path(room4,s,room5).
path(room5,n,room4).
path(room5,s,room7).
path(room7,n,room5).
path(room7,e,room8) :- at(key, in_hand).
path(room7,e,room8) :- write('The door appears to be locked.'), nl, fail.
path(room8,w,room7).
path(room2,s,room9).
path(room9,n,room2).
path(room9,s,room10).
path(room10,n,room9).
path(room10,w,room11).
path(room11,e,room10).
path(room11,s,room12).
path(room12,n,room11).

now i want to check if i one room is connected to another and then perhaps display one of the possible paths. i tried to do something like this connected(X,Y):-path(X,_,Z),path(Z,_,Y). but it keeps giving me false, what am i doing wrong.

Answer 1

Try something like

connect(Bgn,End,Path) :-    % to find a path between two nodes in the graph
  connected(Bgn,End,[],P) , % - we invoke the helper, seeding the visited list as the empty list
  reverse(P,Path)           % - on success, since the path built has stack semantics, we need to reverse it, unifying it with the result
  .

connected(X,Y,V,[X,Y|V]) :- % can we get directly from X to Y?
  path(X,_,Y)               % - if so, we succeed, marking X and Y as visited
  .                         %
connected(X,Y,V,P) :-       % otherwise...
  path(X,_,T) ,             % - if a path exists from X to another room (T)
  T \= Y ,                  % - such that T is not Y
  \+ member(X,V) ,          % - and we have not yet visited X
  connected(T,Y,[X|V],P)    % - we continue on, marking X as visited.
  .

You'll note the test for having visited the room before. If your graph has cycles, if you don't have some sort of test for having previously visited the node, you'll wind up going around and around in circles...until you get a stack overflow.