Starter Files

Download lab06.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.

• To receive full credit for this lab, all questions must be attempted.

When you are ready to submit, run ok with the --submit option:

python3 ok --submit

After submitting, ok will display a submission URL, with which you can view your submission on okpy.org.

Recursion

Question 1: Sum

Using recursion, write a function sum that takes a single argument n and computes the sum of all integers between 0 and n inclusive. Do not write this function using a while or for loop. Assume n is non-negative.

def sum(n):
    """Using recursion, computes the sum of all integers between 1 and n, inclusive.
    Assume n is positive.

    >>> sum(1)
    1
    >>> sum(5)  # 1 + 2 + 3 + 4 + 5
    15
    """
    "*** YOUR CODE HERE ***"
    if n == 1:
        return 1
    return n + sum(n - 1)

Toggle Solution

Use OK to test your code:

python3 ok -q sum

Question 2: Has Seven

Write a function has_seven that takes a positive integer n and returns whether n contains the digit 7. Do not use any assignment statements - use recursion instead:

def has_seven(k):
    """Returns True if at least one of the digits of k is a 7, False otherwise.

    >>> has_seven(3)
    False
    >>> has_seven(7)
    True
    >>> has_seven(2734)
    True
    >>> has_seven(2634)
    False
    >>> has_seven(734)
    True
    >>> has_seven(7777)
    True
    """
    "*** YOUR CODE HERE ***"
    if k % 10 == 7:
        return True
    elif k < 10:
        return False
    else:
        return has_seven(k // 10)

Toggle Solution

Use OK to test your code:

python3 ok -q has_seven

Question 3: Filter

Write the recursive version of the function filter which takes

• filter - a one-argument function that returns True if the argument passed in should be included in the list, and False otherwise.
• s - a sequence of values

def filter(f, seq):
    """Filter a sequence to only contain values allowed by filter.

    >>> def is_even(x):
    ...     return x % 2 == 0
    >>> def divisible_by5(x):
    ...     return x % 5 == 0
    >>> filter(is_even, [1,2,3,4])
    [2, 4]
    >>> filter(divisible_by5, [1, 4, 9, 16, 25, 100])
    [25, 100]
    """
    "*** YOUR CODE HERE ***"
    
    if seq == []:
        return seq
    if f(seq[0]):
        return [seq[0]] + filter(f, seq[1:])
    return filter(f, seq[1:])

Toggle Solution

Use OK to test your code:

python3 ok -q filter

Question 4: Decimal

Write the recursive version of the function decimal which takes in n, a number, and returns a list representing the decimal representation of the number.

def decimal(n):
    """Return a list representing the decimal representation of a number.

    >>> decimal(55055)
    [5, 5, 0, 5, 5]
    >>> decimal(-136)
    ['-', 1, 3, 6]
    """
    "*** YOUR CODE HERE ***"
    if n < 0:
        return ['-'] + decimal(-1 * n)
    elif n < 10:
        return [n % 10]
    else:
        return decimal(n // 10) + [n % 10]

Toggle Solution

Use OK to test your code:

python3 ok -q decimal

Question 5: Insect Combinatorics

Consider an insect in an M by N grid. The insect starts at the bottom left corner, (0, 0), and wants to end up at the top right corner, (M-1, N-1). The insect is only capable of moving right or up. Write a function paths that takes a grid length and width and returns the number of different paths the insect can take from the start to the goal. (There is a closed-form solution to this problem, but try to answer it procedurally using recursion.)

For example, the 2 by 2 grid has a total of two ways for the insect to move from the start to the goal. For the 3 by 3 grid, the insect has 6 different paths (only 3 are shown above). Note that this problem uses tree recursion.

def paths(m, n):
    """Return the number of paths from one corner of an
    M by N grid to the opposite corner.

    >>> paths(2, 2)
    2
    >>> paths(5, 7)
    210
    >>> paths(117, 1)
    1
    >>> paths(1, 157)
    1
    """
    "*** YOUR CODE HERE ***"
    if m == 1 or n == 1:
        return 1
    return paths(m - 1, n) + paths(m, n - 1)

Toggle Solution

Use OK to test your code:

python3 ok -q paths

Submit

Make sure to submit this assignment by running:

python3 ok --submit