Solution Files

You can find solutions for all questions in hw03.py.

Required Questions

Higher-Order Functions

Q1: Product

Write a function called product that returns the product of the first n terms of a sequence. Specifically, product takes in an integer n and term, a single-argument function that determines a sequence. (That is, term(i) gives the ith term of the sequence.) product(n, term) should return term(1) * ... * term(n).

def product(n, term):
    """Return the product of the first n terms in a sequence.

    n: a positive integer
    term: a function that takes an index as input and produces a term

    >>> product(3, identity)  # 1 * 2 * 3
    6
    >>> product(5, identity)  # 1 * 2 * 3 * 4 * 5
    120
    >>> product(3, square)    # 1^2 * 2^2 * 3^2
    36
    >>> product(5, square)    # 1^2 * 2^2 * 3^2 * 4^2 * 5^2
    14400
    >>> product(3, increment) # (1+1) * (2+1) * (3+1)
    24
    >>> product(3, triple)    # 1*3 * 2*3 * 3*3
    162
    """
    prod, k = 1, 1
    while k <= n:
        prod, k = term(k) * prod, k + 1
    return prod

Use Ok to test your code:

python3 ok -q product
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Q2: Funception

Implement funception, which takes in a function func1 and a number begin and returns a function func2. func2 should take a single argument, end, and apply func1 on all the numbers from begin (inclusive) up to end (exclusive) and return the product.

If begin is greater than or equal to end, the range of numbers is invalid, and func2 should return 1.

Note 1: The function returned by funception merges terms over the range [begin, end).

Note 2: If you attempt to modify begin in func2, you will get an UnboundLocalError. You're not allowed to rebind variables defined outside of our current frame! However, you can still access the value of begin in func2 and set it equal to a new variable. An example is shown below:

Unbound Local Error:

>>> def f():
...     x = 5
...     def g():
...         x += 1 # UnboundLocalError: cannot modify variable X outside current frame
...         print(x)
...     return g
...
>>> h = f()
>>> h()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 4, in g
UnboundLocalError: local variable 'x' referenced before assignment

No Error:

>>> def f():
...     x = 5
...     def g():
...         i = x # No Error: accesses value X from parent frame, binds it to I in the current frame
...         i += 1
...         print(i)
...     return g
...
>>> h = f()
>>> h()
6

def funception(func1, begin):
    """ Takes in a function (func1) and a begin value.
    Returns a function (func2) that will find the product of
    func1 applied to the range of numbers from
    begin (inclusive) to end (exclusive)

    >>> def increment(num):
    ...     return num + 1
    >>> def double(num):
    ...     return num * 2
    >>> g1 = funception(increment, 0)
    >>> g1(3)    # increment(0) * increment(1) * increment(2) = 1 * 2 * 3 = 6
    6
    >>> g1(0)    # Returns 1 because begin >= end
    1
    >>> g1(-1)   # Returns 1 because begin >= end
    1
    >>> g2 = funception(double, 1)
    >>> g2(3)    # double(1) * double(2) = 2 * 4 = 8
    8
    >>> g2(4)    # double(1) * double(2) * double(3) = 2 * 4 * 6 = 48
    48
    >>> g3 = funception(increment, -3)
    >>> g3(-1)   # increment(-3) * increment(-2) = -2 * -1 = 2
    2
    """
    def func2(end):
        i = begin
        product = 1
        while i < end:
            product *= func1(i)
            i += 1
        return product
    return func2

Use Ok to test your code:

python3 ok -q funception
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Q3: Make Repeater

Implement the function make_repeater which takes a one-argument function f and a positive integer n. It returns a one-argument function so that make_repeater(f, n)(x) returns the value of f(f(...f(x)...)), in which f is applied n times to x. For example, make_repeater(square, 3)(5) squares 5 three times and returns 390625, just like square(square(square(5))).

def make_repeater(f, n):
    """Returns the function that computes the nth application of f.

    >>> add_three = make_repeater(increment, 3)
    >>> add_three(5)
    8
    >>> make_repeater(triple, 5)(1) # 3 * (3 * (3 * (3 * (3 * 1))))
    243
    >>> make_repeater(square, 2)(5) # square(square(5))
    625
    >>> make_repeater(square, 3)(5) # square(square(square(5)))
    390625
    """
    def repeater(x):
        k = 0
        while k < n:
            x, k = f(x), k + 1
        return x
    return repeater

Use Ok to test your code:

python3 ok -q make_repeater
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Check Your Score Locally

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

python3 ok --score

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