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, positive
integers have a recursive structure of which we can take advantage
when we write direct-recursion procedures. A positive integer, n,
is either (a) zero, or (b) the successor of a smaller positive integer,
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-check)
(base-case val)
(combine (partof val)
(recursive-proc (simplify val))))))
For lists, the check 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 positive integers, 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 positive integer 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 positive
integer into the appropriate return value for a given non-zero integer.
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 positive integer,
number, computes the result of adding together all of the positive
integers up to and including number. At Grinnell, this result is
traditionally called the “termial” of the number, a play on the term
“factorial”..
;;; Procedure:
;;; termial
;;; Parameters:
;;; number, a positive integer
;;; Purpose:
;;; Compute the sum of the positive integers not greater than a given integer
;;; Produces:
;;; sum, an integer
;;; 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 positive integers 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 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 integers is reducing
the number by 1.
termial in actionHere’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
termialThe 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. 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)))))
=> ...
Similarly, 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.
(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)))))))
=> ...
Hence, we might use a husk-and-kernel strategy to protect our procedure.
(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
(termial-kernel number)])))
(define termial-kernel
(lambda (number)
(if (zero? number)
0
(+ number (termial-kernel (- number 1))))))
termialNote that our “sum all the values” algorithm is not the only way to compute the termial of an integer. 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 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. (Not an issue in our implementation of
;;; Scheme, but we mention ;;; it anyway.)
;;; 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.)
Our base case need not be a small number. We can, for example, use
direct recursion to approach the base case from below by repeated
additions of 1, provided we are sure that our starting point is
less than or equal to that target. Here’s a variant of termial
that sums the values from a to b (rather than t to n),
inclusive.
;;; Procedure:
;;; sum-between
;;; Parameters:
;;; a, an integer
;;; b, an integer
;;; Purpose:
;;; Compute the sum of the numbers from a to b, inclusive.
;;; Produces:
;;; sum, an integer
;;; Preconditions:
;;; a <= b.
;;; Postconditions:
;;; sum = a + (a+1) + (a+2) + ... + (b-1) + b.
(define sum-between
(lambda (a b)
(if (= a b)
b
(+ a (sum-between (+ a 1) b)))))
> (sum-between 1 3)
6
> (+ 1 2 3)
6
> (sum-between 7 11)
45
> (+ 7 8 9 10 11)
45
> (sum-between -5 5)
0
In the examples above, our recursive procedures returned numbers. But
we can also have numeric recursive procedures return lists. For example,
here’s one way we might write make-list if it did not already exist.
(define my-make-list
(lambda (n val)
(if (zero? n)
null
(cons val (my-make-list (- n 1) val)))))
The reading shows you how the basic patterns of
recursion we have seen apply
to numbers just as well as lists. Identify the standard pieces of the
recursive procedure sum-between by filling in the blanks below.