Homework 7

Due by 11:59pm on Saturday, 7/25

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 1 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

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 2

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(s):
    """Return whether Link s 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
    """
    "*** YOUR CODE HERE ***"

Use OK to test your code:

python3 ok -q has_cycle

Question 3

The ordered_sequence_to_tree function takes a sorted linked list and converts it into a balanced BinaryTree. A BinaryTree is balanced if

  • the number of entries in its left branch differs from the number of entries in its right branch by at most 1, and
  • its non-empty branches are also balanced trees.

In order to write ordered_sequence_to_tree, implement partial_tree(s, n), which converts the first n elements of the sorted linked list s into a balanced BinaryTree. The return value is a two-element tuple: the resulting balanced tree; and the rest of the linked list.

Hint: This function requires two recursive calls. The first call builds a left branch out of the first left_size elements of s; Then, the next element is used as the entry of the returned tree. Finally, the second recursive call builds the right branch out of the next right_size elements. In total, (left_size + 1 + right_size) = n, where 1 is for the entry:

def partial_tree(s, n):
    """Return a balanced tree of the first n elements of Link s, along with
    the rest of s. A tree is balanced if

      (a) the number of entries in its left branch differs from the number
          of entries in its right branch by at most 1, and

      (b) its non-empty branches are also balanced trees.

    Examples of balanced trees:

    BinaryTree(1)                                # branch difference 0 - 0 = 0
    BinaryTree(1, BinaryTree(2), None)           # branch difference 1 - 0 = 1
    BinaryTree(1, None, BinaryTree(2))           # branch difference 0 - 1 = -1
    BinaryTree(1, BinaryTree(2), BinaryTree(3))  # branch difference 1 - 1 = 0

    Examples of unbalanced trees:

    # branch difference 2 - 0 = 2
    BinaryTree(1, BinaryTree(2, BinaryTree(3)), None)
    # Unbalanced right branch
    BinaryTree(1, BinaryTree(2, BinaryTree(3), None),
            BinaryTree(4, BinaryTree(5, BinaryTree(6), None), None))

    >>> s = Link(1, Link(2, Link(3, Link(4, Link(5)))))
    >>> partial_tree(s, 3)
    (BinaryTree(2, BinaryTree(1), BinaryTree(3)), Link(4, Link(5)))
    >>> t = Link(-2, Link(-1, Link(0, s)))
    >>> partial_tree(t, 7)[0]
    BinaryTree(1, BinaryTree(-1, BinaryTree(-2), BinaryTree(0)), BinaryTree(3, BinaryTree(2), BinaryTree(4)))
    >>> partial_tree(t, 7)[1]
    Link(5)
    """
    if n == 0:
        return BinaryTree.empty, s
    left_size = (n-1)//2
    right_size = n - left_size - 1
    "*** YOUR CODE HERE ***"

def ordered_sequence_to_tree(s):
    """Return a balanced tree containing the elements of ordered Link s.

    Note: this implementation is complete, but the definition of partial_tree
    above is not complete.

    >>> ordered_sequence_to_tree(Link(1, Link(2, Link(3))))
    BinaryTree(2, BinaryTree(1), BinaryTree(3))
    >>> elements = Link(1, Link(2, Link(3, Link(4, Link(5, Link(6, Link(7)))))))
    >>> ordered_sequence_to_tree(elements)
    BinaryTree(4, BinaryTree(2, BinaryTree(1), BinaryTree(3)), BinaryTree(6, BinaryTree(5), BinaryTree(7)))
    """
    return partial_tree(s, len(s))[0]

Use OK to test your code:

python3 ok -q partial_tree
python3 ok -q ordered_sequence_to_tree

Extra question

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

Question 4

Implement has_cycle_constant, which has the same behavior as has_cycle but requires only constant 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(s):
    """Return whether Link s 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