Homework 8
Due at 11:59:59 pm on Thursday, 4/13/2023.

hw08.zip

Instructions

Download hw08.zip. Inside the archive, you will find starter files for the questions in this homework, 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. Check that you have successfully submitted your code on okpy.org.

Readings: This homework relies on following references:

Recall that the order of growth of a function expresses how long it takes for the function to run, and is defined in terms of the function's input sizes.

For example, let's say that we have the function get_x which is defined as follows:

def get_x(x):
    return x

get_x has one expression in it. That one expression takes the same amount of time to run, no matter what x is, or more importantly, how large x gets. This is called constant time, or O(1).

The main two ways that a function in your program will get a running time different than just constant time is through either iteration or recursion. Let's start with some iteration examples!

The (simple) way you figure out the running time of a particular while loop is to simply count the cost of each operation in the body of the while loop, and then multiply that cost by the number of times that the loop runs. For example, look at the following method with a loop in it:

def foo(n):
    i = 1
    sum = 0
    while i <= n:
        sum = sum + i
        i = i + 1
    return sum

This loop has two statements in it sum = sum + i and i = i + 1. Each statement is a constant time operation, since the amount of time each statement takes does not depend on the input to the function n. In C88C, we are not concerned with how long primitive functions such as addition, multiplication, and variable assignment take to run. Rather we are concerned with how many more times a loop is executed or how many more recursive calls occur as the input increases. In this example, we execute the loop n times, and for each iteration, we only execute constant time operations, so we get an order of growth of O(n).

Here are a couple of basic functions, along with their running times. Try to understand why they have the given running time.

O(n)

def bar(n):
    i = 1
    a = 1
    b = 0
    while i <= n:
        temp = a
        a = a + b
        b = temp
        i = i + 1
    return a

O(n^2)

def bar(n):
    sum = 0
    a, b = 0, 0
    while a < n:
        while b < n:
            sum += (a*b)
            b += 1
        b = 0
        a += 1
    return sum

Efficiency

There is nothing to submit for this part. But doing these problems will be good practice. The solutions are given right below the question and to see them you must highlight them.

For each question find the asymptotic runtime in big theta notation.

Question 1

What is the asymptotic run time of the baz function.

def baz(n):
    i = 1
    sum = 0
    while i <= n:
        sum += bam(i)
        i += 1
    return sum

def bam(n):
    i = 1
    sum = 0
    while i <= n:
        sum += i
        i += 1
    return sum
Highlight the text on the line below this line to see the solution:
Answer: O(n2)).

Question 2

def bonk(n):
    sum = 0
    while n >= 2:
        sum += n
        n = n / 2
    return sum
Highlight the text on the line below this line to see the solution:
Answer: O(log(n)).

Inheritance

Question 3: Errors

It is often said that nothing in life is certain but death and taxes. For a programmer or data scientist, however, nothing is certain but encountering errors.

In Python, there are two primary types of errors, both of which you are likely familiar with: syntax errors and exceptions. Syntax errors occur when the proper structure of the language is not followed, while exceptions are errors that occur during the execution of a program. These include errors such as ZeroDivisionError, TypeError, NameError, and many more!

Under the hood, these errors are based in the concepts of object orientation, and all exceptions are class objects. If you're interested in more detailed explanations of the structure of exceptions as well as how to create your own, check out this article from the Python documentation! In the meantime, we'll implement our own version of an Error class

Complete the Error, SyntaxError, and ZeroDivisionError classes such that they create the correct messages when called.

  • The SyntaxError and ZeroDivisionError classes inherit from the Error class and add functionality that is unique to those particular errors. Their code is partially implemented for you.
  • The add_code method adds a new helpful message to your error, while the write method should print the output that you see when an error is raised. Do not worry if that code already is already defined; for this problem, it is safe to overwrite it.
  • You can access the parent class methods using the super() function
class Error:
    """
    >>> err1 = Error(12, "error.py")
    >>> err1.write()
    error.py:12
    """
    def __init__(self, line, file):
        "*** YOUR CODE HERE ***"

    def format(self):
        return self.file + ':' + str(self.line)

    def write(self):
        print(self.format())

class SyntaxError(Error):
    """
    >>> err1 = SyntaxError(17, "lab10.py")
    >>> err1.write()
    lab10.py:17 SyntaxError : Invalid syntax
    >>> err1.add_code(4, "EOL while scanning string literal")
    >>> err2 = SyntaxError(18, "lab10.py", 4)
    >>> err2.write()
    lab10.py:18 SyntaxError : EOL while scanning string literal
    """
    type = 'SyntaxError'
    msgs = {0 : "Invalid syntax", 1: "Unmatched parentheses", 2: "Incorrect indentation", 3: "missing colon"}

