; Recursion with integers
; recursion when the input is natural number (nonnegative integer)
;
; the principle is similar... we can recurse on problems for which there is
;
; 1) base case, an easy case to answer right away
;
; 2) we can break other problems down into simpler problems (each time we will 
; reduce the input integer

; example: factorial
; (factorial n)  ->  n*(n-1)*(n-2)* ... * 2*1
; you may have seen this written n! in math class
; to write a procedure for factorial we need 
; 1) base case:  0! = 1, (factorial 0) --> 1
;
; 2) (factorial n) -> n*(n-1)*(n-2)* ... * 2*1 -> (* n (factorial (- n 1)))
;

(define factorial
  (lambda (n)
    (if  (= n 0)
         1
         (* n (factorial (- n 1))))))
 
; write a procedure: sum-of-digits, which takes a natural number and
; returns the sum of its digits
; break this into two cases:
; 1) base case: input n is a single digit number, test with (< n 10)
;                           in this case we wnat to return n
; 2) try to break down more difficult cases
;       given a number n, is there a way I can just take off a digit?
;       (sum-of-digits n)  -->  (+ (remainder n 10) (sum-of-digits (quotient n 10)))

; Procedure: SUM-OF-DIGITS returns the sum of the digits of the input number
;
; Given: N, a natural number
;
; Result: SUM-OF-DIGITS, a natural number
;
; Preconditions: none
;
; Postconditions: SUM-OF-DIGITS is the sum of the digits of N

(define sum-of-digits
  (lambda (n)
    (if (<  n 10)
        n
        (+ (remainder n 10) (sum-of-digits (quotient n 10))))))