Lab 2: Lambdas, HigherOrder Functions, and Recursion
Due at 11:59pm on Friday, 02/03/2017.
Starter Files
Download lab02.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the OK autograder.
Submission
By the end of this lab, you should have submitted the lab with
python3 ok submit
. You may submit more than once before the
deadline; only the final submission will be graded.
 Questions 1  4 must be completed in order to receive credit for this lab. Starter code for questions 3 and 4 is in lab02.py.
 Questions 5 and 6 (Environment Diagrams) are optional. It is recommended that you work on these should you finish the required section early.
 Questions 7  11 are also optional. It is recommended that you complete these problems on your own time. Starter code for the questions are in lab02_extra.py.
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.
Lambdas
Lambda expressions are oneline functions that specify two things: the parameters and the return value.
lambda <parameters>: <return value>
While both lambda
and def
statements are related to functions, there are some differences.
lambda  def  

Type  lambda is an expression 
def is a statement 
Description  Evaluating a lambda expression does not create or modify any variables.
Lambda expressions just create new function values. 
Executing a def statement will create a new function value and bind it to a variable in the current environment. 
Example 


A lambda
expression by itself is not very interesting. As with any values
such as numbers, Booleans, strings, we usually:
 assign lambdas to variables (
foo = lambda x: x
)  pass them in to other functions (
bar(lambda x: x)
)
Higher Order 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 be exploring many applications of higher order functions.
Required Questions
What Would Python Display?
Question 1: WWPD: Lambda the Free
Use OK to test your knowledge with the following "What Would Python Display?" questions:
python3 ok q lambda u
Hint: Remember for all WWPD questions, input
Function
if you believe the answer is<function...>
,Error
if it errors, andNothing
if nothing is displayed.
>>> lambda x: x
______<function <lambda> at ...>
>>> a = lambda x: x
>>> a(5) # x is the parameter for the lambda function
______5
>>> b = lambda: 3
>>> b()
______3
>>> c = lambda x: lambda: print('123')
>>> c(88)
______<function <lambda> at ...>
>>> c(88)()
______123
>>> d = lambda f: f(4) # They can have functions as arguments as well.
>>> def square(x):
... return x * x
>>> d(square)
______16
>>> t = lambda f: lambda x: f(f(f(x)))
>>> s = lambda x: x + 1
>>> t(s)(0)
______3
>>> bar = lambda y: lambda x: pow(x, y)
>>> bar()(15)
______TypeError: <lambda>() missing 1 required positional argument: 'y'
>>> foo = lambda: 32
>>> foobar = lambda x, y: x // y
>>> a = lambda x: foobar(foo(), bar(4)(x))
>>> a(2)
______2
>>> b = lambda x, y: print('summer')
______# Nothing gets printed by the interpreter
>>> c = b(4, 'dog')
______summer
>>> print(c)
______None
>>> a = lambda b: b * 2
______# Nothing gets printed by the interpreter
>>> a
______Function
>>> a(a(a(2)))
______16
>>> a(a(a()))
______TypeError: <lambda>() missing 1 required positional argument: 'b'
>>> def d():
... print(None)
... print('whoo')
>>> b = d()
______None
whoo
>>> b
______# Nothing gets printed by the interpreter
>>> x, y, z = 1, 2, 3
>>> a = lambda b: x + y + z
>>> x += y
>>> y = z
>>> a('b')
______5
>>> z = 3
>>> e = lambda x: lambda y: lambda: x + y + z
>>> e(0)(1)()
______4
Question 2: WWPD: Higher Order Functions
Use OK to test your knowledge with the following "What Would Python Display?" questions:
python3 ok q hof u
Hint: Remember for all WWPD questions, input
Function
if you believe the answer is<function...>
,Error
if it errors, andNothing
if nothing is displayed.
