Homework 5 Solutions

hw05.zip

Solution Files

You can find the solutions in hw05.py.

Required Questions

Q1: Double Eights

Write a recursive function that takes in a positive integer n and determines if its digits contain two adjacent 8s (that is, two 8s right next to each other).

Hint: Start by coming up with a recursive plan: the digits of a number have double eights if either (think of something that is straightforward to check) or double eights appear in the rest of the digits.

Important: Use recursion; the tests will fail if you use any loops (for, while), the in operator, or the str function.

def double_eights(n: int) -> bool:
    """Returns whether or not n has two digits in row that
    are the number 8.

    >>> double_eights(1288)
    True
    >>> double_eights(880)
    True
    >>> double_eights(538835)
    True
    >>> double_eights(284682)
    False
    >>> double_eights(588138)
    True
    >>> double_eights(78)
    False
    >>> # ban iteration, in operator and str function 
    >>> from construct_check import check
    >>> check(SOURCE_FILE, 'double_eights', ['While', 'For', 'In', 'Str'])
    True
    """
    last, second_last = n % 10, n // 10 % 10
    if last == 8 and second_last == 8:
        return True
    elif n < 100:
        return False
    return double_eights(n // 10)

    # Alternate solution
    last, second_last = n % 10, n // 10 % 10
    if n < 10:
        return False
    return (last == 8 and second_last == 8) or double_eights(n // 10)

    # Alternate solution with helper function:
    def helper(num, prev_eight):
        if num == 0:
            return False
        if num % 10 == 8:
            if prev_eight:
                return True
            return helper(num // 10, True)
        return helper(num // 10, False)
    return helper(n, False)

Use Ok to test your code:

python3 ok -q double_eights

Q2: Add Characters

Given two words, w1 and w2, we say w1 is a subsequence of w2 if all the letters in w1 appear in w2 in the same order (but not necessarily all together). That is, you can add letters to any position in w1 to get w2. For example, "sing" is a substring of "absorbing" and "cat" is a substring of "contrast".

Implement add_chars, which takes in w1 and w2, where w1 is a substring of w2. This means that w1 is shorter than w2. It should return a string containing the characters you need to add to w1 to get w2. Your solution must use recursion.

In the example above, you need to add the characters "aborb" to "sing" to get "absorbing", and you need to add "ontrs" to "cat" to get "contrast".

The letters in the string you return should be in the order you have to add them from left to right. If there are multiple characters in the w2 that could correspond to characters in w1, use the leftmost one. For example, add_words("coy", "cacophony") should return "acphon", not "caphon" because the first "c" in "coy" corresponds to the first "c" in "cacophony".

def add_chars(w1, w2):
    """
    Return a string containing the characters you need to add to w1 to get w2.

    You may assume that w1 is a subsequence of w2.

    >>> add_chars("owl", "howl")
    'h'
    >>> add_chars("want", "wanton")
    'on'
    >>> add_chars("rat", "radiate")
    'diae'
    >>> add_chars("a", "prepare")
    'prepre'
    >>> add_chars("resin", "recursion")
    'curo'
    >>> add_chars("fin", "effusion")
    'efuso'
    >>> add_chars("coy", "cacophony")
    'acphon'
    >>> from construct_check import check
    >>> # ban iteration and sets
    >>> check(SOURCE_FILE, 'add_chars',
    ...       ['For', 'While', 'Set', 'SetComp']) # Must use recursion
    True
    """
    if not w1:
        return w2
    elif w1[0] == w2[0]:
        return add_chars(w1[1:], w2[1:])
    return w2[0] + add_chars(w1, w2[1:])

Use Ok to test your code:

python3 ok -q add_chars

Q3: Sublist

Write a function has_sublist that takes two lists, lst and sublist, and returns whether the elements of sublist appear consecutively in order anywhere within lst.

def has_sublist(lst, sublist):
    """Returns whether the elements of sublist appear consecutively in order anywhere within lst.
    >>> has_sublist([], [])
    True
    >>> has_sublist([3, 3, 2, 1], [])
    True
    >>> has_sublist([], [3, 3, 2, 1])
    False
    >>> has_sublist([3, 3, 2, 1], [3, 2, 1])
    True
    >>> has_sublist([3, 2, 1], [3, 2, 1])
    True
    >>> has_sublist([4, 3, 2, 1], [4, 2, 1])
    False
    >>> has_sublist([9, 3, 2, 1, 9], [3, 2, 1])
    True
    """
    sublist_length = len(sublist)
    lst_length = len(lst)
    if sublist_length > lst_length:
        return False
    elif lst[:sublist_length] == sublist:
        return True
    else:
        return has_sublist(lst[1:], sublist)

Use Ok to test your code:

python3 ok -q has_sublist

Just For Fun Questions

The questions below are optional and not representative of exam questions. You can try them if you want an extra challenge, but they're just puzzles that are not required for the course. Almost all students will skip them, and that's fine. We will not be prioritizing support for these questions on Ed or during office hours.

Q4: Towers of Hanoi

A classic puzzle called the Towers of Hanoi is a game that consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with n disks in a neat stack in ascending order of size on a start rod, the smallest at the top, forming a conical shape.

The objective of the puzzle is to move the entire stack to an end rod, obeying the following rules:

Only one disk may be moved at a time.
Each move consists of taking the top (smallest) disk from one of the rods and sliding it onto another rod, on top of the other disks that may already be present on that rod.
No disk may be placed on top of a smaller disk.

