Lab 8: Midterm Review

Due at 11:59pm on Friday, 10/19/2018.

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 through 4 and submit through Ok. Questions 1, 3, and 4 can be found in
  • The remaining questions are extra practice. They can be found in the file. It is recommended that you complete these problems after finishing the required portion. There are many optional questions on this lab: it is recommended you identify the topics you struggle with most and complete those questions first.
  • In order to facilitate midterm studying, solutions to this lab will be released on Monday, October 15th. We encourage you to try out the problems and struggle for a while before looking at the solutions!

Required Questions

Linked Lists

Q1: Deep Linked List Length

A linked list that contains one or more linked lists as elements is called a deep linked list. Write a function deep_len that takes in a (possibly deep) linked list and returns the deep length of that linked list. The deep length of a linked list is the total number of non-link elements in the list, as well as the total number of elements contained in all contained lists. See the function's doctests for examples of the deep length of linked lists.

Hint: Use isinstance to check if something is an instance of an object.

def deep_len(lnk):
    """ Returns the deep length of a possibly deep linked list.

    >>> deep_len(Link(1, Link(2, Link(3))))
    >>> deep_len(Link(Link(1, Link(2)), Link(3, Link(4))))
    >>> levels = Link(Link(Link(1, Link(2)), \
            Link(3)), Link(Link(4), Link(5)))
    >>> print(levels)
    <<<1 2> 3> <4> 5>
    >>> deep_len(levels)
"*** YOUR CODE HERE ***"
if lnk is Link.empty: return 0 elif not isinstance(lnk, Link): return 1 else: return deep_len(lnk.first) + deep_len( Video walkthrough:

Use Ok to test your code:

python3 ok -q deep_len

Orders of Growth

Q2: Finding Orders of Growth

Use Ok to test your knowledge with the following questions:

python3 ok -q growth -u

Be sure to ask a lab assistant or TA if you don't understand the correct answer!

What is the order of growth of is_prime in terms of n?

def is_prime(n):
    for i in range(2, n):
        if n % i == 0:
            return False
    return True

Explanation: The body of the for loop is executed n - 2 times. Each iteration takes constant time (one conditional check and one return statement). Therefore, the total time is (n - 2) x θ(1), or simply θ(n).

What is the order of growth of bar in terms of n?

def bar(n):
    i, sum = 1, 0
    while i <= n:
        sum += biz(n)
        i += 1
    return sum

def biz(n):
    i, sum = 1, 0
    while i <= n:
        sum += i**3
        i += 1
    return sum

Explanation: The body of the while loop in bar is executed n times. Each iteration, one call to biz(n) is made. Note that n never changes, so this call takes the same time to run each iteration. Taking a look at biz, we see that there is another while loop. Be careful to note that although the term being added to sum is cubed (i**3), i itself is only incremented by 1 in each iteration. This tells us that this while loop also executes n times, with each iteration taking constant time , so the total time of biz(n) is n x θ(1), or θ(n). Knowing the runtime of biz(n), we can conclude that each iteration of the while loop in bar takes θ(n). Therefore, the total runtime of bar(n) is n x θ(n).

What is the order of growth of foo in terms of n, where n is the length of lst? Assume that slicing a list and calling len on a list can both be done in constant time.

def foo(lst, i):
    mid = len(lst) // 2
    if mid == 0:
        return lst
    elif i > 0:
        return foo(lst[mid:], -1)
        return foo(lst[:mid], 1)

Explanation: A single recursive call is made in the body of foo on half the input list (either the first half or the second half depending on the input flag i). The base case is executed when the list either is empty or has only one element. We start with an n element list and halve the list until there is at most 1 element, which means there will be log(n) total calls. Each call, constant work is done if we ignore the recursive call. The total runtime is then log(n) * θ(1).

Note: We simplified this problem by assuming that slicing a list takes constant time. In reality, this operation is a bit more nuanced and may take linear time. As an additional exercise, try determining the order of growth of this function if we assuming slicing takes linear time.

