Due by 11:59pm on Monday, 6/29

## Instructions

Download hw02.zip. Inside the archive, you will find a file called hw02.py, along with a copy of the OK autograder.

Submission: When you are done, submit with ```python3 ok --submit```. You may submit more than once before the deadline; only the final submission will be scored. See Lab 1 for instructions on submitting assignments.

Using OK: If you have any questions about using OK, please refer to this guide.

Readings: You might find the following references useful:

## Composition questions (required)

One of the most important aspects of writing code is readability, or composition; it is often the case that a fellow programmer needs to read and understand your code. A program is composed well if it is

• concise
• uses well-named variables
• understandable and easy to follow

Programs that have poor composition are often difficult to understand, needlessly verbose, or difficult to maintain. For more details, see this article about composition.

For each of the following questions, decide which version of the question's code has better composition explain why. See each question for further instructions.

### Question 1: Composition: summation

Consider these two versions of the `summation` function:

Version 1
``````def summation(term, n):
current, total = 1, 0
while current <= n:
total += term(current)
current += 1
Version 2
``````def summation(f, n):
x, y = 0, 1
while y < n:
x += f(y)
y += 1
x += f(n)
return x``````

Decide which of version (1 or 2) has better composition:

• assign the variable `better_summation` to the number of your choice (either `1` or `2`)
• assign the variable `better_summation_explanation` to a (multi-line) string explaining your decision

``````better_summation = None  # REPLACE None WITH 1 or 2

better_summation_explanation = """
"""``````

### Question 2: Composition: is_positive

Consider these two versions of the `is_positive` function:

Version 1
``````def is_positive(n):
if (n > 0) == True:
return True
else:
return False``````
Version 2
``````def is_positive(n):
return n > 0``````

Decide which of version (1 or 2) has better composition:

• assign the variable `better_is_positive` to the number of your choice (either `1` or `2`)
• assign the variable `better_is_positive_explanation` to a (multi-line) string explaining your decision

``````better_is_positive = None  # REPLACE None WITH 1 or 2

better_is_positive_explanation = """
"""``````

## Required questions

### Doctest functions

Several doctests refer to these one-argument functions:

``````def square(x):
return x * x

def triple(x):
return 3 * x

def identity(x):
return x

def increment(x):
return x + 1``````

### Question 3: Piecewise

Implement `piecewise`, which takes two one-argument functions, `f` and `g`, along with a number `b`. It returns a new function that takes a number `x` and returns either `f(x)` if `x` is less than `b`, or `g(x)` if `x` is greater than or equal to `b`.

``````def piecewise(f, g, b):
"""Returns the piecewise function h where:

h(x) = f(x) if x < b,
g(x) otherwise

>>> def negate(x):
...     return -x
>>> abs_value = piecewise(negate, identity, 0)
>>> abs_value(6)
6
>>> abs_value(-1)
1
"""

Use OK to test your code:

``python3 ok -q piecewise``

### Question 4: Product

The `summation(term, n)` function from lecture adds up `term(1) + ... + term(n)` Write a similar `product(n, term)` function that returns ```term(1) * ... * term(n)```. Show how to define the factorial function in terms of `product`. Hint: try using the `identity` function for `factorial`.

``````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

>>> 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
"""

def factorial(n):
"""Return n factorial for n >= 0 by calling product.

>>> factorial(4)
24
>>> factorial(6)
720
"""

Use OK to test your code:

``````python3 ok -q product
python3 ok -q factorial``````

### Question 5: Accumulate

Show that both `summation` and `product` are instances of a more general function, called `accumulate`, with the following signature:

``````from operator import add, mul

def accumulate(combiner, base, n, term):
"""Return the result of combining the first n terms in a sequence.

>>> 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(combiner, base, n, term)` takes the following arguments:

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

For example, `accumulate(add, 11, 3, square)` is

``11 + square(1) + square(2) + square(3)``

Implement `accumulate` and show how `summation` and `product` can both be defined as simple calls to `accumulate`:

``````def summation_using_accumulate(n, term):
"""An implementation of summation using accumulate.

>>> summation_using_accumulate(5, square)
55
>>> summation_using_accumulate(5, triple)
45
"""

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

>>> product_using_accumulate(4, square)
576
>>> product_using_accumulate(6, triple)
524880
"""

Use OK to test your code:

``````python3 ok -q accumulate
python3 ok -q summation_using_accumulate
python3 ok -q product_using_accumulate``````

### Question 6: Repeated

Implement `repeated(f, n)`:

• `f` is a one-argument function that takes a number and returns another number.
• `n` is a positive integer

`repeated` returns another function that, when given an argument `x`, will compute `f(f(....(f(x))....))` (apply `f` a total `n` times). For example, `repeated(square, 3)(42)` evaluates to `square(square(square(42)))`.

``````def repeated(f, n):
"""Return the function that computes the nth application of f.

8
>>> repeated(triple, 5)(1) # 3 * 3 * 3 * 3 * 3 * 1
243
>>> repeated(square, 2)(5) # square(square(5))
625
>>> repeated(square, 4)(5) # square(square(square(square(5))))
152587890625
"""

Hint: You may find it convenient to use `compose1` from the textbook:

``````def compose1(f, g):
"""Return a function h, such that h(x) = f(g(x))."""
def h(x):
return f(g(x))
return h``````

Use OK to test your code:

``python3 ok -q repeated``

## Extra questions

### Question 7: 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)"""

def two(f):
"""Church numeral 2: same as successor(successor(zero))"""

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
"""

"""Return the Church numeral for m + n, for Church numerals m and n.

5
"""

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
"""

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
"""
``````python3 ok -q church_to_int