>>> def even(f):
... def odd(x):
... if x < 0:
... return f(x)
... return f(x)
... return odd
>>> stevphen = lambda x: x
>>> stewart = even(stevphen)
>>> stewart
______<function ...>
>>> stewart(61)
______61
>>> stewart(4)
______4
>>> def cake():
... print('beets')
... def pie():
... print('sweets')
... return 'cake'
... return pie
>>> a = cake()
______beets
>>> a
______Function
>>> a()
______sweets
'cake'
>>> x, b = a(), cake
______sweets
>>> def snake(x):
... if cake == b:
... x += 3
... return lambda y: y + x
... else:
... return y  x
>>> snake(24)(23)
______50
>>> cake = 2
>>> snake(26)
______Error
>>> y = 50
>>> snake(26)
______24
Coding Practice
Question 3: Lambdas and Currying
We can transform multipleargument functions into a chain of singleargument, higher order functions by taking advantage of lambda expressions. This is useful when dealing with functions that take only singleargument functions. We will see some examples of these later on.
Write a function lambda_curry2
that will curry any two argument function using
lambdas. See the doctest or refer to the
textbook
if you're not sure what this means.
def lambda_curry2(func):
"""
Returns a Curried version of a twoargument function FUNC.
>>> from operator import add
>>> curried_add = lambda_curry2(add)
>>> add_three = curried_add(3)
>>> add_three(5)
8
"""
"*** YOUR CODE HERE ***"
return ______
return lambda arg1: lambda arg2: func(arg1, arg2)
Use OK to test your code:
python3 ok q lambda_curry2
Question 4: 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.
You may use the compose1
function defined below.
def compose1(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 = compose1(square, add_one) # (x + 1)^2
>>> a1(4)
25
>>> mul_three = lambda x: x * 3 # multiplies 3 to x
>>> a2 = compose1(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
"""
"*** YOUR CODE HERE ***"
def identity(x):
return compose1(f, g)(x) == compose1(g, f)(x)
return identity
# Alternative solution
return lambda x: f(g(x)) == g(f(x))
Use OK to test your code:
python3 ok q composite_identity
Optional Questions
Environment Diagrams
Question 5: Lambda the Environment Diagram
Try drawing an environment diagram for the following code and predict what Python will output.
You do not need to submit or unlock this question through Ok. Instead, you can check your work with the Online Python Tutor, but try drawing it yourself first!
>>> a = lambda x: x * 2 + 1
>>> def b(b, x):
... return b(x + a(x))
>>> x = 3
>>> b(a, x)
______21
Question 6: Make Adder
Draw the environment diagram for the following code:
n = 9
def make_adder(n):
return lambda k: k + n
add_ten = make_adder(n+1)
result = add_ten(n)
There are 3 frames total (including the Global frame). In addition, consider the following questions:
 In the Global frame, the name
add_ten
points to a function value. What is the intrinsic name of that function value, and what frame is its parent?  In frame
f2
, what name is the frame labeled with (add_ten
or λ)? Which frame is the parent off2
?  What value is the variable
result
bound to in the Global frame?
You can try out the environment diagram at tutor.cs61a.org.
 The intrinsic name of the function object that
add_ten
points to is λ (specifically, the lambda whose parameter isk
). The parent frame of this lambda isf1
. f2
is labeled with the name λ the parent frame off2
isf1
, since that is where λ is defined. The variable
result
is bound to 19.
Recursion
A recursive function is a function that calls itself in its body, either directly or indirectly. Recursive functions have three important components:
 Base case(s), the simplest possible form of the problem you're trying to solve.
 Recursive case(s), where the function calls itself with a simpler argument as part of the computation.
 Using the recursive calls to solve the full problem.
Let's look at the canonical example, factorial
:
def factorial(n):
if n == 0:
return 1
return n * factorial(n  1)
We know by its definition that 0! is 1. So we choose n == 0
as our base case.
The recursive step also follows from the definition of factorial, i.e., n
! =
n
* (n
1)!.
