Lab 5: Data Abstraction, Trees

Due by 11:59pm on Friday, October 4.

Starter Files

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


By the end of this lab, you should have submitted the lab with python3 ok --submit. You may submit more than once before the deadline; only the final submission will be graded. Check that you have successfully submitted your code on


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.

Data Abstraction

Data abstraction is a powerful concept in computer science that allows programmers to treat code as objects -- for example, car objects, chair objects, people objects, etc. That way, programmers don't have to worry about how code is implemented -- they just have to know what it does.

Data abstraction mimics how we think about the world. When you want to drive a car, you don't need to know how the engine was built or what kind of material the tires are made of. You just have to know how to turn the wheel and press the gas pedal.

An abstract data type consists of two types of functions:

  • Constructors: functions that build the abstract data type.
  • Selectors: functions that retrieve information from the data type.

Programmers design ADTs to abstract away how information is stored and calculated such that the end user does not need to know how constructors and selectors are implemented. The nature of abstract data types allows whoever uses them to assume that the functions have been written correctly and work as described.


A tree is a data structure that represents a hierarchy of information. A file system is a good example of a tree structure. For example, within your cs61a folder, you have folders separating your projects, lab assignments, and homework. The next level is folders that separate different assignments, hw01, lab01, hog, etc., and inside those are the files themselves, including the starter files and ok. Below is an incomplete diagram of what your cs61a directory might look like.


As you can see, unlike trees in nature, the tree abstract data type is drawn with the root at the top and the leaves at the bottom.

Some tree terminology:

  • root: the node at the top of the tree
  • label: the value in a node, selected by the label function
  • branches: a list of trees directly under the tree's root, selected by the branches function
  • leaf: a tree with zero branches
  • node: any location within the tree (e.g., root node, leaf nodes, etc.)

Our tree abstract data type consists of a root and a list of its branches. To create a tree and access its root value and branches, use the following constructor and selectors:

  • Constructor

    • tree(label, branches=[]): creates a tree object with the given label value at its root node and list of branches. Notice that the second argument to this constructor, branches, is optional - if you want to make a tree with no branches, leave this argument empty.
  • Selectors

    • label(tree): returns the value in the root node of tree.
    • branches(tree): returns the list of branches of the given tree.
  • Convenience function

    • is_leaf(tree): returns True if tree's list of branches is empty, and False otherwise.

For example, the tree generated by

number_tree = tree(1,

would look like this:

 / | \
2  3  6
  / \  \
 4   5  7

To extract the number 3 from this tree, which is the label of the root of its second branch, we would do this:


The print_tree function prints out a tree in a human-readable form. The exact form follows the pattern illustrated above, where the root is unindented, and each of its branches is indented one level further.

def print_tree(t, indent=0):
    """Print a representation of this tree in which each node is
    indented by two spaces times its depth from the root.

    >>> print_tree(tree(1))
    >>> print_tree(tree(1, [tree(2)]))
    >>> numbers = tree(1, [tree(2), tree(3, [tree(4), tree(5)]), tree(6, [tree(7)])])
    >>> print_tree(numbers)
    print('  ' * indent + str(label(t)))
    for b in branches(t):
        print_tree(b, indent + 1)

Required Questions

City Data Abstraction

Say we have an abstract data type for cities. A city has a name, a latitude coordinate, and a longitude coordinate.

Our ADT has one constructor:

  • make_city(name, lat, lon): Creates a city object with the given name, latitude, and longitude.

We also have the following selectors in order to get the information for each city:

  • get_name(city): Returns the city's name
  • get_lat(city): Returns the city's latitude
  • get_lon(city): Returns the city's longitude

Here is how we would use the constructor and selectors to create cities and extract their information:

>>> berkeley = make_city('Berkeley', 122, 37)
>>> get_name(berkeley)
>>> get_lat(berkeley)
>>> new_york = make_city('New York City', 74, 40)
>>> get_lon(new_york)

All of the selector and constructor functions can be found in the lab file, if you are curious to see how they are implemented. However, the point of data abstraction is that we do not need to know how an abstract data type is implemented, but rather just how we can interact with and use the data type.

