Fundamentals of Computer Science 1 (CS151 2003S)

Recursion with Natural Numbers

Summary: Our first explorations of recursion involved procedures that take lists as parameters. However, recursive procedures can also take numbers as parameters. In this reading, we consider these kinds of procedures.



While the recursive procedures we've written so far have used lists as arguments, we can also write recursive procedures with numbers as arguments. Like lists, natural numbers have a recursive structure of which we can take advantage when we write direct-recursion procedures. A natural number is either (a) zero, or (b) the successor of a smaller natural number, which we can obtain by subtracting 1.

Numeric Base Cases

Recall that the standard format of a recursive procedure is

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

To write recursive procedures with numeric arguments, we first need a technique for identifying the base case. Standard Scheme provides the predicate zero? to distinguish between the base and recursive cases, so we can again use an if-expression to ensure that only the expression for the appropriate case is evaluated. So we can 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.

An Example, Termial

For instance, here is a procedure that computes the termial of any natural number, number. That is, it computes the result of adding together all of the natural numbers up to and including 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)
        (+ 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.

Watching termial in Action

Here's a summary of what actually happens during the evaluation of a call to the termial procedure, say, (termial 5):

   (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

Preconditions for termial

The restriction that termial takes only non-negative integers as arguments is an important one: If we gave it a negative number or a non-integer, we'd have a runaway recursion, because we cannot get to zero by subtracting 1 repeatedly from a negative number or from a non-integer, and so the base case would never be reached. For example,

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

Similariy, if we gave the termial procedure an approximation rather than an exact number, we might or might not be able to reach zero, depending on how accurate the approximation is and how much of that accuracy is preserved by the subtraction procedure.

A Note on the Implementation of termial

Note that our sum all the values algorithm is not the only way to compute the termial of a natural number. Many of you may have learned a more efficient (or at least more elegant) algorithm. We'll return to this algorithm later.

Other Numeric Base Cases

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 1, rather than 0, as the base case.

;;; Procedure:
;;;   factorial
;;; Parameters:
;;;   number, a positive integer
;;; Purpose:
;;;   Compute number!, the product of positive integers not
;;;   greater than a given positive integer.
;;; Produces:
;;;   product, an integer
;;; Preconditions:
;;;   number 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 not
;;;   greater than number.  That is,
;;;     product = 1 * 2 * ... * number
(define factorial
  (lambda (number)
    (if (= number 1)
        (* number (factorial (- number 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.)

Similarly, 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, a natural number
;;;   upper, a natural number
;;; Purpose:
;;;   Construct a list of the natural numbers from lower to upper, 
;;;   inclusive, in ascending order.
;;; Produces:
;;;   ls, a list
;;; Preconditions:
;;;   lower <= upper
;;;   Both lower and upper are numbers, exact, integers, and non-negative.
;;; Postconditions:
;;;   The length of ls is upper - lower + 1.
;;;   Every natural number between lower and upper, inclusive, appears
;;;     in the list.
;;;   Every value in the list with a successor is smaller than its
;;;     successor.
;;;   For every natural number 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)))))




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The source to the document was last modified on Sun Feb 16 22:26:08 2003.
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Samuel A. Rebelsky,