Lab: Numeric recursion

Held

Friday, 5 April 2019

Writeup due

Monday, 8 April 2019

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 natural numbers (non-negative integers) as the basis of recursion.

Reference

Here is a template for the simplest kind of numeric recursive procedures.

(define RECURSIVE-PROC
  (lambda (n)
    (if (zero? n)
        BASE-CASE
        (COMBINE n (RECURSIVE-PROC (- n 1))))))

We don’t always subtract 1 from n. We might also divide by 2 or make other changes that get us closer to a target.

(define RECURSIVE-PROC
  (lambda (n)
    (if (zero? n)
        BASE-CASE
        (COMBINE n (RECURSIVE-PROC (quotient n 2))))))

You can also use helper recursion to achieve a variety of other goals. For example, helper recursion can let you count up from 1 to n. (A slight change to the following code will allow you to count up from 0 to n.)

(define HELPER
  (lambda (so-far i n)
    (if (> i n)
        so-far
        (HELPER (UPDATE so-far i) (+ i 1) n))))

(define PRIMARY
  (lambda (n)
    (HELPER STARTING-VALUE 1 n)))

We can also use this approach if we are approach n in some way other than adding one.

(define HELPER
  (lambda (so-far i n)
    (if (> i n)
        so-far
        (HELPER (UPDATE so-far i) (* i 2) n))))

Preparation

a. Make a copy of numeric-recursion-lab.rkt, which contains important code from the reading.

b. Do the normal lab setup. That is

• Start DrRacket.
• Make sure that the loudhum package is up to date.
• Add (require loudhum) to the top of the definitions pane.

Exercises

Exercise 1: Counting down

Define and test a recursive 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)

Note: You should use cons to build up the list.

Note: You are better off writing this with direct recursion (the first pattern above), rather than using a helper procedure.

When you are finished, you may want to read the notes on this exercise.

Exercise 2: Filling lists

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

> (value-replicate 5 "sample")
'("sample" "sample" "sample" "sample" "sample")
> (value-replicate 3 10)
'(10 10 10)
> (value-replicate 1 null)
'(())
> (value-replicate 2 null)
'(() ())
> (value-replicate 0 "hello")
'()

You should not call the built-in make-list procedure. Instead, implement value-replicate recursively.

When you are finished writing this procedure, compare it to the notes on this exercise.

Exercise 3: Counting to

As you may recall, range is a 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. As you’ve seen, the range procedure can be quite useful.

Implement and test your own version of range, which you should call my-iota. (Before Racket included range, we wrote our own version, which we call iota for reasons lost to the dustbin of history.)

For example,

> (my-iota 3)
'(0 1 2)
> (my-iota 5)
'(0 1 2 3 4)
> (my-iota 1)
'(0)

Note that you will probably need to use a helper of some sort to write my-iota. You might use the traditional form of helper, which adds an extra parameter. You might also use a helper that simply computes the list in the reverse order. (Most students write a backwards version in the first attempt; instead of throwing it away, rename it and call it from my-iota.)

Exercise 4: Counting digits

Here is the definition of a procedure that computes the number of digits in the base-ten representation of the whole portion 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?

Exercise 5: Powers of …

a. Write a procedure, (powers-of-two n), that produces a list of all of the powers of two less than or equal to n.

> (powers-of-two 5)
'(1 2 4)
> (powers-of-two 10)
'(1 2 4 8)
> (powers-of-two 100)
'(1 2 4 8 16 32 64)

b. Write a procedure, (powers-of-three n), that produces a list of all of the powers of three less than or equal to n.

> (powers-of-three 100)
'(1 3 9 27 81)
> (powers-of-three 1000)
'(1 3 9 27 81 243 729)

c. You’ve likely found that powers-of-two and powers-of-three look remarkably the same. When you find procedures that look the same, one approach is to write a template. But another is to add an extra parameter.

Write a procedure, (powers-of k n), that produces a list of all the powers of k less than or equal to n.

> (powers-of 2 100)
'(1 2 4 8 16 32 64)
> (powers-of 3 100)
'(1 3 9 27 81)
> (powers-of pi 100)
'(1 3.141592653589793 9.869604401089358 31.006276680299816 97.40909103400242)

d. Rewrite powers-of-two and powers-of-three in terms of powers-of.

Hint: Use section.

