Homework 2: Higher Order Functions

Due by 11:59pm on Thursday, September 7


Download hw02.zip. Inside the archive, you will find a file called hw02.py, along with a copy of the ok autograder.

Submission: When you are done, submit the assignment by uploading all code files you've edited to Gradescope. You may submit more than once before the deadline; only the final submission will be scored. Check that you have successfully submitted your code on Gradescope. See Lab 0 for more instructions on submitting assignments.

Using Ok: If you have any questions about using Ok, please refer to this guide.

Readings: You might find the following references useful:

Grading: Homework is graded based on correctness. Each incorrect problem will decrease the total score by one point. There is a homework recovery policy as stated in the syllabus. This homework is out of 2 points.

Required Questions

Getting Started Videos

These videos may provide some helpful direction for tackling the coding problems on this assignment.

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YouTube link

Several doctests refer to these functions:

from operator import add, mul

square = lambda x: x * x

identity = lambda x: x

triple = lambda x: 3 * x

increment = lambda x: x + 1

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 one argument to produce the term

    >>> product(3, identity)  # 1 * 2 * 3
    >>> product(5, identity)  # 1 * 2 * 3 * 4 * 5
    >>> product(3, square)    # 1^2 * 2^2 * 3^2
    >>> product(5, square)    # 1^2 * 2^2 * 3^2 * 4^2 * 5^2
    >>> product(3, increment) # (1+1) * (2+1) * (3+1)
    >>> product(3, triple)    # 1*3 * 2*3 * 3*3
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q product

Q2: Accumulate

Let's take a look at how product is an instance of a more general function called accumulate, which we would like to implement:

def accumulate(merger, start, n, term):
    """Return the result of merging the first n terms in a sequence and start.
    The terms to be merged are term(1), term(2), ..., term(n). merger is a
    two-argument commutative function.

    >>> accumulate(add, 0, 5, identity)  # 0 + 1 + 2 + 3 + 4 + 5
    >>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 5
    >>> accumulate(add, 11, 0, identity) # 11
    >>> accumulate(add, 11, 3, square)   # 11 + 1^2 + 2^2 + 3^2
    >>> accumulate(mul, 2, 3, square)    # 2 * 1^2 * 2^2 * 3^2
    >>> # 2 + (1^2 + 1) + (2^2 + 1) + (3^2 + 1)
    >>> accumulate(lambda x, y: x + y + 1, 2, 3, square)
    >>> # ((2 * 1^2 * 2) * 2^2 * 2) * 3^2 * 2
    >>> accumulate(lambda x, y: 2 * x * y, 2, 3, square)
    >>> accumulate(lambda x, y: (x + y) % 17, 19, 20, square)
    "*** YOUR CODE HERE ***"

accumulate has the following parameters:

  • merger: a two-argument function that specifies how the current term is merged with the previously accumulated terms
  • start: value at which to start the accumulation
  • n: an integer indicating the number of terms to merge
  • term: a single-argument function that determines a sequence; term(i) is the ith term of the sequence

accumulate should merge the first n terms of the sequence defined by term with the start value according to the merger function.

For example, the result of accumulate(add, 11, 3, square) is

11 + square(1) + square(2) + square(3) = 25

Note: You may assume that merger is commutative. That is, merger(a, b) == merger(b, a) for all a and b. However, you may not assume merger is chosen from a fixed function set and hard-code the solution.

After implementing accumulate, show how summation and product can both be defined as function calls to accumulate.

Important: You should have a single line of code (which should be a return statement) in each of your implementations for summation_using_accumulate and product_using_accumulate, which the syntax check will check for. You must delete the ***YOUR CODE HERE*** placeholder under each function.

def summation_using_accumulate(n, term):
    """Returns the sum: term(1) + ... + term(n), using accumulate.

    >>> summation_using_accumulate(5, square)
    >>> summation_using_accumulate(5, triple)
    >>> # You aren't expected to understand the code of this test.
    >>> # Check that the bodies of the functions are just return statements.
    >>> # If this errors, make sure you have removed the "***YOUR CODE HERE***".
    >>> import inspect, ast
    >>> [type(x).__name__ for x in ast.parse(inspect.getsource(summation_using_accumulate)).body[0].body]
    ['Expr', 'Return']
    "*** YOUR CODE HERE ***"

def product_using_accumulate(n, term):
    """Returns the product: term(1) * ... * term(n), using accumulate.

