Homework 5: Object-Oriented Programming, Linked Lists, Iterators and Generators

Due by 11:59pm on Thursday, March 18

Instructions

Download hw05.zip. Inside the archive, you will find a file called hw05.py, along with a copy of the ok autograder.

Submission: When you are done, submit with python3 ok --submit. You may submit more than once before the deadline; only the final submission will be scored. Check that you have successfully submitted your code on okpy.org. See Lab 0 for more instructions on submitting assignments.

Using Ok: If you have any questions about using Ok, please refer to this guide.

Readings: You might find the following references useful:

Grading: Homework is graded based on correctness. Each incorrect problem will decrease the total score by one point. There is a homework recovery policy as stated in the syllabus. This homework is out of 2 points.

Required Questions

OOP

Q1: Vending Machine

In this question you'll create a vending machine that only outputs a single product and provides change when needed.

Create a class called VendingMachine that represents a vending machine for some product. A VendingMachine object returns strings describing its interactions.

Fill in the VendingMachine class, adding attributes and methods as appropriate, such that its behavior matches the following doctests:

class VendingMachine:
    """A vending machine that vends some product for some price.

    >>> v = VendingMachine('candy', 10)
    >>> v.vend()
    'Inventory empty. Restocking required.'
    >>> v.add_funds(15)
    'Inventory empty. Restocking required. Here is your $15.'
    >>> v.restock(2)
    'Current candy stock: 2'
    >>> v.vend()
    'You must add $10 more funds.'
    >>> v.add_funds(7)
    'Current balance: $7'
    >>> v.vend()
    'You must add $3 more funds.'
    >>> v.add_funds(5)
    'Current balance: $12'
    >>> v.vend()
    'Here is your candy and $2 change.'
    >>> v.add_funds(10)
    'Current balance: $10'
    >>> v.vend()
    'Here is your candy.'
    >>> v.add_funds(15)
    'Inventory empty. Restocking required. Here is your $15.'

    >>> w = VendingMachine('soda', 2)
    >>> w.restock(3)
    'Current soda stock: 3'
    >>> w.restock(3)
    'Current soda stock: 6'
    >>> w.add_funds(2)
    'Current balance: $2'
    >>> w.vend()
    'Here is your soda.'
    """
    "*** YOUR CODE HERE ***"

You may find Python's formatted string literals, or f-strings useful. A quick example:

>>> feeling = 'love'
>>> course = '61A!'
>>> f'I {feeling} {course}'
'I love 61A!'

Use Ok to test your code:

python3 ok -q VendingMachine

If you're curious about alternate methods of string formatting, you can also check out an older method of Python string formatting. A quick example:

>>> ten, twenty, thirty = 10, 'twenty', [30]
>>> '{0} plus {1} is {2}'.format(ten, twenty, thirty)
'10 plus twenty is [30]'

Linked Lists

Q2: Store Digits

Write a function store_digits that takes in an integer n and returns a linked list where each element of the list is a digit of n.

Note: do not use any string manipulation functions like str and reversed

def store_digits(n):
    """Stores the digits of a positive number n in a linked list.

    >>> s = store_digits(1)
    >>> s
    Link(1)
    >>> store_digits(2345)
    Link(2, Link(3, Link(4, Link(5))))
    >>> store_digits(876)
    Link(8, Link(7, Link(6)))
    >>> # a check for restricted functions
    >>> import inspect, re
    >>> cleaned = re.sub(r"#.*\\n", '', re.sub(r'"{3}[\s\S]*?"{3}', '', inspect.getsource(store_digits)))
    >>> print("Do not use str or reversed!") if any([r in cleaned for r in ["str", "reversed"]]) else None
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q store_digits

Trees

Q3: Yield Paths

Define a generator function path_yielder which takes in a Tree t, a value value, and returns a generator object which yields each path from the root of t to a node that has label value.

t is implemented with a class, not as the function-based ADT.

Each path should be represented as a list of the labels along that path in the tree. You may yield the paths in any order.

We have provided a (partial) skeleton for you. You do not need to use this skeleton, but if your implementation diverges significantly from it, you might want to think about how you can get it to fit the skeleton.

def path_yielder(t, value):
    """Yields all possible paths from the root of t to a node with the label value
    as a list.

    >>> t1 = Tree(1, [Tree(2, [Tree(3), Tree(4, [Tree(6)]), Tree(5)]), Tree(5)])
    >>> print(t1)
    1
      2
        3
        4
          6
        5
      5
    >>> next(path_yielder(t1, 6))
    [1, 2, 4, 6]
    >>> path_to_5 = path_yielder(t1, 5)
    >>> sorted(list(path_to_5))
    [[1, 2, 5], [1, 5]]

    >>> t2 = Tree(0, [Tree(2, [t1])])
    >>> print(t2)
    0
      2
        1
          2
            3
            4
              6
            5
          5
    >>> path_to_2 = path_yielder(t2, 2)
    >>> sorted(list(path_to_2))
    [[0, 2], [0, 2, 1, 2]]
    """

    "*** YOUR CODE HERE ***"

    for _______________ in _________________:
        for _______________ in _________________:

            "*** YOUR CODE HERE ***"

Hint: If you're having trouble getting started, think about how you'd approach this problem if it wasn't a generator function. What would your recursive calls be? With a generator function, what happens if you make a "recursive call" within its body?

