Sum of the first n numbers in prolog

Question

Hello can anyone help me compute the sum of the first n numbers. For example n=4 => sum = 10. So far I've wrote this

    predicates
  sum(integer,integer)
clauses

  sum(0,0).
   sum(N,R):-
        N1=N-1,
        sum(N1,R1),
        R=R1+N.

This one works but I need another implementation. I don't have any ideas how I could make this differen . Please help

Answer 1

What @mbratch said.

What you're computing is a triangular number . If your homework is about triangular numbers and not about learning recursive thinking, you can simply compute it thus:

triangular_number(N,R) :- R is N * (N+1) / 2 .

If, as is more likely, you're learning recursive thought, try this:

 sum(N,R) :-    % to compute the triangular number n,
   sum(N,1,0,R) % - invoke the worker predicate with its counter and accumulator properly seeded
   .

 sum(0,_,R,R).     % when the count gets decremented to zero, we're done. Unify the accumulator with the result.
 sum(C,X,T,R) :-   % otherwise,
   C > 0 ,         % - assuming the count is greater than zero
   T1 is T+X ,     % - increment the accumulator
   X1 is X+1 ,     % - increment the current number
   C1 is C-1 ,     % - decrement the count
   sum(C1,X1,T1,R) % - recurse down
   .               % Easy!

Edited to add:

Or, if you prefer a count down approach:

 sum(N,R) :- sum(N,0,R).

 sum(0,R,R).       % when the count gets decremented to zero, we're done. Unify the accumulator with the result.
 sum(N,T,R) :-     % otherwise,
   N > 0 ,         % - assuming the count is greater than zero
   T1 is T+N ,     % - increment the accumulator
   N1 is N-1 ,     % - decrement the count
   sum(N1,T1,R)    % - recurse down
   .               % Easy!

Both of these are tail-recursive , meaning that the prolog compiler can turn them into iteration (google "tail recursion optimization" for details).

If you want to eliminate the accumulator, you need to do something like this:

sum(0,0).
sum(N,R) :-
  N > 0 ,
  N1 is N-1 ,
  sum(N1,R1) ,
  R is R1+N
  .

A little bit simpler, but each recursion consumes another stack frame: given a sufficiently large value for N, execution will fail with a stack overflow.

Answer 2

sum(N, Sum) :-
    Sum is (N + 1) * N / 2 .

Answer 3

Since you already got plenty of advice about your code, let me throw in a snippet (a bit off-topic).

Counting, and more generally, aggregating, it's an area where Prolog doesn't shine when compared to other relational,declarative languages (read SQL). But some vendor specific library make it much more pleasant:

?- aggregate(sum(N),between(1,4,N),S).
S = 10.

Answer 4

This is the "heart" of your program:

sum(N,R):-
    R=R+N,
    N=N-1,
    sum(N,R).

The =/2 predicate (note the /2 means it accepts 2 arguments) is the instantiation predicate, not an assignment, or logical equal. It attempts to unify its arguments to make them the same. So if N is anything but 0 , then R=R+N will always fail because R can never be the same as R+N . Likewise for N=N-1 : it will always fail because N and N-1 can never be the same.

In the case of =/2 (unification), expressions are not evaluated. They are just terms. So if Y = 1 , then X = Y + 1 unifies X with 1+1 as a term (equivalently written +(1,1) ).

Because of the above issues, sum will always fail.

Numerical assignment of an arithmetic expression is done in Prolog with the is/2 predicate. Like this:

X is Y + 1.

This operator unifies the value of X to be the same as the value of the evaluated expression Y+1 . In this case, you also cannot have X is X+1 for the same reason given above: X cannot be made the same as X+1 and Prolog does not allow "re-instantiation" of a variable inside of a clause. So you would need something like, X1 is X + 1 . Also note that for is/2 to work, everything in the expression on the right must be previously instantiated. If any variables in the expression on the right do not have a value, you will get an instantiation error or, in the case of Turbo Prolog, Free variable in expression... .

So you need to use different variables for expression results, and organize the code so that, if using is/2 , variables in the expression are instantiated.

---

EDIT

I understand from Sergey Dymchenko that Turbo Prolog, unlike GNU or SWI, evaluates expressions for =/2 . So the = will work in the given problem. However, the error regarding instantiation (or "free variable") is still caused by the same issue I mentioned above.

Answer 5

sum(N, N, N).  
sum(M, N, S):-
  N>M,
  X is M+1,
  sum(X, N, T),
  S is M+T.


?- sum(1,5,N).
   N = 15 .