Lab 8: Linked Lists, Mutable Trees

Due by 11:59pm on Wednesday, October 19.

Starter Files

Download lab08.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the Ok autograder.

Topics

Consult this section if you need a refresher on the material for this lab. It's okay to skip directly to the questions and refer back here should you get stuck.


Linked Lists

We've learned that a Python list is one way to store sequential values. Another type of list is a linked list. A Python list stores all of its elements in a single object, and each element can be accessed by using its index. A linked list, on the other hand, is a recursive object that only stores two things: its first value and a reference to the rest of the list, which is another linked list.

We can implement a class, Link, that represents a linked list object. Each instance of Link has two instance attributes, first and rest.

class Link:
    """A linked list.

    >>> s = Link(1)
    >>> s.first
    1
    >>> s.rest is Link.empty
    True
    >>> s = Link(2, Link(3, Link(4)))
    >>> s.first = 5
    >>> s.rest.first = 6
    >>> s.rest.rest = Link.empty
    >>> s                                    # Displays the contents of repr(s)
    Link(5, Link(6))
    >>> s.rest = Link(7, Link(Link(8, Link(9))))
    >>> s
    Link(5, Link(7, Link(Link(8, Link(9)))))
    >>> print(s)                             # Prints str(s)
    <5 7 <8 9>>
    """
    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_repr = ', ' + repr(self.rest)
        else:
            rest_repr = ''
        return 'Link(' + repr(self.first) + rest_repr + ')'

    def __str__(self):
        string = '<'
        while self.rest is not Link.empty:
            string += str(self.first) + ' '
            self = self.rest
        return string + str(self.first) + '>'

A valid linked list can be one of the following:

  1. An empty linked list (Link.empty)
  2. A Link object containing the first value of the linked list and a reference to the rest of the linked list

What makes a linked list recursive is that the rest attribute of a single Link instance is another linked list! In the big picture, each Link instance stores a single value of the list. When multiple Links are linked together through each instance's rest attribute, an entire sequence is formed.

Note: This definition means that the rest attribute of any Link instance must be either Link.empty or another Link instance! This is enforced in Link.__init__, which raises an AssertionError if the value passed in for rest is neither of these things.

To check if a linked list is empty, compare it against the class attribute Link.empty. For example, the function below 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!')


Mutable Trees

We define a tree to be a recursive data abstraction that has a label (the value stored in the root of the tree) and branches (a list of trees directly underneath the root).

Previously we implemented trees by using a functional data abstraction, with the tree constructor function and the label and branches selector functions. Now we implement trees by creating the Tree class. Here is part of the class included in the lab.

class Tree:
    """
    >>> t = Tree(3, [Tree(2, [Tree(5)]), Tree(4)])
    >>> t.label
    3
    >>> t.branches[0].label
    2
    >>> t.branches[1].is_leaf()
    True
    """
    def __init__(self, label, branches=[]):
        for b in branches:
            assert isinstance(b, Tree)
        self.label = label
        self.branches = list(branches)

    def is_leaf(self):
        return not self.branches

Even though this is a new implementation, everything we know about the functional tree data abstraction remains true. That means that solving problems involving trees as objects uses the same techniques that we developed when first studying the functional tree data abstraction (e.g. we can still use recursion on the branches!). The main difference, aside from syntax, is that tree objects are mutable.

Here is a summary of the differences between the tree data abstraction implemented as a functional abstraction vs. implemented as class:

- Tree constructor and selector functions Tree class
Constructing a tree To construct a tree given a label and a list of branches, we call tree(label, branches) To construct a tree object given a label and a list of branches, we call Tree(label, branches) (which calls the Tree.__init__ method).
Label and branches To get the label or branches of a tree t, we call label(t) or branches(t) respectively To get the label or branches of a tree t, we access the instance attributes t.label or t.branches respectively.
Mutability The functional tree data abstraction is immutable because we cannot assign values to call expressions The label and branches attributes of a Tree instance can be reassigned, mutating the tree.
Checking if a tree is a leaf To check whether a tree t is a leaf, we call the convenience function is_leaf(t) To check whether a tree t is a leaf, we call the bound method t.is_leaf(). This method can only be called on Tree objects.