Recursion & Tree Recursion

Q3: Subsequences

A subsequence of a sequence S is a sequence of elements from S, in the same order they appear in S, but possibly with elements missing. Thus, the lists [], [1, 3], [2], and [1, 2, 3] are some (but not all) of the subsequences of [1, 2, 3]. Write a function that takes a list and returns a list of lists, for which each individual list is a subsequence of the original input.

In order to accomplish this, you might first want to write a function insert_into_all that takes an item and a list of lists, adds the item to the beginning of nested list, and returns the resulting list.

def insert_into_all(item, nested_list):
    """Assuming that nested_list is a list of lists, return a new list
    consisting of all the lists in nested_list, but with item added to
    the front of each.

    >>> nl = [[], [1, 2], [3]]
    >>> insert_into_all(0, nl)
    [[0], [0, 1, 2], [0, 3]]
"*** YOUR CODE HERE ***"
return [[item] + lst for lst in nested_list]
def subseqs(s): """Assuming that S is a list, return a nested list of all subsequences of S (a list of lists). The subsequences can appear in any order. >>> seqs = subseqs([1, 2, 3]) >>> sorted(seqs) [[], [1], [1, 2], [1, 2, 3], [1, 3], [2], [2, 3], [3]] >>> subseqs([]) [[]] """
"*** YOUR CODE HERE ***"
if not s: return [[]] else: subset = subseqs(s[1:]) return insert_into_all(s[0], subset) + subset

Use Ok to test your code:

python3 ok -q subseqs

Q4: Increasing Subsequences

Now, we wish to find subsequences subject to another condition: we only want the subsequences for which consecutive elements are nondecreasing. For example, [1, 3, 2] is a subsequence of [1, 3, 2, 4], but since 2 < 3, this subsequence would not be included in our result.

Fill in the blanks to complete the implementation of the inc_subseqs function. You may assume that the input list only contains positive elements.

You may use the helper function insert_into_all you defined in the previous part.

def inc_subseqs(s):
    """Assuming that S is a list, return a nested list of all subsequences
    of S (a list of lists) for which the elements of the subsequence
    are strictly nondecreasing. The subsequences can appear in any order.

    >>> seqs = inc_subseqs([1, 3, 2])
    >>> sorted(seqs)
    [[], [1], [1, 2], [1, 3], [2], [3]]
    >>> inc_subseqs([])
    >>> seqs2 = inc_subseqs([1, 1, 2])
    >>> sorted(seqs2)
    [[], [1], [1], [1, 1], [1, 1, 2], [1, 2], [1, 2], [2]]
    def subseq_helper(s, prev):
        if not s:
return ____________________
return [[]]
elif s[0] < prev:
return ____________________
return subseq_helper(s[1:], prev)
a = ______________________
a = subseq_helper(s[1:], s[0])
b = ______________________
b = subseq_helper(s[1:], prev)
return insert_into_all(________, ______________) + ________________
return insert_into_all(s[0], a) + b
return subseq_helper(____, ____)
return subseq_helper(s, 0)

Use Ok to test your code:

python3 ok -q inc_subseqs

Optional Questions


Q5: Keyboard

We'd like to create a Keyboard class that takes in an arbitrary number of Buttons and stores these Buttons in a dictionary. The keys in the dictionary will be ints that represent the postition on the Keyboard, and the values will be the respective Button. Fill out the methods in the Keyboard class according to each description, using the doctests as a reference for the behavior of a Keyboard.

class Keyboard:
    """A Keyboard takes in an arbitrary amount of buttons, and has a
    dictionary of positions as keys, and values as Buttons.

