Lab 2: Higher-Order Functions, Lambda Expressions lab02.zip

Due by 11:59pm on Wednesday, September 7.

Q1: WWPD: Lambda the Free

Use Ok to test your knowledge with the following “What Would Python Display?” questions:
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python3 ok -q lambda -u✂️
As a reminder, the following two lines of code will not display anything in the Python interpreter when executed:
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2
>>> x = None
>>> x

>>> lambda x: x  # A lambda expression with one parameter x
______
>>> a = lambda x: x  # Assigning the lambda function to the name a
>>> a(5)
______
>>> (lambda: 3)()  # Using a lambda expression as an operator in a call exp.
______
>>> b = lambda x: lambda: x  # Lambdas can return other lambdas!
>>> c = b(88)
>>> c
______
>>> c()
______
>>> d = lambda f: f(4)  # They can have functions as arguments as well.
>>> def square(x):
...     return x * x
>>> d(square)
______
>>> z = 3
>>> e = lambda x: lambda y: lambda: x + y + z
>>> e(0)(1)()
______
>>> f = lambda z: x + z
>>> f(3)
______
>>> x = None # remember to review the rules of WWPD given above!
>>> x
______
>>> 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?
______
>>> higher_order_lambda(g)(2)
______
>>> call_thrice = lambda f: lambda x: f(f(f(x)))
>>> call_thrice(lambda y: y + 1)(0)
______
>>> print_lambda = lambda z: print(z)  # When is the return expression of a lambda expression executed?
>>> print_lambda
______
>>> one_thousand = print_lambda(1000)
______
>>> one_thousand # What did the call to print_lambda return?
______

Q2: WWPD: Higher Order Functions

Use Ok to test your knowledge with the following “What Would Python Display?” questions:
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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
______
>>> stewart(61)
______
>>> stewart(-4)
______
>>> def cake():
...    print('beets')
...    def pie():
...        print('sweets')
...        return 'cake'
...    return pie
>>> chocolate = cake()
______
>>> chocolate
______
>>> chocolate()
______
>>> more_chocolate, more_cake = chocolate(), cake
______
>>> more_chocolate
______
>>> def snake(x, y):
...    if cake == more_cake:
...        return chocolate
...    else:
...        return x + y
>>> snake(10, 20)
______
>>> snake(10, 20)()
______
>>> cake = 'cake'
>>> snake(10, 20)
______

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
    """
    "*** YOUR CODE HERE ***"
    return lambda x:lambda y :func(x,y)

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
    """
    "*** YOUR CODE HERE ***"
    def dd(n):
        i=1
        count=0
        while i<=n:
            if condition(n,i):
                count+=1
            i += 1
        return count
    return lambda n: dd(n)

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
    """
    "*** YOUR CODE HERE ***"
    return lambda x:composer(f, g)(x) == composer(g, f)(x)

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
    """
    "*** YOUR CODE HERE ***"
    def dd(n,x):

        if(n==0):
            return x
        if(n%3==0):
            return f3(f2(f1(dd(n-3,x))))
        if(n%3==2):
            return f2(f1(dd(n-2,x)))
        if(n%3==1):
            return f1(dd(n-1,x))
        
    return lambda n: lambda x: dd(n,x)