Homework 7

Due by 11:59pm on Sunday, 7/24

Instructions

Download hw07.zip. Inside the archive, you will find a file called hw07.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:

Linked Lists

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: Cycles

The Link class can represent lists with cycles. That is, a list may contain itself as a sublist.

>>> s = Link(1, Link(2, Link(3)))
>>> s.rest.rest.rest = s
>>> s.rest.rest.rest.rest.rest.first
3

Implement has_cycle,that returns whether its argument, a Link instance, contains a cycle.

Hint: Iterate through the linked list and try keeping track of which Link objects you've already seen.

def has_cycle(link):
    """Return whether link contains a cycle.

    >>> s = Link(1, Link(2, Link(3)))
    >>> s.rest.rest.rest = s
    >>> has_cycle(s)
    True
    >>> t = Link(1, Link(2, Link(3)))
    >>> has_cycle(t)
    False
    >>> u = Link(2, Link(2, Link(2)))
    >>> has_cycle(u)
    False
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q has_cycle

Extra question: This question is not worth extra credit and is entirely optional. It is designed to challenge you to think creatively! The questions after this are still required.

Implement has_cycle_constant with only constant space. (If you followed the hint above, you will use linear space.) The solution is short (less than 20 lines of code), but requires a clever idea. Try to discover the solution yourself before asking around:

def has_cycle_constant(link):
    """Return whether link contains a cycle.

    >>> s = Link(1, Link(2, Link(3)))
    >>> s.rest.rest.rest = s
    >>> has_cycle_constant(s)
    True
    >>> t = Link(1, Link(2, Link(3)))
    >>> has_cycle_constant(t)
    False
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q has_cycle_constant

Trees

Question 3: Cumulative Sum

Write a function cumulative_sum that mutates the Tree t, where each node's entry becomes the sum of all entries in the subtree rooted at the node.

def cumulative_sum(t):
    """Mutates t where each node's entry becomes the sum of all entries in the
    corresponding subtree rooted at t.

    >>> t = Tree(1, [Tree(3, [Tree(5)]), Tree(7)])
    >>> cumulative_sum(t)
    >>> t
    Tree(16, [Tree(8, [Tree(5)]), Tree(7)])
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q cumulative_sum

Question 4: Is BST

Write a function is_bst, which takes a Tree t and returns True if, and only if, t is a valid binary search tree, which means that:

  • Each node has at most two children (a leaf is automatically a valid binary search tree)
  • The children are valid binary search trees
  • For every node, the entries in that node's left child are less than or equal to the entry of the node
  • For every node, the entries in that node's right child are greater than the entry of the node

Note that, if a node has only one child, that child could be considered either the left or right child. You should take this into consideration. Do not use the BST constructor or anything from the BST class.

Hint: It may be helpful to write helper functions bst_min and bst_max that return the minimum and maximum, respectively, of a Tree if it is a valid binary search tree.

def is_bst(t):
    """Returns True if the Tree t has the structure of a valid BST.

    >>> t1 = Tree(6, [Tree(2, [Tree(1), Tree(4)]), Tree(7, [Tree(7), Tree(8)])])
    >>> is_bst(t1)
    True
    >>> t2 = Tree(8, [Tree(2, [Tree(9), Tree(1)]), Tree(3, [Tree(6)]), Tree(5)])
    >>> is_bst(t2)
    False
    >>> t3 = Tree(6, [Tree(2, [Tree(4), Tree(1)]), Tree(7, [Tree(7), Tree(8)])])
    >>> is_bst(t3)
    False
    >>> t4 = Tree(1, [Tree(2, [Tree(3, [Tree(4)])])])
    >>> is_bst(t4)
    True
    >>> t5 = Tree(1, [Tree(0, [Tree(-1, [Tree(-2)])])])
    >>> is_bst(t5)
    True
    >>> t6 = Tree(1, [Tree(4, [Tree(2, [Tree(3)])])])
    >>> is_bst(t6)
    True
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q is_bst

Python Sets

A set is an unordered collection of distinct objects that supports membership testing, union, intersection, and adjunction. The main differences between sets and lists are that sets are unordered and contain no duplicates. Other than that, almost everything is the same.

>>> a = [1, 1, 2, 2, 3, 3]
>>> a = set(a)
>>> a  # No duplicates
{1, 2, 3}
>>> a = {3, 1, 2}
>>> a  # Not necessarily in same order
{1, 2, 3}

The Python documentation on sets has more details. The main things you will use with sets include: in, union (|), intersection (&), and difference (-).

One really convenient thing about Python sets is that many operations on sets (adding elements, removing elements, checking membership) run in θ(1) (constant) time (usually).

Some of the problems use a utility method called timeit, which takes a parameterless function as argument, executes it, and returns the time required to do so. It's a variation on the function timeit.timeit function in the Python3 library.

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

Question 5: Add Up

Write the following function so it (usually) runs in θ(m) time, where m is the length of lst.

def add_up(n, lst):
    """Returns True if any two non identical elements in lst add up to n.

    >>> add_up(100, [1, 2, 3, 4, 5])
    False
    >>> add_up(7, [1, 2, 3, 4, 2])
    True
    >>> add_up(10, [5, 5])
    False
    >>> add_up(151, range(0, 200000, 2))
    False
    >>> timeit(lambda: add_up(151, range(0, 200000, 2))) < 1.0
    True
    >>> add_up(50002, range(0, 200000, 2))
    True
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q add_up

Now you have the chance to improve your coding style for this question. Once you have passed all the tests, invoke AutoStyle by running the command python3 ok --style -q add_up. AutoStyle will provide to you hints that improve the style of your code.

Question 6: Missing Value

Write the following function so it (usually) runs in θ(n) time, where n is the length of lst0.

def missing_val(lst0, lst1):
    """Assuming that lst0 contains all the values in lst1, but lst1 is missing
    one value in lst0, return the missing value.  The values need not be
    numbers.

    >>> from random import shuffle
    >>> missing_val(range(10), [1, 0, 4, 5, 7, 9, 2, 6, 3])
    8
    >>> big0 = [str(k) for k in range(15000)]
    >>> big1 = [str(k) for k in range(15000) if k != 293 ]
    >>> shuffle(big0)
    >>> shuffle(big1)
    >>> missing_val(big0, big1)
    '293'
    >>> timeit(lambda: missing_val(big0, big1)) < 1.0
    True
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q missing_val