Implementing trees as a class gives us another advantage: we can specify how we want them to be output by the interpreter by implementing the __repr__ and __str__ methods.

Here is the __repr__ method:

def __repr__(self):
    if self.branches:
        branch_str = ', ' + repr(self.branches)
    else:
        branch_str = ''
    return 'Tree({0}{1})'.format(self.label, branch_str)

With this implementation of __repr__, a Tree instance is displayed as the exact constructor call that created it:

>>> t = Tree(4, [Tree(3), Tree(5, [Tree(6)]), Tree(7)])
>>> t
Tree(4, [Tree(3), Tree(5, [Tree(6)]), Tree(7)])
>>> t.branches
[Tree(3), Tree(5, [Tree(6)]), Tree(7)]
>>> t.branches[0]
Tree(3)
>>> t.branches[1]
Tree(5, [Tree(6)])

Here is the __str__ method. You do not need to understand how this function is implemented.

def __str__(self):
    def print_tree(t, indent=0):
        tree_str = '  ' * indent + str(t.label) + "\n"
        for b in t.branches:
            tree_str += print_tree(b, indent + 1)
        return tree_str
    return print_tree(self).rstrip()

With this implementation of __str__, we can pretty-print a Tree to see both its contents and structure:

>>> t = Tree(4, [Tree(3), Tree(5, [Tree(6)]), Tree(7)])
>>> print(t)
4
  3
  5
    6
  7
>>> print(t.branches[0])
3
>>> print(t.branches[1])
5
  6

Required Questions


Getting Started Videos

These videos may provide some helpful direction for tackling the coding problems on this assignment.

To see these videos, you should be logged into your berkeley.edu email.

YouTube link

Linked Lists

Q1: WWPD: Linked Lists

Read over the Link class in lab08.py. Make sure you understand the doctests.

Use Ok to test your knowledge with the following "What Would Python Display?" questions:

python3 ok -q link -u

Enter Function if you believe the answer is <function ...>, Error if it errors, and Nothing if nothing is displayed.

If you get stuck, try drawing out the box-and-pointer diagram for the linked list on a piece of paper or loading the Link class into the interpreter with python3 -i lab08.py.

>>> from lab08 import *
>>> link = Link(1000)
>>> link.first
______
1000
>>> link.rest is Link.empty
______
True
>>> link = Link(1000, 2000)
______
AssertionError
>>> link = Link(1000, Link())
______
TypeError
>>> from lab08 import *
>>> link = Link(1, Link(2, Link(3)))
>>> link.first
______
1
>>> link.rest.first
______
2
>>> link.rest.rest.rest is Link.empty
______
True
>>> link.first = 9001 >>> link.first
______
9001
>>> link.rest = link.rest.rest >>> link.rest.first
______
3
>>> link = Link(1) >>> link.rest = link >>> link.rest.rest is Link.empty
______
False
>>> link.rest.rest.rest.rest.first
______
1
>>> link = Link(2, Link(3, Link(4))) >>> link2 = Link(1, link) >>> link2.first
______
1
>>> link2.rest.first
______
2
>>> from lab08 import *
>>> link = Link(5, Link(6, Link(7)))
>>> link                  # Look at the __repr__ method of Link
______
Link(5, Link(6, Link(7)))
>>> print(link) # Look at the __str__ method of Link
______
<5 6 7>

Write a function convert_link that takes in a linked list and returns the sequence as a Python list. You may assume that the input list is shallow; that is none of the elements is another linked list.

Try to find both an iterative and recursive solution for this problem!

Challenge: You may NOT assume that the input list is shallow. Hint: use the type built-in.

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

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

Use Ok to test your code:

python3 ok -q convert_link

Write a function duplicate_link that takes in a linked list link and a value. duplicate_link will mutate link such that if there is a linked list node that has a first equal to value, that node will be duplicated. Note that you should be mutating the original link list link; you will need to create new Links, but you should not be returning a new linked list.

