Lab 13: Final Review

Due at 11:59pm on Friday, 04/27/2018.

Starter Files

Download Inside the archive, you will find starter files for the questions in this lab, along with a copy of the Ok autograder.


By the end of this lab, you should have submitted the lab with python3 ok --submit. You may submit more than once before the deadline; only the final submission will be graded. Check that you have successfully submitted your code on

  • To receive credit for this lab, you must complete Questions 1-3 in and lab13.scm and submit through OK.
  • Question 4-8 are considered extra practice. They can be found in the and lab13_extra.scm files. It is recommended that you complete them on your own time.

Required Questions


For a quick refresher on Scheme, see Lab 09.

This question is to be done in lab13.scm.

Q1: Compose All

Implement compose-all, which takes a list of one-argument functions and returns a one-argument function that applies each function in that list in turn to its argument. For example, if func is the result of calling compose-all on a list of functions (f g h), then (func x) should be equivalent to the result of calling (h (g (f x))).

scm> (define (square x) (* x x))
scm> (define (add-one x) (+ x 1))
scm> (define (double x) (* x 2))
scm> (define composed (compose-all (list double square add-one)))
scm> (composed 1)
scm> (composed 2)
(define (compose-all funcs)
(lambda (x) (if (null? funcs) x ((compose-all (cdr funcs)) ((car funcs) x))))

Use Ok to test your code:

python3 ok -q compose-all

Tail Recursion

For a quick refresher on tail recursion, see Discussion 8.

This question is to be done in lab13.scm.

Q2: Replicate

Write a tail-recursive function that returns a list with x repeated n times.

scm> (tail-replicate 3 10)
(3 3 3 3 3 3 3 3 3 3)
scm> (tail-replicate 5 0)
scm> (tail-replicate 100 5)
(100 100 100 100 100)
(define (tail-replicate x n)
(define (helper n so-far) (if (= n 0) so-far (helper (- n 1) (cons x so-far)))) (helper n '())

Use Ok to test your code:

python3 ok -q tail-replicate


For a quick refresher on generators, see Lab 11.

This question is to be done in

Q3: Generate Permutations

Given a list of unique elements, a permutation of the list is a reordering of the elements. For example, [2, 1, 3], [1, 3, 2], and [3, 2, 1] are all permutations of the list [1, 2, 3].

Implement permutations, a generator function that takes in a lst and outputs all permutations of lst, each as a list (see doctest for an example). The order in which you generate permutations is irrelevant.

Hint: If you had the permutations of lst minus one element, how could you use that to generate the permutations of the full lst?

Note that in the provided code, the return statement acts like a raise StopIteration. The point of this is so that the returned generator doesn't enter the rest of the body on any calls to next after the first if the input list is empty. Note that this return statement does not affect the fact that the function will still return a generator object because the body contains yield statements.

def permutations(lst):
    """Generates all permutations of sequence LST. Each permutation is a
    list of the elements in LST in a different order.

    The order of the permutations does not matter.

    >>> sorted(permutations([1, 2, 3]))
    [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
    >>> type(permutations([1, 2, 3]))
    <class 'generator'>
    >>> sorted(permutations((10, 20, 30)))
    [[10, 20, 30], [10, 30, 20], [20, 10, 30], [20, 30, 10], [30, 10, 20], [30, 20, 10]]
    >>> sorted(permutations("ab"))
    [['a', 'b'], ['b', 'a']]
    if not lst:
        yield []
"*** YOUR CODE HERE ***"
for perm in permutations(lst[1:]): for i in range(len(lst)): yield perm[:i] + [lst[0]] + perm[i:]

Use Ok to test your code:

python3 ok -q permutations

Optional Questions


For a quick refresher on streams, see Discussion 9.

This question is to be done in lab13_extra.scm.

Q4: Run-Length Encoding

Run-length encoding is a very simple data compression technique, whereby runs of data are compressed and stored as a single value. A run is defined to be a contiguous sequence of the same number. For example, in the (finite) sequence

1, 1, 1, 1, 1, 6, 6, 6, 6, 2, 5, 5, 5

there are four runs: one each of 1, 6, 2, and 5. We can represent the same sequence as a sequence of two-element lists:

(1 5), (6 4), (2 1), (5 3)

Notice that the first element of each list is the number in a run, and the second element is the number of of times that number appears in the run.

We will extend this idea to streams. Write a function called rle that takes in a stream of data, and returns a corresponding stream of two-element lists, which represents the run-length encoded version of the stream. You do not have to consider compressing infinite runs.

scm> (define s (cons-stream 1 (cons-stream 1 (cons-stream 2 nil))))
scm> (define encoding (rle s))
scm> (car encoding)  ; Run of number 1 of length 2
(1 2)
scm> (car (cdr-stream encoding))  ; Run of number 2 of length 1
(2 1)
scm> (stream-to-list (rle (list-to-stream '(1 1 2 2 2 3))))  ; See functions at bottom of lab13.scm
((1 2) (2 3) (3 1))
(define (rle s)
(define (track-run elem st len) (cond ((null? st) (cons-stream (list elem len) nil)) ((= elem (car st)) (track-run elem (cdr-stream st) (+ len 1))) (else (cons-stream (list elem len) (rle st)))) ) (if (null? s) nil (track-run (car s) (cdr-stream s) 1))

Use Ok to test your code:

python3 ok -q rle

More Tail Recursion

This question is to be done in lab13_extra.scm.

