Homework 2: Higher-Order Functions

Q1: Product

The summation(n, term) function from the higher-order functions lecture adds up term(1) + ... + term(n). Write a similar function called product that returns term(1) * ... * term(n).

def product(n, term):
    """Return the product of the first n terms in a sequence.
    n -- a positive integer
    term -- a function that takes one argument to produce the term

    >>> product(3, identity)  # 1 * 2 * 3
    6
    >>> product(5, identity)  # 1 * 2 * 3 * 4 * 5
    120
    >>> product(3, square)    # 1^2 * 2^2 * 3^2
    36
    >>> product(5, square)    # 1^2 * 2^2 * 3^2 * 4^2 * 5^2
    14400
    >>> product(3, increment) # (1+1) * (2+1) * (3+1)
    24
    >>> product(3, triple)    # 1*3 * 2*3 * 3*3
    162
    """
    "*** YOUR CODE HERE ***"

from operator import add, mul, sub

square = lambda x: x * x

identity = lambda x: x

triple = lambda x: 3 * x

increment = lambda x: x + 1


HW_SOURCE_FILE=__file__


def product(n, term):
    """Return the product of the first n terms in a sequence.
    n -- a positive integer
    term -- a function that takes one argument to produce the term

    >>> product(3, identity)  # 1 * 2 * 3
    6
    >>> product(5, identity)  # 1 * 2 * 3 * 4 * 5
    120
    >>> product(3, square)    # 1^2 * 2^2 * 3^2
    36
    >>> product(5, square)    # 1^2 * 2^2 * 3^2 * 4^2 * 5^2
    14400
    >>> product(3, increment) # (1+1) * (2+1) * (3+1)
    24
    >>> product(3, triple)    # 1*3 * 2*3 * 3*3
    162
    """
    sum=1
    while n>0:
        sum*=term(n)
        n-=1
    return sum

Q2: Accumulate

Let's take a look at how summation and product are instances of a more general function called accumulate:

def accumulate(combiner, base, n, term):
    """Return the result of combining the first n terms in a sequence and base.
    The terms to be combined are term(1), term(2), ..., term(n).  combiner is a
    two-argument commutative function.

    >>> accumulate(add, 0, 5, identity)  # 0 + 1 + 2 + 3 + 4 + 5
    15
    >>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 5
    26
    >>> accumulate(add, 11, 0, identity) # 11
    11
    >>> accumulate(add, 11, 3, square)   # 11 + 1^2 + 2^2 + 3^2
    25
    >>> accumulate(mul, 2, 3, square)    # 2 * 1^2 * 2^2 * 3^2
    72
    >>> accumulate(lambda x, y: x + y + 1, 2, 3, square)
    19
    >>> accumulate(lambda x, y: 2 * (x + y), 2, 3, square)
    58
    >>> accumulate(lambda x, y: (x + y) % 17, 19, 20, square)
    16
    """
    "*** YOUR CODE HERE ***"

accumulate has the following parameters:

• term and n: the same parameters as in summation and product
• combiner: a two-argument function that specifies how the current term is combined with the previously accumulated terms.
• base: value at which to start the accumulation.

For example, the result of accumulate(add, 11, 3, square) is

11 + square(1) + square(2) + square(3) = 25

Note: You may assume that combiner is commutative. That is, combiner(a, b) == combiner(b, a) for all a, b, and c. However, you may not assume combiner is chosen from a fixed function set and hard-code the solution.

After implementing accumulate, show how summation and product can both be defined as simple calls to accumulate:

def summation_using_accumulate(n, term):
    """Returns the sum of term(1) + ... + term(n). The implementation
    uses accumulate.

    >>> summation_using_accumulate(5, square)
    55
    >>> summation_using_accumulate(5, triple)
    45
    >>> from construct_check import check
    >>> # ban iteration and recursion
    >>> check(HW_SOURCE_FILE, 'summation_using_accumulate',
    ...       ['Recursion', 'For', 'While'])
    True
    """
    "*** YOUR CODE HERE ***"

def product_using_accumulate(n, term):
    """An implementation of product using accumulate.

