Sum of even and odd numbers of a list in Prolog

Question

What I need to do is to write a predicate that takes a list of numbers and returns a list consisting of two elements, the first one is the sum of the even numbers and the second one the sum of the odd numbers.

For example:

?- sum([5,4,9,8,1,7], L).
L = [12, 22].

So far I have written:

iseven(N) :-
    0 is mod(N,2). 

Answer 1

Since you've defined iseven/2 you could use it like:

sum([],[0,0]).
sum([H|T],[N1,N2]):- 
             sum(T,[N3,N4]),
             ( iseven(H) 

              ->  N1 is N3+H, N2 is N4
              ;   N2 is N4+H, N1 is N3

             ).

Example:

?-sum([5,4,9,8,1,7], L).
L = [12, 22].

A non if-then-else version using different clauses:

sum([],[0,0]).
sum([H|T],[N1,N2]):- sum(T,[N3,N2]), iseven(H), N1 is N3+H.
sum([H|T],[N1,N2]):- sum(T,[N1,N3]), \+ iseven(H), N2 is N3+H.

Answer 2

You could also write this predicate using accumulators and if_/3 . Furthermore you can incorporate the single goal of iseven/1 into the predicate:

list_sums(L,S) :-
   list_sums_(L,S,0-0).

list_sums_([],[E,O],E-O).
list_sums_([X|Xs],S,E0-O0) :-
   M is X mod 2,
   if_(M=0,(E1 is E0+X, O1 is O0),(E1 is E0, O1 is O0+X)),
   list_sums_(Xs,S,E1-O1).

Note how the accumulators are written as a pair ( EO ). If you are free to choose a representation for the two sums, this pair notation would be an alternative to a list with two elements. Your example query yields the desired result:

?- list_sums([5,4,9,8,1,7],L).
L = [12, 22].

And the example from the comments terminates considerably faster:

?- length(L,100),maplist(=(1),L),time(list_sums(L,[A,B])).
% 703 inferences, 0.000 CPU in 0.000 seconds (100% CPU, 3426895 Lips)
L = [1, 1, 1, 1, 1, 1, 1, 1, 1|...],
A = 0,
B = 100.

However, due to the use of is/2 this is only working in one direction. If you'd like to use the predicate in the other direction as well, you could opt to use CLP(FD). In that case include the line

:- use_module(library(clpfd)).  

in your source file and replace all occurrences of is/2 in list_sums_/3 by #=/2 :

list_sums_([],[E,O],E-O).
list_sums_([X|Xs],S,E0-O0) :-
   M #= X mod 2,
   if_(M=0,(E1 #= E0+X, O1 #= O0),(E1 #= E0, O1 #= O0+X)),
   list_sums_(Xs,S,E1-O1).

Your example query still yields the same result and the example from the comments terminates in an acceptable time:

?- length(L,100),maplist(=(1),L),time(list_sums(L,[A,B])).
% 18,928 inferences, 0.004 CPU in 0.004 seconds (99% CPU, 4888152 Lips)
L = [1, 1, 1, 1, 1, 1, 1, 1, 1|...],
A = 0,
B = 100.

But the predicate works in both directions now. In some cases Prolog can find concrete answers without further ado:

?- list_sums([A,B],[2,1]).
A = 2,
B = 1 ;
A = 1,
B = 2 ;
false.

In other cases you get residual goals as an answer:

?- L=[A,B,C,D,E,F], list_sums(L,[12,22]).
L = [A, B, C, D, E, F],
A+B#=_G3306,
A mod 2#=0,
B mod 2#=0,
_G3306+C#=_G3342,
C mod 2#=0,
_G3342+D#=12,
D mod 2#=0,
E+F#=22,
E mod 2#=_G3402,
F mod 2#=_G3414,
_G3414 in 0..1,
dif(_G3414, 0),
_G3402 in 0..1,
dif(_G3402, 0) ;
...

In these cases you can restrict the elements of the list to some interval and use label/1 to get concrete numbers as solution. For example, you can ask for solutions with six numbers from zero to seven and Prolog will give you all 300 solutions:

?- length(L,6), L ins 0..7, list_sums(L,[12,22]), label(L).
L = [6, 6, 1, 7, 7, 7] ;
L = [6, 6, 3, 5, 7, 7] ;
...

Answer 3

You can use a functionnal way :

one_sum(X, [SE,SO], [SE1, SO1]) :-
    (   X mod 2 =:= 0 ->
        SE1 is SE+X, SO1 = SO
    ;   SE1 = SE, SO1 is SO+X).

sum(L, S) :-
    foldl(one_sum, L, [0,0], S).