Prolog sum list elements at even positions

Question

I stucked at this point. Can't find out how to sum elements at even positions.

domains
slist=integer*
pos,sum=integer
predicates
even(pos).
even_summ(slist,pos,sum).
clauses
even(X):- X mod 2 = 0.
even_summ([],0,0).
even_summ([H|T],P,S):- even_summ(T,P1,S1),
P=P1+1,
S=S1+H.
goal
even_summ([1,2,3,4,5,6,7],P,S).

Answer 1

Here's a solution involving even and odd which doesn't require the auxiliary variable. It treats sum_even and sum_odd as "co-predicates".

sum_even([], 0).
sum_even([_|T], S) :- sum_odd(T, S).

sum_odd([], 0).
sum_odd([X|T], S) :- S #= S1 + X, sum_even(T, S1).

Since you're using Visual (or Turbo or PDC) Prolog, this would need to be rewritten slightly as the following, which is not tail recursive or relational:

sum_even([], 0).
sum_even([_|T], S) :- sum_odd(T, S).

sum_odd([], 0).
sum_odd([X|T], S) :- sum_even(T, S1), S = S1 + 1.

Answer 2

A simple sumEven/2 (a sumOdd/2 come for free)

sum1of2([], _, 0).

sum1of2([H | T], 0, S1) :-
  sum1of2(T, 1, S0),
  S1 is S0 + H.

sum1of2([_ | T], 1, S) :-
  sum1of2(T, 0, S).

sumEven(L, S) :-
  sum1of2(L, 1, S).

sumOdd(L, S) :-
  sum1of2(L, 0, S).