Q5: Insert

Write a tail-recursive function that inserts a number n into a non-empty sorted list of numbers, s.

Hint: Use the built-in Scheme function append to concatenate two lists together.

scm> (append '(1 2) '(3 4))
(1 2 3 4)
scm> (append '(2 4 6) '())
(2 4 6)
scm> (append '() '(5 3 1))
(5 3 1)
scm> (insert 1 '(2))
(1 2)
scm> (insert 5 '(2 4 6 8))
(2 4 5 6 8)
scm> (insert 1000 '(1 2 3 4 5 6))
(1 2 3 4 5 6 1000)
(define (insert n s)
(define (insert-help s so-far) (cond ((null? s) (append so-far (cons n s))) ((< n (car s)) (append so-far (cons n s))) (else (insert-help (cdr s) (append so-far (list (car s))))))) (insert-help s nil)

Use Ok to test your code:

python3 ok -q insert

The Ok tests for this problem don't check whether your solution is tail-recursive! In order to check your understanding, you can run some manual tests with large input such as the following to ensure that your function use a constant number of frames:

scm> (define big-list (tail-replicate 3 1000))
scm> (define result (insert 4 big-list))
scm> (define expected (append big-list (list 4)))
scm> (equal? result expected)

More Scheme

These questions are to be done in lab13_extra.scm.

Q6: Deep Map

Write the function deep-map, which takes a function fn and a nested list s. A nested list is a list where each element is either a number or a list (e.g. (1 (2) 3) where 1, (2), and 3 are the elements). It returns a list with identical structure to s, but replacing each non-list element by the result of applying fn on it, even for elements within sub-lists. For example:

scm> (define (double x) (* 2 x))
scm> (deep-map double '(2 (3 4)))
(4 (6 8))

Assume that the input has no dotted (malformed) lists.

Hint: You can use the function list?, which checks if a value is a list.

(define (deep-map fn s)
(cond ((null? s) s) ((list? (car s)) (cons (deep-map fn (car s)) (deep-map fn (cdr s)))) (else (cons (fn (car s)) (deep-map fn (cdr s)))))

Use Ok to test your code:

python3 ok -q deep-map

Q7: Tally

Implement tally, which takes a list of names and returns a list of pairs, one pair for each unique name in names. Each pair should contain a name and the number of times that the name appeared in names. Each name should appear only once in the output, and the names should be ordered by when they first appear in names.

Hint: Use the eq? procedure to test if two names are the same.

scm> (tally '(james jen jemin john))
((james . 1) (jen . 1) (jemin . 1) (john . 1))
scm> (tally '(billy billy bob billy bob billy bob))
((billy . 4) (bob . 3))
scm> (tally '())
(define (tally names)
(map (lambda (name) (cons name (count name names))) (unique names))

Hint: If you find the procedure getting too complicated, you may want to try implementing the count and unique helper procedures to use in your solution. You may also want to use map and filter in your solution.

; Implementing and using these helper procedures is optional. You are allowed
; to delete them.
(define (unique s)
(if (null? s) nil (cons (car s) (unique (filter (lambda (x) (not (eq? (car s) x))) (cdr s)))))
) (define (count name s)
(if (null? s) 0 (+ (if (eq? name (car s)) 1 0) (count name (cdr s))))

Use Ok to test your code:

python3 ok -q tally

More Generators

This question is to be done in

Q8: Generators generator

Write the generator function make_generators_generator, which takes a zero-argument generator function g and returns a generator that yields generators. For each element e yielded by the generator object returned by calling g, a new generator object is yielded that will generate items 1 through e yielded by the generator returned by g.

def make_generators_generator(g):
    """Generates all the "sub"-generators of the generator returned by
    the generator function g.

    >>> def ints_to(n):
    ...     for i in range(1, n + 1):
    ...          yield i
    >>> def ints_to_5():
    ...     for item in ints_to(5):
    ...         yield item
    >>> for gen in make_generators_generator(ints_to_5):
    ...     print("Next Generator:")
    ...     for item in gen:
    ...         print(item)
    Next Generator:
    Next Generator:
    Next Generator:
    Next Generator:
    Next Generator:
"*** YOUR CODE HERE ***"
def gen(i): for item in g(): if i <= 0: return yield item i -= 1 i = 1 for item in g(): yield gen(i) i += 1

Use Ok to test your code:

python3 ok -q make_generators_generator