Deleting first N prime numbers from a list (Prolog)

Question

Basically I want to remove the first N numbers from a list, the function that checks whether a number is prime or not seems to work well but the program itself doesn't

For example for input [2,4,5,7,6,9,11] and N = 3 I should get [4, 6, 9, 11] but I get only [4, 6, 9].

divisible(X,Y) :- 
    0 is X mod Y, !.
divisible(X,Y) :- 
    X > Y+1, 
    divisible(X, Y+1).

%isPrime function check whether or not the argument is a prime number
isPrime(2) :- true,!.
isPrime(X) :- X < 2,!,false.
isPrime(X) :- not(divisible(X, 2)).

%delFunction (input_list, N, output_list)
delFunction([],_,_).
delFunction(_,0,_).
delFunction([H|T], N, [H|Res]):-
    not(isPrime(H)), !,
    delFunction(T, N, Res).
delFunction([_|T], N, Res):-
    N1 is N-1,
    delFunction(T,N1,Res).

delFunction([2,4,5,7,6,9,11],3,X) -> [4,6,9] (which is not the correct answer)

Honestly, I don't know where I went wrong, the implementation idea seems pretty easy and straight forward, so is the code.

Also, when I run it it stops at [4] and I have to keep on clicking next to get me to the end of the execution (thus the result). Any idea how to fix it? I'm thinking maybe I need some cuts but not sure where.

P.S: I'd rather not use built-in functions (if there are any that would help in this scenario)

Yea, well, those are my problems. Final output isn't correct and it gives me multiple answers (when it should give only one, obviously). Still think it's a misuse of ! somewhere... — , Jan 13 '17 at 22:08

score 2 · Answer 1 · edited Jan 14 '17 at 00:19

Fists of all, instead of

delFunction([],_,_).

you should write

delFunction([],_,[]).

because when the input list (the left one) is empty, you have to construct the base for the output list: an empty list; with delFunction([], _, _) you don't unify the output list with the empty list so the result end with an not unified variable.

Second. Instead of

delFunction(_,0,_).

you should write

delFunction(L,0,L).

The problem is the same: when number is zero you can "copy" the input in the output; that is, you can unify they; that is delFunction(L,0,L). With delFunction(_,0,_) you don't unify and the result ends with a not unified variable.

Third. In the is-prime clause

delFunction([_|T], N, Res):-
    N1 is N-1,
    delFunction(T,N1,Res).

you should check that N is greater than zero

delFunction([_|T], N, Res):-
    N > 0,
    N1 is N-1,
    delFunction(T,N1,Res).

or, if you prefer, you should add a cut (!) in the zero clause

delFunction(L,0,L) :- !.

Just to avoid that Prolog, via backtracking, can give multiple undesired answers.

--- EDIT ---

As Guy Coder, I avoid cuts like a plague; so I propose the following solution to avoid unwanted repetitions.

delFunction([], _, []).

delFunction([H | T], 0, [H | T]).

delFunction([H | T], N, [H | Res]) :-
    N > 0,
    not(isPrime(H)),
    delFunction(T, N, Res).

delFunction([H | T], N, Res):-
    N > 0,
    isPrime(H),
    N1 is N-1,
    delFunction(T, N1, Res).

@GuyCoder - with wich call? With `delFunction([2,4,5,7,6,9,11],3,X)`? — max66, Jan 13 '17 at 22:18
Thanks man, Really helped out, it works now and gives only 1 answer. — , Jan 13 '17 at 22:19
@GuiCoder - with swi-prolog I get only one answer but with gprolog I get 3. There is to add the `N > 0` che before `not(isPrime())` — max66, Jan 13 '17 at 22:23

Deleting first N prime numbers from a list (Prolog)

1 Answers1