Lab 8: Midterm Review

Due at 11:59pm on Monday, 3/20/2017.

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.

  • To receive credit for this lab, you must complete Questions 1, 2, 3, 4 in lab08.py and submit through OK.
  • The remaining questions are extra practice. They can be found in the lab08_extra.py file. It is recommended that you complete these problems on your own time.

Required Questions

Linked Lists

A linked list is either an empty linked list (Link.empty) or a Link object containing a first value and the rest of the linked list. Here is a part of the Link class. The entire class can be found in the assignment's files:

class Link:
    """A linked list.

    >>> s = Link(1, Link(2, Link(3)))
    >>> s.first
    1
    >>> s.rest
    Link(2, Link(3))
    """
    empty = ()

    def __init__(self, first, rest=empty):
        assert rest is Link.empty or isinstance(rest, Link)
        self.first = first
        self.rest = rest

    def __repr__(self):
        if self.rest is Link.empty:
            return 'Link({})'.format(self.first)
        else:
            return 'Link({}, {})'.format(self.first, repr(self.rest))

To check if a Link is empty, compare it against the class attribute Link.empty. For example, the below function prints out whether or not the link it is handed is empty:

def test_empty(link):
    if link is Link.empty:
        print('This linked list is empty!')
    else:
        print('This linked list is not empty!')

Note: Linked lists are recursive data structures! A linked list contains the first element of the list (first) and a reference to another linked list (rest) which contains the rest of the values in the list.

Question 1: Linked Lists as Strings

Stan and Michelle like different ways of displaying the linked list structure in Python. While Stan likes box and pointer diagrams, Michelle 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.

    >>> stans_to_string = make_to_string("[", "|-]-->", "", "[]")
    >>> michelles_to_string = make_to_string("(", " . ", ")", "()")
    >>> lst = Link(1, Link(2, Link(3, Link(4))))
    >>> stans_to_string(lst)
    '[1|-]-->[2|-]-->[3|-]-->[4|-]-->[]'
    >>> stans_to_string(Link.empty)
    '[]'
    >>> michelles_to_string(lst)
    '(1 . (2 . (3 . (4 . ()))))'
    >>> michelles_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

Use OK to test your code:

python3 ok -q make_to_string

Question 2: Mutate Reverse

Implement mutate_reverse, which takes a linked list instance and mutates it so that its values appear in reverse order. For an extra challenge, find an implementation that requires only linear time in the length of the list (big-Theta n).

def mutate_reverse(link):
    """Mutates the Link so that its elements are reversed.

    >>> link = Link(1)
    >>> mutate_reverse(link)
    >>> link
    Link(1)

    >>> link = Link(1, Link(2, Link(3)))
    >>> mutate_reverse(link)
    >>> link
    Link(3, Link(2, Link(1)))
    """

    "*** YOUR CODE HERE ***"

    if link is Link.empty or link.rest is Link.empty:
        return
    mutate_reverse(link.rest)
    while link.rest is not Link.empty:
        link.first, link.rest.first = link.rest.first, link.first
        link = link.rest

# O(n) solution
def mutate_reverse(link):
    """Mutates the Link so that its elements are reversed.

    >>> link = Link(1)
    >>> mutate_reverse(link)
    >>> link
    Link(1)

    >>> link = Link(1, Link(2, Link(3)))
    >>> mutate_reverse(link)
    >>> link
    Link(3, Link(2, Link(1)))
    """
    link.rest = reverse_in_place(link.rest)
    while link.rest is not Link.empty:
        link.first, link.rest.first = link.rest.first, link.first
        link = link.rest

def reverse_in_place(link):
    result = Link.empty
    while link is not Link.empty:
        rest = link.rest
        link.rest = result
        result = link
        link = rest
    return result

Use OK to test your code:

python3 ok -q mutate_reverse

Trees

Question 3: 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 tree 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(tree_map(lambda x: 2**x, numbers))
    2
      4
        8
        16
      32
        64
          128
        256
    """

    "*** 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])

Use OK to test your code:

python3 ok -q tree_map

Question 4: Add trees

Define the function add_trees, which takes in two trees and returns a new tree where each corresponding node from the first tree is added with the node from the second tree. If a node at any particular position is present in one tree but not the other, it should be present in the new tree as well.