Complete the definition of move_stack, which prints out the steps required to move n disks from the start rod to the end rod without violating the rules. The provided print_move function will print out the step to move a single disk from the given origin to the given destination.

Hint: Draw out a few games with various n on a piece of paper and try to find a pattern of disk movements that applies to any n. In your solution, take the recursive leap of faith whenever you need to move any amount of disks less than n from one rod to another. If you need more help, see the following hints.

See the following animation of the Towers of Hanoi, found on Wikimedia by user Trixx.

The strategy used in Towers of Hanoi is to move all but the bottom disc to the second peg, then moving the bottom disc to the third peg, then moving all but the second disc from the second to the third peg.

One thing you don't need to worry about is collecting all the steps. print effectively "collects" all the results in the terminal as long as you make sure that the moves are printed in order.

def print_move(origin, destination):
    """Print instructions to move a disk."""
    print("Move the top disk from rod", origin, "to rod", destination)

def move_stack(n, start, end):
    """Print the moves required to move n disks on the start pole to the end
    pole without violating the rules of Towers of Hanoi.

    n -- number of disks
    start -- a pole position, either 1, 2, or 3
    end -- a pole position, either 1, 2, or 3

    There are exactly three poles, and start and end must be different. Assume
    that the start pole has at least n disks of increasing size, and the end
    pole is either empty or has a top disk larger than the top n start disks.

    >>> move_stack(1, 1, 3)
    Move the top disk from rod 1 to rod 3
    >>> move_stack(2, 1, 3)
    Move the top disk from rod 1 to rod 2
    Move the top disk from rod 1 to rod 3
    Move the top disk from rod 2 to rod 3
    >>> move_stack(3, 1, 3)
    Move the top disk from rod 1 to rod 3
    Move the top disk from rod 1 to rod 2
    Move the top disk from rod 3 to rod 2
    Move the top disk from rod 1 to rod 3
    Move the top disk from rod 2 to rod 1
    Move the top disk from rod 2 to rod 3
    Move the top disk from rod 1 to rod 3
    """
    assert 1 <= start <= 3 and 1 <= end <= 3 and start != end, "Bad start/end"
    if n == 1:
        print_move(start, end)
    else:
        other = 6 - start - end
        move_stack(n-1, start, other)
        print_move(start, end)
        move_stack(n-1, other, end)

Use Ok to test your code:

python3 ok -q move_stack

To solve the Towers of Hanoi problem for n disks, we need to do three steps:

Move everything but the last disk (n-1 disks) to someplace in the middle (not the start nor the end rod).
Move the last disk (a single disk) to the end rod. This must occur after step 1 (we have to move everything above it away first)!
Move everything but the last disk (the disks from step 1) from the middle on top of the end rod.

We take advantage of the fact that the recursive function move_stack is guaranteed to move n disks from start to end while obeying the rules of Towers of Hanoi. The only thing that remains is to make sure that we have set up the playing board to make that possible.

Since we move a disk to end rod, we run the risk of move_stack doing an improper move (big disk on top of small disk). But since we're moving the biggest disk possible, nothing in the n-1 disks above that is bigger. Therefore, even though we do not explicitly state the Towers of Hanoi constraints, we can still carry out the correct steps.

Video walkthrough:

YouTube link

Q5: Anonymous Factorial

This question demonstrates that it's possible to write recursive functions without assigning them a name in the global frame.

The recursive factorial function can be written as a single expression by using a conditional expression.

>>> fact = lambda n: 1 if n == 1 else mul(n, fact(sub(n, 1)))
>>> fact(5)
120

However, this implementation relies on the fact (no pun intended) that fact has a name, to which we refer in the body of fact. To write a recursive function, we have always given it a name using a def or assignment statement so that we can refer to the function within its own body. In this question, your job is to define fact recursively without using a def or assignment statement to give it a name.

Write an expression that computes n factorial using only call expressions, conditional expressions, and lambda expressions (no assignment or def statements).

Note: You are not allowed to use make_anonymous_factorial in your return expression.

The sub and mul functions from the operator module are the only built-in functions required to solve this problem.

Hint: In order to recursively compute the factorial, you need a name to use in the recursive calls. What other ways have we learned to give something a name, besides assignment statements and def statements?

from operator import sub, mul

def make_anonymous_factorial():
    """Return the value of an expression that computes factorial.

    >>> make_anonymous_factorial()(5)
    120
    >>> from construct_check import check
    >>> # ban any assignments or recursion
    >>> check(SOURCE_FILE, 'make_anonymous_factorial',
    ...     ['Assign', 'AnnAssign', 'AugAssign', 'NamedExpr', 'FunctionDef', 'Recursion'])
    True
    """
    return (lambda f: lambda k: f(f, k))(lambda f, k: k if k == 1 else mul(k, f(f, sub(k, 1))))
    # Alternate solution:
    return (lambda f: f(f))(lambda f: lambda x: 1 if x == 0 else x * f(f)(x - 1))

Use Ok to test your code:

python3 ok -q make_anonymous_factorial

Check Your Score Locally

You can locally check your score on each question of this assignment by running

python3 ok --score

This does NOT submit the assignment! When you are satisfied with your score, submit the assignment to Gradescope to receive credit for it.

Submit Assignment

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