Homework 10
Due at 11:59:59 pm on Sunday, 11/22/2020.

hw10.zip

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 1: 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.empty)
    []
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q link_to_list

Question 2: 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.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)
    >>> s
    Link(1, Link(3))
    >>> odd_length = Link(5, Link(3, Link(1)))
    >>> every_other(odd_length)
    >>> odd_length
    Link(5, Link(1))
    >>> singleton = Link(4)
    >>> every_other(singleton)
    >>> singleton
    Link(4)
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q every_other

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

Use OK to test your code:

python3 ok -q deep_map

Question 4: Mutable Mapping

Implement deep_map_mut(fn, link), which applies a function fn onto all elements in the given linked list link. If an element is itself a linked list, apply fn to each of its elements, and so on.

Your implementation should mutate the original linked list. Do not create any new linked lists.

Hint: The built-in isinstance function may be useful.

>>> s = Link(1, Link(2, Link(3, Link(4))))
>>> isinstance(s, Link)
True
>>> isinstance(s, int)
False
def deep_map_mut(fn, link):
    """Mutates a deep link by replacing each item found with the
    result of calling fn on the item.  Does NOT create new Links (so
    no use of Link's constructor)

    Does not return the modified Link object.

    >>> link1 = Link(3, Link(Link(4), Link(5, Link(6))))
    >>> deep_map_mut(lambda x: x * x, link1)
    >>> print_link(link1)
    <9 <16> 25 36>
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q deep_map_mut

Question 5: Slice

Implement a function slice_link that slices a given link. slice_link should slice the link starting at start and ending one element before end, as with a normal Python list.

def slice_link(link, start, end):
    """Slices a Link from start to end (as with a normal Python list).

    >>> link = Link(3, Link(1, Link(4, Link(1, Link(5, Link(9))))))
    >>> new = slice_link(link, 1, 4)
    >>> print_link(new)
    <1 4 1>
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q slice_link

Submit

Make sure to submit this assignment by running:

python3 ok --submit