The next few questions in lab will have you writing recursive functions. Here are some general tips:
 Consider how you can solve the current problem using the solution to a simpler version of the problem. Remember to trust the recursion: assume that your solution to the simpler problem works correctly without worrying about how.
 Think about what the answer would be in the simplest possible case(s). These will be your base cases  the stopping points for your recursive calls. Make sure to consider the possibility that you're missing base cases (this is a common way recursive solutions fail).
 It may help to write the iterative version first.
Note: The following questions are in lab02_extra.py.
Question 7: Common Misconception
Find the bug in the following recursive function.
def factorial(n):
"""Return n * (n  1) * (n  2) * ... * 1.
>>> factorial(5)
120
"""
if n == 0:
return 1
else:
return factorial(n1)
Fix the code in lab02_extra.py
and run:
python3 ok q factorial
The result of the recursive calls is not combined into the correct solution.
def factorial(n):
if n == 0:
return 1
else:
return n * factorial(n1)
Question 8: Common Misconception
Find the bug with this recursive function.
def skip_mul(n):
"""Return the product of n * (n  2) * (n  4) * ...
>>> skip_mul(5) # 5 * 3 * 1
15
>>> skip_mul(8) # 8 * 6 * 4 * 2
384
"""
if n == 2:
return 2
else:
return n * skip_mul(n  2)
Fix the code in lab02_extra.py
and run:
python3 ok q skip_mul
Consider what happens when we choose an odd number for n
. skip_mul(3)
will
return 3 * skip_mul(1)
. skip_mul(1)
will return 1 * skip_mul(1)
. You
may see the problem now. Since we are decreasing n
by two at a time, we've
completed missed our base case of n == 2
, and we will end up recursing
indefinitely. We need to add another base case to make sure this doesn't
happen.
def skip_mul(n):
if n == 1:
return 1
elif n == 2:
return 2
else:
return n * skip_mul(n  2)
Question 9: GCD
The greatest common divisor of two positive integers a
and b
is the
largest integer which evenly divides both numbers (with no remainder).
Euclid, a Greek mathematician in 300 B.C., realized that the greatest
common divisor of a
and b
is one of the following:
 the smaller value if it evenly divides the larger value, or
 the greatest common divisor of the smaller value and the remainder of the larger value divided by the smaller value
In other words, if a
is greater than b
and a
is not divisible by
b
, then
gcd(a, b) = gcd(b, a % b)
Write the gcd
function recursively using Euclid's algorithm.
def gcd(a, b):
"""Returns the greatest common divisor of a and b.
Should be implemented using recursion.
>>> gcd(34, 19)
1
>>> gcd(39, 91)
13
>>> gcd(20, 30)
10
>>> gcd(40, 40)
40
"""
"*** YOUR CODE HERE ***"
a, b = max(a, b), min(a, b)
if a % b == 0:
return b
else:
return gcd(b, a % b)
# Iterative solution, if you're curious
def gcd_iter(a, b):
"""Returns the greatest common divisor of a and b, using iteration.
>>> gcd_iter(34, 19)
1
>>> gcd_iter(39, 91)
13
>>> gcd_iter(20, 30)
10
>>> gcd_iter(40, 40)
40
"""
if a < b:
return gcd_iter(b, a)
while a > b and not a % b == 0:
a, b = b, a % b
return b
Use OK to test your code:
python3 ok q gcd
More Coding Practice
Question 10: 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."""
i, count = 1, 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."""
i, count = 1, 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 counts all the numbers
from 1 to n
that satisfy condition
.
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.
>>> 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
"""
"*** YOUR CODE HERE ***"
def counter(n):
i, count = 1, 0
while i <= n:
if condition(n, i):
count += 1
i += 1
return count
return counter
Use OK to test your code:
python3 ok q count_cond
Question 11: 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
"""
"*** YOUR CODE HERE ***"
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
Use OK to test your code:
python3 ok q cycle