    def __init__(self, line, file, code=0):
        "*** YOUR CODE HERE ***"

    def format(self):
        "*** YOUR CODE HERE ***"

    def add_code(self, code, msg):
        "*** YOUR CODE HERE ***"

class ZeroDivisionError(Error):
    """
    >>> err1 = ZeroDivisionError(273, "lab10.py")
    >>> err1.write()
    lab10.py:273 ZeroDivisionError : division by zero
    """
    type = 'ZeroDivisionError'

    def __init__(self, line, file, message='division by zero'):
        "*** YOUR CODE HERE ***"

    def format(self):
        end = self.type + ' : ' + self.message
        "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q Error

Use OK to test your code:

python3 ok -q SyntaxError

Use OK to test your code:

python3 ok -q ZeroDivisionError

Linked Lists

A linked list is either an empty linked list (Link.empty) or a first value and the rest of the linked list.

class Link:
    """
    >>> s = Link(1, Link(2, Link(3)))
    >>> s
    Link(1, Link(2, Link(3)))
    """
    empty = ()

    def __init__(self, first, rest=empty):
        assert rest is Link.empty or isinstance(rest, Link)
        self.first = first
        self.rest = rest

    def __repr__(self):
        if self.rest is not Link.empty:
            rest_str = ', ' + repr(self.rest)
        else:
            rest_str = ''
        return 'Link({0}{1})'.format(repr(self.first), rest_str)

To check if a Link is empty, compare it against the class attribute Link.empty. For example, the below function prints out whether or not the link it is handed is empty:

def test_empty(link):
    if link is Link.empty:
        print('This linked list is empty!')
    else:
        print('This linked list is not empty!')

Note: Linked lists are recursive data structures! A linked list contains the first element of the list (first) and a reference to another linked list (rest) which contains the rest of the values in the list.

Question 4: Link to List

Write a function link_to_list that converts a given Link to a Python list.

def link_to_list(link):
    """Takes a Link and returns a Python list with the same elements.

    >>> link = Link(1, Link(2, Link(3, Link(4))))
    >>> link_to_list(link)
    [1, 2, 3, 4]
    >>> link_to_list(Link(88))
    [88]
    >>> link_to_list(Link.empty)
    []
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q link_to_list

Question 5: Every Other

Implement every_other, which takes a linked list s. It mutates s such that all of the odd-indexed elements (using 0-based indexing) are removed from the list. For example:

>>> s = Link('a', Link('b', Link('c', Link('d'))))
>>> every_other(s)
>>> s
Link('a', Link('c'))
>>> s.first
'a'
>>> s.rest.first
'c'
>>> s.rest.rest is Link.empty
True

If s contains fewer than two elements, s remains unchanged.

Do not return anything! every_other should mutate the original list.

def every_other(s):
    """Mutates a linked list so that all the odd-indiced elements are removed
    (using 0-based indexing).

    >>> s = Link(1, Link(2, Link(3, Link(4))))
    >>> every_other(s) # removes 2, 4
    >>> s
    Link(1, Link(3))
    >>> odd_length = Link(5, Link(3, Link(1)))
    >>> every_other(odd_length) # removes 3
    >>> odd_length
    Link(5, Link(1))
    >>> two_items = Link(6, Link(7))
    >>> every_other(two_items) # removes 7
    >>> two_items
    Link(6)
    >>> singleton = Link(4)
    >>> every_other(singleton) # doesn't remove anything
    >>> singleton
    Link(4)
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q every_other

Question 6: Deep Map

Implement deep_map, which takes a function f and a link. It returns a new linked list with the same structure as link, but with f applied to any element within link or any Link instance contained in link.

The deep_map function should recursively apply fn to each of that Link's elements rather than to that Link itself.

Hint: You may find the built-in isinstance function useful.

def deep_map(f, link):
    """Return a Link with the same structure as link but with fn mapped over
    its elements. If an element is an instance of a linked list, recursively
    apply f inside that linked list as well.

    >>> s = Link(1, Link(Link(2, Link(3)), Link(4)))
    >>> print_link(s)
    <1 <2 3> 4>
    >>> print_link(deep_map(lambda x: x * x, s))
    <1 <4 9> 16>
    >>> print_link(s) # unchanged
    <1 <2 3> 4>
    >>> t = Link(s, Link(Link(Link(5))))
    >>> print_link(t)
    <<1 <2 3> 4> <<5>>>
    >>> print_link(deep_map(lambda x: 2 * x, t))
    <<2 <4 6> 8> <<10>>>
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q deep_map

Submit

Make sure to submit this assignment by running:

python3 ok --submit