Summary:
In many algorithms, you want to do things again and again and again.
For example, you might want to do something with each value in a list.
In general, the term used for doing things again and again is called
repetition. In Scheme, the primary technique used for repetition is
called recursion, and involves having procedures call themselves.
As we've already seen, it is commonplace for the body of a procedure to
include calls to another procedure, or even to several others. For example,
we might write our find one root of the quadratic equation
as
(define root1
(lambda (a b c)
(/ (+ (- 0 b)
(sqrt (- (* b b)
(* 4 a c))))
(* 2 a))))
Here, there are calls to addition, subtraction, division, multiplication, and
square root in the definition of root1.
Direct recursion is the special case of this construction in
which the body of a procedure includes one or more calls to the very
same procedure -- calls that deal with simpler or smaller arguments.
For instance, let's define a procedure called sum that takes
one argument, a list of numbers, and returns the result of adding all of
the elements of the list together:
> (sum (list 91 85 96 82 89))
443
> (sum (list -17 17 12 -4))
8
> (sum (list 19/3))
19/3
> (sum null)
0
Because the list to which we apply sum may have any number of
elements, we can't just pick out the numbers using list-ref
and add them up -- there's no way to know in general whether an element
even exists at the position specified by the second argument to
list-ref. One thing we do know about lists, however, is that
every list is either (a) empty, or (b) composed of a first
element and a list of the rest of the elements, which we can obtain with
the car and cdr procedures.
Moreover, we can use the predicate null? to distinguish
between the (a) and (b) cases, and conditional evaluation to
make sure that only the expression for the appropriate case is chosen. So
the structure of our definition is going to look something like this:
(define sum
(lambda (numbers)
(if (null? numbers)
; The sum of an empty list
; The sum of a non-empty list
)))
The sum of the empty list is easy -- since there's nothing to add, the
total is 0.
And we know that in computing the sum of a non-empty list, we can use
(car numbers), which is the first element, and (cdr
numbers), which is the rest of the list.
So the problem is to find the sum of a non-empty list, given
the first element and the rest of the list. Well, the rest of the list is
one of those simpler or smaller
arguments mentioned above.
Since Scheme supports direct recursion, we can invoke the sum
procedure within its own definition to compute the sum of the elements of
the rest of a non-empty list. Add the first element to this sum, and we're
done!
;;; Procedure:
;;; sum
;;; Parameters:
;;; numbers, a list of numbers.
;;; Purpose:
;;; Find the sum of the elements of a given list of numbers
;;; Produces:
;;; total, a number.
;;; Preconditions:
;;; All the elements of numbers must be numbers.
;;; Postcondition:
;;; total is the result of adding together all of the elements of numbers.
;;; total is a number.
;;; If all the values in numbers are exact, total is exact.
;;; If any values in numbers are inexact, total is inexact.
(define sum
(lambda (numbers)
(if (null? numbers)
0
(+ (car numbers) (sum (cdr numbers))))))
At first, this may look strange or magical, like a circular definition: If
Scheme has to know the meaning of sum before it can process
the definition of sum, how does it ever get started?
The answer is what Scheme learns from a procedure definition is not so much
the meaning of a word as the algorithm, the step-by-step method, for
solving a problem. Sometimes, in order to solve a problem, you have to
solve another, somewhat simpler problem of the same sort. There's no
difficulty here as long as you can eventually reduce the problem to one
that you can solve directly.
Another way to think about it is in terms of the way we normally write
instructions. We often say go back to the beginning and do the
steps again
. Given that we've named the steps in the algorithm,
the recursive call is, in one sense, a way to tell the computer to go
back to the beginning.
The repeatedly solving simpler problems strategy is how
Scheme proceeds when it deals with a call to a recursive
procedure -- say, (sum (cons 38 (cons 12 (cons 83 null)))).
First, it checks to find out whether the list it is given is empty. In
this case, it isn't. So we need to determine the result of adding together
the value of (car ls), which in this case is 38, and the sum
of the elements of (cdr ls) -- the rest of the given list.
