Lab 8: Midterm Review

Due at 11:59pm on Friday, 10/20/2017.

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. Check that you have successfully submitted your code on okpy.org.

  • To receive credit for this lab, you must complete Questions 1, 2, 3, and 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

Note: For your reference, the Link and Tree classes can be found at the bottom of lab08.py.

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, which is the sum of the deep length of all linked lists contained in a deep linked list.

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)

Use Ok to test your code:

python3 ok -q deep_len

Q2: 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

Use Ok to test your code:

python3 ok -q make_to_string

Trees

Q3: 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))
    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

Q4: 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(add_trees(numbers, numbers))
    2
      4
        6
        8
      10
        12
          14
        16
    >>> print(add_trees(Tree(2), Tree(3, [Tree(4), Tree(5)])))
    5
      4
      5
    >>> print(add_trees(Tree(2, [Tree(3)]), Tree(2, [Tree(3), Tree(4)])))
    4
      6
      4
    >>> print(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.copy_tree() if not t2: return t1.copy_tree() 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

Q5: 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

Q6: 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

Q7: 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.typing([0, 1])
    'HI'
    >>> k.typing([1, 0])
    'IH'
    >>> b1.pressed
    2
    >>> b2.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.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: def __init__(self, pos, key): self.pos = pos self.key = key self.pressed = 0

Use Ok to test your code:

python3 ok -q Keyboard

Nonlocal

Q8: 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

Q9: 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

Q10: Zap (Orders of Growth)

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

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 contant work, so the overall runtime will be log n.