Simple recursive prolog program to add elements

Question

I am trying to calculate the sum of values given. My attempt:

element(0, 988).
element(1, 5434).
element(2, 5433).
element(3, 4543).
element(4, 827).

addElements(5, 0).

addElements(INDEX, SUM):-     %sums from given index to the end of array
    element(INDEX, VALUE),
    addElements(INDEX+1, SUM-VALUE).

My query:

addElements(0,X).

This is not working. Is it a syntactic error?

Answer 1

To evaluate arithmetic expressions in Prolog, you need to use the ISO built-in predicate is/2:

?- Index = 1, NewIndex is Index + 1.
Index = 1,
NewIndex = 2.

?- Accumulator = 10, NewAccumulator is Accumulator + 5.
Accumulator = 10,
NewAccumulator = 15.

Thus, assuming the indices are consecutive integers, a possible solution is:

add_elements(Index, Sum) :-
    add_elements_loop(Index, 0, Sum).

add_elements_loop(Index, Accumulator, Sum) :-
    not(element(Index, _)),              % index out of range!
    Sum = Accumulator.

add_elements_loop(Index, Accumulator, Sum) :-
    element(Index, Value),
    NewIndex is Index + 1,
    NewAccumulator is Accumulator + Value,
    add_elements_loop(NewIndex, NewAccumulator, Sum).

element(0, 70).
element(1, 30).
element(2, 10).
element(3, 20).
element(4, 80).

Examples:

?- add_elements(0, S).
S = 210 ;
false.

?- add_elements(3, S).
S = 100 ;
false.

An improved version of this code, which avoids the redundant call of the predicate element/2 and eliminates the spurious choice point, is as follows:

add_elements(Index, Sum) :-
    add_elements_loop(Index, 0, Sum).

add_elements_loop(Index, Accumulator, Sum) :-
    (   element(Index, Value)                             % if
    ->  NewIndex is Index + 1,                            % then 
        NewAccumulator is Accumulator + Value,
        add_elements_loop(NewIndex, NewAccumulator, Sum)
    ;   Sum = Accumulator ).                              % else

?- add_elements(0, S).
S = 210.

?- add_elements(3, S).
S = 100.