Due at 11:59pm on Friday, 02/02/2018.

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. Check that you have successfully submitted your code on okpy.org.

• Questions 1-3 must be completed in order to receive credit for this lab. Starter code for question 3 is in lab02.py.
• Questions 4 and 5 (Environment Diagrams) are optional, but highly recommended. Try to work on at least one of these if you finish the required section early.
• Questions 6, 7, and 8 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.

Lambda Expressions

Lambda expressions are one-line functions that specify two things: the parameters and the return expression.

lambda <parameters>: <return expression>

While both lambda and def 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.

# A lambda expression by itself does not alter
# the environment
lambda x: x * x

# We can use lambda expressions in assignment
# statements to give the function a name
square = lambda x: x * x
square(3)

# We can pass lambda expressions as arguments
# into call expressions
negate = lambda f, x: -f(x)
negate(lambda x: x * x, 3)

def square(x):
    return x * x

# A function created by a def statement
# can be referred to by its intrinsic name
square(3)

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.

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 functions 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, you can also return them in other functions! 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!

Here's what multiply_by would look like if we wrote it with a lambda expression:

def multiply_by(m):
    return lambda n: n * m

As you can see, lambda expressions make writing higher order functions much cleaner.

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. Here are the rules as a refresher.

def Statements

1. 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.
2. Write the intrinsic name of the function in the current frame and bind it to the newly created function object.

Call expressions

1. Evaluate the operator, whose value should be a function.
2. Evaluate the operands left to right.
3. Open a new frame. Label it with the sequential frame number, the intrinsic name of the function, and its parent.
4. Bind the formal parameters of the function to the arguments whose values you found in step 2.
5. Execute the body of the function in the new environment.

Assignment Statements

1. Evaluate the expression on the right hand side of the = sign.
2. Write the name found on the left hand side of the = sign in the current frame and bind the value obtained in step 1 to this name.

Required Questions

What Would Python Display?

Q1: WWPD: Lambda the Free

Use Ok to test your knowledge with the following "What Would Python Display?" questions:

python3 ok -q lambda -u

For all WWPD questions, type Function if you believe the answer is <function...>, > Error if it errors, and Nothing if nothing is displayed.

>>> 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

>>> 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
______
# 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 hof -u

For all WWPD questions, type Function if you believe the answer is <function...>, > Error if it errors, and Nothing if nothing is displayed.

>>> 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 lambda y: x + y
...    else:
...        return x + y
>>> snake(10, 20)
______
Function
>>> snake(10, 20)(30)
______
40
>>> cake = 'cake'
>>> snake(10, 20)
______
30

Coding Practice

Q3: Lambdas and Currying

We can transform multiple-argument functions into a chain of single-argument, higher order functions by taking advantage of lambda expressions. This is useful when dealing with functions that take only single-argument 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 two-argument 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

Optional Questions

Environment Diagram Practice

Q4: 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

Q5: 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:

1. In the Global frame, the name add_ten points to a function object. What is the intrinsic name of that function object, and what frame is its parent?
2. In frame f2, what name is the frame labeled with (add_ten or λ)? Which frame is the parent of f2?
3. What value is the variable result bound to in the Global frame?

More Coding Practice

Note: The following questions are in lab02_extra.py.

Q6: Composite Identity Function

Write a function that takes in two single-argument 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

Q7: 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 two-argument predicate function condition(n, i). count_cond returns a one-argument 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 two-argument 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

Q8: 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, return x
• n = 1, apply f1 to x, or return f1(x)
• n = 2, apply f1 to x and then f2 to the result of that, or return f2(f1(x))
• n = 3, apply f1 to x, f2 to the result of applying f1, and then f3 to the result of applying f2, or f3(f2(f1(x)))
• n = 4, start the cycle again applying f1, then f2, then f3, then f1 again, or f1(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 higher-order 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