scheme divide witohut using /

[feedmeh8]

Hey guys. Im new to these forums and was hoping i could get some help. Trying to figure out how to divide a number by another number in scheme without using /. I have written these functions that i can use in creating the divide function:

(define pwr
  (lambda (a b) 
    (if
     (= b 0)
     1
     (* a(pwr a(- b 1))))))

(define (add c d)
  (if
   (= d 0)
   c
   (add (+ c 1) (- d 1))))

(define (subt e f)
  (if
   (= f 0)
   e
   (subt (- e 1) (- f 1))))

(define (multi g h)
  (if
   (= h 0)
   0
   (add g(multi g(- h 1)))))

It is supposed to be done like how the multi function above was done. (by using add in this case). I am thinking there is some way to do it by using multi, but i cant figure it out : /

Any help appreciated!
ty :)

[tuffff]

From my first days with scheme… it’s not pretty but it works. Please ignore the markup, “divide-2” is just a function name and “d” just a variable.

; divides two numbers until cutoff is reached
(define (divide n d cutoff)
  (let ((step 10)) ; any value >1 will work
    (define (loop i p sum)
      (letrec ((sp (expt step p))
               (dp (* d sp))
               (q (quotient i dp))
               (r (remainder i dp))
               (new-sum (+ sum (* q sp))))
        (cond
          ((= r 0)
           new-sum)
          ((< (abs r) cutoff)
           (exact->inexact new-sum))
          (else
            (loop r (- p 1) new-sum)))))
    (loop n 0 0)))

; divides two numbers until cutoff is reached -- without quotient/remainder
(define (divide-2 n d cutoff)
  (let ((step 10)) ; any value >1 will work
    (define (loop i ud p sum)
      (cond
        ((= i 0)
         sum)
        ((< (abs i) cutoff)
         (exact->inexact sum))
        (else
          (letrec ((sp (expt step p))
                   (dp (* ud sp))
                   (new-sum (+ sum sp)))
              (cond
                ((< dp i)
                 (loop (- i dp) ud p new-sum))
                (else
                  (loop i ud (- p 1) sum)))))))
    (if (< n 0)
      (if (< d 0)
        (loop (- 0 n) (- 0 d) 0 0)
        (- 0 (loop (- 0 n) d 0 0)))
      (if (< d 0)
        (- 0 (loop n (- 0 d) 0 0))
        (loop n d 0 0)))))

[feedmeh8]

Thanks! :D did not even occur to me to do it like that :P .... i came up with:

(define (divide x y)
  (if
    (< x y)
    0
    (+ 1(divide (- x y) y))))

[tuffff]

You should have said that you only want integer division :)

But for that task, you solution is much cleaner, but not tail-recursive :S