Homework 3

Due by 11:59pm on Monday, 7/4

Instructions

Download hw03.zip. Inside the archive, you will find a file called hw03.py, along with a copy of the OK autograder.

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. See Lab 0 for instructions on submitting assignments.

Using OK: If you have any questions about using OK, please refer to this guide.

Readings: You might find the following references useful:

Required questions

Question 1: G function

A mathematical function G on positive integers is defined by two cases:

G(n) = n,                                       if n <= 3
G(n) = G(n - 1) + 2 * G(n - 2) + 3 * G(n - 3),  if n > 3

Write a recursive function g that computes G(n). Then, write an iterative function g_iter that also computes G(n):

def g(n):
    """Return the value of G(n), computed recursively.

    >>> g(1)
    1
    >>> g(2)
    2
    >>> g(3)
    3
    >>> g(4)
    10
    >>> g(5)
    22
    >>> from construct_check import check
    >>> check(HW_SOURCE_FILE, 'g', ['While', 'For'])
    True
    """
    "*** YOUR CODE HERE ***"

def g_iter(n):
    """Return the value of G(n), computed iteratively.

    >>> g_iter(1)
    1
    >>> g_iter(2)
    2
    >>> g_iter(3)
    3
    >>> g_iter(4)
    10
    >>> g_iter(5)
    22
    >>> from construct_check import check
    >>> check(HW_SOURCE_FILE, 'g_iter', ['Recursion'])
    True
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q g
python3 ok -q g_iter

Question 2: Ping pong

The ping-pong sequence counts up starting from 1 and is always either counting up or counting down. At element k, the direction switches if k is a multiple of 7 or contains the digit 7. The first 30 elements of the ping-pong sequence are listed below, with direction swaps marked using brackets at the 7th, 14th, 17th, 21st, 27th, and 28th elements:

1 2 3 4 5 6 [7] 6 5 4 3 2 1 [0] 1 2 [3] 2 1 0 [-1] 0 1 2 3 4 [5] [4] 5 6

Implement a function pingpong that returns the nth element of the ping-pong sequence. Do not use any assignment statements; however, you may use def statements.

Hint: If you're stuck, try implementing pingpong first using assignment and a while statement. Any name that changes value will become an argument to a function in the recursive definition.

def pingpong(n):
    """Return the nth element of the ping-pong sequence.

    >>> pingpong(7)
    7
    >>> pingpong(8)
    6
    >>> pingpong(15)
    1
    >>> pingpong(21)
    -1
    >>> pingpong(22)
    0
    >>> pingpong(30)
    6
    >>> pingpong(68)
    2
    >>> pingpong(69)
    1
    >>> pingpong(70)
    0
    >>> pingpong(71)
    1
    >>> pingpong(72)
    0
    >>> pingpong(100)
    2
    >>> from construct_check import check
    >>> check(HW_SOURCE_FILE, 'pingpong', ['Assign', 'AugAssign'])
    True
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q pingpong

You may use the function has_seven, which returns True if a number k contains the digit 7 at least once.

def has_seven(k):
    """Returns True if at least one of the digits of k is a 7, False otherwise.

    >>> has_seven(3)
    False
    >>> has_seven(7)
    True
    >>> has_seven(2734)
    True
    >>> has_seven(2634)
    False
    >>> has_seven(734)
    True
    >>> has_seven(7777)
    True
    """
    if k % 10 == 7:
        return True
    elif k < 10:
        return False
    else:
        return has_seven(k // 10)

Question 3: Count change

Once the machines take over, the denomination of every coin will be a power of two: 1-cent, 2-cent, 4-cent, 8-cent, 16-cent, etc. There will be no limit to how much a coin can be worth.

A set of coins makes change for n if the sum of the values of the coins is n. For example, the following sets make change for 7:

  • 7 1-cent coins
  • 5 1-cent, 1 2-cent coins
  • 3 1-cent, 2 2-cent coins
  • 3 1-cent, 1 4-cent coins
  • 1 1-cent, 3 2-cent coins
  • 1 1-cent, 1 2-cent, 1 4-cent coins

Thus, there are 6 ways to make change for 7. Write a function count_change that takes a positive integer n and returns the number of ways to make change for n using these coins of the future:

def count_change(amount):
    """Return the number of ways to make change for amount.

    >>> count_change(7)
    6
    >>> count_change(10)
    14
    >>> count_change(20)
    60
    >>> count_change(100)
    9828
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q count_change

Question 4: 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.

Towers of Hanoi

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.

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"
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q move_stack

Question 5: Flatten

This problem will also be used as part of the AutoStyle study, which you can find more information about on Piazza. Participation in the study will be rewarded with extra credit.

Write a function flatten that takes a (possibly deep) list and "flattens" it. For example:

>>> lst = [1, [[2], 3], 4, [5, 6]]
>>> flatten(lst)
[1, 2, 3, 4, 5, 6]

Hint: you can check if something is a list by using the built-in type function. For example,

>>> type(3) == list
False
>>> type([1, 2, 3]) == list
True
def flatten(lst):
    """Returns a flattened version of lst.

    >>> flatten([1, 2, 3])     # normal list
    [1, 2, 3]
    >>> x = [1, [2, 3], 4]      # deep list
    >>> flatten(x)
    [1, 2, 3, 4]
    >>> x = [[1, [1, 1]], 1, [1, 1]] # deep list
    >>> flatten(x)
    [1, 1, 1, 1, 1, 1]
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q flatten

Question 6: Merge

Write a function merge that takes 2 sorted lists lst1 and lst2, and returns a new list that contains all the elements in the two lists in sorted order.

def merge(lst1, lst2):
    """Merges two sorted lists.

    >>> merge([1, 3, 5], [2, 4, 6])
    [1, 2, 3, 4, 5, 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]
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q merge

Question 7: Mergesort

Mergesort is a type of sorting algorithm. It follows a naturally recursive procedure:

  • Break the input list into equally-sized halves
  • Recursively sort both halves
  • Merge the sorted halves.

Using your merge function from the previous question, implement mergesort.

Challenge: Implement mergesort itself iteratively, without using recursion.

def mergesort(seq):
    """Mergesort algorithm.

    >>> mergesort([4, 2, 5, 2, 1])
    [1, 2, 2, 4, 5]
    >>> mergesort([])     # sorting an empty list
    []
    >>> mergesort([1])   # sorting a one-element list
    [1]
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q mergesort

Extra questions

Extra questions are not worth extra credit and are entirely optional. They are designed to challenge you to think creatively!

Question 8: Y combinator

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 giving it a name!

There's actually a general way to do this that uses a function Y defined

def Y(f):
    return f(lambda: Y(f))

Using this function, you can define fact with an assignment statement like this:

 fact = Y(?)

where ? is an expression containing only lambda expressions, conditional expressions, function calls, and the functions mul and sub. That is, ? contains no statements (no assignments or def statements in particular), and no mention of fact or any other identifier defined outside ? other than mul or sub from the operator module. You are also allowed to use the equality (==) operator. Find such an expression to use in place of ?.

from operator import sub, mul

def Y(f):
    """The Y ("paradoxical") combinator."""
    return f(lambda: Y(f))

def Y_tester():
    """
    >>> tmp = Y_tester()
    >>> tmp(1)
    1
    >>> tmp(5)
    120
    >>> tmp(2)
    2
    """
    "*** YOUR CODE HERE ***"
    return Y(________)  # Replace

Use OK to test your code:

python3 ok -q Y_tester