The rest of the list at this point is the value of (cons 12 (cons 83
null)). How do we compute its sum? We call the sum
procedure again. This list of two elements isn't empty either, so again we
wind up in the alternate of the if-expression. This time we
want to add 12, the first element, to the sum of the rest of the list. By
rest of the list
, this time, we mean the value of (cons 83
null) -- a one-element list.
To compute the sum of this one-element list, we again invoke the
sum procedure. A one-element list still isn't empty,
so we head once more into the alternate of the if-expression,
adding the car, 83, to the sum of the elements of the cdr,
null. The rest of the list
this time around is empty, so
when we invoke sum yet one more time, to determine the sum of
this empty list, the test in the if-expression succeeds and
the consequent, rather than the alternate, is selected. The sum of
null is 0.
We now have to work our way back out of all the procedure calls that have
been waiting for arguments to be computed. The sum of the one-element
list, you'll recall, is 83 plus the sum of null, that is, 83 +
0, or just 83. The sum of the two-element list is 12 plus the sum of the
(cons 83 null), that is, 12 + 83, or 95. Finally, the sum of
the original three-element list is 38 plus the sum of (cons 12 (cons
83 null)) that is, 38 + 95, or 133.
Here's a summary of the steps in the evaluation process.
(sum (cons 38 (cons 12 (cons 83 null))))
=> (+ 38 (sum (cons 12 (cons 83 null)))))
=> (+ 38 (+ 12 (sum (cons 83 null))))
=> (+ 38 (+ 12 (+ 83 (sum null))))
=> (+ 38 (+ 12 (+ 83 0)))
=> (+ 38 (+ 12 83))
=> (+ 38 95)
=> 133
Talk about delayed gratification! That's a while to wait before we can do the
first addition.
The process is exactly the same, by the way, regardless of whether we
construct the three-element list using cons, as in the example
above, or as (list 38 12 83) or '(38 12 83).
Since we get the same list in each case, sum takes it apart in
exactly the same way no matter what mechanism was used to build it.
The method of recursion works in this case because each time we invoke the
sum procedure, we give it a list that is a little shorter and
so a little easier to deal with, and eventually we reach the base
case of the recursion -- the empty list -- for which the answer can be
computed immediately.
If, instead, the problem became harder or more complicated on each
recursive invocation, or if it were impossible ever to reach the base case,
we'd have a runaway recursion -- a programming error that shows up
in DrScheme not as a diagnostic message printed in red, but as an endless
wait for a result. The designers of DrScheme's interface provided a
Break button above the definition window so that you can
interrupt a runaway recursion: Move the mouse pointer onto it and click the
left mouse button, and DrScheme will abandon its attempt to evaluate the
expression it's working on.
As you may have noted, there are three basic parts to these kinds
of recursive functions.
- A recursive case in which the function calls itself with
a simpler or smaller parameter.
- A base case in which the function does not call itself.
- A test that decides which case holds.
You'll come back to these three parts for each function you write.
As you may have noted, an odd thing about the sum procedure
is that it works from right to left. Traditionally, we sum from left
to right. Can we rewrite sum to work from left to right?
Certainly, but we may need a helper procedure (another procedure
whose primary purpose is to assist our current procedure) to do so.
If you think about it, when you're summing a list of numbers from left
to write, you need to keep track of two different things:
- The sum of the values seen so far.
- The remaining values to add.
Hence, we'll build our helper procedure with two parameters,
sum-so-far and remaining. We'll start
the body with a template for recursive procedures (a test to
determine whether to use the base case or recursive case, the base
case, and the recursive case). We'll then fill in each part.
(define new-sum-helper
(lambda (sum-so-far remaining)
(if (test)
base-case
recursive-case)))
The recursive case is fairly easy. We add the first element of
remaining to sum-so-far and continue
with the new sum-so-far and the rest of remaining.
To continue
, we simply call new-sum-helper
again.
(define new-sum-helper
(lambda (sum-so-far remaining)
(if (test)
base-case
(new-sum-helper (+ sum-so-far (car remaining))
(cdr remaining)))))
The recursive case then gives us a clue as to what to use for the test.
We need to stop when there are no elements left in the list.