    >>> product_using_accumulate(4, square)
    576
    >>> product_using_accumulate(6, triple)
    524880
    >>> from construct_check import check
    >>> # ban iteration and recursion
    >>> check(HW_SOURCE_FILE, 'product_using_accumulate',
    ...       ['Recursion', 'For', 'While'])
    True
    """
    "*** YOUR CODE HERE ***"

It should be noted that:The order of operation of two numbers is: base->term(1)->...->term(n)

Implementation: So I usek=1 (starting from the first item) to traverse upward to ensure the correct order of operations

(Before starting the operation from n, the order was wrong)

def accumulate(combiner, base, n, term):
    """Return the result of combining the first n terms in a sequence and base.
    The terms to be combined are term(1), term(2), ..., term(n).  combiner is a
    two-argument commutative function.

    >>> accumulate(add, 0, 5, identity)  # 0 + 1 + 2 + 3 + 4 + 5
    15
    >>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 5
    26
    >>> accumulate(add, 11, 0, identity) # 11
    11
    >>> accumulate(add, 11, 3, square)   # 11 + 1^2 + 2^2 + 3^2
    25
    >>> accumulate(mul, 2, 3, square)    # 2 * 1^2 * 2^2 * 3^2
    72
    >>> accumulate(lambda x, y: x + y + 1, 2, 3, square)
    19
    >>> accumulate(lambda x, y: 2 * (x + y), 2, 3, square)
    58
    >>> accumulate(lambda x, y: (x + y) % 17, 19, 20, square)
    16
    """
    "*** YOUR CODE HERE ***"
    k=1
    while n>=k:
        base=combiner(base, term(k))
        k+=1
    return base

def summation_using_accumulate(n, term):
    """Returns the sum of term(1) + ... + term(n). The implementation
    uses accumulate.

    >>> summation_using_accumulate(5, square)
    55
    >>> summation_using_accumulate(5, triple)
    45
    >>> from construct_check import check
    >>> # ban iteration and recursion
    >>> check(HW_SOURCE_FILE, 'summation_using_accumulate',
    ...       ['Recursion', 'For', 'While'])
    True
    """
    "*** YOUR CODE HERE ***"
    return accumulate(add,0,n,term)

def product_using_accumulate(n, term):
    """An implementation of product using accumulate.

    >>> product_using_accumulate(4, square)
    576
    >>> product_using_accumulate(6, triple)
    524880
    >>> from construct_check import check
    >>> # ban iteration and recursion
    >>> check(HW_SOURCE_FILE, 'product_using_accumulate',
    ...       ['Recursion', 'For', 'While'])
    True
    """
    "*** YOUR CODE HERE ***"
    return accumulate(mul,1,n,term)

Q3: Make Repeater

Implement the function make_repeater so that make_repeater(func, n)(x) returns func(func(...func(x)...)), where func is applied n times. That is, make_repeater(func, n) returns another function that can then be applied to another argument. For example, make_repeater(square, 3)(42) evaluates to square(square(square(42))).

def make_repeater(func, n):
    """Return the function that computes the nth application of func.

    >>> add_three = make_repeater(increment, 3)
    >>> add_three(5)
    8
    >>> make_repeater(triple, 5)(1) # 3 * 3 * 3 * 3 * 3 * 1
    243
    >>> make_repeater(square, 2)(5) # square(square(5))
    625
    >>> make_repeater(square, 4)(5) # square(square(square(square(5))))
    152587890625
    >>> make_repeater(square, 0)(5) # Yes, it makes sense to apply the function zero times!
    5
    """
    "*** YOUR CODE HERE ***"

For an extra challenge, try defining make_repeater using compose1 and your accumulate function in a single one-line return statement.

def compose1(func1, func2):
    """Return a function f, such that f(x) = func1(func2(x))."""
    def f(x):
        return func1(func2(x))
    return f

 Requires n nesting times (accumulate implementation)

Architecture: compose1

( provides a framework to indicate how parameters are related to each other)

Parameter 1: identity (original function)

Number of times: n times

Parameter two: lambda _:func()

(由于 base=combiner(base, term(k))

A number returned by func(x) conflicts with the parameters of compose

So func must be returned to a function)

Implementation: identity(lambda _:func(x))

def accumulate(combiner, base, n, term):
    """Return the result of combining the first n terms in a sequence and base.
    The terms to be combined are term(1), term(2), ..., term(n).  combiner is a
    two-argument commutative function.

