Due by 11:59pm on Wednesday, July 8.

Starter Files

Download lab04.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the Ok autograder.

Submission

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 okpy.org.

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.

>>> list_of_nums = [1, 2, 3, 4]
>>> list_of_bools = [True, True, False, False]
>>> nested_lists = [1, [2, 3], [4, [5]]]

>>> lst = [6, 5, 4, 3, 2, 1]
>>> lst[0]
6
>>> lst[3]
3

>>> len([])
0
>>> len([2, 4, 6, 8, 10])
5

>>> lst = [True, False, True, True, False]
>>> lst[1:4]
[False, True, True]
>>> lst[:3]  # Start index defaults to 0
[True, False, True]
>>> lst[3:]  # End index defaults to len(lst)
[True, False]
>>> lst[:]  # Creates a copy of the whole list
[True, False, True, True, False]

Required Questions

Lists Practice

Q1: List Indexing

Use Ok to test your knowledge with the following "List Indexing" questions:

python3 ok -q list-indexing -u

For each of the following lists, what is the list indexing expression that evaluates to 7? For example, if x = [7], then the answer would be x[0]. You can use the interpreter or Python Tutor to experiment with your answers. If the code would cause an error, type Error.

>>> x = [1, 3, [5, 7], 9]
______
x[2][1]
>>> x = [[3, [5, 7], 9]]
______
x[0][1][1]

What would Python display? If you get stuck, try it out in the Python interpreter!

>>> lst = [3, 2, 7, [84, 83, 82]]
>>> lst[4]
______
Error
>>> lst[3][0]
______
84

Q2: Reverse (iteratively)

Write a function reverse_iter that takes a list and returns a new list that is the reverse of the original. Use iteration! Do not use lst[::-1], lst.reverse(), or reversed(lst)!

def reverse_iter(lst):
    """Returns the reverse of the given list.

    >>> reverse_iter([1, 2, 3, 4])
    [4, 3, 2, 1]
    >>> import inspect, re
    >>> cleaned = re.sub(r"#.*\\n", '', re.sub(r'"{3}[\s\S]*?"{3}', '', inspect.getsource(reverse_iter)))
    >>> print("Do not use lst[::-1], lst.reverse(), or reversed(lst)!") if any([r in cleaned for r in ["[::", ".reverse", "reversed"]]) else None
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q reverse_iter

Q3: Reverse (recursively)

Write a function reverse_recursive that takes a list and returns a new list that is the reverse of the original. Use recursion! You may also use slicing notation. Do not use lst[::-1], lst.reverse(), or reversed(lst)!

def reverse_recursive(lst):
    """Returns the reverse of the given list.

    >>> reverse_recursive([1, 2, 3, 4])
    [4, 3, 2, 1]
    >>> import inspect, re
    >>> cleaned = re.sub(r"#.*\\n", '', re.sub(r'"{3}[\s\S]*?"{3}', '', inspect.getsource(reverse_recursive)))
    >>> print("Do not use lst[::-1], lst.reverse(), or reversed(lst)!") if any([r in cleaned for r in ["[::", ".reverse", "reversed"]]) else None
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q reverse_recursive

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)
'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.

Q4: 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

Q5: 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, city_a, city_b):
    """
    Returns the name of either city_a or city_b, 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)
    '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

Q6: 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.

Submit

Make sure to submit this assignment by running:

python3 ok --submit

Optional Questions

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

Q7: 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. This means that w1 is shorter than 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")
    'h'
    >>> add_chars("want", "wanton")
    'on'
    >>> add_chars("rat", "radiate")
    'diae'
    >>> add_chars("a", "prepare")
    'prepre'
    >>> add_chars("resin", "recursion")
    'curo'
    >>> add_chars("fin", "effusion")
    'efuso'
    >>> add_chars("coy", "cacophony")
    'acphon'
    >>> from construct_check import check
    >>> # ban iteration and sets
    >>> check(LAB_SOURCE_FILE, 'add_chars',
    ...       ['For', 'While', 'Set', 'SetComp']) # Must use recursion
    True
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q add_chars