    >>> b1 = Button(0, "H")
    >>> b2 = Button(1, "I")
    >>> k = Keyboard(b1, b2)
    >>> k.buttons[0].key
    >>> #No button at this position
    >>> k.typing([0, 1])
    >>> k.typing([1, 0])
    >>> b1.times_pressed
    >>> b2.times_pressed

    def __init__(self, *args):
"*** YOUR CODE HERE ***"
self.buttons = {} for button in args: self.buttons[button.pos] = button
def press(self, info): """Takes in a position of the button pressed, and returns that button's output"""
"*** YOUR CODE HERE ***"
if info in self.buttons.keys(): b = self.buttons[info] b.times_pressed += 1 return b.key return ''
def typing(self, typing_input): """Takes in a list of positions of buttons pressed, and returns the total output"""
"*** YOUR CODE HERE ***"
accumulate = '' for pos in typing_input: return accumulate
class Button: def __init__(self, pos, key): self.pos = pos self.key = key self.times_pressed = 0

Use Ok to test your code:

python3 ok -q Keyboard


Q6: Advanced Counter

Complete the definition of make_advanced_counter_maker, which creates a function that creates counters. These counters can not only update their personal count, but also a shared count for all counters. They can also reset either count.

def make_advanced_counter_maker():
    """Makes a function that makes counters that understands the
    messages "count", "global-count", "reset", and "global-reset".
    See the examples below:

    >>> make_counter = make_advanced_counter_maker()
    >>> tom_counter = make_counter()
    >>> tom_counter('count')
    >>> tom_counter('count')
    >>> tom_counter('global-count')
    >>> jon_counter = make_counter()
    >>> jon_counter('global-count')
    >>> jon_counter('count')
    >>> jon_counter('reset')
    >>> jon_counter('count')
    >>> tom_counter('count')
    >>> jon_counter('global-count')
    >>> jon_counter('global-reset')
    >>> tom_counter('global-count')
"*** YOUR CODE HERE ***"
global_count = 0 def make_counter(): count = 0 def counter(msg): nonlocal global_count, count if msg == 'count': count += 1 return count elif msg == 'reset': count = 0 elif msg == 'global-count': global_count += 1 return global_count elif msg == 'global-reset': global_count = 0 return counter return make_counter

Use Ok to test your code:

python3 ok -q make_advanced_counter_maker

Mutable Lists

Q7: Environment Diagram

Draw an environment diagram for the following program.

Some things to remember:

  • When you mutate a list, you are changing the original list.
  • When you concatenate two lists, you are creating a new list.
  • When you assign a name to an existing object, you are creating another reference to that object rather than creating a copy of that object.
def got(lst, el, f):
    welcome = []
    for e in lst:
        if e == el:
            el = f(lst[1:], 2, welcome)
    return lst[3:] + welcome

def avocadis(lst, i, lst0):
    return len(lst0)

bananis = [1, 6, 1, 6]
n = bananis[3]
we = got(bananis, n, avocadis)

You can check your solution here. If you get stuck, ask a Lab Assistant or TA for help before checking the solution! There is nothing to submit for this problem.

Q8: Trade

In the integer market, each participant has a list of positive integers to trade. When two participants meet, they trade the smallest non-empty prefix of their list of integers. A prefix is a slice that starts at index 0.

Write a function trade that exchanges the first m elements of list first with the first n elements of list second, such that the sums of those elements are equal, and the sum is as small as possible. If no such prefix exists, return the string 'No deal!' and do not change either list. Otherwise change both lists and return 'Deal!'. A partial implementation is provided.

Hint: You can mutate a slice of a list using slice assignment. To do so, specify a slice of the list [i:j] on the left-hand side of an assignment statement and another list on the right-hand side of the assignment statement. The operation will replace the entire given slice of the list from i inclusive to j exclusive with the elements from the given list. The slice and the given list need not be the same length.

>>> a = [1, 2, 3, 4, 5, 6]
>>> b = a
>>> a[2:5] = [10, 11, 12, 13]
>>> a
[1, 2, 10, 11, 12, 13, 6]
>>> b
[1, 2, 10, 11, 12, 13, 6]

Additionally, recall that the starting and ending indices for a slice can be left out and Python will use a default value. lst[i:] is the same as lst[i:len(lst)], and lst[:j] is the same as lst[0:j].

def trade(first, second):
    """Exchange the smallest prefixes of first and second that have equal sum.

