This lab is also available in PDF.
Summary:
In this laboratory, we consider the various techniques for creating
local recursive procedures, particularly letrec and named
let. We also review related issues, such as husk-and-kernel
programming.
Contents:
a. If any of the following procedures are not in your library, add them to
your library.
;;; Procedure:
;;; rgb.light?
;;; Parameters:
;;; color, an RGB color
;;; Purpose:
;;; Determine if the color appears light.
;;; Produces:
;;; light?, a Boolean
(define rgb.light?
(lambda (color)
(< 66 (rgb.brightness color))))
;;; Procedure:
;;; rgb.dark?
;;; Parameters:
;;; color, an RGB color
;;; Purpose:
;;; Determine if the color appears dark.
;;; Produces:
;;; dark?, a Boolean
(define rgb.dark?
(lambda (color)
(> 34 (rgb.brightness color))))
;;; Procedure:
;;; rgb.brightness
;;; Parameters:
;;; color, an RGB color
;;; Purpose:
;;; Computes the brightness of color on a 0 (dark) to 100 (bright) scale.
;;; Produces:
;;; b, an integer
;;; Preconditions:
;;; color is a valid RGB color. That is, each component is between
;;; 0 and 255, inclusive.
;;; Postconditions:
;;; If color1 is likely to be perceived as brighter than color2,
;;; then (brightness color1) > (brightness color2).
(define rgb.brightness
(lambda (color)
(round (* 100 (/ (+ (* 0.30 (rgb.red color))
(* 0.59 (rgb.green color))
(* 0.11 (rgb.blue color)))
255)))))
;;; Procedure:
;;; rgb.brighter?
;;; Parameters:
;;; color1, a color
;;; color2, a color
;;; Purpose:
;;; Determine if color1 is strictly brighter than color 2.
;;; Produces:
;;; brighter?, a Boolean
;;; Preconditions:
;;; [No additional preconditions.]
;;; Postconditions:
;;; If (rgb.brightness color1) > (rgb.brightness color2)
;;; then brighter is true (#t)
;;; Otherwise
;;; brighter is false (#f)
(define rgb.brighter?
(lambda (color1 color2)
(> (rgb.brightness color1) (rgb.brightness color2))))
b. Create a new window in which you load your library file.
As the reading suggested,
local kernels are particularly appropriate for checking preconditions.
Local kernels are also appropriate for writing the helpers for the
recursive helper procedures that take an extra parameter, one that
accumulates a result.
For example, here's the standard helper-based definition of
rgb.brightest.
(define rgb.brightest
(lambda (colors)
(rgb.brightest-helper (car colors) (cdr colors))))
(define rgb.brightest-helper
(lambda (brightest-so-far remaining-colors)
(cond
((null? remaining-colors)
brightest-so-far)
((rgb.brighter? (car remaining-colors) brightest-so-far)
(rgb.brightest-helper (car remaining-colors) (cdr remaining-colors)))
(else
(rgb.brightest-helper brightest-so-far (cdr remaining-colors))))))
a. Rewrite this to make the helper local and to name the helper
kernel.
b. Test your procedure on a few simple lists of colors. For example,
you might try some of the following.
(rgb->string (rgb.brightest (list color.red color.green color.blue)))
(rgb->string (rgb.brightest (list color.black)))
(rgb->string (rgb.brightest (list color.red. color.black color.green color.yellow)))
(rgb->string (rgb.brightest (list color.yellow color.red. color.black color.green)))
c. When you are done, you may want to compare
your answer to the sample solution in
the notes on this problem.
The procedure given above, and your rewrite of that procedure, will fail
miserably if given an inappropriate parameter, such as an empty list, a
list that contains values that are not colors, or a non-list. Rewrite
rgb.brightest so that it checks its preconditions (including
that the list contains only colors) before calling the kernel.
Rewrite rgb.brightest-helper using a named let rather
than letrec. (The goal of this problem is to give you experience
using named let, so you need not check preconditions for this
exercise.)
When you are done, you may want to compare
your answer to the sample solution in
the notes on this problem.
A list of spots is a light-dark-alternator if its elements
are alternately light spots and dark spots, beginning with a light spot.
A list of spots is a dark-light-alternator if its
elements are alternately dark and light, beginning with a
dark spot. (We say that the empty list is both a light-dark-alternator
and a dark-light-alternator.)
a. Write a predicate, spots.alternating-brightness?,
that determines whether a non-empty list of spots is either
a light-dark-alternator or a dark-light-alternator. Your
definition should include a a letrec expression
in which the identifiers light-dark-alternator?
and dark-light-alternator? are bound to mutually
recursive predicates, each of which determines whether a given list
has the indicated characteristic.
(define spots.alternating-brightness?
(lambda (spots)
(letrec ((light-dark-alternator?
(lambda (spots) ...))
(dark-light-alternator?
(lambda (spots) ...)))
...)))
When you are done, you may want to compare your answer to the sample
solution in the notes on this problem.
b. Here are a few lists of spots. For which do you expect
spots.alternating-brightness? to hold?