Use Ok to test your code:

python3 ok -q path_yielder

Extra Questions

OOP

Q4: Mint

A mint is a place where coins are made. In this question, you'll implement a Mint class that can output a Coin with the correct year and worth.

  • Each Mint instance has a year stamp. The update method sets the year stamp to the current_year class attribute of the Mint class.
  • The create method takes a subclass of Coin and returns an instance of that class stamped with the mint's year (which may be different from Mint.current_year if it has not been updated.)
  • A Coin's worth method returns the cents value of the coin plus one extra cent for each year of age beyond 50. A coin's age can be determined by subtracting the coin's year from the current_year class attribute of the Mint class.
class Mint:
    """A mint creates coins by stamping on years.

    The update method sets the mint's stamp to Mint.current_year.

    >>> mint = Mint()
    >>> mint.year
    2020
    >>> dime = mint.create(Dime)
    >>> dime.year
    2020
    >>> Mint.current_year = 2100  # Time passes
    >>> nickel = mint.create(Nickel)
    >>> nickel.year     # The mint has not updated its stamp yet
    2020
    >>> nickel.worth()  # 5 cents + (80 - 50 years)
    35
    >>> mint.update()   # The mint's year is updated to 2100
    >>> Mint.current_year = 2175     # More time passes
    >>> mint.create(Dime).worth()    # 10 cents + (75 - 50 years)
    35
    >>> Mint().create(Dime).worth()  # A new mint has the current year
    10
    >>> dime.worth()     # 10 cents + (155 - 50 years)
    115
    >>> Dime.cents = 20  # Upgrade all dimes!
    >>> dime.worth()     # 20 cents + (155 - 50 years)
    125
    """
    current_year = 2020

    def __init__(self):
        self.update()

    def create(self, kind):
        "*** YOUR CODE HERE ***"

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

class Coin:
    def __init__(self, year):
        self.year = year

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

class Nickel(Coin):
    cents = 5

class Dime(Coin):
    cents = 10

Use Ok to test your code:

python3 ok -q Mint

Generators/Trees

Q5: Is BST

Write a function is_bst, which takes a Tree t and returns True if, and only if, t is a valid binary search tree, which means that:

  • Each node has at most two children (a leaf is automatically a valid binary search tree)
  • The children are valid binary search trees
  • For every node, the entries in that node's left child are less than or equal to the label of the node
  • For every node, the entries in that node's right child are greater than the label of the node

An example of a BST is:

bst

Note that, if a node has only one child, that child could be considered either the left or right child. You should take this into consideration.

Hint: It may be helpful to write helper functions bst_min and bst_max that return the minimum and maximum, respectively, of a Tree if it is a valid binary search tree.

def is_bst(t):
    """Returns True if the Tree t has the structure of a valid BST.

    >>> t1 = Tree(6, [Tree(2, [Tree(1), Tree(4)]), Tree(7, [Tree(7), Tree(8)])])
    >>> is_bst(t1)
    True
    >>> t2 = Tree(8, [Tree(2, [Tree(9), Tree(1)]), Tree(3, [Tree(6)]), Tree(5)])
    >>> is_bst(t2)
    False
    >>> t3 = Tree(6, [Tree(2, [Tree(4), Tree(1)]), Tree(7, [Tree(7), Tree(8)])])
    >>> is_bst(t3)
    False
    >>> t4 = Tree(1, [Tree(2, [Tree(3, [Tree(4)])])])
    >>> is_bst(t4)
    True
    >>> t5 = Tree(1, [Tree(0, [Tree(-1, [Tree(-2)])])])
    >>> is_bst(t5)
    True
    >>> t6 = Tree(1, [Tree(4, [Tree(2, [Tree(3)])])])
    >>> is_bst(t6)
    True
    >>> t7 = Tree(2, [Tree(1, [Tree(5)]), Tree(4)])
    >>> is_bst(t7)
    False
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q is_bst

Q6: Preorder

Define the function preorder, which takes in a tree as an argument and returns a list of all the entries in the tree in the order that print_tree would print them.

The following diagram shows the order that the nodes would get printed, with the arrows representing function calls.

preorder

Note: This ordering of the nodes in a tree is called a preorder traversal.

def preorder(t):
    """Return a list of the entries in this tree in the order that they
    would be visited by a preorder traversal (see problem description).

    >>> numbers = Tree(1, [Tree(2), Tree(3, [Tree(4), Tree(5)]), Tree(6, [Tree(7)])])
    >>> preorder(numbers)
    [1, 2, 3, 4, 5, 6, 7]
    >>> preorder(Tree(2, [Tree(4, [Tree(6)])]))
    [2, 4, 6]
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q preorder

Q7: Generate Preorder

Similarly to preorder in Question 3, define the function generate_preorder, which takes in a tree as an argument and now instead yields the entries in the tree in the order that print_tree would print them.

Hint: How can you modify your implementation of preorder to yield from your recursive calls instead of returning them?

def generate_preorder(t): 
   """Yield the entries in this tree in the order that they
   would be visited by a preorder traversal (see problem description).

   >>> numbers = Tree(1, [Tree(2), Tree(3, [Tree(4), Tree(5)]), Tree(6, [Tree(7)])])
   >>> gen = generate_preorder(numbers)
   >>> next(gen)
   1
   >>> list(gen)
   [2, 3, 4, 5, 6, 7]
   """
   "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q generate_preorder