Numeric Recursion

We have written a wide variety of recursive procedures so far. We have written recursive procedures that return lists (e.g., a variety of procedures that select or filter elements from lists), numbers (e.g., procedures that tally elements in a list, as well as things like sum and product), and even Boolean values (e.g., the all-___? and any-___? predicates). Yet the procedures have had one thing in common: All of them took lists as parameters.

While the recursive procedures we've written so far have used lists as the basis of recursion, we can also write recursive procedures with other types as the basis of recursion. All we really need to do recursion are (a) a way to determine if a value is simple enough that we can compute an answer directly and (b) a way to simplify the value.

Natural numbers provide a nice basis of recursion. Like lists, natural numbers have a recursive structure of which we can take advantage when we write direct-recursion procedures. A natural number, n, is either (a) zero, or (b) the successor of a smaller natural number, which we can obtain by subtracting 1 from n.

Recall that the common format of a recursive procedure is

(define recursive-proc
  (lambda (val)
    (if (base-case-test)
        (base-case val)
        (combine (partof val)
                 (recursive-proc (simplify val))))))

For lists, the test for a base case was typically “is the list empty” or “does the list have only one value”, which we would express as (empty? lst) and (empty? (cdr lst)), respectively. We typically simplify a list by taking the cdr of the lst. Hence, the simplest form of a recursive procedure for lists is

(define recursive-proc
  (lambda (lst)
    (if (null? lst)
        (base-case)
        (combine (car lst)
                 (recursive-proc (cdr lst))))))

Clearly, with other data types, we'll have different tests for the base case and different mechanisms for simplifying values.

To write recursive procedures with numeric arguments, we first need a technique for identifying the base case. With natural numbers, 0 often provides an appropriate base case. Standard Scheme provides the predicate zero? to distinguish between the base and recursive cases, which permits us to use an if-expression to ensure that only the expression for the appropriate case is evaluated. We can potentially write a procedure that applies to any natural number if we know (a) what value it should return when the argument is 0 and (b) how to convert the value that the procedure would return for the next smaller natural number into the appropriate return value for a given non-zero natural number.

Hence, a typical numeric recursive procedure will look something like

(define recursive-proc
  (lambda (val)
    (if (zero? val)
        (base-case)
        (combine val (recursive-proc (- val 1))))))

In this sample code, we subtract 1 to simplify the number. However, one can also subtract more than 1, or divide the number by 2, or do anything else that gives a result that is closer to zero.

For instance, here is a procedure that, given a natural number, number, computes the result of adding together all of the natural numbers up to and including number. This result is traditionally called the termial of the number.

;;; Procedure:
;;;   termial
;;; Parameters:
;;;   number, a natural number
;;; Purpose:
;;;   Compute the sum of natural numbers not greater than a given
;;;   natural number
;;; Produces:
;;;   sum, a natural number
;;; Preconditions:
;;;   number is a number, exact, an integer, and non-negative.
;;;   The sum is not larger than the largest integer the language
;;;     permits.
;;; Postconditions:
;;;   sum is the sum of natural numbers not greater than number.
;;;   That is, sum = 0 + 1 + 2 + ... + number
(define termial
  (lambda (number)
    (if (zero? number)
        0
        (+ number (termial (- number 1))))))

Whereas in a list recursion, we called the cdr procedure to reduce the length of the list in making the recursive call, the operation that we apply in recursion with natural numbers is reducing the number by 1.

