Due by 11:59pm on Tuesday, July 11.

Starter Files

Download lab05.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.

Required Questions

Trees

Q1: Finding Berries!

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 berries. Define the function berry_finder, which takes in a tree and returns True if the tree contains a node with the value 'berry' and False otherwise.

Hint: To iterate through each of the branches of a particular tree, you can consider using a for loop to get each branch.

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

    >>> scrat = tree('berry')
    >>> berry_finder(scrat)
    True
    >>> sproul = tree('roots', [tree('branch1', [tree('leaf'), tree('berry')]), tree('branch2')])
    >>> berry_finder(sproul)
    True
    >>> numbers = tree(1, [tree(2), tree(3, [tree(4), tree(5)]), tree(6, [tree(7)])])
    >>> berry_finder(numbers)
    False
    >>> t = tree(1, [tree('berry',[tree('not berry')])])
    >>> berry_finder(t)
    True
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q berry_finder
Copy
✂️

Q2: Replace Loki at Leaf

Define replace_loki_at_leaf, which takes a tree t and a value lokis_replacement. replace_loki_at_leaf returns a new tree that's the same as t except that every leaf label equal to "loki" has been replaced with lokis_replacement.

If you want to learn about the Norse mythology referenced in this problem, you can read about it here.

def replace_loki_at_leaf(t, lokis_replacement):
    """Returns a new tree where every leaf value equal to "loki" has
    been replaced with lokis_replacement.

    >>> yggdrasil = tree('odin',
    ...                  [tree('balder',
    ...                        [tree('loki'),
    ...                         tree('freya')]),
    ...                   tree('frigg',
    ...                        [tree('loki')]),
    ...                   tree('loki',
    ...                        [tree('sif'),
    ...                         tree('loki')]),
    ...                   tree('loki')])
    >>> laerad = copy_tree(yggdrasil) # copy yggdrasil for testing purposes
    >>> print_tree(replace_loki_at_leaf(yggdrasil, 'freya'))
    odin
      balder
        freya
        freya
      frigg
        freya
      loki
        sif
        freya
      freya
    >>> laerad == yggdrasil # Make sure original tree is unmodified
    True
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q replace_loki_at_leaf
Copy
✂️

Data Abstraction

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

Our data abstraction 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)
'Berkeley'
>>> get_lat(berkeley)
122
>>> new_york = make_city('New York City', 74, 40)
>>> get_lon(new_york)
40

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.

Q3: 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(city_a, city_b):
    """
    >>> city_a = make_city('city_a', 0, 1)
    >>> city_b = make_city('city_b', 0, 2)
    >>> distance(city_a, city_b)
    1.0
    >>> city_c = make_city('city_c', 6.5, 12)
    >>> city_d = make_city('city_d', 2.5, 15)
    >>> distance(city_c, city_d)
    5.0
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q distance
Copy
✂️

Q4: 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 you use your distance function to find the distance between the given location and each of the given cities?

def closer_city(lat, lon, city_a, city_b):
    """
    Returns the name of either city_a or city_b, whichever is closest to
    coordinate (lat, lon). If the two cities are the same distance away
    from the coordinate, consider city_b to be the closer city.

    >>> berkeley = make_city('Berkeley', 37.87, 112.26)
    >>> stanford = make_city('Stanford', 34.05, 118.25)
    >>> closer_city(38.33, 121.44, berkeley, stanford)
    '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)
    'Bucharest'
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q closer_city
Copy
✂️

Q5: Don't violate the abstraction barrier!

Note: this question has no code-writing component (if you implemented the previous two questions correctly).

When writing functions that use an data abstraction, we should use the constructor(s) and selector(s) whenever possible instead of assuming the data abstraction'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 the previous questions 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_city_abstraction
Copy
✂️

The check_city_abstraction function exists only for the doctest, which swaps out the implementations of the original 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 data abstraction shouldn't affect the functionality of any programs that use that data abstraction, 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 with the appropriate constructor or selector.

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

Check Your Score Locally

You can locally check your score on each question of this assignment by running

python3 ok --score

This does NOT submit the assignment! When you are satisfied with your score, submit the assignment to Gradescope to receive credit for it.

Submit

Make sure to submit this assignment by uploading any files you've edited to the appropriate Gradescope assignment. For a refresher on how to do this, refer to Lab 00.

Optional Questions

These questions are optional, but you must complete them in order to be checked off before the end of the lab period. They are also useful practice!

Trees

Q6: Recursion on Trees

Define a function dejavu, which takes in a tree of numbers t and a number n. It returns True if there is a path from the root to a leaf such that the sum of the numbers along that path is n and False otherwise.

def dejavu(t, n):
    """
    >>> my_tree = tree(2, [tree(3, [tree(5), tree(7)]), tree(4)])
    >>> dejavu(my_tree, 12) # 2 -> 3 -> 7
    True
    >>> dejavu(my_tree, 5) # Sums of partial paths like 2 -> 3 don ’t count
    False
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q dejavu
Copy
✂️

Q7: Hailstone Tree

We can represent the hailstone sequence as a tree in the figure below, showing the route different numbers take to reach 1. Remember that a hailstone sequence starts with a number n, continuing to n/2 if n is even or 3n+1 if n is odd, ending with 1. Write a function hailstone_tree(n, h) which generates a tree of height h, containing hailstone numbers that will reach n.

Hint: A node of a hailstone tree will always have at least one, and at most two branches (which are also hailstone trees). Under what conditions do you add the second branch?

def hailstone_tree(n, h):
    """Generates a tree of hailstone numbers that will reach N, with height H.
    >>> print_tree(hailstone_tree(1, 0))
    1
    >>> print_tree(hailstone_tree(1, 4))
    1
      2
        4
          8
            16
    >>> print_tree(hailstone_tree(8, 3))
    8
      16
        32
          64
        5
          10
    """
    if _________________________________:
        return _________________________________
    branches = _________________________________
    if ___________ and ___________ and ___________:
        branches += _________________________________
    return tree(n, branches)

Use Ok to test your code:

python3 ok -q hailstone_tree
Copy
✂️