Lab 8: Midterm Review

Due by 11:59pm on Tuesday, March 16.

Starter Files

Download lab08.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the Ok autograder.

Submission

In order to facilitate midterm studying, solutions to this lab were released with the lab. We encourage you to try out the problems and struggle for a while before looking at the solutions! Note: You do not need to run python ok --submit to receive credit for this assignment.

Required Questions

Q1: All Questions Are Optional

The questions in this assignment are not graded, but they are highly recommended to help you prepare for the upcoming midterm. You will receive credit for this lab even if you do not complete these questions.

This question has no Ok tests.

Suggested Questions

Recursion and Tree Recursion

Q2: Subsequences

A subsequence of a sequence S is a subset of elements from S, in the same order they appear in S. Consider the list [1, 2, 3]. Here are a few of it's subsequences [], [1, 3], [2], and [1, 2, 3].

Write a function that takes in a list and returns all possible subsequences of that list. The subsequences should be returned as a list of lists, where each nested 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 each nested list, and returns the resulting list.

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

    >>> nl = [[], [1, 2], [3]]
    >>> insert_into_all(0, nl)
    [[0], [0, 1, 2], [0, 3]]
    """
    "*** YOUR CODE HERE ***"

def subseqs(s):
    """Return a nested list (a list of lists) of all subsequences of S.
    The subsequences can appear in any order. You can assume S is a list.

    >>> seqs = subseqs([1, 2, 3])
    >>> sorted(seqs)
    [[], [1], [1, 2], [1, 2, 3], [1, 3], [2], [2, 3], [3]]
    >>> subseqs([])
    [[]]
    """
    if ________________:
        ________________
    else:
        ________________
        ________________

Use Ok to test your code:

python3 ok -q subseqs

Q3: Non-Decreasing Subsequences

Just like the last question, we want to write a function that takes a list and returns a list of lists, where each individual list is a subsequence of the original input.

This time we have 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 contains no negative elements.

You may use the provided helper function insert_into_all, which takes in an item and a list of lists and inserts the item to the front of each list.

def non_decrease_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 = non_decrease_subseqs([1, 3, 2])
    >>> sorted(seqs)
    [[], [1], [1, 2], [1, 3], [2], [3]]
    >>> non_decrease_subseqs([])
    [[]]
    >>> seqs2 = non_decrease_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 ____________________
        elif s[0] < prev:
            return ____________________
        else:
            a = ______________________
            b = ______________________
            return insert_into_all(________, ______________) + ________________
    return subseq_helper(____, ____)

Use Ok to test your code:

python3 ok -q non_decrease_subseqs

Q4: Number of Trees

A full binary tree is a tree where each node has either 2 branches or 0 branches, but never 1 branch.

Write a function which returns the number of unique full binary tree structures that have exactly n leaves.

For those interested in combinatorics, this problem does have a closed form solution):

def num_trees(n):
    """Returns the number of unique full binary trees with exactly n leaves. E.g.,

    1   2        3       3    ...
    *   *        *       *
       / \      / \     / \
      *   *    *   *   *   *
              / \         / \
             *   *       *   *

    >>> num_trees(1)
    1
    >>> num_trees(2)
    1
    >>> num_trees(3)
    2
    >>> num_trees(8)
    429

    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q num_trees

Generators

Q5: Merge

Implement merge(incr_a, incr_b), which takes two iterables incr_a and incr_b whose elements are ordered. merge yields elements from incr_a and incr_b in sorted order, eliminating repetition. You may assume incr_a and incr_b themselves do not contain repeats, and that none of the elements of either are None. You may not assume that the iterables are finite; either may produce an infinite stream of results.

You will probably find it helpful to use the two-argument version of the built-in next function: next(incr, v) is the same as next(incr), except that instead of raising StopIteration when incr runs out of elements, it returns v.

See the doctest for examples of behavior.

def merge(incr_a, incr_b):
    """Yield the elements of strictly increasing iterables incr_a and incr_b, removing
    repeats. Assume that incr_a and incr_b have no repeats. incr_a or incr_b may or may not
    be infinite sequences.

