Prolog相互递归谓词

Question

Good evening,

I am newbie in Prolog and I have a problem to solve. I tried to do it myself, but it does not work for me. I searched my Prolog books for any help, but didn't find anything either. This question is really confusing me.

The question is:

We need to calculate the early start time for each task (earliest possible time that it could start given its prerequisites must be accomplished first).

• The early start time of a task is given by the latest early finish of its prerequisites.

• The latest early finish of any list of tasks can be calculated by starting at zero and successively taking the maximum with the early finish time of each task.

• The early finish time of a task is given by adding its duration to its early start time.

Write definitions in Prolog for the predicates e_start, l_e_finish, e_finish. These predicates are mutually recursive.

The tasks and their times are defined as follows:

duration(Task, Time),
duration(b, 10),
duration(k,5),
duration(a,2).

The prerequisites are defined as follows:

prerequisites(Task, PreqTask),
prerequisites(k,[b]),
prerequisites(b,[]),
prerequisites(a,[k]).

I tried to solve it but I think I need more explanation on how to do it properly as I just cannot get it right.

My solution is:

e_start(Task,Start):-
   prereqs(Task, X),
   l_e_finish(X,Start).

l_e_finish(Task,Finish) :-
    e_finish(Task,Finish),
    l_e_finish(Task,Finish1),
    duration(Task, Finish),
    max(Finish,Task,Finish1).

e_finish(Task,Finish):-
    duration(Task, Time),
    e_start(Task,Finish),
    Finish is Time+Finish.

Any help would me much appreciate. Thank you fellow coders!

Answer 1

Let's start by considering the case where you have no prerequisites. In that case, your earliest finish is simply your duration:

early_finish(T, Finish) :- 
    duration(T, Finish).

What if you have prerequisites? Then your early finish is your duration plus the latest of those, which would look like this:

early_finish(T, Finish) :-
    prerequisites(T, Prerequisites),
    maplist(early_finish, Prerequisites, PrerequisiteFinishes),
    maxlist(PrerequisiteFinishes, LastFinish),
    duration(T, Duration),
    Finish is LastFinish + Duration.

I assume you have a maxlist/2 predicate here that gives you the largest item in a list. If you define maxlist/2 to give 0 for the empty list and you guarantee that all tasks have a prerequisite/2 definition, then you're done with just the one clause. If not, you will have to find a way to combine the logic of these two clauses.