Lab 2 Solutions
Solution Files
Topics
Consult this section if you need a refresher on the material for this lab. It's okay to skip directly to the questions and refer back here should you get stuck.
HigherOrder Functions
Variables are names bound to values, which can be primitives like 3
or
'Hello World'
, but they can also be functions. And since functions can take
arguments of any value, other functions can be passed in as arguments. This is
the basis for higherorder functions.
A higher order function is a function that manipulates other functions by taking in functions as arguments, returning a function, or both. We will introduce the basics of higher order functions in this lab and will be exploring many applications of higher order functions in our next lab.
Functions as arguments
In Python, function objects are values that can be passed around. We know that one
way to create functions is by using a def
statement:
def square(x):
return x * x
The above statement created a function object with the intrinsic name square
as
well as binded it to the name square
in the current environment. Now let's try
passing it as an argument.
First, let's write a function that takes in another function as an argument:
def scale(f, x, k):
""" Returns the result of f(x) scaled by k. """
return k * f(x)
We can now call scale
on square
and some other arguments:
>>> scale(square, 3, 2) # Double square(3)
18
>>> scale(square, 2, 5) # 5 times 2 squared
20
Note that in the body of the call to scale
, the function object with the intrinsic
name square
is bound to the parameter f
. Then, we call square
in the body of
scale
by calling f(x)
.
As we saw in the above section on lambda
expressions, we can also pass
lambda
expressions into call expressions!
>>> scale(lambda x: x + 10, 5, 2)
30
In the frame for this call expression, the name f
is bound to the function
created by the lambda
expression lambda x: x + 10
.
Functions that return functions
Because functions are values, they are valid as return values! Here's an example:
def multiply_by(m):
def multiply(n):
return n * m
return multiply
In this particular case, we defined the function multiply
within the body of multiply_by
and then returned it. Let's see it in action:
>>> multiply_by(3)
<function multiply_by.<locals>.multiply at ...>
>>> multiply(4)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
NameError: name 'multiply' is not defined
A call to multiply_by
returns a function, as expected. However, calling
multiply
errors, even though that's the name we gave the inner function. This
is because the name multiply
only exists within the frame where we evaluate
the body of multiply_by
.
So how do we actually use the inner function? Here are two ways:
>>> times_three = multiply_by(3) # Assign the result of the call expression to a name
>>> times_three(5) # Call the inner function with its new name
15
>>> multiply_by(3)(10) # Chain together two call expressions
30
The point is, because multiply_by
returns a function, you can use its return
value just like you would use any other function.
Lambda Expressions
Lambda expressions are expressions that evaluate to functions by specifying two things: the parameters and a return expression.
lambda <parameters>: <return expression>
While both lambda
expressions and def
statements create function objects,
there are some notable differences. lambda
expressions work like other
expressions; much like a mathematical expression just evaluates to a number and
does not alter the current environment, a lambda
expression
evaluates to a function without changing the current environment. Let's take a
closer look.
lambda  def  

Type  Expression that evaluates to a value  Statement that alters the environment 
Result of execution  Creates an anonymous lambda function with no intrinsic name.  Creates a function with an intrinsic name and binds it to that name in the current environment. 
Effect on the environment  Evaluating a lambda expression does not create or
modify any variables. 
Executing a def statement both creates a new function
object and binds it to a name in the current environment. 
Usage  A lambda expression can be used anywhere that
expects an expression, such as in an assignment statement or as the
operator or operand to a call expression. 
After executing a def statement, the created
function is bound to a name. You should use this name to refer to the
function anywhere that expects an expression. 
Example 


Currying
We can transform multipleargument functions into a chain of singleargument, higher order functions. For example, we can write a function f(x, y)
as a different
function g(x)(y)
. This is known as currying.
For example, to convert the function add(x, y)
into its curried form:
def curry_add(x):
def add2(y):
return x + y
return add2
Calling curry_add(1)
returns a new function which only performs the addition once the returned function is called with the second addend.
>>> add_one = curry_add(1)
>>> add_one(2)
3
>>> add_one(4)
5
Refer to the textbook for more details about currying.
Environment Diagrams
Environment diagrams are one of the best learning tools for understanding
lambda
expressions and higher order functions because you're able to keep
track of all the different names, function objects, and arguments to functions.
We highly recommend drawing environment diagrams or using Python
tutor if you get stuck doing the WWPD problems below.
For examples of what environment diagrams should look like, try running some
code in Python tutor. Here are the rules:
Assignment Statements
 Evaluate the expression on the right hand side of the
=
sign.  If the name found on the left hand side of the
=
doesn't already exist in the current frame, write it in. If it does, erase the current binding. Bind the value obtained in step 1 to this name.
