Summary: Although most of our prior experiments with recursion have emphasized recursion over lists, it is also possible to use other values as the basis of recursion. In this laboratory, you will explore the use of natural numbers (non-negative integers) as the basis of recursion.
Here is a template for the simplest kind of numeric recursive procedures.
(define recursive-proc (lambda (n) (if (zero? n) base-case (combine n (recursive-proc (- n 1))))))
If you choose to use a helper, here is one common form.
(define recursive-proc
(letrec ((kernel (lambda (so-far n)
(if (zero? n)
so-far
(kernel (update so-far n)
(- n 1))))))
(lambda (n)
(kernel starting-value n))))
We can also use a kernel to count up from 1 to n.
(A slight change lets you count up from 0 to n.)
(define recursive-proc
(letrec ((kernel (lambda (so-far i n)
(if (> i n)
so-far
(kernel (update so-far i) (+ i 1) n)))))
(lambda (n)
(kernel starting-value 1 n))))
a. Make sure that your library includes the following spot procedures.
;;; Procedure:
;;; image-render-spot!
;;; Parameters:
;;; image, an image
;;; spot, a spot
;;; Purpose:
;; Draw the spot on the image.
;;; Produces:
;;; [Nothing; Called for the side effect]
(define image-render-spot!
(lambda (image spot)
(image-set-pixel! image
(spot-col spot) (spot-row spot)
(spot-color spot))))
;;; Procedure:
;;; image-render-stamp!
;;; Parameters:
;;; image, an image
;;; stamp, a list of spots
;;; Purpose:
;; Draw the stamp on the image.
;;; Produces:
;;; image, the same image
(define image-render-stamp!
(lambda (image stamp)
(for-each (lambda (spot) (image-render-spot! image spot)) stamp)
image))
b. Make sure that your library includes the count-from
procedure from the reading.
;;; Procedure:
;;; count-from
;;; Parameters:
;;; lower, a natural number
;;; upper, a natural number
;;; Purpose:
;;; Construct a list of the natural numbers from lower to upper,
;;; inclusive, in ascending order.
;;; Produces:
;;; ls, a list
;;; Preconditions:
;;; lower <= upper
;;; Both lower and upper are numbers, exact, integers, and non-negative.
;;; Postconditions:
;;; The length of ls is upper - lower + 1.
;;; Every natural number between lower and upper, inclusive, appears
;;; in the list.
;;; Every value in the list with a successor is smaller than its
;;; successor.
;;; For every natural number k less than or equal to the length of
;;; ls, the element in position k of ls is lower + k.
(define count-from
(lambda (lower upper)
(if (= lower upper)
(list upper)
(cons lower (count-from (+ lower 1) upper)))))
c. Open a new window, and make the first line an instruction to load your library.
d. Create a new 200x200 and name it canvas.
e. Pick your favorite image, load it, and name it picture.
Here is the final definition of termial from the
reading.
(define termial
(lambda (number)
(cond
((not (number? number))
(error "termial expects a number, received" number))
((not (integer? number))
(error "termial expects an integer, received" number))
((< number 1)
(error "termial expects a positive integer, received" number))
((inexact? number)
(error "termial expects an exact integer, received" number))
(else
(termial-kernel number)))))
(define termial-kernel
(lambda (number)
(if (zero? number)
0
(+ number (termial-kernel (- number 1))))))
Rewrite it so that the kernel is local.
Define and test a recursive Scheme
procedure, (, that takes a natural number as
argument and returns a list of all the natural numbers less than or
equal to that number, in descending order.
count-down
val)
>(count-down 5)(5 4 3 2 1 0)>(count-down 0)(0)
Note that you should use cons to build up the list.
Note also that you are better off writing this with direct recursion (the first pattern above), rather than using a helper procedure.
When you are finished, you may want to read the notes on this exercise.
When you are finished writing this procedure, add
count-down to your Scheme library.
As you may recall from some of the earlier labs, sometimes we want to
draw sequences of spots. In those past exercises, we generated the list
of spot positions by hand. One of the great benefits of procedures like
count-down is that they can help us automate the generation
of a large number of spots.
a. Consider the following expression. Summarize the list of spots it describes.
(map (lambda (n) (spot-new n 5 (rgb-new 255 0 0))) (count-down 100))
b. Check your answer by rendering the spots, using an expression like the following.
(image-render-stamp! canvas (map (lambda (n) (spot-new n 5 (rgb-new 255 0 0))) (count-down 100)))
c. Consider the following expression. Summarize the list of spots it describes.
(map (lambda (n) (spot-new 3 n (rgb-new 0 0 255))) (count-down 50))
d. Check your answer by rendering the spots.
e. Consider the following expression. Summarize the list of spots it describes.
(map (lambda (n) (spot-new n n (rgb-new 0 (* 2 n) 0))) (count-down 128))
f. Check your answer by rendering the spots.
g. Consider the following expression. Summarize the list of spots it describes.
(map (lambda (n) (spot-new (quotient n 10) (modulo n 10) (rgb-new 0 0 0)))
(count-down 200))
h. Check your answer by rendering the spots.
Define and test a Scheme procedure,
(, that takes two arguments, the
second of which is a natural number, and returns a list consisting of
the specified number of repetitions of the first argument.
value-replicate value
count)
>(value-replicate "sample" 5)("sample" "sample" "sample" "sample" "sample")>(value-replicate 10 3)(10 10 10)>(value-replicate null 1)(())>(value-replicate null 2)(() ())>(value-replicate "hello" 0)()
Even if you know a built-in procedure to do this task, please
implement value-replicate recursively.
