recursion - Sum numbers up to N-M in Prolog -

recursion - Sum numbers up to N-M in Prolog -

hey need create function countnm(n,m,s) sums numbers n-m (and returns 0 if n equal or less m). have worked out , have makes sense when work through in head, uninstantiated...

sumnm(0,0,0). sumnm(n,m,s):- n=<m, s 0. sumnm(n,m,s):- n>m, n-m, b s+a, c m+1, sumnm(n,c,b).

the problem on line

b s+a,

at point, s not bound, causing arithmetic look fail.

other that, logic behind look wrong. you're saying sum of 1 fewer number (b) equal total sum (s) plus current number (a). should other way around.

you can replace lastly predicate code:

sumnm(n,m,s):- n>m, n-m, c m+1, sumnm(n,c,b), s b+a.

here first calculate sum b, after add together it, resulting in total sum s.

recursion prolog