    >>> a = [1, 1, 3, 2, 1, 1, 4]
    >>> b = [4, 3, 2, 7]
    >>> trade(a, b) # Trades 1+1+3+2=7 for 4+3=7
    >>> a
    [4, 3, 1, 1, 4]
    >>> b
    [1, 1, 3, 2, 2, 7]
    >>> c = [3, 3, 2, 4, 1]
    >>> trade(b, c)
    'No deal!'
    >>> b
    [1, 1, 3, 2, 2, 7]
    >>> c
    [3, 3, 2, 4, 1]
    >>> trade(a, c)
    >>> a
    [3, 3, 2, 1, 4]
    >>> b
    [1, 1, 3, 2, 2, 7]
    >>> c
    [4, 3, 1, 4, 1]
    m, n = 1, 1

"*** YOUR CODE HERE ***"
equal_prefix = lambda: sum(first[:m]) == sum(second[:n]) while m < len(first) and n < len(second) and not equal_prefix(): if sum(first[:m]) < sum(second[:n]): m += 1 else: n += 1
if False: # change this line!
if equal_prefix():
first[:m], second[:n] = second[:n], first[:m] return 'Deal!' else: return 'No deal!'

Use Ok to test your code:

python3 ok -q trade

Iterators and Generators

Q9: Generate Permutations

Given a sequence of unique elements, a permutation of the sequence is a list containing the elements of the sequence in some arbitrary order. For example, [2, 1, 3], [1, 3, 2], and [3, 2, 1] are some of the permutations of the sequence [1, 2, 3].

Implement permutations, a generator function that takes in a sequence seq and returns a generator that yields all permutations of seq.

Permutations may be yielded in any order. Note that the doctests test whether you are yielding all possible permutations, but not in any particular order. The built-in sorted function takes in an iterable object and returns a list containing the elements of the iterable in non-decreasing order.

Your solution must fit on the lines provided in the skeleton code.

Hint: If you had the permutations of all the elements in lst not including the first element, how could you use that to generate the permutations of the full lst?

def permutations(seq):
    """Generates all permutations of the given sequence. Each permutation is a
    list of the elements in SEQ in a different order. The permutations may be
    yielded in any order.

    >>> perms = permutations([100])
    >>> type(perms)
    <class 'generator'>
    >>> next(perms)
    >>> try:
    ...     next(perms)
    ... except StopIteration:
    ...     print('No more permutations!')
    No more permutations!
    >>> sorted(permutations([1, 2, 3])) # Returns a sorted list containing elements of the generator
    [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
    >>> 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 ____________________:
if not seq:
yield ____________________
yield []
for perm in _____________________:
for perm in permutations(seq[1:]):
for _____ in ________________:
for i in range(len(seq)):
yield perm[:i] + [seq[0]] + perm[i:]

Use Ok to test your code:

python3 ok -q permutations

Recursive objects

Q10: Linked Lists as Strings

Kevin and Jerry like different ways of displaying the linked list structure in Python. While Kevin likes box and pointer diagrams, Jerry prefers a more futuristic way. Write a function make_to_string that returns a function that converts the linked list to a string in their preferred style.

Hint: You can convert numbers to strings using the str function, and you can combine strings together using +.

>>> str(4)
>>> 'cs ' + str(61) + 'a'
'cs 61a'
def make_to_string(front, mid, back, empty_repr):
    """ Returns a function that turns linked lists to strings.

    >>> kevins_to_string = make_to_string("[", "|-]-->", "", "[]")
    >>> jerrys_to_string = make_to_string("(", " . ", ")", "()")
    >>> lst = Link(1, Link(2, Link(3, Link(4))))
    >>> kevins_to_string(lst)
    >>> kevins_to_string(Link.empty)
    >>> jerrys_to_string(lst)
    '(1 . (2 . (3 . (4 . ()))))'
    >>> jerrys_to_string(Link.empty)
"*** YOUR CODE HERE ***"
def printer(lnk): if lnk is Link.empty: return empty_repr else: return front + str(lnk.first) + mid + printer( + back return printer Video walkthrough:

Use Ok to test your code:

python3 ok -q make_to_string

Q11: Tree Map

Define the function tree_map, which takes in a tree and a one-argument function as arguments and returns a new tree which is the result of mapping the function over the entries of the input tree.

def tree_map(fn, t):
    """Maps the function fn over the entries of t and returns the
    result in a new tree.