Q1: Distance

We will now implement the function distance, which computes the distance between two city objects. Recall that the distance between two coordinate pairs (x1, y1) and (x2, y2) can be found by calculating the sqrt of (x1 - x2)**2 + (y1 - y2)**2. We have already imported sqrt for your convenience. Use the latitude and longitude of a city as its coordinates; you'll need to use the selectors to access this info!

from math import sqrt
def distance(city1, city2):
    >>> city1 = make_city('city1', 0, 1)
    >>> city2 = make_city('city2', 0, 2)
    >>> distance(city1, city2)
    >>> city3 = make_city('city3', 6.5, 12)
    >>> city4 = make_city('city4', 2.5, 15)
    >>> distance(city3, city4)
"*** YOUR CODE HERE ***"
lat_1, lon_1 = get_lat(city1), get_lon(city1) lat_2, lon_2 = get_lat(city2), get_lon(city2) return sqrt((lat_1 - lat_2)**2 + (lon_1 - lon_2)**2) # Video walkthrough:

Use Ok to test your code:

python3 ok -q distance

Q2: Closer city

Next, implement closer_city, a function that takes a latitude, longitude, and two cities, and returns the name of the city that is relatively closer to the provided latitude and longitude.

You may only use the selectors and constructors introduced above and the distance function you just defined for this question.

Hint: How can use your distance function to find the distance between the given location and each of the given cities?

def closer_city(lat, lon, city1, city2):
    Returns the name of either city1 or city2, whichever is closest to
    coordinate (lat, lon).

    >>> berkeley = make_city('Berkeley', 37.87, 112.26)
    >>> stanford = make_city('Stanford', 34.05, 118.25)
    >>> closer_city(38.33, 121.44, berkeley, stanford)
    >>> bucharest = make_city('Bucharest', 44.43, 26.10)
    >>> vienna = make_city('Vienna', 48.20, 16.37)
    >>> closer_city(41.29, 174.78, bucharest, vienna)
"*** YOUR CODE HERE ***"
new_city = make_city('arb', lat, lon) dist1 = distance(city1, new_city) dist2 = distance(city2, new_city) if dist1 < dist2: return get_name(city1) return get_name(city2) # Video walkthrough:

Use Ok to test your code:

python3 ok -q closer_city

Q3: Don't violate the abstraction barrier!

Note: this question has no code-writing component (if you implemented distance and closer_city correctly!)

When writing functions that use an ADT, we should use the constructor(s) and selector(s) whenever possible instead of assuming the ADT's implementation. Relying on a data abstraction's underlying implementation is known as violating the abstraction barrier, and we never want to do this!

It's possible that you passed the doctests for distance and closer_city even if you violated the abstraction barrier. To check whether or not you did so, run the following command:

Use Ok to test your code:

python3 ok -q check_abstraction

The make_check_abstraction function exists only for the doctest, which swaps out the implementations of the city abstraction with something else, runs the tests from the previous two parts, then restores the original abstraction.

The nature of the abstraction barrier guarantees that changing the implementation of an ADT shouldn't affect the functionality of any programs that use that ADT, as long as the constructors and selectors were used properly.

If you passed the Ok tests for the previous questions but not this one, the fix is simple! Just replace any code that violates the abstraction barrier, i.e. creating a city with a new list object or indexing into a city, with the appropriate constructor or selector.

Make sure that your functions pass the tests with both the first and the second implementations of the City ADT and that you understand why they should work for both before moving on.


Q4: Acorn Finder

The squirrels on campus need your help! There are a lot of trees on campus and the squirrels would like to know which ones contain acorns. Define the function acorn_finder, which takes in a tree and returns True if the tree contains a node with the value 'acorn' and False otherwise.

def acorn_finder(t):
    """Returns True if t contains a node with the value 'acorn' and
    False otherwise.

    >>> scrat = tree('acorn')
    >>> acorn_finder(scrat)
    >>> sproul = tree('roots', [tree('branch1', [tree('leaf'), tree('acorn')]), tree('branch2')])
    >>> acorn_finder(sproul)
    >>> numbers = tree(1, [tree(2), tree(3, [tree(4), tree(5)]), tree(6, [tree(7)])])
    >>> acorn_finder(numbers)
    >>> t = tree(1, [tree('acorn',[tree('not acorn')])])
    >>> acorn_finder(t)
"*** YOUR CODE HERE ***"
if label(t) == 'acorn': return True for b in branches(t): if acorn_finder(b): return True return False # Alternative solution def acorn_finder(t): if label(t) == 'acorn': return True return True in [acorn_finder(b) for b in branches(t)]

Use Ok to test your code:

python3 ok -q acorn_finder

Q5: Sprout leaves

Define a function sprout_leaves that takes in a tree, t, and a list of values, vals. It produces a new tree that is identical to t, but where each old leaf node has new branches, one for each value in vals.

