CS61A Lab 2

更好的阅读体验

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.

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

______
>>> 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)

______
>>> x = None # remember to review the rules of WWPD given above!
>>> x
>>> lambda x: x

______
>>> z = 3
>>> e = lambda x: lambda y: lambda: x + y + z
>>> e(0)(1)()

______
>>> f = lambda z: x + z
>>> f(3)

______
>>> 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

______

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

______
>>> 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)

______

Parsons Problems

To work on these problems, open the Parsons editor:

python3 parsons

Q3: A Hop, a Skip, and a Jump

Complete hop, which implements a curried version of the function f(x, y) = y.

def hop():
    """
    Calling hop returns a curried version of the function f(x, y) = y.
    >>> hop()(3)(2) # .Case 1
    2
    >>> hop()(3)(7) # .Case 2
    7
    >>> hop()(4)(7) # .Case 3
    7
    """
    "*** YOUR CODE HERE ***"
    return lambda x: lambda y: y

Q4: Digit Index Factory

Complete the function digit_index_factory, which takes in two integers k and num as input and returns a function. The returned function takes no arguments, and outputs the offset between k and the rightmost digit of num. The offset between two numbers is defined to be the number of steps between the two numbers. For example, in 25, there is an offset of 1 between 2 and 5.

Note that 0 is considered to have no digits (not even 0).

def digit_index_factory(num, k):
    """
    Returns a function that takes no arguments, and outputs the offset
    between k and the rightmost digit of num. If k is not in num, then
    the returned function returns -1. Note that 0 is considered to
    contain no digits (not even 0).
    >>> digit_index_factory(34567, 4)() # .Case 1
    3
    >>> digit_index_factory(30001, 0)() # .Case 2
    1
    >>> digit_index_factory(999, 1)() # .Case 3
    -1
    >>> digit_index_factory(1234, 0)() # .Case 4
    -1
    """
    "*** YOUR CODE HERE ***"

    def digit_index_factory(num, k):1
        index = 0
        while num:
            if num % 10 == k:
                return lambda: index
            index += 1
            num //= 107
        return lambda: -1

Coding Practice

Q5: Lambdas and Currying

Write a function lambda_curry2 that will curry any two argument function using lambdas.

Your solution to this problem should fit entirely on the return 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)

Use Ok to test your code:

python3 ok -q lambda_curry2✂️

Q6: 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.

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 counter(n):
        i = 1
        cnt = 0
        while i <= n:
            if condition(n, i):
                cnt = cnt + 1
            i = i + 1
        return cnt
    return counter

Use Ok to test your code:

python3 ok -q count_cond✂️

Submit

Make sure to submit this assignment by running:

python3 ok --submit

Optional Questions

Q7: 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 ***"
    def identity(x):
        return composer(f, g)(x) == composer(g, f)(x)
    return identity

Use Ok to test your code:

python3 ok -q composite_identity✂️

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 ref_fn(n):
        def ret(x):
            if n == 0:
                return x
            return cycle(f2, f3, f1)(n - 1)(f1(x))
        return ret
    return ref_fn

Use Ok to test your code:

python3 ok -q cycle