    >>> numbers = Tree(1,
    ...                [Tree(2,
    ...                      [Tree(3),
    ...                       Tree(4)]),
    ...                 Tree(5,
    ...                      [Tree(6,
    ...                            [Tree(7)]),
    ...                       Tree(8)])])
    >>> print(tree_map(lambda x: 2**x, numbers))
    >>> print(numbers)
"*** YOUR CODE HERE ***"
if t.is_leaf(): return Tree(fn(t.label), []) mapped_subtrees = [tree_map(fn, b) for b in t.branches] return Tree(fn(t.label), mapped_subtrees) # Alternate solution def tree_map(fn, t): return Tree(fn(t.label), [tree_map(fn, b) for b in t.branches]) Video walkthrough:

Use Ok to test your code:

python3 ok -q tree_map

Q12: Long Paths

Implement long_paths, which returns a list of all paths in a tree with length at least n. A path in a tree is a linked list of node values that starts with the root and ends at a leaf. Each subsequent element must be from a child of the previous value's node. The length of a path is the number of edges in the path (i.e. one less than the number of nodes in the path). Paths are listed in order from left to right. See the doctests for some examples.

def long_paths(tree, n):
    """Return a list of all paths in tree with length at least n.

    >>> t = Tree(3, [Tree(4), Tree(4), Tree(5)])
    >>> left = Tree(1, [Tree(2), t])
    >>> mid = Tree(6, [Tree(7, [Tree(8)]), Tree(9)])
    >>> right = Tree(11, [Tree(12, [Tree(13, [Tree(14)])])])
    >>> whole = Tree(0, [left, Tree(13), mid, right])
    >>> for path in long_paths(whole, 2):
    ...     print(path)
    <0 1 2>
    <0 1 3 4>
    <0 1 3 4>
    <0 1 3 5>
    <0 6 7 8>
    <0 6 9>
    <0 11 12 13 14>
    >>> for path in long_paths(whole, 3):
    ...     print(path)
    <0 1 3 4>
    <0 1 3 4>
    <0 1 3 5>
    <0 6 7 8>
    <0 11 12 13 14>
    >>> long_paths(whole, 4)
    [Link(0, Link(11, Link(12, Link(13, Link(14)))))]
"*** YOUR CODE HERE ***"
paths = [] if n <= 0 and tree.is_leaf(): paths.append(Link(tree.label)) for b in tree.branches: for path in long_paths(b, n - 1): paths.append(Link(tree.label, path)) return paths

Use Ok to test your code:

python3 ok -q long_paths

More Orders of Growth

Q13: Boom

What is the order of growth in time for the following function boom? Use big-θ notation.

def boom(n):
    sum = 0
    a, b = 1, 1
    while a <= n*n:
        while b <= n*n:
            sum += (a*b)
            b += 1
        b = 0
        a += 1
    return sum

Use ok to test your understanding:

python3 ok -q boom -u


Q14: Zap

What is the order of growth in time for the following function zap? Use big-θ notation.

def zap(n):
    i, count = 1, 0
    while i <= n:
        while i <= 5 * n:
            count += i
            print(i / 6)
            i *= 3
    return count

Use ok to test your understanding:

python3 ok -q zap -u

θ(log n)

Here, the stopping condition of both loops rely on the same variable i. You might notice that completion of the inner loop will guarantee completion of the outer loop; after all, if i is greater than 5 * n, then it will be greater than n. Therefore, the overall runtime is just the runtime of the inner loop. Since i begins at 1 and is multiplied by 3 at every iteration of the inner loop, the inner loop will have log n iterations overall. Each iteration does constant work, so the overall runtime will be log n.