For example, say we have the tree t = tree(1, [tree(2), tree(3, [tree(4)])]):

 / \
2   3

If we call sprout_leaves(t, [5, 6]), the result is the following tree:

     /   \
    2     3
   / \    |
  5   6   4
         / \
        5   6
def sprout_leaves(t, vals):
    """Sprout new leaves containing the data in vals at each leaf in
    the original tree t and return the resulting tree.

    >>> t1 = tree(1, [tree(2), tree(3)])
    >>> print_tree(t1)
    >>> new1 = sprout_leaves(t1, [4, 5])
    >>> print_tree(new1)

    >>> t2 = tree(1, [tree(2, [tree(3)])])
    >>> print_tree(t2)
    >>> new2 = sprout_leaves(t2, [6, 1, 2])
    >>> print_tree(new2)
"*** YOUR CODE HERE ***"
if is_leaf(t): return tree(label(t), [tree(v) for v in vals]) return tree(label(t), [sprout_leaves(s, vals) for s in branches(t)])

Use Ok to test your code:

python3 ok -q sprout_leaves

Optional Questions

While "Add Characters" is optional, it is good practice for the Cats project and is thus highly recommended!

Q6: Add Characters

Given two words, w1 and w2, we say w1 is a subsequence of w2 if all the letters in w1 appear in w2 in the same order (but not necessarily all together). That is, you can add letters to any position in w1 to get w2. For example, "sing" is a substring of "absorbing" and "cat" is a substring of "contrast".

Implement add_chars, which takes in w1 and w2, where w1 is a substring of w2. It should return a string containing the characters you need to add to w1 to get w2. Your solution must use recursion.

In the example above, you need to add the characters "aborb" to "sing" to get "absorbing", and you need to add "ontrs" to "cat" to get "contrast".

The letters in the string you return should be in the order you have to add them from left to right. If there are multiple characters in the w2 that could correspond to characters in w1, use the leftmost one. For example, add_words("coy", "cacophony") should return "acphon", not "caphon" because the first "c" in "coy" corresponds to the first "c" in "cacophony".

def add_chars(w1, w2):
    Return a string containing the characters you need to add to w1 to get w2.

    You may assume that w1 is a subsequence of w2.

    >>> add_chars("owl", "howl")
    >>> add_chars("want", "wanton")
    >>> add_chars("rat", "radiate")
    >>> add_chars("a", "prepare")
    >>> add_chars("resin", "recursion")
    >>> add_chars("fin", "effusion")
    >>> add_chars("coy", "cacophony")
    >>> from construct_check import check
    >>> # ban iteration and sets
    >>> check(LAB_SOURCE_FILE, 'add_chars',
    ...       ['For', 'While', 'Set', 'SetComp']) # Must use recursion
"*** YOUR CODE HERE ***"
if not w1: return w2 elif w1[0] == w2[0]: return add_chars(w1[1:], w2[1:]) return w2[0] + add_chars(w1, w2[1:])

Use Ok to test your code:

python3 ok -q add_chars

Q7: Add trees

Define the function add_trees, which takes in two trees and returns a new tree where each corresponding node from the first tree is added with the node from the second tree. If a node at any particular position is present in one tree but not the other, it should be present in the new tree as well.

Hint: You may want to use the built-in zip function to iterate over multiple sequences at once.

Note: If you feel that this one's a lot harder than the previous tree problems, that's totally fine! This is a pretty difficult problem, but you can do it! Talk about it with other students, and come back to it if you need to.

def add_trees(t1, t2):
    >>> numbers = tree(1,
    ...                [tree(2,
    ...                      [tree(3),
    ...                       tree(4)]),
    ...                 tree(5,
    ...                      [tree(6,
    ...                            [tree(7)]),
    ...                       tree(8)])])
    >>> print_tree(add_trees(numbers, numbers))
    >>> print_tree(add_trees(tree(2), tree(3, [tree(4), tree(5)])))
    >>> print_tree(add_trees(tree(2, [tree(3)]), tree(2, [tree(3), tree(4)])))
    >>> print_tree(add_trees(tree(2, [tree(3, [tree(4), tree(5)])]), \
    tree(2, [tree(3, [tree(4)]), tree(5)])))
"*** YOUR CODE HERE ***"
if not t1: return t2 if not t2: return t1 new_label = label(t1) + label(t2) t1_children, t2_children = branches(t1), branches(t2) length_t1, length_t2 = len(t1_children), len(t2_children) if length_t1 < length_t2: t1_children += [None for _ in range(length_t1, length_t2)] elif len(t1_children) > len(t2_children): t2_children += [None for _ in range(length_t2, length_t1)] return tree(new_label, [add_trees(child1, child2) for child1, child2 in zip(t1_children, t2_children)])

Use Ok to test your code:

python3 ok -q add_trees

Fun Question!