   (termial 5)
=> (+ 5 (termial 4))
=> (+ 5 (+ 4 (termial 3)))
=> (+ 5 (+ 4 (+ 3 (termial 2))))
=> (+ 5 (+ 4 (+ 3 (+ 2 (termial 1)))))
=> (+ 5 (+ 4 (+ 3 (+ 2 (+ 1 (termial 0))))))
=> (+ 5 (+ 4 (+ 3 (+ 2 (+ 1 0)))))
=> (+ 5 (+ 4 (+ 3 (+ 2 1))))
=> (+ 5 (+ 4 (+ 3 3)))
=> (+ 5 (+ 4 6))
=> (+ 5 10)
=> 15

   (termial -5)
=> (+ -5 (termial -6))
=> (+ -5 (+ -6 (termial -7)))
=> (+ -5 (+ -6 (+ -7 (termial -8))))
=> (+ -5 (+ -6 (+ -7 (+ -8 (termial -9)))))
=> ...

   (termial 4.1)
=> (+ 4.1 (termial 3.1))
=> (+ 4.1 (+ 3.1 (termial 2.1)))
=> (+ 4.1 (+ 3.1 (+ 2.1 (termial 1.1))))
=> (+ 4.1 (+ 3.1 (+ 2.1 (+ 1.1 (termial 0.1)))))
=> (+ 4.1 (+ 3.1 (+ 2.1 (+ 1.1 (+ 0.1 (termial -0.9))))))
=> (+ 4.1 (+ 3.1 (+ 2.1 (+ 1.1 (+ 0.1 (+ -0.9 (termial -1.9)))))))
=> ...

(define termial
  (lambda (number)
    (cond
     [(not (number? number))
      (error "termial expects a number, received" number)]
     [(not (integer? number))
      (error "termial expects an integer, received" number)]
     [(< number 1)
       (error "termial expects a positive integer, received" number)]
     [(inexact? number)
      (error "termial expects an exact integer, received" number)]
     [else
      (let kernel ([current number])
        (if (zero? current)
            0
            (+ current (kernel (- current 1)))))])))

The important part of getting recursion to work is making sure that the base case is inevitably reached by performing the simplification operation enough times. For instance, we can use direct recursion on exact positive integers with a base case of 1, rather than 0.

;;; Procedure:
;;;   factorial
;;; Parameters:
;;;   n, a positive integer
;;; Purpose:
;;;   Compute n!, the product of positive integers less than or
;;;   equal to a given positive integer.
;;; Produces:
;;;   product, an integer
;;; Preconditions:
;;;   n is a number, exact, an integer, and positive.
;;;   The product is not larger than the largest integer the
;;;     language permits.
;;; Postconditions:
;;;   product is the product of the positive integers less than or
;;;   equal to n.  That is,
;;;     product = 1 * 2 * ... * n
(define factorial
  (lambda (n)
    (if (= n 1)
        1
        (* n (factorial (- n 1))))))

We require the invoker of this factorial procedure to provide an exact, strictly positive integer. (Zero won't work in this case, because we can't reach the base case, 1, by repeated subtractions if we start from 0.)

But our base case need not be a small number. We can use direct recursion to approach the base case from below by repeated additions of 1, if we know that our starting point is less than or equal to that base case. Here's an example.

;;; Procedure:
;;;   count-from
;;; Parameters:
;;;   lower, an integer
;;;   upper, an integer
;;; Purpose:
;;;   Construct a list of the integers from lower to upper,
;;;   inclusive, in ascending order.
;;; Produces:
;;;   ls, a list
;;; Preconditions:
;;;   lower <= upper
;;;   Both lower and upper are numbers, exact, and integers.
;;; Postconditions:
;;;   The length of ls is 1 + upper - lower.
;;;   Every integer between lower and upper, inclusive, appears
;;;     in the list.
;;;   Every value in the list with a successor is smaller than its
;;;     successor.
;;;   For every integer k less than or equal to the length of
;;;       ls, the element in position k of ls is lower + k.
(define count-from
  (lambda (lower upper)
    (if (= lower upper)
        (list upper)
        (cons lower (count-from (+ lower 1) upper)))))

Why is this useful? Well, it acts much like a generalized version of iota, and we've already seen that iota is useful in many different situations.

You may also recall much more manually intensive ways of making such lists in the past, such as listing all the integers between 0 and 16.