Lab 14: Final Review (Optional Lab)

Due at 11:59pm on Friday, 8/6/2019.

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

  • This lab is not worth any credit, and is just here for you to get practice. Feel free to attempt any questions you want for practice. The solutions have also been released for you to look at


For a quick refresher on Trees, see Lab 09.

This question is to be done in

Q1: 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 n branches
    of each node with the smallest label.

    >>> 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)])])
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 10.

This question is to be done in lab14.scm.

Q2: 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

Tree Recursion

For a quick refresher on tree recursion, see Discussion 03.

This question is to be done in

Q3: Num Splits

Given a list of numbers s and a target difference d, how many different ways are there to split s into two subsets such that the sum of the first is within d of the sum of the second? The number of elements in each subset can differ.

You may assume that the elements in s are distinct and that d is always non-negative.

Note that the order of the elements within each subset does not matter, nor does the order of the subsets themselves. For example, given the list [1, 2, 3], you should not count [1, 2], [3] and [3], [1, 2] as distinct splits.

Hint: If the number you return is too large, you may be double-counting somewhere. If the result you return is off by some constant factor, it will likely be easiest to simply divide/subtract away that factor.

def num_splits(s, d):
    """Return the number of ways in which s can be partitioned into two
    sublists that have sums within d of each other.

    >>> num_splits([1, 5, 4], 0)  # splits to [1, 4] and [5]
    >>> num_splits([6, 1, 3], 1)  # no split possible
    >>> num_splits([-2, 1, 3], 2) # [-2, 3], [1] and [-2, 1, 3], []
    >>> num_splits([1, 4, 6, 8, 2, 9, 5], 3)
"*** YOUR CODE HERE ***"
def difference_so_far(s, difference): if not s: if abs(difference) <= d: return 1 else: return 0 element = s[0] s = s[1:] return difference_so_far(s, difference + element) + difference_so_far(s, difference - element) return difference_so_far(s, 0)//2

Use Ok to test your code:

python3 ok -q num_splits

Linked Lists

For a quick refresher on Linked Lists, see Lab 07.

This question is to be done in

Q4: Insert

Implement a function insert that takes a Link, a value, and an index, and inserts the value into the Link at the given index. You can assume the linked list already has at least one element. Do not return anything -- insert should mutate the linked list.

Note: If the index is out of bounds, you can raise an IndexError with:

raise IndexError
def insert(link, value, index):
    """Insert a value into a Link at the given index.

    >>> link = Link(1, Link(2, Link(3)))
    >>> print(link)
    <1 2 3>
    >>> insert(link, 9001, 0)
    >>> print(link)
    <9001 1 2 3>
    >>> insert(link, 100, 2)
    >>> print(link)
    <9001 1 100 2 3>
    >>> insert(link, 4, 5)
"*** YOUR CODE HERE ***"
if index == 0: = Link(link.first, link.first = value elif is Link.empty: raise IndexError else: insert(, value, index - 1) # iterative solution def insert(link, value, index): while index > 0 and is not Link.empty: link = index -= 1 if index == 0: = Link(link.first, link.first = value else: raise IndexError

Use Ok to test your code:

python3 ok -q insert


For a quick refresher on streams, see Discussion 10.

Q5: Cycles

In Scheme, it's possible to have a stream with cycles. That is, a stream may contain itself as part of the stream definition.

scm> (define s (cons-stream 1 (cons-stream 2 s)))
scm> (car s)
scm> (car (cdr-stream (cdr-stream s)))
scm> (eq? (cdr-stream (cdr-stream s)) s)

Implement has-cycle?, which returns whether a stream contains a cycle. You may assume that the input is either a stream of some unknown finite length, or is one that contains a cycle. You should implement and use the contains? procedure in your solution. We have provided a skeleton for has-cycle?; your solution must fit on the lines provided.

Hint: A stream has a cycle if you see the same pair object more than once. The built-in procedure eq? may be used to check if two expressions evaluate to the same object in memory.

  scm> (define lst1 '(1 2 3))
  scm> (define lst2 lst1)
  scm> (define lst3 '(1 2 3))
  scm> (eq? lst1 lst2)
  scm> (eq? lst1 lst3)
(define (has-cycle? s)
(define (pair-tracker seen-so-far curr)
(define (pair-tracker seen-so-far curr)
(cond (_________________ ____________)
(cond ((null? curr) #f)
(_________________ ____________)
((contains? seen-so-far curr) #t)
(else _________________________))
(else (pair-tracker (cons curr seen-so-far) (cdr-stream curr))))
(pair-tracker nil s)
) (define (contains? lst s)
(cond ((null? lst) #f) ((eq? s (car lst)) #t) (else (contains? (cdr lst) s)))

Use Ok to test your code:

python3 ok -q has-cycle