CS 61A: Quiz 2

Due by 11:59pm on Thursday, 3/5

Instructions

Download quiz02.zip. Inside the archive, you will find a file called quiz02.py, along with a copy of the OK autograder.

Complete the quiz and submit it before 11:59pm on Thursday, 3/5. You must work alone, but you may talk to the course staff (see Asking Questions below). You may use any course materials, including an interpreter, course videos, slides, and readings. Please do not discuss these specific questions with your classmates, and do not scour the web for answers or post your answers online.

Your submission will be graded automatically for correctness. Your implementations do not need to be efficient, as long as they are correct. We will apply additional correctness tests as well as the ones provided. Passing these tests does not guarantee a perfect score.

Asking Questions: If you believe you need clarification on a question, make a private post on Piazza. Please do not post publicly about the quiz contents. If the staff discovers a problem with the quiz or needs to clarify a question, we will email the class via Piazza. You can also come to office hours to ask questions about the quiz or any other course material, but no answers or hints will be provided in office hours.

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.

Using OK

The ok program helps you test your code and track your progress. The first time you run the autograder, you will be asked to log in with your @berkeley.edu account using your web browser. Please do so. Each time you run ok, it will back up your work and progress on our servers. You can run all the doctests with the following command:

python3 ok

To test a specific question, use the -q option with the name of the function:

python3 ok -q <function>

By default, only tests that fail will appear. If you want to see how you did on all tests, you can use the -v option:

python3 ok -v

If you do not want to send your progress to our server or you have any problems logging in, add the --local flag to block all communication:

python3 ok --local

When you are ready to submit, run ok with the --submit option:

python3 ok --submit

Readings: You might find the following references useful:

Table of Contents

Question 1

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.

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 ***"

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

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.

Question 2

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.

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 ***"

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