    >>> m = merge([0, 2, 4, 6, 8, 10, 12, 14], [0, 3, 6, 9, 12, 15])
    >>> type(m)
    <class 'generator'>
    >>> list(m)
    [0, 2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
    >>> def big(n):
    ...    k = 0
    ...    while True: yield k; k += n
    >>> m = merge(big(2), big(3))
    >>> [next(m) for _ in range(11)]
    [0, 2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
    """
    iter_a, iter_b = iter(incr_a), iter(incr_b)
    next_a, next_b = next(iter_a, None), next(iter_b, None)
    "*** YOUR CODE HERE ***"

Watch the hints video below for somewhere to start:


Use Ok to test your code:

python3 ok -q merge

Objects

Minilecture Video: OOP

Q6: 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 Button:
    """
    Represents a single button
    """
    def __init__(self, pos, key):
        """
        Creates a button
        """
        self.pos = pos
        self.key = key
        self.times_pressed = 0

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
    'H'
    >>> k.press(1)
    'I'
    >>> k.press(2) #No button at this position
    ''
    >>> k.typing([0, 1])
    'HI'
    >>> k.typing([1, 0])
    'IH'
    >>> b1.times_pressed
    2
    >>> b2.times_pressed
    3
    """

    def __init__(self, *args):
        ________________
        for _________ in ________________:
            ________________

    def press(self, info):
        """Takes in a position of the button pressed, and
        returns that button's output"""
        if ____________________:
            ________________
            ________________
            ________________
        ________________

    def typing(self, typing_input):
        """Takes in a list of positions of buttons pressed, and
        returns the total output"""
        ________________
        for ________ in ____________________:
            ________________
        ________________

Use Ok to test your code:

python3 ok -q Keyboard

Q7: Bank Account

Implement the class Account, which acts as a a Bank Account. Account should allow the account holder to deposit money into the account, withdraw money from the account, and view their transaction history. The Bank Account should also prevents a user from withdrawing more than the current balance.

Transaction history should be stored as a list of tuples, where each tuple contains the type of transaction and the transaction amount. For example a withdrawal of 500 should be stored as ('withdraw', 500)

Hint: You can call the str function on an integer to get a string representation of the integer. You might find this function useful when implementing the __repr__ and __str__ methods.

Hint: You can alternatively use fstrings to implement the __repr__ and __str__ methods cleanly.

class Account:
    """A bank account that allows deposits and withdrawals.
    It tracks the current account balance and a transaction
    history of deposits and withdrawals.

    >>> eric_account = Account('Eric')
    >>> eric_account.deposit(1000000)   # depositing paycheck for the week
    1000000
    >>> eric_account.transactions
    [('deposit', 1000000)]
    >>> eric_account.withdraw(100)      # make a withdrawal to buy dinner
    999900
    >>> eric_account.transactions
    [('deposit', 1000000), ('withdraw', 100)]
    >>> print(eric_account) #call to __str__
    Eric's Balance: $999900
    >>> eric_account.deposit(10)
    999910
    >>> eric_account #call to __repr__
    Accountholder: Eric, Deposits: 2, Withdraws: 1
    """

    interest = 0.02

    def __init__(self, account_holder):
        self.balance = 0
        self.holder = account_holder
        "*** YOUR CODE HERE ***"

    def deposit(self, amount):
        """Increase the account balance by amount, add the deposit
        to the transaction history, and return the new balance.
        """
        "*** YOUR CODE HERE ***"

    def withdraw(self, amount):
        """Decrease the account balance by amount, add the withdraw
        to the transaction history, and return the new balance.
        """
        "*** YOUR CODE HERE ***"

    def __str__(self):
        "*** YOUR CODE HERE ***"

    def __repr__(self):
        "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q Account

Mutable Lists

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
    'Deal!'
    >>> 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)
    'Deal!'
    >>> a
    [3, 3, 2, 1, 4]
    >>> b
    [1, 1, 3, 2, 2, 7]
    >>> c
    [4, 3, 1, 4, 1]
    """
    m, n = 1, 1

    equal_prefix = lambda: ______________________
    while _______________________________:
        if __________________:
            m += 1
        else:
            n += 1

    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

Q9: Shuffle

Define a function shuffle that takes a sequence with an even number of elements (cards) and creates a new list that interleaves the elements of the first half with the elements of the second half.

def card(n):
    """Return the playing card numeral as a string for a positive n <= 13."""
    assert type(n) == int and n > 0 and n <= 13, "Bad card n"
    specials = {1: 'A', 11: 'J', 12: 'Q', 13: 'K'}
    return specials.get(n, str(n))

def shuffle(cards):
    """Return a shuffled list that interleaves the two halves of cards.