If there is more than one name/expression in the statement, evaluate all the expressions first from left to right before making any bindings.
def Statements
 Draw the function object with its intrinsic name, formal parameters, and parent frame. A function's parent frame is the frame in which the function was defined.
 If the intrinsic name of the function doesn't already exist in the current frame, write it in. If it does, erase the current binding. Bind the newly created function object to this name.
Call expressions
Note: you do not have to go through this process for a builtin Python function like
max
or
 Evaluate the operator, whose value should be a function.
 Evaluate the operands left to right.
 Open a new frame. Label it with the sequential frame number, the intrinsic name of the function, and its parent.
 Bind the formal parameters of the function to the arguments whose values you found in step 2.
 Execute the body of the function in the new environment.
Lambdas
Note: As we saw in the
lambda
expression section above,lambda
functions have no intrinsic name. When drawinglambda
functions in environment diagrams, they are labeled with the namelambda
or with the lowercase Greek letter λ. This can get confusing when there are multiple lambda functions in an environment diagram, so you can distinguish them by numbering them or by writing the line number on which they were defined.
 Draw the lambda function object and label it with λ, its formal parameters, and its parent frame. A function's parent frame is the frame in which the function was defined.
This is the only step. We are including this section to emphasize the fact that
the difference between lambda
expressions and def
statements is that
lambda
expressions do not create any new bindings in the environment.
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.
What Would Python Display?
Important: For all WWPD questions, type
Function
if you believe the answer is<function...>
,Error
if it errors, andNothing
if nothing is displayed.
Q1: WWPD: Lambda the Free
Use Ok to test your knowledge with the following "What Would Python Display?" questions:
python3 ok q lambda u
As a reminder, the following two lines of code will not display anything in the Python interpreter when executed:>>> x = None >>> x
>>> lambda x: x # A lambda expression with one parameter x
______<function <lambda> at ...>
>>> a = lambda x: x # Assigning the lambda function to the name a
>>> a(5)
______5
>>> (lambda: 3)() # Using a lambda expression as an operator in a call exp.
______3
>>> b = lambda x: lambda: x # Lambdas can return other lambdas!
>>> c = b(88)
>>> c
______<function <lambda> at ...
>>> c()
______88
>>> d = lambda f: f(4) # They can have functions as arguments as well.
>>> def square(x):
... return x * x
>>> d(square)
______16
>>> z = 3
>>> e = lambda x: lambda y: lambda: x + y + z
>>> e(0)(1)()
______4
>>> f = lambda z: x + z
>>> f(3)
______NameError: name 'x' is not defined
>>> x = None # remember to review the rules of WWPD given above!
>>> x
______# x evaluates to None, so nothing gets displayed
>>> higher_order_lambda = lambda f: lambda x: f(x)
>>> g = lambda x: x * x
>>> higher_order_lambda(2)(g) # Which argument belongs to which function call?
______Error
>>> higher_order_lambda(g)(2)
______4
>>> call_thrice = lambda f: lambda x: f(f(f(x)))
>>> call_thrice(lambda y: y + 1)(0)
______3
>>> print_lambda = lambda z: print(z) # When is the return expression of a lambda expression executed?
>>> print_lambda
______Function
>>> one_thousand = print_lambda(1000)
______1000
>>> one_thousand # What did the call to print_lambda return?
______# print_lambda returned None, so nothing gets displayed
Q2: WWPD: Higher Order Functions
Use Ok to test your knowledge with the following "What Would Python Display?" questions:
python3 ok q hofwwpd u
>>> def even(f):
... def odd(x):
... if x < 0:
... return f(x)
... return f(x)
... return odd
>>> steven = lambda x: x
>>> stewart = even(steven)
>>> stewart
______<function ...>
>>> stewart(61)
______61
>>> stewart(4)
______4
>>> def cake():
... print('beets')
... def pie():
... print('sweets')
... return 'cake'
... return pie
>>> chocolate = cake()
______beets
>>> chocolate
______Function
>>> chocolate()
______sweets
'cake'
>>> more_chocolate, more_cake = chocolate(), cake
______sweets
>>> more_chocolate
______'cake'
>>> def snake(x, y):
... if cake == more_cake:
... return chocolate
... else:
... return x + y
>>> snake(10, 20)
______Function
>>> snake(10, 20)()
______30
>>> cake = 'cake'
>>> snake(10, 20)
______30
Coding Practice
Q3: Lambdas and Currying
Write a function lambda_curry2
that will curry any two argument function
using lambdas.
Your solution to this problem should only be one line. You can try first writing a solution without the restriction, and then rewriting it into one line after.
If the syntax check isn't passing: Make sure you've removed the line containing
"***YOUR CODE HERE***"
so that it doesn't get treated as part of the function for the syntax check.
def lambda_curry2(func):
"""
Returns a Curried version of a twoargument function FUNC.