    >>> product_using_accumulate(4, square)
    >>> product_using_accumulate(6, triple)
    >>> # You aren't expected to understand the code of this test.
    >>> # Check that the bodies of the functions are just return statements.
    >>> # If this errors, make sure you have removed the "***YOUR CODE HERE***".
    >>> import inspect, ast
    >>> [type(x).__name__ for x in ast.parse(inspect.getsource(product_using_accumulate)).body[0].body]
    ['Expr', 'Return']
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q accumulate
python3 ok -q summation_using_accumulate
python3 ok -q product_using_accumulate

Takeaway: Notice how quick it is now to create accumulator functions with different merger functions! This is because we abstracted away the logic of product and summation into the accumulate function. Without this abstraction, our code for a summation function would be just as long as our code for the product function from Question 1, and the logic would be highly redundant!

Q3: 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: While similar to accumulate, the function returned by funception merges terms over the range [begin, end) (rather than [1, n] alongside some arbitrary begin term).

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()

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
    >>> g1(0)    # Returns 1 because begin >= end
    >>> g1(-1)   # Returns 1 because begin >= end
    >>> g2 = funception(double, 1)
    >>> g2(3)    # double(1) * double(2) = 2 * 4 = 8
    >>> g2(4)    # double(1) * double(2) * double(3) = 2 * 4 * 6 = 48
    >>> g3 = funception(increment, -3)
    >>> g3(-1)   # increment(-3) * increment(-2) = -2 * -1 = 2
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q funception

Lambda Expressions

Q4: Lambda Math

To get familiar with writing lambda expressions, let's see how we can use them to help us do some math.

Important: For each of the following problems, your solution should have only a single line of code (which should be a return statement). The syntax check will check for this.

We can start by using a lambda to complete the mul_by_num function.

def mul_by_num(num):
    """Returns a function that takes one argument and returns num
    times that argument.

    >>> x = mul_by_num(5)
    >>> y = mul_by_num(2)
    >>> x(3)
    >>> y(-4)
    return ______

mul_by_num takes in a single number as an argument, and returns a one argument function that multiplies any value passed to it by the original number.

Use Ok to test your code:

python3 ok -q mul_by_num

The next thing we want to do is write a function that takes in two functions, f1 and f2, and returns another function that takes in a single argument x. The returned function should compute f1(x) + f2(x). You can assume both f1 and f2 take in one argument, and their result can be added together.

def add_results(f1, f2):
    Return a function that takes in a single variable x, and returns
    f1(x) + f2(x). You can assume the result of f1(x) and f2(x) can be
    added together, and they both take in one argument.

    >>> identity = lambda x: x
    >>> square = lambda x: x**2
    >>> a1 = add_results(identity, square) # x + x^2
    >>> a1(4)
    >>> a2 = add_results(a1, identity)     # (x + x^2) + x
    >>> a2(4)
    >>> a2(5)
    >>> a3 = add_results(a1, a2)           # (x + x^2) + (x + x^2 + x)
    >>> a3(4)
    return ______

Use Ok to test your code:

python3 ok -q add_results

Finally, let's use a lambda expression to create a function that takes in two integers, x and y, and returns whether or not x is evenly divisible by y. Complete the function mod_maker, which has no input but will return a function that, when called on two integers, will return True if x is divisible by y. Otherwise, it should return the remainder of x % y. You may not use an if or ...if...else... statement in your lambda expression.

def mod_maker():
    """Return a two-argument function that performs the modulo operation and
    returns True if the numbers are divisble, and the remainder otherwise.

    >>> mod = mod_maker()
    >>> mod(7, 2) # 7 % 2
    >>> mod(4, 8) # 4 % 8
    >>> mod(8,4) # 8 % 4
    >>> from construct_check import check
    >>> check(HW_SOURCE_FILE, 'mod_maker', ['If', 'IfExp']) # no if / if-else statements
    return ______

Note: You are allowed (and expected) to use the modulo operator itself in your solution. The goal of this function is not to recreate the operator from scratch, but to create an alternate way of calling it.

Hint: Recall that Python implements short-circuiting. Additionally, recall that Python values like False, 0, None, and "" are considered false values. Can we utilize an and or or statement in our lambda to solve this problem?

Use Ok to test your code:

python3 ok -q mod_maker

Use Ok to run the local syntax checker (which checks that each of your solutions only contains one line):

python3 ok -q lambda_math_syntax_check

Takeaway: Although lambda functions have more restrictions about what they can contain in their bodies than regular functions, they are still useful, especially for making minor tweaks and additions to pre-existing code that doesn't have quite the right structure for what we are trying to do.

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.


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.

Exam Practice

Homework assignments will also contain prior exam questions for you to try. These questions have no submission component; feel free to attempt them if you'd like some practice!

Note that exams from Spring 2020, Fall 2020, and Spring 2021 gave students access to an interpreter, so the question format may be different than other years. Regardless, the questions below are good problems to try without access to an interpreter.

  1. Fall 2019 MT1 Q3: You Again [Higher Order Functions]
  2. Spring 2021 MT1 Q4: Domain on the Range [Higher Order Functions]
  3. Fall 2021 MT1 Q1b: tik [Functions and Expressions]