Homework 2: Higher Order Functions
Due by 11:59pm on Thursday, September 7
Instructions
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.
To see these videos, you should be logged into your berkeley.edu email.
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 i
th 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
"""
"*** 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
15
>>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 5
26
>>> accumulate(add, 11, 0, identity) # 11
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
>>> # ((2 * 1^2 * 2) * 2^2 * 2) * 3^2 * 2
>>> accumulate(lambda x, y: 2 * x * y, 2, 3, square)
576
>>> accumulate(lambda x, y: (x + y) % 17, 19, 20, square)
16
"""
"*** 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 termsstart
: value at which to start the accumulationn
: an integer indicating the number of terms to mergeterm
: a single-argument function that determines a sequence;term(i)
is thei
th 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 alla
andb
. However, you may not assumemerger
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)
55
>>> summation_using_accumulate(5, triple)
45
>>> # 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)
576
>>> product_using_accumulate(6, triple)
524880
>>> # 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 byfunception
merges terms over the range[begin, end)
(rather than[1, n]
alongside some arbitrarybegin
term).
Note 2: If you attempt to modify
begin
infunc2
, you will get anUnboundLocalError
. You're not allowed to rebind variables defined outside of our current frame! However, you can still access the value ofbegin
infunc2
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
"""
"*** 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)
15
>>> y(-4)
-8
"""
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)
20
>>> a2 = add_results(a1, identity) # (x + x^2) + x
>>> a2(4)
24
>>> a2(5)
35
>>> a3 = add_results(a1, a2) # (x + x^2) + (x + x^2 + x)
>>> a3(4)
44
"""
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
1
>>> mod(4, 8) # 4 % 8
4
>>> mod(8,4) # 8 % 4
True
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'mod_maker', ['If', 'IfExp']) # no if / if-else statements
True
"""
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 anand
oror
statement in ourlambda
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.
Submit
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.
- Fall 2019 MT1 Q3: You Again [Higher Order Functions]
- Spring 2021 MT1 Q4: Domain on the Range [Higher Order Functions]
- Fall 2021 MT1 Q1b: tik [Functions and Expressions]