(define sample0 null)
(define sample1
(list (spot.new 0 0 color.white)))
(define sample2
(list (spot.new 0 0 color.black)))
(define sample3
(list (spot.new 0 0 color.white)
(spot.new 0 1 color.black)))
(define sample4
(list (spot.new 0 0 color.black)
(spot.new 0 1 color.white)))
(define sample5
(list (spot.new 0 0 color.black)
(spot.new 0 1 color.black)))
(define sample6
(list (spot.new 0 0 color.white)
(spot.new 0 1 color.black)
(spot.new 0 2 color.white)))
(define sample7
(list (spot.new 0 0 color.white)
(spot.new 0 1 color.black)
(spot.new 0 2 color.white)
(spot.new 0 3 color.black)
(spot.new 0 4 color.white)))
(define sample8
(list (spot.new 0 0 color.red)
(spot.new 1 0 color.yellow)
(spot.new 2 0 color.blue)
(spot.new 3 0 color.white)
(spot.new 4 0 color.black)))
c. Check your answers experimentally.
Define and test a procedure, (list.take lst n),
returns a list consisting of the first n elements of
the list, lst, in their original order.
You might also think of take as returning all the values
that appear before index n.
For example,
> (list.take (list "a" "b" "c" "d" "e") 3)
("a" "b" "c")
> (list.take (list 2 3 5 7 9 11 13 17) 2)
(2 3)
> (list.take (list "here" "are" "some" "words") 0)
()
> (list.take (list null null) 2)
(() ())
> (map rgb->string (list.take (list rgb.black rgb.white rgb.green) 1))
("0/0/0")
The procedure should signal an error if lst is not a list, if
n is not an exact integer, if n is negative,
or if n is greater than the length of lst.
Note that in order to signal such errors, you may want to take advantage
of the husk-and-kernel programming style.
Rewrite list.take to use whichever of named let
and letrec you didn't use in the previous exercise.
You've now seen two examples in which you've written two different
solutions, one using letrec and one use named let.
Reflect on which of the two strategies you prefer and why.
As you may recall, the iota procedure takes a natural
number as a parameter and returns a list of all the lesser
natural numbers in ascending order. For example,
a. Define and test a version of the iota procedure that uses
letrec to pack an appropriate kernel inside a husk. The
husk should do precondition testing and the kernel should build the list.
This version of iota should
look something like
(define iota
(lambda (num)
(letrec ((kernel (lambda (...) ...)))
(cond
((fails-precondition) (error ...))
...
(else (kernel num))))))
b. Define and test a version of the iota procedure that uses
a named let. This version of iota should
look something like
(define iota
(lambda (num)
(cond
((fails-precondition) (error ...))
...
(else
(let kernel (...)
...)))))
Here's one possible solution.
(define rgb.brightest
(lambda (colors)
(letrec ((kernel
(lambda (brightest-so-far remaining-colors)
(cond
((null? remaining-colors)
brightest-so-far)
((rgb.brighter? (car remaining-colors) brightest-so-far)
(kernel (car remaining-colors)
(cdr remaining-colors)))
(else
(kernel brightest-so-far
(cdr remaining-colors)))))))
(kernel (car colors) (cdr colors)))))
Here's one possible solution.
(define rgb.brightest
(lambda (colors)
(let kernel ((brightest-so-far (car colors))
(remaining-colors (cdr colors)))
(cond
((null? remaining-colors)
brightest-so-far)
((rgb.brighter? (car remaining-colors) brightest-so-far)
(kernel (car remaining-colors)
(cdr remaining-colors)))
(else
(kernel brightest-so-far
(cdr remaining-colors)))))))
Once light-dark-alternator? and dark-light-alternator? are written, the definition is fairly straightforward. A non-empty list of
spots has alternating brightness if it's not empty and it's either a
light-dark-alternator or a dark-light-alternator.
(and (not (null? spots))
(or (light-dark-alternator? spots)
(dark-light-alternator? spots)))
Each definition is also fairly straightforward. A list is a
light-dark-alternator if it's empty or if the car is light and
the cdr is a dark-light-alternator.
(light-dark-alternator?
(lambda (spots)
(or (null? spots)
(and (rgb.light? (spot.color (car spots)))
(dark-light-alternator? (cdr spots))))))
The definition of dark-light-alternator? is so similar
that we will not give it separately.
Putting everything together, we get
(define spots.alternating-brightness?
(lambda (spots)
(letrec ((light-dark-alternator?
(lambda (spots)
(or (null? spots)
(and (rgb.light? (spot.color (car spots)))
(dark-light-alternator? (cdr spots))))))
(dark-light-alternator?
(lambda (spots)
(or (null? spots)
(and (rgb.dark? (spot.color (car spots)))
(light-dark-alternator? (cdr spots)))))))
(and (not (null? spots))
(or (light-dark-alternator? spots)
(dark-light-alternator? spots))))))
Note that there's a hidden moral here: The procedures defined in a
letrec can be mutually recursive.
;
