SWI-Prolog: Sum-List

2
votes

I'm starting with Prolog, and i'm a bit confused...

I have a simple program:

sum(0, []).
sum(Total, [Head|Tail]) :- sum(Sum, Tail), Total is Head + Sum.

When i debug, i can see that Prolog first splits the list with Head and Tail, so the result is 0 + empty list, and AFTER THAT it starts to sum the numbers and adds it again to the list.

Can someone explain why it doesn't come to Total is Head + Sum. first and then splits the list to Head and Tail again?

EDIT: Here is the trace:

[trace]  ?- sum(X, [1,2,3]).
Call: (6) sum(_G345, [1, 2, 3]) ? creep
Call: (7) sum(_G424, [2, 3]) ? creep
Call: (8) sum(_G424, [3]) ? creep
Call: (9) sum(_G424, []) ? creep
Exit: (9) sum(0, []) ? creep
Call: (9) _G430 is 3+0 ? creep
Exit: (9) 3 is 3+0 ? creep
Exit: (8) sum(3, [3]) ? creep
Call: (8) _G433 is 2+3 ? creep
xit: (8) 5 is 2+3 ? creep
Exit: (7) sum(5, [2, 3]) ? creep
Call: (7) _G345 is 1+5 ? creep
Exit: (7) 6 is 1+5 ? creep
Exit: (6) sum(6, [1, 2, 3]) ? creep
X = 6.
2
listprolog

2 Answers

5
votes

Your definition puts addition on the stack. The optimization that avoids putting away the addition would be a special case of a general technique known as tail recursion.

The following definition can use tail recursion:

sum(X,L):-sum(0,L,X).
sum(X,[],X).
sum(N, [Head|Tail],Y) :- N1 is Head + N, sum(N1,Tail,Y).

It introduces an accumulator for the values of the partial sum and carries it with the tail of the list. Here is the trace of the execution of the sum(X,[1,2,3]) query.

?- trace, sum(S,[1,2,3]),notrace,nodebug.
   Call: (7) sum(_G584, [1, 2, 3]) ? creep
   Call: (8) sum(0, [1, 2, 3], _G584) ? creep
^  Call: (9) _G792 is 1+0 ? creep
^  Exit: (9) 1 is 1+0 ? creep
   Call: (9) sum(1, [2, 3], _G584) ? creep
^  Call: (10) _G795 is 2+1 ? creep
^  Exit: (10) 3 is 2+1 ? creep
   Call: (10) sum(3, [3], _G584) ? creep
^  Call: (11) _G798 is 3+3 ? creep
^  Exit: (11) 6 is 3+3 ? creep
   Call: (11) sum(6, [], _G584) ? creep
   Exit: (11) sum(6, [], 6) ? creep
   Exit: (10) sum(3, [3], 6) ? creep
   Exit: (9) sum(1, [2, 3], 6) ? creep
   Exit: (8) sum(0, [1, 2, 3], 6) ? creep
   Exit: (7) sum(6, [1, 2, 3]) ? creep
S = 6 .
-2
votes

Here is another version doing that. I have been using concept mapping software to help in designing code, I cannot do complicated stuff in my head. 

sum(A, [], A).
sum(Total, [Head|Tail], AuxNum) :-
    Sum is Head+AuxNum,
    sum(Total, Tail, Sum).


   1 ?- trace,sum(Total,[1,2,3],0).
   Call: (7) sum(_G2090, [1, 2, 3], 0) ? creep
   Call: (8) _G2221 is 1+0 ? creep
   Exit: (8) 1 is 1+0 ? creep
   Call: (8) sum(_G2090, [2, 3], 1) ? creep
   Call: (9) _G2224 is 2+1 ? creep
   Exit: (9) 3 is 2+1 ? creep
   Call: (9) sum(_G2090, [3], 3) ? creep
   Call: (10) _G2227 is 3+3 ? creep
   Exit: (10) 6 is 3+3 ? creep
   Call: (10) sum(_G2090, [], 6) ? creep
   Exit: (10) sum(6, [], 6) ? creep
   Exit: (9) sum(6, [3], 3) ? creep
   Exit: (8) sum(6, [2, 3], 1) ? creep
   Exit: (7) sum(6, [1, 2, 3], 0) ? creep
Total = 6.