CODEWITHSUNDEEP
recursionschemeracketperfect-numbers
user1661660
user1661660

Reputation:

Perfect number recursion in SCHEME. (beginner)

Hey so I am creating a function (divides n), which is supposed to calculate the number of divisors in a number n with the help of a modulo function and a function that acts as a counter going down from the number n. My issue is that the modulo function should output a true or false depending on whether the number is whole however my if statement 

(if (= (divides n k) #f)
    0

Im not sure why but the code wont evaluate the if statement as true or false.. it just skips it. Also im not sure that 0 should be the correct output I want it to just skip that number and not count it.

Heres my code:

(define (divides a b) (= 0 (modulo b a)))
(define (divisors-upto n k)
  (if (= (divides n k) #f)
      0
      (+ k (divisors-upto n (- k 1)))))
(define (divisors n) (divisors-upto n n))

(divisors 4) ;for example should produce the result 3

Upvotes: 1

Views: 1239

Answers (1)

Óscar López
Óscar López

Reputation: 236004

Start by fixing the divides procedure, you reversed the arguments to modulo. This is how it should look:

(define (divides a b)
  (= 0 (modulo a b)))

The above tests if b divides a, that's how you're using it in the divisors-upto procedure. Also you should replace this:

(= (divides n k) #f)

With this:

(equal? (divides n k) #f)

Or even better, this:

(not (divides n k))

Apart from that, isn't this the same question that you posted before? I told you there that you're missing a case in the recursion, look at my previous answer in the link.

If it's not the same procedure, then I'm not really sure of what you want to do: in this question you state that the procedure "is supposed to calculate the number of divisors in a number", but that's not what the procedure is doing - you're adding the actual divisors (the k parameter in the procedure) and not the number of divisors. And again, you'd be missing a case - what happens if the current k is not a divisor? the recursion would exit prematurely! Try to work on this a bit, fill-in the blanks:

(define (divisors-upto n k)
  (cond ((zero? k)
         <???>) ; how many divisors are there if k is zero?
        ((not (divides n k))
         <???>) ; if k is not a divisor of n, we must proceed without incrementing
        (else   ; if k is a divisor of n, by how many should the count be incremented?
         (+ <???> (divisors-upto n (- k 1))))))

Upvotes: 2

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