CS61A Homework 2
easterling

Homework 2: Higher Order Functions hw02.zip

Due by 11:59pm on Thursday, September 8

Required questions

Several doctests refer to these functions:

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from operator import add, mul

square = lambda x: x * x

identity = lambda x: x

triple = lambda x: 3 * x

increment = lambda x: x + 1

Q1: Product

Write a function called product that returns term(1) * ... * term(n).

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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=term(1)
for i in range(n-1):
sum*=term((i+2))
return sum

Q2: Accumulate

Let’s take a look at how product is an instance of a more general function called accumulate, which we would like to implement:

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def accumulate(merger, start, n, term):
"""Return the result of merging the first n terms in a sequence and start.
The terms to be merged are term(1), term(2), ..., term(n). merger 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
>>> # 2 + (1^2 + 1) + (2^2 + 1) + (3^2 + 1)
>>> accumulate(lambda x, y: x + y + 1, 2, 3, square)
19
>>> # ((2 * 1^2 * 2) * 2^2 * 2) * 3^2 * 2
>>> accumulate(lambda x, y: 2 * x * y, 2, 3, square)
576
>>> accumulate(lambda x, y: (x + y) % 17, 19, 20, square)
16
"""
"*** YOUR CODE HERE ***"
sum = start
for i in range(n):
sum = merger(sum,term(i+1))
return sum

accumulate has the following parameters:

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

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

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

Note: You may assume that merger is commutative. That is, merger(a, b) == merger(b, a) for all a and b. However, you may not assume merger 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 function calls to accumulate.

Important: You should have a single line of code (which should be a return statement) in each of your implementations for summation_using_accumulate and product_using_accumulate, which the syntax check will check for.

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def summation_using_accumulate(n, term):
"""Returns the sum: term(0) + ... + term(n), using accumulate.

>>> summation_using_accumulate(5, square)
55
>>> summation_using_accumulate(5, triple)
45
>>> # You aren't expected to understand the code of this test.
>>> # Check that the bodies of the functions are just return statements.
>>> # If this errors, make sure you have removed the "***YOUR CODE HERE***".
>>> import inspect, ast
>>> [type(x).__name__ for x in ast.parse(inspect.getsource(summation_using_accumulate)).body[0].body]
['Expr', 'Return']
"""
return accumulate(add, term(0), n, term)

def product_using_accumulate(n, term):
"""Returns the product: term(1) * ... * term(n), using accumulate.

>>> product_using_accumulate(4, square)
576
>>> product_using_accumulate(6, triple)
524880
>>> # You aren't expected to understand the code of this test.
>>> # Check that the bodies of the functions are just return statements.
>>> # If this errors, make sure you have removed the "***YOUR CODE HERE***".
>>> import inspect, ast
>>> [type(x).__name__ for x in ast.parse(inspect.getsource(product_using_accumulate)).body[0].body]
['Expr', 'Return']
"""
return accumulate(mul, 1, n, term)
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