CODEWITHSUNDEEP
prolog
Angelrina
Angelrina

Reputation: 1

How to make a list of pairs from facts in SWI-Prolog?

Assuming I have some facts like the following

person(jessica,19,usa).
person(james,18,uk).
person(eric,34,italy).
person(jake,24,france).

how can I create a predicate that creates a large list of pairs of all the names and their corresponding country like so:

?-filter(L).
L=[(jessica,usa),(james,uk),(eric,italy),(jake,france)]

Upvotes: 0

Views: 1779

Answers (2)

lambda.xy.x
lambda.xy.x

Reputation: 4998

Assuming person/3 is ground and terminates, you can implement it without setof as:

notin(_, []).
notin(X, [Y|Ys]) :-
    dif(X,Y),
    notin(X,Ys).

lt_list(_, []).
lt_list(X, [Y|Ys]) :-
    X @< Y,
    lt_list(X,Ys).

f( [ Name-Location | Rest], Acc) :-
    person(Name, _, Location),
    lt_list( Name-Location, Acc ),
    f(Rest, [Name-Location | Acc]).
f( [], Acc) :-
    \+ (person(Name,_,Location), notin(Name-Location,Acc)).

When we query f, we get our solutions:

?- f(Xs,[]).
Xs = [jessica-usa, james-uk, jake-france, eric-italy] ;
false.

I used X-Y instead of (X,Y) for better readability. The predicate notin describes an element that is not contained in a list and lt_list describes an element that is smaller than anything in the list by the standard term order.

The idea is that the first rule generates persons that I have not seen yet. Using the term order makes sure that we don't generate all permutations of the list (try replacing lt_list by notin to see what happens). The second rule makes sure we only terminate if there are no more solutions to generate. Be aware that the rule contains negation, which can have some unwanted side-effects. Most of them are filtered out by only looking at ground terms, but I have not thought well, how bad the impact is in this solution.

Upvotes: 3

Daniel Lyons
Daniel Lyons

Reputation: 22803

The best solution is this one:

?- bagof((P,C), Age^person(P,Age,C), People).
People = [(jessica, usa),  (james, uk),  (eric, italy),  (jake, france)].

This gives you the same result as findall/3, because findall/3 implicitly assumes existential quantification on all variables not present in the template ((P,C) is the template). In your case you only have one, the age variable. Notice what happens if you don't include that:

?- bagof((P,C), person(P,_,C), People).
People = [(james, uk)] ;
People = [(jessica, usa)] ;
People = [(jake, france)] ;
People = [(eric, italy)].

What happened here? The value of the second parameter was the same across each solution because we didn't inform bagof/3 that we didn't care what it was bound to or even if it was bound to just one thing. This property of bagof/3 and setof/3 (but not findall/3) sometimes turns out to be surprisingly useful, so I tend to prefer using bagof/3 over findall/3 if I only need to mark a variable or two.

It's more obvious if we add another person the same age to the database:

person(janet,18,australia).

?- bagof((P,C), person(P,Age,C), People).
Age = 18,
People = [(james, uk),  (janet, australia)] .

?- bagof((P,C), person(P,_,C), People).
People = [(james, uk),  (janet, australia)] ;

Upvotes: 3

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