CODEWITHSUNDEEP
listfilterprolog
Wirsing
Wirsing

Reputation: 45

Prolog - Filter List

So I have this homework due tommorow. I have to filter every nth element of a list and return it as a list. So for example:

?- everyNth(3,[a,b,c,d,e,f],Rs). Rs = [c,f].

My Idea was basically:

everynth(N, [X|Xs], L) :- everynth(N, [X|Xs], N, L).
everynth(N, [], C, L).
everynth(N, [X|Xs], 0, [X]) :- everynth(N, Xs, N, [X]).
everynth(N, [X|Xs], C, L) :- C1 is C -1,
  everynth(N,Xs,C1,L).

But it does not work this way, because in the third row it tries to match X and the return X and the Count 0 the second time it goes there.

Upvotes: 2

Views: 606

Answers (2)

applicative_functor
applicative_functor

Reputation: 4976

everynth(_, _, [], R, R).
everynth(1, M, [X|Xs], Z, R) :- append(Z, [X], Z1),  everynth(M, M, Xs, Z1, R).              
everynth(N, M, [_|Xs], Z, R) :- N > 1, N1 is N - 1, everynth(N1, M, Xs, Z, R).




?- everynth(3, 3, [a,b,c,d,e,f], [], Rs).
Rs = [c, f] .

Upvotes: 1

gusbro
gusbro

Reputation: 22585

You are almost there. Check these modifications:

everynth(N, L, NL) :- everynth(N, L, N, NL).

everynth(_, [], _, []).
everynth(N, [X|Xs], 1, [X|NXs]) :- everynth(N, Xs, N, NXs).
everynth(N, [_|Xs], C, NXs) :- C1 is C-1, C1>0,
  everynth(N,Xs,C1,NXs).

The first clause of everynth/4 is the termination of the recursion. It should give an empty list when there are no more items in the input list.

The second clause of everynth/4 deals with the nth item, it has to put the input item in the output list and keep processing the remaining items restarting your item counter.

And the third clause of everynth/4 deals with items which are not the nth element, so you have to skip the item, decrement the counter and continue with the remaining items.

Upvotes: 3

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