CS88 | Lab 7: Object-Oriented Programming

Starter Files

Download lab07.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.

Submit the lab07.py file to ok.

OOP terminology

Object-oriented programming (OOP) is a style of programming that allows you to think of code in terms of "objects." Here's an example of a Car class:

class Car(object):

num_wheels = 4
gas = 30
headlights = 2
size = 'Tiny'

def __init__(self, make, model):
    self.make = make
    self.model = model
    self.color = 'No color yet. You need to paint me.'
    self.wheels = Car.num_wheels
    self.gas = Car.gas

def paint(self, color):
    self.color = color
    return self.make + ' ' + self.model + ' is now ' + color

def drive(self):
    if self.wheels < Car.num_wheels or self.gas <= 0:
        return 'Cannot drive!'
    self.gas -= 10
    return self.make + ' ' + self.model + ' goes vroom!'

def pop_tire(self):
    if self.wheels > 0:
        self.wheels -= 1

def fill_gas(self):
    self.gas += 20
    return 'Gas level: ' + str(self.gas)

Here's some terminology:

class: a blueprint for how to build a certain type of object. The Car class (shown above) describes the behavior and data that all Car objects have.
instance: a particular occurrence of a class. In Python, we create instances of a class like this:
```
>>> my_car = Car('Tesla', 'Model S')
```
my_car is an instance of the Car class.
attribute or field: a variable that belongs to the class. Think of an attribute as a quality of the object: cars have wheels and size, so we have given our Car class self.wheels and self.size attributes. We can access attributes using dot notation:
```
>>> my_car.size
'Tiny'
>>> my_car.wheels
4
```
method: Methods are just like normal functions, except that they are tied to an instance or a class. Think of a method as a "verb" of the class: cars can drive and also pop their tires, so we have given our Car class the methods drive and pop_tire. We call methods using dot notation:
```
>>> my_car = Car('Tesla', 'Model S')
>>> my_car.drive()
'Tesla Model S goes vroom!'
```
constructor: As with data abstraction, constructors describe how to build an instance of the class. Most classes have a constructor. In Python, the constructor of the class defined as __init__. For example, here is the Car class's constructor:
```
def __init__(self, make, model):
        self.make = make
        self.model = model
        self.color = 'No color yet. You need to paint me.'
        self.wheels = Car.num_wheels
        self.gas = Car.gas
```
The constructor takes in two arguments, make and model. As you can see, the constructor also creates the self.color, self.wheels and self.gas attributes.
self: in Python, self is the first parameter for many methods (in this class, we will only use methods whose first parameter is self). When a method is called, self is bound to an instance of the class. For example:
```
>>> my_car = Car('Tesla', 'Model S')
>>> my_car.drive()
```
Notice that the drive method takes in self as an argument, but it looks like we didn't pass one in! This is because the dot notation implicitly passes in car as self for us.

Car WWPD

Question 1: Car

Use OK to test your knowledge with the following What would Python print questions:
python3 ok -q car -u

If you get stuck try typing these in the interpreter yourself

python3 -i

Make Change

Question 2

Implement make_change, which takes a positive integer amount and a dictionary of coins. The coins dictionary keys are positive integer denominations and its values are positive integer coin counts. For example, {1: 4, 5: 2} represents four pennies and two nickels. The make_change function returns a list of coins that sum to amount, where the count of any denomination k in the return value is at most coins[k].

If there are multiple ways to make change for amount, prefer to use as many of the smallest coins available and place the smallest coins first in the returned list.

Hint: Try using the smallest coin to make change. If it turns out that there is no way to make change using the smallest coin, then try making change without the smallest coin.

Hint: The simplest solution does not involve defining any local functions, but you can define additional functions if you wish.

def make_change(amount, coins):
    """Return a list of coins that sum to amount, preferring the smallest coins
    available and placing the smallest coins first in the returned list.

    The coins argument is a dictionary with keys that are positive integer
    denominations and values that are positive integer coin counts.

