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)
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 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 ***"
f = func
if n == 0:
return identity
for i in range(n-1):
f = compose1(func, f)
return f
For an extra challenge, try defining make_repeater using compose1 and your accumulate function in a single one-line return statement.
return accumulate(compose1, identity, n, lambda _: func)
Just for fun Question
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 zero(f):
return lambda x: x
def successor(n):
return lambda f: lambda x: f(n(f)(x))
def one(f):
"""Church numeral 1: same as successor(zero)"""
"*** YOUR CODE HERE ***"
return lambda x: f(x)
def two(f):
"""Church numeral 2: same as successor(successor(zero))"""
"*** YOUR CODE HERE ***"
return lambda x: f(f(x))
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 ***"
return n(increment)(0)
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 ***"
return lambda f: lambda x: m(f)(n(f)(x))
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 ***"
return lambda f: m(n(f))
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 ***"
return n(m)
