# Recursion Basics

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. While you have learned a few techniques for repeating instructions, these techniques only work with specific data structure, such as images and lists. Hence, you should also know a more general technique, that will work with any instance in which you need to repeat code. In Scheme, the primary technique used for repetition is called recursion, and involves having procedures call themselves.

## Introduction

In your work so far, you've seen how to repeat actions using two basic types: You can do something for each position in an image, or you can do something for each element of a list. You've also seen two general strategies for what you do at each step: You can either update something (such as an image), as is the case for `image-transform!`, `image-compute-pixels!`, or `for-each`. Alternately, you can build a new compound value by applying a function to each value in the original compound value, as in the case of `image-variant` or `map`.

But what if you want to do other kinds of repetition? For example, what if you want to do something for every integer between 1 and 100, of if you want to do something for every third element of a list? What if you want to compute a list that contains only some elements of an original list? What if you want to compute a value that is based on all of the elements of an original list, such as the average of a list? Since it is not possible to predict every way that a programmer will want to repeat code, most languages provide a general mechanism for repeating instructions. In Scheme, the most general mechanism for repeating instructions is called recursion, and involves having a procedure call itself.

## Rethinking Procedure Calls

But what does it mean to have a procedure “call itself”?

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 instructions to find the average component as

```(define rgb-average-component
(lambda (color)
(/ (+ (rgb-red color) (rgb-green color) (rgb-blue color))
3)))
```

Here, there are calls to division, addition, and the three rgb component procedures in the definition of `rgb-average-component`.

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.

## An Example: Summation

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 9.3))`
`9.3`
`>` `(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 empty or 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 two 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! We'd express that portion as follows.

```        (+ (car numbers) (sum (cdr numbers)))
```

As we put it together into a complete procedure, we'll also add some documentation.

```;;; 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.
;;;   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 that 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.

### Watching Sum Work

The strategy of repeatedly solving simpler problems 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 x `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 is a summary of some of the key steps in the evaluation process. (We've left out the evaluation of the conditionals, since they are fairly straightforward.)

```   (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.

## Base Cases

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 MediaScript not as a diagnostic message printed in red, but as an apparently endless wait for a result. The good news is that you can hit the Stop button to stop the computation. While MediaScript should behave normally after you click Stop, we do recommend that you save your code after stopping a recursion, just in case something else goes wrong.

As you may have noted, there are three basic parts to the kind of recursive functions you have learned about.

• A recursive case in which the function calls itself with a simpler or smaller parameter. For `sum`, this was the call to `sum` on the cdr of the list.
• A base case in which the function does not call itself. For `sum`, this was simply the value 0. In writing other recursion procedures, you may find that you need to do a computation in the base case.
• A test that decides which case holds. For `sum`, the test was whether or not the list we want to sum is empty.

You'll need to consider each of these three parts for each recursive function you write.

## Filtering Lists

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 colors and “filters out” the dark ones. Let's assume that we've already defined a predicate, `rgb-dark?`, that determines whether or not a color is dark. For most definitions of `rgb-dark?`, if we started with the list of colors red, yellow, dark grey, green, blue, black, and white, we would end up with red, yellow, green, and white. 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 dark colors 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 dark, 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:

```;;; Procedure:
;;;   rgb-filter-out-dark
;;; Parameters:
;;;   colors, a list of RGB colors.
;;; Purpose:
;;;   Create a new list of colors, consiting only of non-dark colors.
;;; Produces:
;;;   not-dark, a list of RGB colors.
;;; Preconditions:
;;;   rgb-dark? is defined.
;;; Postconditions:
;;;   Every element of not-dark appears in colors.
;;;   If (not (rgb-dark? (list-ref colors i))) holds for some i,
;;;     then the corresponding color appears in not-dark.
(define rgb-filter-out-dark
(lambda (colors)
(if (null? colors)
null
(if (rgb-dark? (car colors))
(rgb-filter-out-dark (cdr colors))
(cons (car colors) (rgb-filter-out-dark (cdr colors)))))))
```

Copyright (c) 2007-9 Janet Davis, Matthew Kluber, Samuel A. Rebelsky, and Jerod Weinman. (Selected materials copyright by John David Stone and Henry Walker and used by permission.)

This material is based upon work partially supported by the National Science Foundation under Grant No. CCLI-0633090. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.

This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License. To view a copy of this license, visit `http://creativecommons.org/licenses/by-nc/2.5/` or send a letter to Creative Commons, 543 Howard Street, 5th Floor, San Francisco, California, 94105, USA.