Hint: You may want to use the built-in zip function to iterate over multiple sequences at once.

def add_trees(t1, t2):
    """
    >>> numbers = Tree(1,
    ...                [Tree(2,
    ...                      [Tree(3),
    ...                       Tree(4)]),
    ...                 Tree(5,
    ...                      [Tree(6,
    ...                            [Tree(7)]),
    ...                       Tree(8)])])
    >>> print_tree(add_trees(numbers, numbers))
    2
      4
        6
        8
      10
        12
          14
        16
    >>> print_tree(add_trees(Tree(2), Tree(3, [Tree(4), Tree(5)])))
    5
      4
      5
    >>> print_tree(add_trees(Tree(2, [Tree(3)]), Tree(2, [Tree(3), Tree(4)])))
    4
      6
      4
    >>> print_tree(add_trees(Tree(2, [Tree(3, [Tree(4), Tree(5)])]), \
    Tree(2, [Tree(3, [Tree(4)]), Tree(5)])))
    4
      6
        8
        5
      5
    """

    "*** YOUR CODE HERE ***"

    if not t1:
        return t2 # Could replace with copy_tree(t2)
    if not t2:
        return t1 # Could replace with copy_tree(t1)
    new_label = t1.label + t2.label
    t1_branches, t2_branches = list(t1.branches), list(t2.branches)
    length_t1, length_t2 = len(t1_branches), len(t2_branches)
    if length_t1 < length_t2:
        t1_branches += [None for _ in range(length_t1, length_t2)]
    elif length_t1 > length_t2:
        t2_branches += [None for _ in range(length_t2, length_t1)]
    return Tree(new_label, [add_trees(branch1, branch2) for branch1, branch2 in zip(t1_branches, t2_branches)])

Use OK to test your code:

python3 ok -q add_trees

Optional Questions

Objects

Question 5: WWPP: Methods

Use OK to test your knowledge with the following "What Would Python Print?" questions:

python3 ok -q foobar -u

Hint: Remember for all WWPP questions, enter Function if you believe the answer is <function ...> and Error if it errors.

>>> class Foo:
...     def print_one(self):
...         print('foo')
...     def print_two():
...         print('foofoo')
>>> f = Foo()
>>> f.print_one()

______
foo

>>> f.print_two()

______
Error

>>> Foo.print_two()

______
foofoo

>>> class Bar(Foo):
...     def print_one(self):
...         print('bar')
>>> b = Bar()
>>> b.print_one()

______
bar

>>> Bar.print_two()

______
foofoo

>>> Bar.print_one = lambda x: print('new bar')
>>> b.print_one()

______
new bar

Question 6: WWPP: Attributes

Use OK to test your knowledge with the following "What Would Python Print?" questions:

python3 ok -q attributes -u

Hint: Remember for all WWPP questions, enter Function if you believe the answer is <function ...> and Error if it errors.

>>> class Foo:
...     a = 10
...     def __init__(self, a):
...         self.a = a
>>> class Bar(Foo):
...     b = 1
>>> a = Foo(5)
>>> b = Bar(2)
>>> a.a

______
5

>>> b.a

______
2

>>> Foo.a

______
10

>>> Bar.b

______
1

>>> Bar.a

______
10

>>> b.b

______
1

>>> Foo.c = 15
>>> Foo.c

______
15

>>> a.c

______
15

>>> b.c

______
15

>>> Bar.c

______
15

>>> b.b = 3
>>> b.b

______
3

>>> Bar.b

______
1

Nonlocal

Question 7: 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

Lists

Question 8: 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.

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

Orders of Growth

Question 9: Boom (Orders of Growth)

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

θ(n4)

Use OK to test your code:

python3 ok -q boom -u