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At times, 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 form 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
sqaure 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 (ls) (if (null? ls) ; The sum of an empty list ; The sum of a non-empty list ) ; if ) ; lambda ) ; sum

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 ls)`

, which is the first element, and ```
(cdr
ls)
```

, 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!

;;; sum: find the sum of the elements of a given list of numbers ;;; Given: ;;; LS, a list of numbers. ;;; Result: ;;; TOTAL, a number. ;;; Preconditions: ;;; None. ;;; Postcondition: ;;; TOTAL is the result of adding together all of the elements of LS. (define sum (lambda (ls) (if (null? ls) 0 (+ (car ls) (sum (cdr ls))) ) ; if ) ; lambda ) ; sum

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.

That's 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.

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:

;;; greatest-of-list: find the greatest element of a given list of real ;;; numbers ;;; Given: ;;; LS, a list of real numbers. ;;; Result: ;;; GREATEST, a real number. ;;; Precondition: ;;; LS is not empty. ;;; Postconditions: ;;; (1) GREATEST is an element of LS. ;;; (2) GREATEST is greater than or equal to every element of LS.

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 (ls) (if (null? (cdr ls)) (car ls) (max (car ls) (greatest-of-list (cdr ls))))))

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:

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:

;;; all-even?: determine whether all of the elements of a list of ;;; integers are even ;;; Given: ;;; LS, a list of integers. ;;; Result: ;;; RESULT, a Boolean. ;;; Preconditions: ;;; None. ;;; Postconditions: ;;; RESULT is #T if all of the elements of LS are even, #F if any ;;; of them is not even. (define all-even? (lambda (ls) (or (null? ls) (and (even? (car ls)) (all-even? (cdr ls))))))

When `ls`

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 `ls`

is
odd, the test fails, so `all-even?`

stops, returning
`#f`

. However, if the first element of `ls`

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.

September 2, 1997 (John David Stone)

- Created

March 17, 2000 (John David Stone)

- Last revised

Monday, 5 September 2000 (Samuel A. Rebelsky)

- Rewritten and reformatted

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**Disclaimer** Often, these pages were created "on the fly" with little, if any, proofreading. Any or all of the information on the pages may be incorrect. Please contact me if you notice errors.

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