Fundamentals of Computer Science I: Media Computing (CS151.01 2008S)
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Summary: We've learned a number of basic techniques for writing recursive functions over lists, including the common pattern of recursion. But these techniques aren't always the clearest or most elegant for every case. Here, we extend your recursion toolbox to include a few more techniques, particularly the identification of special base cases and ways to write recursive predicates.
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 leftmost element of a given non-empty list of spots. (The list must be non-empty because there is no “leftmost element” of an empty list.)
>
(spots-leftmost (list (spot-new 10 0 color-white) (spot-new 3 5 color-white) (spot-new 6 2 color-white) (spot-new 1 50 color-white) (spot-new 8 5 color-white)))
(1 50 -1)
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. A precondition is a requirement
that must be met in order for your procedure to work correctly.
You should form the habit of figuring out when such preconditions
are appropriate. With the six-P technique for documenting procedures,
you should also make it a habit to document such preconditions as you
write the initial comment for a procedure:
;;; Procedure:
;;; spots-leftmost
;;; Parameters:
;;; spots, a list of spots.
;;; Purpose:
;;; Find the leftmost spot in spots.
;;; Produces:
;;; leftmost, a spot.
;;; Preconditions:
;;; spots is not empty.
;;; All the values in spots are spots. That is, each is a list of
;;; three values, two integers and an RGB color. The first
;;; integer represents the column, the second the row, and the
;;; third the color.
;;; Postconditions:
;;; leftmost is an element of spots (and, by implication, is a spot).
;;; For each spot in spots, leftmost is either in the same column as
;;; the spot, or to the left.
If a list of spots is a singleton, the answer is trivial -- its only
element is its leftmost. Otherwise, we can take the list apart
into its car and its cdr, invoke the procedure recursively to find the
leftmost of the cdr, and then try to figure out which comes first.
How do we figure whether or not one spot is to the left of another?
We compare their columns. Let's use a helper procedure. (This is not
a recursive helper procedure. Rather, like rgb-darker
,
it is a relatively straightforward procedure that simplifies our
recursive definitions.)
;;; Procedure: ;;; spot-leftmost ;;; Parameters: ;;; spot1, a spot ;;; spot2, a spot ;;; Purpose: ;;; Determine the leftmost of spot1 and spot2. ;;; Produces: ;;; leftmost, a spot. ;;; Preconditions: ;;; spot1 and spot2 have the correct form for spots. ;;; Postconditions: ;;; leftmost is equal to either spot1 or spot2. ;;; leftmost is either in the same column as both spot1 and spot2 or ;;; has the same column as one, and is to the left of the other. (define spot-leftmost (lambda (spot1 spot2) (if (< (spot-col spot1) (spot-col spot2)) spot1 spot2)))
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
spots))
is #t
if spots
is a singleton,
#f
if color
has two or more elements.
Here, then, is the procedure definition:
(define spots-leftmost (lambda (spots) (if (null? (cdr spots)) (car spots) (spot-leftmost (car spots) (spots-leftmost (cdr spots))))))
If someone who uses this procedure happens to violate its precondition, applying the procedure to the empty list, DrFu notices the error and prints out a diagnostic message:
>
(spots-leftmost null)
Error: cdr: argument 1 must be: pair
If you think back to the tail-recursive version of difference
,
you may note another time that we had a special singleton case. When we
compute
v
_{1} -
v
_{2} -
v
_{3} -
v
_{k},
the base case is not “we have nothing to subtract”, but rather
“we have nothing to subtract from
v
_{1}”.
We didn't note the need for a singleton base case until we tried to write a tail-recursive version, but the need was there. Of course, that means that we might consider rewriting the non-tail-recursive version, but that version gave us the wrong answer, anyway.
If you consider the examples you've studied over the past few days, you will see that there is a common form for most of the procedures. The form goes something like this:
(define recursive-proc (lambda (val) (if (base-case-test?) (base-case-computation val) (combine (partof val) (recursive-proc (simplify val))))))
For example, for the spots-leftmost
procedure,
spots-leftmost
.
spots
, our
list of numbers.
(null? (cdr
spots))
, which checks whether spots
has only one element.
car
, which extracts
the one spot left in spots
.
car
, which extracts the first spot in
spots
.
cdr
, which drops the first element, thereby
giving us a simpler (well, smaller) list.
spot-leftmost
.
Similarly, consider the first complete version of sum
.
;;; 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))))))
In the sum
procedure,
sum
.
numbers
, a list
of numbers.
(null? numbers)
, which checks if we have no numbers.
0
.
(This computation does not quite match the form above, since we
don't apply the 0 to numbers
. As this example
suggests, sometimes the base case does not involve the parameter.)
car
,
which extracts the first value in numbers
.
cdr
, which drops the the first element.
When you write your own recursive procedures, it's often useful
to start with the general structure and then to fill in the
pieces. When you are recursing over lists (as you have in our
first explorations of recursion), partof is
almost always car
and simplify
is almost always cdr
. There's a bit more to
the base-case-test, since we've used both
(null? ___)
and (null? (cdr? ___))
. We may
also find other techniques.