>>> from operator import add, mul, mod
>>> curried_add = lambda_curry2(add)
>>> add_three = curried_add(3)
>>> add_three(5)
8
>>> curried_mul = lambda_curry2(mul)
>>> mul_5 = curried_mul(5)
>>> mul_5(42)
210
>>> lambda_curry2(mod)(123)(10)
3
"""
return lambda arg1: lambda arg2: func(arg1, arg2)
Use Ok to test your code:
python3 ok q lambda_curry2
To curry a two argument function, we want to only call func
once we have
received its two arguments on separate calls to our curry function. We can
do so by using lambdas. The outermost lambda will receive the first argument
that we will later pass into func
, and since we haven't yet received both
arguments that we need, we want this outermost lambda to return a function
itself. This function (which we can implement using another lambda) will take
in a single parameter, so when we call it, then we will have had both
arguments that we need to call func
.
Q4: Count van Count
Consider the following implementations of count_factors
and count_primes
:
def count_factors(n):
"""Return the number of positive factors that n has.
>>> count_factors(6)
4 # 1, 2, 3, 6
>>> count_factors(4)
3 # 1, 2, 4
"""
i = 1
count = 0
while i <= n:
if n % i == 0:
count += 1
i += 1
return count
def count_primes(n):
"""Return the number of prime numbers up to and including n.
>>> count_primes(6)
3 # 2, 3, 5
>>> count_primes(13)
6 # 2, 3, 5, 7, 11, 13
"""
i = 1
count = 0
while i <= n:
if is_prime(i):
count += 1
i += 1
return count
def is_prime(n):
return count_factors(n) == 2 # only factors are 1 and n
The implementations look quite similar! Generalize this logic by writing a
function count_cond
, which takes in a twoargument predicate function
condition(n, i)
. count_cond
returns a oneargument function that takes
in n
, which counts all the numbers from 1 to n
that satisfy condition
when called.
Note: When we say
condition
is a predicate function, we mean that it is a function that will returnTrue
orFalse
based on some specified condition in its body.
def count_cond(condition):
"""Returns a function with one parameter N that counts all the numbers from
1 to N that satisfy the twoargument predicate function Condition, where
the first argument for Condition is N and the second argument is the
number from 1 to N.
>>> count_factors = count_cond(lambda n, i: n % i == 0)
>>> count_factors(2) # 1, 2
2
>>> count_factors(4) # 1, 2, 4
3
>>> count_factors(12) # 1, 2, 3, 4, 6, 12
6
>>> is_prime = lambda n, i: count_factors(i) == 2
>>> count_primes = count_cond(is_prime)
>>> count_primes(2) # 2
1
>>> count_primes(3) # 2, 3
2
>>> count_primes(4) # 2, 3
2
>>> count_primes(5) # 2, 3, 5
3
>>> count_primes(20) # 2, 3, 5, 7, 11, 13, 17, 19
8
"""
def counter(n):
i = 1
count = 0
while i <= n:
if condition(n, i):
count += 1
i += 1
return count
return counter
One question that might be nice to ask is:
in what ways is the logic for count_factors
and count_primes
similar,
and in what ways are they different?
The answer to the first question can tell us the logic that we want to
include in our count_cond
function, while the answer to the second
question can tell us where in count_cond
we want to be able to have
the difference in behavior observed between count_factors
and
count_primes
.
It'll be helpful to also keep in mind that we want count_cond
to return
a function that, when an argument n
is passed in, will behave similarly
to count_factors
or count_primes
. In other words, count_cond
is a
higher order function that returns a function, that then contains the
logic common to both count_factors
and count_primes
.
Use Ok to test your code:
python3 ok q count_cond
Submit
Make sure to submit this assignment by running:
python3 ok submit
Environment Diagram Practice
There is no Ok submission for this component.
However, we still encourage you to do this problem on paper to develop familiarity with Environment Diagrams, which might appear in an alternate form on the exam. To check your work, you can try putting the code into PythonTutor.
Q5: HOF Diagram Practice
Draw the environment diagram that results from executing the code below.
n = 7
def f(x):
n = 8
return x + 1
def g(x):
n = 9
def h():
return x + 1
return h
def f(f, x):
return f(x + n)
f = f(g, n)
g = (lambda y: y())(f)
Optional Questions
Q6: Composite Identity Function
Write a function that takes in two singleargument functions, f
and g
, and
returns another function that has a single parameter x
. The returned
function should return True
if f(g(x))
is equal to g(f(x))
. You can
assume the output of g(x)
is a valid input for f
and vice versa.
Try to use the composer
function defined below for more HOF practice.
def composer(f, g):
"""Return the composition function which given x, computes f(g(x)).