Note: in order to insert a link into a linked list, you need to modify the .rest of certain links. We encourage you to draw out a doctest to visualize!

def duplicate_link(link, val):
    """Mutates `link` such that if there is a linked list
    node that has a first equal to value, that node will
    be duplicated. Note that you should be mutating the
    original link list.

    >>> x = Link(5, Link(4, Link(3)))
    >>> duplicate_link(x, 5)
    >>> x
    Link(5, Link(5, Link(4, Link(3))))
    >>> y = Link(2, Link(4, Link(6, Link(8))))
    >>> duplicate_link(y, 10)
    >>> y
    Link(2, Link(4, Link(6, Link(8))))
    >>> z = Link(1, Link(2, (Link(2, Link(3)))))
    >>> duplicate_link(z, 2) #ensures that back to back links with val are both duplicated
    >>> z
    Link(1, Link(2, Link(2, Link(2, Link(2, Link(3))))))
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q duplicate_link

Trees

Q4: WWPD: Trees

Read over the Tree class in lab08.py. Make sure you understand the doctests.

Use Ok to test your knowledge with the following "What Would Python Display?" questions:

python3 ok -q trees-wwpd -u

Enter Function if you believe the answer is <function ...>, Error if it errors, and Nothing if nothing is displayed. Recall that Tree instances will be displayed the same way they are constructed.

>>> from lab08 import *
>>> t = Tree(1, Tree(2))
______
Error
>>> t = Tree(1, [Tree(2)]) >>> t.label
______
1
>>> t.branches[0]
______
Tree(2)
>>> t.branches[0].label
______
2
>>> t.label = t.branches[0].label >>> t
______
Tree(2, [Tree(2)])
>>> t.branches.append(Tree(4, [Tree(8)])) >>> len(t.branches)
______
2
>>> t.branches[0]
______
Tree(2)
>>> t.branches[1]
______
Tree(4, [Tree(8)])

Q5: Cumulative Mul

Write a function cumulative_mul that mutates the Tree t so that each node's label becomes the product of its label and all labels in the subtrees rooted at the node.

Hint: Consider carefully when to do the mutation of the tree and whether that mutation should happen before or after processing the subtrees.

def cumulative_mul(t):
    """Mutates t so that each node's label becomes the product of all labels in
    the corresponding subtree rooted at t.

    >>> t = Tree(1, [Tree(3, [Tree(5)]), Tree(7)])
    >>> cumulative_mul(t)
    >>> t
    Tree(105, [Tree(15, [Tree(5)]), Tree(7)])
    >>> otherTree = Tree(2, [Tree(1, [Tree(3), Tree(4), Tree(5)]), Tree(6, [Tree(7)])])
    >>> cumulative_mul(otherTree)
    >>> otherTree
    Tree(5040, [Tree(60, [Tree(3), Tree(4), Tree(5)]), Tree(42, [Tree(7)])])
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q cumulative_mul

Submit

Make sure to submit this assignment by running:

python3 ok --submit

Optional Questions

Q6: 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

Q7: Prune Small

Complete the function prune_small that takes in a Tree t and a number n and prunes t mutatively. If t or any of its branches has more than n branches, the n branches with the smallest labels should be kept and any other branches should be pruned, or removed, from the tree.

def prune_small(t, n):
    """Prune the tree mutatively, keeping only the n branches
    of each node with the smallest labels.

    >>> t1 = Tree(6)
    >>> prune_small(t1, 2)
    >>> t1
    Tree(6)
    >>> t2 = Tree(6, [Tree(3), Tree(4)])
    >>> prune_small(t2, 1)
    >>> t2
    Tree(6, [Tree(3)])
    >>> t3 = Tree(6, [Tree(1), Tree(3, [Tree(1), Tree(2), Tree(3)]), Tree(5, [Tree(3), Tree(4)])])
    >>> prune_small(t3, 2)
    >>> t3
    Tree(6, [Tree(1), Tree(3, [Tree(1), Tree(2)])])
    """
    while ___________________________:
        largest = max(_______________, key=____________________)
        _________________________
    for __ in _____________:
        ___________________

Use Ok to test your code:

python3 ok -q prune_small