(define new-sum-helper
(lambda (sum-so-far remaining)
(if (null? remaining)
base-case
(new-sum-helper (+ sum-so-far (car remaining))
(cdr remaining)))))
We're almost done. What should the base case be? In the previous version,
it was 0. However, in this case, we've been keeping a running sum. When
we run out of things to add, the value of the complete sum is the value
of the running sum.
(define new-sum-helper
(lambda (sum-so-far remaining)
(if (null? remaining)
sum-so-far
(new-sum-helper (+ sum-so-far (car remaining))
(cdr remaining)))))
Now we're ready to write the primary procedure whose responsibility it is
to call new-sum-helper. Like sum,
new-sum will take a list as a parameter. That list will
become remaining. What value should sum-so-far
begin with? Since we have not yet added anything when we start, it
begins at 0.
(define new-sum
(lambda (numbers)
(new-sum-helper 0 numbers)))
Putting it all together, we get the following.
;;; Procedure:
;;; new-sum
;;; Parameters:
;;; numbers, a list of numbers.
;;; Purpose:
;;; Find the sum of the elements of a given list of numbers
;;; Produces:
;;; total, a number.
;;; Preconditions:
;;; All the elements of numbers must be numbers.
;;; Postcondition:
;;; total is the result of adding together all of the elements of numbers.
;;; total is a number.
;;; If all the values in numbers are exact, total is exact.
;;; If any values in numbers are inexact, total is inexact.
(define new-sum
(lambda (numbers)
(new-sum-helper 0 numbers)))
;;; Procedure:
;;; new-sum-helper
;;; Parameters:
;;; sum-so-far, a number.
;;; remaining, a list of numbers.
;;; Purpose:
;;; Add sum-so-far to the sum of the elements of a given list of numbers
;;; Produces:
;;; total, a number.
;;; Preconditions:
;;; All the elements of remaining must be numbers.
;;; sum-so-far must be a number.
;;; Postcondition:
;;; total is the result of adding together sum-so-far and all of the
;;; elements of remaining.
;;; total is a number.
;;; If both sum-so-far and all the values in remaining are exact,
;;; total is exact.
;;; If either sum-so-far or any values in remaining are inexact,
;;; total is inexact.
(define new-sum-helper
(lambda (sum-so-far remaining)
(if (null? remaining)
sum-so-far
(new-sum-helper (+ sum-so-far (car remaining))
(cdr remaining)))))
Does this change make a difference in the way in which the sum is
evaluated? Let's watch.
(new-sum (cons 38 (cons 12 (cons 83 null))))
=> (new-sum-helper 0 (cons 38 (cons 12 (cons 83 null))))
=> (new-sum-helper (+ 0 38) (cons 12 (cons 83 null)))
=> (new-sum-helper 38 (cons 12 (cons 83 null)))
=> (new-sum-helper (+ 38 12) (cons 83 null))
=> (new-sum-helper 50 (cons 83 null))
=> (new-sum-helper (+ 50 83) null)
=> (new-sum-helper 133 null)
=> 133
Note that the intermediate results for new-sum were different,
primarily because new-sum operates from left to right.
Often the computation for a non-empty list involves making another test.
Suppose, for instance, that we want to define a procedure that takes a list
of integers and filters out
the negative ones, so that if, for
instance, we give it a list consisting of -13, 63, -1, 0, 4, and -78, it
returns a list consisting of 63, 0, and 4. We can use direct recursion to
develop such a procedure:
- If the given list is empty, there are no elements to filter out and also no
elements to keep, so the correct result is the empty list.
- If the given list is not empty, we examine its car and its cdr. We can use
a call to the very procedure that we're defining to filter negative
elements out of the cdr. That gives a list comprising all of its
non-negative elements.
- If the car of the given list -- that is, its first element -- is negative,
we ignore the car and just return the result of the recursive procedure
call, without change.
- Otherwise, we invoke
cons to attach the car to the new list.
