Question

I need a predicate routing which gives all the cities between start & end. For example:

path(chicago,atlanta).
path(chicago,milwaukee).
path(milwaukee,detroit).
path(milwaukee,newyork).
path(chicago,detroit).
path(detroit, newyork).
path(newyork, boston).
path(atlanta,boston).
path(atlanta, milwaukee).

?- routing(chicago,newyork,X).
X=[chicago,milwaukee,newyork];
X=[chicago,detroit,newyork];
X=[chicago,milwaukee,detroit,newyork];
X=[chicago,atlanta,milwaukee,newyork];
X=[chicago,atlanta,milwaukee,detroit,newyork]

I have tried this, and keep coming back to it.

routing(FromCity,ToCity,[FromCity|ToCity]) :-
  path(FromCity,ToCity).

routing(FromCity,ToCity,[FromCity|Connections]) :- 
  path(FromCity,FromConnection), 
  path(FromConnection,ToConnection),
  path(ToConnection,ToCity),
  routing(ToConnection,ToCity,Connections).

routing(FromCity,ToCity,[]).

but it just keeps giving

X=[chicago,milwaukee,newyork];
X=[chicago,chicago,newyork];
X=[chicago,chicago,chicago,newyork]
...
..

Can some one please point me in the right direction …

Answer 1

If you are sure (by definition) that your graph is acyclic, you can simplify your rule, exploiting Prolog depth first search:

routing(FromCity, ToCity, [FromCity, ToCity]) :-
  path(FromCity, ToCity).

routing(FromCity, ToCity, [FromCity|Connections]) :- 
  path(FromCity, ToConnection),
  routing(ToConnection, ToCity, Connections).

This find all availables paths on backtracking:

?- routing(chicago,newyork,X).
X = [chicago, atlanta, milwaukee, newyork] ;
X = [chicago, atlanta, milwaukee, detroit, newyork] ;
X = [chicago, milwaukee, newyork] ;
X = [chicago, milwaukee, detroit, newyork] ;
X = [chicago, detroit, newyork] ;
false.

Note the difference between the first and second pattern of list construction: [FromCity, ToCity] vs [FromCity|Connections]. This because Connections will be a list, while ToCity will be an atom, when the rule will succeed.

If your graph contains cycles, this code will loop. You can refer to another answer for a simple schema that handles this problem.