Summary:
In this laboratory, you will ground your understanding of the basic
techniques for locally naming values and procedures in Scheme,
let and let*.
a. Write a nested
let-expression that binds a total of
five names, alpha,
beta, gamma, delta,
and epsilon. With
alpha bound to
(irgb 0 0 255), and each subsequent value to a redder
version of the previous name.
That is,
beta should be a a redder version of alpha
(computed by applying irgb-redder to alpha),
gamma a redder version of beta (computed by
applying irgb-redder to beta), and
so on and so forth. The body of the innermost let
should make a color swatch from those colors.
Your result will look something like
(let ([___ ___])
(let ([___ ___])
(let ([___ ___])
(let ([___ ___])
(let ([___ ___])
(image-show (color-swatch alpha beta gamma delta epsilon)))))))
b. Write a similar expression, this time with
alpha bound to
(irgb 0 0 0). The remaining names should still be bound
to subsequently redder versions of alpha.
Write a let*-expression equivalent to the
let-expression in the previous exercise, but using a
different starting color.
When you are done, you may want to read the notes on this problem.
Consider the following expression, which is a potential solution to the previous problem.
(let* ([alpha (irgb 128 128 128)]
[beta (irgb-redder alpha)]
[gamma (irgb-redder beta)]
[delta (irgb-redder gamma)]
[epsilon (irgb-redder delta)])
(image-show (color-swatch alpha beta gamma delta epsilon)))
a. Open a new pane, enter the traditional (require
gigls/unsafe) at the top of the definitions pane, click
, and then
copy and paste the code into the interactions
pane and make sure that it does what it is supposed to do.
b. What do you expect to have happen if we enter the following in the interactions pane and then rerun the code?
(define irgb-redder
(lambda (rgb)
(rgb-new (- (irgb-red rgb) 32)
(- (irgb-green rgb) 32)
(+ (irgb-blue rgb) 32))))
c. Check your answer experimentally. That is, copy and paste the
new definition of irgb-redder into the interactions pane,
then copy and paste the expression that creates contain progressively
redder values of grey.
d. Click run to restore the old definition of
irgb-redder.
e. You may have noted that someone can make our code misbehave by
redefining irgb-redder. Can you tell why?
Suggest a way to update the expression so that it is not affected by
changes to the definition of irgb-redder.
If you're not sure of your answer, you can check the notes on this problem.
Recall that sometimes it's useful to see values as they are being computed. Here's a procedure that makes it easy to tell when an expression is being evaluated. It prints the value it is called with and then returns the value.
(define value
(lambda (val)
(display "Computed: ")
(display val)
(newline)
val))
The value procedure should be familiar, since it's much like
the yield procedure we explored earlier.
a. What do you expect to happen when you execute the following command?
(+ (value 5) (value 7))
b. Check your answer experimentally.
c. What do you expect to happen when you execute the following command?
(* (value (+ (value 2) (value 3))) (value (+ (value 1) (value 1))))
d. Check your answer experimentally.
e. What do you expect to happen when you execute the following command?
(define tmp (value (* 3 4 5)))
In the reading, we
noted that it is possible to move bindings outside of the lambda in
a procedure definition. In particular, we noted that the first of
the two following versions of years-to-seconds required
recomputation of seconds-per-year every time it was called
while the second required that computation only once.
(define years-to-seconds
(lambda (years)
(let* ([days-per-year 365.24]
[hours-per-day 24]
[minutes-per-hour 60]
[seconds-per-minute 60]
[seconds-per-year (* days-per-year hours-per-day
minutes-per-hour seconds-per-minute)])
(* years seconds-per-year))))
(define years-to-seconds
(let* ([days-per-year 365.24]
[hours-per-day 24]
[minutes-per-hour 60]
[seconds-per-minute 60]
[seconds-per-year (* days-per-year hours-per-day
minutes-per-hour seconds-per-minute)])
(lambda (years)
(* years seconds-per-year))))
a. Rename the first version years-to-seconds-a and
the second years-to-seconds-b.