    >>> make_change(2, {2: 1})
    [2]
    >>> make_change(2, {1: 2, 2: 1})
    [1, 1]
    >>> make_change(4, {1: 2, 2: 1})
    [1, 1, 2]
    >>> make_change(4, {2: 1}) == None
    True

    >>> coins = {2: 2, 3: 2, 4: 3, 5: 1}
    >>> make_change(4, coins)
    [2, 2]
    >>> make_change(8, coins)
    [2, 2, 4]
    >>> make_change(25, coins)
    [2, 3, 3, 4, 4, 4, 5]
    >>> coins[8] = 1
    >>> make_change(25, coins)
    [2, 2, 4, 4, 5, 8]
    """
    if not coins:
        return None
    smallest = min(coins)
    rest = remove_one(coins, smallest)
    "*** YOUR CODE HERE ***"
    
    if amount == smallest:
        return [smallest]
    result = make_change(amount-smallest, rest)
    if result:
        return [smallest] + result
    else:
        return make_change(amount, rest)

You can use the remove_one function in your implementation:

def remove_one(coins, coin):
    """Remove one coin from a dictionary of coins. Return a new dictionary,
    leaving the original dictionary coins unchanged.

    >>> coins = {2: 5, 3: 2, 6: 1}
    >>> remove_one(coins, 2) == {2: 4, 3: 2, 6: 1}
    True
    >>> remove_one(coins, 6) == {2: 5, 3: 2}
    True
    >>> coins == {2: 5, 3: 2, 6: 1} # Unchanged
    True
    """
    copy = dict(coins)
    count = copy.pop(coin) - 1
    if count:
        copy[coin] = count
    return copy

Use OK to test your code:

python3 ok -q make_change

Question 3

Complete the change method of the ChangeMachine class. A ChangeMachine instance holds some coins, which are initially all pennies. The change method takes a positive integer coin, adds that coin to its coins, and then returns a list that sums to coin. The machine prefers to return as many of the smallest coins available, ordered from smallest to largest. The coins returned by change are removed from the machine's coins.

Hint: Call the make_change function in order to compute the result of change, but update self.coins before returning that result.

Hint: To remove key-value pairs from a dictionary, you can use use .pop(<key>). For example, d.pop("first_key") will remove the key-value pair associated with "first_key" from d.

class ChangeMachine:
    """A change machine holds a certain number of coins, initially all pennies.
    The change method adds a single coin of some denomination X and returns a
    list of coins that sums to X. The machine prefers to return the smallest
    coins available. The total value in the machine never changes, and it can
    always make change for any coin (perhaps by returning the coin passed in).

    The coins attribute is a dictionary with keys that are positive integer
    denominations and values that are positive integer coin counts.

    >>> m = ChangeMachine(2)
    >>> m.coins == {1: 2}
    True
    >>> m.change(2)
    [1, 1]
    >>> m.coins == {2: 1}
    True
    >>> m.change(2)
    [2]
    >>> m.coins == {2: 1}
    True
    >>> m.change(3)
    [3]
    >>> m.coins == {2: 1}
    True

    >>> m = ChangeMachine(10) # 10 pennies
    >>> m.coins == {1: 10}
    True
    >>> m.change(5) # takes a nickel & returns 5 pennies
    [1, 1, 1, 1, 1]
    >>> m.coins == {1: 5, 5: 1} # 5 pennies & a nickel remain
    True
    >>> m.change(3)
    [1, 1, 1]
    >>> m.coins == {1: 2, 3: 1, 5: 1}
    True
    >>> m.change(2)
    [1, 1]
    >>> m.change(2) # not enough 1's remaining; return a 2
    [2]
    >>> m.coins == {2: 1, 3: 1, 5: 1}
    True
    >>> m.change(8) # cannot use the 2 to make 8, so use 3 & 5
    [3, 5]
    >>> m.coins == {2: 1, 8: 1}
    True
    >>> m.change(1) # return the penny passed in (it's the smallest)
    [1]
    >>> m.change(9) # return the 9 passed in (no change possible)
    [9]
    >>> m.coins == {2: 1, 8: 1}
    True
    >>> m.change(10)
    [2, 8]
    >>> m.coins == {10: 1}
    True

    >>> m = ChangeMachine(9)
    >>> [m.change(k) for k in [2, 2, 3]]
    [[1, 1], [1, 1], [1, 1, 1]]
    >>> m.coins == {1: 2, 2: 2, 3: 1}
    True
    >>> m.change(5) # Prefers [1, 1, 3] to [1, 2, 2] (more pennies)
    [1, 1, 3]
    >>> m.change(7)
    [2, 5]
    >>> m.coins == {2: 1, 7: 1}
    True
    """
    def __init__(self, pennies):
        self.coins = {1: pennies}

    def change(self, coin):
        """Return change for coin, removing the result from self.coins."""
        "*** YOUR CODE HERE ***"
        
        self.coins[coin] = 1 + self.coins.get(coin, 0)
        result = make_change(coin, self.coins)
        for coin in result:
            count = self.coins.pop(coin) - 1
            if count:
                self.coins[coin] = count
        return result