Question

Sample database:

 any(a,b,2). 
 any(b,c,2). 
 any(c,d,3). 
 any(d,e,1). 
 any(e,f,3).

Key: (Station1,Station2,Time).

The question:

enter (station1,time) into prolog and have it return all of the stations within reach.

For example:

If you enter: (b,2).

The output should be: ‘a’ and ‘c’

This is because from station ‘b’ in ‘2’ minutes, you will be able to get to stations ‘a’ and ‘c’ as they are within reach of the time entered.

---

I have tired using lists and recursion but no luck, any help/suggestions?

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reachable(Station1, Limit, Result) :-
    reachable(Station1, Limit, 0, 0, Result).

reachable(Station1, Visited, TimeSpent, Limit, Result) :-
    overground(Station1,Station2,Time),
    Visited is Limit - Time,
    Limit =< TimeSpent,
    Result = [Station2];
    overground(Station1, Waypoint, Time),
    NewVisited is Visited - Limit,
    NewTimeSpent is TimeSpent - NewVisited,
    reachable(Waypoint,Station2, NewTimeSpent, NewVisited,Result).

<<< something I tried but doesnt seem to work X_X

Answer 1

So, let’s look at the problem :

1. you need to be able to find the station in “both ways” from your predicate, ie any(MyStation, Y) or any(Y, MyStation), so you’ll need a disjunction when using any/3.
2. you need to find a station that isn’t directly in reach of the departure station if necessary, through recursion.
3. since you have to use recursion, you have to keep track of time already spent at a given point of the recursion with an accumulator, in order to know if a station at a given point still satisfies the time limit.
4. since you have to use recursion, you have to keep track of already visited stations, in order not to fall into an infinite loop, or to give twice the same result just through one more loop in a circuit this will be done thanks to an accumulator this time again.

General look of the solution :

First step, calling the working predicate :

reachable(From, Limit, Result) :-
    ...

Where ... should be a call to reachable/5 (this one + the 2 accumulators).

Second step, the working predicate :

reachable(From, Visited, TimeSpent, Limit, Result) :-
    ...

Where ... should satisfy those guidelines :

• first you have to find a station that is related to From thanks to any/3
• then you have to check if this station is in Visited or not
• then you have to write a disjunction with two cases (or write two clauses) :
  • the final case ie you check that time spent is under the limit and unify Result with the appropriate variable
  • the recursive case ie you call reachable with a new TimeSpent that you calculate, a new Visited that you update and a new departure station.

If you want more precise advice, I’d advise you to publish the work you’ve already done so that we can work from there !