b. Using value, confirm that years-to-seconds-a does, in fact,
recompute the values each time it is called. You might, for example, replace
(seconds-per-year (* days-per-year hours-per-day
minutes-per-hour seconds-per-minute)))
with
(seconds-per-year (value (* days-per-year hours-per-day
minutes-per-hour seconds-per-minute))))
c. Confirm that years-to-seconds-b does not recompute the
values each time it is called. Again, make changes like those reported above.
d. Given that years-to-seconds-b does not recompute each
time, when does it do the computation? (Consider when you see the
messages.)
You may recall that we defined a procedure to compute a grey with the same brightness as a given color.
(define irgb-greyscale
(lambda (color)
(let ([component (+ (* 0.30 (irgb-red color))
(* 0.59 (irgb-green color))
(* 0.11 (irgb-blue color)))])
(irgb component component component))))
You might be tempted to move the let clause outside the
lambda, just as we did in the previous exercise. However, as we noted
in this reading, this reordering will fail.
a. Verify that it will not work to move the let before
the lambda, as in the following. (Note that we have renamed the
procedure because irgb-greyscale is part of
Mediascheme.)
(define irgb-greyscale-1
(let ([component (+ (* 0.30 (irgb-red color))
(* 0.59 (irgb-green color))
(* 0.11 (irgb-blue color)))])
(lambda (color)
(irgb component component component))))
b. Explain, in your own words, why this fails (and why it should fail).
If you find that you have some extra time, you can try any or all of these problems, in any order you prefer.
Some programmers find anonymous procedures a bit too anonymous. Hence, even when they only want to use a procedure once, they still name it. However, because they want to limit the impact of creating that procedure (e.g., they don't want to conflict with someone else's procedure with a similar name), because they don't want to bother writing the six-P documentation, or because they find their code is more readable if they name procedures, they write a local definition.
For example, here's an alternate way to transform an image by dropping all but the green component.
>(let ([only-green (lambda (color) (irgb 0 (irgb-green color) 0))]) (image-show (image-variant picture only-green)))
a. Load an image of your choice and call it picture.
b. Verify that the code above works as described.
c. Write a similar expression that computes a variant in which the blue component of each pixel is 255 minus the blue component of the corresponding pixel in the original.
Consider the following procedure
(define drawing-munge
(lambda (drawing)
(let ([d0 drawing])
(let ([d1 (hshift-drawing 5 d0)])
(let ([d2 (hshift-drawing 5 d1)])
(let ([d3 (hshift-drawing 5 d2)])
(let ([d4 (hshift-drawing 5 d3)])
(let ([d5 (hshift-drawing 5 d4)])
(drawing-group d0 d1 d2 d3 d4 d5)))))))))
a. Explain, in your own words, what drawing-munge
does.
b. Check your answer experimentally.
c. Rewrite the expression to be more concise.
In previous labs and homework assignments, you've written instructions
that use the GIMP tools to create complex pictures (e.g., of a smiley
face, a house, or a scene of your choice). Turn those instructions
into a procedure, using let or let* to name
the image and any other values that are used repeatedly.
a. Here's the form of answer we expected.
(let* ([alpha (irgb 128 128 128)]
[beta (irgb-redder alpha)]
[gamma (irgb-redder beta)]
[delta (irgb-redder gamma)]
[epsilon (irgb-redder delta)])
(image-show (color-swatch alpha beta gamma delta epsilon)))
Since we're relying on a particular definition of irgb-redder, we may want to put it in our let* statement.
(let* ([irgb-redder (lambda (color)
(irgb (+ 32 (irgb-red color))
(irgb-green color)
(irgb-blue color)))]
[alpha (irgb 128 128 128)]
[beta (irgb-redder alpha)]
[gamma (irgb-redder beta)]
[delta (irgb-redder gamma)]
[epsilon (irgb-redder delta)])
...)