    >>> shuffle(range(6))
    [0, 3, 1, 4, 2, 5]
    >>> suits = ['♡', '♢', '♤', '♧']
    >>> cards = [card(n) + suit for n in range(1,14) for suit in suits]
    >>> cards[:12]
    ['A♡', 'A♢', 'A♤', 'A♧', '2♡', '2♢', '2♤', '2♧', '3♡', '3♢', '3♤', '3♧']
    >>> cards[26:30]
    ['7♤', '7♧', '8♡', '8♢']
    >>> shuffle(cards)[:12]
    ['A♡', '7♤', 'A♢', '7♧', 'A♤', '8♡', 'A♧', '8♢', '2♡', '8♤', '2♢', '8♧']
    >>> shuffle(shuffle(cards))[:12]
    ['A♡', '4♢', '7♤', '10♧', 'A♢', '4♤', '7♧', 'J♡', 'A♤', '4♧', '8♡', 'J♢']
    >>> cards[:12]  # Should not be changed
    ['A♡', 'A♢', 'A♤', 'A♧', '2♡', '2♢', '2♤', '2♧', '3♡', '3♢', '3♤', '3♧']
    """
    assert len(cards) % 2 == 0, 'len(cards) must be even'
    half = _______________
    shuffled = []
    for i in _____________:
        _________________
        _________________
    return shuffled

Use Ok to test your code:

python3 ok -q shuffle

Linked Lists

Q10: 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 should raise an IndexError with:

raise IndexError('Out of bounds!')
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>
    >>> other_link = link
    >>> insert(link, 9001, 0)
    >>> print(link)
    <9001 1 2 3>
    >>> link is other_link # Make sure you are using mutation! Don't create a new linked list.
    True
    >>> insert(link, 100, 2)
    >>> print(link)
    <9001 1 100 2 3>
    >>> insert(link, 4, 5)
    Traceback (most recent call last):
        ...
    IndexError: Out of bounds!
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q insert

Q11: 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))))
    3
    >>> deep_len(Link(Link(1, Link(2)), Link(3, Link(4))))
    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)
    5
    """
    if ______________:
        return 0
    elif ______________:
        return 1
    else:
        return _________________________

Use Ok to test your code:

python3 ok -q deep_len

Q12: 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)
'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)
    '[1|-]-->[2|-]-->[3|-]-->[4|-]-->[]'
    >>> kevins_to_string(Link.empty)
    '[]'
    >>> jerrys_to_string(lst)
    '(1 . (2 . (3 . (4 . ()))))'
    >>> jerrys_to_string(Link.empty)
    '()'
    """
    def printer(lnk):
        if ______________:
            return _________________________
        else:
            return _________________________
    return printer

Use Ok to test your code:

python3 ok -q make_to_string

Trees

Q13: 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
    Tree(6)
    >>> 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 ___________________________:
        largest = max(_______________, key=____________________)
        _________________________
    for __ in _____________:
        ___________________

Use Ok to test your code:

python3 ok -q prune_small

Q14: 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 list of node labels that starts with the root and ends at a leaf. Each subsequent element must be from a label of a branch 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 ordered in the output list from left to right in the tree. See the doctests for some examples.

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

    >>> long_paths(Tree(1), 0)
    [[1]]
    >>> long_paths(Tree(1), 1)
    []
    >>> 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])
    >>> print(whole)
    0
      1
        2
        3
          4
          4
          5
      13
      6
        7
          8
        9
      11
        12
          13
            14
    >>> 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)
    [[0, 11, 12, 13, 14]]
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q long_paths

Complexity

Q15: Determining Complexity

Use Ok to test your knowledge with the following questions:

python3 ok -q wwpd-complexity -u

Be sure to ask a member of course staff 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
Linear Θ(n).

Explanation: In the worst case, n is prime, and we have to execute the loop n - 2 times. Each iteration takes constant time (one conditional check and one return statement). Therefore, the total time is (n - 2) x constant, or simply linear.

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
Quadratic Θ(n^2).

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 constant, or linear. Knowing the runtime of linear, we can conclude that each iteration of the while loop in bar is linear. Therefore, the total runtime of bar(n) is quadratic.