Homework 5 Solutions

Solution Files

You can find the solutions in hw05.py.

Required Questions


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Q1: Count Dollars

Given a positive integer total, a set of dollar bills makes change for total if the sum of the values of the dollar bills is total. Here we will use standard US dollar bill values: 1, 5, 10, 20, 50, and 100. For example, the following sets make change for 15:

  • 15 1-dollar bills
  • 10 1-dollar, 1 5-dollar bills
  • 5 1-dollar, 2 5-dollar bills
  • 5 1-dollar, 1 10-dollar bills
  • 3 5-dollar bills
  • 1 5-dollar, 1 10-dollar bills

Thus, there are 6 ways to make change for 15. Write a recursive function count_dollars that takes a positive integer total and returns the number of ways to make change for total using 1, 5, 10, 20, 50, and 100 dollar bills.

Use next_smaller_dollar in your solution: next_smaller_dollar will return the next smaller dollar bill value from the input (e.g. next_smaller_dollar(5) is 1). The function will return None if the next dollar bill value does not exist.

Important: Use recursion; the tests will fail if you use loops.

Hint: Refer to the implementation of count_partitions for an example of how to count the ways to sum up to a final value with smaller parts. If you need to keep track of more than one value across recursive calls, consider writing a helper function.

def next_smaller_dollar(bill):
    """Returns the next smaller bill in order."""
    if bill == 100:
        return 50
    if bill == 50:
        return 20
    if bill == 20:
        return 10
    elif bill == 10:
        return 5
    elif bill == 5:
        return 1

def count_dollars(total):
    """Return the number of ways to make change.

    >>> count_dollars(15)  # 15 $1 bills, 10 $1 & 1 $5 bills, ... 1 $5 & 1 $10 bills
    6
    >>> count_dollars(10)  # 10 $1 bills, 5 $1 & 1 $5 bills, 2 $5 bills, 10 $1 bills
    4
    >>> count_dollars(20)  # 20 $1 bills, 15 $1 & $5 bills, ... 1 $20 bill
    10
    >>> count_dollars(45)  # How many ways to make change for 45 dollars?
    44
    >>> count_dollars(100) # How many ways to make change for 100 dollars?
    344
    >>> count_dollars(200) # How many ways to make change for 200 dollars?
    3274
    >>> from construct_check import check
    >>> # ban iteration
    >>> check(HW_SOURCE_FILE, 'count_dollars', ['While', 'For'])
    True
    """

    def constrained_count(total, largest_bill):
        if total == 0:
            return 1
        if total < 0:
            return 0
        if largest_bill == None:
            return 0
        without_dollar_bill = constrained_count(total, next_smaller_dollar(largest_bill))
        with_dollar_bill = constrained_count(total - largest_bill, largest_bill)
        return without_dollar_bill + with_dollar_bill
    return constrained_count(total, 100)

Use Ok to test your code:

python3 ok -q count_dollars

This is remarkably similar to the count_partitions problem, with a few minor differences:

  • A maximum partition size is not given, so we need to create a helper function that takes in two arguments: current total and dollar bill value.
  • Partition size is not linear. To get the next partition you need to call next_smaller_dollar.

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python3 ok --score

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Submit Assignment

Submit this assignment by uploading any files you've edited to the appropriate Gradescope assignment. Lab 00 has detailed instructions.

Optional Questions

These questions are optional. If you don't complete them, you will still receive credit for this assignment. They are great practice, so do them anyway!

Q2: Count Dollars Upward

Write a recursive function count_dollars_upward that is just like count_dollars except it uses next_larger_dollar, which returns the next larger dollar bill value from the input (e.g. next_larger_dollar(5) is 10). The function will return None if the next dollar bill value does not exist.

Important: Use recursion; the tests will fail if you use loops.

def next_larger_dollar(bill):
    """Returns the next larger bill in order."""
    if bill == 1:
        return 5
    elif bill == 5:
        return 10
    elif bill == 10:
        return 20
    elif bill == 20:
        return 50
    elif bill == 50:
        return 100

def count_dollars_upward(total):
    """Return the number of ways to make change using bills.

    >>> count_dollars_upward(15)  # 15 $1 bills, 10 $1 & 1 $5 bills, ... 1 $5 & 1 $10 bills
    6
    >>> count_dollars_upward(10)  # 10 $1 bills, 5 $1 & 1 $5 bills, 2 $5 bills, 10 $1 bills
    4
    >>> count_dollars_upward(20)  # 20 $1 bills, 15 $1 & $5 bills, ... 1 $20 bill
    10
    >>> count_dollars_upward(45)  # How many ways to make change for 45 dollars?
    44
    >>> count_dollars_upward(100) # How many ways to make change for 100 dollars?
    344
    >>> count_dollars_upward(200) # How many ways to make change for 200 dollars?
    3274
    >>> from construct_check import check
    >>> # ban iteration
    >>> check(HW_SOURCE_FILE, 'count_dollars_upward', ['While', 'For'])
    True
    """

    def constrained_count(total, smallest_bill):
        if total == 0:
            return 1
        if total < 0:
            return 0
        if smallest_bill == None:
            return 0
        without_dollar_bill = constrained_count(total, next_larger_dollar(smallest_bill))
        with_dollar_bill = constrained_count(total - smallest_bill, smallest_bill)
        return without_dollar_bill + with_dollar_bill
    return constrained_count(total, 1)

Use Ok to test your code:

python3 ok -q count_dollars_upward

This is remarkably similar to the count_partitions problem, with a few minor differences:

  • A maximum partition size is not given, so we need to create a helper function that takes in two arguments: current total and dollar bill value.
  • Partition size is not linear. To get the next partition you need to call next_larger_dollar.