However, when you work with other kinds of information (as you will do soon), you'll have different techniques for extracting a piece of information, for simplifying the argument, and for deciding when you're done.
Note, also, that examples like filtering suggest a similar, but more complex structure for recursive procedures.
(define recursive-proc (lambda (val) (cond ((one-base-case-test?) (one-base-case-computation val)) ((another-base-case-test?) (another-base-case-computation val)) ... ((special-recursive-case-test-1?) (combine-1 (partof-1 val) (recursive-proc (simplify-1 val)))) ((special-recursive-case-test-2?) (combine-2 (partof-2 val) (recursive-proc (simplify-2 val)))) ... (else (combine (partof val) (recursive-proc (simplify val)))))))
However, in practice you will find that you rarely have more than two base-case tests (and mostly one) and rarely more than two recursive cases.
When we write tail-recursive procedures, we simply use the result of the recursive call, and don't combine it with anything. Here's a simple tail recursive pattern.
(define procedure (lambda (val) (procedure-helper initial-result val))) (define procedure-helper (lambda (so-far remaining) (if (base-case-test? remaining) (final-computation so-far) (procedure-helper (combine (part-of remaining) so-far) (simplify remaining)))))
Of course, these common forms are not the only way to define recursive
procedures. In particular, 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 rgb-all-dark?
that takes a given
list of colors and determines whether all of them are dark. As usual,
we consider the cases of the empty list and non-empty lists separately:
rgb-all-dark?
should return #t
when given the empty list.
and
to make sure that they are both satisfied.
Thus, rgb-all-dark?
should return #t
when the given list either is empty or has a dark first element and
all dark elements after that. This yields the following definition:
;;; Procedure: ;;; rgb-all-dark? ;;; Parameters: ;;; colors, a list of rgb colors. ;;; Purpose: ;;; Determine whether all of the elements of a list of colors ;;; represent dark colors. ;;; Produces: ;;; all-dark?, a Boolean. ;;; Preconditions: ;;; All the values in the list are rgb colors. ;;; Postconditions: ;;; all-dark? is #t if all of the elements of values are dark. ;;; all-dark? is #f if at least one element is not dark. (define rgb-all-dark? (lambda (colors) (or (null? colors) (and (rgb-dark? (car colors)) (rgb-all-dark? (cdr colors))))))
When colors
is the empty list,
rgb-all-dark?
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 rgb-all-dark?
proceeds to evaluate
the first test in the and
-expression. If the first
element of colors
is not dark, the test fails,
so rgb-all-dark?
stops, returning #f
.
However, if the first element of colors
is dark,
the test succeeds, so rgb-all-dark?
goes on to the
recursive procedure call, which checks whether all of the remaining
elements are dark, and returns the result of this recursive call,
however it turns out.
Here's a template for that solution.
(define all-____? (lambda (lst) (or (null? lst) (and (____? (car lst)) (all-____? (cdr lst))))))
We can use a similar technique to ask if any value in a list is dark. In this case, if there are no values in the list, we know that no values are dark. Otherwise, we check if the first value is dark. If it is, then some value must be dark.
The complicated part is getting the base case right (particularly
if we want to avoid using if
). The standard technique
is to require that the list not be null (using not
and and
). If the list is null,
(not (null? lst))
returns #f
. And, since
(and #f ...)
is #f
, we get false back for
the empty list, just as we wanted.
;;; Procedure: ;;; rgb-any-dark? ;;; Parameters: ;;; colors, a list of rgb colors. ;;; Purpose: ;;; Determine whether any of the elements of a list of colors ;;; represent dark colors. ;;; Produces: ;;; any-dark?, a Boolean. ;;; Preconditions: ;;; All the values in the list are rgb colors. ;;; Postconditions: ;;; any-dark? is #t if at least one of the elements of values are dark. ;;; any-dark? is #f if all of the elements are not dark. (define rgb-any-dark? (lambda (colors) (and (not (null? colors)) (or (rgb-dark? (car colors)) (rgb-any-dark? (cdr colors))))))
And, once again, we can generalize.
(define any-____? (lambda (lst) (and (not (null? lst)) (or (____? (car lst)) (any-____? (cdr lst))))))
Primary: [Front Door] [Syllabus] - [Academic Honesty] [Instructions]
Current: [Outline] [EBoard] [Reading] [Lab] [Assignment]
Groupings: [Assignments] [EBoards] [Examples] [Exams] [Handouts] [Labs] [Outlines] [Projects] [Readings]
References: [A-Z] [Primary] [Scheme Report (R5RS)] [Scheme Reference] [DrScheme Manual]
Related Courses: [CSC151.02 2008S (Davis)] [CSC151 2007F (Rebelsky)] [CSC151 2007S (Rebelsky)] [CSCS151 2005S (Stone)]
Copyright (c) 2007-8 Janet Davis, Matthew Kluber, and Samuel A. Rebelsky. (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.
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