>>> add_one = lambda x: x + 1 # adds one to x
>>> square = lambda x: x**2
>>> a1 = composer(square, add_one) # (x + 1)^2
>>> a1(4)
25
>>> mul_three = lambda x: x * 3 # multiplies 3 to x
>>> a2 = composer(mul_three, a1) # ((x + 1)^2) * 3
>>> a2(4)
75
>>> a2(5)
108
"""
return lambda x: f(g(x))
def composite_identity(f, g):
"""
Return a function with one parameter x that returns True if f(g(x)) is
equal to g(f(x)). You can assume the result of g(x) is a valid input for f
and vice versa.
>>> add_one = lambda x: x + 1 # adds one to x
>>> square = lambda x: x**2
>>> b1 = composite_identity(square, add_one)
>>> b1(0) # (0 + 1)^2 == 0^2 + 1
True
>>> b1(4) # (4 + 1)^2 != 4^2 + 1
False
"""
def identity(x):
return composer(f, g)(x) == composer(g, f)(x)
return identity
# Alternative solution
return lambda x: f(g(x)) == g(f(x))
Solution using composer
:
Calling composer
will return us a function that takes in a single parameter
x
.
Here, the order in which we pass in the two functions f
and g
from
composite_identity
matters. composer
will give us a function that first
calls the second argument to composer
on the input x
(let's call this
return value to be y
), and we will then call the first argument to composer
on this return value (aka on y
), which is what we finally return.
We want to compare the results of f(g(x))
with g(f(x))
, so we will want
to call composer
and then pass in (as a separate argument) x
to these
composed functions in order to get a value to actually compare them.
Solution not using composer
:
We can also directly call f(g(x))
and g(f(x))
instead of calling
composer
, and then compare the results of these two function calls.
Use Ok to test your code:
python3 ok q composite_identity
Q7: I Heard You Liked Functions...
Define a function cycle
that takes in three functions f1
, f2
,
f3
, as arguments. cycle
will return another function that should
take in an integer argument n
and return another function. That
final function should take in an argument x
and cycle through
applying f1
, f2
, and f3
to x
, depending on what n
was. Here's what the final function should do to x
for a few
values of n
:
n = 0
, returnx
n = 1
, applyf1
tox
, or returnf1(x)
n = 2
, applyf1
tox
and thenf2
to the result of that, or returnf2(f1(x))
n = 3
, applyf1
tox
,f2
to the result of applyingf1
, and thenf3
to the result of applyingf2
, orf3(f2(f1(x)))
n = 4
, start the cycle again applyingf1
, thenf2
, thenf3
, thenf1
again, orf1(f3(f2(f1(x))))
 And so forth.
Hint: most of the work goes inside the most nested function.
def cycle(f1, f2, f3):
"""Returns a function that is itself a higherorder function.
>>> def add1(x):
... return x + 1
>>> def times2(x):
... return x * 2
>>> def add3(x):
... return x + 3
>>> my_cycle = cycle(add1, times2, add3)
>>> identity = my_cycle(0)
>>> identity(5)
5
>>> add_one_then_double = my_cycle(2)
>>> add_one_then_double(1)
4
>>> do_all_functions = my_cycle(3)
>>> do_all_functions(2)
9
>>> do_more_than_a_cycle = my_cycle(4)
>>> do_more_than_a_cycle(2)
10
>>> do_two_cycles = my_cycle(6)
>>> do_two_cycles(1)
19
"""
def ret_fn(n):
def ret(x):
i = 0
while i < n:
if i % 3 == 0:
x = f1(x)
elif i % 3 == 1:
x = f2(x)
else:
x = f3(x)
i += 1
return x
return ret
return ret_fn
# Alternative solution
def ret_fn(n):
def ret(x):
if n == 0:
return x
return cycle(f2, f3, f1)(n  1)(f1(x))
return ret
return ret_fn
There are three main pieces of information we need in order to calculate the value that we want to return.
 The three functions that we will be cycling through, so
f1
,f2
,f3
.  The number of function applications we need, namely
n
. Whenn
is 0, we want our function to behave like the identity function (i.e. return the input without applying any of our three functions to it).  The input that we start off with, namely
x
.
The functions are the parameters passed into cycle
. We want the return
value of cycle
to be a function ret_fn
that takes in n
and outputs
another function ret
. ret
is a function that takes in x
and then will
cyclically apply the three passed in functions to the input until we have
reached n
applications. Thus, most of the logic will go inside of ret
.
Solution:
To figure out which function we should next use in our cycle, we can use the
mod operation via %
, and loop through the function applications until we
have made exactly n
function applications to our original input x
.
Use Ok to test your code:
python3 ok q cycle