Exercise 6: Exponentiation

Using recursion over natural numbers, define a recursive procedure, (two-to-the n), that takes a non-negative integer as its input and returns the result of raising 2 to the power of that number.

For example,

> (two-to-the 3)
8
> (two-to-the 0)
1
> (two-to-the 10)
1024

It is is possible to implement the procedure non-recursively, using Scheme’s primitive expt procedure. The point of this exercise is to use recursion.

For those who finish early

Extra 1: The nth element

Write a procedure, (my-list-ref lst n), that extracts element n of a list. For example,

> (my-list-ref (list "red" "orange" "yellow" "green" "blue" "indigo" "violet") 5)
"indigo"
> (my-list-ref (list "red" "orange" "yellow" "green" "blue" "indigo" "violet") 0)
"red"

Even though this procedure does the same thing as list-ref, you should not use list-ref to implement it. Instead, your goal is to figure out how list-ref works, which means that you will need to implement this procedure using direct recursion.

Hint: When recursing, you will need to simplify the numeric parameter (probably by subtracting 1) and the list parameter (probably by taking its cdr).

Extra 2: Computing list prefixes

Define and test a recursive procedure, (list-prefix lst n), that returns a list consisting of the first n elements of the list, lst, in their original order. You might also think of list-prefix as returning all the values that appear before index n.

For example,

> (list-prefix (list "a" "b" "c" "d" "e") 3)
("a" "b" "c")
> (list-prefix (list 2 3 5 7 9 11 13 17) 2)
(2 3)
> (list-prefix (list "here" "are" "some" "words") 0)
()
> (list-prefix (list null null) 2)
(() ())
> (map rgb->color-name (list-prefix (map color-name->rgb (list "black" "white" "green") 1)) 
("black")

While your procedure has much the same purpose as take, you should not call take. Rather, your goal is to implement the procedure recursively.

Extra 3: A more efficient exponentiation procedure

The typical definition of two-to-the looks something like the following.

(define two-to-the 
  (lambda (n)
    (if (zero? n)
        1
        (* 2 (two-to-the (- n 1))))))

This definition requires approximately n recursive calls. Can we do better? It turns out we can, if we take advantage of a simple rule about exponentiation: 2 to the 2*k power is the same as the square of 2 to the kth power. 2^(2k) = (2^k)*(2^k) or square(2^k).

Using that technique for even powers, come up with a more efficient version of two-to-the.

Notes

Notes on exercise 1: Counting down

Here’s a possible solution to the problem. The base case is easy. If the number is zero, then the list of all non-negative numbers less than or equal to zero is the list that contains only zero.

    (if (zero? n)
        (list 0)

In the recursive case, we assume that we can compute the list of all numbers less than or equal to n-1. To get the complete list, we simply add n to the front.

        (cons n (count-down (- n 1)))

Putting it all together, we get

;;; Procedure:
;;;   count-down
;;; Parameters:
;;;   n, a non-negative integer
;;; Purpose:
;;;   Create a list of the form (n n-1 n-2 ... 3 2 1 0).
;;; Produces:
;;;   nums, a list
;;; Preconditions:
;;;   [No additional.]
;;; Postconditions:
;;;   (length nums) = n+1
;;;   (list-ref nums i) = n-i for all i, 0 <= i <= n.
(define count-down
  (lambda (n)
    (if (zero? n)
        (list 0)
        (cons n (count-down (- n 1))))))

Return to the problem.

Notes on exercise 2: Filling lists

We begin by considering the base case. The last example may give a hint: If you want zero copies, you end up with the empty list.

    (if (zero? n)
        null

Now, on to the recursive case. If we can create a list of n-1 copies, we can then create a list of n copies by prepending one more copy.

        (cons val (value-replicate val (- n 1)))

Putting it all together, we get

;;; Procedure:
;;;   value-replicate
;;; Parameters:
;;;   n, a non-negative integer
;;;   val, a Scheme value
;;; Purpose:
;;;   Create a list of n copies of val.
;;; Produces:
;;;   val-lst
;;; Preconditions:
;;;   [No additional]
;;; Postconditions:
;;;   (length val-lst) is n
;;;   (list-ref val-list i) is val for all i, 0 <= i < n
(define value-replicate
  (lambda (n val)
    (if (zero? n)
        null
        (cons val (value-replicate (- n 1) val)))))

Return to the problem.