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.
Contents:
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.
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.
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)
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.
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
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.
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.
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)
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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