Shakespeare and Dictionaries

We will use dictionaries to approximate the entire works of Shakespeare! We're going to use a bigram language model. Here's the idea: We start with some word -- we'll use "The" as an example. Then we look through all of the texts of Shakespeare and for every instance of "The" we record the word that follows "The" and add it to a list, known as the successors of "The". Now suppose we've done this for every word Shakespeare has used, ever.

Let's go back to "The". Now, we randomly choose a word from this list, say "cat". Then we look up the successors of "cat" and randomly choose a word from that list, and we continue this process. This eventually will terminate in a period (".") and we will have generated a Shakespearean sentence!

The object that we'll be looking things up in is called a "successor table", although really it's just a dictionary. The keys in this dictionary are words, and the values are lists of successors to those words.

Q8: Successor Tables

Here's an incomplete definition of the build_successors_table function. The input is a list of words (corresponding to a Shakespearean text), and the output is a successors table. (By default, the first word is a successor to "."). See the example below.

Note: there are two places where you need to write code, denoted by the two "*** YOUR CODE HERE ***"

def build_successors_table(tokens):
    """Return a dictionary: keys are words; values are lists of successors.

    >>> text = ['We', 'came', 'to', 'investigate', ',', 'catch', 'bad', 'guys', 'and', 'to', 'eat', 'pie', '.']
    >>> table = build_successors_table(text)
    >>> sorted(table)
    [',', '.', 'We', 'and', 'bad', 'came', 'catch', 'eat', 'guys', 'investigate', 'pie', 'to']
    >>> table['to']
    ['investigate', 'eat']
    >>> table['pie']
    >>> table['.']
    table = {}
    prev = '.'
    for word in tokens:
        if prev not in table:
"*** YOUR CODE HERE ***"
table[prev] = []
"*** YOUR CODE HERE ***"
table[prev] += [word]
prev = word return table

Use Ok to test your code:

python3 ok -q build_successors_table

Q9: Construct the Sentence

Let's generate some sentences! Suppose we're given a starting word. We can look up this word in our table to find its list of successors, and then randomly select a word from this list to be the next word in the sentence. Then we just repeat until we reach some ending punctuation.

Hint: to randomly select from a list, import the Python random library with import random and use the expression random.choice(my_list)

This might not be a bad time to play around with adding strings together as well. Let's fill in the construct_sent function!

def construct_sent(word, table):
    """Prints a random sentence starting with word, sampling from

    >>> table = {'Wow': ['!'], 'Sentences': ['are'], 'are': ['cool'], 'cool': ['.']}
    >>> construct_sent('Wow', table)
    >>> construct_sent('Sentences', table)
    'Sentences are cool.'
    import random
    result = ''
    while word not in ['.', '!', '?']:
"*** YOUR CODE HERE ***"
result += word + ' ' word = random.choice(table[word])
return result.strip() + word

Use Ok to test your code:

python3 ok -q construct_sent

Putting it all together

Great! Now let's try to run our functions with some actual data. The following snippet included in the skeleton code will return a list containing the words in all of the works of Shakespeare.

Warning: Do NOT try to print the return result of this function.

def shakespeare_tokens(path='shakespeare.txt', url=''):
    """Return the words of Shakespeare's plays as a list."""
    import os
    from urllib.request import urlopen
    if os.path.exists(path):
        return open('shakespeare.txt', encoding='ascii').read().split()
        shakespeare = urlopen(url)

Uncomment the following two lines to run the above function and build the successors table from those tokens.

# Uncomment the following two lines
# tokens = shakespeare_tokens()
# table = build_successors_table(tokens)

Next, let's define a utility function that constructs sentences from this successors table:

>>> def sent():
...     return construct_sent('The', table)
>>> sent()
" The plebeians have done us must be news-cramm'd."

>>> sent()
" The ravish'd thee , with the mercy of beauty!"

>>> sent()
" The bird of Tunis , or two white and plucker down with better ; that's God's sake."

Notice that all the sentences start with the word "The". With a few modifications, we can make our sentences start with a random word. The following random_sent function (defined in your starter file) will do the trick:

def random_sent():
    import random
    return construct_sent(random.choice(table['.']), table)

Go ahead and load your file into Python (be sure to use the -i flag). You can now call the random_sent function to generate random Shakespearean sentences!

>>> random_sent()
' Long live by thy name , then , Dost thou more angel , good Master Deep-vow , And tak'st more ado but following her , my sight Of speaking false!'

>>> random_sent()
' Yes , why blame him , as is as I shall find a case , That plays at the public weal or the ghost.'