Lab 8: Midterm Review

lab08.zip

Due by 11:59pm on Friday, October 25.

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

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 okpy.org.

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!

Topics

Consult this section if you need a refresher on the material for this lab. It's okay to skip directly to the questions and refer back here should you get stuck.

Orders of Growth

Recall that the order of growth of a function expresses how long it takes for the function to run, and is defined in terms of the function's input sizes.

For example, let's say that we have the function get_x which is defined as follows:

def get_x(x):
    return x

get_x has one expression in it. That one expression takes the same amount of time to run, no matter what x is, or more importantly, how large x gets. This is called constant time.

The main two ways that a function in your program will get a running time different than just constant time is through either iteration or recursion. Let's start with some iteration examples!

The (simple) way you figure out the running time of a particular while loop is to simply count the cost of each operation in the body of the while loop, and then multiply that cost by the number of times that the loop runs. For example, look at the following method with a loop in it:

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

This loop has one statement in it sum, i = sum + i, i + 1. This statement is considered to run in constant time, as none of its operations rely on the size of the input. Individually, sum = sum + 1 and i = i + 1 are both constant time operations. However, when we're looking at order of growth, we take the maximum of those 2 values and use that as the running time. In 61A, we are not concerned with how long primitive functions, such as addition, multiplication, and variable assignment, take in order to run - we are mainly concerned with how many more times a loop is executed or how many more recursive calls occur as the input increases. In this example, we execute the loop n times, and for each iteration, we only execute constant time operations, so we get an order of growth of linear.

Here are a couple of basic functions, along with their running times. Try to understand why they have the given running time.

Linear

def bar(n):
    i, a, b = 1, 1, 0
    while i <= n:
        a, b, i = a + b, a, i + 1
     return a

Quadratic

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

Required Questions

Linked Lists

Q1: 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)
    IndexError
    """
    "*** YOUR CODE HERE ***"
    if index == 0:
        link.rest = Link(link.first, link.rest)
        link.first = value
    elif link.rest is Link.empty:
        raise IndexError
    else:
        insert(link.rest, value, index - 1)

# iterative solution
def insert(link, value, index):
    while index > 0 and link.rest is not Link.empty:
        link = link.rest
        index -= 1
    if index == 0:
        link.rest = Link(link.first, link.rest)
        link.first = value
    else:
        raise IndexError

Use Ok to test your code:

python3 ok -q insert

Orders of Growth

Q2: Finding Orders of Growth

Use Ok to test your knowledge with the following questions:
python3 ok -q wwpd-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

Linear.

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.

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.

Recursion and 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

In Lab 4, we examined the Subsequences problem. 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. For example, the lists [], [1, 3], [2], and [1, 3, 2] are subsequences of [1, 3, 2]. Again, 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 insert_into_all(item, lsts):
    """Assuming that lsts is a list of lists, return a new list consisting of
    all the lists in lsts, but with item added to the front of each.

    >>> lsts = [[], [1, 2], [3]]
    >>> insert_into_all(0, lsts)
    [[0], [0, 1, 2], [0, 3]]
    """
    return [[item] + lst for lst in lsts]

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)        else:
            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_subsequence

Optional Questions

Objects

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
    '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):
        "*** 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:
            accumulate+=self.press(pos)
        return accumulate
class Button:
    """
    Represents a single button
    """
    def __init__(self, pos, key):
        """
        Creates a button
        """
        self.pos = pos
        self.key = key
        self.times_pressed = 0

Use Ok to test your code:

python3 ok -q Keyboard

Nonlocal

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')
    1
    >>> tom_counter('count')
    2
    >>> tom_counter('global-count')
    1
    >>> jon_counter = make_counter()
    >>> jon_counter('global-count')
    2
    >>> jon_counter('count')
    1
    >>> jon_counter('reset')
    >>> jon_counter('count')
    1
    >>> tom_counter('count')
    3
    >>> jon_counter('global-count')
    3
    >>> jon_counter('global-reset')
    >>> tom_counter('global-count')
    1
    """
    "*** 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: 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

    "*** 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

Q8: 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'
    "*** YOUR CODE HERE ***"
    half = len(cards) // 2
    shuffled = []
    for i in range(half):
        shuffled.append(cards[i])
        shuffled.append(cards[half+i])
    return shuffled

Use Ok to test your code:

python3 ok -q shuffle

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)
    [100]
    >>> 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 []    else:
        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: 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
    """
    "*** 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(lnk.rest)

    # Video Walkthrough: https://youtu.be/pbMeCRUU7yw?t=2m28s

Use Ok to test your code:

python3 ok -q deep_len

Q11: 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)
    '()'
    """
    "*** YOUR CODE HERE ***"
    def printer(lnk):
        if lnk is Link.empty:
            return empty_repr
        else:
            return front + str(lnk.first) + mid + printer(lnk.rest) + back
    return printer

    # Video Walkthrough: https://youtu.be/pbMeCRUU7yw?t=14m22s

Use Ok to test your code:

python3 ok -q make_to_string

Q12: 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 ___________________________:
    while len(t.branches) >  n:
        largest = max(_______________, key=____________________)
        largest = max(t.branches, key=lambda x: x.label)
        _________________________
        t.branches.remove(largest)
    for __ in _____________:
    for b in t.branches:
        ___________________
        prune_small(b, n)

Use Ok to test your code:

python3 ok -q prune_small

Recursion / Tree Recursion

Q13: Number of Trees

How many different possible full binary tree (each node has two branches or 0, but never 1) structures exist that have exactly n leaves?

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

def num_trees(n):
    """How many full binary trees have 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 ***"
    if n == 1:
        return 1
    return sum(num_trees(k) * num_trees(n-k) for k in range(1, n))

Use Ok to test your code:

python3 ok -q num_trees