When you are finished writing this procedure, compare it to
the notes on
this exercise and then add value-replicate
to your Scheme library.
Define and test a recursive Scheme procedure that takes a natural number
as argument and returns a list of all the natural numbers that are
strictly less than the argument, in ascending order. (The traditional
name for this procedure is iota, a Greek letter.)
For example,
>(iota 3)(0 1 2)>(iota 5)(0 1 2 3 4)>(iota 1)(0)
Note that you will probably need to use a helper of some sort to
write iota. You might use the traditional form
of helper, which adds an extra parameter. You might also use a helper
that simply computes iota in the reverse order. (Most students write
a backwards iota in the first attempt; instead of throwing it away,
rename it and call it from iota.)
When you are done, add iota to your library.
You may recall the count-from procedure from
the
reading on recursion
over natural numbers which you added to your library at the
beginning of this lab.
a. What is the value of the call (count-from -10 10)?
b. Check your answer experimentally.
c. It is possible to implement count-from in terms
of iota. The implementation looks something like
the following.
(define count-from
(lambda (lower upper)
(map (lambda (n) (+ _____ n)) (iota _____))))
Finish this definition.
d. What do you see as the advantages and disadvantages of each way of
defining count-from?
Write a procedure, (, that
extracts element list-nth
n lst)n of a list. For example,
>(list-nth 5 (list "red" "orange" "yellow" "green" "blue" "indigo" "violet">))"indigo">(list-nth 0 (list "red" "orange" "yellow" "green" "blue" "indigo" "violet">))"red"
Even though this procedure does the same thing as
list-ref, you should not use
list-ref to implement it. Instead, your goal
is to figure out how list-ref works, which means
that you will need to implement this procedure using direct recursion.
Hint: When recursing, you will need to simplify the numeric parameter (probably by subtracting 1) and the list parameter (probably by taking its cdr).
Add the following procedures to your library.
;;; Procedure:
;;; image-get-spot
;;; Parameters:
;;; image, an image
;;; position, a position represented as a list of the form (col row)
;;; Purpose:
;;; Get a spot from the image
;;; Produces:
;;; spot, a spot
;;; Preconditions:
;;; col and row are integers.
;;; 0 <= col < (image-width image)
;;; 0 <= row < (image-height image)
;;; Postconditions:
;;; (spot-col spot) = col
;;; (spot-row spot) = row
;;; (spot-color spot) = (image-get-pixel image col row)
(define image-get-spot
(lambda (image position)
(let ((col (car position))
(row (cadr position)))
(spot-new col row (image-get-pixel image col row)))))
a. Write a procedure, (image-get-row , that extracts a row of
spots from image
row)image. You should be able
to write this procedure by mapping an anonymous procedure that
includes image-get-spot onto a call to
count-down, as in
(define image-get-row
(lambda (image row)
(map (lambda (n) (image-get-spot image ____ ____)) (count-down ____))))
b. Use this procedure to copy a row from picture to
canvas. For example,
(image-render-stamp! canvas (image-get-row picture 5))
a. Write a procedure, (,
that builds a list of positions that correspond to a rectangle of the
given width and height with an upper-left-corner of 0,0. For example,
if rectangle
width height)width is 3 and height
is 2, the points will be some permutation of (0,0), (0,1), (0,2),
(1,0), (1,1), and (1,2).
Hint: Look at problem 3.g.
b. Write a procedure, (, that builds a list of positions that
correspond to a rectangular region as described.
rectangular-region left top width height)
c. Write a procedure, (,
that extracts a rectangular image from
the given image and returns it as a list of spots.
image-get-region
image left top
width height)
Here's a possible solution to the problem. The base case is easy. If the number is zero, then the list of all non-negative numbers less than or equal to zero is the list that contains only zero.
(if (zero? n)
(list 0)
In the recursive case, we assume that we can compute the list of all
numbers less than or equal to n-1. To get the
complete list, we simply add n to the front.
(cons n (count-down (- n 1)))
Putting it all together, we get
;;; Procedure:
;;; count-down
;;; Parameters:
;;; n, a non-negative integer
;;; Purpose:
;;; Create a list of the form (n n-1 n-2 ... 3 2 1).
;;; Produces:
;;; nums, a list
;;; Preconditions:
;;; [No additional.]
;;; Postocnditions:
;;; (length nums) = n+1
;;; (list-ref nums i) = n-i for all i, 0 <= i <= n.
(define count-down
(lambda (n)
(if (zero? n)
(list 0)
(cons n (count-down (- n 1))))))
We begin by considering the base case. The last example gives a hint: If you want zero copies, you end up with the empty list.
(if (zero? n)
null
Now, on to the recursive case. If we can create a list of
n-1 copies, we can then create a list of
n copies by prepending one more copy.
(cons val (value-replicate val (- n 1)))
Putting it all together, we get
;;; Procedure:
;;; value-replicate
;;; Parameters:
;;; val, a Scheme value
;;; n, a non-negative integer
;;; Purpose:
;;; Create a list of n copies of val.
;;; Produces:
;;; val-lst
;;; Preconditions:
;;; [No additional]
;;; Postconditions:
;;; (length val-lst) is n
;;; (list-ref val-list i) is val for all i, 0 <= i < n
(define value-replicate
(lambda (val n)
(if (zero? n)
null
(cons val (value-replicate val (- n 1))))))