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.

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

# We can assign lambda functions to a name
# with an assignment statement
square = lambda x: x * x
square(3)

# Lambda expressions can be used as an operator
# or operand
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)

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Currying

We can transform multiple-argument functions into a chain of single-argument, 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

  1. Evaluate the expression on the right hand side of the = sign.
  2. 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

  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. 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 built-in Python function like max or print.

  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.

Lambdas

Note: As we saw in the lambda expression section above, lambda functions have no intrinsic name. When drawing lambda functions in environment diagrams, they are labeled with the name lambda 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.

  1. 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.

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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.

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What Would Python Display?

Important: For all WWPD questions, type Function if you believe the answer is <function...>, Error if it errors, and Nothing 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 hof-wwpd -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 two-argument 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 two-argument predicate function condition(n, i). count_cond returns a one-argument 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 return True or False 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 two-argument 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 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. 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, 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
    """
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.

  1. The three functions that we will be cycling through, so f1, f2, f3.
  2. The number of function applications we need, namely n. When n 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).
  3. 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