CS61A 学习笔记 Homework 2: Higher Order Functions

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__

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 ***"
    result = 1
    for i in range(1, n+1):
        result *= term(i)
    return result


def square(x):
    return x * x

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 ***"
    result = base
    for i in range(1, n+1):
        result = combiner(result, term(i))
    return result

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 ***"
    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)
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