Homework 2 Solutions

Solution Files

You can find solutions for all questions in hw02.py.

Required Questions


Getting Started Videos

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

To see these videos, you should be logged into your berkeley.edu email.

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
    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

The prod variable is used to keep track of the product so far. We start with prod = 1 since we will be multiplying, and anything multiplied by 1 is itself. We then initialize the counter variable k to use in the while loop to ensures that we get through all values 1 through k.

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(fuse, start, n, term):
    """Return the result of fusing together the first n terms in a sequence 
    and start.  The terms to be fused are term(1), term(2), ..., term(n). 
    The function fuse is a two-argument commutative & associative function.

    >>> accumulate(add, 0, 5, identity)  # 0 + 1 + 2 + 3 + 4 + 5
    15
    >>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 5
    26
    >>> accumulate(add, 11, 0, identity) # 11 (fuse is never used)
    11
    >>> accumulate(add, 11, 3, square)   # 11 + 1^2 + 2^2 + 3^2
    25
    >>> accumulate(mul, 2, 3, square)    # 2 * 1^2 * 2^2 * 3^2
    72
    >>> # 2 + (1^2 + 1) + (2^2 + 1) + (3^2 + 1)
    >>> accumulate(lambda x, y: x + y + 1, 2, 3, square)
    19
    """
total, k = start, 1 while k <= n: total, k = fuse(total, term(k)), k + 1 return total # Alternative solution def accumulate_reverse(fuse, start, n, term): total, k = start, n while k >= 1: total, k = fuse(total, term(k)), k - 1 return total

accumulate has the following parameters:

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

Implement accumulate, which fuses the first n terms of the sequence defined by term with the start value using the fuse function.

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

add(11,  add(square(1), add(square(2),  square(3)))) =
    11 +     square(1) +    square(2) + square(3)    =
    11 +     1         +    4         + 9            = 25

Assume that fuse is commutative, fuse(a, b) == fuse(b, a), and associative, fuse(fuse(a, b), c) == fuse(a, fuse(b, c)).

Then, implement summation (from lecture) and product as one-line calls to accumulate.

Important: Both summation_using_accumulate and product_using_accumulate should be implemented with a single line of code starting with return.

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

    >>> summation_using_accumulate(5, square) # square(1) + square(2) + ... + square(4) + square(5)
    55
    >>> summation_using_accumulate(5, triple) # triple(1) + triple(2) + ... + triple(4) + triple(5)
    45
    >>> # This test checks that the body of the function is just a return statement.
    >>> import inspect, ast
    >>> [type(x).__name__ for x in ast.parse(inspect.getsource(summation_using_accumulate)).body[0].body]
    ['Expr', 'Return']
    """
return accumulate(add, 0, n, term)
def product_using_accumulate(n, term): """Returns the product: term(1) * ... * term(n), using accumulate. >>> product_using_accumulate(4, square) # square(1) * square(2) * square(3) * square() 576 >>> product_using_accumulate(6, triple) # triple(1) * triple(2) * ... * triple(5) * triple(6) 524880 >>> # This test checks that the body of the function is just a return statement. >>> import inspect, ast >>> [type(x).__name__ for x in ast.parse(inspect.getsource(product_using_accumulate)).body[0].body] ['Expr', 'Return'] """
return accumulate(mul, 1, n, term)

Use Ok to test your code:

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

We want to abstract the logic of product and summation into accumulate. The differences between product and summation are:

  • How to fuse terms. For product, we fuse via * (mul). For summation, we fuse via + (add).
  • The starting value. For product, we want to start off with 1 since starting with 0 means that our result (via multiplying with the start) will always be 0. For summation, we want to start off with 0.

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, where 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

There are many correct ways to implement make_repeater. This solution repeatedly applies h.

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 Assignment

Submit this assignment by uploading any files you've edited to the appropriate Gradescope assignment. Lab 00 has detailed instructions.

[Optional] Exam Practice

Here are some related questions from past exams for you to try. These are optional. There is no way to submit them.

  1. Fall 2019 MT1 Q3: You Again [Higher-Order Functions]
  2. Fall 2021 MT1 Q1b: tik [Functions and Expressions]