Prolog list and filtering

Question

I have a problem with a Prolog exercise. I'm trying to study Computer Science alone at my home, and I'm doing a series of exercises I've taken from the Internet.

The question is the following;

Construct a predicate called filteringPairsAtoms/3 so that given an atom (first argument), and a list of pairs, unify a third parameter with the filtered list of pairs by selecting only the pairs that have the first component as the atom of the first argument.

Example

filteringPairsAtoms(sA,[[basA,absAb],[ab,bbsA],[sA,abbsB],[bsA,sAsB],[sA,bb]],X)

must result in

X = [[sA,abbsB],[sA,bb]]

I'm trying to solve it but I don't have anyone to ask, because I'm learning alone, and don't have a professor to write to or something.

Any help is welcome.

Thanks!!!

What have you tried so far? Please provide the steps that you have tried and any specific questions that you may have from that. — Seeta Somagani
I've used recursive search on the list, but i dont have any positive results yet. — Emily_1990
you can show us anything you got. any code, any attempt is OK. so we can see where's your difficulty. otherwise it's shooting in the dark. — Will Ness
@Emily_1990 It's a bit off-topic, but since you study on your own, you can first spend time on more structured and simple tasks like the classical Ninety-Nine Prolog Problems. Then you can add helpful details to your questions like "I don't understand this problem, but I do understand these and these simpler related problems" (and in this case even a rubber duck may help you no less than a professor.) — Kirill Bulygin
Adding to Kirill, there are a lot of great Prolog books for self-study, and the language hasn't changed much in a few decades, so even fairly old books can be helpful still. I would recommend Programming in Prolog by Clocksin & Mellish or Art of Prolog by Sterling & Shapiro. — Daniel Lyons

Will Ness Will Ness · Accepted Answer · 2018-07-05T16:21:02

You can start from what you already have:

filteringPairsAtoms(sA, [[basA,absAb],[ab,bbsA],[sA,abbsB],[bsA,sAsB],[sA,bb]], X) :-
    X = [[sA,abbsB],[sA,bb]].

That's as good a start as any. But we know more: it must be the case that

filteringPairsAtoms(sA, [[sA,abbsB],[bsA,sAsB],[sA,bb]], X) :-
    X = [[sA,abbsB],[sA,bb]].

%% and
filteringPairsAtoms(sA, [[bsA,sAsB],[sA,bb]], X) :-
    X = [[sA,bb]].

%% and
filteringPairsAtoms(sA, [[sA,bb]], X) :-
    X = [[sA,bb]].

%% and
filteringPairsAtoms(sA, [], X) :-
    X = [].

Comparing the sequential clauses, we also notice that it must be

filteringPairsAtoms(sA, [[sA,abbsB] | [[bsA,sAsB],[sA,bb]] ], X) :-
    sA = sA,
    X = [[sA,abbsB] | Y],
    filteringPairsAtoms(sA, [[bsA,sAsB],[sA,bb]], Y).

(this is just the first two clauses from above, combined); and

filteringPairsAtoms(sA, [[bsA,sAsB] | [[sA,bb]] ], X) :-
    dif( sA, bsA),
    X = Y,
    filteringPairsAtoms(sA, [[sA,bb]], Y).

All that's left for you to do, is generalize this by replacing concrete terms with logical variables. Like this:

filteringPairsAtoms(SA, [Head | Tail ], X) :-
    Head = [BsA, _]
    dif( SA, BsA),
    X = Y,
    filteringPairsAtoms(SA, Tail, Y).

and carry this out fully to the end.

Prolog list and filtering

1 Answers