Longest sequence of even numbers in a list with prolog

user_18543953 0
Viewed 83

I want to implement a predicate even_sequence(X,Y,Z) where X is the given list, Y is a counter of the max length of even numbers sequence, and Z is the counter that stores the length of the current even number subsequence. For example for the given list 2,4,6,3,5,2,2 i want to return 3 because 2,4,6 is the longest sequence.

Here is the code that i tried and i don't know how to make it work(i am a total beginner in Prolog).If i run even_sequence([2,2,2,3,3],C,R) i get the error: Arguments are not sufficiently instantiated In: [1] even_sequence([2,2|...],_1730,_1732)

even_sequence([],_,_).
even_sequence([H|T],GlobalMax,LocalMax):-
    H mod 2 =:= 0,
    LocalMax1 is LocalMax+1,
    even_sequence(T,GlobalMax,LocalMax1).
even_sequence([H|T],GlobalMax,LocalMax):-
    H mod 2 =\= 0,
    GlobalMax1 is LocalMax,
    LocalMax1 is LocalMax-LocalMax,
    even_sequence(T,GlobalMax1,LocalMax1).
prolog
1 Answers
longest_even_sequence(Lst, EvenSeqLen) :-
    % Keep running total of current sequence, and best-so-far
    longest_even_sequence_(Lst, 0, 0, EvenSeqLen).

longest_even_sequence_([], Cur, Best, EvenSeqLen) :-
    % Check final sequence, and assign EvenSeqLen
    num_num_max(Cur, Best, EvenSeqLen).

longest_even_sequence_([H|T], Cur, Best, EvenSeqLen) :-
    0 is H mod 2, !,
    % Is even, so sequence gets longer
    Cur1 is Cur + 1,
    longest_even_sequence_(T, Cur1, Best, EvenSeqLen).

longest_even_sequence_([_H|T], Cur, Best, EvenSeqLen) :-
    % Is odd (because not even), so check the finished sequence
    num_num_max(Cur, Best, Best1),
    % Starting new sequence
    longest_even_sequence_(T, 0, Best1, EvenSeqLen).


% Convenient wrapper
num_num_max(X, Y, Max) :-
    Max is max(X, Y).

Result in swi-prolog:

?- longest_even_sequence([2,4,6,3,5,2,2], N).
N = 3.

?- longest_even_sequence([1,2,3], N).
N = 1.
user_176283360
Related