Fastest way between two cities - prolog

Question

I need to find the fastest way to travel from one city to another. I have something like
way(madrid, barcelona, 4).
way(barcelona, paris, 5).
way(madrid, londres, 3).
way(londres,paris,1).
I have come up with a predicate shortway(A,B,C,D) where C is the list of towns between A and B and D the distance.
so I have
shortway(A,B,C,D):-
way(A,B,_,_) , (A,_,C,D). D<C.
shortway(A,_,C).
I trying my best but I really cant get it to work!

Answer 1

You have a bunch of problems with your code! First of all, way/3 has arity 3, not 4, so calling way(A,B,_,_,) is clearly not going to do what you think. Second, I have no idea what you're trying to do with (A,_,C,D). The period after this signifies the end of the predicate! So the next line, D<C. is just a free-floating query that cannot be fulfilled. And then shortway(A,_,C) is basically a fact, with three singletons, but it would define a shortway/3 clause when the previous one is a shortway/4 clause.
There really isn't enough that's on the right track here to try and recover. It looks here like you are extremely confused about even the basics of Prolog. I would strongly encourage you to go back to the beginning and start over. You can't rush Prolog! And this code looks like you're trying to make a combustion engine by banging rocks together.

Answer 2

I wrote some line of code to help you, but as https://stackoverflow.com/users/812818/daniel-lyons said, it's better than you learn something easier before.
To solve your problem I advice you to read, at least, the first 3 chapters of this book: http://www.learnprolognow.org/lpnpage.php?pageid=online and do the practical session at paragraph 3.4.
Then, you could take a look at my code (you can find some explenation of it here:Out of local stack error in Prolog route planner .
Here the code
way(madrid, barcelona, 4).
way(barcelona, paris, 5).
way(madrid, londres, 3).
way(londres,paris,1).
shortway(From, To):- findall(Journey, travel(From, To, Journey, _) , Travels_list),
findall(Total_distance, travel(From, To, _, Total_distance) , Distances_list),
min_member(Y, Distances_list), find_minimum_index(Y, Distance_list, 1, Distance_index),
find_journey(Distance_index, Travels_list, 0, Shortest_path),
format('The shortest path is ~w', [Shortest_path]).
travel(From, To, Journey, Total_distance) :- dif(From, To),
AccDistance is 0,
path(From, To, [From], Journey, AccDistance, Total_distance).
path(From, To, Passed_cities, go(From, To), AccDistance, Total_distance) :- way(From, To, Way_distance),
Total_distance is AccDistance + Way_distance.
path(From, To, Passed_cities, go(From, Intermediate, GO), AccDistance, Total_distance) :- way(From, Intermediate, Way_distance),
dif(Intermediate, To),
\+ member(Intermediate, Passed_cities),
NewAccDistance is AccDistance + Way_distance,
path(Intermediate, To, [Intermediate|Passed_cities], GO, NewAccDistance, Total_distance).
min_member(Min, [H|T]) :- min_member_(T, H, Min).
min_member_([], Min, Min).
min_member_([H|T], Min0, Min) :-
( H >= Min0
-> min_member_(T, Min0, Min)
; min_member_(T, H, Min)
).
find_minimum_index(X, [], N, I) :- fail.
find_minimum_index(X, [X|T], N, I) :- I is N, !.
find_minimum_index(X, [H|T], N, I) :- H \= X, increment(N, N1), find_minimum_index(X, T, N1, I).
find_journey(I, [H|T], N, Elemento) :- N = I, Elemento = H, !.
find_journey(I, [H|T], N, Elemento) :- N \= I, increment(N, N1), find_journey(I, T, N1, Elemento).
increment(X, X1) :- X1 is X+1.
Then you call, for example
?:- shortway(madrid,paris).
and it will return
"The shortest path is go(madrid, londres, go(londres,paris))"
which distance is, 4
rather than
go(madrid, barcelona, go(barcelona, madrid)
which distance is 9.
Summing up: calling shortway/2, with the predicates findall/3 you'll find the lists of all possible pathes and their relative distances, respectively, then you'll browse the list of the distances to find the index of the minimum element and so using it to find the shortest path from the list of all pathes found previously.