Translating this algorithm into Scheme yields the following definition:
(define filter-out-negatives
(lambda (ls)
(if (null? ls)
null
(if (negative? (car ls))
(filter-out-negatives (cdr ls))
(cons (car ls) (filter-out-negatives (cdr ls)))))))
Sometimes the problem that we need an algorithm for doesn't apply to the
empty list, even in a vacuous or trivial way, and the base case for a
direct recursion instead involves singleton lists -- that is,
lists with only one element. For instance, suppose that we want an
algorithm that finds the greatest element of a given non-empty list of real
numbers.
> (greatest-of-list (list -17 38 62/3 -14/9 204/5 26 19))
204/5
The assumption that the list is not empty is a precondition for
the meaningful use of this procedure, just as a call to Scheme's built-in
quotient procedure requires that the second argument, the
divisor, be non-zero. You should form the habit of noting and detailing
such preconditions as you write the initial comment for a procedure:
;;; Procedure:
;;; greatest-of-list
;;; Parameters:
;;; numbers, a list of real numbers.
;;; Purpose:
;;; Find the greatest element of a given list of real numbers
;;; Produces:
;;; greatest, a real number.
;;; Preconditions:
;;; numbers is not empty.
;;; All the values in numbers are real numbers. That is, numbers
;;; contains only numbers, and none of those numbers are complex.
;;; Postconditions:
;;; greatest is an element of numbers (and, by implication, is real).
;;; greatest is greater than or equal to every element of numbers.
If a list of real numbers is a singleton, the answer is trivial -- its only
element is its greatest element. Otherwise, we can take the list apart
into its car and its cdr, invoke the procedure recursively to find the
greatest element of the cdr, and use Scheme's built-in procedure
max to compare the car to the greatest element of the cdr,
returning whichever is greater.
We can test whether the given list is a singleton by checking whether its
cdr is an empty list. The value of the expression (null? (cdr
ls)) is #t if ls is a singleton,
#f if ls has two or more elements.
Here, then, is the procedure definition:
(define greatest-of-list
(lambda (numbers)
(if (null? (cdr numbers))
(car numbers)
(max (car numbers) (greatest-of-list (cdr numbers))))))
If someone who uses this procedure happens to violate its precondition,
applying the procedure to the empty list, DrScheme notices the error and
prints out a diagnostic message:
> (greatest-of-list null)
cdr: expects argument of type <pair>; given ()
When we define a predicate that uses direct recursion on a given list, the
definition is usually a little simpler if we use and- and
or-expressions rather than if-expressions. For
instance, consider a predicate all-even? that takes a given
list of integers and determines whether all of them are even. As usual, we
consider the cases of the empty list and non-empty lists separately:
- Since the empty list has no elements, it is (as mathematicians say)
vacuously true
that all of its elements are even -- there is certainly
no counterexample that one could use to refute the assertion. So
all-even? should return #t when given the empty
list.
- For a non-empty list, we separate the car and the cdr. If the list is to
count as
all even
, the car must clearly be even, and in addition the
cdr must be an all-even list. We can use a recursive call to determine
whether the cdr is all-even, and we can combine the expressions that test
the car and cdr conditions with and to make sure that they are
both satisfied.
Thus all-even? should return #t when the given
list either is empty or has an even first element and all even elements
after that. This yields the following definition:
;;; Procedure:
;;; all-even?
;;; Parameters:
;;; values, a list of integers.
;;; Purpose:
;;; Determine whether all of the elements of a list of integers
;;; are even.
;;; Produces:
;;; result, a Boolean.
;;; Preconditions:
;;; All the values in the list are integers.
;;; Postconditions:
;;; result is #t if all of the elements of values are even.
;;; result is #f if at least one element is not even.
(define all-even?
(lambda (values)
(or (null? values)
(and (even? (car values))
(all-even? (cdr values))))))
When values is the empty list, all-even? applies the
first test in the or-expression, finds that it succeeds, and
stops, returning #t. In any other case, the first test fails,
so all-even? proceeds to evaluate the first test in the
and-expression. If the first element of values is
odd, the test fails, so all-even? stops, returning
#f. However, if the first element of values is even,
the test succeeds, so all-even? goes on to the recursive
procedure call, which checks whether all of the remaining elements are
even, and returns the result of this recursive call, however it turns out.
;
;
Check with Bobby