Recursion on Natural Numbers

Summary: Although most of our prior experiments with recursion have emphasized recursion over lists, it is also possible to use other values as the basis of recursion. In this laboratory, you will explore the use of natura l numbers (non-negative integers) as the basis of recursion.

Contents:

• Exercises
  • Exercise 0: Preparation
  • Exercise 1: Power of two
  • Exercise 2: Counting down
  • Exercise 3: Filling lists
  • Exercise 4: Counting To
  • Exercise 5: Counting Between
  • Exercise 6: Counting To, Revisited
  • Exercise 7: How Many Digits?
• For Those Who Finish Early
• Notes
  • count-from

Exercises

Exercise 0: Preparation

a. Make sure you understand the first reading on recursion and the second reading on recursion.

b. Start DrScheme.

Exercise 1: Power of two

Using recursion over natural numbers, define and test a recursive Scheme procedure, (power-of-two power) that takes a natural number (an integer greater than or equal to 0) as its argument and returns the result of raising 2 to the power of that number. For example,

> (power-of-two 3)
8
> (power-of-two 10)
1024
> (power-of-two 1)
2

It is possible to define this procedure non-recursively, using Scheme's primitive expt procedure, but the point of the exercise is to use recursion.

Exercise 2: Counting down

Define and test a Scheme procedure, (count-down val), that takes a natural number as argument and returns a list of all the natural numbers less than or equal to that number, in descending order:

> (count-down 5)
(5 4 3 2 1 0)
> (count-down 0)
(0)

Exercise 3: Filling lists

Define and test a Scheme procedure, (fill-list value count), that takes two arguments, the second of which is a natural number, and returns a list consisting of the specified number of repetitions of the first argument:

> (fill-list 'sample 5)
(sample sample sample sample sample)
> (fill-list (list 'left 'right) 3)
((left right) (left right) (left right))
> (fill-list null 1)
(())
> (fill-list null 2)
(() ())

Exercise 4: Counting To

Define and test a recursive Scheme procedure that takes a natural number as argument and returns a list of all the natural numbers that are strictly less than the argument, in ascending order. (The traditional name for this procedure is iota -- another Greek letter.)

For example,

> (iota 3)
(0 1 2)
> (iota 5)
(0 1 2 3 4)
> (iota 1)
(0)

Exercise 5: Counting Between

You may recall the count-from procedure from the reading on recursion over natural numbers. That procedure is also reproduced at the end of this lab.

What is the value of the call (count-from -10 10)?

a. Write down what you think that it should be.

b. Copy the definition of count-from into DrScheme and use it to find out what the call actually returns.

Exercise 6: Counting To, Revisited

Using count-from, define and test a Scheme procedure that takes a natural number as argument and returns a list of all the natural numbers that are strictly less than the argument, in ascending order. Note that your procedure must use count-from as a helper.

Exercise 7: How Many Digits?

Here is the definition of a procedure that computes the number of digits in the decimal representation of number:

(define number-of-decimal-digits
  (lambda (number)
    (if (< number 10)
        1
        (+ (number-of-decimal-digits (quotient number 10)) 1))))

a. Test this procedure.

The definition of number-of-decimal-digits uses direct recursion.

b. Describe the base case of this recursion.

c. Identify and describe the way in which a simpler instance of the problem is created for the recursive call. That is, explain what problem is solved recursively and why you know that that problem is simpler.

d. Explain how the procedure correctly determines that the decimal numeral for the number 2000 contains four digits.

e. What preconditions does number-of-decimal-digits impose on its argument?

For Those Who Finish Early

1. Finish your work on the previous lab.

2. Rewrite power-of-two to permit negative exponents.

Notes

count-from

For those of you unable to find the reading on recursion over natural numbers and for completeness, here is the count-from procedure.

;;; 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)))))