Aggregate solution over multiple facts

Question

Trying to create a predicate (timePeriod/2) that calculates the time period between two dates for a specific fact. I've managed to do this by myself, but face issues when 'other answers' exist in the same list (i.e. easier to explain with examples).

I have the following knowledge-base facts;

 popStar('Jackson',1987,1991).
 popStar('Jackson',1992,1996).
 popStar('Michaels',1996,2000).
 popStar('Newcastle',2000,2007).
 popStar('Bowie',2008,2010).

And the following function, calculates the time between dates for a specific fact (as per below).

Predicate (timePeriod/2) -

 timePeriod(PS,X) :-
    bagof((Name,Start,End),popStar(Name,Start,End),PSs),X is End-Start+1)

Using Bowie as an example; it returns X=3 (which is correct).

However, when there is repetition in the list, with more than one answer available, the predicate just states 'false'. Using the facts 'Jackson' as an example, I want to be able to calculate both of the time periods for both facts; at the same time.

So, if the predicate would work for both of the Jackson facts, the predicate timePeriod would state X=10.

Would really appreciate if anyone could suggest what to change in order for this to work correctly.

Thanks.

Answer 1

SWI-Prolog has a nice library to handle aggregation: it builds upon standard 'all solutions' predicates like findall/3,setof/3,bagof/3, so you should first grasp the basic of these (as Boris explained in his answer). With the library, a single query solves your problem:

timePeriod(PS,X) :-
    aggregate(sum(P), B^E^(popStar(PS,B,E),P is E-B+1), X).

Answer 2

You probably don't quite understand what foreach/3 does. I don't think I fully understand foreach/3 either. I know for sure that it is not the same as say:

for each x in xs:
    do foo(x)

Another thing: "tuples" in Prolog are not what you might expect, coming from a language like Python or Haskell. This: (a,b,c) is actually this: ','(a,','(b,c)). Much better is to use a flat term, the generic form would be triple(a,b,c). For a pair, the idiom is First-Second.

So, you can simplify your call to bagof/3 to this:

?- bagof(From-To, pop_star(Name, Start, End), Ts).
Name = 'Bowie',
Ts = [2008-2010] ;
Name = 'Jackson',
Ts = [1987-1991, 1992-1996] ;
Name = 'Michaels',
Ts = [1996-2000] ;
Name = 'Newcastle',
Ts = [2000-2007].

Once you have a list as above, you need to sum the differences, which would be maybe something like:

periods_total(Ps, T) :-
    maplist(period_length, Ps, Ls),
    sum_list(Ls, T).

period_length(From-To, Length) :-
    Length is To - From + 1.

And then you can query like this:

?- bagof(From-To, pop_star('Jackson', From, To), Ps), periods_total(Ps, T).
Ps = [1987-1991, 1992-1996],
T = 10.

?- bagof(From-To, pop_star(Name, From, To), Ps), periods_total(Ps, T).
Name = 'Bowie',
Ps = [2008-2010],
T = 3 ;
Name = 'Jackson',
Ps = [1987-1991, 1992-1996],
T = 10 ;
Name = 'Michaels',
Ps = [1996-2000],
T = 5 ;
Name = 'Newcastle',
Ps = [2000-2007],
T = 8.