Homework 2 Solutions

Solution Files

You can find solutions for all questions in hw02.py.

Required Questions


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Recursion

Q1: Num Eights

Write a recursive function num_eights that takes a positive integer n and returns the number of times the digit 8 appears in n.

Important: Use recursion; the tests will fail if you use any assignment statements or loops. (You can define new functions, but don't put assignment statements there either.)

def num_eights(n):
    """Returns the number of times 8 appears as a digit of n.

    >>> num_eights(3)
    0
    >>> num_eights(8)
    1
    >>> num_eights(88888888)
    8
    >>> num_eights(2638)
    1
    >>> num_eights(86380)
    2
    >>> num_eights(12345)
    0
    >>> num_eights(8782089)
    3
    >>> from construct_check import check
    >>> # ban all assignment statements
    >>> check(SOURCE_FILE, 'num_eights',
    ...       ['Assign', 'AnnAssign', 'AugAssign', 'NamedExpr', 'For', 'While'])
    True
    """

    if n % 10 == 8:
        return 1 + num_eights(n // 10)
    elif n < 10:
        return 0
    else:
        return num_eights(n // 10)

Use Ok to test your code:

python3 ok -q num_eights

The equivalent iterative version of this problem might look something like this:

total = 0
while n > 0:
    if n % 10 == 8:
        total = total + 1
    n = n // 10
return total

The main idea is that we check each digit for a eight. The recursive solution is similar, except that you depend on the recursive call to count the occurences of eight in the rest of the number. Then, you add that to the number of eights you see in the current digit.

Q2: Interleaved Sum

Write a function interleaved_sum, which takes in a number n and two one-argument functions: f_odd and f_even. It returns the sum of applying f_odd to every odd number and f_even to every even number from 1 to n inclusive.

For example, executing interleaved_sum(5, lambda x: x, lambda x: x * x) returns 1 + 2*2 + 3 + 4*4 + 5 = 29.

Important: Implement this function without using any loops or directly testing if a number is odd or even (no using %). Instead of directly checking whether a number is even or odd, start with 1, which you know is an odd number.

Hint: Introduce an inner helper function that takes an odd number k and computes an interleaved sum from k to n (including n). Alternatively, you can use mutual recursion.

def interleaved_sum(n, f_odd, f_even):
    """Compute the sum f_odd(1) + f_even(2) + f_odd(3) + ..., up
    to n.

    >>> identity = lambda x: x
    >>> square = lambda x: x * x
    >>> triple = lambda x: x * 3
    >>> interleaved_sum(5, identity, square) # 1   + 2*2 + 3   + 4*4 + 5
    29
    >>> interleaved_sum(5, square, identity) # 1*1 + 2   + 3*3 + 4   + 5*5
    41
    >>> interleaved_sum(4, triple, square)   # 1*3 + 2*2 + 3*3 + 4*4
    32
    >>> interleaved_sum(4, square, triple)   # 1*1 + 2*3 + 3*3 + 4*3
    28
    >>> from construct_check import check
    >>> check(SOURCE_FILE, 'interleaved_sum', ['While', 'For', 'Mod']) # ban loops and %
    True
    >>> check(SOURCE_FILE, 'interleaved_sum', ['BitAnd', 'BitOr', 'BitXor']) # ban bitwise operators, don't worry about these if you don't know what they are
    True
    """

    def sum_from(k):
        if k > n:
            return 0
        elif k == n:
            return f_odd(k)
        else:
            return f_odd(k) + f_even(k+1) + sum_from(k + 2)
    return sum_from(1)

Use Ok to test your code:

python3 ok -q interleaved_sum

Q3: 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(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.

Sequences and Python Lists

Q4: Shuffle

Implement shuffle, which takes a sequence s (such as a list or range) with an even number of elements. It returns a new list that interleaves the elements of the first half of s with the elements of the second half.

To interleave two sequences s0 and s1 is to create a new list containing the first element of s0, the first element of s1, the second element of s0, the second element of s1, and so on. For example, if s = [1, 2, 3, 4, 5, 6] then s0 = [1, 2, 3] and s1 = [4, 5, 6], and interleaving s0 and s1 would result in [1, 4, 2, 5, 3, 6].

Challenge: Try to implement shuffle in less lines using a list comprehension!

def shuffle(s):
    """Return a shuffled list that interleaves the two halves of s.

    >>> shuffle(range(6))
    [0, 3, 1, 4, 2, 5]
    >>> letters = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
    >>> shuffle(letters)
    ['a', 'e', 'b', 'f', 'c', 'g', 'd', 'h']
    >>> shuffle(shuffle(letters))
    ['a', 'c', 'e', 'g', 'b', 'd', 'f', 'h']
    >>> letters  # Original list should not be modified
    ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
    """
    assert len(s) % 2 == 0, 'len(seq) must be even'

    half = len(s) // 2
    shuffled = []
    for i in range(half):
        shuffled = shuffled + [s[i]]
        shuffled = shuffled + [s[half + i]]
    return shuffled

    # Alternative List Comprehension Solution
    half = len(s) // 2
    return [x for i in range(half) for x in (s[i], s[half + i])]

Use Ok to test your code:

python3 ok -q shuffle

Q5: Merge

Implement merge, which takes 2 sorted lists of numbers s and t in ascending order. It returns a new list that contains all the elements from both s and t in sorted order. Do not remove duplicates; the length of the returned list should be len(s) + len(t). Use only recursion; while or for loops are not allowed.

def merge(s, t):
    """Merges two sorted lists.

    >>> s1 = [1, 3, 5]
    >>> s2 = [2, 4, 6]
    >>> merge(s1, s2)
    [1, 2, 3, 4, 5, 6]
    >>> s1
    [1, 3, 5]
    >>> s2
    [2, 4, 6]
    >>> merge([], [2, 4, 6])
    [2, 4, 6]
    >>> merge([1, 2, 3], [])
    [1, 2, 3]
    >>> merge([5, 7], [2, 4, 6])
    [2, 4, 5, 6, 7]
    >>> merge([2, 3, 4], [2, 4, 6])
    [2, 2, 3, 4, 4, 6]
    >>> from construct_check import check
    >>> check(SOURCE_FILE, 'merge', ['While', 'For'])    # ban iteration
    True
    """

    # Recursive Solution
    if not s or not t:
        return s + t
    elif s[0] < t[0]:
        return [s[0]] + merge(s[1:], t)
    else:
        return [t[0]] + merge(s, t[1:])

    # Alternative Recursive Solution
    if not s or not t:
        return s + t
    elif s[0] <= t[0]:
        return [s[0]] + merge(s[1:], t)
    else:
        return merge(t, s)
    # For the last line, instead of concatenating [t[0]] at the front like we did for the first solution,
    # we instead call merge again with the arguments switched, so that in the next call to merge, s
    # is now t, and t is s.
    # Note that in order for this to work successfully, we have to modify the elif case to have <= rather
    # than just <, otherwise we run into infinite recursion. (Try out the last doctest to see why)

Use Ok to test your code:

python3 ok -q merge

Check Your Score Locally

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python3 ok --score

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Submit Assignment

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[Optional] Exam Practice

Here are some related questions from past exams for you to try. These are optional. There is no way to submit them.

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