    >>> accumulate(add, 0, 5, identity)  # 0 + 1 + 2 + 3 + 4 + 5
    15
    >>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 5
    26
    >>> accumulate(add, 11, 0, identity) # 11
    11
    >>> accumulate(add, 11, 3, square)   # 11 + 1^2 + 2^2 + 3^2
    25
    >>> accumulate(mul, 2, 3, square)    # 2 * 1^2 * 2^2 * 3^2
    72
    >>> accumulate(lambda x, y: x + y + 1, 2, 3, square)
    19
    >>> accumulate(lambda x, y: 2 * (x + y), 2, 3, square)
    58
    >>> accumulate(lambda x, y: (x + y) % 17, 19, 20, square)
    16
    """
    "*** YOUR CODE HERE ***"
    k=1
    while n>=k:
        base=combiner(base, term(k))
        k+=1
    return base

def compose1(func1, func2):
    """Return a function f, such that f(x) = func1(func2(x))."""
    def f(x):
        return func1(func2(x))
    return f
def make_repeater(func, n):
    """Return the function that computes the nth application of func.

    >>> add_three = make_repeater(increment, 3)
    >>> add_three(5)
    8
    >>> make_repeater(triple, 5)(1) # 3 * 3 * 3 * 3 * 3 * 1
    243
    >>> make_repeater(square, 2)(5) # square(square(5))
    625
    >>> make_repeater(square, 4)(5) # square(square(square(square(5))))
    152587890625
    >>> make_repeater(square, 0)(5) # Yes, it makes sense to apply the function zero times!
    5
    """
    "*** YOUR CODE HERE ***"
    return  accumulate(compose1,identity,n,lambda _: func)

Just for fun Question

This question is out of scope for 61a. Do it if you want an extra challenge or some practice with HOF and abstraction!

Q4: Church numerals

The logician Alonzo Church invented a system of representing non-negative integers entirely using functions. The purpose was to show that functions are sufficient to describe all of number theory: if we have functions, we do not need to assume that numbers exist, but instead we can invent them.

Your goal in this problem is to rediscover this representation known as Church numerals. Here are the definitions of zero, as well as a function that returns one more than its argument:

def zero(f):
    return lambda x: x

def successor(n):
    return lambda f: lambda x: f(n(f)(x))

First, define functions one and two such that they have the same behavior as successor(zero) and successsor(successor(zero)) respectively, but do not call successor in your implementation.

Next, implement a function church_to_int that converts a church numeral argument to a regular Python integer.

Finally, implement functions add_church, mul_church, and pow_church that perform addition, multiplication, and exponentiation on church numerals.

def one(f):
    """Church numeral 1: same as successor(zero)"""
    "*** YOUR CODE HERE ***"

def two(f):
    """Church numeral 2: same as successor(successor(zero))"""
    "*** YOUR CODE HERE ***"

three = successor(two)

def church_to_int(n):
    """Convert the Church numeral n to a Python integer.

    >>> church_to_int(zero)
    0
    >>> church_to_int(one)
    1
    >>> church_to_int(two)
    2
    >>> church_to_int(three)
    3
    """
    "*** YOUR CODE HERE ***"

def add_church(m, n):
    """Return the Church numeral for m + n, for Church numerals m and n.

    >>> church_to_int(add_church(two, three))
    5
    """
    "*** YOUR CODE HERE ***"

def mul_church(m, n):
    """Return the Church numeral for m * n, for Church numerals m and n.

    >>> four = successor(three)
    >>> church_to_int(mul_church(two, three))
    6
    >>> church_to_int(mul_church(three, four))
    12
    """
    "*** YOUR CODE HERE ***"

def pow_church(m, n):
    """Return the Church numeral m ** n, for Church numerals m and n.

    >>> church_to_int(pow_church(two, three))
    8
    >>> church_to_int(pow_church(three, two))
    9
    """
    "*** YOUR CODE HERE ***"