I\'m trying to create a program in Prolog which takes two numbers - A and B and finds the sum of the numbers from A to B, including them.
To sum up: sum(1, 5, C) should
Available in SWI-Prolog, library(aggregate) is a full fledged interface:
?- aggregate(sum(N), between(1,5,N), S).
S = 15.
it does internally a similar work as you can read in Kaarel' answer. Anyway, to learn Prolog, follow C.B. detailed explanation.
In SWI-Prolog, the simplest way to implement it is:
?- numlist(1, 5, List), sum_list(List, Sum).
List = [1, 2, 3, 4, 5],
Sum = 15.
Your code tries to evaluate C+A
but C
is not known yet, therefore the error message.
Btw, you can also calculate this sum directly, without any iteration or list generation, see https://math.stackexchange.com/questions/50485/sum-of-n-consecutive-numbers
I suppose you could use brute force to compute the sum of a finite arithmetic sequence, but wouldn't it be easier just to use your maths?
sum(A,B,X) :-
A =< B ,
N is 1+B-A ,
X is N*(A+B) / 2
.
Or, more generally, for computing the sum of a finite sequence of n terms, starting with a, with a "step size" (relative difference) of d, you get:
sum(A,D,N,S) :- S is (N*(2*A+(N-1)*D))/2 .
But if you're going to use brute force, make it tail recursive:
sum(A,B,S) :- A =< B , sum(A,B,0,S) .
sum(A,B,S,S) :-
A > B
.
sum(A,B,T,S) :-
A =< B ,
T1 is T+A ,
A1 is A+1 ,
sum(A1,B,T1,S)
.
You can't instantiate C1
as C+A
because C
is not yet an integer.
Let's call sum(1, 5, C)
sum(1, 5, C)
---> 1 =< 5 ... true
---> A1 is 1 + 1 ... A1 = 2 (fine)
---> C1 is C + 2, C = something like _G232, not yet instantiated.
Let's take a look at your logic.
We would have a base case, when A and B are equal, that will be our 'starting' sum
sum(X,X,X).
Then we have our general case. each recursive call will give us the sum from k + 1 to n where we call sum(k,n,sum).
sum(A, B, C):- A =< B, A1 is A + 1, sum(A1,B,C1), C is C1 + A.