Lab 13: Final Review

Due at 11:59pm on Friday, 08/03/2018.

Lab Check-in 7 questions here.

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

Mutable Functions

For a quick refresher on mutable functions, see Discussion 9.

This question is to be done in

Q1: Efficient Count Change

Recall the count_change function from earlier in the semester that takes in some amount and returns the number of ways to make change for amount using coins with denominations that are powers of two. We can simplify this problem by introducing an additional parameter, coins, a tuple of coin values with which to count change:

def count_change(amount, coins=(50, 25, 10, 5, 1)):
    >>> count_change(7, (1, 2, 4, 8))
    if amount == 0:
        return 1
    elif amount < 0 or len(coins) == 0:
        return 0
    return count_change(amount, coins[1:]) + count_change(amount - coins[0], coins)

This function is quite slow on larger inputs, since it repeats the same computation many times. For example, to find count_change(7, (1, 2, 4, 8)), we make recursive calls to count_change(7, (2, 4, 8)) and count_change(6, (1, 2, 4, 8)). Both these calls make recursive calls to count_change(6, (2, 4, 8)), which will get computed twice! Draw out the recursive call tree to convice yourself of this repeated call with the same arguments.

To speed things up, complete the implementation below for make_count_change, which returns a function that is equivalent to count_change but is more efficient. This more efficient version should only perform any computation it has not seen before, i.e. for arguments that have never been passed in previously. If it is called with a pair of arguments previously seen, then it should just return the previously computed value. Once you are done, compare the two versions on a large number such as 500 to make sure that your code is faster than the original version.

def make_count_change():
    """Return a function to efficiently count the number of ways to make

    >>> cc = make_count_change()
    >>> cc(500, (50, 25, 10, 5, 1))
    computed = {}
    def count_change(amount, coins=(50, 25, 10, 5, 1)):
if _________________________:
if amount == 0:
return 1
elif _________________________:
elif amount < 0 or len(coins) == 0:
return 0 elif (amount, coins) in computed:
return ____________________
return computed[(amount, coins)]
count = _____________________________ + _____________________________
count = count_change(amount, coins[1:]) + count_change(amount - coins[0], coins)
computed[(amount, coins)] = count
return count return count_change

Use Ok to test your code:

python3 ok -q make_count_change


For a quick refresher on streams, see Discussion 10.

This question is to be done in lab13.scm.

Q2: 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 in 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

Tail Recursion

For a quick refresher on tail recursion, see Discussion 10.

This question is to be done in lab13.scm.

Q3: 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

Optional Questions


For a quick refresher on trees, see Lab 9.

This question is to be done in

Q4: Prune Small

Complete the function prune_small that takes in a Tree t and a number n and prunes t mutatively. If t or any of its branches has more than n branches, the n branches with the smallest labels should be kept and any other branches should be pruned, or removed, from the tree.

def prune_small(t, n):
    """Prune the tree mutatively, keeping only the smallest
    n branches of every tree node.

    >>> t1 = Tree(6)
    >>> prune_small(t1, 2)
    >>> t1
    >>> t2 = Tree(6, [Tree(3), Tree(4)])
    >>> prune_small(t2, 1)
    >>> t2
    Tree(6, [Tree(3)])
    >>> t3 = Tree(6, [Tree(1), Tree(3, [Tree(1), Tree(2), Tree(3)]), Tree(5, [Tree(3), Tree(4)])])
    >>> prune_small(t3, 2)
    >>> t3
    Tree(6, [Tree(1), Tree(3, [Tree(1), Tree(2)])])
    if t.is_leaf():
while ___________________________:
while len(t.branches) > n:
largest = max(_______________, key=____________________)
largest = max(t.branches, key=lambda x: x.label)
for __ in _____________:
for b in t.branches:
prune_small(b, n)

Use Ok to test your code:

python3 ok -q prune_small


For a quick refresher on Scheme, see Lab 07.

These questions are to be done in lab13_extra.scm.

Q5: 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

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 Tail Recursion

This question is to be done in lab13_extra.scm.

Q8: 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)